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k 

COURSE 

OF 

MATHEMATICS; 

FOR THB 

USE OF ACADEMIES 

AS WILL AS 

PRIVATE TUITION. 



IN TWO VOLUMES* 



CHARLj^g; IWTTON,. LUR JFJtS. 

fcATS PROFESSOR OF ajA&XMATICS IN THE WfAL VlMTART ACADEMY . 

THE FIFTH AMENlfeKlV FROM THE NINTH 

WITH MANY COR.RjplC£i0r;S 4 AND IMPROVEMENTS. 

RV OUNTHUS GREGORY, LL.D. 

-CorretponSing Associate of the Academy of Dijon. Honorary Member of the Literary 
and Philosophical Society of New- York, of the New- York Historical Society, of the 
Literary and Philosophical, and the Antiquarian Societies of Newcastle opoa Tyoa, 
of the Cambridge Philosophical Society, of tho Institution of Civil Engineers, ate. fce. 
Secretary to the Astronomical Society of London, and Professor of TTithamatioa in 
wfaa Royal Military Academy. 



WITH THE ADDITIONS 
or 

ROBERT ADRAIN, LL.D. F A. P S. F.A.A.S, lie. 

And Professor of Mathematics and Natural Philosophy. 

THE WHOLE 

CORRECTED AND IMPROVED. 



VOL. I. 



NEW-YORK: . 

W. JL DZAJf, PR UTTER.' 
r. AMD I. SW0RD8; T. A. RONALDS ; COLLINS AND CO. ; COLLINS iNb RAN* 
NAT J H. AND C. AND H. CARVILL; WHITE, OALLARERj AND WHITE f 
a A. ROORBACHf AND M'CLRATB AND BANGS. 

1831. 




SiutiUr* JHrtrict o/ JrVw-Ter*, to wit I 
BE IT REMEMBERED, That oatb« 224 day of February, Ana© Domini ISSt, W. 
E. DEAN, of the Mid district, bath j i r w i ti4 U «kU edbce the title of a book, the title 
ef which U in the word* following, to wit: 

• A Count of Mathematics ; for the nse of Academies as well eg private toitioo. la 
Two Volomes. By Charles Hottoo, LL.D. F.R.S„ Uto Profeieor of Mathematics in 
the Royal Military Academy. The Fifth American from the Ninth London Edition, 
with many corrections and improvements. By Olinthus Gregory, LL.D. Correspond* 
Ing Associate of the Academy of Dijon, Honorary Member of the Literary and Philo- 
sophical Society of New- Fork, of the New- York Historical Society, of the Literary 
and Philosophical, and the Antiquarian Societies of Newcastle upon Tyno, of the 
Cambridge Philosophical Society, of the Institution of Clril Engineers, dec dec Secre- 
tary to the Astronomical Society of London, and Professor of Mathematics in the Ro- 
yal Military Academy. With the Additions of Robert Adrain, LL.D. F.A.P.8. F.A. 
AA, dec and Professor of Mathematics and Natural Philosophy. The whole correct* 
od and improved." 

Om right whereof ho claim as proprietor. In conformity with an Aat of Congress, e*> 
ttttt « A* Act to sjmb4 the atreral Acts respecting eopy-rights." 

FRED. J. BETTS, 
Cfcr* «/ we fiaHws Dirtrtrt e/AWFerfc . 



PREFACE. 



The present American edition is in part a re- 
print of the Ninth English edition by Dr. Oun- 
thus Gregory, with 'mQst.pf, the improvements 
introduced into former Al^ejricaa editions by Dr. 
Adrain, together with mich modifications of the 
English editions as appeared calculated to in- 
crease the general usefulness of the work. At 
the same time two or three Chapters, devoted 
to subjects of no great value at present to the 
American student, have been omitted, to leave 
room for matter of more interest and importance. 



CONTENTS 



OF 
VOL. I. 



Gin iRAt. Preliminary Principles I ] 

ARITHMETIC, 
ffotaiion and Numeration *j 
Roman Notation - . 7 
Addition g 
Subtraction - . - ][ 
Multiplication - 13 
Division - . - IS 

Reduction - 

Compound Addition • 32 

Commissioned Officers' Regimen 

talPay . 
Compound Subtraction * 

Multiplication 

Division - 
1 Rule, or Rule of Thr** ■ 



r _ jnd Proportion 
Vulgar Fractions 
Reduction of Vulgar Fractions 
Addition of Vulgar Fractions 
Subtra ct ion of VwTgar Fractions 
Multiplication t*f Vol fat Frtfst&ns'. ib. 
Di v i s ion of Vu *r*c«loas ? (ft 



Roto of Three ia Vulgar Fractions 65 
imnl Fractions * .«HSe 



Involution by Logarithms 
Evolution by Logvrtthntt 

ALGEBRA,. 

Definitions and ] 
Addition 
Subtraction 
Multiplication 
Division - 
Fractious 
Involution 
Evolution 
Surds * 
Arithmaticiil 



Pa. 159 
160 



Decimal . . 

Su1*ra™m ofDectmftlf. *- : " A i . ^ 
Multiplication of Decimal*- - . ,.ib* 
Division of Decimals* m -l *' ^.55 

Duodecimals - - 77 

Involution - - - 70 

Evolution - - 60 
To extract the Square Root - ft I 
To extract »he Cube Root - 85 
VTo extract any Root whatever • 8ti 
Table nf Powers and Roots - 9] 
Ratios, Proportions, end Progres- 
sion* - - . - Hi 
Ariihmetical Proportion - 112 
Geometrical Proportion - - 116 
Harmonical Proportion - 181 
Fellowship, or Partnership - 132 
Single Fellowship * - ib. 
Double Fellowship * - 125 
Simple Interest - - 127 
Compound Interest - - 13Q 
13* 

- 119 
136 

- las 

141 



- 16* 
166 

- 170 
172 

- 175 
179 

- 189 
192 

- 196 
id Pro- 

- 203 
207 

Geometrical Proportion and Pro- 
gression - - - 212 
Infinite Series, and their Summa- 
tion * 214 
Simple. Equations * • 231 
Q r vulca(jc Equal iona - . 249 
Xu&cTiftitl Higher Equations- 256 
. Sixfiih inlerest - - - 266 
,Coni pound Interest * - 267 

- 270 



gression 
Piles of Shot and Shells 



frfinhloas 



- 



Alligation Alternate 
Single Position 
Double Position * 
Practical Questions 

LOGARITHM*. 
Definition and Properties of Loga- 
rithms • - - .146 
To compute Logarithms 149 
Description and Use of Logarithms 15 
Multiplication by Logarithms - 157 
Division by Logarithms 15* 



GEOtt£TRT* 

- 275 
281 

- ib. 
Defi- 

.318 

Theorems <- 320 
Of Plane* and Sollds^Deftriitions 336 
Theorems - , . 340 

Problems * . 355 

Application of Algebra to Geome- 

iry - , . .371 
Problems » - 372 

Plane Trigonometry * - 378 
Trigonometrical Formula - 393 
Heights and Distances - - 396 
Mensuration otPianes or Areas 405 
Mensuration of Solids - * 420 
Land Surveying - • 430 
ArtifuerV Worls - - 459 

Timber Measuring - . 468 
Conic SecUoM - - -472 
Of the Ellipse - - 476 

Of the Hyperbole . -494 

Of lhe Parabola - - 518- 
Problems, &c, in Conic Sections - 534 
Equations of the Curve - 536 
Element* of Isoperiim try * - 539 
Surfaces - - - 541 
Solids • - - .551 
PracticaT Qiie^cmiinMantiiratlOB 562 
Logarithms of Numbers * • 571 
Table of Lorarithznic Sines, and 

590 



A 



COURSE 



OF 



MATHEMATICS, & c . 



GENERAL PRINCIPLES. 

1. Quantity, or Magnitude, is any thing that will 
admit of increase or decrease ; or that is capable of any sort 
of calculation or mensuration ; such as numbers, lines, space, 
time, motion, weight, &c. 

- 2. Mathematics is the science which treats of all kinds 
of quantity whatever, that can be numbered or measured.— 
That part which treats of numbering is called Arithmetic ; 
and that which concerns measuring, or figured extension, 
is called Geometry. — Not only these two, but Algebra and 
Fluxions, which are conversant about multitude, magnitude, 
form, and motion, being the foundation of all the other 
parts, are called Pure or Abstract Mathematics ; because 
they investigate and demonstrate the properties of abstract 
numbers and magnitudes of all sorts. And when these two 
parts are applied to particular or practical subjects, they 
constitute the branches or parts called Mixed Mathematics* 
— Mathematics is also distinguished into Speculative and 
Practical: viz. Speculative, when it is concerned in dis- 
covering properties and relations ; and Practical, when 
applied to practice and real use concerning physical objects. 

The peculiar topics of investigation in the four prmci^ %1 
departments of pure mathematics may be indicated Vw fo\u 

Vol. L 2 



2 



OEIfEKAL PRINCIPLES. 



words : viz. arithmetic by number, geometry by form, algtbrm 
by generality, fluxions by motion. 

3. In mathematics are several general terms or principles ? 
such as, Definitions, Axioms, Propositions, Theorems, Pro- 
blems, Lemmas, Corollaries, Scholia, &c. 

4. A Definition is the explication of any term or word in a 
science ; showing the 'sense and meaning in which the term 
is employed. — Every Definition ought to be clear, and ex* 
pressed in words that are common and perfectly well under* 
stood. 

5. A Proposition is something proposed to be demon- 
strated, or something required to be done ; and is accordingly 
either a Theorem or a Problem. 

6. A Theorem is a demonstrative Proposition ; in which 
some property is asserted, and the truth of it required to be 
proved. Thus, wpea R is said that, The sum of the three 
angles of a plane triangle is equal to two right angles, that is 
a Theorem, the truth of which is demonstrated by Geometry* 
— A set or collection of such Theorems constitutes a Theory, 

7. A Problem is a proposition or a question requiring 
something to be done ; either to investigate some truth or 
property, or to perform some operation. As, to find out the 
quantity or sum of all the three angles of any triangle, or to 
draw one line perpendicular to another. — A Limited Pro* 
blem is that which has but one answer or solution. An Un- 
limited Problem is that which has innumerable answers. 
And a Determinate Problem is that which has a certain nun*, 
her of answer*. 

8. Solution of a Problem, is the resolution or answer 
given to it. A Numerical or Numeral Solution, is the an. 
89 ex given in numbers. A Geometrical Solution, is the an- 
swer even by tl*e principles of Geometry. And a Mechani- 
cal Solution, is one whicji is gained by trials. 

9. A Lemma is a preparatory proposition, lajd down in 
ojrder U> shorten the demonstration of the main proposition 
which follows it. 

ip. A Corollary, or Conseetary, is a consequence draws 
imq^aWy from some proposition or other premises. 

11. A Scholium is a remark or observation nade upon 
some foregping proposition or premises. 

12. MMim,Qr Masim, is a self-evident prejwwuon ; 
requiring no formal deopnsteation to prove its truth ; but 
received and assented to as soon as mentioned. Such as, 
Tb* wlpole of any thing is. greater than a part of it ; or, The 
wjhoJ* i* equal to a)| its parts taken together ; or, Two quan. 
tides that are each of them equal to a third quantity, are 
equal to eaqh other, 



OSNK&AL AINCIPLE8. 



3 



13. A Postulate, or Petition, is something required to be 
done, which is so easy and evident that no person will hesi- 
tate to allow it. 

14. An Hypothesis is * stipulation assumed to be true, 
in order to argue from, or to found upon it the reasoning and 
demonstration of some proposition. 

19k Demonstration is the* collecting the several torments 
alri proofs, 4nd ttryiilg them together in proper order to shbifr 
the trtitt of the proposition under consideration, 

10» A Ihtreet, TbsUloe, of Affirmative D&ntviishiitiun', is 
thai *hr6h cotidrides with the direct and certain pttibf df th* 
prtjpoeitioh in hand. 

IT. 4ji Indira*, or irt&trtttxr Beinortstrdtioh, is that whicti 
shows a proposition to be true, by proving that some ab- 
aofdlry would hecessanly fblloW if the proposition advanced 
were false* This is also sometitties called Redbctio ad Afr- 
HftNAdSi ; because it shows the absurdity arid falsehood of 
alt suppositions cbritrarjr to that contained in the proposi- 
tion. 

18. Method is the art of disposing a train of arguments in 
a proper order, to investigate either the truth or falsity of 
a proposition, df to demons tra te it to others when it has been 
(bund oat.— 'this is either Analytical or Synthetical. 

19. Analysis or the Analytic Method, is the art or mode of 
finding out the truth of a proposition, by first supposing 
the thing to be done, and then reasoning back, step by step, 
till we arrive at some known truth. This is also called the 
Method of Invention, or Resolution ; and is that which is com- 
monly used in Algebra. 

20. Synthesis, or the Synthetic Method, is the searching 
out truth, by first laying down some simple and easy prin- 
ciples, and then pursuing the consequences flowing from 
them till we arrive at the conclusion. — This is also called 
the Method of Composition ; and is the reverse of the Ana* 
lytic method, as this proceeds from known principles to an 
unknown conclusion ; while the other goes in a retrograde 
order, from the thing sought, considered as if it were true, 
to some known principle or fact. Therefore, when any 
troth has been found out by the Analytic method, it may be 
demonstrated by a process in the contrary order, by Syn- 
thesis : and in the solution of geometrical propositions, it is 
Vety instructive to carry through both the analysis and the 
synthesis* 



4 



ARITHMETIC. 

Arithmetic is the art or science of numbering ; being 
that branch of Mathematics which treats of the nature and 
properties of numbers. — When it treats of whole numbers, 
it is called Vulgar, or Common Arithmetic ; but when of 
broken numbers, or parts of numbers, it is called Fractions. 

Unity, or an Unit, is that by which every thing is called 
one ; being the beginning of number ; as, one man, one 
ball, one gun. 

Number is either simply one, or a compound of several 
units ; as, one man, three men, ten men. 

An Integer, or Whole Number, is some certain precise 
quantity of units ; as, one, three, ten. — These are so called 
as distinguished from Fractions, which are broken num- 
bers, or parts of numbers ; as, one-half, two-thirds, or three- 
fourths. 

A Prime Number is one which has no other divisor than 
unity ; as 2, 3, 5, 7, 17, 19, &c. A Composite Number is 
one which is the product of two or more numbers ; as, 4* 6, 
8, 9, 28, &c. 



NOTATION AND NUMERATION. 

These rules teach how to denote or express any pro* 
posed number, either by words or characters : or to read 
and write down any sum or number. 

The Numbers in Arithmetic are expressed by the follow- 
ing ten digits, or Arabic numeral figures, which were intro- 
duced into Europe by the Moors, about eight or nine 
hundred years since ; viz. 1 one, 2 two, 3 three, 4 four, 
5 five, 6 six, 7 seven, 8 eight, 9 nine, cipher, or nothing. 
These characters or figures were formerly all called by the 
general name of Ciphers ; whence it came to pass that the 
art of Arithmetic was then often called Ciphering. The 
first nine are called Significant Figures, as distinguished 
from the cipher, which is of itself quite insignificant. 

Besides this value of those figures, they have also another, 
which depends on the place they stand in when joined to- 
gether ; as in the following table : • 



NOTATION AND KUXBBAITON. 



s 



-3 




6c. '9 8 7 6 5 4 3 2 1 
98 765432 
9 8 7 6 5 4 3 
9 8 7 6 5 4 
9 8 7 6 5 
9 8 7 6 
9 8 7 
9 8 
9 



Here, any figure in the first place, reckoning from right 
to left, denotes only its own simple value ; but that in the 
second place, denotes ten times its simple value ; and that in 
the third place, a hundred times its simple value ; and so 
on : the value of any figure, in each successive place, be- 
ing always ten times its former value. 

Thus, in the number 1796, the 6 in the first place denotes 
only six units, or simply six ; 9 in the second place signifies 
nine tens, or ninety ; 7 in the third place, seven hundred ; 
and the 1 in the fourth place, one thousand : so that the 
whole number is read thus, one thousand seven hundred and 
ninety- six. 

As to the cipher, 0, though it signify nothing of itself, yet 
being joined on the right-hand side to other figures, it in- 
creases their value in the same ten-fold proportion : thus, 5 
signifies only five ; but 50 denotes 5 tens, or fifty ; and 500 
is five hundred ; and so on. 

For the more easily reading of large numbers, they are 
divided into periods and half-periods, each half-period con. 
sisting of three figures ; the name of the first period being 
units ; of the secoud, millions ; of the third, millions of 
millions, or bi-millions, contracted to billions ; of the fourth, 
millions of millions of millions, or tri-millions, contracted to 
trillions, and so on. Also the first part of any period is so 
many units of it, and the latter part so many thousands. 



6 AUOTonnno. 

The following Table contains a summary of the whole 
doctrine. 



Periods. Quadril. ; Trillions ; Billions ; Millions ; Units. 



Half-per. 



th. un. th. un. th. un. th. un. th. un. 



Figures. (123,456 ; 789,098 ; 765,432 ; 101,234 ; 567,890. 



Numeration is the reading of any number in words that 
is proposed or set down in figures ; which will be easily done 
by help of the following rule, deduced from the foregoing 
tables and observations — viz. 

Divide the figures in the proposed number, as in the sum- 
mary, above, into periods and half-periods ; then begin at the 
left-hand side, and read the figures with the names set to 
them in the two foregoing tables. 



EXAMPLXfi. 



Express in words the following numbers ; viz. 



34 
96 
380 
704 
6134 
9028 



15080 
72003 
109026 
483500 
2500639 
7523000 



13405670 
47050023 
309025600 
4723507689 
274856390000 
6578600307024 



Notation is the setting down in figures any number 
proposed in words ; which is done by setting down the figures 
instead of the words or names belonging to them in the sum- 
mary above ; supplying the vacant places with ciphers 
where any words do not occur. 



EXAMPLES. 



Set down in figures the following numbers : 
Fifty-seven. 

Two hundred eighty-six. 
Nine thousand two hundred and ten. 
Twenty-seven thousand five hundred and ninety-four. 
Six hundred and forty thousand, four hundred and eighty-one. 
Three millions, two hundred sixty thousand, one hundred 
and six. 



NOTATION AND NUMXBATION. 



Four hundred and eight millions, two hundred and fifty-five 
thousand, one hundred and ninety-two. 

Twenty-seven thousand and eight millions, ninety-six thou- 
sand two hundred and four. 

Two hundred thousand and five hundred and fifty millions, 
one hundred and ten thousand, and sixteen. 

Twenty-one billions, eight hundred and ten millions, sixty, 
four thousand, one hundred and fifty. 

OF THE BOKAN NOTATION. 

The Romans, like several other nations, expressed their 
numbers by certain letters of the alphabet. The Romans 
used only seven numeral letters, being the seven following 
capitals : viz. i for one ; v for five ; x for ten ; l for fifty ; 
c for an hundred ; d for five hundred ; m for a thousand ; The 
other numbers they expressed by various repetitions and 
combinations of these, after the following manner : 



1 = 


i 




2 = 


ii 


As often as any character is re- 


3 = 


iii 


peated, so many times is its 






value repeated. 


4 = 


mi or iv 


A less character before a greater 


5 = 


V 


diminishes its value. 


6 = 


VI 


A less character after a greater 


7 = 


VII 


increases its value. 


8 = 


VIII 




9 = 


IX 




10 = 


X 




60 = 


L 




100 = 


C 




500 = 


d or io 


For every o annexed, this be- 






comes 10 times as many. 


1000 = 


m or cu 


For every c and o, placed one 


2000 = 


MM 


at each end, it becomes 10 






times as much. 


6000 = 


v or loo 


A bar over any number in- 


6000 = 


VI 


creases it 1000 fold. 


10000 = 


x or ccioo 




50000 = 


| l or 1003 




00000 = 


f- 




100000 = 


■ < c or ccciooo 





1000000 = nor CCCCI0O33 
2000000 = ra 



9 



ARITHMETIC. 



EXPLANATION OF CERTAIN CHARACTERS* 

There are various characters or marks used in Arithmetic, 
and Algebra, to denote several of the operations and propo- 
sitions ; the chief of which are as follow : 

+ signifies plus, or addition. 
— - . minus, or subtraction. 
X or - multiplication. 
H- - division, 
t :: : - proportion. 
«=* - T equality. 
«v/ - - square root. 
%/ - - cube root, &c. 

*r . . diff. between two numbers when it is not known 
which is the greater. 

Thus, 

5 + 3, denotes that 3 is to be added to 5. 

6 — 2, denotes that 2 is to be taken from 6. 

7 X 3, or 7 • 3, denotes that 7 is to be multiplied by 3. 

8 -r- 4, denotes that 8 is to be divided by 4. 

2 : 3 : : 4 : 6, shows that 2 is to 3 as 4 is to 6. 

6 + 4 = 10, shows that the sum of 6 and 4 is equal to 10. 

or 3^, denotes the square root of the number 3. 

$/5, or 5^, denotes the cube root of the number 5. 
7 s , denotes that the number 7 is to be squared. 
S 3 , denotes that the number 8 is to be cubed, 
dec. 



OF ADDITION. 

Addition is the collecting or pitting of several numbers 
together, in order to find their sum, or the total amount of the 
whole. This is done as follows : 

Set or place the numbers under each other, so that each 
figure may stand exactly under the figures of the same value, 



ADDITION* 



9 



that is, units under units, tens under tens, hundreds under 
hundreds, dec. and draw a line under the lowest number, to 
separate the given numbers from their sum, when it is found. 
— Then add up the figures in the column or row of units, 
and find how many tens are contained in that sum. — Set 
down exactly below, what remains more than those tens, or 
if nothing remains, a cipher, and carry as many ones to the 
next row as there are tens. — Next add up the second row, 
together with the number carried, in the same manner as the 
first. And thus proceed till the whole is finished, setting 
down the total amount of the last row. 



TO PROVE ADDITION. 



First Method. — Begin at the top, and add together all the 
rows of numbers downwards, in the same manner as they 
were before added upwards ; then if the two sums agree, it 
may be presumed the work is right. — This method of proof 
is only doing the same work twice over, a little varied. 

Second Method. — Draw a line below the uppermost num. 
ber, and suppose it cut off. — Then add all the rest of the 
numbers together in the usual way, and set their sum under 
the number to be proved. — Lastly, add this last found num- 
ber and the uppermost line together ; then if their sum be 
the same as that found by the first addition, it may be pre- 
sumed the work is right. — This method of proof is founded 
on the plain axiom, that " The whole is equal to all its parts 
taken together." 

Third Method. — Add the figures in 
the uppermost line together, and find example i. 
how many nines are contained in 
their sum. — Reject those nines, and 3497 g 5 
set down the remainder towards the 6512 .g 5 
right hand directly even with the 8295 £ 6 

figures in the line, as in the annexed — — o 

example. — Do the same with each of 18304 8 7 
the proposed lines of numbers, set- — — g — 
ting all these excesses of nines in a co- W 
lumn on the right-hand, as here 5, 5, 0. Then, if the excess 
of 9's in this sum, found as before, be equal to the excess 
of 9's in the total sum 18304, the work is probably right. — 
Thus, the sum of the right-hand column, 5, 5, 6, is 16, the 
excess of which above 9 is 7. Also the sum of the figure* V& 

Vol. I. 3 



10 



ABZTHMSTXC. 



the sum total 18304, is 16, the excess of which above 9 m 
also 7> the same as die former*. 



OTHER EXAHPLES. 



2. 


3. 


4. 


12345 


12345 


12345 


87890 


67890 


876 


98765 


9876 


9087 


432)0 


543 


56 


12345 


21 


234 


67890 


9 


1012 


302445 


90684 


23610 


290100 


78339 


11265 


302445 


90684 


23610 



Ex. 5. Add 3426 ; 9024 ; 5106 ; 8890 ; 1204, together. 

Ans. 27650. 

6. Add 509267; 235809; 72920 ; 8392; 420 ; 21; and 
9, together. Ans. 82683a. 



* This method of proof depends on a property of the number 9, 
which, except the number 3 ; belongs to no other digit whatever ; 
namely* that 44 any number divided by 9, will leave the same remain- 
der as the sum of its figures are digits divided by 9:" which may be 
demonstrated in this manner. 

Demonstration. Let there be any number proposed, as 4668. This, 
separated into its several parts, becomes, 4000 + 600+60 4- 8. Bnt 
4000 = 4 X 1000 = 4 X (999 + 1) = (4 X 999) + 4. In like man- 
ner 600 = (6 X 99) +6 ; and 50 = (5 X 9) 4-5. Therefore the gi- 
ven number 4658 = (4 X 999) + 4 + (6 X 99) +6 +(5 X9) + 6 + 
8 = (4 X 999) + (6 X 99) + (5 X 9) + 4 + 6 + 5 + 8; and 
4658 + 9 = (4 X 999+6 X 99 + 5 X 9 + 4+ 6 + 5 +8) + 9. But 
(4 X 999) + (6 X 99; + (5 X 9) is evidently divisible by 9, without 
a remainder ; therefore if the given number 4658 be divided by 9, 
it will leave the same remainder as4+6+6+6 divided by 9. And 
the same, it is evident, will hold for any other number whatever. 

In like manner, the same property may be shown to belong to the 
number 3 ; but the preference is usually given to the number 9, on ac- 
count of its being more convenient in practice. 

Now, from the demonstration above given, the reason of the rale it- 
self is evident : for the eicess of 9's in two or more numbers being 
taken separately, and the excess of 9's taken also out of the sum of the 
former excesses, it is plain that this last excess must be equal to the ex- 
cess of 9's contained in the total sum of all these numbers ; all the parts 
taken together being equal to the whole.— This rule was first given by 
Dr. Waflis in his Arithmetic, published in the year 1657. 



•mmuonoN. 11 

7. Add 2; 19; 817; 4298 ; 50916 ; 730205 ; 91806% 
together. Ans. 9966891. 

8. How many days are in the twelve calendar months ? 

Ana. 965. 

9. How many days are there from the. 15th day of April to 
die 24th day of November, both days included f Ans, 224 % 

10. An army consisting of 52714 infantry*, or fbot, 51 HI 
horse, 6250 dragoons, 3927 light-horse, 928 artillery, of 
pinners, 1410 pioneers, 250 sappers, and 406 miners : what 
is the whole number of men ? Ans. 70995. 



OF SUBTRACTION. 

Subtraction teaches to find how much one number ex- 
ceeds another, called their difference, or the remainder, by 
taking the less from the greater. The method of doing which 
is as follows: 

Place the less number under the greater, in the same man- 
ner as in Addition, that is, units under units, tens under tens, 
and so on ; and draw a line below them. — Begin at the right 
hand, and take each figure in the lower line, or number, 
from the figure above it, setting down the remainder below 
it. — But if the figure in the lower line be greater than that 
above it, first borrow, or add, 10 to the upper one, and then 
take the lower figure from that sum, setting down the remain- 
der, and carrying 1, for what was borrowed, to the next 
lower figure, with which proceed as before ; and so on till the 
whole is finished. 



* The whole body of foot soldiers is denoted by the word htfanlrg; 
and all those that charge on horseback by the word Cavalry. — Some 
authors conjecture that the term infantry is derived from a certain In- 
fanta of Spain, who, finding that the army commanded by the king 
her lather had been defeated by the Moors, assembled n body of the 
people together on foot, with which she engaged and totally rooted the 
enemy. In honour of this event, and to distinguish the foot soldiers, 
who were not before held In much estimation, they received the name 
of Infantry, from her own title of Infanta. 



4 



12 



AxrrBXxnc. 



TO PSOVE SUBTRACTION. 

Add the remainder to the less number, or that which is 
just above it ; and if the sum be equal to the greater or up- 
permost number, the work is right*. 



EXAMPLES. 



From 5386427 
Take 2164315 



From 5386427 
Take 4258792 



From 1234567 
Take 702973 



Rem. 3222112 



Rem. 1127635 



Rem. 531594 



Proof. 5366427 



Proof. 5386427 



Proof. 1234567 



4. From 5331806 take 5073918. Ans. 257888. 

5. From 7020974 take 2766809. Ans. 4254 165. 

6. From 8503402 take 574271. Ans. 7929131. 

7. Sir Isaac Newton was born in the year 1642, and he 
died in 1727 : how old was he at the time of his decease ? 

Ans. 85 years. 

8. Homer was born 2560 years ago, and Christ 1827 years 
ago : then how long before Christ was the birth of Homer ? 

Ans. 733 years. 

9. Noah's flood happened about the year of the world 1656, 
and the birth of Christ about the year 4000 : then how long 
was the flood before Christ ? Ans. 2344 years. 

10. The Arabian or Indian method of notation was first 
known in England about the year 1150: then how long is 
it since to this present year 1827 1 Ans. 677 years. 

11. Gunpowder was invented in the year 1330 : how long 
was that before the invention of printing, which was in 1441 ? 

Ans. Ill years. 

12. The mariner's compass was invented in Europe in the 
year 1302 : how long was that before the discovery of Ame- 
rica by Columbus, which happened in 1492 ? 

Ans. 190 years. 



* The reason of this method of proof is evident; for if the difference 
of two jpombers be added to the less, it most manifestly make up a sum 
equal to the greater. 



jcttltolicatioh. 18 



OF MULTIPLICATION. 

Multiplication is a compendious method of Addition, 
teaching how to find the amount of any given number when 
repeated a certain number of times ; as, 4 times 6, which 
is 24. 

The number to be multiplied, or repeated, is called the 
Multiplicand. — The number you multiply by, or the number 
of repetitions, is the Multiplier. — And the number found, 
being the total amount, is called the Product. — Also, both 
the multiplier and multiplicand are, in general, named the 
Terms or Factors. 

Before proceeding to any operations in this rule, it is 
necessary to learn off very perfectly the following Table, of 
all the products of the first 12 numbers, commonly called 
the Multiplication Table, or sometimes Pythagoras's Table, 
from its inventor. 



MULTIPLICATION TABLE. 



JJ 


2| 


3 


4 


5 


6 


71 


8 




3 


11 


12 


2 


4 


6 


8 


JO 


12 


14 


LO 


18 


20 


S3 


24 


3 


6 


9 




15 


18 


21 


34 


27 


30 


33 


361 


4 


8 


12 


16 


20 


24 


28 


32 


36 


40 


44 


48 


5 


10 


15 1 


20 


25 


30 


35 


40 


45 


50 


55 


60 


6 


12 


18 


24 


30 


36 


42 


48 


54 


60 66 


72 


7 


14 


21 


28 


35 


42 


49 


56 


63 


70 


77 


84 


8 


16 


24 


32 


40 


48 


56 


64 


72 


80 


88 


96 


9 


18 


27 


36 


45 


54 


63 


72 


81 


130 


99 


108 


10 


20 


30 


40 


50 


60 


70 


80 


00 


100 


110 


120 


11 


22 


33 


44 


55 


66 


77 


88 


99 


110 


121 


132 


12 


24 


36 


48 


00 


72 


84 


96 


10s|l20 


132 


144; 



14 



ARITHMETIC* 



To multiply any Given Number by a Single Figure, or by any 
Number not exceeding 12. 

* Set the multiplier under the unite 9 figure or right-hand 
place, of the multiplicand, and draw a line below it. — Then, 
beginning at the right-hand, multiply every figure in this 
by the multiplier. — Count how many tens there are in the 
product of every single figure, and set down the remainder 
directly under the figure that is multiplied ; and if nothing 
remains, set down a cipher. — Carry as many units or ones as 
mere are tens ' counted, to the product of the next figures ; 
and proceed in the same manner till the whole is finished. 



EXAMPLE* 

Multiply 9876543210 the Multiplicand. 
By 2 the Multiplier. 



19753066420 



To multiply by a Number consisting of Several Figures. 

f Set the multiplier below the multiplicand, placing them 
as in Addition, namely, units under units, tens under tens, &c. 
drawing a line below it. —Multiply the whole of the multi- 
plicand by each figure of the multiplier, as in the last article ; 



6678 

* The reason of this rale is the same as for 4 
the process in Addition, in which 1 is car- — — 
ried for every 10, to the next place, gradu- 32 = 8X4 
ally as the several products are produced 280 = 70 X 4 
one after another, instead of setting them all 2400 = 600 X 4 
down below each other, as in the annexed ex- 20000 = 6000 X 4 

ample. ■ 

22712 = 5678 X 4 



t After having found the product of the multiplicand by the first 
figure of the multiplier, as in the former case, the multiplier is supposed 
to be divided into parts, and the product is found for the second figure 
in the same manner: but as this figure stands in the place of tens, the 
product must be ten times <jts simple value ; and therefore the first 
figure of this product must be set in the place of tens ; or, which is the 
same thing, directly under the figure multiplying by. And proceeding 



MULTIPLICATION. 



15 



setting down a line of products for each figure in the multi- 
plier, oo as that the first figure of each line may stand straight 
under the figure multiplying by. Add all the lines of pro* 
ducts together, in the order in which they stand, and their 
sum will be the answer or whole product required. 

TO PROVE MULTIPLICATION. 

There are three different ways of proving multiplication, 
which are as below : 

First Method. — Make the multiplicand and multiplier 
change places, and multiply the latter by the former in the 
same manner as before. Then if the product found in this 
way be the same as tho former, the number is right. 

Second Method.—* Cast all the 9's out of the sum of the 
figures in each of the two factors, as in Addition, and set 
down the remainders. Multiply these two remainders to- 
gether, and cast the 9's out of the product, as also out of the 
whole product or answer of the question, reserving the re- 
mainders of these last two, which remainders must be equal 
when the work is right. — Note, It is common to set the four 
remainders within the four angular spaces of a cross, as in 
the example below. 



In this manner separately with all the 1234567 the multiplicand, 
figures of the multiplier, it is evident 4567 

that we shall multiply all the parts of 

the multiplicand by all the parts of 8641969— 7 times the mult, 
the multiplier, or the whole of the 7407402 = 60 times ditto, 
multiplicand by the whole of the mul- 6172b35 = 500 times ditto, 
tiplier: therefore these several pro- 4938268 =4000 times ditto. 

ducts being added together, will be 

equal to the whole required product ; 6638267489— 4667 times ditto, 
as in the example annexed. 

* This method of proof is derived from the peculiar property of the 
number 9, mentioned in the proof of Addition, and the reason for the 
one includes that of the other. Another more ample demonstration of 
this rule may, however, be as follows : — Let p and q denote the number 
of 9's in the factors to be multiplied, and a and 6 what remain ; then 9p 
-4- a and 9q + 6 will be the numbers themselves, and their product ia 
(9r X 9q) + (9p X b) + (9q X fl) + (« v 6 J ; but the first three of these 
products are each a precise number of 9's, because their factors are so, 
either one or both : these therefore being cast away, there remains only 
II X 6; and if the 9's also be cast out of this, the excess is the excess of 
9's in the total product : but a and b are the eicesses in the factors 
themselves, and a X b is their product ; therefore the rule is true. This 
mode of proof, however, is not an ample check against the errors that 
night arise from a transposition of figures. 



16 



ARITHMETIC. 



Third Method. — Multiplication is also very naturally prov- 
ed by Division ; for the product divided by either of the fac- 
tors, will evidently give the other. But this cannot be prac- 
tised till the rule of division is learned. 



EXAMPLES* 



Mult 3542 or Mult. 6190 

by 6196 Proof. by 3542 



21252 \ / 12392 

31878 \2/ 24784 

3542 30980 

21252 X *\ 18588 



21946232 21946232 Proof! 



OTHER EXAMPLES. 



Multiply 123456789 
Multiply 123456789 
Multiply 123456789 
Multiply 123456789 
Multiply 123456789 
Multiply 123456789 
Multiply 1234567b9 
Multiply 123456789 
Multiply 123456789 
Multiply 302914603 
Multiply 273580961 
Multiply 402097316 
Multiply 82164973 
Multiply 7564900 
Multiply 8496427 
Multiply 2760625 



by 3. 
by 4. 
by 5. 
by 6. 
by 7. 
by 8. 
by 9. 
by 11. 
by 12. 
by 16. 
by 23. 
by 195. 
by 3027. 
by 579. 
by 874359. 
by 37072. 



Ans. 370370367. 
Ans. 493827156. 
Ans. 617288945. 
Ans. 740740734; 
Ans. 864197523. 
Ans. 987654312. 
Ans. 1111111101. 
Ans. 1358024679. 
Ans. 1481481468. 
Ans. 4846633648. 
Ans. 6292362103. 
Ans. 78408976620. 
Ans. 248713373271. 
Ans. 4380077100. 
Ans. 7428927415293. 
Ans. 102330768400. 



CONTRACTIONS IN MULTIPLICATION. 

I. When there are Ciphers in the Factors. 

If the ciphers be at the right-hand of the numbers ; mul- 
tiply the other figures only, and annex as many ciphers to 
the right-hand of the whole product, as are in both the fac- 
tors. — When the ciphers are in the middle parts of the mul- 
tiplier ; neglect them as before, oniy taking care to place 



MULTIPLICATION. 



17 



the first figure of every line of products exactly under the 
figure you are multiplying with. 



EXAMPLES. 



1. 

Mult. 9001035 
by - 70100 



9001635 
63011445 



2. 

Mult. 390720400 
by . 406000 

23443224 
15628816 



631014613500 Products 158632482400000 



3. Multiply 81503600 by 7030. Ans. 572970308000. 

4. Multiply 9030100 by 2100. Ans. 18963210000. 

5. Multiply 8057069 by 70050. Ans. 564397683450. 

II. When the Multiplier is the Product of iwo or more Num- 
bers in the Table ; then 

* Multiply by each of those parts separately, instead of 
the whole number at once. 



EXAMPLES. 

1. Multiply 51307298 by 56, or 7 times 8. 
51307298 
7 



359151086 
8 



2873208088 



2. Multiply 31704592 by 

3. Multiply 29753804 by 

4. Multiply 7128368 by 

5. Multiply 160430800 by 

6. Multiply 61835720 by 



36. Ans. 1141365312. 

72. Ans. 2142273888. 

96. Ans. 684323328. 

108. Ans. 17326526400. 

1320. Ans. 81623150400. 



* The reason of this rule is obvious enough ; for any number multi- 
plied by the component parts of another, must give the same product 
as if it were multiplied by that number at once. Thus, in the 1st ex- 
ample, 7 times the product of 8 by the given number, makes 66 time! 
the same number, as plainly as 7 times 8 make 5o\ 

Vol. I. 4 



is 



ARITHMETIC. 



7. There was an army composed of 104 4 battalions, eacb 
consisting of 500 men ; what was the number of men con- 
tained in the whole ? Ans. 52000. 

8. A convoy of ammunition f bread, consisting of 250 
waggons, and each waggon containing 320 loaves, having 
been intercepted and taken by the enemy, what is the num- 
ber of loaves lost ? Ans. 80000. 



OF DIVISION. 

Division is a kind of compendious method of Subtrac- 
tion, teaching to find how often one number is contained in 
another, or may be taken out of it : which is the same thing. 

The number to be divided is called the Dividend*— 
The number to divide by, is the Divisor. — And the number 
of times the dividend contains the divisor, is called the 
Quotient. — Sometimes 'there is a Remainder left, after the 
division is finished. 

The usual manner of placing the terms, is, the dividend 
in the middle, having the divisor on the left hand, and the 
quotient on the right, each separated by a curve line; as, 
to divide . 12 by 4, the quotient is 3, 

Dividend 12 
Divisor 4) 12 (3 Quotient; 4subtr. 
showing that the number 4 is 3 times — 
contained in 12, or may be 3 times 8 
subtracted out of it, as in the margin. 4 subtr. 

% Ride.— Having placed the divisor — 
before the dividend, as above directed, 4 
find how often the divisor is contained 4 subtr. 

in as many figures of the dividend as — 
are just necessary, and place the num- 
ber on the right in the quotient. Mul- — 



* A battalion is a body of foot, consisting of 500, or 000, or 700 men, 
more or less. 

t The ammunition bread, is that which is provided for, and distribut- 
ed to, the soldiers ; the usual allowance being a loaf of 6 pounds to 
•vary soldier, once in 4 days. 

X In this way the dividend is resolved into parts, and by trial is found 
how often the divisor is contained in each of those parti, one after an- 
other, arranging the several figures of the quotient one after another, 
into one number. 



19 



tiply the divisor by this number, and set the product under 
the figures of the dividend, before-mentioned. — Subtract this 
product from that part of the dividend under which it stands, 
and bring down the next figure of the dividend, or more if 
necessary, to join oh the right of the remainder.— Divide this 
numbef, so increased, in the same manner as before ; and so 
on, till all the figures are brought down and used. 

Note* If it be necessary to bring down more figures than 
one to any remainder, in order to make it as large as the di- 
visor, or larger, a cipher must be set in the quotient for every 
figure so brought down more than one. 

TO PROVE DIVISION. 

* Multiply the quotient by the divisor ; to this product 
add the remainder, if there bo any ; then the sum will be 
equal to the dividend, when the work is right. 



When there is no remainder to a division, the quotient is the whole 
and perfect answer to the question. But when there is a remainder, it 

res so much towards another time, as it approaches to the divisor : so, 
the remainder be half the divisor, it will go the half of a time more ; 
if the 4th part of the divisor, it will go one-fourth of a time more ; and 
so on. Therefore, to complete the quotient, set the remainder at (he 
end of it, above a small line, and the divisor below it, thus forming a 
fractional part of the whole quotient. 

* This method of proof is plain enough: for since the quotient is the 
number of times the dividend contains the divisor, the quotient multi- 
plied by the divisor must evidently be equal to the dividend. 

There are several other methods sometimes used for proving Divi- 
sion, some of the most useful of which are as follow: 

Second Method Subtract the remainder from the dividend, and di- 
vide what is left by the quotient ; so shall tbe new quotient from this 
last division be equal to the former divisor, when the work is right. 

Third Method. — Add together the remainder and all the products of 
the several quotient figures by the divisor, according to the order in 
which they stand in the work ; and the sum will be equal to the divi- 
dend, when the work is right 



to 



abithottic. 



examples. 



1 QuoU 
3) 1234567 ( 411522 
12 mult. 3 



3 
3 



15 
15 



1234566 
add 1 



4 1234567 
3 



Proof. 



2 Quot, 
37) 12345678 ( 333666 
111 37 



124 
111 

135 
111 



2335662 
1000996 
rem. 36 

12345G78 



246 Proof. 
222 



6 
6 



247 
222 



7 
6 



258 
222 



Rem. 1 



Rem. 36 



Divide 73146085 by 4. 
Divide 53179b6027 by 7. 
Divide 570196382 by 12. 
Divide 74638105 
Divide 137896254 
Divide 35821649 
Divide 72091365 



Ana. 18286521J. 
Ans. 759712289$. 
Ans. 47516365/,. 
Ans. 2017246,^. 
Ans. 142I6l6f$. 
Ans. 46886}Jf . 



An*. 13*61 § y±\, 



by 37. 
by 97. 
by 764. 
by 5201. 

10. Divide 4637064283 by 57606. Ans. 80496ftf£f 

11. Suppose 471 men are formed into ranks of 3 deep, 
what is the number in each rank? Ans. 157. 

12. A party, at the distance of 378 miles .from the head 
quarters, receive orders to join their corps in 18 days : what 
number of miles must they march each day to obey their 
orders? Ans. 21. 

13. The annual revenue of a nobleman being 37960Z. ; 
how much per day is that equivalent to, there being 365 days 
in the year ? Ans. 104Z. 



CONTRACTIONS IN DIVISION. 



There are certain contractions in Division, by which the 
operation in particular cases may be performed in a shorter 
manner : as follows : 



DIVISION. 



21 



I. Division by any Small Number, not greater than 12, may- 
be expeditiously performed, by multiplying and subtracting 
mentally, omitting to set down the work except only the quo- 
tient immediately below the dividend. 

BXAMPLXS. 

3) 56103061 4)52610675 5) 1370192 



Quot. 18701320$ 



6) 38672040 7) 81306627 8) 23718020 



9) 43081062 11) 576 14230 12) 27080373 



II. * When Ciphers are annexed to the Divisor'; cut off 
those ciphers from it, and cut off the same number of figures 
from the right-hand of the dividend ; then divide with the re- 
maining figures, as usual. And if there be any thing remain- 
ing after this division, place the figures cut off from the di- 
vidend to the right of it, and the whole will be the true re- 
mainder ; otherwise, the figures cut off only will be the re- 
mainder. 

EXAMPLES. 

1. Divide 3704196 by 20. 2. Divide 31086901 by 7100, 
2,0) 3*/04l9,6 . 71,00) 310869,01 (4378$Hft. 

284 

Quot. 185209$} 

268 
213 

556 
497 

599 

568 

31 



* This method serves to avoid a needless repetition of ciphers, wVncb 
would happen in the common way. And the truth of the pnivdpYt ou 



82 



ARITHMETIC 



3. Divide 7380064 by 23000. 

4. Divide 2304109 by 5800. 



Ana. 320}f£f 
Ant. 397ifiJ. 



III. When the Divisor is the exact Product of two or more 
of the small Numbers not greater than 12 : * Divide by each 
of those numbers separately, instead of the whole divisor at 
once. 

Note. There are commonly several remainders in work- 
ing by this rule, one to each division ; and to find the true or 
whole remainder, the same as if the division had been per- 
formed all at once, proceed as follows : Multiply the last re. 
mainder by the preceding divisor, or last but one, and to the 
product add the preceding remainder ; multiply this sum by 
the next preceding divisor, and to the product add the next 
preceding remainder ; and so on till you have gone backward 
through all the divisors and remainders to the first. As in 
the example following : 



EXAMPLES. 



1. Divide 31046835 by 56 or 7 times 8. 



7) 31046835 

8) 4435262—1 first rem. 

554407 — 6 second rem. 



Ans. 554407}} 



6 the last rem. 
mult. 7 preced. divisor. 

42 

add 1 to the 1st rem. 
43 whole rem. 



2. Divide 7014596 by 72. 

3. Divide 5130652 by 132. 

4. Divide 83016572 by 240. 



Ans. 974244f. 
Ans. 38808 T W- 
Ans. 345902,VV- 



which it is founded, is evident ; for, cutting off the same number of 
ciphers, or figures, from each, is (he same as dividing each of them by 
10, or 100, or 1000, &c. according to the number of ciphers cut off; 
and it is evident, that as often as the whole divisor is contained in the 
whole dividend, so often must any part of the former be contained in a 
like part of the latter. 

* This follows from the second contraction in Multiplication, being 
only the converse of it ; for the half of the third part of any thing, is 
evidently the same as the sixth part of the whole ; and so of any other 
numbers. — The reason of the method of finding the whole remainder 
from the Several particular ones, will best appear from the nature of 
Vulgar Fractions. Thus, in the first example above, the first remainder 
being 1, when the divisor is 7, makes ; this must be added to the se- 
cond remainder, 6, making 6+ to the divisor 8, or to be divided by 8. 

Bat 6f =^|±?= y; and this divided by 8 gives ^ g -^- 



REDUCTION. 



28 



• IV. Common Division may be performed more concisely, 
by omitting the several products, and setting down only the 
remainders; namely, multiply the divisor by the quotient 
figures as before, and, without setting down the product, 
subtract each figure of it from the dividend, as it is produced ; 
always remembering to carry as many to the next figure as 
were borrowed before. 



EXAMPLES. 

1. Divide 3104679 by 833. 
833) 3104679 (3727# 5 . 
6050 
2257 
5919 



2. Diyide 79165238 by 238. 

3. Divide 29137062 by 5317. 

4. Divide 62015735 by 7803. 



Ans. 832627,^. 
Ans. 5479f|{J. 
Ans. 79474f f f. 



OF REDUCTION. 

Reduction is the changing of numbers from one name 
or denomination to another, without altering their value. — 
This is chiefly concerned in reducing money, weights, and 
measures. 

When the numbers are to be reduced from a higher name 
to a lower, it is called Reduction descending ; but when, 
contrarywise, from a lower name to a higher, it is Reduction 
ascending. 

Before we proceed to the rules and questions of Reduction, 
it will be proper to set down the usual tables of money, 
weights, and measures, which are as follow : 



OF MONEY, WEIGHTS, AND MEASURES. 



TABLES OF MONET. 



2 Farthings = 1 Halfpenny | 
4 Farthings = 1 Penny d 
12 Pence = 1 Shilling s 
20 Shillings = 1 Pound £ 



qrs d 

4=1 s 
48 = 12 = 1 £ 
960 = 240 = 20 = \ 



24 



ASXTHMBTIC. 



PBNCB TABLE. SHILLINGS TABLE* 



d 




8 


d 


t 




d 


20 


is 


1 


8 


1 


is 


12 


30 




2 


6 


2 




24 


40 




3 


4 


3 




36 


50 




4 


2 


4 




48 


60 




5 





5 




60 


70 




5 


10 


6 




72 


80 




6 


8 


7 




84 


90 




7 


6 


8 




96 


100 




8 


4 


9 




108 


110 




9 


2 


10 




120 


120 




10 





11 




132 



/Vote.— £ denotes pounds, « shillings, and d denotes pence. 

| denotes 1 farthing, or one quarter of any thing. 

£ denotes a halfpenny, or the half of any thing. 

} denotes 3 farthings or three quarters of any thing. 
The fall weight and value of the English gold and silver coin, 
old and new, are as here below. 



both 



Gold. 

Guinea 
Half do. 
Third do. 
Double Sov. 
Sovereign 
Half do. 



Value 
£ $ d 

1 1 
10 

7 

2 

1 
10 



Weight, 
dwtgr 

'6 h 

2 16} 

1 l*t 
10 6+t 

5 6A 

2 13-fr 



Silver. 



Value, 
s d 



A Crown 5 
Half-crown 2 
Shilling 1 
Sixpence 



Old Wt 
dwt gr 
19 8* 
9 I6t 
3 21 
1 22A 



New Wt. 

dwt gr. 

18 4^ 
9 2^ 
3 15* 
119# 



The usual value of gold is nearly 41 an ounce, or 2d a grain ; and 
that of silver is nearly 5* an ounce. Also the value of any quantity of 
gold, was to the value of the same weight of standard silver, as l^rir 
to 1, in the old coin ; bdt in the new coin they are as 14^ to 1. 

Pure gold, free from mixture with other metals, usually called fine 

Sold, is of so pure a nature, that it will endure the fire without wasting, 
lough it be kept continually melted. But silver, not having the purity 
of gold, will not endure the fire like it : yet fine silver will waste but 
very little by being in the fire any moderate time ; whereas copper, 
tin, lead, kc. will n#t only waste, but may be calcined, or burnt to a 
powder. 

Both gold and silver, in their purity, are so soft and flexible (like new 
lead, &c.) that they are not so useful, either in coin or otherwise (ex- 
cept to beat into leaf gold or silver), as when they are alloyed, or mix- 
ed and hardened with copper or brass. And though most nations differ, 
more or less, in the quantity of such alloy, as well as in the same place 
at different times, yet in England the standard for gold and silver coin 
has been for a long time as follows— viz. That 22 parts of fine gold, 
and 2 parts of copper, being melted together, shall be esteemed the 
true standard for sold coin : And that 11 ounces and 2 pennyweights 
of fine silver, and 18 pennyweights of copper, being melted together, 
be esteemed the true standard for silver coin, called Sterling silver. 

In the old coin the pound of sterling gold was coined into 42| gui- 
neas, of 21 shillkip each, of which the pound of sterling silver was di- 
vided into 02. Tie new coin is also of the same quality or degree of 



TABUS OF WEIGHTS. 2$ 



TKOY WEIGHT*. 

Grains - - marked gr 
24 Grains make 1 Pennyweight dwt 
20 Pennyweightsi Ounce oz 
12 Ounces 1 Pound lb 



fr dwt 
4= I oz 
480= 20= 1 lb 
5760=240=12=1 



By this weight are weighed Gold, Silver, and Jewels. 

APOTHECARIES' WEIGHT. 

Grains - - marked gr , 
20 Grains make 1 Scruple sc or 3 
3 Scruples 1 Dram dr or 3 
8 Drams 1 Ounce oz or J 
12 Ounces 1 Pound lb or ft 





8C 








20 = 


1 


dr 






60 = 


3 = 


1 


02? 




480 = 


24 = 


8 = 


1 


lb 


5760 = 


288 = 


96 = 


12 = 


1 



This is the same as Troy weight, only having some dif- 
ferent divisions. Apothecaries make use of this weight in 
compounding their medicines ; but they buy and sell their 
Drugs by Avoirdupois weight. 



AVOIHDUPOIS WEIGHT. 



Drams 
16 Drams 
16 Ounces 
28 Pounds 
4 Quarters 



make 1 Ounce 
- 1 Pound - 
1 Quarter - 
1 Hundred weight 



marked dr 
oz 
lb 

cwt 



20 Hundred Weight 1 Ton 



ton 



fineness with that of the old sterling gold and silver above described, but 
divided into pieces of other names or values ; viz. the pound of the sil- 
ver into 66 shillings, of course each shilling is the 66th part of a pound ; 
and 20 pounds of the gold into 934£ pieces called sovereigns, or the 
pound weight into 4Gf« sovereigns, each equal to 20 of the new shil- 
lings. So that the weight of the sovereign is 46^$Ujs of a pound, 
which is equal to pennyweights, or equal to 5 dwt. 3^ gr. very 
nearly, as stated in the preceding table. And multiples and parts of the 
sovereign and shilling in their several proportions. 

* The original of all weights used in England, was a grain or corn 
of wheat, gathered out of the middle of the ear, and, being well dried, 
SI of them were to make one pennyweight, 20 penny weigjhU out 

Vol. I. 5 



26 



ARITHMETIC. 



dr oz 
16 = 1 lb 

256 = 16 = 1 qr 

7168 = 448 = 28= 1 ewi 

28672 = 1752 = 112 = 4 = 1 Urn 

573440 = 35840 = 2240 = 80 = 20 = 1 

By this weight are weighed all things of a coarse or drossy 
nature, as Corn, Bread, Butter, Cheese, Flesh, Grocery 
Wares, and some Liquids ; also all metals, except Silver 
and Gold. 

oz dwt gr 

Note, that lib Avoirdupois = 14 11 15J Troy 
\oz - - = 18 
Idr . - =013} 



LONG MEASURE. 



2 Barley-corns make 
12 Inches - 

3 Feet 

6 Feet - 
5 Yards and a half 
40 Poles 
8 Furlongs 
3 Miles 

69 iV Miles nearly 



1 Inch - 
1 Foot - 
1 Yard - 
1 Fathom 
1 Pole or Rod 
1 Furlong 
1 Mile 
1 League 
1 Degree - 



In Ft 

12= 1 

36= 3 = 

V98 = 16} = 



In 
Ft 

Yd ' 

Fth 

PI 

Fur 

Mile 

Lea 

Deg or 



Yd 

1 PI 
5} = 1 Far 
7920 = 660" = 220 = 40= 1 Mile 
63360 = 5280 = 1760 = 320 = 8 = 1 



CLOTH MEASURE. 



2 Inches and a quarter make 
4 Nails 

3 Quarters 

4 Quarters 

6 Quarters - 

4 Quarters l£ Inch 



1 Nail - - Nl 
1 Quarter of a Yard Qr 
1 Ell Flemish - EF 
1 Yard . - Yd 
1 Ell English - EE 
1 Ell Scotch - E S 



ounce, and 12 ounces one pound. But in later times it was thought 
sufficient to divide the same pennyweight into 24 equal parts* still, 
called grains, being the least weight now in common use ; and from 
thence the rest are computed, as in the Tables above. 



TABLES OP HEASURE8. 



27 



SQUARE MEA8URB. 

144 Square Inches make 1 Sq Foot - Ft 

9 Square Feet . 1 Sq Yard - Yd 

30$ Square Yards - 1 Sq Pole - Pole 

40 Square Poles - 1 Rood - Rd 

4 Roods - - 1 Acre - Act 

Sq Inc SqFt 

144 = 1 Sq Yd 
1296= 9 = 1 SqPl 
39204 = 272± = 30}= 1 Rd 
1568160 = 10890 = 1210 = 40 = 1 Acr 
6272640 = 43560 = 4840 = 160 = 4 = 1 

By this measure, Land, and Husbandmen and Gardeners' 
work are measured ; also Artificers' work, such as Board, 
Glass, Pavements, Plastering, Wainscoting, Tiling, Floor- 
ing, and every dimension of length and breadth only. 

When three dimensions are concerned, namely, length, 
breadth, and depth or thickness, it is called cubic or solid 
measure, which is used to measure Timber, Stone, dec. 

The cubic or solid Foot, which is 12 inches in length and 
breadth and thickness, contains 1728 cubic or solid inches, 
and 27 solid feet make one solid yard. 



DRY, OR CORN MEASURE* 



2 Pints make 1 Quart - - Qt 

2 Quarts - 1 Pottle - - Pot 

2 Pottles - 1 Gallon - - Gal 

2 Gallons - 1 Peck - - Pec 

4 Pecks - 1 Bushel - - Bu 
8 Bushels 1 Quarter - - Qr 

5 Quarters 1 Wey, Load, or Ton Wey 
2 Weys - 1 Last - - Last 

Put Gal 
8=1 Pec 
16= 2= 1 Bu 

64= 8= 4= 1 Qr 
512= 64 = 32 = 8= 1 Wey 

2560 = 320 = 160 = 40= 5=1 Last 

5120 = 640=320 = 80 = 10 = 2=1 



28 



▲uiTMjcmc* 



By this are measured all dry wares, as, Corn, Seeds, 
Roots, Fruits, Salt, Coals, Sand, Oysters, <Scc. 

The standard Gallon dry-measure contained 268$ cubic or 
solid inches, and the corn or Winchester bushel 2150} cubic 
inches ; for the dimensions of the Winchester bushel, by the 
old Statute, - were 8 inches deep, and 18£ inches wide or in 
diameter. But the Coal bushel was to be 19£ inches in dia- 
meter ; and 36 bushels, heaped up, made a London chaldron 
of coals, the weight of which was 3136 lb Avoirdupois, or 1 
ton 8 cwt nearly. See, however, page 29. 



ALE AND BEER MEASURE. 



2 Pints make 




1 Quart 


Qt 


4 Quarts 




1 Gallon . 


Gal 


36 Gallons - 




1 Barrel 


Bar 


1 Barrel and a half 


1 Hogshead - 


Hhd 


2 Barrels - 




1 Puncheon ~ 


Pun 


2 Hogsheads 




1 Butt 


Butt 


2 Butts 




1 Tun 


Tun 


Pfc Qt 








2= 1 


Gal 






8 = 4 = 


1 


Bar 




288 = 144 = 


36 = 


1 Hhd 




432 = 216 = 


54 = 


1}= 1 Butt 




864= 432 = 


108 = 


3=2=1 





Nate. The Ale Gallon contained 282 cubic or solid inches, 
by which also milk was measured. 



WISE MEASURE. 



2 Pints make 




1 Quart 


Qt 


4 Quarts 




1 Gallon 


Gal 


42 Gallons, . 




1 Tierce 


Tier 


63 Gallons or 1£ Tierces 




1 Hogshead • 


Hhd 


2 Tierces - 




1 Puncheon - 


Pun 


2 Hogsheads 




1 Pipe or Butt 


Pi 


2 Pipes or 4 Hhds 




1 Tun 


Tun 


Pts Qt 








2 = 1 Gal 








8= 4= 1 


Tier 




336 = 168 = 42 = 


1 


Hhd 




504 = 252 = 63 = 


H 


= 1 Pun 




672 = 336 = 84 = 


2 


= 1*= 1 


Pi 


1008 = 504 = 126 = 


3 


-4-3*- 


1 Tun 


2016 = 1008 = 252 = 


6 




2 = 1 



TABLES OF MEASURES AITD TIME. 29 

Note, by this are measured all Wines, Spirits, Strong- 
waters, Cyder, Mead, Perry, Vinegar, Oil, Honey, &c. 

The old Wine Gallon contained 231 cubic or solid inches. 
And it is remarkable that these Wine and Ale Gallons have 
the same proportion to each other, as the Troy and Avoir- 
dupois Pounds have ; that is, as one Pound Troy is to one 
Pound Avoirdupois, so is one Wine Gallon to one Ale 
Gallon. 



OF 



60 Seconds or 60" make 


- 1 Minute - 


ilfor' 


60 Minutes 


- 1 Hour - 


Hr 


24 Hours 


• 1 Day 


Day 


7 Days 


. lWeek - 


»* 


4 Weeks 


. 1 Month - 


Mo 


13 Months 1 Day 6 Hours, ) 
or 365 Days 6 Hours $ 


1 Julian Year 


Yr 



Sec 
60 = 
3600 = 
86400 = 



Min 
1 

60 *= 
1440 = 



Hr 

1 Day 
24 = 1 



Wk 



604800 = 10080 = 168 = 7 = 1 Mo 
2419200 = 40320 = 672 = 28 = 4 = 1 
31557600 == 525960 = 8766 = 365J = 1 Year 



Wk Da Hr Mb 
Or 52 1 6 = 13 

Da Hr M Sec 

But 365 5 48 - 45£ = Solar Year 



Da Hr 
1 6=1 Julian Year 



IMPERIAL MEASURES. 

By the late Act of Parliament for Uniformity of Weights 
and Measures, which commenced its operation on the 1st of 
January, 1826, the chief part of the weightsjand measures 
are allowed to remain as they were ; the Act simply pre- 
scribing scientific modes of determining them, in case they 
should be lost. 

The pound troy contains 5760 grains. 
The pound avoirdupois contains 7000 grains. 
The imperial gallon contains 277*274 cubic inches. 
The corn bushel, eight times the above. 



30 



ARITHMETIC. 



Henco, with respect to Ale, Wine, and Corn, it will be 
expedient to possess a 

TABLE OF FACTORS, 



For converting old measures into new, and the contrary. 





By decimals. 


By vulgar frac- 
tions nearly. 


Corn 
Measure. 


Wine 
Measure. 


Ale 
Measure. 


Corn 
Mea- 
sure. 


Wine 
Mea- 
sure. 


Ale 
Mm. 
sure. 


To convert old j 
measures to neW. 1 


•96943 


•83311 


1-01704 


H 




H 


To convert new ) f ,*« . . 
measures to old. \ 1H)3IM 


1-20032 


•98324 


! H 


* 





N. B. For reducing the prices, these numbers must all 
be reversed. 



RULES FOR REDUCTION. 

I. When the Numbers are to be reduced from a Higher 
Denomination to a Lower : 

Multiply the number in the highest denomination by as 
many of the next lower as make an integer, or 1, in that 
higher ; to this product add the number, if any, which was 
in this lower denomination before, and set down the amount. 

Reduce this amount in like manner, by multiplying it by 
as many of the next lower as make an integer of this, taking 
in the odd parts of this lower, as before. And so proceed 
through all the denominations to the lowest ; so shall the 
number last found be the value of all the numbers which 
were in the higher denominations, taken together*. 



* The reason of this rule is very evident; for pounds are brought 
into shillings by multiplying them by 20 ; shillings into pence, by mul- 
tiplying them by 12 ; and pence into farthings, by multiplying* by 4; 
and the reverse of this rule by division.— And the same, it is evident, 
will be true in the reduction of numbers consisting of any denomina- 
tions whatever. 



HOLES FOR SEDUCTION. 



31 



EXAMPLE. 

1. Ill 12342 15* 7d, how many farthings ? 
I 8 d 
1234 15 7 



24095 Shillings 
12 



206347 Pence 
4 



Answer 1 1 85388 Farthings. 



II. When the Numbers are to be reduced from a Lower De- 
nomination to a Higher : 

Divide the given number by as many of that denomina- 
tion as make 1 of the next higher, and set down what re- 
mains, as well as the quotient. 

Divide the quotient by as many of this denomination as 
make 1 of the next higher ; setting down the now quotient, 
and remainder, as before. 

Proceed in the same manner through all the denomina- 
tions to the highest ; and the quotient last found, together 
with the several remainders, if any, will be of the same value 
as the first number proposed. 

EXAMPLES. 

2. Reduce 1185388 farthings into pounds, shillings, and 
pence. 

4) 1185388 
12) 2963474 



2,0) 2469,5*— Id 



Answer 1234Z 15* Id 



3. Reduce 24Z to farthings. Ans. 23040. 

4. Reduce 337587 farthings to pounds, &c. 

Ans. 3511 \3s 



83 



.ARITHMETIC* 



5. How many farthings are in 36 guineas ? Ans. 36288. 

6. In 36288 farthings how many guineas ? Ans. 36. 

7. In 69 lb 13 dwts 5 gr. how many grains ? Ans. 340157. 

8. In 8012131 grains how many pounds, d&c. 

Ans. 1390 lb 11 oz 18 dwfc 19 gr. 

9. In 35 ton 17 cwt 1 qr 23 lb 7 oz 13 dr how many drams ? 

Ans. 20571005. 

10. How many barley-corns will reach round the earth, 
supposing it, according to the best calculations, to be 25000 
miles? Ans. 4752000000. 

11. How many seconds are in a solar year, or 365 days 
5 hrs 48 min. 45} sec ? Ans. 31556925}. 

12. In a lunar month, or 29 ds 12 hrs 44 min 3 sec, how 
many seconds? Ans. 2551443. 



COMPOUND ADDITION. 

Compound Addition shows how to add or collect several 
numbers of different denominations into one sum. 

Rule. — Place the numbers so, that those of the same de- 
nomination may stand directly under each other, and draw a 
line below them. Add up the figures in the lowest deno- 
mination, and find, by Reduction, how many units, or ones, 
of the next higher denomination are contained in their sum. 
—Set down the remainder below its proper column, and 
carry those units or ones to the next denomination, which 
add up in the same manner as before. — Proceed thus through 
all the denominations, to the highest, whose sum, together 
with the several remainders, will give the answer sought. 

The method of proof is the same as in Simple Addition. 



COMPOUND ADDITION. 



SI 



EXAMPLES OF MONEY*. 





1. 




2. 






3. 






4. 




I 


s 


d 


/ 1 


<* 


/ 


8 


d 


Z 




d 


7 


13 


3 


14 7 


5 


15 


17 


10 


53 


14 


8 


3 


5 


lO-i 


8 19 


2f 


3 


14 


6 


5 


10 


Sf 


6 


18 


7 


7 8 


1} 


23 


6 


*1 


93 


11 


6 





2 


5j 


21 2 


9 


14 


9 


*i 


7 


5 





4 





3 


7 16 


8J- 


15 


6 


4 


13 


2 


5 


17 


15 


*i 


4 


3 


6 


12 







18 


7 


99 


15 


9 ? 


















32 


2 


6j 


















39 


15 


9J 



















5. 6. 

I s d lad 

14 7J 37 15 8 

8 15 3 14 12 9; 

62 4 7 17 14 9 

4 17 8 23 10 9J 

23 4f 8 6 

6 6 7 14 5J 

91 10{ 54 2 l\ 



7. 


8. 




1 8 d 


/ 8 


d 


61 3 2£ 


472 15 


3 


7 16 8 


9 2 




29 13 lOf 


27 12 


si 


12 16 2 


370 16 


2 1 


7 5J 


13 7 


4 


24 13 


6 10 


5f 


5 10? 


30 


11* 



Exam. 9. A nobleman, going out of town, is informed by 
his steward, that his butcher's bill comes to 197/ 13* 7\d ; 
his baker's to 592 5* 2Jd ; his brewer's to 85/ ; his wine* 
merchant's to 103/ 13* ; to his corn chandler is due 75/ 3d ; 
to his tallow-chandler and cheesemonger, 27/ 15* 11{</; and 
to his tailor 55/ 3* 5f d ; also for rent, servants' wages, and 
other charges, 127/ 3* : Now, supposing he would take 100/ 
with him, to defray his charges on the road, for what sum 
must he send to -bis bankor ? Ans. 8301 14s $Vd 

Vol. I. 6 



34 



ARITHMETIC. 



10. The strength of a regiment of foot, of 10 companies, 
and the amount of their subsistence*, for a month of SO 
days, according to the annexed Table, are required ? 



Numb. 


Rank. 


Subsistence for a Month. 
• 


1 


Colonel 


£27 








1 


Lieutenant Colonel 


19 


10 





1 


Major 


17 


5 





7 


Captains 


78 


15 





11 


Lieutenants 


57 


15 





9 


Ensigns 


40 


10 





1 


Chaplain 


7 


10 





1 


Adjutant 


4 


10 





1 


Quarter-Master 


5 


5 





1 


Surgeon 


4 


10 





1 


Surgeon's Mate 


4 


10 





30 


Serjeants 


45 








30 


Corporals 


30 








20 


Drummers 


20 








2 


Fifers 


2 








390 


Private Men 


292 


10 





507 


Total. 


£656 10 






* Subsistence Money, is the money paid to the soldiers weekly ; 
which is short of their fall pay, because their clothes, accoutrements, 
Ac. are to be accounted for. It is likewise the money advanced to 
officers till their accounts are made up, which is commonly once a 
year, when they are paid their arrears. The following table shows the 
foil pay and subsistence of each rank on the English establishment. 



COMPOUND ADDITION. 



35 



1*4 
a in 



7 4 « — « _ o ~ no< d _ t t- ' 
| o as " o '<= o * - » 



if 3 



i 



£3 



CS=J CI Q w 2 . , O Q . , , 

1W - Q 3 O O . .QQ - . . 

5*^> - - - -» cor-' + vo | 

2* * n cc*fl ,«t^ . , — l- 



*"- QQ ' 5 '3000 , ( QO 

« o o a v o ^ a, 6 -a »~ — 

^ £ S£ ** *o *^ » <c . . — * r* 

O w i fi ^ g o " 

: " ii ■ ^ , 



Ira Jtuoq fg 

"TV r»™ <wj w» n 



.irfj Pin .>i!.|..„[ *a 
Pi Jiff 4#J .uinj.ttdE 
-jy ptrm liv j wajQ 



U mi Lii.ifi r 
ail ,rjtu)*Min 



3IJ] Mtl JUJCM] STT 

-(V pin Vq,[ twijij 



2 * *S ■ ■ & « c ^ r«S 

* « « 7J . <£> . T . O i c 



.3. K 



2 - "2 • : ' a ■ 

r 9 - O O . O O 



3* 



1 



f s 1 1 

■i|f! 

■ c 



2U 

- c 



- £ = 



in 

ill 
ill 

if 5 
HI 




■ 

Mi 



* 111 



5 I : 
1 t5S 

c - — 

*^ _ — 

| • 

I. 1 ? 

ti: 

* r- 51 



36 



ARITHMETIC. 



EXAMPLES OF WEIGHTS. MEASURES, <f . 



TROT WEIGHT. APOTHEC ARIES' WEIGHT. 





1. 






2. 




3. 






4. 






lb 


oz 


dwt 


oz 


dwt gr 


lb 


oz dr 


sc 


oz 


dr 


sc 




17 


3 


15 


37 


9 3 


3 


5 7 


2 


3 


5 


1 


17 


7 


9 


4 


9 


5 3 


13 


7 3 





7 


3 


2 


5 





10 


7 


8 


12 12 


19 


10 6 


2 


16 


7 





12 


9 


5 





17 


7 8 





9 1 


2 


7 


3 


2 


9 


176 


2 


17 


5 


9 


36 


3 5 





4 


1 


2 


18 


23 


11 


12 


3 


19 


5 


8 6 


1 


36 


4 


1 


14 



AVOIRDUPOIS WEIGHT. LONG MEASURE. 

5. 6. 7. 8. 

lb oz dr cwt qr lb mis fur pis yds feet inc 

17 10 13 15 2 15 29 3 14 127 1 5 

5 14 8 6 3 2 4 19 6 29 12 2 9 
12 9 18 9 1 14 7 24 10 10 
27 1 6 9 1 17 9 1 37 54 1 11 

040 10 2 6 703 527 

6 14 10 3 3 4 5 9 23 5 



CTOTH MEASURE. LAND MEASURE. 

9. 10 11. 12. 

yds qr nls el en qrs nls ac ro |> ac ro p 

26 3 1 270 1 225 3 37 19 16 

13 1 2 57 4 3 16 1 25 270 3 29 

9 1 2 18 1 2 7 2 18 6 3 13 

217 3 3 2 4 2 9 23 34 

9 1 10 1 42 1 19 7 2 16 

55 3 1 4 4 1 7 6 75 23 



WINE .MEASURE. ALF. AND BEER MEASURE. 

13. 14. 15. 16. 

t hds gal hds gal pts hds gal pts hds gal pts 

13 3 15 15 61 5 17 37 3 29 43 5 
8 1 37 17 14 13 9 10 15 12 19 7 

14 1 20 29 23 7 3 6 2 14 16 6 
25 12 3 15 1 5 14 6 8 1 

3 1 9 16 8 12 .9 6 57 13 4 

72 3 21 4 96 (5 8 42 4 5 6 



COMPOUND SUBTRACTION. 



37 



COMPOUND SUBTRACTION. 



Compound Subtraction shows how to find the difference 
between any two numbers of different denominations. To 
perform which, observe the following Rule. 

* Place the less number below the greater, so that the 
parts of the same denomination may stand directly under 
each other ; and draw a line below them. — Begin at the 
right-hand, and subtract each number or part in the lower 
line, from the one just above it, and set the remainder 
straight below it. — But if any number in the lower line.be 
greater than that above it, add as many to the upper number 
as make 1 of the next higher denomination ; then take the 
lower number from the upper one thus increased, and set 
down the remainder. Carry the unit borrowed to the next 
number in the lower line ; after which subtract this number 
from the one above it, as before ; and so proceed till the 
whole is finished. Then the several remainders, taken to- 
gether, will be the whole difference sought. 

The method of proof is the same as in Simple Subtraction. 



EXAMPLES OP MONKF. 



1. 2. 3. 4. 

1 s d 1 * d 1 s d 1 s d 

From 79 17 8} 103 3 2£ 81 10 11 254 12 

Take 35 12 4' 71 12 5j 29 13 3| 37 9 4f. 

Rem. 44 5 4£ 31 10 83 

Proof 79 17 8J- 103 3 2£ 



5. What is the difference between 73/ 5irf and 19/ 13s 10J ? 

Ans. 53Zt>*7irf. 



* The reason of this Rule will easily appear from what has been said 
in Simple Subtraction ; for the borrowing depends on the same princi- 
ple, and is only different as the numbers to be subtracted are of differ- 
ent denominations. 



38 ARITHMETIC 

Ex. 6. a lends to b 100/, how much is b in debt after a 
has taken goods of him to the amount of 73/ 12* 4|c/ ? 

Ans. 26/ 7s 7J<*. 

7. Suppose that my rent for half a year is 20/ 12s, and 
that I have laid out for the land-tax 14s 6</, and for several 
repairs 1/3* 3*d, what have I to pay of ray half- year's rent? 

Ans. \81Us2$d. 

8. A trader, failing, owes to a 35/ 7s 6d, to b 91/ 13s \d, 
to c 53/ 7» cZ, to d 87/ 5*, and to e 111/ 2s 5$d. When 
this happened, he had by him in cash 23/ 7s bd, in wares 
53/11* 10{d f in household furniture 63/ 17* 7£</, and in 
recoverable book-debts 25/ 7s 5d. What will his creditors 
lose by him, supposing these things delivered to them ? 

Ans. 212/ 5* 3}<*. 

EXAMPLES OF WEIGHTS, MEASURES, 6fC. 

TROT WEIGHT. APOTHECARIES 1 WEIGHT. 

1. 2. 3. 

lb oz dwt gr lb oz dwt gr lb oz dr scr gr 

From 9 2 12 10 7 10 4 17 73 4 7 14 

Take 5 4 6 17 3 7 16 12 29 5 3 4 19 



Rem. 
Proof 



AVOIRDUPOIS WEIGHT. LONG MEASURE. 

4. 5. 6. 7. 

c qrs lb lb oz dr m fu pi vd ft in 

From 5 17 71 5 9 14 3 17 96 4 

Take 2 3 10 17 9 18 7 6 11 72 2 9 



Rem. 



Proof 



CLOTH MEASURE. LAND MEASURE. 

8. ». 10. 11. 

yd qr nl vd qr nl ac ro p ac ro p 

From 17 2 1 9 2 17 1 14 57 1 16 

Take 9 2 7 2 1 16 2 8 22 3 29 

Rem. 



Proof 



COMPOUND MULTIPLICATION. 



39 



WI5* MEASURE. ALE AND BEER MEASURE. 

12. 13. 14. 15. 

t hd gal hd gal pt hd gal pt hd gal pt 

From 17 2 23 5 4 14 29 3 71 16 5 

Take 9 1 36 2 12 6 9 35 7 19 7 I 

Rem. 

Proof 



DRY MEASURE. TIME.. 

16. 17. 18. 19. 

la qr bu bu gal pt mo we da ds hrs min 

From 9 4 7 13 7 1 71 2 5 114 17 26 

Take 6 3 5 9 2 7 17 1 6 72 10 37 

Rem. 

Proof 



20. The line of defence in a certain polygon being 236 
yards, and that part of it which i9 terminated by the curtain 
and shoulder being 146 yards 1 foot 4 inches ; what then was 
the length of the face of the bastion ? Ans. 89 yds 1 ft 8 in. 



COMPOUND MULTIPLICATION. 

Compound Multiplication shows how to find the amount 
of any given number of different denominations repeated a 
certain proposed number of times ; which is performed by 
the following rule. 

Set the multiplier under the lowest denomination of the 
multiplicand, and draw a line below it. — Multiply the num- 
ber in the lowest denomination by the multiplier, and find 
how many units of the next higher denomination arc con- 
tained in the product, setting down what remains. — In like 
manner, multiply the number in the next denomination, and 
to the product carry or add the units, before found, and dud 
how many units of the next higher denomination arc m VVvvs 



40 



ARITHMETIC 



amount, which carry in like manner to the next product, 
setting down the overplus. — Proceed thus to the highest de- 
nomination proposed : so shall the last product, with the se- 
veral remainders, taken as one compound number, be the 
whole amount required. — The method of Proof, and the rea- 
son of the Rule, are the same as in Simple Multiplication. 



EXAMPLES OF MONF.Y. 



1. To find the amount of 8 lb of Tea, at 5*. 8hd. per lb. 
s d 

5 Si 



£2 5 8 Answer. 

/ -v d 

2. 4 lb of Tea, at 7s 8d per lb. Ans. 1 10 8 

3. 6 lb of Butter, at $\d per lb. Ans. 4 9 

4. 7 lb of Tobacco, at 1* 8$d per lb. Ans. 11 1H 

5. 8 stone of Beef, at 2s l\d per st. Ans. 110 

6. 10 cwt cheese, at 2Z 17* lOcZ per cwt. Ans. 28 18 4 

7. 12 cwt of Sugar, at 3Z Is 4d per cwt. Ans. 40 8 



CONTRACTIONS. 



I. If the multiplier exceed 12, multiply successively by its 
component parts, instead of the whole number at once. 



EXAMl'l.KS. 



1. 15 cwt of Cheese, at 17* tod per cwt. 

/ * d 
17 6 
3 



2 12 

5 



13 t 6 Answer. 



I s d 

2. 20 cwt of Hops, at 41 7* 2d per cwt. Ans. 87 3 4 

3. 24 tons of Hay, at 3J 7* 672 per ton. Ans. 81 

4. 46 eUs of Cloth, at 1* M per ell. Ans. 3 7 6 



COMPOUND HULTOUCATION. 



41 



I sd 

Ex. 5. 63 gallons of Oil, at 2s 3d per gall. Ana. 7 19 

& 70 barrels of Ale, at 11 4* per barrel. Ans. 84 

7. 84 quarters of Oats, at 1/ 12* Sd per qr. Ans. 137 4 

8. 96quartersofBarley,atH3*4<2perqr. Ans.112 

9. 120 days' Wages, at 5s 9d per day. Ans. 34* 10 

10. 144 reams of Paper, at 13s 4d per ream. Ans. 96 

11. If the multiplier cannot be exactly produced by the 
multiplication of simple numbers, take the nearest number 
to it, either greater or less, which can be so produced, and 
multiply by its parts, as before. — Then multiply the given 
multiplicand by the difference between this assumed number 
and the multiplier, and add the product to that before found, 
when the assumed number is less than the multiplier, but 
subtract the same when it is greater. 

EXAMPLES. 

1. 26 yards of Cloth, at 3* Of cZ per yard. 
I s d 
3 0; 
5 



15 3J 

5 



3 16 Of 
3 0| add 

£3 19 7* Answer. 



EXAMPLES OF WEIGHTS AND MEASURES. 

2. 29 quarters of Corn, at 21 5s 3{d per qr. 

Ans. 65 12 lOj 

3. 53 loads of Hay, at 3Z 15* 2d per Id. Ans. 199 3 10 

4. 79 bushels of Wheat, at 11* 5fd per bush. 

Ans. 45 6 10* 

5. 97 casks of Beer, at 12* 2d per cask. Ans. 59 2 
C. 114 stone of Meat, at 15* 3jrf per st. Ans. 87 5 7} 

1. 2. 3. 

lb oz dwt gr lb oz dr sc gr cwt qr lb oz 

28 7 14 10 2 6 3 2 10 29 2 16 14 

5 8 12 



Vol. L 



7 



48 



rote fit . pis yds 
22 5 20 .6 
4 



5. 

yds qra na 
126 3 1 
7 



6. 

EC 10 pO 

28 8 27 
9 



7, . 8. 9. 

luns hhd gal pts we qr bu pe mo we da ho mm 

20 2 26 2 24 2 5 3 172 3 6 16 49 

3 6 10 



COMPOUND DIVISION. 

Compound Division teaches how to divide a number of 
several denominations by any given number, or into any 
number of equal parts ; as follows : 

Place the divisor on the left of the dividend, as in Simple 
Divis ion. — Begin at the left-hand, and divide the number of 
jflfehighest denomination by the divisor, setting down the 
^Hrent in its proper place. — If there be any remainder after 
this division, reduce it to the next lower denomination, which 
add to the number, if any, belonging to that denomination, 
and* divide the sum by the divisor. — Set down again this quo- 
tient, reduce its remainder to the next lower denomination 
again, and so on through all the denominations to the last. 



EXAMPLE* OF MONEY. 

Divide 237Z 8* 6d by 2. 

I 8 d 

2) 237 8 6 



£118 14 3 the Quotient. 



conform Division. 



I 9 d I 9 d 

2. Divide 482 12 1} by 3. Ans. 144 4 Oj 

3. Divide 507 3 5 by 4. Ana. 126 15 10' 

4. Divide 032 7 0} by 5. Ans. 126 9 6 

5. Divide 090 14 3fby6. Ana. 115 2 4* 

6. Divide 705 10 2 by 7. Ans. 100 15 8} 

7. Divide 760 5 6 by a Ans. 95 s\ 

8. Divide 761 5 7} by 9. Ans. 84 11 8} 

9. Divide 829 17 10 by 10. Ans. 82 19 9£ 

10. Divide 937 8 8} by 11. Ans. 85 4 5 

11. Divide 1145 11 4} by 12. Ans. 95 9 3J 



COATKACTlOlfS. 

i 

I. If the divisor eieeed 12, find what simple numbers, 
multiplied together, will produce it, and divide by them 
separately, as in Simple Division, as below. 



EXAXFLES. 



1. What is Cheese per cwt, if 16 cwt cost 252 14i Sdl 
I 9 d 

4) 25 14 8 



4) 6 8 8 



£ 1 12 2 the Answer. 



I $ d 

2. If 20 cwt of Tobacco come to ) k 7 in ± 

1501 6s 8rf, what is that per cwt ? \ 

3. Divide 982 8s by 36. Ans. 2 14 8 

4. Divide 712 13s \0d by 56. Ans. 1 5 1{ 

5. Divide 442 4s by 96. Ans. 9 2} 

6. At 312 10s per cwt, how much per lb ? Ans. 5 7} 

II. If the divisor cannot be produced by the multiplica- 
tion of small numbers, divide by the whole divisor at once, 
after the mariner oT Long division, as follows. 



44 



ARITHMETIC* 



EXAMPLES. 



1. Divide 59Z 6* 3fd by 19. 
I s d ltd 
19) 59 6 3} '(3 2 5j Ana. 
57 




09 (5 
95 

' ~4 
4 

19 (i 

I s d I s d 

2. Divide 89 14 5} by 57. Ans. 13 llj 

3. Divide 125 4 9 by 43. Ans. 2 18 3 

4. Divide 542 7 10 by 97. Ans. 5 11 10 

5. Divide 123 11 2 J by 127. Ans. 19 5* 



EXAMPLES OF WEIGHTS AND XEASUBS8. 

1. Divide 17 lb 9 oz dwts 2 gr by 7. 

Ans. 2 lb 6 oz 8 dwts 14 gr. 

2. Divide 17 lb 5 oz 2 dr 1 scr 4 gr by 12. 

Ans. 1 lb 5 oz 3 dr 1 scr 12 gr. 

3. Divide 178 cwt 3 qrs 14 lb by 53. Ans. 3 cwt 1 qr 14 lb. 

4. Divide 144 mi 4 fur 20po 1 yd 2ft in by 39. 

Ans. 3 mi 5 fur 26 po yds 2 ft 8 in. 

5. Divide 534 yds 2 qrs 2 na by 47. Ans. 11 yds 1 qr 2 na. 

6. Divide 77 ac 1 ro 33 po by 51. Ans. 1 ac 2 ro 3 po. 

7. Divide 2 tu hhds 47 gal 7 pi by 65. Ans. 27 gal 7 pi. 

8. Divide 387 la 9 qr by 72. Ans. 5 la 3 qrs 7 bu. 

9. Divide 206 mo 4 da by 26. Ans. 7 mo 3 we 5 ds. 



BULK OF THREE. 



45 



THE GOLDEN RULE, OR RULE OF THREE. 

The Rule of Thus- teaches how to* find a fourth propor- 
tional to three numbers given : for which reason it is some, 
times called the Rule of Proportion. It is called the Rule 
of Three, because three terms or numbers are given, to find 
a fourth. And because of its great and extensive usefulness, 
it is often called the Golden Rule. This Rule is usually by 
practical men considered as of two kinds, namely, Direct 
and Inverse. The distinction, however, as well as the man- 
ner of stating, though retained here for practical purposes, 
does not well accord with the principles of proportion ; as 
will be shown farther on. 

The Rule of Three Direct is that in which more requires 
more, or less requires less. As in this ; if three men dig 21 
yards of trench in a certain time, how much will six men dig 
in the same time ? Here more requires more, that is, 6 men, 
which are more than three men, will also perform more work, 
in the same time. Or when it is thus : if 6 men dig 42 yards, 
how much will 3 men dig in the same time ? Here then, less 
requires less, or 3 men will perform proportionately less work 
than 6 men, in the same time. In both these cases then, 
the Rule, or the Proportion, is Direct ; and the stating must 
be 

thus, as 3 : 21 : : 6 : 42, or as 3 : 6 : : 21 : 42. 

And, as 6 : 42 : : 3 : 21, or as 6 : 3 : : 42 : 21. 
But the Rule of Three Inverse, is when more requires less, 
or less requires more. As in this : if 3 men dig a certain 
quantity of trench in 14 hours, in how many hours will 6 
men dig the like quantity ? Here it is evident that 6 men, 
being more than 3, will perform an equal quantity of work 
in less time, or fewer hours. Or thus : if 6 men perform a 
certain quantity of work in 7 hours, in how many hours will 
3 men perform the same ? Here less requires more, for 3 
men will take more hours than 6 to perform the same work. 
In both these cases then the Rule, or the Proportion, is In- 
verse ; and the stating must be 

thus, as 6 : 14 : : 3 : 7, or as 6 : 3 : : 14 : 7. 

And, as 3 : 7 : : 6 : 14, or as 3 : 6 : : 7 : 14. 
And in all these statings, the fourth term is found, by mul- 
tiplying the 2d and 3d terms together, and dividing the pro* 
duct by the 1st term. 

Of the three given numbers : two of them contain the sup- 
position, and the third a demand. And for stating and work- 
ing questions of these kinds, observe the following general 
Rule: 



46 



ARITHMETIC. 



State the question by setting down in a straight line the 
three given numbers, in the following manner, viz. so that 
the 2nd term be that number of supposition which is of the 
same kind that the answer or 4th term is to be ; majtiog the 
other number of supposition the 1st term, and the demanding 
number the 3d term, when the question is in direct propor- 
tion ; but contrariwise, the other number of supposition the 
3d term, and the demanding number the 1st term, when the 
question has inverse proportion. 

Thon, in both cases, multiply the 2d and 3d terms together, 
and divide the product by the 1st, which will give the answer, 
or 4th term sought, viz. of the same denomination as the 
second term. 

Note, If the first and third terms consist of different deno- 
minations, reduce them both to the same : and if the second 
term be a compound number, it is mostly convenient to re* 
duce it to the lowest denomination mentioned. — If, after 
division, there be any remainder, reduce it to the next lower 
denomination, and divide by the same divisor as before, and 
the quotient will be of this last denomination. Proceed in 
the same manner with all the remainders, till they be. reduc- 
ed to the lowest denomination which the second admits of, 
and the several quotients taken together will be the answer 
required. 

Note also, The reason for the foregoing Rules will appear, 
when we come to treat of the nature of Proportions. — Some- 
times two or more statings are necessary, which may always 
be known from the nature of the question. 

EXAMPLES. 

1. If 8 yards of Cloth cost 11 4*, what will 96 yards cost? 
yds 1 s yds 1 s 
As 8 : 1 4 : : 96 : 14 8 the Answer. 
20 

24 
96 



144 
216 

8)2304 

2,0) 28,8* 

£14 8 Answer. 



BULB OF THREE. 



47 



Ex. 2. An engineer having raised 100 yards of a certain 
work in 34 days with 5 men ; how many men must he em* 
ploy to finish a like quantity of work in 15 days ? 



ds men ds men 
As 15 : 5 : : 24 : 8 Ans. 
5 

15) 120 (8 Answer. 
120 



3. What will 72 yards of cloth cost, at the rate of 9 yards 
for 52 12* ? Ans. 44/ 16*. 

4. A person's annual income being 146/ ; how much is 
that per day ? Ans. 8*. 

5. If 3 paces or common steps of a certain person be equal 
to 2 yards, how many yards will 160 of his paces make ? 

Ans. 106 yds 2 ft. 

6. What length must be cut off a board, that is 9 inches 
broad, to make a square foot, or as much as 12 inches in 
length and 12 in breadth contains ? Ans. 16 inches. 

7. If 750 men require 22500 rations of bread for a 
month ; how many rations will a garrison of 1000 men re. 
quire ? Ans. 36000. 

8. If 7 cwt 1 qr. of sugar cost 26/ 10* 4d ; what will be 
the price of 43 cwt 2 qrs ? Ans. 159/ 2*. 

9. The clothing of a regiment of foot of 750 men amount- 
ing to 28312 5* ; what will the clothing of a body of 3500 
men amount to ? Ans. 13212/ 10*. 

10. How many yards of matting, that is 3 ft broad, will 
cover a floor that is 27 feet long and 20 feet broad ? 

Ans. 60 yards. 

11. What is the value of six bushels of coals, at the rate 
of 1/ 14*. 6d the chaldron ? Ans. 5* \)d. 

12. If 6352 stones of 3 feet long complete a certain quan- 
tity of walling ; how many stones of 2 feet long will raise a 
like quantity ? Ans. 9528. 

13. What must be given for a piece of silver weighing 
73 lb 5 oz 15 dwts, at the rate of 5* 9d per ounce ? 

Ans. 253/ 10* 0f<Z. 

14. A garrison of 536 men having provision for 12 
months ; how long will those provisions last, if the garrison 
be increased to 1124 men ? Ans. 174 days and T f | T . 

15. What will be the tax upon 763/ 15* at the rate of 
3t Qd per pound sterling ? Ans. 1331 \3s \\d. 



46 



ARITHMETIC. 



16. A certain work being raised in 12 days, by working 4 
hours each day ; how long would it nave been in raising by 
working 6 hours per day ? Ans. 8 days. 

17. What quantity of corn can I buy for 90 guineas, at 
the rate of 6* the bushel ? Ans. 39 qra 3 bu. 

18. A person, failing in trade, owes in all 9772 ; at which 
time he has, in money, goods, and recoverable debts, 420Z 6* 
3JcZ ; now supposing these things delivered to his creditors, - 
how much will they get per pound? Ans. 8* 7\d. 

19. A plain of a certain extent having supplied a body of 
3000 horse with forage for 18 days ; then how many days 
would the same plain have supplied a body of 2000 horse ? 

Ans. 27 days. 

20. Suppose a gentleman's income is 600 guineas a year, 
and that he spends 25* 6d per day, one day with another ; 
how much will he have saved at the year's end ? 

Ans. 164/ 12* 6c*. 

21. What cost 30 pieces of lead, each weighing 1 cwt 
121b. at the rate of 16* 4d the cwt ? Ans. 27Z 2s 6d, 

22. The governor of a besieged place having provision for 
54 days, at the rate of l£lb of bread ; but being desirous to 
prolong the siege to 80 days, in expectation of succour, in 
that case what must the ration of bread be ? Ans. 1 ^ ¥ lb. 

23. At half-a-guinea per week, how long can I be boarded 
for 20 pounds ? Ans. 38 T <& wks. 

24. How much will 75 chaldrons 7 bushels of coals come 
to, at the rate of 11 13* 6d per chaldron ? 

Ans. 125Z 19* 0j<*. 

25. If the penny loaf weigh 8 ounces when the bushel of 
wheat costs 7* 3d, what ought the penny loaf to weigh when 
the wheat is at 8* 4d ? Ans. 6 oz 15 ffo dr. 

26. How much a year will 173 acres 2 roods 14 poles of 
land give, at the rate of 11 Is 8d per acre ? 

Ans. 240Z 2* 7&d. 

27 To how much amounts 73 pieces of lead, each weigh- 
ing 1 cwt 3 qrs 7 lb, at 10Z 4* per fother of 19| cwt ? 

Ans. 69Z4*2d l^fq. 

28. How many yards of stuff, of 3 qrs wide, will line a 
cloak that is 1 J yards in length and 3J yards wide ? 

Ans. 8 yds Oqrs 2$ nl. 

29. If 5 yards of cloth cost 14* 2d, what must be given for 
9 pieces, containing each 21 yards 1 quarter ? 

Ans. 27Z 1* 

30. If a gentleman's estate be worth 2107Z 12* a year ; 
what may he spend per day, to save 500Z in the year ? 

Ans. 4Z 8s ljftd. 



RULE OF TIIHKE. 49 

31. Wanting just an acre of land cut off from a piece 
which is 131 poles in breadth, what length must the picco 
be ? Ans. 11 po 4 yds 2 ft Otf in. 

32. At 7s 9' d per yard, what is the value of a piece of 
cloth containing 53 ells English 1 qr ? Ans. 25/ ISs 1 J<J. 

33. If the carriage of 5 cwt 11 lb for 90 miles he 1/ 12s (yd; 
how fur may I have 3 cwt 1 qr carried for the same money? 

Ans. 151 m 3 fur 3,-^ pol. 
3-1. Bought a silver tankard, weighing 1 lb 7 oz 14 dwts ; 
what did it cost me at Hs 4d the ounce ? Ans. (U 4s 9jd. 

35. What is tho half 3 T ear's rent jf 547 acres of land, at 
15s Oct the acre ? Ans. 211/ 19* 3d. 

36. A wall that is to be built to the height of 30 feet, was 
raised feet high by 10 men in days ; then how many men 
must be employed to finish the wall in 4 days, at the same 
rate of working / Ans. 72 men. 

37. What will be the charge of keeping 20 horses for a 
year, at the rate of 14 id pur dav for each horse? 

Ans. 441/ 0* lOd. 

38. If 18 ells of stuff that is J yard wide, cost 39s M ; 
what will 50 oils, of the same goodness, cost, being yard 
wide? Ans. 11 i\s 3J{<#. 

39. How many yards of paper that is 30 inches wide, will 
hang a room that is 20 yards in circuit and 9 feet high ? 

Ans. 72 yards. 

40. If a gentleman's estate be worth 384/ 1 0* a year, and 
the Jand-ta.x be assessed at 2s \)} A d per pound, what is his net 
annual income ? Ans. 331/ Is 9£<Z. 

41. The circumference of the earth is about 25000 miles ; 
at what rate per hour is a person at the middle of its surface 
carried round, one whole rotation being made in 23 hours 
56 minutes ? Ans. 1044 T s Afe miles. 

42. If a person drink 20 bottles of wine per month, when 
it costs 8s. a gall ; how many bottles per month may ho 
drink, without increasing the expense, when wine costs 10s 
he gallon ? Ans. 16 bottles. 

43. What cost 43 nrs 5 bushels of corn, at 1/ 8s 6*2 the 
quarter ? * Ans. 62/ 3s 3J<Z. 

44. How T many yards of canvas that is ell wide will line 
50 yards of say that is 3 quartern wide ? Ans. 30 yds. 

45. If an ounce of gold cost 4 guineas, what is the value 
of a grain ? Ans. 2 T Vd. 

40. If 3 cwt of tea cost 10/ 12s ; at how much a pound 
must it L9 rolailH, to gain 10/ by the whole \ Ans. 3^$. 



Vol. I. 



50 



COMPOUND PROPORTION. 

Compound Pbofostion is a rule by means of which the 
student may resolve such questions as require two or more 
stsiings in simple proportion. 

The general rule for questions of this kind may be ex. 
hibited in the following precepts : viz. 

1. Set down the terms that express the conditions of the 
question in one line. 

2. Under each conditional term, set its corresponding one, 
in another line, putting the letter a in the (otherwise) blank 
place of the term required. 

3. Multiply the producing terms of one line, and the pro* 
duced terms of the other line, continually, and take the re. 
suit for a dividend. 

4. Multiply the remaining terms continually, and let the 
product be a divisor. 

5. The quotient of this division will be q, the term re. 
quired.** 

Note. By producing terms are here meant whatever ne- 
cessarily and jointly produce any effect ; as the cause and 
the time ; length, breadth, and depth ; buyer and his mo- 
ney ; things carried, and their distance, dec. all necessarily 
inseparable in producing their several effects. 

In a question where a term is only understood, and not ex- 
pressed, that term may always be expressed by unity. 

A quotient is represented by the dividend put above a line, 
and the divisor put below it. 



EXAMPLES. 

1. How many men can complete a trench of 135 yards 
long in 8 days, when 16 men can dig 54 yards of the same 
trench in 6 days ? 

M D Yds 

16 • ... 6 ... • 54 
a .... 8 .... 135 



* This rale, which is as applicable to Simple as to Compmmd Propor- 
tion, was given, in 1706, by W. Jones, Esq. F.R.S., the father of the 
late Sir W. Jones. 



COMPOUND PROPORTION. 



51 



Here 16 men and 6 days, are the producing terms of the 
first line, and 185 yards, the produced term of the other. 
Therefore, by the rule, 

16X6X135 2X135 m 

UCa -8X54- 9 * 

the number of men required. 



ANOTHER question. 

If a garrison of 3600 men have bread for 35 days, at 
94 os each a day : How much a day must be allowed to 
4800 men, each for 45 days, that the same quantity of bread 
may serve? 

men ok days bread 

3600 . . 24 . . 35 . . 1 
4800 . . a . . 45 . . 1 

3600X24X35 



AN EXAJfPLE IN SIMPLE PROPORTION. 

If 14 yards of cloth cost 21Z, how many yards may be 
bought for 73/ 10f? 

man £ yds. 

1 .... 21 .... 14 
1 .... 73* .... q 

a = = { of 73* = 49 yards, Answer. 

«1 

2. If 1002 in one year gain 51 interest, what will be the 
interest of 7501 for seven years ? Ans. 2622 10*. 

3. If a family of 8 peraons expend 200/ in 9 months ; 
how much will serve a family of 18 people 12 months ? 

Ans. 600/. 

4. If 27# be the wages of 4 men for 7 days ; what will be 
the wages of 14 men for 10 days ? Ans. 6/ 15*. 

If a footman travel 130 miles in 3 days, when the days 
are 12 hours long ; in how many days, of 10 hours each, 
may he travel 360 miles 7 Ans. 9j j days. 

G. If 120 bushels of corn can serve 14 horses 56 days ; 
how many days will 94 bushels serve 6 horses ? 

Ans. VXL\\ day*. 



52 



ARITHMETIC. 



7. If 8000 lbs of beef serve 340 men 15 days ; how many 
lbs will serve 120 men for 25 days ? Ans. 1764 lb 1 1 ^ oz. 

8. If a barrel of beer be sufficient to last a family of 8 
persons 12 days ; how many barrels will be drank by 16 
persons in the space of a year 7 Ans. 60 J barrels. 

9. If 180 men, in six days, of 10 hours each, can dig a 
trench 200 yards long, 3 wide, and 2 deep ; in how many 
days of 8 hours long, will 100 men dig a trench of 360 yards 
long, 4 wide, and 3 deep ? Ans. 48} days. 



OP VULGAR FRACTIONS. 

A Fraction, or broken number, is an expression of a 
part, or some parts, of something considered as a whole. 

It is denoted by two numbers, placed one below tho other, 
with a line between them : 

Thus, JL I j umer * tor \ which is named 3-fourths. 
4 denominator > 

The denominator, or number placed below the line, shows 
how many equal parts the whole quantity is divided into ; 
and it represents the Divisor in Division. — And the Nu- 
merator, or number set above the line, shows how many of 
these parts are expressed by the Fraction : being the re- 
mainder after division. — Also, both these numbers are in 
general named the Terras of the Fraction. 

Fractions are either Proper, Improper, Simple, Compound, 
Mixed, or Complex. 

A Proper Fraction, is when the numerator is less than the 
denominator ; as, £, or |, or £, &c. 

An Improper Fraction, is when the numerator is equal to, 
or exceeds, the denominator ; as, f , or f , or }, &c. In 
these cases the fraction is called Improper, because it is equal 
to, or exceeds unity. 

A Simple Fraction, is a single expression, denoting any 
number of parts of the integer ; as, f , or |. 

A Compound Fraction, is the fraction of a fraction, or 
two or more fractions connected with the word of between 
them ; as, j of §, or $ of £ of 3, dec. 

A Mixed Number, is composed of a whole number and a 
fraction together ; as, 3|, or 12f , dec. 

A Complex Fraction, is one that has a fraction or a mixed 
number for its numerator, or its denominator, or both ; 

i 2 I 3 * jt 
**' "t"' or 7' 01 4 ' or 4 9 &c ' 



REDUCTION OF VULGAR FRACTIONS. 



53 



A whole or integer number may be expressed like a frac- 
tion, by writing 1 below it, as a denominator ; so 3 is f , or 4 
is f , Ac. 

A fraction denotes division ; and its value is equal to the 
quotient obtained by dividing the numerator by the deno- 
minator : so y is equal to 3, and V ' 8 equal to 4}. 

Hence then, if the numerator be less than the denominator, 
the value of the fraction is less than 1. But if the numerator 
be the same as the denominator, the fraction is just equal to 
1. And. if the numerator be greater than the denominator, 
the fraction is greater than 1. 



REDUCTION OF VULGAR FRACTIONS. 

Reduction of Vulgar Fractions, is the bringing them out 
of one form or denomination into another ; commonly to pre- 
pare them for the operations of Addition, Subtraction, dec. ; 
of which there are several cases. 

PROBLEM. 

To find the Greatest Common Measure of Two or more 
Numbers. 

The Common Measure of two or more numbers, is that 
number which will divide them all without remainder ; so, 3 
is a common measure of 18 and 24; the quotient of the 
former being 6, and of the latter 8. And the greatest num- 
ber that will do this, is the greatest common measure : so 6 
is the greatest common measure of 18 and 24 ; the quotient 
of the former being 3, and of the latter 4, which will not 
both divide further. 

RULE. 

If there be two numbers only, divide the greater by the 
less ; then divide the divisor by the remainder ; and so on, 
dividing always the last divisor by the last remainder, till 
nothing remains ; so shall the last divisor of all be the great, 
est common measure sought. 

When there are more than two numbers, find the greatest 
common measure of two of them, as before ; then do tho 
same for that common measure and another of lYifc uronfofct*% 



54 ARITHMETIC 

and so on, through all the numbers ; so will the greatest com- 
mon measure last found be the answer. 

If it happen that the common measure thus found is 1 ; 
then the numbers are said to be incommensurable, or not to 
have any common measure, or they are said to be prime to 
each other. 

examples. 

1. To find the greatest common measure of 1908, 086, 
and 030. 

036 ) 1908 ( 2 So that 36 is the greatest common 

1872 measure of 1908 and 930. 

36 ) 936 ( 26. Hence 36 ) 630 ( 17 
7© 36 

216 270 
216 252 

18) 36 (2 
36 

m 

Hence 18 is the answer required. 

2. What is the greatest common measure of 246 and 372 ? 

An?. 6. 

3. What is the greatest common measure of 324, 612, 
and 1032? Ans. 12. 

CASE I. 

To Abbreviate or Reduce Fractions to their Lowest Terms. 

* Divide the terms of the given fraction by any number 
that will divide them without a remainder ; then divide these 



* That dividing both the terms of the fraction by the same number, 
whatever it be, will give another fraction equal to the former, is evi- 
dent. And when these divisions are performed as often as can be done, 
or when the common divisor is the greatest possible, the terms of the 
resulting fraction must be the least possible 

Note. 1. Any number ending with an even number, or a cipher, is 
divisible, or can be divided, by 2. 

2. Any number, ending with 6, or 0, is divisible by 6. 

3. If the right-hand place of any number be 0, the whole is divisible 
by 10 ; if there be two ciphers, it is divisible by 100 ; if three ciphers, by 
1000 : and so on ; which if only cutting off those ciphers. 



REDUCTION OF WLQAJL FRACTIONS* 



35 



quotients again in the same manner ; and so on, till it appears 
that there is no number greater than 1 which will divide 
them ; then the fraction will be in its lowest terms. 

Or, divide both the terms of the fraction by their greatest 
common measure at once, and the quotients will be the terms 
of the fraction required, of the same value as at first. 

EXAMPLES. 

1. Reduce }jf to its least terms. 

ttt = tt = » = tt = f = h the answer. 
Or thus : 

216) 288 (1 Therefore 72 is the greatest common 

216 measure ; and 72) |jj = J the 

Answer, the same as before. 

72) 216 (3 
216 

2. Reduce t0 iis ^west terms. Ans. J. 

3. Reduce J$f to its lowest terms. Ans. §. 

4. Reduce £f$ to its lowest terms. Ans. |. 



4. If the two right-hand figures of any number be divisible by 4, the 
whole is divisible by 4. And if ihe three right-hand figures be divisible 
by 8, the whole is divisible by 8. And so on. 

5. If the sum of the digits in any number be divisible by 3, or by 9, 
the whole is divisible by 3, or by 9. 

6. If the right-hand digit be even, and the sum of all the digits be di- 
visible by 6, then the whole is divisible by 6. 

7. A number is divisible by 11, when the sum of the 1st, 3d, 5th, &c. 
or all the odd places, is equal to the sum of the 2d, 4th, 6th, Ac. or of 
all the even places of digits. 

8. If a number cannot be divided by some quantity less than the 
square root of the same, that number is a prime, or cannot be divided 
by any number whatever. 

9. All prime numbers, except 2 and 5, have either 1, 3, 7, or 9, in the 
place of units; and all other numbers are composite, or can be divided. 

10. When numbers, with the sign of addition or subtraction between 
them, are to be divided by any number, then each of those numbers 

must be divided by it. Thus l^iAlli = 5-1-4 — 2 = 7. 

11. But if the numbers have the sign of multiplication between them, 
only one of them must be divided. Thus, 

10X8X3 _ 10 X 4 X 3 _ 10 X 4 X 1 = 10 X 2 X 1 20 

6X2" 6X1 2X1 1X1 1 



56 



AKITHIUETIC. 



CASE II. 

To Reduce a Mixed Number to its Equivalent Improper Frac- 
tion* 

* Multiply the integer or whole number by the deno- 
minator of the fraction, and to the product add the numera- 
tor ; then set that sum above the denominator for the fraction 
required. 

EXAMPLES. 

1. Reduce 23} to a fraction. 
23 
5 



115 Or, thus, 

2 (23X5)+2 117 t . 
_ ~ = the Answer. 

117 

5 

2. Reduce 12 J to a fraction. Ans. 1 j 5 . 

3. Reduce 14 T \ to a fraction. Ans. y 7 7 . 

4. Reduce 183 f s r to a fraction. Ans. 3 £-f ». 

CASE III. 

To Reduce an Improper Fraction to its Equivalent Whole or 
Mixed Number. 

f Divide the numerator by the denominator, and the quo- 
tient will be the whole or mixed number sought. 

EXAMPLES. 

1. Reduce ^ to its equivalent number. 
Here y or 12-7-3=4, the Answer. 



* This is no more than first multiplying a Quantity by some number, 
and then dividing the result back again by the same : which it is evi- 
dent does not alter the value ; for any fraction represents a division of 
the numerator by the denominator. 

t This rule is evidently the reverse of the former ; aod the reason of 
it is manifest from the nature of Common Division. 



.msBUonozr or vulgab fkactiojci. 57 

3. Reduce y to its equivalent number. 

Here y or 15-4-7=24, the Answer. 

& Reduce 7 T y to its equivalent number. 

Thus, 17 ) 749 ( 4 T V4 
68 

69 So that \y =44 jV, the Answer. 
68 

1 

4. Reduce y to its equivalent number. Ans. 8. 

5. Reduce to its equivalent number. Ans. 54}}. 

6. Reduce *ff • to its equivalent number. Ans. 171ff. 

CASE IV. 

To Reduce a Whole Number to an Equivalent Fraction, hav- 
ing a Given Denominator. 

* Multiply the whole number by the given denominator ; 
then set the product over the said denominator, and it will 
form the fraction required. 

EXAMPLES. 

1. Reduce 9 to a fraction whose denominator shall be 7. 

Here 9X7=63: then V is the Answer; 
For y =63-r-7=9, the Proof. 

2. Reduce 12 to a fraction whose denominator shall be 13. 

Ans. yj. 

3. Reduce 27 to a fraction whose denominator shall be 11. 

Ans. yp. 

CASE V. 

7b Reduce a Compound Fraction to an Equivalent Simple 
one. 

f Multiply all the numerators together for a numerator, 
. and all the denominators together for a denominator, and 
they will form the simple fraction sought. 



• Multiplication and Division being here equally osetf, the result 
mst be the same as the quantity first proposed. 

f The troth of this rule may be shown as follows : Let the compound 
faction be | of Now } of ^ is which is ^ \ tonaftqtMutty 
Vol, I 9 



58 AR I T H METIC, 

When part of the compound fraction is a whole or mixed 
number, it must first be reduced to a fraction by one of the 
former cases. 

And, when it can be done, any two terms of the fraction 
may be divided by the same number, and the quotients used 
instead of them. Or, when there are terms that are com* 
mon, they may be omitted, or cancelled. 

EXAMPLES. 

1. Reduce J of f of f to a simple fraction. 

_ 1X2X3 6 1 , . 

Here 2^3X4 = 24 = 4* *• An8Wer ' 

♦ 

Or, ^^| = 1, by cancelling the 2*s and 3's. 

2. Reduce | of J of H to a simple fraction. 

2X3X10 60 12 4 . . 
Here 3 X5X11 = 165 = 33 = TP the Answer ' 
2 

_ 2X£X20 4 . . _ . 

^XjgXll = IP same as tofore, by cancelling the 

3's, and dividing by 5's. 

3. Reduce ^ of f to a simple fraction. Ana. 

4. Reduce $ of } of $ to a simple fraction. Ans. }• 

5. Reduce f pf f of 3} to a simple fraction. Ans. }. 

6. Reduce f of \ of } of 4 to a simple fraction. Ans. f • 

7. Reduce 2 and f of £ to a fraction. Ans. 2. 

CASE vi. 

Tb Reduce Fraction* of Different Denominations to Equivalent 
Fractions having a Common Denominator. 

* Multiply each numerator by all the denominators ex* 
cept its own for the new numerators : and multiply all the 
denominators together for a common denominator. 



f of f will be ^X2 or ^ ; that is, the numerators are multiplied to- 
gether, and also the denominators, as in the Role. When the compound 
fraction consists of more than two single ones ; having first reduced 
two of them as above, then the resulting fraction and a third will be the 
same as a compound fraction of two parts ; and so on to the last of all. 

* This is evidently no more than multiplying each numerator and its- 
denominator by the same quantity r and consequently the value of the* 
fraction ti not altered. 



mttcnoK or vttloae fractions. 



59 



Male, It it evident, that in this and several other operations, 
'when any of the proposed quantities are integers, or mixed 
tratnbersy or oompoond fractions, they must first be reduced, 
by their proper Rules, to the form of simple fractions. 



EXAMPLES. 



1. Reduce *, }, and }, to a common denominator, 

1 X 3 X 4 = 12 the new numerator for £. 
2X2X4 = 16 ditto |. 
3 X 2 X 3 =* 18 ditto f . 

2 X 3 X 4 = 24 the common denominator. 
Therefore the equivalent fractions are £}, £f , and J }. 

Or the whole operation of multiplying may often be per- 
formed mentally, only setting down the results and given 
^ 0M ««*»^tt|» = «,Jf,if = ^A.A»by abbre. 
vianon. 

2. Reduce f and | to fractions of a common denominator. 

Ans - Ih *f • 

3. Reduce |, }, and f to a common denominator, 

Ans. f$, |f, 

4. Reduce |, 2}, and 4 to a common denominator. 

Ans.H,}f,W- 

iVsfe 1, When the denominators of two given fractions 
have a common measure, let them be divided by it ; then 
multiply the terms of each given fraction by the quotient 
arising from the other's denominator. 

Ex. ft and ft = {ft {ft, by multiplying the former 
5 7 by 7 and the latter by 5. 

2. When the less denominator of two fractions exactly 
divides the greater, multiply the terms of that which has the 
less denominator by the quotient. 

Ex. 4 and ft = ft and ft, by mult, the former by 2.* 
2 

3. When more than two fractions are proposed, h is some- 
times convenient, first to reduce two of them to a common 
denominator ; then these and a third ; and so on till they he 
-all reduced to their least common denominator. 

Ex. | and J and J = J and f and \ =» \\ and \\ and \\. 



CASK VII. 

7b reduce Complex Fractions to single ones. 

Reduce the two parts both to simple fractions ; then mul. 
tiply the numerator of each by the denominator of the other ; 
which is in fact only increasing each part by equal multi- 



fft Asrrusnc. 

pUCatjons, which makes no difference in the value of the 
▼hole. 



6' 



And 



= A lso?i = V : 
4 12 4J- f 



17 v 2 _ 34 
~5 * 9 ~ 45' 



CASE Tin. 

3b find the valve of a Fraction in Parte of the Integer. 

Multiply the integer by the numerator, and divide the 
product by the denominator, by Compound Multiplication 
and Division, if the integer be a compound quantity. 

Or, if it be a single integer, multiply the numerator by the 
parts in the next inferior denomination, and divide the pro- 
duct by the denominator. Then, if any thing remains, mul- 
tiply it by the parts in the next inferior denomination, and 
divide by the denominator, as before ; and so on as far as ne- 
cessary ; so shall the quotients, placed in order, be the value 
of the fraction required.* 

EXAMPLES. 



1. What is the \ of 2J6*? 
By the former part of the Rule 
2/6* 
4 



5) 9 4 

11 \QsQd2iq. 



2. What is the value of |of III 
By the 2d part of the Rule, 
2 
20 

3) 40 (13* 4d Ans. 

1 
12 

3) 12 (4d 



3. Find the value of } of a pound sterling. Ans. lit W. 

4. What is the value of } of a guinea ? Ads. As 8d. 
& What is the value of J of a half crown ? 



6. What is the value of} of 4* \0dl 

7. What is the value off lb troy ? 

8. What is the value of ft of a cwt ? 

9. What is the value of J of an acre ? 
10. What is the value of ft of a day? 



Ans 1* lOjrf. 
Ans. Ullfd. 
Ans. 9 oz 12 dwts. 
Ans. 1 qr 7 lb. 
Ans. 3 ro 20 po. 
Ans. 7 hrs 12 min. 



* The numerator of a fraction being considered as a remainder, in 
. Division, and the denominator as the divisor, this rule is of ibe same 
aature as Compound Division, or the valuation of remainders in the 
Hole of Three, before explained. 



mwcTioN op tom^r nuonoics. tt 



cabs is. ft 



7> JMpMtf a JVaefofi ,/r©ii one JDe^onifiafiofi to another. 

* Consider how many of the leas denomination make 
one of the greater ; then multiply the numerator by that 
number, if the reduction be to a leas name, but multiply the 
denominator, if to a greater. 



1. Reduce } of a pound to the fraction of a penny. 

f X Y * Y = T = l Vt Ae Answer. 

2. Reduce 4 of a penny to the fraction of a pound. 

# x tV * 1V 88 ^ Answer. 

$. Reduce ftl to the fraction of a penny. Ana. *fd. 

4. Reduce }g to the fraction of a pound. Ana. t*Vt* 

5. Reduce f cwt to the fraction of a lb. Ana. y . 

6. Reduce ] dwt to the fraction of a lb troy. Ana. T £ r . 

7. Reduce f crown to the fraction of a guinea. Ana. fy, 

8. Reduce { half-crown to the fract. of a shilling. Ana. . 

9. Reduce 2* 6d to the fraction of a £. Ana. \. 
10. Reduce 17* Id 3{? to the fraction of a £. Ans. fiff. 



ADDITION OF VULGAR FRACTIONS. 

If the fractions have a common denominator ; add all the 
numerators together, then place the sum over the common 
denominator, and that will be the sum of the fractions re* 
quired. 

f If the proposed fractions have not a common denomina- 
tor, they must be reduced to one. Also compound fractions 



* This is the same as the Rale of Reduction in whole numbers from 
one denomination to another. 

\ Before fractions are reduced to a common denominator, they are 
quite dissimilar, as much as shillings and pence are, and therefore can- 
not be incorporated with one another, any more than these can. But 
when they are reduced to a common denominator, and made parts of 
the same thing, their sum, or difference, may then be as properly ex- 
pressed by the sum or difference of the numerators, as the sum or dif- 
ference of any two quantities whatever, by the sum or ditfemw* <A 



62 AKrr&wftncr 

must be reduced to simple ones, and fractions of different 
denominations to -those |>f the same denomination. Then 
add the numerators, as before. As to mixed, numbers, they 
may either be reduced to improper fractions, and so added 
with the others ; or else the fractional parts only added, and 
the integers united afterwards. 

' examples. 

1. To add | and £ together. 

Here }+| = J = If, the Answer. 

2. To add } and | together. 

* + * = it + tt = it = Hi* Answer. 

3. To add J and 7\ and | of f together. 

i+7i+i off = i+.V+i « i+V+t = V 

4. To add ^ and 4 together. Ans. If. 

5. To add $ and f together. Ans. 

6. Add \ and ft together. Ans. ft. 



their individuals. Whence the reason of the Role is manifest, both for 
Addition and Sob traction. 

When several fractions are to be collected, it is commonly best first 
to add two of them together that most easily reduce to a common de- 
nominator; then add their sum and a third, and so on. 

Note 2. Taking any two fractions whatever, ft and f-j- t for example, 
after reducing them to a common denominator, we judge whether they 
are equal or unequal, by observing whether the products 35 X 11, and 
7 X 65, which constitute the new numerators, are equal or unequal. 
If, therefore, we have two equal products 35x11=7X56, we may 
compose from them two equal fractions, as -g-f- = ft, or = ^ . 

If, then, we take two equal fractions, such as ft and we shall 
have 36 X 11 = 7 X 56 ; taking from each of these 7 X 11, there will 

OK J 

remain (35 — 7) X 11 = (65 — 11) X 7, whence we have ^ = 

55 — 11 

In like manner, if the terms of ft were respectively added to 
those of ff, we should have = tf = ft. 

Or, generally, if = ^, it may in a similar way be shown, that 
_ tf _ c 

Hence, when two fractions are of equal valuta the fraction formed by ta- 
king the sum (or Ike diffirence) of their numerators respctireJy, and of their 
denominators respecticelij, it a fraction equal in value to each of the original 
fractions. This proposition will be found useful in the doctrine of pro- 
portions. 



MULTlFJJCATIOlf OF VULGAR FRACTIOUS. 68 

• 7. What is the sum of f and f and 4 ! Ans. l}ff . 

& What is the sum of f and | and ? Ans. 3f|. 

9. What is the sum off and \ of |, and 9ft ? Ans. 10^. 

10. What is the sum of } of a pound and $ of a shilling ? 

Ans. l |»t or 13* MM 2}?. 
11: What is the sum of } of a shilling and ft of a penny 1 

Ans. y/dor7d Hftf- 
12. What is the* sum of } of a pound, and f of a shilling, 
and ft of a penny ? Ans. fiff* or 3 * ld Hffl- 



SUBTRACTION OF VULGAR FRACTIONS. 

Prepare the fractions the same as for Addition, when 
necessary ; then subtract the one numerator from the other, 
and set the remainder over the common denominator, for the 
difference of the fractions sought. 

EXAMPLES. 

1. To find the difference between £ and £. 

Here $ — £ = $ = |, the Answer. 

2. To find the difference between J and f . 

? — i = H — H = *V Answer. 

3. What is the difference between ft and ft 1 Ans. 

4. What is the difference between ft and ft ? Ans. ft. 

5. What is the difference between ft and ft ? Ans. T \fr. 

6. What is the diff. between 5} and 4 of 4$ ? Ans. 4^- 

7. What is the difference between £ of a pound, and | of 
I of a shilling ? Ans. 'ft 8 or 10* Id \\q. 

8. What is the difference between $ of 5£ of a pound, and 
} of a shilling. Ans. or 11 Ss llftd. 



MULTIPLICATION OF VULGAR FRACTIONS. 
^ * Reduce mixed numbers, if there be any, to equivalent 



♦ Multiplication of any thing by a fraction, implies the taking some 
put or parts of the thing; it may therefore be truly eipresaadb^ * 



▲RlTHttBTft/* 



fractions ; then multiply all the numerators together for a 
numerator, and all the denominators together for a denomi- 
nator, which will give the product required. 



BXAMPLS8. 



1. Required the product of } and f . 

Here J Xf =^=^, the Answer. 
Or|X*=iX*=*. 

2. Required the continued product of |, 3£, 5, and f of j. 

Heie y x T T T 5 r "4xf""I"'^ 

3. Requited the product of f and f . Ans. 

4. Required the product of <fo and Ana. ^ . 

5. Required the product of f, }|. Ans. 

6. Required the product of }, }, and 3. Ans. 1. 

7. Required the product of }, f , and 4^. Ans. 3^. 

8. Required the product of f , and } of f . Ans. 

9. Required the product of 6, and } of 5. Ans. 20. 

10. Required the product off of j, and f of 3f. Ans. |{. 

11. Required the product of 8} and 4}}. Ans. 14|}f • 

12. Required the product of 5, f, f of}, and 4}. Ans. 2^-. 



DIVISION OP VULGAR FRACTIONS. 

* Prepabb the fractions as before in Multiplication : then 
divide the numerator by the numerator, and the denominator 
by the denominator, if they will exactly divide : but if not, 



compound fraction ; which is resolved by multiplying together the 
numerators and the denominators. 

Ad/a. A Fraction b best multiplied by an integer, by dividing the 
denominator by it ; bat if it will not exactly divide, then multiply the 
numerator by ft. 

• Division being the reverse of Multiplication, the reason of the rale 
is evident. 

Acta, A fraction is best divided by an integer, by dividing the nume- 
rator by It ; bat if it will not exactly divide, then multiply the denorni- 
nstor by it. 



ftULE Or THXE1 IN VULGAR FRACTIONS. 65 

invert the terms of the divisor, and multiply the dividend by 
K, as in Multiplication. • 

EXAMPLES. 

1. Divide V by |. 

Here y 4. f = a = If, by the first method. 

2. Divide f by ft. 
Heref^A=*Xy=|Xf = V-4}. 



3. It is required to divide by |. Ans. f . 

4. It is required to divide ^ by }. Ans. fa 

5. It is required to divide y by J. Ans. 1 J. 

6. It is required to divide { by y . Ans. y^. 

7. It is required to divide by |. Ans. 4. 

8. It is required to divide ij- by }. Ans. . 

9. It is required to divide ft by 3. Ans. fa 

10. It is required to divide } by 2. Ans. fa 

11. It is required to divide 7± by 9|. Ans. }|. 

12. It is required to divide f of } by | of 7|. Ans. jJt. 



RULE OF THREE IN VULGAR FRACTIONS. 

Make the necessary preparations as before directed ; then 
multiply continually together, the second and third terms, 
and the first with its parts inverted as in Division, for the 
answer*. 

EXAMPLES. 

1. If } of a yard of velvet cost f of a pound sterling ; what 
will -f € of a yard cost ? 

2. What will 3f oz. of silver cost, at 6s 4d an ounce ? 

Ans. I/ I* 4J<*. 



* This Is only multiplying the 2d And Sd terms together, and divld- 
bftbe product by the first, as in the Rale of Three in whole nunbtm 
Vol. I. 10 



AJtlTHXBTIC. 



a If A of a ihip be worth 27813* 64; what are A of 
her worth? • An* 8871 12$ Id. 

4. What is the purchase of 12901 bank-stock, at 108f per 
cent.? Ana. 133«1# W. 

5. What is the interest of 2731 15* for a year, at 3£ per 
cent.? Ans. 8Z17* 11J<*. 

6. If | of a ship be worth 731 Is 3d ; what part of her is 
worth 250Z 10* ? Ans. J. 

7. What length must be cut off a board that is 7| inches 
broad, to contain a square foot, or as touch as another piece 
of 12 inches long and 12 broad ? Ans. 18jf inches* 

8. What quantity of shalloon that is J of a yard wide, will 
line flivards of cloth, that is 2} yards wide? Ans. 31 J yds. 

fc' if the penny loaf weigh 6^ 0z. when the price of 
wheat is 5* the bushel ; what ought it to weigh when the 
wheat is 8a 6d the bushel ? Ans. 4^ oz. 

10. How much in length, of a piece of land that is 11{4 
ptoles broad, will make an acre of land, or as much as 40 
pdlei in length and 4 in breadth ? Ans. 13 T y T poles. 

Hi If a courier perform a certain journey in 35$ days, 
travelling 13f hours a day ; how long would he be in per- 
forming the same, travelling only 11 -ft hours a day ? 

Ans. 40f)f days. 

12. A regiment of soldiers, consisting of 976 men, are to 
be new clothed ; each coat to contain 2J yards of cloth that 
is If yard wide, and lined with shalloon J yard wide : how 
many yards of shalloon will line them ? 

Ans. 4531 yds 1 qr 2f nails. 



DECIMAL FRACTIONS. 

A Decimal Fraction is that which has for its deno- 
minator an unit (1), with as many ciphers annexed as the 
numerator has places ; and it is usually expressed by setting 
down thgfiumetator only, with a point before it, on the left, 
hand. Thus, & is -4, and ffr is -24, and rifo is *074, and 
ttVvVt i* '00124 ; where ciphers are prefixed to make up as 
many placet ma are ciphers in the denominator, when there 
is a deficiency in the figures. 

A mixpd numbejr is made up of a whole number with some 
.decimal fraction, the one being separated from the other by 
a fcoint. Thus, 3*25 is the same as or f| f . 

Ciphers on the right-hand of decimals make no alteration 
in their value ; for -4, or '40, or *400 are decimals having all 
IbeMne value, each being = ^ or f . Bur when they are 



ADDlfHMI W MCDIAIS. 



placed m the left-hand, they decrease the value in m ten4bld 
proportion : That, -4 is ^ or 4 tenths ; but -04 is only t4y> 
or4himfw>dths > and -004 is «nly y/^, or 4 thousandths. 

Ia sWrknalfr as well as in whole numbers, the vahief .of 
tfcefiaee* increase towards Joe left-hand, and 4e&*ase to. 
wards die right, both in the same tenfold proportion $ s# in 
the following Scale or Table of Notation. 



3333338*333939 



ADDITION OF DECIMALS. 

Sbt the numbers under each other according to the value 
of their places, as in whole numbers ; in which state the 
decimal separating points will stand all exactly under each 
other. Then, beginning at the right hand, add up all the 
columns of numbers as in integers ; and point off as many 
places for decimals, as are in the greatest number of decimal 
places in any of the lines that are added ; or pla<?e the point 
directly below all the other points. 



1. To add together 29-0146, and 3146*5, and 2 J 09, tod 
03417, and 14-16. 



29 0146 
8146-5 
2109- 

•62417 

1416 




EXAMPLES. 



5299-29877 the Sum. 



6S 



ABXTHXSTIC* 



2. What is the sum of 276, 39-213, 72014-9, 417, -and 
50821 Ans. 77770-113. 

3. What is the sum of 7530, 16 201, 3-0142, 957*13, 
6*72119 and -03014 ? Ans. 8513-09653. 

4. What is the sum of 312-09, 3-5711, 7195-6, 71-498, 
9739-215, 179, and -0027 ? Ans. 17500-9718. 



SUBTRACTION OF DECIMALS. 

Place, the numbers under each other according to the 
value of their places, as in the last Rule. Then, beginning 
at the right-hand, subtract as in whole numbers, and point off 
the decimals as in Addition. 

EXAMPLES. 

1. To find the difference between 91*73 and 2*138. 
91*73 
2*138 



Ans. 89.592 the Difference. 



2. Find the diff. between 1 -9185 and 2*73. Ans. 0*81 15. 

3. To subtract 4*90142 from 214*81. Ans. 209*90858. 

4. Find the diff. between 2714 and -916. Ans. 2713*084. 



MULTIPLICATION OF DECIMALS. 

* Place the factors, and multiply them together the same 
as if they were whole numbers. — Then point off in the pro- 
duct just as many places of decimals as there are decimals in 
both the factors. But if there be not so many figures in the 
product, then supply the defect by prefixing ciphers. 



* The rule will be evident from this example Let it be required to 
multiply -12 by '961 ; these numbers are equivalent to -jfo and ftftfc ; 
the product of which is v \ g g g = 04432, by the nature of Notation, 
which consists of as many places as there are ciphers, that is v of as 
many places as there are in both numbers. And in like manner for any 
other numbers. 



HULTIPLICATfOll OF DECIMAL!. 



69 



EXAMPLES* 

1. Multiply -321006 
by -2465 



1605480 
1926576 
1284384 
642192 



Ana.' •0791501640 the Product. 



2. Multiply 79-347 by 23-15. Ans, 1836-88305. 

3. Multiply -63478 by -8204. Ana. -520773512. 

4. Multiply -385746 by -00464. Ana. -00178986144. 

CONTRACTION I. 

To multiply Decimals by 1 with any Number of Ciphers, as 
by 10, or 100, or 1000, $c. 

This is done by only removing the decimal point so many 
places farther to the right-hand, as there are ciphers in the 
multiplier ; and subjoining ciphers if need be. 

EXAMPLES. 

1. The product of 51-3 and 1000 is 51300. 

2. The product of 2-714 and 100 is 
8. The product of -916 and 1000 is 
4. The product of 2i»31 and 10000 is 

CONTRACTION II. 

To contract the Operation so as to retain only as many Deci- 
mals in the Product as may be thought necessary, when the 
Product would naturally contain several more Places. 

Set the unit's place of the multiplier under the figure of 
the multiplicand whose place is the same as is to be retained 
for the last in the product ; and dispose of the rest of the 
figures in the inverted or contrary order to what they are 
usually placed in. — Then, in multiplying, reject all the 
figures that are more to the right-hand than each multiplying 
figure, and set down the products, so that their njVv\-\i«iA 



70 



figures may fall in a column straight below each other ; but 
observe to increase the first figure of every line with what 
would arise from the figures omitted, in this manner namely 
1 from 5 to 14, 2 from 15 to 24, 3 from 25 to 34, dec. ; and 
the sum of all the lines will be the product as required, com- 
monly to the nearest unit in the last figure. 

EXAMPLES. 



1. To multiply 27-14986 by 92-41035, so as to retain only 
four places of decimals in the product 



Contracted Way. 
27 14986 
53014-29 



Common Way. 
27-14986 
92-41035 



24434874 


13 


574930 


542997 


81 


44958 


108599 


2714 


986 


2715 


108599 


44 


81 


542997 


2 


14 


24434874 




2508-9280 


2508-9280 


650510 



2. Multiply 480*14936 by 2-72416, retaining only four 
decimals in the product. 

3. Multiply 2490-3048 by -5*73286, retaining only five 
decimals in the product. 

4. Multiply 325-701428 by -7218393, retaining only three 
decimals in the product. 



DIVISION OF DECIMALS. 



Divide as in whole numbers ; and point off in the quo- 
tient as many places for decimals, as the decimal places in 
the dividend exceed those in the divisor*. 



* The mason of this Rale is evident; for, since the divisor multiplied 
by the quotient gives the dividend, therefore the Dumber of deoimal 
places iu the dividend, is equal to those in the divisor and quotient, 
taken together, by the nature of Multiplication ; and consequently 
the quotient itself mast contain as many as the dividend eieeeds the 
divisor. 



DIVISION OF 9MQULLL8. 



71 



Another way to know the place for the decimal point is 
this : The first figure of the quotient must be made to occupy 
the same place, of integers or decimals, as that figure of the 
dividend which stands over the unit's figure of the first pro- 

When the places of the quotient are not so many as) the 
Rule requires, the defect is to be supplied by prefixing 
ciphers. 

When there happens to be a remainder after the division ; 
or when the decimal places in the divisor are more than those 
in the dividend ; then ciphers may be annexed to the divi- 
dend, and the quotient carried on as far as required. 



EXAMPLES. 



00272589 -2639 



1. 

178) "48580998 ( 
1392 
460 
1049 
1599 
1758 
156 



3. Divide 123-70536 by 54 25. 

4. Divide 12 by -7854. 

5. Divide 4195-68 by 100. 

6. Divide -8297592 by -153. 



1. 

n 27-00000 (102-3114 
6100 

8220 
3030 
3910 
12710 
2154 

Ans. 2-2802. 
Ans. 15-278. 
Ans. 41-9568. 
Ans. 5-4232. 



CONTRACTION I. 



When the divisor is an integer, with any number of ciphers 
annexed : cut off those ciphers, and remove the decimal 
point in the dividend as many places farther to the left as 
there are ciphers cut off, prefixing ciphers, if need be ; then 
proceed as before. 

EXAMPLES. 

1. Divide 45-5 by 2100. 

21-00) -455 ( 0216, <&c. 
35 
140 
14 



3. Divide 41020 by 32000. 
& Divide 953 by 21600. 

4. Divide 61 by 79000. 



73 



ARITHMETIC. 



CONTRACTION II. 



Hence, if the divisor be 1 with ciphers, as 10, 100, or 
1000, &c ; then the quotient will be found by merely mov- 
ing the decimal point in the dividend so many places farther 
to the left, as the divisor hath ciphers ; prefixing ciphers if 
need be. 



EXAMPLES. 

So, 217-3 -r 100 = 2 173 Ans. 419 10 = 

And 5-16 -f- 100= Ans. -21 -f- 1000 = 

CONTRACTION HI. 

When there are many figures in the divisor ; or when only 
a certain number of decimals are necessary to be retained 
in the quotient ; then take only as many figures of the divi- 
sor as will be equal to the number of figures, both integers 
and decimals, to be in the quotient, and find how many times 
they may be contained in the first figures of the dividend, as 
usual. 

Let each remainder be a new dividend ; and for every such 
dividend, leave out one figure more on the right-hand side 
of the divisor ; remembering to carry for the increase of the 
figures cut off, as in the 2d contraction in Multiplication. 

Note. When there are not so many figures in the divisor 
as are required to be in the quotient, begin the operation with 
all the figures, and continue it as usual till the number of 
figures in the divisor be equal to those remaining to be found 
in the quotient ; after which begin the contraction. 

examples. 

1. Divide 2508-92800 by 92*41035, so as to have only 
four decimals in the quotient, in which case the quotient will 
contain six figures. 



Contracted. 
92-4103,5) 2608-928,09(971498 
660721 
13849 
4608 
912 
80 
6 • 



Coitittum. 
92.4103,5) 2608-928,06 (27-1498 
66072106 
18848610 
46075750 
91116100 
79467850 
5639570 



2. Divide 4109*2351 by 230-409, so that the quotient may 
contain only four decimals. Ans. 17*8345. 



REDUCTION OP DECIMALS. 



78 



3. Divide 37- 10438 by 5713-96, that the quotient may 
contain only five decimals. Ans. '00649. 

4. Divide 913 08 by 2137-2, that the quotient may contain 
only three decimals. 



REDUCTION OF DECIMALS. 



CASE I, 

To reduce a Vulgar Fraction to its equivalent Decimal. 

Divide the numerator by the denominator, as in Division 
ef Decimals, annexing ciphers to the numerator as far as 
necessary ; so shall the quotient be the decimal required*. 



* The following method of throwing a vulgar fraction, whose de- 
nominator is a prime number, into a decimal consisting of a great num- 
ber of figures, is given by Mr. Cohan in page 162 of Sir Isaac Newton's 
Fluxions. 

[ EXAMPLE. 

Let ^ be the fraction which is to be converted into an equivalent 
decimal. 

Then, by dividing in the common way till the remainder becomes a 
tingle figure, we shall have -fy — -03448^ for the complete quotient^ 
and this equation being multiplied by the numerator 8, wiJJ give -f^ == 
275g4ji4. f ( ,r rather u % ~ 27586\/ ff : and if this be substituted instead 
of the friction in the first equation, it will make ^ = -0344827586^. 
Again, let this eqaation be multiplied by 6, and it will give A = 
'206*8965517^ ; and then by substituting as before 
-if = -034482758610689605175^ ; 

and so on, as far as may he thought proper; each fresh multiplication 
doubling the number of figures in the decimal value of the fraction. 

In the present instance the decimal circulates in a complete period of 
28 figures, i. e. one less than the denominator of the fraction. This, 
again, may be divided into equal periods, each ot 14 figures, as below : 
•03448275862068 
•96551724137931 

in which it will be found that each figure with the figure vertically be- 
low it makes 9; + 9 = 9; 3 -f 6 ^ 9 ; and so on. This circulate also 
comprehends all the separate values of &c. in correspond- 

ing circulates of 28 figures, only each beginning in a distinct place, easi- 
ly ascertainable. Thus, ^ — 06896, &c. beginning at the 12th place 
of the primitive circulate. ^ = 103448, &c. beginning at the 28th 
plnce. So that, in fact, this circle includes 28 complete circles. 

8ee, on this curious subject, Mr. Goodwyn's Tables of Decimal Cir- 
cles, and the Ladies' Diary for 1824. 
Vol. I. 11 



AKTEOtBTICV 



BX AMPLE** 



I. Reduce yV to a decimal. 
24 = 4 X 6. Then 4) 7- 



6) 1*750000 
•291606 dec. 



2. Reduce J, and J, and J, to decimals. 



3» Reduce f to a decimal. 
4. Reduce ^ to a decimal. 
5* Reduce T J ¥ to a decimal. 



6\ Reduce to a decimah. 



Ans* *25, and # 5, and *75v 
Ana. -625. 
Ana. -12. 
Ana. -08135. 

Ana. -14*154 IK- 



CASE n. 



To Jmd the Yakut of a Decimal in terms of the Inferior 
Denominations. 

Multiply the decimal by the number of parts in the 
next lower denomination ; and cut off as many places for a 
remainder to the right-hand, as there are places in the given 
decimal. 

Multiply that remainder by the parts in the next lower 
denomination again, cutting off for another remainder as 
before. 

Proceed in the same manner through all the parts of the 
integer ; then the several denominations separated on the 
left-hand will make up the answer. 

Note, This operation is the same as Reduction Descending 
in whole numbers. 



1. Required to find the value of -775 pounds sterling. 



EXAMPLE?. 



•775 
20 



s 15-500 
12 



4 6-0Q0 



Ans. 15* 6rf» 



■SDUCTlOlf OF BECfMALS. 76 

% What is the value of -625 shil ? fc Ans. 74*. 

3. What is the value of -86352 ? Ans. 17s 3-242. 

4. What is the value of -0125 lb troy? Ans. 3 dwts. 
& What is the value of -4694 lb troy ? 

Ans. 5 or. 12 dwts 15*744 gr. 
& What is the value of -625 cwt ? Ans. 2 qr 14 lb. 

7. What is the value of -009943 miles? 

Ans. 17 yd 1 ft 5-98848 inc. 
& What is the value of -6875 yd ? Ans. 2 qr 3 nls. 
9. What is the value of -3375 acr ? Ans. 1 rd 14 poles* 
19. What is the value of -2083 hhd of wine? 

Ans. 13-4229 gaL 

CASS III. 

lb reduce Integer* or Decimals to Equivalent Decimal* of 
Higher Denominations. 

Divide by the number of parts in the next higher deno- 
ounation ; continuing; the operation to as many higher de- 
aominations as may be necessary, the same as in Reduction 
Ascending of whole numbers. 

EXAMPLES. 

1. Reduce 1 dwt to the decimal of a pound troy* 

20 \ 1 dwt 
12 0-05 os 

1 0-004166 dec, lb. Ans. 

2. Reduce 9d to the decimal of a pound. Ans/ 0*3757. 

3. Reduce 7 drams to the decimal of a pound avoird. 

Ans. -027343751b. 

4. Reduce -264 to the decimal of a I. Ans. 0910833 6zc. /. 

5. Reduce 2-15 lb to the decimal of a cwt. 

Ans. -019196 + cwt. 

6. Reduce 24 yards to the decimal of a mile. 

Ans. 013636 &c. mile. 

7. Reduce -056 pole to the decimal of an acre. 

Ans. O0035 ac. 

8. Reduce 1-2 pint of wine to the decimal of a hhd. 

Ans. -00238 + hhd. 

3. Reduce 14 minutes to the decimal of a day. 

Ans. -009722 <fcc. da* 

JO. Reduce -21 pint to the decimal of a peck. 

Ans. -031325 pec. 

11. Reduce 28" 12" to the decimal of a minute. 



79 



AKIfflMlG* 



Hon, When (here 
to Vke ieemdl of the highest ; 

Set the given numbers directly under each other, for di- 
vidends* proceeding orderly from the lowest denomination 
to the highest. 

Opposite to each dividend, on the left-hand, set such a 
number for a divisor as will bring it to the next higher name ; 
drawing a perpendicular line between all the divisors and 
dividends ' 

Begin at the uppermost, and perform all the divisions : 
only observing to set the quotient of each division, as deci- 
mal parts, on the right-hand of the dividend next below it ; 
so shall the last quotient be the decimal required. 



EXAMPLES. 

1. Seduce 17* 9f J to the decimal of a pound. 
4 1 3- 
12 9-75 
20 J 17-8125 

£0 890625 Aits. 



2. Beduce 191 17s 3}d to a I. Ans. 19*86354166 &c. I. 

3. Reduce 15* 6d to the decimal of a I. Ans., -775/. 

4. Reduce l\d to the decimal of a shilling. Ans. *625*. 

5. Reduce 5 oz 12 dwts 16 gr to lb. Ans. -46944 dec. lb. 



RULE OP THREE IN DECIMALS. 

• Pbbpaxe the terms, by reducing the vulgar fractions to 
decimals, and any compound number either to decimals of 
the higher denominations, or. to integers of the lower, also 
the first and third terms to the same name : Then multiply 
and divide as in whole numbers. 

Note. Any of the convenient Examples in the Rule of 
Three or Rule of Five in Integers, or Vulgar Fractions, may 
be taken as proper examples to the same rules in Decimals. 
—The following example, which is the first in Vulgar Frac- 
tions, is wrought out here, to show the method. 



DUODKCTJLAXt* 77 

If I of a yard of relvet cost fJ, what will ^ yd cost? 

yd I yd I 9 d 

} = -375 -375 : -4 : : -3125 : -333 dec. or 6 8 

•4 



J = -4 -375) -12500 (-333333 fcc. 

1250 20 
125 



#6-60666 &c. 
^ = -8125 12 



Ans. 6* Bd. d 7-99099 &c. = 8d. 



DUODECIMALS. 



Duodecimals, or Cross Multiplication, is a rule used 
by workmen and artificers, in computing the contents of 
their works. ♦ 

Dimensions are usually taken in feet, inches, and quarters ; 
any parts smaller than these being neglected as of no con- 
sequence. And the same in multiplying them together, or 
computing the contents. The method is as follows. 

Set down the two dimensions to be multiplied together, 
one under the other, so that feet may stand under feet, inches 
under inches, &c. 

Multiply each term in the multiplicand, beginning at the 
lowest, by the feet in the multiplier, and set the result of 
each straight under its corresponding term, observing to car- 
ry 1 for every 12, from the inches to the feet. 

In like manner, multiply all the multiplicand by the inches 
and parts of the multiplier, and set the result of each term 
one place removed to the right-hand of those in the mul- 
tiplicand ; omitting, however, what is below parts of inches, 
only carrying to these the proper numbers of units from the 
lowest denomination. 

Or, instead of multiplying by the inches, take such parts 
of the multiplicand as these are of a foot. 

Then add the two lines together, after the manner of 
Compound Addition, carrying 1 to the feet for every 12 
inches, when these come to so many. 



78 



▲JEtlTHMBTIC. 



EXAMPLES. 

1. Multiply 4 f 7 inc. 2. Multiply 14 f 9 inc. 

by 6 4 by 4 6 



27 


6 


59 





1 


<** 


7 




Ans. 29 


°* 


Ans. 66 


4* 



3. Multiply 5 feet 7 inches by 9 f 6 inc. Ans. 43 f 6 J. inc. 

4. Multiply 12 f 5 inc by 6 f 8 inc. Ans. 82 9} 

5. Multiply 35 f 4£ inc by 12 f 3 inc. Ans. 433 4| 

6. Multiply 64 f 6 inc by 8 f 9i inc. Ans. 565 8f 

Nate, The denomination which occupies the place of 
inches in these products, means not square inches, but recf- 
angles of an inch broad and a foot long. Thus, the answer 
to the first example is 29 sq. feet, 4 sq. inches ; to the second 
66 sq. feet, 54 sq. inches* 



INVOLUTION. 

Involution is the raising of Powers from any given num. 
ber, as a root. 

A Power is a quantity produced by multiplying any given 
number, called the Root, a certain number of times conti- 
nually by itself. Thus, 

2 = 2 is the root, or 1st power of 2. 
2X2 = 4 is the 2d power, or square of 2. 
2X2X2= 8 is the 3d power, or cube of 2. 
' 2X2X2X2 = 16 is the 4th power of 2, &c. 

And in this manner may be calculated the following Table of 
the first nine powers of the first 9 numbers* 



nrroLtmozr. 79 



TABLE OT THE FIRST NINE POWERS OF NUMBERS. 



1 


5 


3d 


4th 


5 th 


6th 


7th 


8th 


9ih " 


1 

2 
3 
4 
5 
8 
7 
8 
9 


I 


1 


1 


1 


1 


1 


1 


1 


1 


8 


16 


32 


64 


128 


256 


512 





27 


81 


243 


729 


2187 


6561 


19683 


16 


64 


256 


1024 


409G 


16384 


65536 


262144 


25 


125 


625 


31^5 


15620 


78125 


390625 


1953125 


m 


216 


1296 


7776 


46656 


279936 


1679616 


10077696 


id 


343 


2401 16807 


117649 


823543 


5764801 


40353607 


H 
si 


512 

721) 


4006(32768 
6561 59049 


262144 


2097152 


16777216 


134217728 


[6814*] 


4782969 


43046721 


387420489 



The Index or Exponent of a Pnwer, is the number de- 
noting the height or degree of that power ; and it is 1 more 
than the number of multiplications used in producing the 
same. So 1 is the index or exponent of the 1st power or 
root, 2 of the 2d power or square, 8 of the third power or 
cube, 4 of the 4th power, and so on. 

Powers, that are to be raised, are usually denoted by 
placing the index above the root or first power. 

So 2 s = 4 is the 2d power of 2. 

2 3 = 8 is the 3d power of 2. 

2 4 = 16 is the 4th power of 2. 

540* is the 4th power of 540, ozc. 

When two or more powers are multiplied together, their 
.product is that power whose index is the sum of the expo- 
nent of the factors or powers multiplied. Or the multiplica- 
tion of the powers, answers to the addition of the indices. 
Thus, in the following powers of 2, 

1st 2d 3d 4th 5th 6th 7th 8th 9th 10th 
2 4 8 16 32 64 128 256 512 1024 
orS 1 2* 2 s 2* 2 8 2 s 2 7 2 s 2 B * 10 



80 



▲XtTHXETIC. 



Here, 4 X 4 = 16, and 2 + 2 = 4 its index ; 

and 8 x 16= 128, and 3 + 4= 7 its index; 
also 16 X 64 — 1024, and 4 + 6 = 10 its index. 



OTHER EXAMPLES. 

1. What is the 2d power of 45 1 Ans. 2025. 

2. What is the square of 4*16 ? Ans. 17*8056. 

3. What is the 3d power of 3-5 ? Ans. 42-575, 

4. What is the 5th power of -029 ? Ans. -00000002051 1149. 

5. What is the square of f ? Ans. f . 

6. What is the 3d power of f ? Ans. 

7. What is the 4th power of } ? Ans. ,VV- 



EVOLUTION. 



Evolution, or the reverse of Involution, is the extracting 
or finding the roots of any given powers. 

The root of any number, or power, is such a number, as 
being multiplied into itself a certain number of times, will 
produce that power. Thus, 2 is the square root, or 2d root 
of 4, because 2 a = 2 x 2 == 4 ; and 3 is the cube root or 3d 
root of 27, because 3 3 == 3 X 3 X 3 = 27. 

Any power of a given number or root may be found ex- 
actly, namely, by multiplying the number continually into 
itself. But there are many numbers of which a proposed 
root can never be exactly found. Yet, by means of deci- 
mals, we may approximate or approach towards the root, to 
any degree of exactness. 

Those roots which only approximate, are called Surd 
Roots ; but those which can be found quite exact, are called 
Rational Roots. Thus, the square root of 3 is a surd root ; 
but the square root of 4 is a rational root, being equal to 2 : 
also the cube root of 8 is rational, being equal to 2 ; but the 
cube root of 9 is surd or irrational. 

Roots' are sometimes denoted by writing the character •/ 
before the power, with the index of the root against it. 
Thus, the 3d root of 20 is expressed by f/JJO ; and the square 



SQUARE ROOT. 



SI 



root or 2d root of it is ^/20, the index 2 being always omit- 
ted, when only the square root is designed. 

When the power is expressed *»y several numbers, with 
the sign + or — beiwxeu them, a line is drawn from the top 
of the sign over all the parts of it : thus the third root of 
-15 — 12 is \/ 45 — 12, or thus, y(45— - 12), inclosing the 
numbers in parentheses. 

But all roots arc now often designed like powers, with 

i 

fractional indices ; thus, the square root of 8 is 8 3 , the cube 
root of 25 is 25», and the 1th root of 4i> — 18 is (45 - 18)*. 

TO EXTRACT THE SQUARE ROOT. 

* Divide the given number into period" of two figures 
each, by setting a point over the place of units, another over 
the place of hundreds, a.id so on, over every second figure, 
both to the left-hand in integers, and to the right in deci- 
mals. 



* The reason for separating the figures of the dividend into periods 
or portions of two places each, U, that the square of any single figure 
never consists of more than two places; t tic square of a number of two 
figures, <if nmi more than four places, and so on. So that there will he 
as many figures in the root us the given number contains periods so di- 
vided or parted off. 

And the reason of the several steps hi the operation appears from the 
ilgchraic form of the -quare of any number of terms, whether two or 
three or more. Thus 

(«- h)2 . a-* -•- 2ntt-\- fta _- aa \-( % la f ft) ft, the square of two terms ; 
wuerc it appears that a i* the first term of the root, and ft the second 
term ; al«*o a the first divisor, and 'lie new divisor i« 2a j-ft, or double 
the first term i». creased by the second. And hence the manner of ex- 
traction is thus : 

h: divisor a) »*2 -f Sift -|- ft (a j- ft the root. 

?.a di-;$or°.a-j-ft .^aft-j-fts 
ft I ft :-fta 



Again, for a root of three P'.Im, ft. r f thus. 

(a \-b-\-c)i -■•|3"-2rtft bi -' r %ar-\-%bc j /8 

a- -i ( m 2a ; h)h ■: (2/*' r *2ft j r)c, the 
square of three terms, where a is too fuM term of the root, ft the «»rond, 
and c the third term ; also a the fu?t divisor, 2.i- r ft the second, and &r 
-| 'lb --c the third, each consi.-ting of the double of the root increased 
by thu next term of 'lie -»»m. . And the mode of extr.ictifui agrees with 
iiilu. r?«.e far'.iier, Case 2. of Kvolution ii. ire. Algebra. 

hi' -■; "»h , . 

I* or an approximation observe that Vul± v - a. -— ---nearly in 

All cases where <i it small in respect of a. 
Vol. J. l'J 



82 



Find the greatest square in the first period en the left-hand, 
and set its root on the right-hand of the given number, after 
the manner of a quotient figure in Division. 

Subtract the square thus found from the said period, and 
to the remainder annex the two figures of the next following 
period, for a dividend. 

Double the root above mentioned for a divisor ; and find 
how often it is contained in the said dividend, exclusive of its 
right-hand figure ; and set that quotient figure both in the 
quotient and divisor. 

Multiply the whole augmented divisor by this last quotient 
figure, and subtract the product from the said dividend, bring- 
ing down to it the next period of the given number, for a new 
dividend. 

Repeat the same process over again, viz. find another new 
divisor, by doubling; all the figures now found in the root ; 
from which, and the last dividend, find the next figure of 
the root as before ; and so on through all the periods, to the 
last. 

iVbfe, The best way of doubling the root, to form the new 
divisors, is by adding the last figure always to the last divi- 
sor, as appears in the following examples. — Also, after the 
figures belonging to the given number are all exhausted, the 
operation may be continued into decimals at pleasure, by add- 
ing any number of periods of ciphers, two in each period. 

EXAMPLES. 

1. To find the square root of 20506624. 

20506624 (5432 the root. 
25 



104 


450 


4 


416 


1083 


3466 


3 


3249 



10862 21724 
2 21724 



SQUABR ROOT* W 

Nero, When the ml it to he extracted to many places of 
figures, the work may he considerably shortened, thus y 

. Having proceeded in the extractioa after the common 
method, till there be found half the required number of 
figures in the root, or one figure more ; then, for the retf, 
divide the last remainder by its corresponding divisor, after 
the manner of the third contraction in Division of Deci- 
mals; thus, 

2. To find the root of 2 to nine places of figures. 
2 (1-41421356 the root. 
1 



24 
4 



100 



281 | 
1 I 



400 

281 



2824 
4 



11900 
11206 



60400 
56564 



3. What 

4. What 

5. What 

6. What 

7. What 

8. What 
0. What 

10. What 

11. What 

12. What 



28284) 



is the square 
is the square 
is the square 
is the square 
is the square 
is the square 
is the square 
is the square 
is the square 
is the square 



31936 (1356 
1008 
160 
19 
2 

root of 2025? 
root of 17-3056? 
root of -000729? 
root of 3? 
root of 5? 
root of 6 ? 
root of 7 ? 
root of 10? 
root of 11? 
root of 12? 



Ads. 45« 
Ads. 416. 
Ans. -027. 

Ans. 1-732050. 

Ans. 2*236068. 

Ans. 2*449489. 

Ans. 2-645751. 

Ans. 3-162277. 

Ads. 3-316624. 

Ans. 3-464101. 



RULES FOR THE SQUARE ROOTS OF VULGAR FRACTIONS AND 
MIXED NUMBERS. 



First prepare all vulgar fractions, by reducing \Y&m to 
their least terms, both for this and all other roots. Then 



84 ARITHMETIC* 

1. Take the root of the numerator and of the denominator 
for the respective terms of the root required ; which is the 
best way if the denominator be a complete power : but if it 
be not, then 

2. Multiply the numerator and denominator together ; 
take the root of the product : this root being made the nu. 
merator to the denominator of the given fraction, or made 
the denominator to the numerator of it, will form the frac- 
tional root required. 

That is, ^ « ^ = / flft ---L- 



b ~~ x/b "~ b ~~ i/ab 

This rule will serve, whether the root be finite or infinite. 

3. Or reduce the vulgar fraction to a decimal, and extract 
its root. 

4. Mixed numbers may be either reduced to improper 
fractions, and extracted by the first or second rule, or the 
vulgar fraction may be reduced to a decimal, then joined to 
the integer, and the root of the whole extracted. 



EXAMPLES. 



1. What is the root of |§? 

2. What is the root of T y T ? 

3. What is the root of 7 \ ? 

4. What is the root of ? 

5. What is the root of 17} ? 



Ans. | 
Ans. if. 
Ans. 0-866025. 
Ans. 0-645497 
Ans. 4-16S3&3. 



By means of the square root also may readily be found the 
4th root, or the 8th root, or the 16th root, <fcc. that is, the 
root of any power whose index is some power of the number 
2 ; namely, by extruding so oleen the square root as is de- 
noted by that power of 2 ; that is ; two extractions for the 4th 
root, three for the 8th root, and so on. 

So, to find the 4th root of the number 21035*8, extract the 
square root two times as follows : 



CUBE BOOT* 



85 



21035-8000 ( 145 037237 ( 12 0431407 the 4th root. 
1 1 



24 
4 



110 
06 



45 
44 



285 
5 



1435 
1425 



2404 
4 



10372 
0610 



29003 



108000 24083 
87009 3 



20991 (7237 
6*7 
107 



75637 
72249 

3388 ( 1407 
980 
17 



Ex. 2. What is the 4th root of 97-41 ? 



TO EXTRACT THB CUBE ROOT. 



I. By tint Common Rule*. 

1. Having divided the given number into periods of three 
figures each (by setting a point over the place of units, and 
also over every third figure, from thence, to the left hand in 
whole numbers, and to the right in decimals), find the nearest 
less cube to the first period ; set its root in the quotient, and 
subtract the said cube from the first period ; to the remainder 
bring down the second period, and cull this the resolvend. 

2. To three times the square of the root, just found, add 
three times the root itself, setting this one place more to the 
right than the former, and call this sum the divisor. Then 
divide the resolvend, wanting the last figure, by the divisor, 
for the next figure of the root, which annex to the former ; 



* The reason for pointing the given number into periods of three fi- 
gures each, is because the cube of one figure never amounts to more 
than three places. And, for a similar reason, a given number is point- 
ed into periods of four figures for the 4th root, of five figures for the 5th 
root, and so on. 

The reason for the other parts of the rule depends on the algebraic 
formation of a cube : for, if the root consist of the two parts a r I 
then its cube is as follows: («-{-fc)3 - aa -j- 3a 6-f- 3a6« -f b* ; where 
a \n the root of the first part a3 ; the resolvend is 3a»6-f 3a^2 -f 63 ; 
which is also the same as the three parts of the subtrahend ; also the 
divisor is 3a -7- 3a, by which dividing the first two terms of the resolv- 
end 3aa 6 + ab* , gives b for the second part of the rool \ and w> on. 



86 ARITHMETIC. 

calling this last figure e, and the part of the root before found 
let be called a. 

3. Add all together these three products, namely, thrice a 
square multiplied by c, thrice a multiplied by e square, and 
e cube, setting each of them one place more to the right than 
the former, and call the sum the subtrahend ; which must 
not exceed the resolvend ; but if it does, then make the last 
figure e less, and repeat the operation for finding the subtra- 
hend, till it be less than the resolvend. 

4. From the resolvend take the subtrahend, and to the 
remainder join the next period of the given number for a new 
resolvend ; to which form a new divisor from the whole root 
now found ; and from thence another figure of the root, as 
directed in article 2, and so on. 



EXAMPLE* 



To extract the cube root of 48228*544. 



3 X 3» = 27 
3 X 3 = 09 

Divisor 279 



48228-544 ( 36-4 root. 
27 



21228 resolvend. 



3 X 3* X 6 =162 
3X3 X6>= 324 ) add 
6» = 216 f 



3 x 36 J =; 
3 X 36 = 108 



38988 



19656 subtrahend. 



1572544 resolvend. 



3 X 36» X 4 = 15552 ) 
3 X 36 X 4* = 1728 > add 



4 3 = 64 J 



1572544 subtrahend. 



0000000 remainder. 



Ex. 2. Extract the cube root of 571482-19. 
Ex. 3. Extract the cube root of 1628-1582. 
Ex. 4. Extract the cube root of 1332. 



CUBE SOOT. 



87 



II. Ib extract the Cube Root by a short Way*. 

1. By trials, or by the table of roots at p. 93, dec. take 
the nearest rational cube to the given number, whether it be 
greater or less ; and call it the assumed cube. 

2. Then say, by the Rule of Three, As the sum of the 
given number, and double the assumed cube, is to the sum of 
the assumed cube and double the given number, so is the 
root of the assumed cube, to the root required, nearly. Or, 
As the first sum is to the difference of the given and assumed 
cube, so is the assumed root to the difference of the roots 
nearly. 

3. Again, by using, in like manner, the cuhe root of the 
last found as a new assumed cube, another root will be ob- 
tained still nearer. And so on as far as we please ; using 
always the cube of the last found root, for the assumed 
cube. 

EXAMPLE. 

To find the cube root of 21034-8. 

Here we soon find that the root lies between 20 and 30, 
and then between 27 and 28. Taking therefore 27, its cube 
is 19683, which is the assumed cube. Then 

19683 21035-8 
2 2 



39366 42071-6 
21035-8 19083 



As 60401-8 : 61754-6 : : 27 : 27-6047. 
27 



4322822 
1235092 



60401-8) 1667374-2 (27-6047 the root nearly, 
459338 
36525 
284 
42 



* The method usually given for extracting the cube root, Vi so ev 
ceedingly tedious, and difficult to be remembered, that \w\ov\* oXtat 



88 ARITHMETIC. 

Again, for a second operation, the cube of this root is 
21035-318645155823, and the process by the* latter method 
will be thus : 

21035-118645 &c. 



42070-637290 21035-8 
,21035-8 21035-318645 &c. 



As 63106-43729 : difT. -481355 : : 27-6017 : 

thediff. -0002^ 0560. 



conseq. the root req. is 27-604910560. 

Ex. 2. To extract the cube root of -67. 
Ex. 3. To extract the cube root of -01. 



TO EXTRACT ANY ROOT WJIATKVER • 

Let p be the given power or number, ?* the index of the 
power, a the assumed power, r its root, r the required root 
of p. Then say, 

As the sum of n + 1 times a and n — 1 times p, 
is to the sum of n + 1 times p and n — 1 times a ; 
so is the assumed root r, to the required root r, 

Or, as half the said sum of n + 1 times a and n — 1 times 
r, is to the difference between the given and assumed powers, 



approximating rules have been invented, viz. by Newton, Rnphson, 
Halley, De Lagny, Simpson, Emerson, and several other mathemati- 
cians ; but no one that I have yet seen is so simple in it form, or seems 
so well adapted for general use, as that above given. This rule is the 
same in effect as Dr. Halley's rational formula, but more commodious- 
ly expressed ; and the first investigation of it was given in my Tracts, 
p. 49. The algebraic form of it is this: 

As p j - 2a : a -|- 2r : : r : jr. Or, 
A* p -(- 2a : p *r a : : r : k *r r ; 

where p is the given number, a is the assumed nearest cube, r the cube 
root of a, and n the root of p sought. 

* This is a very general approximating rule, of which that for the 
cube root is a particular ca*e, and is the best adapted for practice, and 
for memory, of any that I have yet seen. It was first discovered in this 
form by myself, and the investigation and use of it were given at large 
in my Tracts, p. 45, &c 



GENERAL ROOTS. 



80 



so is tlss^sjhjuMd root r, to the difference between the true 
end iMMn roots ; which difference, added or subtracted, 
ee the case requires, gives the true root nearly. 
That ie, (n 4-1) a + (n-l)r: (w+1) p. H-(n-l) a: :r:a- 
Or, (n + 1) Aa + (n — 1) : p^a : : r : it ^ r. 

And the operation may be repeated as often as we please, 
by using always the lust found root for the assumed root, and 
its nth power for the assumed power a. 



example. 
To extract the 5th root of 21035& 

Here it appears that the 5th root is between 7 3 and 7*4* 
Taking 7-3, its 5th power is 20730 71503. Hence we have 
r « 21035-8, n = 5, r = 7-3, and a = 20730-71593 ; then 
» + 1 . £a -f- n — 1 . £p : p ^ a : : r : r «^ r, that is, 
8 X20730-71593 + 2 x 21053-8 : 305-0*4 : : 7 3 : -0213005 
3 2 7-3 



62192 14779 
42071-6 

104263 74779 



42071 -(J 915252 
2135568 



2227 1 13 1( -0213605=11^ 
7 3=r, add 



7 321360- r, true 
to the last figure. 



OTHER EXAMPLES. 



1. What is tho 3d root of 2 ? Ans. 1-259921. 

2. What is tho 3d root of 3214 ? Ans. 14-75758. 
8. What is the 4th root of 2 ? Ans. 1 189207. 

1 4. Whnt is the 4th root of 97-41 ? Ans. 3 1415909. 

5. Whnt is the 5th root of 2 ? Ans. 1-148699. 

6. What is the 6th mot of 21035-8 ? Ans. 5-254037. 

7. What is the 6th root of 2 ? Ans. 1-122462. 

8. What is tho 7th root of 21035-8 ? Ans. 4-145392. 
0. What is the 7th root of 2 ? Ans. 1-104080. 

10. What is the 8ih root of 21035-8 ? Ans. 3-470328. 
Vol. I. 13 



00 AStTHWCTIC. 

1 1. What is the 8th root of 2 ? • • An. l^HK 

12. What is the 9th root of 21035*8 1 Ana. S^tHS9. 

13. What is the 9th root of 2 ? Ans. 1-080059. 



The following is a Table of squares and cubes, and also the 
square roots and cube roots, of all numbers from 1 to 1000, 
which will be found very useful on many occasions, in nu- 
meral calculations, when roots or powers are concerned. 

The use of this table may be greatly extended, either by 
the addition of ciphers, or by changing the places of the 
separating points. The following examples will suffice to 
suggest the method. 

Root. Square. Cube. 

36- 12<J6* 46656- 

360- 129600- 46656000- 

3600- 12960000- 46656600000- 

546- 298116- 162771336- 

54-6 2981-16 162771-330 

'546 -298116 -162771336 

For a simple and ingenious method of constructing tables 
of square and cube roots, and the reciprocals of numbers, 
see Dr. Hutton's Tracts on Matheirtotical and Philosophical 
Subjects, vol, i. Tract 24, pa. 459. 



A TABU OF 4QUAAU, CUBES, AND ROOTS. 



91 





square. 




Square ttooi. 


ouce ttoot* 


1 


" 1 


1 


1 0000000 


1 000000 


i 


4 


8 


14142136 


1 259921 




9 


27 


1-7320508 


1-442250 


4 


16 


64 


2 0000000 


1-587401 


5 


25 


125 


2 2360680 


1 709976 


6 


36 


216 


24494397 


1 817121 


7 


49 


343 


2 6457513 


1-912931 


8 


64 


512 


28284271 


2 000000 


9 


81 


729 


30000000 


2 080084 


10 


100 


1000 


3 1622777 


2 154435 


11 


121 


1331 


33166248 


2 223990 


12 


144 


1728 


3 4641016 


2289428 


13 


169 


2197 


3 6055513 


2 351335 


14 


196 


2744 


37416574 


2410142 


15 


225 


3375 


38729833 


2 466212 


16 


256 


4096 


40000000 


2519842 


17 


289 


4913 


4 1231056 


2 571282 


IS 


324 


5832 


4 2426407 


2 620741 


10 


861 


6359 


4 35SS9S9 


2 668402 


30 


400 


SO 00 


4 4721360 


2714418 


21 


44 1 


9261 


4 5825757 


2 758924 


22 


434 


10643 


4 6904158 


2 802039 


23 


529 


12107 


4 795S315 


2-S43867 


24 


576 


13S24 


4 8989795 


2 884499 


25 


625 


15625 


50000000 


2 924018 


SO 


676 


17576 


5 0990195 


29G2496 


27 


729 


19683 


5 1961524 


3 000000 


28 


7S4 


21952 


5 2915026 


3036589 


29 


841 


24389 


5 385164S 


307231 7 


30 


900 


27000 


54772256 


3 107232 


31 


9G1 


29791 


5 5677644 


3 141381 


39 


1024 


32768 


5656S542 


3 174802 


33 


1089 


35937 


5 7445626 


3207534 


» j'l 


1 1 'iJi 

1 1<JD 








35 


1225 


42875 


59160798 


3271066 


36 


12D6 


46656 


60000000 


3 301927 


37 


1369 


50653 


60827G25 


3-332222 


38 


1444 


54S72 


6 16 14140 


3 361975 


39 


1521 


59319 


62449980 


3 391211 


40 


1600 


61000 


6-3245553 


3 419952 


41 


1681 


69921 


64031242 


3 448217 


42 


1764 


7408* 


6-4807407 


3476027 


43 


1849 


79507 


6 5574385 


3 503398 


44 


i<m 


85184 


6 0332496 


3 530348 


45 


2025 


91125 


670S2039 


3 556893 


46 


2116 


97336 


0-7823300 


3 583043 


47 


2209 


103823 


68556546 


3603826 


48 


2304 


110592 , 


6 9282032 






M02 


117649 


70000000 




^ si 


2500 / 


125000 


70710678 


\ 3 



AJUIBMETtC. 



Number. 


Square. 


Cube. 


Square EooL 

, 


Cube Root. 


51 


260] 


132651 


7 1414284 


3-708430 


52 


2704 


1-1 CO US 


72111026 


3 732511 


53 


2300 


14S877 


7-2801099 


3756286 


54 


2916 


157464 


73484692 


3779763 


55 


3025 


16.G375 


7 4161985 


3802953 


56 


3136 


i 175616 


74833148 


3 825862 


57 


3249 


185193 


75498344 


3 848501 


58 


3364 


195)12 


7 6157731 


3870877 


50 


34*1 


205379 


7 681)457 


3 892996 


60 


3600 


216C0D 


77459667 


3 914863 


61 


3721 


226981 


78102497 


3936497 


62 


3844 


238328 


76740079 


3-957883 


63 


3969 


250047 


7*9372539 


3979057 


64 


4096 


262144 


800C00OO 


4 000000 


w 


4225' 


274625 


80622577 


4 020726 


60 


4356 


287496 


S12403S4 


4041240 


67 


4489 


300763 


8 1 85352S 


4061548 


6S 


4624 


314432 


8 2462113 


4-081655 


69 


4761 


3285G9 


83066239 


4101566 


70' 


4900 


343000 


83666003 


4 121285 


71 


son 


357911 


842G149S 


4 140818 


72 


5184 


373248 


8-4852814 


4 160153 


73 


5329 


399017 


8*5440037 


4 179339 


74 


5476 


405224 


86023253 


4 198336 


75 


5625 


421875 


86602540 


4217163 


76 


5776 


438976 


87177979 


4 235824 


77 


5929 


456533 


S 7749644 


4 25432J 


78 


6084 


474552 


SS317609 


4 272659 


79 


6241 


493039 


8 8881944 


4290841 


80 


6400 


512000 


S 9442719 


4 308870 


61 


6561 


531441 


90000000 


4 326749 


82 


6724 


551368 


90553S51 


4 344431 


83 


0889 


571787 


& 11 04336 


4 362071 


84 


7C56 


51)2704 


Ai 1 PSl £ 1 4 


4*379519 


65 


7225 


614125 


92195445 


4396830 


86 


7396 


636056 


2736185 


4414005 


87 


7569 


658503 


9 3273791 


4431047 


88 


7744 


6S1472 


936CS315 


4447960 


89 


7921 


704969 


9 4339811 


4464745 


90 


8100 


720000 


9486S:^30 


4481405 


91 


£281 


753571 


95393920 


4-497941 


92 


64JG4 


77S6SS 


9 5916630 


4-514357 


93 


S649 


S04357 


9-6436508 


4530655 


94 


8836 


8305S4 


96953597 


4 546836 


95 


9025 


857375 


97467943 


4 562903 


96 


9216 


884736 


9*7979590 


4 578857 


97 


9409 


912673 


98488578 


4-594701 


98 


9604 


941192 


9S994949 


4-610436 


99 / 


9S01 


970299 


L 9949S744 


. 4626065 


100 1 


10000 


10000GO 


\ 10 moggou 





•WAftlS, CUUS, AlfD BOOTf . 93 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


101 


10201 


1030301 


10' 04 98756 


4657010 


102 


10404 


106120S 


100995049 


4 672329 


103 


lCGGO 


1092727 


10 1488916 


4 687548 


104" 


10813 


1124>G4 


1 /> i AC) Art 

10' 1930390 


4 702t)b9 


105 


1 1 02 


1157625 


1 J"l J Aft t J lO 

10 2469508 


4-717694 


100 


I12CG 


11010IG 


10 23o6301 


4*732624 


107 


11449 


1225043 


JO 3 I40J50* 


4 747459 


loa 


11064 


1250712 


10 3923048 


4762203 


109 


11 SSI 


1295029 


104403065 


4 77685b 


110 


12100 


1331000 


10 4880885 


4 791420 


111 


12321 


1367631 


10 5356536 


4 HO 5896 


112 


12544 


1404928 


105830052 


j .a tin ao * 

4 82 0284 


113 


12700 


1442S97 


10 6301459 


A Q'li CQQ 

4 Wo45S9 


114 


12996 


1481544 


10 67707S3 


4o4SS0S 


115 


13225 


1520875 


10 7239053 


4 00^944 


116 


13456 


1560896 


1 A^mv aaa nit 

1 07 703 2 06 


J V ""t" ■ 1 1 1 f 1 


117 


13698 


1601613 


10 8166538 


4 3905173 


11S 


13924 


1643032 


10 8627805 


4 yu4obo 


119 


14161 


1685159 


10'90S7121 


j,ni bjjqe 

4 y 1 ooSo 


120 


14400 


1728000 


109544512 




121 


14641 


1771561 


i rooooooo 


4 946088 


122 


149S4 


181584S 
186CS67 


11 0453610 


4 959676 


123 ; 


15120 


I 1 .Anne 'ic r 

I I 0905365 


j A^At AA 

4 973 19U 
4 98663 1 


124 


15376 


1 A A** Cm 

1906624 


11 1 r\r rnriH 

11 1355287 




15625 


1953125 


T 1 . i cm a a a a 

1 1 1803399 


o uuuuuu 


126 


15&7G 


2000376 


1 I 2249722 


o U ] a/Ho 


127 


16129 


2048383 


1 1 2694277 


t o OZOfr^b 


129 


16331 


2007152 


11 '3137085 


U3Ubb4 


129 


16641 


2146689 


I IOC TO i c T 

I I 3o7SI67 


o 05^774 


139 


16900 


2107000 


11 4017543 


065797 


131 


17161 


2248091 


11 j j *~ ^ A A 1 

11*4455231 


5 078753 


132 


17424 


2299908 


1 1 4891253 


c a a l i* m a 

5 091643 


133 


17639 


2352637 


1 lo325626 


5 104469 


134 


17956 


2406104 


11-5758369 


5 117230 


135 


18225 


2460375 


11'6189500 


5 129928 


136 


18496 


251545G 


11-6619038 


5 142563 


137 


18760 


2571353 


11 7046999 


5 155137 


138 


19044 


2628072 


11 7473444 


5 167649 


130 


19321 


2685610 


11*7898261 


5 1S0101 


140 


19600 


2744000 


11 8321596 


5 192494 


111 


198S1 


2803221 


U 8743421 


5 204828 


142 


20164 


2863283 , 


11 9163753 


5217103 


143 


20440 


2924207 i 


11 9582607 


5 220321 


144 


20736 


23859S4 


12 0000000 


5241483 


145 


21025 


3048625 


V2 0415946 


5 253583 


146 


21316 


3112136 


120830460 


5 265637 


147 


21609 


3176523 


12 1243557 


5277632 


148 


21904 


3241792 


12 1655251 


5-2S<ft&n 


149 


22201 


3307949 


122065556 




150 J 


22500 j 


3375000 j 


12-24744S7 





^» 1 rTYiltar 

Jl ^ LI 114 Ucl . 


Souafe 


Cube, 


Snuare Root 


CuhA Rrmt 

VUMC HWl. 


151 


22801 


3442951 


12 2882037 


5325074 


15^ 


23104 


35 11808 


12-3288280 


5336303 


153 


23409 


3581577 


12 3693169 


5 348481 


154 


23716 


3652264 


124006736 


5-360108 l 


155 


24025 


3723875 


12449S996 


5 371685 


156 


24336 


3796416 


12 4899960 


5383213 


157 


24649 


3809S93 


12 5299641 


5 394691 


158 


24964 


3944312 


125693051 


5406120 


159 


25281 


4019679 


12 6095202 


5 417501 


160 


25600 


4096000 


126491106 


5428835 


Ml 


26921 


4173281 


126885776 


5440122 


16a 


26244 


425152S 


12 7279221 


5 451362 


163 


26569 


4330747 


127671453 


5462556 


164 


26896 


4410944 


12 8062485 


5473704 


165 


27225 


4492125 


128452326 


5494806 


166 


27566 


4574296 


12 8840987 


5495865 


167 


27SS9 


46*7463 


12 922S480 


5506879 


168 




4741632 


129614814 


5517S4S 


169 


2S561 


4826309 


13 0000000 


552S775 


170 


28900 


4913300 


13 0384048 


5 539658 


171 


29241 


5000211 


130766968 


5550499 


172 


29584 


5088448 


13 1148770 


5 561298 


173 


29929 


5177717 


13 1529464 


5572055 


174 


30276 


5268024 


13 1909060 


5-592770 


175 


30625 


6359375 


132287566, 


5 593445 


176 


30976 


5451776 


13 2664992 


5604079 


177 


31329 


5545233 


133041347 


5 614673 


178 


31684 


5639752 


13 3416641 


5625226 


179 


32041 


6735339 


13-3790882 


5 635741 


180 


32*00 


5932000 


13 4164079 


5 646216 


181 


,32761 


5929741 


134536240 


5656653 


182 


33124 


6028568 


134907376 


5667051 


183 


33439 


6128487 


13 5277493 


5 677411 


194 




0^29,0 04 


Jo Ot>4bbUU 


5 687734 


185 


34225 


6331625 


13 6014705 


5699019 


186 


34596 


6434856 


136381817 


5 708267 


187 


34969 


6539203 


136747943 


5718479 


188 


35344 


6644672 


13 7113092 


5 729654 


189 


35721 


6751269 


13 7477271 


5738794 


190 


3G100 


6959000 


137840488 


5749S07 


191 


36481 


6967871 


138202750 


5758965 


192 


36864 


70778S8 


13 8564065 


5768998 


193 


37249 


7189057 


13 8924440 


5 778996 


194 


37636 


7301384 


139283883 


5 789960 


195 


38025 


7414875 


139642400 


5 798890 


196 


38416 


7529536 


1400000(10 


5S08786 


197 


3S809 


7645373 


140356688 


5 818648 


198 | 


39204 


7762302 


140712473 


5828476 


199 * 


39601 


7880599 




\ vmm 


2W / 


40000 j 


8000000 







■40AAIS, CUBES, AND ROOT*. 



/Number. 



201 
202 
203 
204 
205 
206 
207 
208 
209 
210 
211 
212 
213 
214 
215 
216 
217 
218 
219 
220 
221 
222 
223 
224 
225 
226 
227 
228 
229 
230 
231 
232 
233 
234 
235 
236 
237 
238 
239 
240 
241 
242 
243 
244 
245 
246 
247 
248 
349 



Square. 



/ 



40401 
40804 
41209 
41616 
42025 
42436 
42849 
43264 
43661 
44100 
44521 
44944 
45369 
45796 
46225 
46656 
47089 
47524 
47961 
48400 
48841 
49264 
49729 
50176 
50625 
51076 
51529 
51984 
52441 
52900 
53361 
53S24 
54289 
54756 
55225 
55696 
56169 
56644 
57121 
57600 
58061 
58564 
59049 
59536 
60025 
60516 
61009 
61504 
62001 
62500 



Cube. 



8120601 
8242408 
8365427 
8489664 
8615125 
8741816 
8869743 
8998912 
9123329 
9261000 
9393931 
9528128 
9663597 
9800344 
9938375 
10077696 
10218313 
10360232 
10503459 
10648000 
10793861 
10941048 
11089567 
11239424 
11390625 
11543176 
11697063 
11852352 
12008989 
12167000 
12326391 
12487168 
12649337 
12S12904 
12977875 
13144256 
13312053 
13481272 
13651919 
13824000 
13997521 
1417248S 
14348907 
14526789 
14706125 
14886936 
15069223 
15252992 
15438249 
15625000 1 



Square Root. Cube Rogt 



141774469 
142126704 
142478068 
142828569 
14 3178211 
143527001 
143874946 
144222051 
14 4568323 
144913767 
145258390 
14 5602198 
14 5945195 
146287388 
146628783 
146969385 
14 7309199 
147648231 
147986486 
14 8323970 
148660687 
148996644 

14 9331845 
149666295 

15 0000000 
15 0332964 
150665192 
15 0996689 
15 1327460 
15- 1657509 
15 1986842 
152315462 
15 2643375 
15-2970585 
153297097 
15-3622915 
15-3948043 
15 4272486 
15 4596248 
15-4919334 
15-5241747 
155563492 
15 5884573 
156204994 
156524758 
156843871 
157162336 
157480157 
157797338 
15-3113883 



5857766 
5 867464 
5877130 
5 886765 
5896368 
5 905941 
5915482 
5924992 
5 934473 
5 943922 
5953342 
5962731 
5972091 

5 981426 
5990727 
6000000 
6009244 
6018463 
6027650 

6 036811 
6045943 
6055048 
6064126 
6073178 
6082201 
6 091199 
6 100170 
6 109115 
6 118033 
.6 126925 
6 135792 
6 144634 
6 153449 
6 162239 
6 171005 
6 179747 
6- 188463 
6 197154 
6205822 
6 214465 
6223084 
6231679 
6240251 
6248800 
6257325 
6265826 



M 



Number. 


Square, 


Cube. 


Square Root. 1 


Cube Root. 


251 


■ 1 

63001 


15813251 


15 r S429 795 1 


b o07994 


252 


63504 


16003G08 


15 8745079 


6'31 6359 


253 


64009 


1 C -1 " 1 t\ J a 1 -*t*f 

16134277 


1 5' 059737 


024704 


254 


64516 


1 1 f r>fi A a j 

16387064 


15 9373775 


b 3 302 6 


net 


65025 


1 CCDl 'ITS 

lbo3 137o 


159b37194 


b 44 1*3 2 b 


256 


65536 


lb77721b 


lb 0000000 


b <H9b04 


257 


06049 


loy7459o 


i c . ao 1 a 1 a el 
1 b Oo 1 2) 9o 


b o57ool 


2 DO 


66564 




1 C AC ill 1 2 4 

lb Ug JJ734 


o t5bouyo 


259 


67081 


17d*3y79 


lb 0934769 


b i>74o 1 1 


260 


67600 


17576000 


lb 1245155 


332504 


261 


68121 


17779591 


16 1554944 


6 390676 


262 


68644 


17934729 


161964141 


6 39S829 


263 


69169 


1S191447 


16 2172747 


6 406959 


264 


69696 


18399744 


162480768 


6 415003 


265 


70225 


1360962:> 


16*2788206 


6 423153 


266 


70756 


18821096 


^ n a a p a n a 

163095064 


a , t a t n ftn 

b 43122S 


267 


nr 1 nan 

71289 


19034163 


16 3401346 


G 4392 / * 


268 


71824 


1 A A *. O a !i »i 

19249332 


16 p37070o5 


6 447305 


269 


72361 


19465109 


te 1 Ai i rtr 

164012195 


6 455315 


OTA 




lybSJOOO 


164316767 


( 6 463304 


271 


73441 


iyyo25i i 


16 4620776 


6 4712i4 


272 


73 934 


20123643 


16 4924225 


6 479224 


273 


74529 


20046417 


16 52271 16 


6 4S7154 


274 


75076 


2O570S24 


lb 5529454 


6 495065 


275 


75625 


207 9 6,3 7o 


16-5831240 


6 5029^6 


J76 


76176 


21 024576 


lb'6 132477 


6 510^30 


277 


76729 




it,/; IQ'JtTn 
iD'O-ldiJl iKJ 


b a 1 ob34 


275 


772a 4 


4I-io4yD2 


1 D O r ooo20 


b *>2b519 


279 


77341 


Ol Tl 7KQQ 


ID / JJ 


b Qd4iJo5 


230 


• 7Q 4 A A 

7&4i>u 


41 4 J4UUU 


1 ifi . "i " J fl A A tl 

1 (W*5d2O0O 


542133 


43 1 




99 1 QCtHi 1 


ID f OoUO-^O 


i? - c j* A,A 1 n 


- ~ 


70*194 




1 ,fi 7Q9Q^1A 


d Do 4 b7dt 


283 


80089 


226J51S7 


168226038 


6 565415 


234 


80656 


22906304 


16-8522995 


6573139 


285 


81225 


23149125 


16 8819430 


6-580844 


286 


81796 


23393656 


16-9115345 


6588532 


287 


82369 


23639903 


16-9410743 


6-596202 


288 


82944 


23987872 


169705627 


6603354 


289 


53521 


24137560 


17-0000000 


6 611489 


290 


84100 


24389000 


17-0293864 


6 6191 00 


291 


84631 


24642171 


17-0587221 


6 626705 


292 


85264 


24897088 


1708S0075 


6634287 


293 


85849 


25153757 


171172428 


6641852 


294 


86436 


25412184 


1T1464282 


6649399 


2&5 


87025 


25672375 


171755640 




296 


87616 


25934336 


172046505 




297 


SS209 


26198073 


172336S79 


6671940 


S&S / 


3SS04 


26463592 






at* 


89401 


26730S99 






300 / 


90000 


27000000 





SQ.UARE8, CUBES, AND ROOTS. 



97 



Number. 


Square. 


Cube. j 




Cubo Root. 


oUX 


one a 1 

yUbUl 


0*70^1 IftAl 1 


■ *> 1 a •* "-\ i ^ 1 


1 u 1 / oy 


aiKS 


1* J ZVJ4 


Z 7 04obUS 


1 / »5 / ■*» I -1 / — 


l*«-7AQ 1 7*i 

b tyiji to 


QUO' 


Ql A AO 


•>7tt 1 4il 07 1 


17 lA<".^iQ'i , > 1 


n-7 1 a !%7n 


0114 


9Z4XO 


*r>Uy4-l )4 ! 




b /-o;*.jl 


OUO 






1 7'J.H !•> l0-> ' 
1 l 4b i^-ii'Z - 


b / »3 1 1 b 


OUU 


0OO0O 


z*>b^-5i.> I b 


1 7- IOOk'%%7 


b / Oobbo 


QAT 


94Z4;f 


OwO'i 4 i t M 


1 1 •)£ 1 •! 1 O'J 


b HOyy/ 


OAO 
•JUO 


y4oo4 


2'JJISI 12 


i*.r: ((HIOh:Q 
1 / t\t — • ^ 


b / o»5o 1 


QAA 

ouy 


OI% 4 Q 1 

y»481 


zl'olMojy 




b /b0bi4 


olO 


96100 


2979 1 00(1 


1 / bUOSJ oil 


b /b/syy 


Ql 1 

ol 1 




o(JU802.>l 


1 / 00 ) 1 y^i 1 


rfi^T^ 1 ftft 


oiZ 


y/o44 


o\K57132S 


J 7 bbo >J1 / . 


b /S24zd 


«ll *1 
Old 




oUob4297 


17 oU LNUbU 


ioybbl 


ol4 


QDSnA 

yoosib 


*J A ft X 1 1 1 < ■< 

ousjoy 14 * 


1 / / JUO l.ll 


b #ub.-?v± 


QIC 

olo 


yy^zo 


OlJOO.S70 


17 /4->Jo:M | 


b olMUyj 


olo 


yyoob 


31554496 


1 T-T^TlIM wCki 1 
1 / / / Oo-T^*) ' 


/••CI 1 .'JC 4 

boll Jh4 




1 An 4 co 


Qlti': 1 •> 

0I000U 1 .3 


1 7-ViA < iO*>tt 


/< vim 
b -^4b J 


010 


1U1 iz4 


o.)i KT 1 -JO 


1 / OOJO.) 1.) 


b oj..)bJ4 


olU 


1 A I T/l 1 
101 / Ol 


tSJ In 17o9 


1 "1 \ 1 1 1 
J / nUH.) /II 


b 80 J / 7 1 


oon 

o^u 




/ b800l» 






0^1 


1 A'<AJ. 1 


00U / b 1 u L 


1 / f 1 bt / — J 


17 no 1 
b n-i tv)jz\ 


OZJ 


i 'MU-.H 


od-*r*bi - s 


1*7 fl « 1 '> \<< » 


b r».J'l 1 J4 


100 
O40 


1 A 1 "*Oft 


odbii^vtb # 


1 / .» / iil.'U? 


b nO \£l & 


oZ4 


1 A i A"? ft 


'1 1 A 1 OOO 1 


1 QlUli'lfiMllM 




*io % 
0^0 


l v'OO JtO 


> •.••iOj.il-,-) 


1 ~ « J«- # I ... 1 1 


r: w7 1 .4 


'•ton 


1 nfW7<: 

A UU^ f U 


O iU-JO.I / b 




b *"io(33 


f 






1 w As;:; Hl'l 


ft-K^QJ 1 O 
b o^y*i 1 ;| 




1 rt7%ftJ. 

1 V/ 1 »JO-± 




tc-ii ii/70'? 
l»p IM'// yjt* 


b «M/b-joo 


ozy 




'\~xfl 1 1 OCA 




b JU-HOQ 




1 option 


-*>017( win 


ivirrVK]')' ' 


1 A IOQ 

b i' 1 u-j^o 




109561 


Ml'i'^fi 0"»<i 1 


] s* 1934054 


#VA 1 7*-lQA 


oo<s 


1 1022 l 


yvf'/J I • J li ~ 




fi QO fJ^X 


333 


1 lOSsO 




18'24S2S76 • 


i;n*i 1 oni 
«'•)[ 1 


334 


111556 


37259704 


18 2756669 ' 


6 938232 


335 


11222.3 


37595375 


18 303U052 , 


6 945149 


336 


112896 


37933056 


1S 330302S ; 


6 952053 


337 


113.i69 


3S272753 


18 357559S ; 


6 958943 


338 


11 1214 


38614472 


jS 3s 17763 ! 


6 965819 


339 


114921 


3.S95S219 


i s 4 H 9526 ! 


6 972683 


340 


115600 


39304000 


18-4390SS9 i 


6 97"532 


341 


116281 


3965 1S21 


18 4661S53 


6986368 


342 


116964 


40001 6 ss 


Is : 93:120 ! 


6993191 


343 


11 76-1 9 


40353**0? 


18 52;:::592 j 


7O00000 


344 


118.336 


4070 : .^4 


18 5172370 i 


7006796 


345 


119025 


41063i>25 


18 5741756 | 


7013579 


346 


119716 


41421736 


18 6010752 j 


7020349 


347 


120409 


4 178 1923 


18 6279360 j 


7 02710ft 


348 


121104 




18-6547581 1 




849 121*01 


4250S549 


18 6S15417 


\ 7-^V)o^\ 


850 I 


122600 : 


42875000 


18 70S2S69 




Vol. I 











Number. 




Cube. 


Square Root. 


Cube Root. ! 


351 


123201 


43243551 


18*7349940 


7054004 


352 


123904 


43614208 


187616630. 


7060696 


353 


124609 


43936977 


18-7882942 


7067376 


354 


, 125316 


44361964 


188148S77 


7074044 


355 


126025 


44739875 


18 8414437 


7080609 


356 


126736 


45118016 


1B8679623 


7 087341 


357 


127449 


45499293 


1SS944436 


7093971 


35S 


128164 


45882712 


IS 9208879 


7100588 


359 


128881 


46268279 


189472953 


7 107194 


360 


129600 


46656000 


18 9736660 


7 113786 


361 


i 130321 


47045881 


19 0000000 


7 120367 


362 


131044 


4743792S 


19 0262976 


7 126936 


363 


131769 


47832147 


19 0525589 


7133492 


364 


132496 


48228544 


19 0787840 


7*140037 


365 


133225 


4S627125 


19' 1049732 


7146569 


366 


133956 


49027896 


19 1311265 


7 153090 


367 


134689 


49430863 


19 1572441 


7 159599 


363 


135424 


49836032 


19 1333261 


7166096 


369 


136161 


50243409 


19 2093727 


7172580 


370 


136900 


50653000 


19*2353841 


7179054 


371 


137641 


51064811 


19 2613603 


7185516 


372 


138384 


51478S48 


19 2873015 


7 191966 


373 


139129 


51895117 


193132079 


7 198405 


374 


139S7G 


52313624 


19 3390796 


7 204832 


375 


140625 


52734375 


193649167 


7-311448 


376 


141376 


53157376 


19 3907194 


7 217652 


377 


* 142129 


535S2633 


19*4164878 


7 224045 ; 


378 


142884 


54010152 


19 4422221 


7230427 


379 


143641 


54439939 


194679223 


7236797 


380 


144400 


54872000 


194935887 


7 243156 


381 


145161 


55306341 


19 5192213 


7-240504 


3S2 


145924 


55742968 


19 5448203 


7255841 


383 


146689 
147456 


561S1S87 
56623104 


195703858 


7 262167 


384 


19 5959179 


T2694&2 


3S5 


14S225 


19 0214169 

* & **** i'lVf 


7-274786 


386 


14S996 


57512456 


196468827 


7281079 


397 


149769 


57960603 


19 6723156 


7287362 


388 


150544 


5841 J 072 


19 6977156 


7293633 


3S9 


151321 


58S63869 


197230829 


7-299894 


390 


152100 


59319000 


197484177 


7 306143 


391 


152S81 


59776471 


19T737199 


7 312383 


392 


153664 


60236289 


19 7989899 


7318611 


393 


154449 


60698457 


198242276 


7324829 


394 


155236 


61162984 


19 8494332 


7-331037 


395 


156025 


61629875 


198746069 


7 337234 


396 


156S16 


62099136 


198997487 


7*343420 


3t)7 


157609 


62570773 


199248588 


7349597 


39? 


158404 


63044792 


19 9499373 


7355762 


399 1 


159201 


6352 1199 


19 9749844 


7361918 


400 j 


leoooo 


64000000 


200000000 





SQUARES, CUBES, AND ROOTS. 



99 



uTTkuer, 


Square. 


v>ube< 


Square Root. 


Cube Root, 


401 


160801 


64481201 


20 0249844 


7 37419S 


401 


161604 


64964808 


200499377 




403 
404 


162409 


65450827 


20 0748599 


7386437 


163216 


05939264 


20 0997512 


7 392542 


405 


164025 


66430125 


20 1246118 


7 39H636 


406 


16493G 


669234 16 


20 1494417 


7-404720 


407 


165649 


67419143 


20 1742410 


7 410795 


408 


166464 


U7''l 1131 2 


20 1 990099 


7 41 6859 


409 


167291 


68417929 


SO 9237481 


7-422 r na. 

| 1 Jd H.' J 7t 


410 


168100 






7-4 28050 


411 


163921 


60426 531 




7-434994 


412 


169744 


69934 52S 


20 2Q77A3I 


7 44 1019 


413 


170569 


70444997 


2032240 14 


7-447034 


414 


171396 


70957944 


20 3469899 


7453040 


415 


172225 


71473H75 


20 3715488 


7 459036 


416 


173056 






7' 4 65022 


417 


173889 


7251 1713 


20 4205779 


7 470999 


418 


174724 


73034632 


20 4450483 


7 476966 


419 


175561 


73560059 


204694895 


7 4S2924 


420 


176400 


740SS000 


20 4939015 


7 4B8H72 


421 


177241 


74618461 


205182845 


7 49481 1 


422 


17S0S4 


75151448 


20 5426380 


7 500741 


423 


17S929 


75G969C7 




7 506 tiG 1 


424 


179776 


76225024 


20-59 1 2603 


7 512571 


425 


180625 


76765625 


20-6)55281 


7 51S473 


426 


181476 


77308776 


20 6397674 


7 524365 


427 


182329 


77854483 


20 6639783 


7-530248 


423 


183184 


78402752 


20 6Stf 1609 


7 536 J 21 


429 


18404 I 


78953589 


20 7123152 


7 541986 


430 


184900 


79507000 


20 7364414 


7-547^42 


431 


185761 


80062991 


20 7605395 


7 553688 


432 


186624 


80621568 


207846097 


7559526 


433 


187489 


81182737 


208086520 


7 565355 


434 


188356 


61746504 


208326667 


7'57 1174 


435 


189225 


82312875 


20S566530 


7 5769*5 


436 


190096 


82381856 


208806130 


7582786 


437 


190969 


83453453 


20 9045450 


75wy579 


438 


191844 


84027672 


20 9284495 


7594363 


439 


192721 


846045 1 9 


209523268 


760013S 


440 


193600 


85184000 


209761770 


7605905 


441 


194481 


85766121 


21 0000000 


7611662 


442 


195364 


863503S8 


21 0237960 


7 617412 


443 


196249 


86938307 


21 0475652 


7623152 


444 


197136 


87528384 


210713075 


7 628884 


445 


198025 


88121125 


21 0950231 


7634607 


446 


198916 


88716536 


211187121 


7640321 


447 


199809 


89314623 


21 1423745 


7646027 


448 


200704 




21 1660105 


\ tmviw 


449 


201601 


90513849 


21-1896201 




450 I 


202500 f 


91125Q0Q | 


21 2132034 \ TWWM 


t-rrx- 



\ 



100 



ARITHMETIC. 



iNJ 11 m nor 




Cube. 






451 


203401 


91733851 


21 2367606 


7668766 


452 


204304 


9234540b 


21-2602916 


7674430 


453 


205209 


92959677 


21 2837967 


7 680086 


454 


206106 


93576664 


21 3072758 


7685733 


455 


207025 


94196375 


21 3307290 


7691372 


456 


207936 


94818S16 


21 3541565 


7697002 


457 


208849 


95443993 


21 3775583 


7-702625 


458 


209764 


96071912 


21 4009346 


7 708239 


459 


210681 


96702579 


21 4242853 


7713845 


460 


211600 


97336000 


214476106 


7 719442 


461 


212521 


97972181 


21 4709106 


7725032 


462 


213444 


9S611128 


21 4941853 


7 730614 


463 


214369 


99252847 


21 517434S 


7-736188 


464 


215296 


99897344 


21 5406592 


7741753 


465 


216225 


100544625 


21 5638587 


7 747311 


466 


217156 


101194696 


21 5870331 


7 752861 


467 


218089 


101847563 


21 6101828 


•7-758402 


468 


219024 


102503232 


21-6333077 


7763936 


469 


219961 


103161709 


21 6564078 


7769462 


470 


220900 


103823000 


21-6794834 


7774980 


471 


221841 


104487111 


21 7025344 


7780490 


472 


222784 


105154048 


21 7255610 


7785993 


473 


223729 


105823817 


217485632 


7791487 


474 


224676 


106496424 


21 7715411 


7796974 


475 


225625 


107171875 


21 7944947 


7-802454 


476 


226576 


107850176 


21 8174242 


7807925 


477 


227529 


108531333 


21 8403297 


7813389 


478 


22*184 


109215352 


21 8632111 


7818846 


479 


229411 


109902239 


21 88606S6 


7S24294 


480 


230400 


110592000 


21-9089023 


7829735 


481 


231361 


111284641 


21 9317122 


7835169 


482 


232324 


111980168 


21 9544984 


7840595 


483 


233289 


11267S587 


21 9772610 


7 846013 


484 


234254 


113379904 


22 0000000 


7851424 


485 


235225 


114084125 


220227155 


7856828 


4S6 


236196 


114791256 


220454077 


7862224 


487 


237169 


115501303 


220680765 


7 867613 


488 


238144 


116214272 


220907220 


7872994 


489 


239121 


116930169 


22 1133444 


7878368 


490 


240100 


117649000 


22 1359436 


7883735 


491 


241081 


118370771 


22- 1585 198 


7889095 


492 


212064 


119095488 


22 1810730 


7894447 


493 


243049 


119823157 


22-2036033 


7899792 


494 


244036 


120553784 


22 2261108 


7905129 


495 


245025 


121287375 


222485955 


7-910460 


496 


246016 


122023936 


222710575 


7-915783 


497 


247009 


122763473 


22-2934968 


7 921100 


498 


24S004 


123505992 


223159 Vifc 




499 / 


249001 


124251499 






oOO l 250000 1 


125000000 





\ 



SQUARES, CUBES, AND ROOTS. 



101 



Number. 


Square. 

— 


Cube. 


Square Root. 


Cube Root. 


u 501 


251001 


125751501 


223330293 


7942293 


502 


252004 


126506008 


224053565 


7947574 


503 


253009 


127263527 


224276615 


7-952848 


504 


254016 


128024064 


224499443 


7-958114 


505 


255025 


12S7t*7625 


224722051 


7963374 


506 


256036 


129554216 


224944438 


7968627 


507 


257049 


130323843 


225166605 


7973873 


50S 


25S064 


131096512 


225333553 


7 979112 


509 


259081 


131872229 


22 5610283 


7984344 


510 


260100 


132651000 


22 5831796 


7989570 


511 


261121 


133432831 


226053091 


7994788 


512 


262144 


134217728 


226274170 


8000000 


513 


263169 


135005697 


226495033 


8005205 


514 


264196 


135796744 


22 6715681 


8010403 


515 


265225 


136590875 


22 6936114 


8015595 


516 


266256 


137388096 


22 7156334 


8020779 


517 


267289 


138188413 


22 7376340 


8025957 


518 


26S324 


138991832 


227596134 


8 031129 


519 


269361 


139798359 


227815715 


8036293 


520 


270400 


140608000 


228035085 


8041451 


521 


271441 


141420761 


228254244 


8046603 


522 


272484 


142236643 


228473193 


8051748 


523 


273529 


143055667 


228691933 


8056886 


524 


274576 


143877824 


22 8910463 


8062018 


525 


275625 


144703125 


22 9128785 


8067143 


526 


276676 


145531576 


229346899 


8072262 


527 


277729 


146363183 


22 9564806 


8077374 


528 


278784 


147197952 


22 9782500 


8082480 


529 


279841 


148035889 


230000006 


8087579 


530 


280900 


148877000 


23 0217289 


8092672 


531 


281961 


149721291 


23 0434372 


8097759 


532 


283024 


15056S768 


230651 252 


8 102839 


533 


284089 


151419437 


230867928 


8 107913 






1 •J^Z / OOUi 




Q.i 1 OOQA 


535 


286225 


153130375 


23- 1300670 


8 118041 


536 


287296 


153990656 


231516738 


8123096 


537 


286369 


154854153 


23 1732605 


8128145 


538 


289444 


155720872 


23 1948270 


8 133187 


539 


290521 


156590S19 


23 2163735 


8138223 


540 


291600 


157464000 


232379001 


8 143253 


541 


292681 


158340121 


232594067 


8-148276 


542 


293764 


159220088 


232808935 


8 153294 


543 


294849 


160103007 


233023604 


8- 158305 


544 


295936 


160989184 


233238076 


8 163310 


545 


297025 


161878625 


23 3452351 


8 168309 


546 


298116 


162771336 


233666429 


8- 173302 


547 


299209. 


163667323 


23 3880311 


8- 178289 


54S 


300304 


164566592 


234093998 




549 


301401 


165469149 


23 4307490 


S-\88*44 


550 1 


302500 I 


166375000 


23-4520788 


8-19**1* 



ARITHMETIC. 



Number. 


Square. 


Cube. 


Square Root* 


Cube Root 


551 




303601 


167284151 


23 4733892 


8 198175 


553 


304704 


168196608 


234946802 


8 203132 


553 


305309 


169112377 


23 5159520 


8208082 


554 


306916 


170031464 


23 5372046 


8 213027 


555 


308025 


170953875 


235584380 


8 217966 


556 


309136 


171879616 


235796522 


8222898 


557 


310249 


172308693 


23-6008474 


8227825 


558 


311364 


173741112 


23 6220236 


8 232746 


559 


312481 


174676879 


23-6431303 


8 237661 


560 


313600 


175616000 


23 6643191 


8242571 


561 


314721 


176553481 


23 6854386 


8247474 


562 


315844 


177504328 


237065392 


8252371 


563 


316969 


178453547 


23 7276210 


8257263 


564 


318096 


179406144 


237486842 


8262149 


565 


319225 


180362125 


237697236 


8 267029 


566 


320356 


181321496 


237907545 


8 271904 


567 


321439 


182284263 


23 8117618 


8 276773 


568 


322624 


133250432 


23-8327506 


8-281635 


569 


323761 


184220009 


23 5537209 


3236493 


570 


324900 


135193000 


23-8746728 


8 291344 


571 


326041 


186169411 


2389560(53 


8 296190 


572 


327184 


187149248 


23 9165215 


8 301030 


573 


323329 


183132517 


23 9374184 


8305865 


574 


329476 


189119224 


239532971 


8 310694 


575 


330625 


190109375 


23 9791576 


8315517 


576 


331776 


191102976 


240000000 


3320335 


577 


332929 


192100033 


240208243 


3 325147 


578 


3340S4 


193100552 


24 04 16306 


8 329954 


579 


335241 


194104539 


24 0624183 


8 334755 


580 


336400 


195112000 


24 0831892 


8339551 


581 


337561 


196122941 


24 1039416 


8 344341 


582 


333724 


197137368 


24 1246762 


8 349126 


583 


339889 


198155287 


24 1453929 


8353905 


584 


341056 




1% lbbU919 


W d&3o78 


585 


342225 


200201625 


24 1867732 


8363446 


586 


343396 


201230056 


24 2074369 


8-368209 


587 


344569 


202262003 


24 2230829 


8372967 


588 


345744 


203297472 


24 2487113 


8 377719 


589 


346921 


204336469 


24 2693222 


8-382465 


590 


348100 


205379000 


242899156 


8387206 


591 


349281 


206425071 


24 3104916 


3-391942 


592 


350464 


207474688 


24 3310501 


8396673 


593 


351649 


208527857 


243515913 


8-401393 


594 


352836 


209584584 


24-3721152 


8406118 


595 


354025 


210644875 


24 3926218 


8 410833 


596 


355216 


2 11 703736 


244131112 


8 415542 


597 


356409 


212776173 


24 4335334 


8420246 


598 


357604 | 


213847192 


244540335 


. 8 424945 


599 f 


358801 


214921799 


24 47447^ 




600 I 360000 \ 


216000000 


24*M4fiOT4 





SQUARES, CUBES, AND ROOTS. 



103 



Numbtf. 


Square. 




Square KooL 


Cube Hoot, 


601 


361201 


217081801 


24 5153013 | 


3 439010 


602 


362404 


218167208 


24 5356333 


8 443688 


603 


363609 


919256227 


24 5560583 


8 448 360 


604 


364816 


220348864 


24 5764115 


8 453028 ! 


605 


366025 


221445L25 


24-5967473 


3 457691 


60S 


367236 


222545016 


24 6170673 


8 462343 


607 


36S449 


223648543 


24 6373700 


8 467000 


60S 


369664 


224755712 


24 6576560 


8 471647 


609 


37088 1 


225366529 


24 6779254 


8 476289 


610 


372 1 00 


226 9S 1000 


24698 1781 


3480426 


611 


373321 


228099131 


24 -7 134 142 


8 43 55 58 


612 


374544 


229220928 


24-7386338 


8490 185 


613 


375769 


230346397 


24-7538363 


8 4 94 806 


614 


376996 


23 1475544 


24-7790234 


8 499423 


615 


378225 


232608375 


247991935 


ft 504035 


616 


379456 


233744396 


24-31 93473 


8508642 


617 


330639 


234885113 


24-8394847 


8-513243 


618 


381924 


236029032 


24 3596053 


8 517840 


619 


3S3161 


237176659 


24 3797106 


3-522432 


620 


384400 


236328000 


24-8997992 


8 527019 


6 + >l 


38564 1 


239483061 


24 9198716 


8 531601 


622 


336884 


240641848 


24 9399278 


8 536178 


623 


388129 


241804367 


24 9599679 


8510750 


624 


3S9376 


24297QG24 


24 9799920 


8 545317 


625 


390625 


244140625 


25 0000000 


8-549379 


626 


391876 


245314376 


25 0199920 


8554437 


627 


393129 


246491833 


25 0399681 


8 558990 


623 


394384 


247673152 


25 0599282 


8563533 


629 


395641 


243858189 


25079S724 


8 563081 


630 


396900 


250047000 


25 0993008 


85726 19 


631 


398161 


251239591 


25 1197134 


3 577152 


632 


399424 


252435963 


25 1396102 


8531631 


633 


400689 


253636137 


25 1594913 


8 536205 


634 


401956 


254840104 


25 1793566 


8-590724 


635 


4032*5 


256047875 


25 1992063 


8595238 


636 


404496 


257259456 


252190404 


8-599747 


637 


405769 


25S474S53 


252388539 


3604252 


638 


407044 


259694072 


25 2536GJ9 


8608753 


639 


408321 


260917119 


252784493 


8613248 


640 


409600 


262144000 


25 2982213 


8 617739 


641 


410881 


263374721 


25 3179778 


8622225 


642 


412164 


2646092S8 


25-3377189 


8626706 


643 


413449 


265347707 


253574447 


8631183 


644 


1 414736 


2670399*4 


253771551 


8635655 


645 


416025 


263336125 


25 3968502 


8 640123 


646 


417316 


269536136 


25 4165301 


8-644585 


647 


418609 


570340023 


25 4361947 


3649044 


648 


419904 


272097792 


25-4558441 




649 


421201 : 


273359449 j 


25-4754784 


I 8657946 


050 j 


422500 j 274625000 j 


25 4950976 


\ 8 Wt 



104 



ARITHMETIC. 



Number. 


Square, 


Cube. 


Square Root. 


Cube Root. 


651 


423S01 


275894451 


25 5147016 


8666831 


652 


425104 


277167808 


25 5342907 


8671266 


653 


426409 


278445077 


25 5 538647 


8-675697* 


654 


427716 


279726264 


(CtC C.TO #n*l»w 

25 5734 237 


S 680 124 


655 


429025 


£\<Cjr 1 111 1 O T £ 

28101 14 7o 


25 5929678 


8 684540 


656 


430336 


282300416 


25 6124969 


8 68y96i> 


657 


431649 




25 63201 12 


8 69337b 


653 


432964 


2848903 12 


256515107 


8 6SJ7784 


659 


434291 


2obl91 1 TSJ 


Ac,;' nfj'i^v ftco 


8 7l>£ loo 


660 


4d5h00 


not i ncAAn 

2a7496000 


256904652 


b r Ubao 7 


661 


436921 


2 8 88 04781 


257099203 


8 7109S3 


662 
663 


43S244 


290117528 


25 7203607 


8 715373 


439569 


fin l ^ n j a j pv 

291434247 


25 7487864 


8 7197^9 


664 


a. incinc 
*4U$Ub 


292754944 
29407962a 


25 7681975 


8724141 


665 


A A a aa r 

442225 


25 7875939 
25S069758 


8 728518 


G66 


443556 


295408296 


8 732892 


667 


444889 


296740963 


25 8263431 


8737260 


GtiH 


446224 


298077632 


258456960 


8 741624 


669 


447561 


299418309 


25 S6 50343 


O wf 4 t A£5 ^ 

8 74 59 So 


670 


A A D AA/\ 

44S900 


300763000 


25 8o43582 


O'7o0o40 


671 


450241 


3021 1 171 1 


25 9036677 


A f E J /■fill 

8 754691 


672 


45158! 


3 034 G 444 8 


25-9229628 


s- ?5903S 


673 


452929 


304821217 


25 9422435 


8 76oo81 


674 


454276 


306182024 


25961 51 00 


8 767719 


675 


455625 


307546875 


25 9307621 


S 772053 


676 


456976 


308915776 


2G 0000000 


S 7/6383 


677 


45S329 


3 1 0288733 


2» 0192237 


8 780708 


67S 


459684 


31 16G5752 


2t> Uo^4*id 1 


Q .FT U & f\ A A 


679 


461041 


h 1 lid i ni'i-lA 

313046839 


2605762S4 


□ .frtfin j j ■ 


680 


462400 


o 1 44 o2trO(i 


ZD u /onUiJb 


S /HoboU 


6S1 


4bi37b 1 


•> 1 COO 1 O IT 






6S2 


465124 


*5 1 72 14568 


nf i i Ei rtfli-f 

2b 1 1 J 2*17 


8 r>02272 


683 


46G489 


31861 1987 


2b 1342GS7 


S MJ6572 








JO L-j.j...ipji 


o OI U^DEs 


685 


469225 


321419125 


261725047 


8S15160 


BSC 


470596 


322828856 


26-1916017 


8819447 


687 


471969 


324242703 


262106848 


8 823731 


683 


473344 


325660672 


262297541 


8S2S009 


689 


474721 


3270827G9 


262488095 


8-832285 


690 


476100 


32850900G ! 


26 267851 1 


8836556 


691 


4774S1 


329939371 


262868789 


S-840823 


692 


478864 


331373S3S 


26 3058929 


8-845085 


693 


480249 


332812557 


263248932 


8849344 


694 


481636 


3342553S4 


26-3438797 


8S53598 


695 


483025 


335702375 


26-3628527 


8-857849 


696 


484416 


33715353G 


26-3818119 


8862095 


697 


485809 


338608873 


26-4007576 


SS66337 


698 


487204 


340068392 


26*4196896 


8870576 


699 


488601 


341632099 






TOO j 


490000 


343000000 


\ MA^bm 





SQUARES, CUBES, AND ROOTS. 



105 







Cube. | Square Foot j CuLe Root. 


701 


491401 


344472101 I 


264764046 ! 


8883266 


702 


492804 


345948088 


26 4052826 , 


8887488 


703 


494209 


347428927 i 


265141472 


8891706 


704 


495616 


348913664 
35041)2625 


265329988 


8895920 


705 


497025 


265518361 | 


8900130 


706 


498436 


351895816 


26 5706605 j 


8904336 


707 


499849 


353303243 


26 5894716 j 


8-908538 i 


i 708 


501264 


354894912 


26 6082694 


8 912737 


709 


502681 


356400*29 


26 6270539 


8916931 


710 


504 1 00 


357911000 


26 6458252 > 


8921121 


711 


505521 


359425431 


26 6645833 


8925308 


712 


506944 


360944 12^ 


266833281 


8929490 


713 


50S369 


362467097 


267020598 


8933668 


714 


509796 


363994344 


267207784 


8*937843 


715 


511225 


365525875 


26 7394839 


8 942014 


716 


512656 


367061696 


267581763 


8946181 


717 


514089 


368601813 


26 7768557 


8950344 


718 


515524 


370146232 


267955220 


8954503 i 


719 


516961 


371694959 


268141754 


8958658 j 


720 


518400 


373248000 


26 8328157 


8962809 | 


721 


519841 


374805361 


26 8514432 


8966957 ! 


722 




37636704 H 


26 8700577 


8 971101 ! 


723 


522729 
524176 


377933067 


26 8886593 


8975240 


724 


379503424 


26-90724H1 1 


8979376 


725 


525625 


381078125 


26 9258240 


8-983509 


726 


527076 


382657 176 


26-9443872 


8 9*7637 


727 


52S529 


384240583 


. 269629375 


S 991762 


728 


529984 


3S5828352 


269814751 


8 995883 


729 


531441 


3S7420489 


270000000 


9 000000 


730 


532900 


389017000 


270185122 


9 004113 


731 


534361 


390617SUI 


27 0370117 


9008223 


732 


535S24 


392223168 


270554985 


9*012328 


733 


537289 




£.1 U tOiJ 1 41 


o.m £1 A M 1 
\i i) I b4ol 


734 


538756 


395446904 


27 0924 344 


9 020529 


735 


540225 


397065375 


27 1108834 


9024624 


736 


541696 


398688256 


27 1293199 


9 028715 


737 


543169 


400315553 


27 1477439 , 


9-032802 


738 


544644 


401947272 


27 1661554 


9036886 


739 


546121 


403583419 


27 1S45514 


9040965 


740 


547600 


405224000 


27 2029410 


9 045041 


741 


549081 


406869021 


27 2213152 


9 049114 


742 


550564 


40851848H 


272396769 


9053183 


743 


552049 


410172407 


27 2580263 


9057248 


744 


553536 


4118307*4 


27 2763634 j 


9061310 


745 


555025 


113403625 


27 2946SS1 


9065367 


746 


556516 


i 415160936 


27 3130006 : 


9 069422 


747 


558009 


416832723 


27 3313007 « 


9073473 


748 


559504 


418508992 


! 273495K87 




/ 749 1 


561001 . 


' 4201S9749 


27 367S644 


\ 9-QSU-&& 


L_750 I 


562500 I 


421875000 j 


27 3861270 




Vol. I. 




15 







106 ARITHMETIC. 
- . — * t ■■ ■ * 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


751 


564001 


42o0b47ol 


274043792 


- — - — ■ 

9 089639 


752 


565504 


42525900H 


27 42261S4 


9 093672 


753 


567009 


426957777 


2744084 55 


9 + 09770 1 


754 


568516 


428661064 


27 4590604 


9 101726 


755 


570025 


430368875 


27 "4772633 


9 1 105748 


756 


571536 


432081216 


27" 4954542 


9109766 


757 


573049 


433798093 


MH jr ill ft A A A #V 

27 5136330 


9113781 


758 


574564 


I ■ I • - 1 , , - 1 l 


L^nif ft 1 PV ft n fi 

27 5317998 


9" 117793 


759 


576081 


437245479 


Afv E A n H E J 

27 5499 546 


9121801 


760 


577600 


43a!J7600U 


Af¥~. E Jf* OftftPV f 

27*5680975 


9125S05 


761 


579121 


4 * ATT 1 jn o i 

440711081 


2 7- 5862284 


9 129606 


762 


580644 


44245Q72S 


27 H 6043475 


9-133803 


763 


582169 


444194947 


27 6224546 


9*137797 


764 


58369b 


445943744 


27-6405499 


9*141788 


765 


585225 


447697125 


27 b 586334 


9' 145774 


766 


586 7a6 


449455096 


27-6767050 


9' 149757 


767 


588289 


Jt(£TA*tWji.jtA 

451217663 


276947648 


9 153737 


76S 


f£!}ft£?A J 

589824 


452984832 


27 7128129 


9157714 


769 


591361 


454756609 


27-7308492 


9161686 


770 


r fill nriA 

592900 


456533000 


27- 7438739 


9' 165656 


771 


594441 


458314011 


27-7668868 


9 1 169622 


772 


595984 


460099648 


27- 7848880 


n, -ft Pwn jp ja. n 

9- 173585 


773 


597529 


461889917 


27 8028775 


9 177544 


774 


599076 


463bS4H24 


27S20S555 


9181500 


775 


600625 


465484375 


27*8388218 


9 185453 


776 


602176 


467288576 


A*V A f nfVM n ^1 

27 8567766 


9- 189402 


777 


603729 


A £3 A J"ki i\M j nil 

469097433 


27 8747197 


9 193347 


778 


605284 


4709 1 0952 


27 8 9265 14 


9' 197239 


779 


bU6341 


47272^139 


£)■* Ai AFrVi r- 

279105715 


9201229 


780 


608400 


47 4 DO 2000 


Ilk ft AO J 4 

2 7 + 9284801 


9 205 164 


781 


bU99bl 


4ro379541 


279463772 


9 209096 


782 


fill 524 


47821 1768 


279642629 


9 213025 


783 


j-fc i A AAA 

613089 


480048687 


279921 372 


1 9216950 


784 


614656 


481890304 


280000000 


9 '220873 


785 


616225 


483736625 


280178515 


9 224791 


786 


617796 


485587656 


28 0356915 


9228707 


787 


619369 


487443403 


280535203 


9 232619 


789 


620944 


489303872 


28 0713377 


9 237528 


789 


622521 


491169069 


28 0891438 


9240433 


790 


624100 


493039000 


28 1069386 


9 244335 


791 


625681 


494913671 


28 1247222 


9243234 


792 


627264 


496793088 


28 1424946 


9252130 


793 


628849 


498677257 


28 1602557 


9256022 


794 


630436 


500566184 


28 1780056 


9 259911 


795 


632025 


502459875 


28 1957444 


9263797 


796 


633616 


504358336 


23 2134720 


9*267680 


797 


635209 


506261573 


28 231 1884 
28 2488938 


9271559 
9 275435 


798 


' 636S04 


5031 69592 


799 


63S401 


510082399 


40b0881 


9279308 


800 j 


640000 | 


512000000 







BQVAftES, CUBES, AND ROOTS. 



107 



Number, 



801 
803 
803 
804 
805 
806 
807 



810 
811 
812 
813 
814 
815 
816 
817 
818 
810 
820 
831 
822 
823 
824 
825 
826 
837 
823 
820 
830 
831 
832 



835 



837 



840 
841 
842 
843 
844 
845 
846 
847 
848 
849 



Square. 



6416Q1 
643304 
644300 
646416 
643025 
649636 
651249 



654481 
656100 
657721 
659344 
660960 
662596 
664225 
665956 
6674S9 
669124 
670761 
672400 
674041 
675684 
677329 
678976 
680625 
682276 
683929 
6S5584 
637241 
688900 
690561 



697225 



848 I 
850 I S 



700569 
702244 
703921 
705600 
707281 
708964 
710649 
712336 
714025 
715716 
717409 
719104 
720601 j 
722500 / 



Cube. 



513922401 
515849608 
517781627 
519718464 
521660125 
623606616 
525557943 
527514112 
529475120 
531441000 
533411731 
635387328 
537366797 
530353144 
541343375 
543338496 
545336513 
547343432 
540353250 
55136S000 
553387661 
555412248 
557441767 
559476224 
561515625 
563559976 
565600283 
567663552 
569722780 
571787000 
573S5G191 
575030368 
578009537 
580093704 
5821S2875 
584277056 
586376253 
5884S0472 
590589719 
592704000 
594823321 
596947683 
599077107 
601211584 
603351125 
605495736 
607645423 
! 609800192 
611960049 
614125000 



Square Root. 



Cube Root. 



28 3019434 
23 3106045 
283372546 
28-3543933 
28 3725210 
233901391 
28 4077454 
23-4253408 
284420253 
284604989 
23 4780617 
28 4956137 
285131549 
285306852 
23 5482048 
28 5657137 
23 5832119 

28- 6006993 
28 6181760 
236356421 
28*6530976 
28 6705424 
23 6370766 
287054002 
28 7228132 
287402157 
28 7576077 
28 7749891 
28 7923601 
238097206 
288270706 
28 8444102 
288617394 
288790582 
28 8063666 
28 9136646 
289309523 
28 9482297 
289654967 

28 9827535 
290000000 
200172363 

29 0344623 
29 0516781 
290688837 
29 0860791 
29 1032644 
29 1204396 
29*1376046 

29- 1547595 



0287044 
0290907 
9294767 
9298624 
9302477 
9 306328 
9310175 
9314019 
317860 
321697 
9325532 
9329363 
9 333192 
337017 
9*340838 
0344657 
0348473 
9352286 
0356095 
9359902 
9 363705 
9367505 
9 371302 
9-375096 
9-37S887 
9382675 
9386460 
9300242 
9394020 
0397796 
9401569 
9405339 
9 409105 
9 412369 
9 416630 
9 420337 
9424142 
9427894 
9431642 
9 435388 
9439131 
9 442870 
944C607 
9450341 
9454072 
945TOQQ 

9-4-WMA 
9-4*ra&9A 



108 



ARITHMETIC. 



rHUIIlUci. 




Cube. 


SnunTP Root 


Cube Root 


851 


724201 


616295051 


29 1719043 


9476395 


852 


725904 


618470208 


29 1690390 


9480106 


853 


727609 


620650477 


292061637 


9-483813 


854 


729316 


622835864 


292232784 


9487518 


855 


731025 


625026375 


292403830 


9491220 


856 


732736 


627222016 


292574777 


9494919 


857 


734449 


629422793 


292745623 


9 498615 


858 


736164 


631628712 


292916370 


9502308 


859 


737881 


633839779 


293087018 


9505998 


860 


739600 


636056000 


293257566 


9509685 


861 


741321 


638277381 


293428015 


9513370 


862 


743044 


640503928 


293598365 


9517051 


863 


744769 


642735647 


293768616 


9520730 


864 


746496 


644972544 


293938769 


9524406 


865 


748225 


647214625 


294108823 


9528079 


866 


749956 


649461896 


294278779 


9531749 


867 


751689 


651714363 


294446637 


9 535417 


868 


753424 


653972032 


29 4618397 


9539082 


869 


755161 


656234909 


294788059 


9542744 


870 


756900 


658503000 


294957624 


9546403 


871 


758641 


660776311 


29-5127091 


9550059 


872 


760384 


663054848 


295296461 


9553712 


873 


762129 


665338617 


295465734 


9557363 


874 


763876 


667627624 


295634910 


9561011 


875 


765625 


669921875 


295803989 


9564656 


876 


767376 


672221376 


295972972 


9568298 


877 


769129 


674526133 


296141858 


9571938 


878 


770884 


676836152 


29 6310648 


9575574 


879 


772641 


679151439 


296479325 


9579208 


880 


774400 


681472000 


296647939 


9-582840 


881 


776161 


683797841 


296616442 


9586468 


S82 


777924 


686128968 


296984848 


9590094 


883 


779689 


688465387 


297153159 


9593716 


884 


781456 


690807104 


297321375 


9597337 


885 


783225 


693154125 


297489496 


9600955 


886 


784996 


695506456 


29 7657521 


9-604570 


8S7 


786769 


697864103 


297825452 


9608182 


888 


788544 


700227072 


297993289 


9611791 


889 


790321 


702595369 


29 8161030 


9615398 


890 


792100 


704969000 


298328678 


9619002 


891 


793881 


707347971 


29 8496231 


9622603 


892 


795664 


7097322S8 


298663690 


9626201 


893 


797449 


712121957 


298831056 


9629797 


894 


799236 


714516984 


298998328 


9633390 


895 


801025 


716917375 


299165506 


9636981 


896 


802816 


719323136 


299332591 


9640569 


897 


804609 


721734273 


299499583 


9 644154 


898 I 


806404 


724150792 




I 9 647737 


899 


808201 


726572699 






900 / 810000 


729000000 \ 30 OOOOWto \ 9 



•QUARKS, CUBES, AND ROOTS. 



109 



Number. Square, Cube 



811801 
813604 
815409 
817316 
819025 
H2083G 



826281 
828100 
829921 
831744 
833569 
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902500 / 



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Square Hoot* Cube Root 



300166620 

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30 0832179 

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30 1164407 

30 1330393 

30 1496269 

30 1662063 

30- 1827765 

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302324329 

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30 5941171 

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30*8220700 



9658469 
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9S1665* 



110 



ARITHMETIC 



951 
952 
955 
954 
955 
956 
957 
958 
959 
960 
961 
962 
963 
964 
965 
966 
967 
968 
969 
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971 
972 
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978 
979 
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981 
982 
983 
984 
985 
986 
987 
988 
989 
990 
991 
992 
993 
994 
995 
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998 

WOO / 



Square. Cube. Square Root. Cube Root 



904401 
906304 



910116 
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! 917764 
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921600 
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( 996004 
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1000000 



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970299000 
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994011992 
997002999 
1000000000 



1 



m 

T2 

30 8706981 
308868904 
309030743 

30 9192497 
309354166 
30*9515751 
309677251 

30- 9838668 

31 0000000 
310161248 
31 0322413 
31 0483494 
310644491 

31- 0805405 
31 0966236 
311126984 
31 1287648 
31 1448230 
31 1608729 
31 1769145 
31 1929479 
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31 2249900 
31 
31 

313729915 
31 2889757 
31 3049517 
31 3209195 
31 3368792 
31 3528303 
31 3687743 
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31 4642654 
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31 4960315 
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31 6753068 



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9 963198 
9-966555 
9-969909 
9 973262 
9 976612 
9979960 
9983305 
9986649 
9989990 



Ill 



OF RATIOS, PROPORTIONS, AND PRO- 
GRESSIONS. 

Numbers are compared to each other in two different 
ways : the one comparison considers the difference of the two 
numbers, and is named Arithmetical Relation ; and the dif- 
ference sometimes the Arithmetical Ratio : the other con- 
siders their quotient, which is called Geometrical Relation ; 
and the quotient is the Geometrical Ratio. So, of these two 
numbers 6 and 3, the difference, or arithmetical ratio is 
6—8 or 8, but the geometrical ratio is f or 2. 

There must be two numbers to form a comparison : the 
number which is compared, being placed first, is called the 
Antecedent : and that to which it is compared, the Con- 
sequent. So, in the two number* above, 6 is the antecedent, 
and 3 the consequent. 

If two or more couplets of numbers have equal ratios, or 
equal differences, the equality is named Proportion, and the 
terms of the ratios Proportionals. So, the two couplets, 4, 2 
and 8, 6, are arithmetical proportionals, because 4-2 = 8 
—6 = 2; and the two couplets 4, 2 and 0, 3, are geome- 
trical proportions, because | = f = 2, the some ratio. 

To denote numbers as being geometrically proportional, a 
colon is set between the terms of each couplet, to denote their 
ratio ; and a double colon, or else a mark of equality, between 
the couplets or ratios. So, the four proportionals, 4, 2, 6, 3 
are set thus, 4 : 2 : : 6 : 3, which means, that 4 is to 2 as 6 
is to 3 ; or thus, 4:2 = 6:3, or thus, | = |, both which 
mean, that the ratio of 4 to 2, is equal to the ratio of 
6 to 3. 

Proportion is distinguished into Continued and Discon- 
tinued. When the difference or ratio of the consequent of 
one couplet, and the antecedent of the next couplet, is not 
the same as the common difference or ratio of the couplets, 
the proportion is discontinued. So, 4, 2, 8, 6, are in discon- 
tinued arithmetical proportion, because 4—2=8 — 6=2, 
whereas 8 — 2=6: and 4, 2, 6, 3 are in discontinued geo- 
metrical proportion, because f = §■ = 2, but -J = 3, which is 
not the same. 

But when the difference or ratio of every two succeeding 
terms is the same quantity, the proportion is said to be Con- 
tinued, and the numbers themselves make a scrisa ot Cwv- 



112 



ARITHMETIC. 



Untied Proportionals, or a progression. So 2, 4, 6, 8 form 
an arithmetical progression, because 4 — 2 = — 4 = 8 — 
6 = 2, all the same common difference ; and 2, 4, 8, 16, a 
geometrical progression, because f = } = Y = 2, all the 
same ratio. 

When the following terms of a progression increase, or 
exceed each other, it is called an Ascending Progression, or 
Series ; but when the terms decrease, it is a descending 
one. 

So, 0, 1, 2, 3, 4, &c. is an ascending arithmetical progression, 
but 9, 7, 5, 3, 1, dec. is a descending arithmetical progression. 
Also 1 ,2, 4, 8, 16, dec. is an ascending geometrical progression, 
and 16, 8, 4, 2, 1 , dec. is a descending geometrical progression* 



ARITHMETICAL PROPORTION AND 
PROGRESSION. 

In Arithmetical Progression, the numbers or terms have 
all the same common difference. Also, the first and last 
terms of a Progression, are called the Extremes ; and the 
other terms, lying between them, the Means. The most 
useful part of arithmetical proportion, is contained in the 
following theorems : 

Theorem 1. When four quantities are in arithmetical 
proportion, the sum of the two extremes is equal to the sum 
of the two means. Thus, of the four 2, 4, 6, 8, here 2 + 
8 = 4 + 6=10. 

Theorem 2. In any continued arithmetical progression, 
the sum of the two extremes is equal to the sum of any two 
means that are equally distant from them, or equal to double 
the middle term when there is an uneven number of terms. 

Thus, in the terms 1, 3, 5, it is 1 + 5 = 3 + 3 = 6. 

And in the series 2, 4, 6, 8, 10, 12, 14, it is 2 + 14 = 4 
+ 12 =6 + 10 = 8 + 8 = 10. 

Theorem 3. The difference between the extreme terms 
of an arithmetical progression, is equal to the common dif- 
ference of the series multiplied by one less than the number 
of the terms. So, of the ten terms, 2, 4, 6, 8, 10, 12, 14, 
16, 18, 20, the common difference is 2, and one less than 
the number of terms 9 ; then the difference of the extremes 
is 20 - 2 = 18, and 2 X 9 = 18 also. 



ARITHMETICAL PHOFORTION. 



113 



Consequently the greatest term is equal to the least term 
added to the product of the commoo difference multiplied by 
1 less than the number of terms. 

Theorem 4. The sum of all the terms, of any arith- 
metical progression, is equal to the sum of the two extremes 
multiplied by the number of terms, and divided by 2 ; or the 
sum of the two extremes multiplied by the number of the 
terms, gives double the sum of all the terms in the series. 

This is made evident by setting the terms of the series in 
an inverted order, under the same series in a direct order, and 
adding the corresponding terms together in that order. Thus, 
in the series 1, 3, 5, 7, 9, 11, 13, 15; 
ditto inverted 15, 13, 11, 9, 7, 5, 3, 1 ; 
the sums are 16 +10 + 16 + 16 + 16 + 16 + 16 -f- 16, 
which must be double the sum of the single series, and is 
equal to the sum of the extremes repeated as often as are the 
number of the terms. 

From these theorems may readily be found any one of 
these five parts ; the two extremes, the number of terms, the 
common difference, and the sum of all the terms, when any 
three of them are given ; as in the following problems : 

PROBLEM I. 

Given the Extremes, and the Number of Terms, to find the Sum 
of all the Terms, 

Add the extremes together, multiply the sum by the 
number of terms, and divide by 2. 

EXAMPLES. 

1. The extremes being 3 and 19, and the number of 
terms 9 ; required the sum of the terms ? 

19 
3 

22 

2) 198 
Ans. 99 

2. It is required to find the number of all the strokes a 
common clock strikes in one whole revolution of the index, 
or in 12 hours. Aua.nfc. 

Vox. 1. 1G 



= ^X9=11 X 9 = 99, 
the same answer. 



114 



▲xinufsnc. 



Ex. 3. How many strokes do the clocks of Venice strike 
in the compass of the day, which go continually on from 1 
to 24 o'clock ? Ana. 300. 

4. What debt can be discharged in a year, by weekly 
payments in arithmetical progression, the first payment being 
U y and the last or 52d payment 5i 3t ? Ans. 1851 4*. 



PROBLRX n. 



Given the Extremes, and the Number of Terms ; t^findthe 
Common Difference. 

Subtract the less extreme from the greater, and divide 
the remainder by 1 less than the number of terms, for the 
common difference. 



EXAMPLES. 



1. The extremes being 3 and 19, and the number of terms 
9 ; required the common difference ? 

19 

Ur ' 9-T ¥ 




2. If the extremes be 10 and 70, and the number of terms 
21 ; what is the common difference, and the sum of the 
series ? Ans. the com. diff. is 3, and the sum is 840/ 

3. A certain debt can be discharged in one year, by weekly 
payments in arithmetical progression, the first payment being 
1*, and the last 51 Ss ; what is the common difference of the 
terms? Ans. 2. 



PROBLEM III. 

v 

Given one of the Extremes, the Common Difference, and the 
Number of Terms ; to find the other Extreme, and the 
Sum of the Series. 

Multiply the common difference by 1 less than the 
number of terms, and the product will be the difference of 
the extremes : Therefore add the product to the less ex- 
treme to give the greater ; or substract it from the greater, 
to give the less extreme. 



ARITHMETICAL PROGRESSION. 



115 



EXAMPLES. 

1. Given the least term 3, the common difference 2, of an 
arithmetical aeries of 9 terms ; to find the greatest term, and 
the sum of the series. 

Here 2 X (9 — 1) + 3 = 19, the greatest terra. Theref. 
(10 +3)| = »}• =99, the sum of the series. 

2. If the greatest term be 70, the common difference 3, 
and the number of terms 21, what is the least term, and the 
sum of the series ? 

Ans. The least term is 10, and the sum is 840. 

8. A debt can be discharged in a year, by paying 1 shilling 
the first week, 8 shillings the second, and so on, always 2 
shillings more every week ; what is the debt, and what will 
the last payment be ? 

Ans. The last payment will be 5* 3*, and the debt is 1357 4s. 



PROBLEM IV. 

To find an Arithmetical Mean Proportional between two 
given terms. 

Add the two given extremes or terms together, and take 
half their sum tor the arithmetical mean required. 

EXAMPLE. 

To find an arithmetical mean between the two numbers 4 
and 14. Here 
14 
4 

2) 18 

Ans. 9 the mean required. 



problem v. 

To find two Arithmetical Means between two given 
Extremes. 

Subtract the less extreme from the greater, and divide 
the dMfeienee by 8, so will the quotient be the coranKfli&i- 



116 



ARITHMETIC. 



ference ; which being continually added to the less extreme, 
or taken from the greater, will give the means. 

EXAMPLE. 

To find two arithmetical means between 2 and 3. 
Here 8 
2 

3) 6 Then 2 + 2 = 4 the one mean. 
— — and 4 + 2 = 6 the other mean, 
com. dif. 2 



PROBLEM VI. 

To find any Number of Arithmetical Means between two 
given Terms or Extremes. 

Subtract the less extreme from the greater, and divide 
the difference by 1 more than the number of means required 
to be found, which will give the common difference ; then this 
being added continually to the least term, or subtracted from 
the greatest, will give the mean terms required. 

EXAMPLE. 

To find five arithmetical means between 2 and 14. 
Here 14 
2 

6) 12 Then by adding this com. dif. continually, 

the means are found 4, 6, 8, 10, 12. 

com. dif. 2 

See more of Arithmetical progression in the Algebra. 



GEOMETRICAL PROPORTION AND PRO- 
GRESSION. 

If there be taken two ratios, as those of 6 to 3, and 14 
to 7, which, by what has been already said (p. 113), may 



GEOMETRICAL tfftOGBBMIO*. 



117 



be expressed fractionally, } and \f ; to judge whether they 
are equal or unequal, we must reduce them to a common 
denominator, and we shall have 6X7, and 14 X 3 for the 
two numerators. If these are equal, the fractions or ratios 
are equal. Therefore, 

Theorem i. If four quantities be in geometrical propor- 
tion, the product of the two extremes will be equal to the 
product of the two means. 

And hence, if the product of the two means be divided 
by one of the extremes, the quotient will give the other ex- 
treme. So, of the above numbers, if the product of the means 
42 be divided by 6, the quotient 7 is the other extreme ; 
and if 42 be divided by 7, the quotient 6 is the first ex- 
treme. This is the foundation of the practice in the Rule 
of Three. 

We see, also, that if we have four numbers, 6, 3, 14, 7, 
such, that the products of the means and of the extremes are 
equal, we may hence infer the equality of the ratios $ =y , 
or the existence of the proportion 6 : 3 : : 14 : 7. Hence 

Theorem n. We may always form a proportion of the 
factors of two equal products. 

If the two means are equal, as in the terms 3, 6, 6, 12, 
their product becomes a square. Hence 

Theorem hi. The mean proportional between two num- 
bers is the square root of their product. 
We may, without destroying the accuracy of a proportion, 

S've to its various terms all the changes which do not affect 
e equality of the products of the means and extremes. 
Thus, with respect to the proportion 6 : 3 : : 14 : 7, 
which gives 6X7 = 3X1 4, we may displace the extremes, 
or the means, an operation which is denoted by the word 
AMcrnando. 

This will give 6 : 14 : : 3:7 

or 7 : 3 : : 14 : 6 

or 7 : 14 : : 3:6 

Or, 2dly, we may put the extremes in the places of the means, 
oalled lnvertendo. 

Thus 3 : 6 : : 7 : 14. 

Or, 3dly, we may multiply or divide the two antecedents, 
or the two consequents, by the same number, when propor- 
tionality will subsist. 



118 ARITHMETIC. 

As 6 X 4 : 3 : : 14 X 4 : 7 ; viz. 24 : 3 : : 66 : 7 
and6^-2:3:: 14-t-2:7; viz. 3:3:: 7:7. 

Also, applying the proposition in note 2, Addition of 
Vulgar Fractions, to the terms of a proportion, such as 
30 : 6 : : 15 : 3, or y = y, we shall have 

30±15 15 . 30+15 30-15 „ 

= — and = . Hence 

6 ± 3 3 6+3 6-3 

Theorem iv. The sum or the difference of the antece- 
dents, is to that of the consequents, as any one of the ante- 
cedents is to its consequent. 

Theorem v. The sum of the antecedents is to their 
difference, as the sum of the consequents is* to their dif- 
ference. 

In like manner, if there he a series of equal ratios, 
1 = V = V 4 = i J 5 we have 
6+10+14+30 = 14 = 30 
3+5+7+15 7 15 *nereiure, 

Theorem vi. In any series of equal ratios, the sum of 
the antecedents is to that of the consequents, as any one an- 
tecedent is to its consequent. 

Theorem vii. If two proportions are multiplied, term by 
term, the products will constitute a proportional. 

Thus, if 30 : 15 :: 6 : 3 
and 2 : 3 : : 4 : 6. 

Then 30 X 2 : 15 X 3 :: 6 X 4 : 3 X 6 
or 60 : 45 :: 24 : 18 ; or f J- = 

Theorem viii. If four quantities are in proportion, their 
squares, cubes, &c. will be in proportion. 

For this will evidently be nothing else than assuming the 
proportionality of the products, term by term, of two, three, 
or more identical proportions. 

The same properties hold with regard to surd or irrational 
expressions, 

Thus, v'TSO : ^80 :: ^567 : ^63 
and -y/12 :-y/3 :: : -y/1. 



y/720 = y'O . 80 = 3 y/567 _ y/9X63 _ 3 
y/80 "~ V ** -v/63 y/63 1 

, 12 v/4 2 

and ^3 c= ^-3 = — = r 



GEOMETRICAL PROGRESSION. 



119 



Theorem ix. The quotient of the extreme terms of a 
• geometrical progression is equal to the common ratio of the 
series raised to the power denoted by 1 less than the number 
of the terms. 

So, of the ten terms 2, 4, 8, 16, 32, 64, 128, 256, 512, 
1024, the common ratio is 2, one less than the number of 
terms 9 ; then the quotient of the extremes is 1 °^ 4 = 512, 
and 2° = 512 also. 

Consequently the greatest term is equal to the least term 
multiplied by the said power of the ratio whose index is 1 
less than the number of terms. 

Theorem x. The sum of all the terms, of any geome- 
trical progression, is found by adding the greatest term to the 
difference of the extremes divided by 1 less than the ratio. 

So, the sum of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 
1024-2 

(whose ratio is 2) is 1024 + -^-y- = 1024 + 1022 =2046. 

This subject will be resumed in the Algebraic part of this 
"work. A few examples may here be added. 



EXAMPLES. 

1. The least of ten terms, in geometrical progression, 
being 1, and the ratio 2 ; what is the greatest term, and the 
sum of all the terms ? 

Ans. The greatest term is 512, and the sum 1023. 

2. What debt may be discharged in a year, or 12 months, 
iy paying 1Z the first month, 21 the second, 4Z the third, and 
40 on, each succeeding payment being double the last ; and 
what will the last payment be ? 

Ans. The debt 4095Z, and the last payment 2048/. 



PROBLEM I. 

To find one Geometrical Mean Proportional between any two 
Numbers. 

Multiply the two numbers together, and extract the 
*C|uare root of the product, which will give the mean propor- 
tional sought. 



120 



ARITHMETIC* 



EXAMPLE. 

To fiad a geometrical mean betweea the two numbers 
3 and 12. 

12 
3 

36 (6 the mean. 
36 



PROBLEM II. 

To find two Geometrical Mean Proportionals between any two 
Numbers. 

Divide the greater number by the less, and extract the 
cube root of the quotient, which will give the common ratio 
of the terms. Then multiply the least given term by the 
ratio for the first mean, and this mean again by the ratio for 
the second mean : or, divide the greater of the two given 
terms by the ratio for the greater mean, and divide this again 
by the ratio for the less mean. 

EXAMPLE. 

To find two geometrical means between 3 and 24. 

Here 3 ) 24 ( 8 ; its cube root 2 is the ratio. 

Then 3 X 2 = 6, and 6 X 2 = 12, the two means. 

Or 24 + 2 = 12, and 12 -r- 2 = 6, the same. 

That is, the two means between 3 and 24, are 6 and 12. 



problem in. 

To find any number of Geometrical Means between two 
Numbers. 

Divide the greater number by the less, and extract such 
root of the quotient whose index is 1 more than the number 
of means required ; that is, the 2d root for one mean, the 3d 
root for two means, the 4th root for three means, and so on ; 
and that root will be the common ratio of all the terms. 



OF HARJf OXICAL PROPORTION* 



121 



Then, with the ratio, multiply continually from the first term, 
or divide continually from the last or greatest term. 

EXAMPLE. 

To find four geometrical means between 3 and 96. 
Here 3) 96 (32 ; the 5th root of which is 2, the ratio. 
Then 3X2=6, dc 6X2=12, & t2 X 2=24, & 24 X2=48. 
Or 96-1-2=48, <fc 48-^2=24, & 24^-2=12, & 12-5-2=6. 
That is, 6, 12, 24, 48, are the four means between 3 and 96. 



f " OF HARMONIC AL PROPORTION. 

There is also a third kind of proportion, called Harmo- 
nical or musical, which being but of little or no common use, 
a very short accouut of it may here suffice. 

Musical Proportion is when, of three numbers, the first 
has the same proportion to the third, as the difference be- 
tween the first and second has to the difference between the 
second and third. 

As in these three, 6, 8, 12 ; 
where 6 : 12 : : 8-6 : 12-8, 
that is 6 : 12 : : 2 : 1. 

When four numbers are in musical proportion ; then the 
first has the same ratio to the fourth, as the difference be- 
tween the first and second has to the difference between the 
third and fourth. 

As in these, 6, 8, 12, 18 ; 
where 6 : 18 : : 8-6 : 18-12, 
that is 6 : 18 : : 2 : 6. 

When numbers are in musical progression, their recipro- 
cals are in arithmetical progression ; and the converse, that 
is, when numbers are in arithmetical progression, their reci- 
procals are in musical progression. 

So in these musicals 6, 8, 12, their reciprocals, £, |, yiy, 
are in arithmetical progression ; for ^ + = = i I 
and i + i = | = I ; that is, the sum of the extremes is 
equal to double the mean, which is the property of arithme- 
ticals. 

Vol. I. 17 



122 



ARITHMETIC. 



The method of finding out numbers in musical proportion 
ii best expressed by letters in Algebra. 



FELLOWSHIP, OR PARTNERSHIP. 

Fellowship is a rule, by which any sum or quantity may 
be divided into any number of parts, which shall be in any 
given proportion to one another. 

By this rule are adjusted the gains or loss or charges of 
partners in company ; or tho effects of bankrupts, or lega- 
cies in case of a deficiency of assets or effects ; or the shares 
of prizes ; or the numbers of men to form certain detach* 
ments ; or the division of waste lands among a number of 
proprietors. 

Fellowship is either Single or Double. It is single, when 
the shares or portions are to be proportional each to one 
single given number only ; as when the stocks of partners 
are all employed for the same time; and Double, when 
each portion is to be proportional to two or more numbers ; 
as when the stocks of partners are employed for different 
times. 



SINGLE FELLOWSHIP. 

GENERAL RULE. 

Add together the numbers that denote tho proportion of 
the shares. Then say, 

As the sum of the said proportional numbers, 

Is to the whole sum to be parted or divided, 

So is each several proportional number, 

To the corresponding share or part. 
Or, as the whole stock, is to the whole gain or loss, 

So is each man's particular stock. 

To his particular share of the gain or loss. 

To prove the Work. Add all the shares or parts to- 
gether, and the sum will be equal to the whole number to be 
shared, when the work is right. 



•XH6UB FBLLOWSHIP. 



138 



EXAMPLES. 

1. To divide the number 240 into three such parts, as 
shall be in proportion to each other as the three numbers 1, 
Sand 3. 

Here 1 + 2 + 3 = 6, the sum of the numbers* 
Then, as 6 : 240 : : 1 : 40 the let part, 

and as 6 : 240 : : 2 : 80 the 2d part, 

also as 6 : 240 : : 8 : 120 the 8d part. 

Sum of all 240, the proof. 

2. Three persons, a, b, c, freighted a ship with 840 tuns of 
wine, of which a loaded 100 tuns, n 97, and c the rest : in a 
storm the seamen were obliged to throw overboard 85 tuns ; 
how much must each person sustain of the loss ? 

Here 110 + 97 « 207 tuns, loaded by a and n ; 
theref. 340 — 207 = 133 tuns, loaded by c. 

Hence, as 840 : 85 : : 110 

or as 4 : 1 : : 1 10 : 27| tuns = a's loss ; 
and as 4 : 1 : : 97 : 24| tuns = b's loss ; 
also as 4 : 1 : : 133 : 33} tuns = c's loss ; 

Sum 85 tuns, the proof. 

8. Two merchants, c and d, made a stock of 120/ ; of 
irbich c contributed 757, and d the rest : by trading they 
gained 30Z ; what must each have of it ? 

Ans. c 18/ 15*, and d 11/ 5*. 

4. Three merchants, b, f, g, make a stock of 700/, of 
^vhich b contributed 128/, r 358/, and o the rest : by trading 
fihey gain 125/ 10* ; what must each have of it ? 

Ans. b must have 22/ Is Od 2fjq. 
p ... 64 3 8 Off. 
g ... 30 5 3 1,V- 

5. A General imposing a contribution* of 700/ on four 



* Contribution it a Ui paid by provinces, towns, village! , Ac. to ex- 
xon them from being; plundered. It is paid in provisions or in money, 
*fcd sometimes in both. 



134 



ARITHMETIC. 



villages, to be paid in proportion to the number of inhabitants 
contained in each ; the first containing 250, the 2d 350, the 
3d 400, and the 4th 500 persons ; what part must each vil- 
lage pay ? Ans. the 1st to pay 116/ 13s Ad 

the 2d - - 163 6 8 
the 3d - - 186 13 4 
the 4th - - 233 ' 6 8 

6. A piece of ground, consisting of 37 ac 2 ro 14 ps, is 
to be divided among three persons, l, m, and n, in propor- 
tion to their estates : now if l's estate be worth 5002 a year, 
m's 320/, and n's 75/ ; what quantity of land must each one 
have ? Ans. l must have 20 ac 3 ro 39^} {ps. 

m - - - 13 1 30 T \V *. 
n - - - 3 23tf| 

7. A person is indebted to o 57/ 15*, to r 108/ 3s 84, to 
a 22/ 10c/, and to r 73/ ; but at his decease, his effects are 
found to be worth no more than 170/ 14s ; how must it be 
divided among his creditors ? 

Ans. o must have 37/ 15* 5c/ 2^/^c/. 
p ... 70 15 2 2^V- 
q ... 14 8 4 2flftV- 
B - - - 47 14 11 2tWA. 

8. A ship, worth 900/, being entirely lost, of which -J- be- 
longed to s, I to t, and the rest to v ; what loss will each 
sustain, supposing 540/ of her were insured ? 

Ans. s will lose 45/, t 907, and v 225/. 

9. Four persons, w, x, y, and z, spent among them 25*, 
and agree that w shall pay £ of it, x |, y }, and z j ; that 
is, their shares are to be in proportion as £, |, J, and £ : 
what are their shares t Ans. w must pay 9s 8d 3%]q. 

x - - . 6 5 34$. 
y - - - 4 10 14$. 
z - . . 3 10 3,V- 

10. A detachment, consisting of 5 companies, being sent 
into a garrison, in which the duty required 76 men a day ; 
what number of men must be furnished by each company, in 
proportion to their strength ; the 1st consisting of 54 men, 



DOVBU n&LOWlBXP. 136 

» 

the 2d of 51 men, the 3d of 48 men, the 4th of 89, and the 
5th of 36 men? 

Ana. The 1st must furnish 18, the 2d 17, the 3d 16, the 
4th 13, and the 5th 12 men 41 . 



DOUBLE FELLOWSHIP. 

Doubub Fkllowbhip, aa has been said, is concerned in 
cases in which the stocks of partners are employed or con- 
tinued for different times. 

RuLsf. — Multiply each person's stock by the time of 
its continuance ; then divide the quantity, aa in Single 
Fellowship, into shares, in proportion to these products], oy 
saying, 

As the total sum of all the said products, 

Is to the whole gain or loss, or quantity to be parted, 

So is each particular product 

To the correspondent share of the gain or loss. 



EXAMPLES. 



1. a had in company 50f for 4 months, and b had 60J for 
6 months ; at the end of which time they find 242 gained : 
how roust it be divided between them ? 

Here 50 60 
4 5 

200 + 300 = 500 

Then as 500 : 24 : : 200 : 9} = 91 12* = a's share, 
and as 500 : 24 : : 300 : 14f = 14 8 = b's share. 



* Questions of this nature frequently occurring in military service. 
General Haviland, an officer of great merit, contrived an ingenious in- 
strument, for more expeditiously resolving them ; which is distinguish- 
ed bv the name of I he inventor, being called a Haviland. 

t Th»* proof of this rule is as follows : When the times are equal, 
the slmres of the gain or loss are evidently as the stocks, as in Single 
Fellowship; and when the stocks are equal, the shares are as the 
times; therefore, when neither are equal, the shares must be esthete 
products. 



196 



ABTTH3CKTIC. 



2. c and d hold a piece of ground in common, for which 
they are to pay 542. c put in 23 horses for 27 days, and d 
21 horses for 30 days ; how much ought each man to pay 
of the rent ? . Ans. c must pay 23/ 5* 9d. 

d must pay 30 14 3. 

3. Three persons, e, f, o, hold a pasture in common, 
for which they are to pay 30/ per annum ; into which e put 
7 oxen for 3 months, f put 9 oxen for 5 months, and o put 
in 4 oxen for 12 months ; "how much must each person pay 
of the rent 1 Ans. e must pay 57 10* Gd lftq. 

f .. 11 16 10 0^V 
o - - 12 12 7 2ft. 

4. A ship's company take a prize of 1000/, which they 
agree to divide among them according to thejr pay and the 
time they have been on board : now the officers and midship- 
men have been on board 6 months, and the sailors 3 months ; 
the office re have 40* a month, the midshipmen 30*, and the 
sailors 22* a month ; moreover,' there are 4 officers, 12 mid- 
shipmen, and 110 sailors ; what will each man's share be ? 

Ans. each officer must have 231 2s 5d Y y^q. 
each midshipman - 17 6 9 3^. 
each seaman - - 6 7 2 0fo\. 

5. r, with a capital of 1000/, began trade the first of 
January, and, meeting with success in business, took in i as 
a partner, with a capital of 1500/, on the first of March fol- 
lowing. Three months after that they admit k as a third 
partner, who brought into stock 2K007. After trading toge- 
ther till the end of the year, they find there has been gained 
1776/ 10s ; how must this be divided among the partners ? 

Ans. h must have 475/ 9s A\d j^q* 
i 571 16 8J 

k - . - 747 3 11J iff. 

6. x, y, and z made a joint stock for 12 months ; x at 
first put in 20/, and 4 months after 20/ more ; y put in at 
first 30/, at the end of 3 months he put in 20/ more, and 2 
months after he put in 407 more ; z put in at first 60/, and 
5 months after he put in 10/ more, 1 month after which he 
took out 30/ ; during the 12 months they gained 50/ ; how 
much of it must each have ? 

Ans. x must have 10/ 18s 6d 3} fa. 
y ... 22 8 1 0ft. 
z ... 16 13 4 0. 



SIMPLE onraBST. 



SIMPLE INTEREST. 



Interest is the premium or sum allowed for the loan, or 
forbearance of money. The money lent, or forborn, is 
called the Principal ; and the sum of the principal and its x 
interest added together, is called the Amount. Interest is 
allowed at so much per cent, per annum ; which premium 
per cent, per annum, or interest of 100/ for a year, is 1 called 
the rale of interest : — So, 

When interest is at 3 per cent, the rate is 3 ; 

- 4 per cent. - - - 4 ; 
. 5 per cent. - - - 5 ; 

- 6 per cent. ... 6. 

But, by law, interest ought not to be taken higher than at 
the rate of 5 per cent. 

Interest is of two sorts ; Simple and Compound. 

Simple Interest is that which is allowed for the principal 
lent or forborn only, for the whole time of forbearance* 
As the interest of any sum, for any time, is directly pro- 
portional to the principal sum, and also to the time of con- 
tinuance ; hence arises the following general rule of calcu- 
lation. 

As 100Z is to the rate of interest, so is any given principal 
to its interest for one year. And again, 

As 1 year is to any given time, so is the interest for a 
year, just found, to the interest of the given sum for that 
time. 

Otherwise. Take the interest of 1 pound for a year, 
which multiply by the given principal, and this product 
again by the time of loan or forbearance, in years and parts, 
for the interest of the proposed sum for that time. 

Note. When there are certain parts of years in the time, 
as quarters, or months, or days : they may be worked for, 
either by taking the aliquot or like parts of the interest of a 
year, or by the Rule of Three, in the usual way. Also, to 
divide by 100, is done by only pointing off two figures Cot 
decimals. 



138 



ASfHDtBTXG. 



EXAKPLK8. 



1. To find the interest of 2801 10«, for 1 year, at the rate 
of 4 per cent, per annum.„ 

Here, As 100 : 4 :: 230/ 10* : 97 4* 4Jd. 

4 



100) 0,22 
20 

4-40 
12 

4 80 t Ans. 01 4s 4f<*. 



3*20 



2. To find the interest of 547/ 15*, for 3 years, at 5 per 
cent, per annum. 

As 100 : 5 :: 547-75 

Or 20 : 1 :: 547*75 : 27-3875 interest for 1 year. 

3 



I 82-1625 ditto for 3 years. 
20 



s 3-2500 
12 



<2 3-00 Ans. 82/ 3* 3d. 



3. To find the interest of 200 guineas, for 4 years 7 months 
and 25 days, at 4$ per cent, per annum. 



SIMPLE INTEREST. 



129 



ds 2 ds 
< 8102 As 356: 9-45: : 25: 2 
41 or 73:9-45:: 5 : -6472 

5 

340 

105 73) 47-25 (-6472 

345 

9-45 interest for 1 yr. 530 
4 19 



37*80 ditto 4 years. 
6 mo = \ 4*725 ditto 6 months. 
1 mo = £ -7875 ditto 1 month. 

-6472 ditto 25 days. 

I 43-9597 
20 



* 191940 
12 



d 2-3280 

4 Ans. 432 19* 2\d. 



q 1-3120 



4» To find the interest of 450/, for a year, at 5 per cent, 
per annum. Ans. 222 10*. 

5. To find the interest of 7152 12* 6c2, for a year, at 41 
per cent, per annum. Ans. 322 4* 0}a. 

6. To find the interest of 7202, for 3 years, at 5 per cent 
per annum. Ans. 1082. 

7. To find the interest of 3552 15*, for 4 years, at 4 per 
cent per annum. Ans. 562 18* 4Jd. 

- 8. To find the interest of 322 5* Bd 9 for 7 years, at 4£ per 
cent per annum. Ans. 92 12* Id. 

9. To find the interest of 1702, for 1} year, at. 5 per cent 
per annum. Ans. 122 15** 

10. To find the insurance on 2052 15*, for J of a year, at 
4 per cent, per annum. Ans. 22 1* lf<f. 

11. To find the interest of 3192 6d, for 5? years, at 3| per 
cent, per annum. Ans. 682 14* 9|<2. 

12. To find the insurance on 1072, for 117 days, at 4f per 
cent per annum. Ana. U \fe Id* 

Vol. I. 18 



180 



ARITHMETIC 



13. To find the interest of 172 5*, for 117 days, at 4f per 
cent, per annum. Ans. 5* 3d. 

It To find the insurance on 7122 6s, for 8 months, at 71 
per cent, per annum. Ans. 352 1£# 3J<£ 

Note. The Rules for Simple Interest, serve also to calcu- 
late Insurances, or the Purchase of Stocks, or any thing else 
that is rated at so much per cent. 

See also more on the subject of Interest, with the algebrai- 
cal expression and investigation of the rules at the«adof the 
Algebra. 



COMPOUND INTEREST. 

Compound Interest, called also Interest upon Interest, 
is that which arises from the principal and interest, taken 
together, as it becomes due, at the end of each 'stated time 
of payment. Though it be not lawful to lend money at 
Compound Interest, yet in purchasing annuities, pensions, or 
leases in reversion, it is usual to allow Compound Interest to 
the purchaser for his ready money. 

Rules. — 1. Find the amount of the given principal, for 
the time of the first payment, by Simple Interest. Their 
consider this amount as a new principal for the second pay- 
ment, whose amount calculate as before. And so on through 
all the payments to the last, always accounting the last amount 
as a new principal for the next payment. The reason of 
which is evident from the definition of Compound Interest* 
Or else, 

2. Find the amount of 1 pound for the time of the first 
payment, and raise or involve it to the power whose index 
is denoted by tho number of payments. Then that power 
multiplied by the given principal, will produce the whole 
amount. From which the said principal being subtracted, 
leaves the Compound Interest of the same. As is evident 
from the first Rule. 

EXAMPLES. 

1. To find the amount of 720/, for 4 years, at 5 per cent* 
per annum. 

Here 5 is the 20th part of 100, and the interest of H for a 
year is -fa or *95, and its amount 1*05. Therefore, 



ALLIGATION. 



1. By ike lit Rtde. 2. By the 2d Rule. 

I s d 1*05 amount of 1Z. 

SO) 720 1st yr's princip. 1-05 * 

88 1st yr*s interest. , tnCTe 

J 1 1026 2d power of it. 

20)756 2d yr*s princip. 1-1025 

97 16 2d yr's interest.-———-:^ , 
J 1 -21550625 4th power of it. 

SO) 703 16 3d yr's princip. 720 

30 13 9 h 3d yr's interest y 

20) 838 9 9J 4th yr's princip. 20 
41 13 5} 4thfr' 8 int«e,t. ~~ 

£875 3 3} the whole amou 12 
or ans. required. 



2. To find the amount of 502 in 5 years, at 5 per cent, 
per annum, compound interest. Ans. 032 16* 8jtf. 

3. To find the amount of 50Z in 5 years, or 10 half-years, 
at 5 per cent, per annum, compound interest, the interest 
payable half-yearly. Ans. 64Z 0* Id. 

4 To find the amount of 50Z in 5 years, or 20 quarters, 
at 5 per cent, per annum, compound interest, the interest 
payable quarterly. Ans. 04/ 2s (){d. 

5. To find the compound interest of 370Z forborn for 6 
years, at 4 percent, per annum. Ans. 98Z 3* 4£<Z. 

6. To find the compound interest of 41 OZ forborn for 2* 
years, at 4£ per cent, per annum, the interest payable half, 
yearly. Ans. 48Z 4* 11 {d* 

7. To find the amount, at compound interest, of 217Z, for* 
born at 2{ years, at 5 per cent, per annum, the interest pay- 
able quarterly. Ans. 242/ 13* 4£rf. 



ALLIGATION. 

Alligation teaches how to compound or mix together 
several simples of different qualities, so that the composition 
•nay be of some intermediate quality, or rate. It is com- 
monly distinguished into two cases. Alligation Medial, and 
Alligation Alternate. 



1*2 



AS1THXXTIC. 



^ ALLIGATION MEDIAL. 

Alligation Medial is the method of finding the rate or 
quality of the composition, from having* the quantities and 
rates or qualities of the several simples given. And it is 
thus performed : 

* Multiply the quantity of each ingredient by its rate or 

Siality ; then add all the products together, and add also all 
e Quantities together in another sum ; then divide the 
former sum by the latter, that is, the sum of the products by 
the sum of the quantities, and the quotient will be the rate or 
quality of the composition required. 

EXAMPLES. 

1. If three sorts of gunpowder be mixed together, viz. 
601b at I2d a pound, 441b at 9d, and 261b at 8d a pound ; 
how much a pound is the composition worth ? 
Here 50, 44, 26 are the quantities, 
and 12, 9, 8 the rates or qualities ; 
then 50 X 12 = 600 
44 X 9 = 396 
26 X 8 = 208 

120) 1204 (lOrfr = 10 f V- 

Ans. The rate or price is lO^d the pound. 



• Demonstration. The Rule is thus proved by Algebra. 
Let a, 6, e be the quantities of the ingredients, 
and m, n, p their rates, or qualities, or prices ; 
then am, in, cp are their several values, 
and am + bn -\- cp the sum of their values, 
also « + b + e is the sum of the quantities, 
and if r denote the rate of the whole composition, 
then (ff-f-6-fc)Xr will be the value of the whole, 
conseq. {a -f b 4- c) X r = am + bn + cp, 
and r = (am -f bn + cp) (a + b -f- c), u hich is the Rule. 

Note. If an ounce or any other quantity of pure gold be reduced 
into 94 equal parts, these parts are called Caracts ; but gold is often 
mixed with some base metal, which is called the Alloy, and the mixture 
is said to be of so many caracts fine, according to the proportion of para 

Kid contained in it : thus, if 522 caracts of pure gold, and 2 of a) Joy 
mixed together, it is said to be 32 caracts fine. 
If any one of the simples be of little or no value with respect to the 
rail, its rate is supposed to be nothing ; as water mixed with wine, and 
alloy with gold and silver. 



AU.IGATION ALTERNATE. 



138 



2. A composition being made of 51b of tea at 7s per lb, 
91b at 8* 64 per lb, and 14£lb at 5* lOd per lb ; what is a lb 
of it worth? Ana. 6* I0\d. 

3. Mixed 4 gallons of wine at 4s lOd per gall, with 7 gal- 
Ions at 5# 3d per gall, and 9} gallons at 6* Sd per gall ; what 
is a gallon of this composition worth ? Ans. 5* 4{d» 

4. Having melted together 7 oz of gold of 22 caracts fine, 
12 1 oz of 21 caracts fine, and 17 oz of 19 caracts fine : I 
would know the fineness of the composition ? 

Ans. 2(ty$ caracts fine. 



ALLIGATION ALTERNATE. 

Alligation Alternate is the method of finding what 
quantity of any number of simples, whose rates are given, 
will compose a mixture of a given rate. So that it is the re- 
verse of Alligation Medial, and may be proved by it. 

RULE i*. 

1. Set the rates of the simples in a column under each 
other. — 2. Connect, or link with a continued lino, the rate 



* Demonst. By connecting the less rate with the greater, and placing 
the difference between them and the rate alternately, the quantities re- 
sulting are such, that there is precisely as much gained by one quantity 
as is Inst by the other, and therefore the gain and loss upon the whole 
is equal, and is exactly the proponed rate : and the same will be true of 
any other two simples managed according to the Rule. 

In like manner, whatever the number of simples may be, and with 
how many soever every one is linked, since it is always a le^s with a 
greater than the mean price, there will be an equal balance of loss and 
gain between every two, and consequeutly an equal balance on the 
whole. s. d. 

It is obvious, from the Rule, that questions of this sort admit of a 
great variety of answers ; for, having found one answer, we may find 
as many more as we please, by only multiplying or dividing each of 
the quantities found, by 2, or 3, or 4, fee. : the reason of which is evi- 
dent: for, if two quantities, of two simples, make a balance of loss and 
gain, with respect to the mean price, so must also the double or treble, 
the 1 or | part, or any other ratio of these quantities, and so on ad w- 
Jutiltm. 

These kinds of questions are called by algebraists indeterminate or 
unlimited problems ; and by an analytical process, theorems may be 
raited that will give all the possible answers. 



184 



ASmULETlG. 



of each simple, which is less than that of the compound, with 
one, or any number, of those that are greater than the com- 
pound*; and each greater rate with one or any number of the 
less.— -3. Write the difference between the mixture rate, and 
that of each of the simples, opposite the rate with which they 
are linked. — 4. Then if only one difference stand against 
any rate, it will be the quantity belonging to that rate ; but 
if there be several, their sum will be the quantity. 

The examples may be proved by the rule for Alligation 
Medial. 



EXAMPLES. 

1. A merchant would mix wines at 16*, at 18*, and at 22* 
per gallon, so as that the mixture may be worth 20* the gal- 
lon ; what quantity of each must be taken ? 

1 2 at 16* 
Here 20 ?J8>v J2 at 18* 

(22j/4 + 2 = 6at 22* 

2. How much sugar at 4d, at 6d, and at lid per lb, must 
be mixed together, so that the composition formed by them 
may be worth 7d per lb ? 

Ans. 1 lb, or 1 stone, or 1 cwt, or any other equal quan- 
tity of each sort. 

3. How much corn at 2* 6d, 3* Sd y 4*, and 4* 8d per 
bushel must be mixed together, that the compound may be 
worth 3* lOd per bushel ? 

Ans. 2 at 2* 6d, 3 at 3* 8d, 3 at 4*, and 3 at 4* Sd. 

RULE II. 

When the whole composition is limited to a certain quan- 
tity : Find an answer as before by linking ; then say, as 
the sum of the quantities, or differences thus determined, is 
to the givm quantity ; so is each ingredient, found by link- 
ing, to the required quantity of each. 

EXAMPLE. 

1. How much gold of 15, 17, 18, and 22 caracts fine, mutt 
be mixed together, to form a composition of 40 oz of 20 cap 
racta fino ? 



ALLIGATION AI/TlHUVATE. 



135 




- 2 

Here 20? l^M )- . . 2 

5 + 3+ 2=10 
16 

Then as 10 : 40 : : 2 : 5 
and 16 : 40 : : 10 : 25 
Ans. 5 oz of 15, of 17, and of 18 caracts fine, and 25 oz of 
22 caracts fine*. 

rule inf. t 

When one of the ingredients is limited to a certain quan- 
tity ; Take the difference between each price, and the mean 
rate as before ; then say, As the difference of that simple, 
whose quantity is given, is to the rest of the differences 
severally ; so is the quantity given, to the several quantities 
required. 



* A great number of questions might be here given relating to the 
specific gravities of metals, &c. but one of the most curious may suf- 
fice. 

Hiero, king of Syracuse, gave orders for a crown to be made entire- 
ly of pure gold ; but suspecting the workmen had debased it by mixing 
it with silver or copper, be recommended the discovery of the fraud to 
the famous Archimedes, and desiied to know the exact quantity of alloy 
in the crown. 

Archimedes, in order to detect the imposition, procured two other 
masses, the one of pure gold, the other of. silvei or copper, and each ot 
the same weight with the former; and by putting each separately into 
a vessel full of water, the quantity of water expelled by them deter- 
mined their specific gravities : from which, and their givr»n weights, the 
exact quantities of gold and alloy in the crown may be determined. 

Suppose the weight of each crown to be 101b, and that the water ex- 
pelled by the copper or silver was 921b, by the gold 5*2 lb, and by the 
compound crown C4lb ; what will be the quantities of gold and alloy 
Id the crown ? 

The rates of the simples are 92 and 52, und of the compound 64 ; 
therefore 

CxA I 62 > 12 of copper 

°* I 62... ^28 of gold 
And the sum of these is 12 + 28 — 40, which should have been 10; 
therefore by the Rule, 

40 : 10 :: 12 : 31b of copper „„,„,„, 
40: 10: :28 : 71b of «olJ Jthe answer 
t Id- the very same manner questions may be wrought when several 
•iof the ingredients are limited to certain quantities, by finding first fur 
one limit, and then for another. The two last Rules can need do de- 
monstration, as they evidently result from the first, the reason ut 
has been already explained. 



186 



ABTTHXBTIC 



EXAMPLES. 



1. How much wine at 6*, at 5* 64, and 6* the gallon, must 
be mixed with 3 gallons at 4* per gallon, so that the mixture - 
may be worth 5* 4d per gallon ? 



16+4=20 

Then 10 : 10 : : 3 : 3 
10 : 20 : : 3 : 6 
10:20::3:6 
Ads. 3 gallons at 5*, 6 at 5s 6d 9 and 6 at 6*. 
2. A grocer would mix teas at 12*, 10*, and 6* per lb, with 
201b at 4s per lb : how much of each sort must he take to 
make the composition worth 8* per lb ? 

Ans. 201b at 4*, 101b at 6*, 101b at 10*, and 201b at 12*. 



Position is a rule for performing certain questions, which 
cannot be resolved by the common direct rules. It is some- 
times called False Position, or False Supposition, because 
it makes a supposition of false numbers, to work with the 
same as if they were the true ones, and by their means dis- 
covers the true numbers sought. It is sometimes also called i 
Trial-and-Error, because it proceeds by trials of f ilse num- j 
bers, and thence finds out the true ones by a comparison ^ 
of the errors. — Position is either Single or Double. 



Single Position is that by which a question is resolved 
by means of one supposition only. Questions which have 
their result proportional to their supposition, belong to 
Single Position : such as those which require the multiplier 




POSITION. 



SINGLE POSITION. 



SINGLE POSITION. 



137 



tion or division of the number sought by any proposed num- 
ber ; or when it is to be increased or diminished by itself, 
or any parts of itself, a certain proposed number, of times. 
The rule is as follows : 

Take or assume any number for that which is required, 
and perform the same operations with it, as are described or 
performed in the question. Then say, As the result of the 
said operation, is to the position, or number assumed ; so is 
the result in the question, to a fourth term, which will be 
the number sought*. 



EXAMPLES. 



1. A person after spending J- and { of his money, has yet 
remaining 60/ ; what had he at first ? 

Suppose he had at first 1201. Proof. 
Now { of 120 is 40 £ of 144 is 48 

J of it is 30 i of 144 is 36 

their sum is 70 their sum 84 

which taken from 120 taken from 1 14 

leaves 50 leaves (50 as 

Then, 50 : 120 : : 60 : 144 the Answer. per question. 

2. What number is that, which, being increased by », J, 
and i of itself, the sum shall be 75? Ans. 36. 

3. A general, after sending out a foraging i and £ of his 
men, had yet remaining 1000; what number had he in 
command ? Ans. 6000. 

4. A gentleman distributed 52 pence among a number of 
poor people, consisting of men, women, and children ; to 
each man he gave 6d, to each woman 4d, and to each child 
2d : moreover there were twice as many women as men, and 



* The reason of this Kale is evident, because it is supposed that the 
results are proportional to the suppositions. 

Thus, na : a : : nz : s, 



— h — «fcc. 

n — m 





a 


or 


- : a : 




n 




a a 


or 


- - 




n — m 



and so on. 
Vol. I. 19 



138 



AJCXTHMCTTC. 



thrice as many children as women. How many were them 
of each ? Ana. 2 men, 4 women, and 12 children. 

ft. One being aaked his age, said, if f of the years I have 
lived, be multiplied by 7, and § of them be added to the 
product, the sum will be 210. What was his age ? 

Ana. 45 yeanr. 



DOUBLE POSITION. 

Double Position is the method of resolving certain ques- 
tions by means of two suppositions of false numbers. 

To the Double Rule of Position belong such questions as 
have their results not proportional to their positions : such are 
those in which the numbers sought, or their parts* or their 
multiples, are increased or diminished by some given absolute 
number, which is no known part of the number sought, 

bulk*. 

Take or assume any two convenient numbers, and pro- 
ceed with each of them separately, according to the con- 



* Dcmonstr. The Rule is founded on this supposition, namely, tbet 
the first error is to the second, as the difference between the true and 
first supposed number, is to the difference between the true end second 
supposed number: when that is not the case, the exact answer to the 
question cannot be found by this Rule.— That the Rule is true, accord- 
ing to the assumption, may be thus proved. 

Let a and 6 be the two suppositions, and A and a their results, pro- 
duced by similar operation ; also r and s their errors, or the differences 
between the results a and a from the true result a ; and let x denote 
the number sought, answeriog to the true result a of the question. 

Then is k — a = r, and k — b ~ #, or b — a = r— s. And r according 
to the supposition on which the Rule is founded, r : s :: x — a : x— 6 ; 
hence, by multiplying extremes and means, rx — rb = sx— as; then, by 

transposition, rx — #x = rb — $a ; and, by division, x = r ^~~* g = the 

number sought, which is the rule when the results are both too little. 

If the results be both too great, so that ▲ and a are both mater than 
h ; then h — a = — r, and a — a = — *, or r and s are both negative ; 
hence— r : — t : : x — a : x— b f but — r : — $ t r -f r : -4- *, there- 
fore r : s : : x—a : x—b \ and the rest wHI be exactly as in the for- 
mer case. 

But if one result a only be too little, and the other a too great, or 
one error r positive, and the other » negative, then the theorem be- 
comes x = ^4— t which is the rule in this case, or whea the errors 
are unlike. 



DOVBU POSITION. 



180 



Virions or the question, as in Single Position ; and find how 
enuch each remit is different from the result mentioned in 
the question, calling these differences the errors, noting also 
whether the results are too great or too little. 

Then multiply each of the said errors by the contrary 
supposition, namely, the first position by the second error, 
and the second position by the first error. Then, 

If the . errors are alike, divide the difference of the pro- 
ducts by the difference of the errors, and the quotient will 
lie the answer. 

But if the errors are unlike, divide the sum of the pro- 
ducts by the sum of the errors, for the answer. 

Note, The errors are said to be alike, when they are either 
both too great or both too little ; and unlike, when one is 
too great and the other too little. 



SXAMPLB. 

1* What number is that, which being multiplied by 6, the 
product increased by 18, and the sum divided by 9, the quo- 
tient should be 20 ? 

Suppose the two numbers 18 and 30. Then, 



First Position. 8econd Position. 
18 Suppose 30 
6 mult. 6 


ProoC 
27 
6 




108 
18 


add 


180 
18 


162 
18 


*) 126 


div. 


9) 198 


-0) 180 




14 

20 


results 
true res. 


22 
20 


20 


:2d pos. 


80 


errors unlike 
mult. 


-2 

18 1st pos. 




Er. >2 
jorsyC 


\ 180 

\ 36 




36 




sum 8) 


216 
27 


sum of products 
Answer sought. 







140 



ARITHMETIC. 



RULE II. 

Find, by trial, two numbers, as near the true number as 
convenient, and work with them as in the question ; mark- 
ing the errors which arise from each of them. 

Multiply the difference of the two numbers assumed, or 
found by trial, by one of the errors, and divide the product 
by the difference of the errors, when they are alike, but by 
their sum when they are unlike. Or thus, by proportion : 
As the difference of the errors, or of the results, (which is 
the same thing), is to the difference of the assumed numbers, 
so is either of the errors, to the correction of the assumed 
number belonging to that error. 

Add the quotient, or correction, last found, to the number 
belonging to the said error, when that number is too little, 
but subtract it when too great, and the result will give the 
true quantity sought *. 

EXAMPLES. 

1. So, the foregoing example, worked by this 2d rule, 
will be as follows : 

30 positions 18 ; their diff. 12 

-2 errors +6 ; least error 2 

sum of errors 8 ) 24 ( 3 subtr. 
from the position 30 

leaves the answer 27 

Or, as 22 - 14 : 30 - 18, or as 8 : 12 : : 2 : 3 the cor- 
rection, as above. 

2. A son asking his father how old he was, received this 
answer : Your age is now one-third of mine ; but 5 years 
ago, your age was only one-fourth of mine. What then are 
their two ages ? Ans. 15 and 45. 

3. A workman was hired for 20 days, at 3s per day, for 
every day he worked ; but with this condition, that for every 
day he did not work, he should forfeit 1*. Now it so hap. 



• For since, by the supposition, r : t :: x : — a:x — b, therefore by 
division, r — f : s :: b — a: z — 6, or as b — a : 6 — a : : s :x — 6, for b 
— a if = r — $ : which is the 2d Rule. 



PRACTICAL QUESTIONS. 



141 



pened, that upon the whole he had 27 4* to receive. How 
many of the days did he work ? Ans. 16* 

4. a and b began to play together with equal sums of 
money: a first won 20 guineas, but afterwards lost back} 
of what he then had ; after which b had four times as much 
as a. What sum did each begin with ? Ans. 100 guineas. 

5. Two persons, a and b, have both the same income, 
a saves } of his ; but b, by spending 60/ per annum more 
than a, at the end of 4 years finds himself 100/ in debt. 
"What does each receive and spend per annum ? 

Ans. They receive 125/ per annum ; also a spends 100/, 
and b spends 150/ per annum. 



PRACTICAL QUESTIONS IN ARITHMETIC. 

Quest. 1. The swiftest velocity of a cannon-ball, is 
about 2000 feet in a second of time. Then in what time, 
at that rate, would such a ball move from the earth to the 
sun, admitting the distance to be 100 millions of miles, and 
the year to contain 365 days 6 hours ? Ans. 8 tVtVV years. 

Quest. 2. What is the ratio of the velocity of light to 
that of a cannon-ball, which issues from the gun with a ve- 
locity of 1500 feet per second ; light passing from the sun 
to the earth in 7£ minutes ? Ans. the ratio of 782222$ to 1. 

Quest. 3. The slow or parade-step being 70 paces per 
minute, at 28 inches each pace, it is required to determine 
at what rate per hour that movement is ? Ans. miles. 

Qukht. 4. The quick-time or step, in marching, being 
2 paces per second, or 120 per minute, at 28 inches each ; 
at what rate per hour does a troop march on a rout, and 
how long will they be in arriving at a garrison 20 miles 
distant, allowing a halt of one hour by the way to refresh ? 

. ) the rate is 3^ T miles an hour. 
' \ and the time 7$ hr, or 7h 17| min. 
Quest. 5. A wall was to be built 700 yards long in 29 
days. Now, after 12 men had been employed on it for 11 
dtiys, it was found that they had completed only 229 yards 
of the wall. It is required to determine how many men must 
be* added to the former, that the whole number of them may 
just finish the wall in the time proposed, at the same rate of 
forking. Ans. 4 men to be 



142 



ARITHMETIC* 



Quest. 6. Determine how far 500 millions of guineas will 
reach, when laid down in a strait line touching one an- 
other ; supposing each guinea to be an inch in diameter, as 
it is very nearly. Ans. 7891 miles, 728 yds, 2 ft. 8 in. 

Quest. 7. Two persons, a and b, being on opposite sides 
of a wood, which is 536 yards about, they begin to go round 
it, both the same way, at the same instant of time ; a goes at 
the rate of 1 1 yards per minute, and b 34 yards in 3 mi. 
nutes ; the question is, how many times will the wood be 
gone round before the quicker overtake the slower 1 

Ans. 17 times. 

Quest. 8. a can do a piece of work alone in 12 days, 
and b alone in 14 ; in what time will they both together per- 
form a like quantity of work ? Ans. 6 T T days. 

Quest. 9. A person who was possessed of a | share of a 
copper mine, sold f of his interest in it for 1800/ ; what was 
the reputed value of the whole at the same rate ? Ans. 4000/. 

Quest. 10. A person after spending 20/ more than J of 
his yearly income, had then remaining 30/ more than the 
half of it ; what was his income ? Ans. 200/. 

Quest. 11. The hour and minute hand of a clock are 
exactly together at 12 o'clock ; when are they next together T 

Ans. at l T y hr. or 1 hr. 5> r min. 

Quest. 12. If a gentleman whose annual income is 15002, 
spend 20 guineas a week ; whether will he save or run in 
debt, and how much in the year ? Ans. save 408/, 

Quest 13. A person bought 180 oranges at 2 a penny, 
and 180 more at 3 a penny ; after which, selling them out 
again at 5 for 2 pence, whether did he gain or lose by the 
bargain ? Ans. he lost 6 pence. 

Quest. 14. If a quantity of provisions serves 1500 men 
12 weeks, at the rate of 20 ounces a day for each man ; how 
many men will the same provisions maintain for 20 weeks, at 
the rate of 8 ounces a day for each man ? Ans. 2250 men. 

Quest. 15. In the latitude of London, the distance round 
the earth, measured on the parallel of latitude, is about 15550 
miles ; now as the earth turns round in 23 hours 56 minutes, 
at what rate per hour is the city of London carried by this 
motion from west to east ? Ans. 649$}} miles an hour. 

Quest. 16. A father left his son a fortune, ± of which he 
ran through in 8 months : $ of the remainder lasted him 12 
months longer ; after which he had 820/ left. What sum 
did the father bequeath his son ? Ans. 1913/ 6s 8d. 

Quest. 17. If 1000 men, besieged in a town, with pro* 



PRACTICAL QUESTIONS. 



143 



visions for 5 weeks, allowing each man 16 ounces a day, be 
reinforced with 500 men more ; and supposing that they can- 
not be relieved till the end of 8 weeks, how many ounces a 
day must each man have, that tho provision may last that 
time ? Ans. 6} ounces. 

Qukst. 18. A younger brother received 8400/, which 
was just J of his elder brother's fortune : What was the 
lather worth at his death ? Ans. 19200/. 

Qukst. 19. A person, looking on his watch, was asked 
what was the time of the day, who answered, It is between 
5 and 6 ; but a more particular answer being required, he 
said that the hour and minute hands were then exactly to- 
gether : What was the time ? Ans. 27^ min. past 5. 

Quest. 20. If 20 men can perforin a piece of work in 
12 days, how many men will accomplish another thrice as 
large in one-fifth of the time ? Ans. 300. 

Quest. 21. A father devised T \ of his estate to one of 
his sons, and T \ of the residue to another, and the surplus to 
his relict for life. The children's legacies were found to be 
514/ 6* 8d different : What money did he leave the widow 
the use of? Ans. 1270/ 1* 9^d. 

Quest. 22. A person, making his will, gave to one child 

of his estate, and the rest to another. When these legacies 
came to be paid, the one turned out 1200/ more thun the 
other : What did the testator die worth ? Ans. 4000/. 

Quest. 23. Two persons, a and b, travel between Lon- 
don and Lincoln, distant 100 miles, a from London, and b 
from Lincoln, at the same instant. After 7 hours they meet 
on the road, when it appeared that a had rode l£ miles an 
hour more than b. At what rate per hour then did each of 
the travellers ride ? Ans. a 7§J and b miles. 

Quest. 24. Two persons, a and n, travel between Lon- 
don and Exeter, a leaves Exeter at 8 o'clock in the morn- 
ing, and walks at the rate of 3 miles an hour, without inter- 
naission ; and b sets out from London at 4 o'clock the same 
evening, and walks for Exeter at the rate of 4 miles an hour 
constantly. Now, supposing the distance between the two 
cities to be 130 miles, whereabouts on the road will they 
xtieet ? Ans. 09} miles from Exeter. 

Quest. 25. One hundred eggs being placed on the 
ground, in a straight line, at the distance of a yard from each 
other : How far will a person travel who shall bring them 
one by one to a basket, which is placed at one yard from the 
first egg? Ans. 10100 yards, or 5 miles and 1300 yds. 

Quest. 26. The clocks of Italy go on to Yvoura \ 



144 



ARITHMETIC. 



Then how many strokes do they strike in one complete re- 
volution of the index ? Ans. 300. 

Quest. 27. One Sessa, an Indian, having invented the 
game of chess, showed it to his prince, who was so delighted 
with it, that he promised him any reward he should ask ; on 
which Sessa requested that he might be allowed one grain of 
wheat for the first square on the chess board, 2 for the second, 
4 for the third, and so on, doubling continually, to 64, the 
whole number of squares. Now, supposing a pint to con- 
tain 7680 of these grains, and one quarter or 8 bushels to be 
worth 27^ 6tZ, it is required to compute- the value of all the 
corn ? Ans. 6450468916285/ 17* 3d -Sitffff. 

Quest. 28. A person increased his estate annually by 
100/ more than the j- part of it ; and at the end of 4 years 
found that his estate amounted to 10342/ Ss 9d. What had 
he at first ? Ans. 4000/. 

Qukst. 29. Paid 1012/ 10* for a principal of 750/, taken 
in 7 years before : at what rate per cent, per annum did 1 
pay interest ? Ans. 5 per cent. 

Quest. 30. Divide 1000/ among a, b, c ; so as to give 
a 120 more, and b 95 less than c. 

Ans. a 445, b 230, c 325. 

Quest. 31. A person being asked the hour of the day, 
said, the time past noon is equal to £ths of the time till mid- 
night. What was the time? Ans. 20 min. past 5. 

Quf.8T. 32. Suppose that I have T ^ of a ship worth 
1200/; what part of her have I left after selling J of £ of 
my share, and what is it worth ? Ans. 5 y„, worth 185/. 

Quest. 33. Part 1200 acres of land among a, b, c ; bo 
that b may have 100 more than a, and c (>4 more than b. 

Ans. a 312, b 412, c 476. 

Quest. 34. What number is that, from which if there 
be taken f of }, and to the remainder be added T 8 5 - of \> the 
sum will be 10? Ans. O^J. 

Quest. 35. There is a number which, if multiplied by 
f of | of 1J, will produce 1 : what is the square of that 
number ? Ans. 1 TT . 

Quest. 36. What length must be cut off a board, 8} 
inches broad, to contain a square foot, or as much as 12 
inches in length and 12 in breadth ? Ans. 16 f $ inches. 

Quest. 37. What sum of money will amount to 138/ 2s 
6d, in 15 months, at 5 per cent, per annum simple interest ? 

Ans. 130/. 

Quest. 38. A father divided his fortune among his three 



nucncAft Sjussrnoifs. 



145 



•om, a, a, c, giving i 4 u often as b 3, and o 5 at often as 
b 6 ; what was the whole legacy, supposing a's share was 
40002? Ans. 95002. 

Quest. 39. A young hare starts 40 yards before a grey, 
bound, and is not perceived by him till she has been up 40 
seconds ; she scuds away at the rate of 10 miles an hour, and 
the dog, on view, makes after her at the rate of 18 : how 
long will the course hold, and what ground will be run over, 
counting from the outaetting of the dog ? 

Ans. 60^ sec. and 530 yards run. 

Quest. 40. Two young gentlemen, without private for- 
tune, obtain commissions at the same time, and at the age of 
18. One thoughtlessly spends 102 a year more than his pay ; 
but, shocked at the idea of not paying his debts, gives his 
creditor a bond for the money, at the end of every year, and 
also insures his life for the amount ; each bond costs him 30 
■billings, besides the lawful interest of 5 per cent, and to in- 
sure his life costs him 6 per cent. 

The other, having a proper pride, is determined never to 
ran in debt ; and, that he may assist a friend in need, per. 
•everes in saving 102 every year, for which he obtains an 
interest of 5 per cent, which interest is every year added to 
his savings, and laid out, so as to answer the effect of com- 
pound interest. 

Suppose these two officers to meet at the age of 50, when 
each receives from Government 4002 per annum ; tha the 
one, seeing his past errors, is resolved in future to spend no 
more than he actually has, after paying the interest for what 
he owes, and the insurance on his life. 

The other, having now something beforehand, means in 
future to spend his full income, without increasing his stock* 

It is desirable to know how much each has to spend per 
annum, and what money the latter has by him to assist the 
distressed, or leave to those who deserve it ? 

Ana. The reformed officer has to spend 662 19# 1}*5389& 
per annum. 

The prudent officer has to spend 4372 12s 1 lj«4379cf. 

per annum, and 
The latter has saved, to dispose of, 7522 19s 9-1890& 



▼ax. /. 



20 



146 



LOGARITHMS. 



OF LOGARITHMS *. 



Logarithms are made to ftrinWfe troublesome calcu- 
lations in numbers. This they doy because they perform 
multiplication by only addition, and division by subtraction, 
and raising of powers by multiplying the logarithm by the 
index of the power, and extracting of roots by dividing 
the logarithm of the number by the index of the root 
For, logarithms are numbers so contrived, and adapted to 
other numbers, that the sums and differences of the former 
shall correspond to, and show the products and quotients of 
the latter, dec. 

Or, more generally, logarithms are the numerical expo- 
nents of ratios ; or they are a series of numbers in arith- 



* The invention of Logarithms is due to Lord Napier, Baron of 
Merchiston, in Scotland, and is properly considered as one of the most 
useful inventions of modern times. A table of these numbers was first 
published by the inventor at Edinburgh, in the year 1614, in a treaties 
entitled Canon Mirificum Logarithmorum ; which was eagerly received 
by all the learned throughout Europe. Mr. Henry Bnggs, then pro- 
fessor of geometry at Gresham College, soon after the discovery, 
went to visit the noble inventor ; after which, they jointly undertook 
the arduous task of computing new tables ou this subject, and reducing 
them to a more convenient form than that which was at first thought 
of. But Lord Napier dying soon after, the whole burden fell upon 
Mr. Briggs, who, with prodigious labour and great skill, made an entire 
Canon, according to the new form, for all numbers from 1 to 20000, 
and from 90000 to 101000, to 14 places of figures, and published it at 
London in the year 1624, in a treatise entitled Arithmetica Logarithmic*, 
with directions for supplying the intermediate parts. 

This Canon was again published in Holland by Adrian Vlacq, in the 
year 1628, together with the Logarithms of all the numbers which Mr. 
Briggs had omitted ; but he contracted them down to 10 places of de- 
cimals. Mr. Briggs also computed the Logarithms of the sines, tan- 

Sents, and secants, to every degree, and centestn, or 100th part of a 
egree, of the whole quadrant ; and annexed them to the natural sines* 
tangents, and secants, which lie had before computed, to fifteen placet 
of figures. These tables, with their construction and use, were first 

Eublished in the year 1633, after Air. Briggs 's death, by Mr. Henry Getti* 
rand, under the title of Trigouometria Britomiica. 



LOGARITHMS. 147 

metical progression, answering to another series of numbers 
in geometrical progression. 

rp. $0,1,2,3, 4, 5, 0, Indices, or logarithms, 
inus I 1, 2, 4, 8, 10, 32, 04, Geometric progression, 
ft i 0, 1, 2, 3, 4, 5, 6, Indices, or logarithms, 
w £ 1, 3, 9, 27, 81, 243, 729, Geometric progression. 

n« $ 0, 1, 2, 3, 1, 5, Indices, or logs. 

w $ 1, 10, 100, 1000, 10000, 100000, Geom. progres. 

Where it is evident, that the same indices serve equally 
for any geometric aerie* ; and consequently there may be an 
endless variety of system of logarithms, to the same com- 
mon numbers, by only changing the second term, 2, 3, or 
10, &c. of the geometrical series of whole numbers ; and by 
interpolation the whole system of numbers may be made to 
enter the geometric series, and receive their proportional loga- 
rithms, whether integers or decimals. 

It is also apparent, from the nature of these series, that if 
any two indices be added together, their sum will be the in- 
dex of that number which is equal to the product of the two 
terms, in the geometric progression, to which those in- 
dices belong. Thus the indices 2 and 3, being added toge- 
ther, make 5 ; and the numbers 4 and 8, or the terms cor- 
responding to those indices, being multiplied together, make 
82, wnich is the number answering to the index 5. 

In like manner, if any one index be subtracted from an- 
other, the difference will be the index of that number which 



Benjamin Ursinus also gave "a Table of Napier's Logs, and of sines, 
to every 10 seconds. And Chr. Wolf, in his Mathematical Lexicon, says 
that one Van Loser had computed them to every single second, but 
Ms untimely death prevented their publication. Many other authors 
have treated on this subject ; but as their numbers are frequently in ac- 
curate and incommodiously disposed, they are now generally neglect- 
ed. The Tables in most repute at present, nrc those of Gardiner in 
4to, first published in the* year 1742 ; and my own Tables in 8vo, first 
printed in the year 1785, where the Logarithms of all numbers may be 
easily found from 1 to 10800000 ; and those of the sines, tangents, and 
secants, to any degree of accuracy required. 

Mr. Michael Taylor's Tables in large 4to, containing the common 
logarithms, and the logarithmic sines and tangents to every second of 
the quadrant, are very valuable. And, in France, the new book of 
logarithms by Callet; the 2d edition of which, in 1795, has the 
tables still further extended, and are printed with what are called stereo- 
types, the types in each page beng soldered together into a solid mass 
or block. 

Dodson's Antiiogarithmic Canon is likewise a very elaborate work, 
and used for finding the numbers answering to any given Vogpx\\\k\a, 
neb to 11 places. 



148 



LOGARITHM. 



is equal to the quotient of the two feme to which thoee io« 
dices belong. Thai, the index 6, minus the index 4, is = 2 ; 
and the terms corresponding to those indices are 64 and 16, 
whose quotient is = 4, which is the number answering to the 
index 2. 

For the same reason, if the logarithm of any number be 
multiplied by the index of its power, the product will be equal 
to the logarithm of that power. Thus, the index or loga- 
rithm of 4, in the above series, is 2;. ind if this number be 
multiplied by 3, the product will be «6 j. which is the loga- 
rithm of 64, or the third power-of 4. 

And, if the logarithm of any number be divided by the 
index of its root, the quotient will be equal to the logarithm 
of that root. Thus, the index or logarithm of 64 is 6 ; 
and if this number be divided by 2, the quotient will be 
= 3 ; which is the logarithm of 8, or the square root of 

The logarithms most convenient for practice, are such as 
are adapted to a geometric series increasing in a tenfold pro- 
portion, as in the last of the above forms ; and are those 
which are to be found, at present, in most of the common 
tables on this subject* The distinguishing mark of this 
system of logarithms is, that the index or logarithm of 10 
is 1 ; that of 100 is 2 ; that of 1000 is 3 ; &c. And, in 
decimals, the logarithm of •! is — 1 ; that of *01 is — 2 ; thai 
of "001 is — 3, die. the log. of 1 being in every systesjt 
Whence it follows, that the logarithm of any number beV" 
tween 1 and 10, must be and some fractional parts ; and^ 
that of a number between 10 and 100, will he 1 and some? 
fractional parts ; and so on, for any other number whatever. 
And since the integral part of a logarithm, usually called the 
Index, or Characteristic, is always thus readily found, it is 
commonly omitted in the tables ; being left to be supplied by 
the operator himself) as occasion requires. 

Another Definition of Logarithms is, that the logarithm of 
any number is the index of that power of some other num- 
ber, which is equal to the given number. So, if there be 
»-r", then n is the log. of n ; where n may be either po- 
sitive or negative, or nothing, and the root or base r any 
number whatever, according to the different systems of lo- 
garithms. When n is = 0, then n is = 1, whatever the 
value of r is ; which shows, that the log. of 1 is always 0, in 
every system of logarithms. When n is = 1, then n is = r ; 



140 



to dial the radix r it always that number whose log. is 1, in 
every system. When the radix r is = 2 •718281826459 &c. 
the indices n are the hyperbolic or Napier's log. of the num- 
bers ]t ; so that n is always the hyp. log. of the number if 
or (2.718 &c.)\ 

But when the radix r is = 10, then the index n becomes 
the common or Briggs's log. of the number n : so that the 
common log. of any number 10* or n, is n the index of that 
power of 10 which is equal to the said number. Thus 100, 
being the second power of 10, will have 2 for its logarithm ; 
and 1000, being the third power of 10, will have 3 for its 
logarithm : hence also, if GO be = 10"** 7 , then is 1*69807 
the common log. of 50. And, in general, the following de- 
cuple series of terms, 

to. 1C, 10', 10 9 , Iff, 10°, io-\ 10-», 10- 3 , 10- 4 , 
or 10000, 1000, 100, 10, 1, -1, -01, 001, -0001, 
have 4, 3, 2, 1, 0, -1, -2, -3, -4, 
for their logarithms, respectively. And from this scale of 
numbers and logarithms, the same properties easily follow, 
as above mentioned. 



problem. 

To compute the Logarithm to any of the Natural Number* 
1, 2, 3, 4, 5, 6fc. 

RULE I*. 

Take the geometric series, 1, 10, 100, 1000, 10000, die. 
and apply to it the arithmetic series, 0, 1,2, 3, 4, &c. as 
logarithms. — Find a geometric mean between 1 and 10, or 
/between 10 and 100, or any other two adjacent terms of the 
aeries, between which the number proposed lies. — In like 
manner, between the mean, thus found, and the nearest ex- 
treme, find another geometrical mean ; and so on, till you 
arrive within the proposed limit of the number whose loga- 
rithm is sought. — Find also as many arithmetical means, in 
the same order as you found the geometrical ones, and these 
will be the logarithms answering to the said geometrical 
means. 



* The reader who wishes to inform himself more particularly con- 
cerning the history, nature, and construction of Logarithms, may con- 
ssK my Mathematical Tracts, vol. 1, lately published, where he wit 
sod his cariosity amply gratified. 



150 



LOGARITHMS. 



EXAMPX.B. 

Let it be required to find the logarithm of 9. 

Here the proposed number lies between 1 and 10. 
First, then, the log. of 10 is I, and the log. of 1 is fr; 

theref. (1 + 0) -r- 2 = % = *5 is the arithmetical mean, 
and -v/(10 X 1) = y/10 = 3*1622777 the geom. mean ; 
hence the log. of 0*1022777 is '5. 

Secondly, the log. of 10 is 1, and the log. of 3* 1622777 is -5; 
theref. (1 + -5) 2 = -75 ia the arithmetic a! mean, 
and ^(10 X3-102277?)=5-6234132 is the geom. mean ; 
hence the log. of 5*6234132 is »75. 

Thirdly, the log. of 10 is 1, and the log* of 5*0234132 is -75; 
theref. (1 + -75} 2 = 875 is the arithmetical mean, 
and v/(10 X 5*6234132) = 7*4989422 the geom. mean ; 
hence the log. of 7-4989422 is -875. 

Fourthly, the log.of 10 is 1, and the log. of 7-4989422 is -875; 
theref. (1 + -875) -r- 2 =-9375 is the arithmetical mean, 
and v/(10 X 7-4989422) = 8-6596431 the geom. mean ; 
hence the log. of 8-6596431 is -9375. 

Fifthly, the log. of 10 is 1, and the log. of 8-6596431 is -9375 ; 
theref. (l+-9375)-i-2=-96875 is the arithmetical mean, 
and v/( 10 X 8-6596431) = 9-3057204 the geom. mean ; 
hence the log. of 9-3057204 is -96875. 

Sixthly, the log. oT 8-6596431 is -9375, and the log. of 
9-3057204 is -96875. 
theref. ( -9375+ -96875) -r-2= -953 125 is the arith-mean, 
and ^(8-6596431 X 9-3057204) = 8*9768713 the geo- 
metric mean ; 
hence the log. of 8-9768713 is -953125. 

And proceeding in this manner, after 25 extractions, it 
will be found that the logarithm of 8-9999998 is -9542425 ; 
which may be taken for the logarithm of 9, as it differs so 
little from it, that it is sufficiently exact for all practical pur- 
poses. In this manner were the logarithms of almost all the 
prime numbers at first computed. 

RUIjE II*. 

Let b be the number whose logarithm is required to be 
found ; and a the number next less than b } so that b — a =1, 



* For the demonstration of this rule, see my Mathematical Tablet, 
p. 109, &c. and my Tracts, vol. 1 . 



LOGARITHMS. 



151 



the logarithm of a being known ; and let * denote the sum 
of the two numbers a + b. Then 

1. Divide the constant decimal -8085889038 &c. by *, 
and reserve the quotient : divide the reserved quotient by 
the square of *, and reserve this quotient ; divide this last 
quotient also by the square of s y and again reserve the quo- 
tient : and thus proceed, continually dividing the last quotient 
by the square of s , as long as division can be made. 

2. Then write these quotients orderly under one another, 
the first uppermost, and divide them respectively by the odd 
numbers, 1, 3, 5, 7, 0, && as long as division can be made ; 
that is, divide the first reserved quotient by 1, the second by 
3, the third by 5, the fourth by 7, and so on. 

8. Add all these last quotients together, and the sum will 
be the logarithm of b a ; thorefore to this logarithm add 
also the given logarithm of the said next less number a, so 
will the last sum be the logarithm of the number b proposed. 

That is, 

Log. of b. is log. a + i X ( 1 + J- + -L + -L + &c. 
where n denotes the constant given decimal '8085889638 dec. 



examples. 



Ex. 1. Let it be required to find the log. of the number 2. 
Here the given number b is 2, and the next less number a 
is 1, whose log. is ; also the sum 2 + 1=3 = *, and its 
square * 3 = 9. Then the operation will be as follows : 



3 
9 
9 
9 

9 
9 
9 
O 



•80858890-1 
•289529054 
34109902 
3571410 
397100 
44129 
4903 
515 
01 



1 
3 
5 
7 
9 
11 
13 
15 



•289529054 ( 
32109902 ( 
3574410 ( 
397100 ( 
44129 ( 
4903 ( 
545 ( 
01 ( 

log. of ! { - 
add log. 1 - 



•289529(554 
10723321 
714888 
50737 
4903 
440 
42 
1 

•301029995 
•000000000 



log. of 2 



•301029995 



LOflARITMMB. 



Ex. 3. To compute the logarithm of the number 3. 

Here b = 3, the next less number a = 2, and the sum 
a + b = 5 = «, whose square «* is 25, to divide by which, 
always multiply by -04. Then the operation is as follows : 



5 
25 
25 
25 
25 
25 



•868588964 
173717793 
6948712 
277948 
11118 
445 
18 



1 ) 
3 ) 
5 ) 
7 ) 

11 ) 



173717793 
6948712 
277948 
11118 
445 
18 



( 
[ 

Si 



log. off 
log. of 2 add 



173717798 
2316*37 
55590 
1588 
50 
2 

176091260 
3U1029995 



log. of 3 sought -477121255 



Then, because the sum of the logarithms of numbers, 
gives the logarithm of their product ; and the difference of 
the logarithms, gives the logarithm of the quotient of the 
numbers ; from the above two logarithms, and the logarithm 
of 10, which is 1, we may obtain a great many logarithms, 
as in the following examples : 



EXAMPLE 3. 

Because 2x2= 4, therefore 
to log. 2 . -301029995| 
add log. 2 - -301029995} 



sum is log. 4 -602059991} 



EXAMPLE 4. 

Because 2X3 = 6, therefore 
to log. 2 - -301029995 
add log. 3 - -477121255 



sum is log. 6 -778151250 



KXAMPUB 5. 

Because 2 s = 8, therefore 
log. 2 - -301029995 
mult, by 3 3 



example. 6. 
Because 3" = 9, therefore 
log. 3 - -477121254/, 
mult by 2 2 



gives log. 9 -954242509 



EXAMPLE 7. 

Because y = 5, therefore 
from log. 10 1 -000000000 
take log. 2 -301029995} 



leaves log. 5 -698970004} 



EXAMPLE 8. 

Because 3X4 = 12, therefore 
to log. 3 - -477121255 
add log. 4 * -602059991 



gives log. 8 -903089987 j gives log. 12 1-079181246 



LOGARITHMS. 



158 



And thus, computing by this general rule, the logarithms 
to the other prime numbers, 7, 11, 13, 17, 19, 23, dec. and 
then using composition and division, we may easily find as 
many logarithms as we please, or may speedily examine any 
logarithm in the table*. 

Description a*d Use if the Table of Logarithms. 

Having explained the manner of forming a table of the 
logarithms of numbers, greater than unity ; the next thing to 
be done is, to show how the logarithms of fractional quan- 
tities may be found. In order to this, it may be observed, 
that as in the former case a geometric series is supposed to 
increase towards the left, from unity, so in the latter case 
it is supposed to decrease towards the right hand, still be- 
ginning with unit; as exhibited in the general description, 
page 152, where the indices being made negative, still show 
the logarithms to which they belong. Whence it appears, 
that as + 1 is the log. of 10, so — 1 is the log. of T \ or -1 ; 
and as + 2 is the log. of 100, so — 2 is the log. of T ^ or *01 : 
and so on. 

Hence it appears in general, that all numbers which con- 
sist of the same figures, whether they be integral, or frac- 
tional, or mixed, will have the decimal parts of their loga- 
rithms the same, but differing only in the index, which Mill 
be more or less, and positive or negative, according to the 
place of the first figure of the number. 

Thus, the logarithm of 2051 being 3-423410, the log. of 
tV» or iiv> or T? ! o7> & c - P art °f *t W 'M 06 113 follows : 
Numbers. Logarithms. 



2 6 5 1 
2 0-51 
2 6 5-1 
2-651 
•2 6 5 1 
•0 2 6 5 1 
•0 2 6 5 1 



3-4 2 3 4 1 

2-4 2 3 4 1 

1-4 2 3 4 1 

-4 2 3 4 1 

-1-4 2 3 4 1 

-2-4 2 3 4 1 

-3-4 2 3 4 1 



* There are, besides these, many other ingenious methods, which 
later writers have discovered for finding the logarithms of numbers, 
la a much easier way than by the original inventor ; but, as they cannot 
lie understood without a kno'wledgc; of some of the higher branches of 
Uie mathematics, it is thought proper to omit them, and to refer the 
a-asderto those works which are written e.xpres ly on the subject. It 
Vrould likewise much exceed the limits of this compendium, Vo 
Oat all the peculiar artifices that are made use of fur COn*Vru>cftfe&u& 
Vol. 1. 21 



154 



LOGARITHMS. 



Hence it also appears, that the index of any logarithm, is 
always less by 1 than the number of integer figures which 
the natural number consists of: or it is equal to the distance 
of the first figure from the place of units, or first place of in- 
tegers, whether on the left, or on the right, of it : and this 
index is constantly to be placed on the left-hand side of the 
decimal part of the logarithm. 

When there are integers in the given number, the index 
is always affirmative ; but when there are no integers, the 
index is negative, and is to be marked by a short line drawn 
before it, or else above it. Thus, 

A number having 1, 2, 3, 4, 5, &c. integer places, 

the index of its log. is 0, 1, 2, 3, 4, die. or 1 less than those 
places. 

And a decimal fraction having its first effective figure in the 
1st, 2d, 3d, 4th, &c, place of the decimals, has always 
— 1,-2, — 3, — 4, &c. for the index of its logarithm. 

It may also bo observed, that though the indices of frac- 
tional quantities are negative, yet the decimal parts of their 
logarithms are always affirmative. And the negative mark 
( — ) may be set either before the index or over it. 

I. TO FIND IN T1IE TABLE, THE LOGARITHM TO ANY 
NTMHER*. 

1. If the given Number be less than 100, or consist of only 
two figures ; its log. is immediately found by inspection in 
the first page of the table, which contains all numbers from 
1 to 100, with their logs, and the index immediately annexed 
in the next column. 

So the log. of 5 is 0-G9S970. The log. of 23 is 1 -301728. 
The log. of 50. is 1 -008070. And so on. 

2. If the Number be more than 100 but less than 10000; 
that is, consisting of either three or four figures : the decimal 
part of the logarithm is found by inspection in the other 
pages of the table, standing against the given number in this 
manner ; viz. the first three figures of the given number in the 
first column of the page, and the fourth figure one of those 
along the top line of it ; then in the angle of meeting are the 
last lour figures of the logarithm, and the first two figures 
of the same at the beginning of the same line in the second 



entire table of these numbers ; but any information of this kind, which 
the learner may wish to obtain, may he found in my Tables. See alto 
the article on Logarithms in the 2d volume, p. 340. &c. 
' See the table of Logarithms, at the end of this volume. 



L0GABITHK8. 



156 



column of the page : to which is to be prefixed the proper 
index which is always 1 less than the number of integer 
figures. 

So the logarithm of 251 is 2 399674, that is, the decimal 
•399674 found in the table, with the index 2 prefixed, because 
the given number contains three integers. And the log. of 
94-09 is 1-532627, that is, the decimal -532627 found in the 
table, with the index 1 prefixed, because tho given number 
contains two integers. 

2. But if the given Number contain more than four figures ; 
take out the logarithm of the first four figures by inspection 
in the table, as before, as also the next greater logarithm, 
subtracting the one logarithm from the other, as also their 
corresponding numbers the one from the other. Then say, 

As the difference between the two numbers, 

Is to the difference of their logarithms, 

So is the remaining part of the given number, 

To the proportional part of the logarithm. 

Which part being added to the less logarithm, before taken 
out, gives the whole logarithm sought very nearly. 



EXAMPLE. 

To find the logarithm of the number 34-0926. 
Tho log. of 340900, as before, is 532627. 
And log. of 341000 - - is 532754. 
The ditfs. are 100 and 127. 



Then, as 100 : 127 : : 26 : 33, the proportional part. 

This added to - - - 532627, the first log. 

Gives, with the index, 1-532660 for the log. of 34-0926. 

4. If the number consist both of integers and fractions, or 
is entirely fractional ; find the decimal part of the logarithm 
the same as if all its figures were integral ; then this, having 
prefixed to it the proper index, will give the logarithm re- 
quired. 

5. And if the given number be a proper vulgar fraction : 
subtract the logarithm of the denominator from the loga- 
rithm of the numerator, and the remainder will be the loga- 
rithm sought ; which, being that of a decimal fraction, must 

*" always have a negative index. 

6. But if it be a mixed number ; reduce it to an impro- 
per fraction, and find the difference of the logarithm of the 
numerator and denominator, in the same manner as before. 



156 



LOGARITHMS. 



EXAMPLES. 



1. To find the log. of J?. 
Log, of 87 - 1-568202 
Log. of 94 - 1-973128 



Dif. log. of» J —1-595074 



Where the index 1 is negative 



2. To find the log. of 17$f • 
First, 17j| = Then, 
Log. of 405 . 2-807455 
Log. of 2ft - 1-361728 

Difl ]p|[. of 17iJ 1-245727 



n. TO FIND THE NATUHAX ITDMBSS TO ANY GIVEN 
LOGARITHM. 

This is to be found in the tables by the reverse method 
to the former, namely, by searching for the proposed loga- 
rithm among those in the table, and taking out the corre- 
sponding number by inspection, in which the proper number 
of integers are to be pointed off, viz. 1 more than the index. 
For, in finding the number answering to any given logarithm, 
the index always shows how far the first figure must be 
removed from the place of units, viz. to the left hand, or in- 
tegers, when the index is affirmative ; but to the right hand, 
or decimals, when it is negative. 



EXAMPLES. 



So, the number to the log. ]h532S82 is 34-11. 
And the number of the log. "l'532b82 is -3111. 

But if the logarithm cannot be exactly found in the table ; 
take out the next greater and the next less, subtracting the 
one of these logarithms from the other, as also their natural 
numbers the one from the other, and the less logarithm from 
the logarithm proposed. Then say, 

As the difference of the first or tabular logarithms, 
Is to the difference of their natural numbers, 
So is the differ, of the given log. and the least tabular log. 
To their corresponding numeral difference. 
Which being annexed to the least natural number above ta- 
ken, gives the natural number sought, corresponding to the 
proposed logarithm. 



EXAMPLE. 



So, to find the natural number answering to the given 
logarithm 1*532708. 



L04AJUTHM8. 157 

Here the next greater and next less tabular logarithms, 
trith their corresponding numbers, are as below : 

Next greater 532754 its num. 341000 ; given log. 532708 
Next less 532627 its num. 340900 ; next less 532627 

— 81 




learly the numeral differ, 
i is the number sought, marking off two 
integers, because the index of th e given logarithm is 1. 
Had the index been negative, thus 1-532708, its com- 
■ number would have been 340964, wholly decimal. 





MULTIPLICATION BY LOGARITHMS. 



RULE. 



Take out the logarithms of the factors from the table, 
then add them together, and their sum will be the logarithm 
of the product required. Then, by means of the table, take 
out the natural number, answering to the sum, for the pro- 
duct sought. 

Take care to add what is to be carried from the decimal 
part of the logarithm to the affirmative index or indices, or 
else subtract it from the negative. 

Also, add the indices together when they are of the same 
kind, both affirmative or both negative ; but subtract the 
less from the greater, when the one is affirmative and the 
other negative, and prefix the sign of the greater to the re- 
mainder. 



EXAMPLES. 



1. To multiply 23-14 by 
5062 
Numbers. Logs. 
23 14 . 1-364363 
5 062 . 0-704322 



2. To multiply 2-581926 
by 8«457291 
Numbers. Logs. 

2- 581920 . 0-411944 

3- 457291 . 0-538786 



**rodoct 117-1347 2 068685 



Prod. 8-92648 . 0-950680 



158 



DIVISION BY &OGABXTHK8. 



3. To mult. 3-902 and 597 16 
and -0314726 all together. 



Numbers. 
3-902 
597 16 
•0314728 



0-59F287 ! 
2-776091 
-2-497935 



Prod. 73-3333 . 1-865313 



Here the — 2 cancels the 2, 
and the 1 to carry from the 
decimals is set down. 



4. To mult. 3-586, and 2-1046, 
and 0-8372, and 0-0294 all 
together. 

Numbers. Logs. 
3 586 - 0-554610 
2-1046 - 323170 
0-8372 , -1-922829 
0-0294 -2-468347 



Prod. 01857618 - 1 -268956 



Here the 2 to carry cancels 
the -2, and there remains 
the —1 to set down. 



DIVISION BY LOGARITHMS. 

RULE. 

From the logarithm of the dividend, subtract the loga- 
rithm of the divisor, and the number answering to the re- 
mainder will be the quotient required. 

Change the sign of the index of the divisor, from affirm- 
ative to negative, or from negative to affirmative ; then take 
the sum of the indices if they be of the same name, or their 
difference when of different signs, with the sign of the greater, 
for the index to the logarithm of the quotient. 

Also, when 1 is borrowed, in the left-hand place of the 
decimal part of the logarithm, add it to the index of the 
divisor when that index is affirmative, but subtract it when 
negative ; then let the sign of the index arising from hence 
be changed, and worked with as before. 



EXAMPLES. 



1. To divide 24163 by 4567. 

Numbers. Logs. 
Dividend 24163 - 4 383151 
Divisor 4567 - 3-659631 



2. To divide 37 -1 49 by 523-76. 

Numbers. Logs. 
Dividend 37149 - 1-569947 
Divisor 523-76 . 2 719132 



Quot. 5-29078 0-723520 



Quot. -0709275-2-85081* 



UfTOLUTIOM BY LOGARITHMS. 



150 



8. Divide -06314 by -007241 

Numbers. Logs. 
Dividend -06314-2-800305 
Divisor -007241-3-859799 



4. Todivide-7438by 12-9476. 

Numbers. Logs* 
Divid. -7438 -1-871456 
Divisor 12-9476 1-112169 



Quo*. 8-71979 0-940506 Quot. -057447 - 2-750267 

. '•' v — «— 

Here 1 ctjjirfcd from • the Here the 1 taken from the 
decimals to pfc 8, jnake* it — 1, makes it become —2, to 
become — % wen from set down, 
the other —2, leaves re* 
maining. 

Note. The RuletaftThree, or Rule of Proportion, is per- 
formed by adding the logarithms of tho 2d and 3d terms, 
jnnd subtracting that of the first term from their sum. In- 
stances will occur in Plain Trigonometry. 



INVOLUTION BY LOGARITHMS. 



RULE. 



Take out the logarithm of the given number from the ta- 
ble. Multiply the logarithm thus found, by the index of the 
power proposed. Find the number answering to the product, 
and it will be the power required. 



Note. In multiplying a logarithm with a negative index, 
by an affirmative number, the product will be negative. But 
what is to bo carried from the decimal part of the logarithm, 
will always be affirmative. And therefore their difference 
wilj be the index of the product, and is always to be made 
of the same kind with the greater. 



EXAMPLES. 



1. To square the number 
2-5791. 
Numb. Log. 
Root 2-5791 - - 0-411468 
The index - - 2 



Power 6-65174 0-822936 



2. To find the cube of 
3-07146. 
Numb. Log. 
Root 3 07146 - - 0-487345 
The index - - 3 



Power 28-9758 1-462035 



■VOLUTION BY LOGARITHMS. 



3. To raise •09163 to the 
4th power. 
Numb. Log. 
Root •09163 —2-962038 
The index - - 4 



Pow. -000070494— 5-848152 



Here 4 times the negative 
index being — 8, and 3 to 
carry, the difference — 5 is 
the index of the product. 



4. To raise 1*0045 to the 
365th power. 
Numb. Log. 
Root 1 '0045 - - 0-001950 
Thejndex - . 365 



9750 
■ 11700 
5850 




0-711750 



EVOLUTION BY LOGARITHMS. 

Take the log. of the given number out of the table. 
Divide the log. thus found by the index of the root. Then 
the number answering to the quotient will be the root. 

Note. When the index of the logarithm, to be divided 
is negative, and does not exactly contain the divisor, without 
some remainder, increase the index by such a number as 
will make it exactly divisible by the index, carrying the units 
borrowed, as so many tens, to the left-hand place of the deci- 
mal, and then divide as in whole numbers. 



EXAMPLES. 



1. To find the square root of 
365. 

Numb. Log. 
Power 365 2) 2-562293 
Root 19-10496 1-281146£ 



2. To find the 3d root of 
12345. 
Numb. Log. 
Power 12345 3)4 091491 
Root 231116 1-363830} 



3. To find the 10th root of 
2. 

Numb. Log. 
Power 2 - 10) 0-301030 
Root 1-071773 0-030103 



4. To find the 365th root of 
1-045. 
Numb. Log. 
Power 1 045 365) 019116 
Root 1-000121 0-000052| 



EVOLUTION BY tOGA 



161 



5- To find </ <m. 

Numb. Log* 
Power 093 2) —2-968483 
Root -304959 —1-48424!* 

Here the divisor 2 is con- 
tained exactly once in the ne- 
gative index — 2, and there- 
fore the index of the quotient 
it — 1. 



6* To find the -00048. 
Numb. Log. 
Power -00048 3)— 4-681241 
Root -0782973 —2-893747 

Here the divisor 3, not being exactly 
contained in —4, it is augmented by 2, 
to make up 6, in which the divisor is 
contained juit 2 timet; then the 2, 
thus borrowed, being carried to the de- 
' * aal figure 6, makes 86, which dividV 
by S, gives 8, &c. 



7. To find 3-1416 X 82 X ff- 

8. To find 029IG X 751-3 X fl T 

9. Afl 7241 : 3-58 : : 20-46 : ? 
10, : -v/ff : : 6-927 : ? 



Vol. I. 



22 



ALGEBRA. 



DEFINITIONS AND NOTATION. 

. 

1. Algebra is the science of investigation by means of 
symbols. It is sometimes also called Analysis ; and is ft 
general kind of arithmetic, or universal way of computa- 
tion. 

2. In this science, quantities of all kinds are represented 
by the letters of the alphabet. And the operations to to 
performed with them, as addition or subtraction, &c. are de- 
noted by certain simple characters, instead of being express- 
ed by words at length. 

3. In algebraical inquiries, some quantities are known or 

E'ven, viz. those whose values are known : and others un- 
iown, or are to be found out, viz. those whose values are 
not known. The former of these are represented by the 
leading letters of the alphabet, a, b, c, d, <3uc. ; and the latter, 
or unknown quantities, by the final letters, *, y, «, tt, &c. 

4. The characters used to denote the operations, are 
chiefly the following : 

+ signifies addition, and is named plus. 

— signifies subtraction, and is named minus. 

X or • signifies multiplication, and is named into. 

signifies division, and is named by. 
y/ signifies the square root ; the cube root ; (/ the 
4th root, dtc. ; and the nth root. 
: : : signifies proportion. 
=s signifies equality, and is named equal to* 
And so on for other operations. 

Thus a + b denotes that the number represented by b is 
to be added to that represented by a. 

<x — 6 denotes that the number represented by b is to be 
subtracted from that represented by a. 

a*b denotes the difference of a and 6, when it is not 
known which is the greater. 



vzrorrnoiai jam HbTATiozr. 163 

Hi, or a X b, or a . b f expresses the product, by multipli- 
cation of the numbers represented by a and 4. 

a -f- ft, or denotes, that the number represented by a 
b 

is to be divided by that which is expressed by b. 

a i b : : e : a\ signifies that a is in the same proportion to 
b> as e is to a\ , .y 

# ■= a — & 4- * is an equation, expressing that * is equal 
to the difference of a and b, added to the quantity c. 

a, or a*, denotes the square root of a.; $/a, or a^, the 

eubexoot of a ; and or a' the cube root of the square of a ; 

JL * 

also ^/a, or a w , is theath root of a; and ^/a n ora m is the 

nth power of the mth root of a, or it is a to the — power. 

fn 

a 1 denotes the square of a ; a 3 the cube of a ; a 4 the fourth 
power of a ; and a n the nth power of a. 
a + b X c, or (a + b) c, denotes the product of the com- 

6wnd quantity a + b multiplied by the simple quantity c. 
sing the bar , or the parenthesis ( ) as a vinculum, to 

connect several simple quantities into one compound. 

a + b -f- a — b 9 or a - ^" , , expressed like a fraction, means 
a — o 

the quotient of a + b divided by a— b. 

i/ab + cd 9 or ^ai+crf)^, is the square root of the com* 

pound quantity ab + cd. And e y/ ab + or c (ab + cd)^ 9 
denotes the product of c into the square root of the com- 
pound quantity ab + cd. 

a + 6 — c 3 , or (a + b — cf denotes the cube, or third 
power, of the compound quantity a + b — c. 

8a denotes that the quantity a is to be taken 3 times, and 
4(a + b) is 4 times a + 6. And these numbers, 3 or 4, 
showing how often the quantities are to be taken, or multi- 
plied, are called Co-efficients. 

Also fx denotes that x is multiplied by { ; thus J Xx or Jx. 

5. Like quantities, are those which consist of the same 
letters, and powers. As a and 3a ; or 2ab and 4ab ; or 30*60 
and — 5a*ftc. 

6. Unlike Quantities, are those which consist of different 
letters, or different powers. As a and b ; or 2a and a 9 ; or 
4aP and 3abc. 

7. Simple Quantities are those which consist of one tettft. 
only. As 3a, or bob, or 6ak?. 



1M .* 



ALGEBRA. 



" 8. Compound Quantities are thoae which consist of two or 
more terms. As a+b, or 2a— 3c, or a+2b - 3c. 

9. And when the compound quantity consists of two terms, 
it is called a Binomial, as a+b ; when of three terms, it is a 
Trinomial, as a+2b — 3c ; when of four terms, a Quad ri no- 
mi al, as 2a — 3b+c—4d ; and so on. Also a Multinomial or 
Polynomial, consists of many terms. 

10. A Residual Quantity, is a binomial having one of the 
terms negative. Asa— 2b. 

11. Positive or affirmative Quantities, are those which are 
to be added, or have the sign +. As a or + a, or obi for 
when a quantity is found without a sign, it is understood to 
be positive, or have the sign + prefixed. 

12. Negative Quantities, are those which are to be sub- 
tracted* As — a, or —2ab, or — Sab\ 

* 13. Like Signs, are cither all positive ( + ), or all nega- 
tive ( - ). 

14. Unlike Signs, are when some are positive ( + ), and 
others negative ( — ). 

15. The Co-efficient of any quantity, as shown above, is 
the number prefixed to it. As 3, in (he quantity Sab. 

16. The power of a quantity (a), is its square (a 8 ), or cube 
(a 3 ), or biquadrate (a 4 ), &c. ; called also, the 2d power, or 
3d power, or 4th power, &c. 

17. The Index or Exponent, is the number which denotes 
the power or root of a quantity. So 2 is the exponent of 
the square or second power a 2 ; and 3 is the index of the 

cube or 3d power ; and £ is the index of the square root, at 

or y/a ; and \ is the index of the cube root, a^, or 

18. A Rational Quantity, is that which has no radical sign 
or index annexed to it. As a, or Sab. 

19. An Irrational Quantity, or Surd, is that of which the 
value cannot be accurately expressed in numbers, as the 
square root of 2, 3^J^> Surds are commonly expressed by 

means of the radical sign </ : as y/2, or y/a, or -J/a", or a$. 

20. The Reciprocal of any quantity, is that quantity in- 
verted, or unity divided by it. So, the reciprocal of a, or 

is i, the reciprocal of % is that of — ~ is ^— -. 
la b a* x + y a 

21. The letters by which any simple quantity is expressed, 
may be ranged according to any order at pleasure. So the 
product of a and b, may be cither expressed by ab, or ba ; 



DEFINITIONS AND NOTATION. * 165 



and the product of a, 6, and c, by either abc f or acb, or hoc, 
or bca> or cab, or cba ; as it matters not which quantities are 
placed or multiplied first. But it will be sometimes found 
convenient in long operations, to place the several letters 
according to their order in the alphabet, as a be, which order 
also occurs most easily or naturally to the mind. 

22. Likewise, the several members, or terms, of which a 
compound quantity is composed, may be disposed in any 
order at pleasure, without altering the value of the significa- 
tion of the whole. Thus, 3a — 2a6+4a6c may also be writ- 
ten 2a+4abc—2ab, or 4abe+oa -2ab 9 or -2ab+3a+4abc, 
fee. ; for all these represent the same thing, namely, the 
quantity which remain^, when the quantity or term iab is 
subtracted from the sum of the terms or quantities 3a and 
4abc» But it is most usual and natural, to begin with a po- 
sitive term, and with the first letters of the alphabet. 



SOXV EXAMPLES FOR PRACTICE. 



In finding the numeral values of various expressions, or com- 
binations, of quantities. 

Supposing a=6, and b =5, and c=4, and a*=l, and e=0. 
Then 

1. Will a a + 3aft-c 3 =36 + 90-16 = 110. 

2. And 2a*-3a 2 b + c 3 = 432-540 + (54 = - 44. 

3. And a 3 x(a+6)-2ate = 36 X 11-240=156. 

a 3 21ft 

4. And +c==^- + 10 = 12+ 16 = 28. 

5. And y/2ac+cr or (2ac + c 3 )^ = y/64 = 8. 

A A 1 / _!_ 2lfC « 0_1_ 40 * 

6 - An(1 ^ c + 7C^H^) = 2 + ¥ = 7 - 

* a j < l3 ~"_v/(^ 3 ~~ ac ) 36 — 1 35 

7 * And Sa-v/^+^T) " VJZ7 " "5 ~ 7 ' 

8. And t/(* 3 -ac)+ v^(2flc+c»)=l + 8=9. 

9. And v'^cT^^+f 1 ) = vT 25 - 2 * + 8) = 3. 

10. And <r& + c-d = 183. 

11. And0a6-106 2 + c=24. 

3? 

12. And — - X d = 45. 

c 

18. Andl±*xj = 13J. 



166 



AMKBRA. 



14. And 



a+b a — b 



= 1|. 



15. And— + e = 45. 

c 

16. And — X « = 0. 

c 

17. And (& — c) X (<*— = L 
ia And (a+ 6) — (c — 4) = 8 

19. And (a + b) — c — d ^ Q. 

20. And afc X (P= 144. 

21. And acd — d = 23. 

22. And o% + Ve + d = V. 



23. 



Aod^X^^lSi. 



<J — e c — d 



24. And v^a 8 + ft 2 — — b' = 4-4936249. 

25. And 3ac» + i/a 3 — V = 292-497942. 

26. And 4a 1 — 3a ^a 8 — \ab = 72. 



ADDITION. 

Addition, in Algebra, is the connecting the quantities 
together by their proper signs, and incorporating or uniting 
into one term or sum, such as are similar, and can be united. 
As 3a -f- 26 - 2a = a + 2o, the sum. 

The rule of addition in algebra, may be divided into three 
cases : one, when the quantities are like, and their signs like 
also ; a second, when the quantities are like, but their signs 
unlike; and the third, when the quantities are unlike. 
Which was performed jftfelows*. 



•The reasons on which these operation! are founded, will readily ap- 
pear, by a little reflection on the nature of the quantities to be added, 
or collected together. For, with regard to the first example, where 
the quantities are 3a and 6a, whatever a represents in the one term, it 
will represent the same thing in the other; so that 3 times any thing 
and 6 times the same thing, collected together, must needs make 8 times 
thatching. As if a denote a shilling ; then 3a is 3 shillings, and 6a is 6 
shillings, and their sum 8 shillings. In like manner, —2ab and — 1mb 9 
or —2 times any thing, and —7 timet the same thing, make — f> times 
that thing. 

* i 



ADDITION. 



• 167 



CASE I. 

When the Quantities are Like, and have Like Signs. 

Add the co-efficients together, and set down the sum ; 
after which set the common letter or letters of the like quan- 
tities, and prefix the common sign + or — . 

Thus, 8a added to 6a, makt a 8a. 

And — 2ab added to — 7ab, makes —dab. 

And 5a + 7b added to 7a + U t makes 12a + 10*. 







; FOR PRACTICE. 






bxy 


9a 


— 5bx 


2bxy 


5a 


— 4bx 


5fey 


12a 


— 2bx 


bxy 


a 


— 7bx 


Zbxy 


2a 


— bx 


Uxy 


32a 


— 22bx 


18bxy 



As to the second case, in which the quantities are like, but the signs 
unlike; the reason of its operation will easily appear, by reflecting, 
that addition means only the uniting of Quantities together by means of 
the arithmetical operations denoted by their signs-)- and — , or of addi- 
tion and subtraction ; which being of contrary or opposite natures, the 
one co-efficient must be subtracted from the other, to obtain the incor- 
porated or united mass. 

As to the third case, where the quantities are unlike, it is plain that 
such quantities cannot be united into one, or otherwise added, than by 
means of their signs : thus, for example, if a be supposed to represent 
a crown, and 6 a shilling ; then the sum of a and u can be neither 2a 
nor 2*, that is, neither 2 crowns nor 2 shillings, but only 1 crown plus 1 
•Billing, that is a +6. 

In this rule, the word addition is not very properly used ; being much 
too limited to express the operation here performed. The business of 
this operation is to incorporate into one mass, or algebraic expression, 
different algebraic quantities, as far as asUStoa) incorporation or union 
is possible ; and to retain the algebraic sajfkslbr doing it, in cases where 
the former is not possible. When wtfjlfo several quantities, some 
affirmative and some negative ; and the relation of these quantities can 
in the whole or In part be - discovered ; such incorporation of two or 
more quantities into one, is plainly effected by the foregoing rules. 

It may seem a paradox, that what is called addition in algebra, should 
sometimes mean addition, and sometimes subtraction. But the para- 
dox wholly arises from the scantiness of the name given to the alge- 
braic process; from employing an old term in a new and more enlarged 
sense. Instead of addition, call it incorporation, or union, or g&iking a 
Islands, or give it any name to which a more extensive idea may be 
annexed, than that which is usually implied by the word addition : and 
the paradox vanishes. 



168 * 



AMBMU. 



3* 


dx*+5xy 


2<i* — 4y 


2* 


**+ xy 


4ax — y 


4s 


2* a +4*y 


ax — 8y 


% 


5x*+2*y 


5ax — 5y 


6x 


4x*+3*y 


7ax — 2y 


15* 







5*y 
14xy 
22xy 
Ylxy 
l\xy 

\xy 




30- 


13**- 


-3*y 


23- 


10**- 


•4xy 


14- 


14**- 


•7xy 


10- 


16**- 


•5xy 


16- 


20**- 


' x 9 



5xy — 3x + 4a& 
8xy— 4x + Sab 
3xy — 5* + bob 
xy— 2* + afc 
4*y— * + 7ab 



CASE II. 



When the Quantities are alike, ha have Unlike Signs. 

Add all the affinnatif^co-efficients into one sum, and all 
the negative ones into'aoother, when there are several of a 
kind. Then subtract the less sum, or the less co-efficient, 
from the greater, and to the remainder prefix the sign of the 
greater, and subjoin the common quantity or letters. 

So + 5fg and — 3a, united, make + 2a. 
And — 5a and — 3a, united, make — 2a. 



ADDITION. 



169 



OTHER EXAMPLES FOE PRACTICE. 




+ Qx* + 3y 

- 5x* + 4y 
— 16x 3 + by 
+ 3s 3 — ly 
+ 2X 3 — 2y 

— 8x* + 3y 



+ 4ab+ 4 

— 4ab + l2 
+ 7ab— 14 
+ ab+ 3 

— bob— 10 



— 3ox* 
+ ax* 
+ 5ax* 

— 6ax* 



H-lOv/ax 

— 3v/ax 
+ Ay/ax 

— Yty/ax 



+ 3y + 4ax* 

— y — bax^ 
+ 4y + 2ax* 

— 2y + 6ax* 



case m. 
When the Quantities arc Unlike. 

Having collected together all the like quantities, as in the 
two foregoing cases, set down those that are unlike, one after 
another, with their proper signs. 

: W - 
EXAMPLES.™ 

Sxy 6xy — 12x* 4ax — 130 + 3x* 

2ax -4X 1 + 3xy Sx 3 + 3ax + 9*" 

-5xy +4x« - 2xy Ixy - 4x* + 90 
fax -3xy + 4r» yx + 40 -Ox 8 

— 2xy+8ax 4xy— 8x" 7ax + 8x* + 7xy 



Vol. I. 



23 



170 ALGEBRA. 

9x Y 1 4ax - 2x* 9 + x — 5y 

-li*y 5ax + Zry 2x + 7-/*? + 5y 

+ 3oxy Sif — tex 5y + 3^/ax — 4y 

— 3x a + 26 10— Ay/ax + 4y 




2-v/xy + 14x 
3x + 2y 
—9 + 3-v/jy 




Add a+6 and 3a — 56 together. 

Add 5a — 8x and 3a — 4x together. 

Add 6x- 56+a+8 to — 5a — 4r+4fc — 3. 

Add a+26-3c-10 to 36 -4a + 5c + 10 and 56 -c. 

Add a +6 and a — 6 together. 

Add 3a + 6-10 to>c-<*-a and — 4c + 2a- 36 -7. 
Add 3a a +6 3 -cto 2rt6-3« 2 +6c— -6. 
Add a 3 +6 3 c-6 2 to a6 3 ~a6c+6 3 . 

Add 9a-86 -f 10x-6a*-7c + 50 to $r-3a-5c +46 +6d 
-10. 



SUBTRACTION. 

Set down in one line the first quantities from which the 
subtraction is to be made ; and underneath them place all the 
other quantities composing the subtrahend ; ranging the like 
quantities under each sifter, as in Addition. 

Then change all the signs (+ and — ) of the lower line, 
or conceive them to be changed ; after which, collect all the 
terms together as in the cases of Addition*. 



* This rule is founded on (he consideration, that addition and sub- 
traction are opposite to each other in their nature and operation, as are 
the signs -{-and — , by which they are expressed and represented. So 
that, since to unite a negative quantity with a positive one of the same 
kind, has the effect of diminishing it, or subducting an equal positive 



•UBTRAOTIO.N. 



1T1 



XXAXPLK3. 

From la % — 36 9x* ~ 4y + 8 8xy — 3 + 6z — y 
Take 2a a — 86 bV+5y-4 4xy — 7 — 6x — 4y 

*l*y + 4 + 12x+3y 



— 20 — 6x— 5xy 
3xy — 9x X 8 — 2ay 

Rem. 7xy— 12 2$f£%fc — 8 -r-28+3x — 8ry+2ay 



From 8x»y + 6 5 > /xy + 2x x /xy 7x 3 + 2^/z- 18+ 36 
Take— 2x=y + 2 7 x /xy + S—2xy Ox 3 — 12 + 56+ ** 

Rem. 



5xy — 30 7x 3 — 2 (a + b) 3xy 7 + 20a + 10) 
7xy — 50 2x 3 ~4(a + 6) Ixy + 12a y/(xy + 10) 



Rem. 4a*+56 3x 2 — 9y+12 



From 5xy— i\5^^ — 3y — 4 
Take— 2xy + 6 -4Qfgjb+4 




From a + 6, take a — b. 
From 4a + 46, take b + a. 
From 4a — 46, take 3a + 56. 
From 8a — 12x, take 4a — 3x* 
From 2x — 4a — 26 + 5, take 8— 56 + a + 6x. 
From3a + 6+c — d — 10, takec + Ua — d. 
From 3a + b + c — d — 10, take 6— 10 + 3a. 
From 2a6 + 6 3 — 4c + be — 6, take 3a 3 — r + 6 3 . 
From a 3 + 36"c + 06* — a6c, take 6 2 + a6 a — a6c. 
From 12x + 6a— 46 +40, take 46 — 3a + 4x + 6d— 10. 
• From 2x — 3a + 46 + 6c — 50, take 9a + x + 66 — 6c 
40. 

From 6a — 46 — 12c + 12x, take 2a — 8a + 46 — 5c. 



^De from it, therefore to subtract a positive (which is the opposite of 
^anitiog or adding) is to add the equal negative quantity. In like map. 
-^«r, to subtract a negative quantity, is the same in effect as to add or 
***rfte an equal positive one. So that, changing the sign of a quantity 
^*om-fto— , or from — to -)-, changes its nature from a subductive 
Quantity to an additive one ; and any quantity is in effect subtracted, by 
<k*raly changing its sign. 



TO 



AL6KBKA. 



MULTIPLICATION. 



This consists of several cases, according as the factors are 
simple or compound quantities. 

cash. When both the Factors are Simple Quantities. 

Fcbst multiply the co-efficienti .of ijie two terms together, 
then to the product annex all the letters in those terms, which 
will give the whole product required. 

Note*. like signs, in the factors, produce +, and unlike 
signs — , in the products. 

EXAMPLES. 

10a —2a 7a — 6* 

2b + 2b -4c — 4a 



20ab —Sab —28ac +24ax 



* That this rule for the signs is true, may be thus shown. 

1. When -f-a is to be multiplied by + e; the meaning is, that + * is 
to be taken as many times as there are units in c ; and since the sum of 
any number of positive terms is positive, it follows that -faX+o 
makes 4- ac 

2. When two quantities are to be multiplied together, the result will 
be exactly the same, in whatever order they are placed ; for a times c 
is the same as c times «, and therefore, when — a is to be multiplied by 
+ $ t or + e by — a: this is the same thing as taking — a as many times 
as there are units in 4-c ; and as the sum of any number of negative 
terms is negative, it follows that — a X + c, or + a X — • « make or pro- 
duce — ac. 

3. When — a is to be multiplied by — c : here — a is to be subtract- 
ed as often as there are units in c: but subtracting negatives is the same 
thing as adding affirmatives, by the demonstration of the rule for sub- 
traction ; consequently the product is c times a, or -f- ac. 

Otherwise. Since a — a = 0, therefore (a — a) X — c is also = 0, be- 
cause multiplied by any quantity, is sUll but ; and since the flrst 
term of the product, or a X — c is = — ac, by the second case ; then* 1 
fore the last term of the product, or — a X — c, must be -f- ac, to make 
the sum = 0, or — tu+ac = 0; that in, , — a X — c = ~|- ac. 

Other demonstrations upon the principles of proportion, or by menus 
of geometrical diagrams, have also been given ; but the above may «tf> 
ice. 



4oc 
-3a6 

— 12a 3 6c 



Sax — ax +3jcy — 5xyz 

4x . yi —6c —4 -5ax 
— >■* ■ " — — 




CASE II. 

When ome of ike Factors is a Compound Quantity. 

Multiply every term of the multiplicand, or compound 
quantity, separately, by the multiplier, as in the former 
case ; placing the products one after another, with the 
proper signs ; and the result will be the whole product re- 
quired* 



XULTIFUGATION. 173 

9a 3 * — 2x*y —4xy 

4x 3*^ — xy 

36aV — 6fy +4*V 



EXAMPLES. 



5a — 3c 3ac — 46 2a 3 — 3c + 5 
2a 3a 6c 



lOa^-Goc Qa^ — 12a* 2a'bc-3bc l + 56c 



12x-2ac 
4a 



25c — 76 
-2a 



4x — 6 + 3a6 
2a6 



3c 3 + i 
4xy 



10s* — 3^ 
— 4X 3 



8a 3 — 2x 3 — 66 
2a! 3 



174 



ALGEBRA. 



CASS III. 

When both the Factors are Compound Quantities : 

Multiply every term of the multiplier by every term of 
the multiplicand separately ; setting down the products one 
after or under another, with their proper signs ; and add the 
several lines of products all together for the whole product 



required. 

.A v 

a+b 3x+2y 2&+*y - 2jr» 

a+b 4x— by 3x— Sy 

a*+ab lib*+8xy 6x 3 + 3x^-6*^ 

+ab+b* - 15xy- I0y 2 —6x 2 y—3xy 3 +6y s 

a*+2ab+b 2 12X 3 — 7xy-l(y 6x 3 - 3x^-9x^+6^ 

a+b x 2 +y a'+ab+b* 

a — b x 2 +y a — b 

a*+ab x x +yx* a 3 +a~b+ab 2 

— ab-b 2 +yx 3 +if — a**— <*>*>'— V 

a* * x 4 +2yx 3 +3T « 3 * * — 6 3 



Note. In the multiplication of compound quantities, it is 
the best way to set them down in order, according to the 
powers and the letters of the alphabet. And in the actual 
operation, begin at the left-hand side, and multiply from the 
left hand towards the right, in the manner that we write, 
which is contrary to the way of multiplying numbers. But 
in setting down the several products, as they arise, in the 
second and following lines, range them under the like terms 
in the lines above, when there are such like quantities ; 
which is the easiest way for adding them up together. 

In many cases, the multiplication of compound quantities 
is only to be performed by setting them down one after 
another, each within or under a vinculum, with a sign 
multiplicat ion be tween them. As {a + b) X (a — b) X 
or a + b . a — b . 3a&. „ 



DTVI8ION. 



175 



EXAMPLES TOR PRACTTCF. 



1 . Multiply 10ac by 2a. 

2. Multiply 3a a — 26 by 36. 

3. Multiply 3a + 26 by 3a — 26. 

4. Multiply X 3 — xy + by x + y. 



Ans. 20aV. 
Ans. 9a*b — 66 2 . 
Ans. 9a a — 4o\ 
Ans. + y\ 



5. Multiply a 3 + a f 6 -f a& 2 + 6 3 by a — 6. Ans. a 4 — 6 4 . 

6. Multiply a 8 + ab + 6* by a 3 — «6 + 6 3 . 

7. Multiply 3x* — 2xy + 5 by x 2 + 2xy — 6. 

8. Multiply 3a 2 — 2** + bx 1 by 3a 2 — 4ax — 7x 2 . 

9. Multiply 3x 3 + 2xy + 3y 3 by 2x 3 — 3xY + 3y\ 
10. Multiply a 3 + a6 + 6 a by a — 26. 



Division in Algebra, like that in numbers, is the converse 
of multiplication ; and it is performed like that of numbers 
also, by beginning at the left-hand side, and dividing all the 
parts of the dividend by the divisor, when they can be so 
divided ; or else by setting them down like a fraction, the 
dividend over the divisor, and then abbreviating the fraction 
as much as can be done. This may naturally be distin- 
guished into the following particular cases. 



When the Divisor and Dividend arc both Simple Quantities : 

Set the terms both down as in division of numbers, 
either the divisor before the dividend, or below it, like the 
denominator of a fraction. Then abbreviate these terms as 
much as can be done, by cancelling or striking out all the 
letters that are common to them both, and also dividing 
the one co-efficient by the other, or abbreviating them after 
the manner of a fraction, by dividing them by their common 
measure. 

Noli* Like signs in the two factors make + in * ne <l uo " 
tient ; and unlike signs make — ; the same as in multipli- 
cation*. 



- * Became the divisor multiplied by the quotient, must product lVv% 
dfridend. Therefore, 



DIVISION. 



CASE I. 



170 



ALOKBBA. 



EXAMPLES. 

1. To divide Gab by 3a. 

Here Gab -e- 3a, or 3a ) Gab ( or ^ = 26. 

~ c , , , • abx a 

2. Also c-t-c = - = 1; and ate wry = 7— = 

c oxy y 

3. Divide lBx 3 by 8x. Ans. 2*. 

4. Divide 12a 8 * 2 by — 3a*x. Ans. — 4x. 

5. Divide— 15ay* by 3ay. Ans. — 5*> 

9sy 

6. Divide — 18ax*y by — Saxx. Ans. 



CASE. II. 

When the Dividend i* a Compound Quantity, and the Divitor 
a Simple one. 

Divide every term of the dividend by the divisor, as in 
the former case. 

EXAMFLE8. 

1. (ai + ft a )H-26,or^ a =^ = *a + i6. 

2. (10*6 + 15a*) -s- 5a, or 10fl6 + 15fl * =26 + 3*. 

3. (30az->48z) -f *, or 30g *~ 48 * = 30a-48. 

4. Divide 6ao— 8ax + a by 2a. 

5. Divide Sx'-lS + 6x + 6a by 3x. 



1. When both the terms are +, tbe quotient must be + ; because + 
in the divisor X + in the quotient, produces -j- in the dividend. 

2. When the terms are both — , the quotient is also + ; because — 
in the divisor X — in the quotient, produces -f- ia the dividend. 

3. When one term is-f and the other—, the quotient must be--; 
because -f in the divisor X — in the quotient produces — in the divi- 
dend, or — in the divisor X + in the quotient gives — in the dividend. 

So that the rule is general ; vis. that like signs give +, and unlike 
**ffn* give — , in the quotient. 



DIVISION. 



6. Divide 6abc + \2abx — 9a*b by Sab. 

7. Divide lOa*x - 25* by 5x. 

8. Divide I5a 2 be — Ibacx* + bad? by — 5ac. 

9. Divide 15a + Say - lSy 8 by 21a. 
10. Divide - 2(W 8 6 8 + OOab 3 by - Qab. 



CASE III* 

TFAen Divisor and Dividend are both Compound 
Quantities. 

1. Set them down as in common division of numbers, 
the divisor before the dividend, with a small curved line be- 
tween them, and range the terms according to the powers of 
aomo one of the letters in both, the higher powers before the 
lower. 

2. Divide the first term of the dividend by the first term 
of the divisor, as in the first case, and set the result in the 
quotient. 

3. Multiply the whole divisor by the term thus found, and 
subtract the result from the dividend. 

4. To this remainder bring down as many terms of the 
dividend as are requisite for the next operation, dividing as 
before ; and so on to the end, as in common arithmetic. 

Note. If the divisor be not exactly contained in the divi- 
dend, the quantity which remains after the operation is 
% finished may be placed over the divisor, like a vulgar frac- 
tion, and set down at the end of the quotient as in arith- 
metic. 



EXAMPLES. 

a - b) a a - %tb + b* (a - b 
a 3 - ab 



-ab + V 
- ab + ^ 



Vol. I. 



24 



178 ALGEBRA. 

a - c) a 3 — 4a 9 c + 4^ - c 3 — (a 3 — 3ac + 



- 3« 3 c + W 

- 3a*c + 3'^ 



oc 3 -c 1 
ac 3 - c 3 



«_ 2) a 3 — 6a 3 + 12a - 8 (a 3 - 4a + 4 
a 3 -2d 3 



- 4a 2 + 12a 

- 4a 3 + 8a 



4a — 8 

4a — 8 



« + *) a * - 3x 4 (a 3 - a 3 * + ax 3 — ~ 3 - 



a 4 + a 3 x a + * 



— a 3 * - 3x* 

— a 3 x — aV 



aV - 3x 4 
a 3 x 3 -f • ax 3 



— ar 3 - 3x* 



- 2x 4 



EXAMPLES FOB PRACTICE. 

1. Divide a 3 + 4ax + 4x 3 by a + 2x. Ans. a + 2*. 

2. Divide a 3 - 3a 3 z + 3a* 3 - z 3 by a - z. 

Ans. a 3 — 2az + z\ 

3. Divide 1 by 1 + a. Ans. 1 - a + a 2 — a 3 + &c. 

4. Divide 12x 4 — 102 by 3x - 6. 

Ans. 4x 3 + Sx 2 + 16x + 32. 

5. Divide a 5 - 5a 4 6 + 10a 3 6 3 - 10a 3 6 3 -f 5a6 4 — b* by 

a 3 - 2a& + b\ Ans. a 3 - 3a 3 6 + Sab 2 - 6'. 



FRACTIONS. 179 

6. Divide 48s 3 - 96a* 3 - 64a 2 z + 150a 3 by 2z - 3a. 

7. Divide b* — 36 V + 36 2 x 4 — x*by 6 3 — 2b'x + 36s 3 

— x 3 . 

8. Divide a 1 — x 7 by a — x. 

■ 9. Divide a 3 + 5a ? x + 5ax 3 + x 3 by a + x, 
10. Divide a 4 + 4a 2 6 3 — 326* by a + 26. 
Ik Divide 24a 4 — 6< by 3a — 26. 



ALGEBRAIC FRACTIONS. 

Algebraic Fractions have the same names and rales of 
operation, as numeral fractions in common arithmetic ; <as 
Appears in the following Rules and Cases. 



To reduce a Mixed Quantity to an Improper Fraction. 

Multiply the integer by the denominator of the fraction, 
and to the product add the numerator, or connect it with 
its proper sign, 4" or — ; then tho denominator being set 
under this sum, will give the improper fraction required. 

EXAMPLES. 

1. Reduce 3}, and a ^- to improper fractions. 

OA (3X5) +4 15 + 4 10 , . 
First, 3| = ~- — = — = — the Answer. 

And, a — - = ( aXar )~ ^_ = — — - the Answer. 
xx x 
a 2 z i a a 

2. Reduce a + -p and a to improper fractions. 

o a 

«. . , a 2 (a X 6) + a 3 a6 + a 2 , A 



. 180 ALGEBRA. 

3. Reduce 5f to an improper fraction. An*. V • 

4. Reduce 1 — ^ to an improper fraction. Ans. 

5. Reduce 2a — ** gx a ^ an improper fraction. 

4x 
4r 18 

6. Reduce 12 -| jr- — to an improper fraction. 

1 — 3a c 

7. Reduce x H — to an improper fraction. 

2r* 3a 

8. Reduce 4 + 2x : — — to an improper fraction. 

CA8k II. 

To reduce an Improper Fraction to a Whole or Mixed 
Quantity. 

Divide the numerator by the denominator, for the inte- 
gral part; and set the remainder, if any, over the denomina- 
tor, for the fractional part ; the two joined together will be 
the mixed quantity required. 



EXAMPLES. 

1. To reduce and to mixed quantities. 

o o 

First, y = 16 -T- 3 = 5J, the answer required. 
And, ^^g— = (ab + a 2 ) -r- b = a + Answer. 

2ac — 3a 3 ,3ax + 4x f 

2. To reduce und , to mixed quanti- 

c a + x 

ties. « 

2ac — So 2 M n nK 3a 2 _ 
First, = (2ac — 3a 2 ) c = 2a . Answer. 

3ax+4x a x 2 

And, Ip_==(3ax+4x 2 Wa+s)=3x+— - Ans. 

a+x x / \ / a+x 

_ , 33 _ 2ax — 3x> . , 

3. Reduce — and to mixed quantities. 

3x* 

Ans. 6|, and 2x — . 

4a a x 2n 2 -f- 26 

4. Reduce and — - — - — to whole or mixed quan- 

dties. 



FRACTIONS. 181 

5. Reduce **** , ^ and ^ ^ t o whole or mixed 

x+y x—y 

quantities. 

a D . 10a 1 — 4a + 6 4 . , 

o. Reduce — to a mixed quantity. 

7. Reduce r- rr~ a « to a mixed quantity. 

Sa 1 + 2a 9 — 2a — 4 n ' 



case in. 



To reduce Fractions to a Common Denominator. 

Multiply every numerator, separately, by all the deno- 
minators except its own, for the new numerators ; and all the 
denominators together, for the common denominator. 

When the denominators have a common divisor, it will be 
better, instead of multiplying by the whole denominators, to 
multiply only by those parts which arise from dividing by 
the common divisor. Observing also tho several rules and 
directions, as in Fractions in the Arithmetic. 



EXAMPLES. 

1. Reduce - and - to a common denominator. 

x z 

Here - and - = a% and — , by multiplying the terms of 
x z tz xz 

the first fraction by z, and the terms of the 2d by x. 

2. Reduce -, and — to a common denominator. 

x o c 

n a x j & abc ex 7 , b*x . 

Here - , - , and — = — and 7 — , by multiplying the 

x b c bcx bcx bcx r J ° 

terms of the 1st fraction by be, of the 2d by cx, and of the 3d 
by bx. 

3. Reduce — and ^ to a common denominator. 

x 2c 

4ac , Sbx 
Ans. — - and — . 
2cx 2cx 

4. Reduce ^ and — - to a common denominator. 

b 2c 

k 4ac , 3ab+2b* 
Ans.^and— b — , 



182 ALGEBRA. 

5a 36 

5. Reduce and and 4 a*, to a common denominator. 

Sx 2c 

lOffc ,06a: J 24cox 
Ans. - — and — and — — • 
hex ocx ocx 

6. Reduce jj and and 26+ ~, to fractions having a com- 

4 206 ,18a6 J 486 2 +72a 

mon denominator. Ans. - — and --r-, and — -rn . 

246 246 246 

1 2a 2 2tt 3 +6 3 

7. Reduce - and — and — r-r~ to a common denomina- 

3 4 o+6 

tor. 

8. Reduce — and and ^- to a common denominator. 

4u s 3a 2a 



CASE IV. 



Tb find the greatest common Measure of tlie Terms of a 
Fraction. 

Divide the greater term by the less, and the last divisor 
by the last remainder, and so on till nothing remains ; then 
the divisor last used will be the common measure required ; 
just the same as in common numbers. 

But note, that it is proper to range the quantities accord- 
ing to the dimensions of some letters, as is shown in division. 
Note also, that all the letters or figures which are common 
to each term of the divisors, must be thrown out of them, or 
must divide them, before they are used in the operation. 



EXAMPLES. 



1. To find the greatest common measure of — ^r-r „» 

ar J +6c J 

a6 + 6 3 ) or 2 + be 2 
or a + 6 ) ac 2 + bc z (c 2 
ac 3 + 6c 2 



Therefore the greatest common measure is a + 6. 



a 3 — n6 a 



2. To find the greatest common measure of — — -r-j-r,. 

a-+2a64-6 1 



FRACTIONS. 188 



a*+2ab+P) a 3 — <io 3 (a 



— tab— 2ab 2 ) a 2 4- 2*6 + 6 3 
or a-f 6 ) <r + 2ab + b* (a + b 
a m 4- a6 



oA + ft 2 
ab + b 1 

Therefore a-\-b is the greatest common divisor. 

a 3 — 4 

3. To find the greatest common divisor of ^q^- 

Ans. a — 2» 

4. To find the greatest common divisor of "r^gr- 



5. Find the greatest com. measure of —-—r.-r.r-r —r-—7-r m • 

CASK V. 

To reduce a Fraction to its lowest Terms, 

Find the greatest common measure, as in the last pro- 
blem. Then divide both the terms of the fraction by the com- 
mon measure thus found, and it will reduce it to its lowest 
terms at once, as was required. Or divide the terms by any 
quantity which it may appear will divide them both as in 
arithmetical fractious. 

EXAMPLES. 

, n , ab+b* . t 
1. Reduce — z~~r~. — :» to its lowest terms. 
ac 2 +bc* 

ab+b 2 ) ar'+bc 1 
or a +6 ) ucP+bc 2 (c 3 
ac-+b(r 



Here ab-\-b 2 is divided by the common factor b. 
Therefore a + b is the greatest common measure, and 

hence a+b) rf ^^", ^r=-? > is the fraction required. 

acr-\-bcr <r 

S-b-c . , 
2. To reduce -r-r-r-. . ... to its least terms. 
cr+%bc+b* 



184 ALOEBRA. 

Hero, by a process similar to that of Ex. 2, Case rv. 9 we 
find c + b is the greatest common measure, and hence 

e + b) -f ° -;.= C . ^ C is the fraction required. 
T ' c*+2bc + b 9 c+b M 

c 3 — ft 3 c*+bc+V 

3. Reduce -r — rrs t0 its lowest terms. Ans. . -?-,-» 

£i J 

4. Reduce n to its lowest terms. Ans. 

a' — o ar-x-or 

a 1 — o 4 . , 

5. Reduce -= — 5-^-7-0—53 — TJ to lts lowest terms. 

a 3 — Serb +3*o a — o J 

6. Reduce - 1 - T - -~ i - r . , , . to its lowest terms. 

a 3 c +3aV+3ac 3 +c* 



CASE VI. 

To aaV2 Fractional Quantities together. 

If the fractions have a common denominator, add all the 
numerators together ; then under their sum set the common 
denominator, and it is done. 

If they have not a common denominator, reduce them to 
one, and then add them as before. 

EXAMPLES. 

1. Let ~ and be given, to find their sum. 

a . a 4a , 3a 7a . 
Here — + — = ^ + «F is the sum required. 

2. Given — , and to find their sum. 

be a 

_.. a , 6 c aco* , 66d ,bcc acd4-6W+occ 

Uere y + 7 + 7 ,= kS + 6^ + 6^ "635" 

the sum required. 

3s 2 2ax 
* 3. Let a — and 6 + be added together. 



* In the addition of mixed quantities, it is best to bring the fractional 
parts only to a common denominator, and to annex their sum to the ram 
of the integers, with the proper sign. And the same rule may be ob- 
served for mixed quantities in subtraction also. 

See, also, the note to Addition of Fractions in the Arithmetic. 



FRACTIONS. 185 

Here a — r- + 6 H = a 7 r + — — 

b c be be 

. . , 2abx-3cx* . . 

= a + H £ 9 the sura required. 

be 

, ...4* . . 20ta+6ax 

4. Ada jr- and -=r together. Ans. - — — — 7 — . 

3a 56 6 15a6 

5. Add ~, ~ and together. Ans. £{a. 

« AJJ 2a— 3 .5a . . 9a -6 

6. Add — - — and — together. Ans. — - — . 

7. Add 2a -| — to 4a -\ Ans. Ga Hf ^ — . 

O 4 *U 

8. Add 6a, and ^ and -^jp together. 

o a^ 5 * j 6a ,3a+2 4 

9. Add — , and — and — — — together. 

10. Add 2a, and ~ and 3 + ^ together. 

11. Add 8a + ^ and 2a — ^ together. 



, CASE VII. 

To subtract one Fractional Quantity from another. 

Reduce the fractions to a common denominator, as in 
addition, if they have not 1 a common denominator. 

Subtract the numerators from each other, and under their 
difference set the commoi denominator, and the work is 
done. 

EXAMPLES. 



3a 4a 

1. To find the difference of — andy. 

„ 3a 4a 21a 16a 5a . . J ir . , 

Here — — y = — - — — - = ^ is the d ffer. lequired. 

2* To find the difference of ^7 -- - and 

4c 36 

Vol. I. 25 



difference of &» ■»* "4* 

4. ^ ^ 4 * 

7. T«k« — 



9 2a 
from 4a c 

cask vin. 



Beto-g-Jt5 40 , ______ 



to* 



FRACTIONS. 



187 



2. Required the product of |, ~, and 5?. 
TxTxT "^-H^e product required. 

3* Required the product of ~ and !^p~* 

„ 2m x (a + b) 2aa + 2ab. _ _ 
H6r6 6— (2T+T) = 23T+-£ ^ produCt 

4a 6a 

4. Required the product of 7 and -r- • 

O DC 

5. Required the product of ^ and —j*. 

6. To multiply y, and ~ 9 and ^ together. 

ao 3a* 

7. Required the product of 2a + -g^and 

o. Required the product of — — and ^ ^ ». 

2a 4-1 2a— 1 
9. Required the product of 8a, and — - — and 

10. Multiply« + ±-^byr-£ + ^. 



CASE IX. 



To oroide one Fractional Quantity by another. 

Divide the numerators by each other, and the deno- 
minators by each other, if they will exactly divide. But, if 
•Hot, then invert the terms of the divisor, and multiply by it 
exactly as in multiplication*. 



* 1. If the fraction! to be divided have a common denominator, take 
the numerator of the dividend for a new numerator, and tfae numerator 
"Of the divisor for the new denominator. 

ft. When a fraction is to be divided by any quantity, it is the tame 
thing whether the numerator be divided by it, or the denominator mul- 
tiplied by it 

8. When the two numerators, or the two denominators, can be divi- 
ded by some common quantity, let that be done, and the ajtttUM&iwA ' 
Instead ef the fractions first propweiL 



188 



ALGEBRA. 



EXAMPLES. 



a 3a 

1. Required to divide - by 

a 3a a 8 8a 2 . 
Here i + T = 4 X 3a = I2S " 3 * e qU0Uent ' 

2. Required to divide ?^ by 

w 3a 5c 3a 4d I2ad 6ad _ 

Here » + 45 = 26 X 5c = TOfc = 56? the qUOUenU 

« m j 2a+6 _ 3a+26 „ 

3. Todmde— - 6 by— Here, 

2a+6 w 4a+6 8a*+6ab+b* _ 
3a--26 X 3a+26 = Tra*- tbe qUOtMmt n ^ md ^ 

4. To divide ^r- 3 by- a 



a 3 +6 3 J a+&' 

H — a+6_3a a X(a+6)_ 3a . 

6 aM-6> X a (a 3 +6 3 ) Xa a* — ab + 6* 18 
quotient required. 

5. To divide — by ~. 

4 J 12 

6. To divide -f by 3x. 

5 

7. To divide — - — by — « 

8. To divide jr^-- Ly^. 

4>X. — 1 u 

9. To divided by I?. 

5 3 56 

10. Todividc^by^-. 

4ca oa 

„ _. . . 5a 4 - to f5a6 

11. Dmde^-^^by 



160 



INVOLUTION. 

Involution is the raising of powers from any proposed 
root ; such as finding the square, cube, biquadrate, &c. of 
any given quantity. The method is as follows. 

* Multiply the root or given quantity by itself, as many 
times as there are units in the index less one, and the last 
product will be the power required. Or, in literals, mul- 
tiply the index of the root by the index of the power, and 
the result will be the power, the same as before. 

Note. When the sign of the root is +, all the powers of 
it will be + ; but when the sign is — , all the even powers 
will be +, and all the odd powers — ; as is evident from mul- 
tiplication. 



examples. 



a, the root 
a 3 3= square 
a 3 = cube 
a 4 = 4th power 
a 1 = 5th power 
6zc. 


a 9 , the root 
a 4 ~ square 
a 6 = cube 
a" = 4th power 
a ,0 = 5th power 
dec. 


— * 2a, the root 
+* 4a a = square 

— 8a 3 =- cube 

+ 16a 4 = 4th power 

— 32a 5 = 5ih power 


— 3ab*, the root 
+ 9c?b* = square 

— 270*6*= cube 

+ 81a 4 6 f = 4th power 
— 243a , 6 ie = 5th power 


2ar 2 
— the root 

, 4aV 

+ .— = square 

8aV 

— = cube 

276* 

+ - Q j- 6 T 4th P ower 


^g, the root 
a a 

4^ = square 

?L =cube 
a 3 

166 4 = bi( l uadrate 



* Any power of the product of two or more quantities, is equal to 
the tame power of each of the factors, multiplied together. 

And any power of a fraction, is enual to the same noyr*T cA tab trap 
aerator, divided by the like power of the denominator* 



190 ALOBBBA. 

x — a =root x+ a=root 

x — a x+a 



x 2 — ax x 9 + ax 

— at + a* +ax + a* 



X s — 2ax + a 2 square x* + 2ax + a 9 

x — a x + a 



x 5 — 2ax a +o a x x 3 + 2a* 8 f a 1 * 

—ax 2 +2a»x— a 3 + ax 3 + 2a a x + «* 



x 3 — 3ax* "4* Sa'x-— a 3 x 3 + Sax 2 + So 3 * + a 3 



the cubes, or third powers, of x — a and x + a. 



KXAMJ LES FOR PRACTICE. 

1. Required the cube or 3d power or 3a a . 

2. Required the 4th power of 2a s b. # 

3. Required the 3d power of — 4a a o 3 . 

Q^X 

4. To find the biquadrate of — ~ a . 

5. Required the 5th power of a — 2x. 

6. To find the 6th power of 2a*. 

Sir Isaac Nekton's Rule for raising a Binomial to*any 
Power whatever* : . # 

1. To find the Terms without, the Co-efficients. The inder. 
of the first, or leading quantity, begins with the index of the 
given power, and in the succeeding terms decreases con- 
tinually by 1, in every term to the last ; and in the 2d or 



Alto, powers or roots of the same quantity, are multiplied by one> 
another, by adding iheir exponents ; or divided, by subtracting their 
exponents. 

aa 

Thus, «3 X«* = aa+a = as. And as -f-aa or — = as— a — a . 

as 

* This rule, expressed in general terms, is as follows: 
<.+x)*=a^». a-ix+n . "^a^a-j-n. 'I" 1 . "^-v ^ 

<«-^)»^-«.a»-ix4-n . *^ l a*~ z"-n. *~ l . n ^ 2 a --y Ac. 
/toil. Xb* ma of the co-efficienU, In every power, it equal Co ther- 



INVOLUTION. 



101 



following quantity, the indices of the terms are t 1,2, 3, 4, 
&c. increasing always by 1. That is, the first term will con- 
tain only the 1st part of the root with the same index, or of 
the same height as the intended power : and the last term- of 
the series wUl contain only the 2d part of the given root, 
when raised also to the same height of the intended power : 
bat all the other or intermediate terms will contain the pro- 
ducts of some powers of both the members of the root, in 
such sort, that the powers or indices of the 1st or leading 
member will always decrease by 1, while those of the 2d 
member always increase by 1. 

2. To find the Co efficients. The first co-efficient is al- 
ways 1, and the second is the same as the index of the in- 
tended power ; to find the 3d co-efficient, multiply that of the 
2d term by the index of the leuding letter in the same term, 
mod divide the product by 2 ; and so on, that is, multiply the 
co-efficient of the term last found by the index of the leading 
quantity in that term, and divide the product by the number 
of terms to that place, and it will give the co-efficient of the 
term next following ; which rule will find all the co-efficients, 
one after another. 

Note. The whole number of terms will be 1 more than 
the index of the given power : and when both terms of the 
root are + , all the terms of the power will be + ; but if the 
second term be — , all the odd terms will be + > and all the 
even terms — , which causes the terms to be + and — alter- 
nately. Also the sum of the two indices, in each term, is 
always the same number, viz. the index of the required 
power ; and counting from the middle of the series, both 
ways, or towards the right and left, the indices of the two 
terms are the same figures at equal distances, but with mutu- 
ally changed places. Moreover, the co-efficients are the 
same numbers at equal distances from the middle of the se- 
ries, towards the right and left ; so by whatever numbers 
they increase to the middle, by the same in the reverse order 
they decrease to the end. 



number 2, when raised to that power. Thus 1— |— 1 = 2 in the first power ; 
1+2+1 =4=2- in the square; 1 +3 + 3 + 1 = 8 =2* in the cube, 
or third power : and so on. 

A. trinomial or a quadrinomial may be expanded in the same manner. 
Thua, to raise a — ft + c — d to the 6th power, put a — b = x, e — d—z, 
mod raise i + z to the 6th power ; after which substitute for the powers 
of x and y their corresponding values in terms of o— b, aad c— d, and 
their powers respectively. 



192 



AXGKBRA. 



EXAMPLES. 

1. Let a + x be involved to the 5th power. 

The terms without the co-efficients, by the 1st rule, 
will be 

a 5 , a 4 *, aV, aV, or 4 , **, 
and the co-efficients, by the 2d rule, will be 
5Xj 10X3 10X2 5X1 

it 5, , — 3—, — — , — — ; 

or, 1, 5, 10, 10, 5, 1 ; % 

Therefore the 5th power altogether is 

a 5 + 5a 4 * + lOaV + lOaV + 5a* 4 + X s . 

But it is best to set down both the co-efficients and the 
powers of the letters at once, in one line, without the inter* 
mediate lines in the above example, as in the example here 
below. The operation is very easily effected by performing 
the division first. 

2. Let a — x be involved to the 6th power. 

The terms with the co-efficients will be 
a 8 — 6a 5 * + 15aV-20aV + 15aV — 6a* 5 + a*. 
8. Required the 4th power of a — x. 

Ans. a 4 — 4a 3 x + 6aV- 4ax 3 + ac< 
And thus any other powers may be set down at once, in 
the same manner, which is the best way. 

4. Involve a — * to the ninth power ; x — y to the tenth 
power, and a + b — c to the fourth power. 



EVOLUTION. 

Evolution is the reverse of Involution, being the method 
of finding the square root, cube root, 6zc. of any given quan- 
tity, whether simple or compound. 

case 1. To find the Boots of Simple Quantities. 

Extract the root of the co-efficient, for the numeral part * r 
and divide the index of the letter or letters, by the index oC 



EVOLUTION. 



193 



the power, and it will give the root of the literal part ; then 
annex this to the former, for the whole root sought*. 



EXAMPLES. 



1. The square root of 4a 3 , is 2a. 

2. The cube root of 8a 3 , is 2a' or 2a. 

3. The square root of -ttt-, or \Z~7Tn~9 is — v" 5. 

16a 4 6 8 2ab 2 

4. The cube root of — -~--z- 9 is ^ - 2/2a. 

27c 3 3c v 

5. To find the square root of 2db\ Ans. a6 3 v/2. 

6. To find the cube root of - 64a <6 8 . Ans. -4ci6 3 . 

~ m ^ i , „8dr6' 2a6 2 

7. To find the square root of -57-7-. Ans. — J - . 

3c J c 3c 

a To find the 4th root of 81aW. Ans. 3aV&. 

9. To find the 5th root of — 32aW. Ans. — 2ab*/b. 



CASE II. 

To find the square root of a Corn-pound Quantity. 

This is performed like as in numbers, thus : 

1. Range the quantities according to the dimensions of 
one of the letters, and set the root of the first term in the 
quotient. 

2. Subtract the square of the root, thus found, from the 
first term, and bring down ihe next two terms to the re- 
mainder for a dividend ; and take double the root for a di- 
visor. 



* Any evpn root of nn affirmative quantity, 1. ay be ml'.ier -j- or — : 
thus the square root of -|- a* is either + a i or — a » because -f- a X + 
a = -f- «a . and — a < — a = -f- a* aUo. 

But an odd root of any rj'ianti'y will have the fame s : gn as the quan- 
tity itsjlt : thus tbe cube root of -f- a* is -J- .i, and tue cub'j root of — a * 
k — a ; for -r- >l X + « X + * - 4~ a * » n,, d — a * — '* X — a — — « 3 • 

Any evon root of a negative quantity ia imp' i«ib'e ; for neithe; -|- a 
X+ «t nor — a v — a can prr.cHce — /?a. 

Any root of a prod ict is cq.ial to the like root o each of l! 1 f -jcto-s 
moltipl-pd together. Tor the root of a fraction, *ake the roM of t'ie 
numerator and the ro-^t of the dciominutor. 
Vol. 1. 



/ 



104 ALGEBRA* 

3. Divide the dividend by the divisor, and annex the re- 
sult both to the quotient and to the divisor. 

4. Multiply the divisor, thus increased, by the term last 
set in the quotient, and subtract the product from the divi- 
dend. 

And so on, always the same, as in common arithmetic. 



EXAMPLES. 

1. Extract the square root of a 4 — 4a 5, &+6a s &» — 4ab*+b*. 
a 4 — 4a*6 + 6a»6« — 4a6» + 6 4 (a" — tab + b* the root. 



2f—2ab)—4a*b + 6a 2 6 3 
— 4a 3 6 + 4a 2 6* 

2a»_4a* + 6 s ) 2a 2 6 a — 4ab> + 6 4 
2a*b 2 — 4a6 3 + b l 



2. Find the root of a 4 + 4tfb + 10a 2 6 s + 12a6 3 + 96*. 
a* + 4a 3 * + lOaV + 12a6 3 + 96 4 (a 2 + 2a£ + 3^ 



2a 2 + 2a6) 4a?b + lOa 2 * 2 
4a 3 6 + 4a a 6 a 



2a a + 4a& + 36 2 ) 6a 8 6 2 + 12aft 3 + 96 4 
6aW + 12a6 3 + 9b* 



3. To find the square root of a 1 + 4a 3 + 6a 2 + 4a + I - 

Ans. a 2 + 2a + 1 - 

4. Extract the square root of a 4 - 2a 3 + 2a 2 — a + J. 

Ans. a 2 — a + ■£ — 

5. It is required to find the square root of a 3 — ab. 



CASE III. 

To find the Boats of any Powers in general. 

This is also done like the same roots in numbers, thus ; 

Find the root of the first term, and set it in the quotie*"***' 
—Subtract its power from that term, and bring down tJ^* 
second term for a dividend. — Involve the root, last found, 
Jbe next lower power, and multiply it by the index of ***** 



EVOLUTION* 



105 



given power, for a Jivisor. — Divide the dividend by the di- 
visor, and set the quotient as the next term of the root. — 
Involve now the whole root to the power to be extracted ; 
then subtract the power thus arising from the given power, 
and divide the first term of the remainder by the divisor first 
found ; and so on till the whole is finished *. 

EXAMPLES. ' 

1. To find the square root of a 4 — 2a' , &+3aV— 2aft 3 +tt 
««_2a b + 3a*V— 2aP + b* {a 3 —ab + &*, 



2a 3 )— 2a b 



** 2a b + a fl 6 > = (a 8 — ab)' 



2a 3 ) 2a a #» 

<t—2ab + 3a a 6 a — 2a6 3 + © 4 = (a* - ab + b 3 f. 

2. Find the cube root of a 9 - 6a 5 + 21a 4 -44a 3 + 63a* — 
54a + 27. 

a 8 — 6a» + 2 la 4 — 44a 3 + 63a 2 — 54a + 27 (a 3 — 2a+3. 



a 9 



3a 1 ) — 6a 5 



a*— 6a 5 + 12a 4 — 6a 3 = (a 3 — 2«) 3 



3a 4 ) + 9a 4 



a t -fa>+*\a*--44a : ±G3a 3 --54a+2rr= (a 3 — 2a+3)\ 



* As this method, in high powers, may be thought too laborious, it 
will not be improper to observe, that the roots of compound quantities 
may sometimes be easily discovered, thus: 

Eitract the roots of some of the most simple terms, and connect 
them together by the sign -f or — , as may be judged most suitable for 
the purpose.— Involve (he compound root, thus found, to the proper 
power ; then, if this be the same with the given quantity, it is the root 
required. — But if it be found to differ only in some of the signs, change 
them from -f to — , or from — to + , till its power agrees with the 
given one throughout. 

Thus, in the 5th example, the root 3a — 26, is the difference of the 
toots of the first and last terms ; and in the 3d example, the mot 
ft*— 1 * + x, is the sum of the roots of the 1st, 4th, and 6th terms. The 
'Ipi nay also be observed of the 6th example, where the KOQttofo^k 
'3HM'Umi first and last terms. t 



106 ALGEBRA- 

3. To find the square root of a' — 2ab + 2ax + P — 
2bx + x* Ans. a — b + x. 

4. Find the cube root of a 6 — 3a 5 + 9a 4 — - l&r + 18a 3 — 
12a + 8. Ans. a 3 — a + 2. 

5. Find the 4th root of 81a 4 — 216a 3 6 + 216a 3 6 3 - 90a6. 
+ 166 4 . Ans. 3a - 26. 

6. Find the 5th root of a s — 10a 4 + 40a 3 — 80a 3 + 80a 
— 32. Ans. a — 2. 

7. Required tho square root of 1 — **• 

8. Required the cube root of 1 — x\ 



SURDS. 

Sum>s are such quantities as have no exact root ; and are 
usually exprcbded by fractional indices, or by means of the 

radical sign y/. Thus, 3^, or ^3, denotes tho square root 

of 3 ; and 2 1 , or ^/2 3 , or ^/4. the cube root of the square of 
2; where »ht numer'or shows tho power to w! 'zh the 
quantity is to be raised, and the denominator its root. 



PROHLFM I. 



To reduce a Rational Quantify to the Farm of a Surd. 

R * ise the piver* quantity to the power denoted by the 
index of the surd ; then over or above the new quantity set 
the radical sign, an I it will be of the form required. 



EXAMPLES. « 

1. To reduce 4 to the form of the square root. 
First, 4 2 — 4 X 4 ■= 10 ; then y/ J 6 is the answer. 

2. To reduce 3<r to the form of the cube root. 
First 3<r X 3i 3 =^ X 3/7 = \3a>) ' = 27 a G ; 

then \y27a* or (27a*)* is the answer. 

3. Reduce C to the form of the cube root. 

Ans. (210)* or ?/216. 

4. Reduce lab to the form of the square root. 



8VXDI. 197 

5. Reduce 2 to the form of the 4th root. Ans. (16)*. 

6. Reduce a 4 to the form of the 5th root. 

7. Reduce a + x to ihe form of the square root. 

8. Reduce a — * to the form of the cube root. 



PROBLEM II. 

To reduce Quantities to a Common Index. 

1. Reduce the indices of the given quantities to a com- 
mon denominator, and involve each of them to the power 
denoted by its numerator; then 1 set over the common de- 
nominator will form the common index. Or, 

2. If the common index be given, divide the indices of the 
quantities by the given index, and the quotients will he the 
new indices for those quantities. Then over the said quan- 
tities, with their new indices, set the given index, and they 
will make the equivalent quantities Bought. 

EXAMPLK8. 

1. Reduce 3* and 5* to a common index. 
Here £ and } = J z and fy. 

Therefore 3^ and 5^ = (3')^ and (5*)^ = V 5 and 1 
= ■ J/243 and l {/25. 

2. Reduce a* and b* to the same common index |. 
Here, f •+ $ = J X \ = f the 1st index, 

and Jr} = iX \ = | the 2d index. 

Therefore (a 6 )' and (6*)', or yAr 5 and are the quanti- 
ties. 

3. Reduce 4* and 5* to the common index \. 

Ans. (256*)* and 25*. 

4. Reduce afc and r* to the common index j-. 

Ans. (a 1 )* and (x')*. 

5. Reduce d 2 and x 3 to the same radical sign. 

Ans. y/a* and y/x*. 

6. Reduce (a + x)* and (a — x)^ to a common index. 

7. Reduce (a + b)^ and (a — t)* to a comumVbfa** 



1G8 



alg::bra. 



l'RORI.r.M nr. 

To reduce. Surds to more. Simple Terms. 

Divide the s'.ird, if possible, into two factors, cne of which 
is a power of the kind that accords wiih the root sought ; as 
a complete square, if it be a square root, a complete cube, 
if it he a cube root ; and so on. Set the root f this com- 
plete power before the surd expression which indicates the 
root of the other factor ; and the quantity is reduced, ad re- 
quired. 

If the surd be a fraction, the reduction is effected by mul- 
tiplying both its numerator and denominator by some number 
that will transform the denominator into a complete square, 
cube, &c. its root will be the denominator to a fraction that 
will stand before the remaining part, or surd. Sec Example 
3, below. 

EXAMPLKS. 

1. To reduce v /32 to simpler terms. 

Here ^32 = ^(10X2) = ^/lO x ^/2~4 X v/2 =4 </2. 

2. To reduce ^/3*20 to simpler terms. 

V320 =y(t)4 x 5) = yoi x y:> = i x y5 =^ 4 ys. 

3. Reduce -/J* to simpler terms. 

44 44 , 4i 4.11 H _ 2 2 .55 _ 

IB'* 5 ' 

4. Reduce v/75 to its simplest terms. Ans. 5^/3. 

5. Reduce ^/189 to its simplest terms. Ans. 3^/7. 
0. Reduce '^/\;\f to its simplest terms. Ans. -JJ/10. 
7. Reduce ^/loarh to its simplest terms. Ans. f>tf v /36. 

Note. There are other cases of reducing algebraic surds 
to simpler forms, that are practised on several occasions ; 
one of which, on account of its simplicity and usefulness, may 
be here noticed, viz. in fractional forms having compound 
surds in the denominator, multiply both numerator and de- 
nominator by the same terms of the denominator, but having 
one sign chanced, from to — or from — to +, which will 
reduce the fraction »■> a rational denominator. 

Ex. To reduce ^ 2 ?- + -^-~, multiply it bv ^ and it 



becomes - - — = 8 + 2 v /l. r >. 

49 



SURDS. 



190 



* „ 3v/15 — 4*'5 ,. , . _ «/lf> — y5 

■ Also, to reduce - _ — ; multiply it bv -- — — and 

y/lr +■ v 'o * ,/la— ^ 

. , 65—7,770 65 -33 v /3 13— 7,/3 

it becomes — - - — — = — . 

15— o 10 2 

And the same method may easily be applied to examples with 
three or more surds. 



PROBLK3I IV. 

To add Suid Quantities together 

1. Bring all fractions to a common denominator, and 
reduce the quantities to their simplest terms, as in the last 
problem. — 2. Reduce also such quantities as have unlike 
indices to other equivalent ones, bavin" a common index. — 
3. Then if the Mini part be the same in them all, annex it 
to the sum of the rational pars. \vi;h the sign of multiplica- 
tion, and it will give the total sum recr-iired. 

But if the surd pan be not ilu> s«ni<' in all the quantities, 
they can only be added by the signs + and — . 

KX AMI* LI-IS. 

1. Required to add ^/IS and X /S2 together. 
FirstV18=V(9X2; = %/2; ™<l — y/{WX2)^4y/2: 
Then, 3y2 +4^/2= (3+4)^/2=7^2= sum required. 

2. It is required to add */375. and yi92 together. 
First,y375=^/( 1 2f> X 3 ; =5y 3:and y 102=^/(0.1x3) =4^/3: 
Then, 5^3 + 4^3 = (5+ l)y.*J 9y3 = sum required. 

3. Required the sum of\/27 and x /4t\ Ans. 7 y/S, 

4. Required the sum of \/f>0 and >/72. Ans. 11^/2. 

5. Required the sum of and v'rs- -^ ns - t^V^' 
G. Required the sum of -y 50 and i[/lbli. Ans. 5|/7. 

7. Required the sum of \/ \ and \/ , ! f Ans. -Jy2. 

8. Required the sum of 3^/a-ft and SyMbV/j. 

fruhlfm v. 
7'ri t //W the. Diijcrvncc of Surd Quantities. 

Pre pa it k th^ quantities the same w:«y as in the last rule : 
then subtract the rational parts, and to the remainder acuiftx. 
the common-surd, for the difference of the. surtta tec^ivt^* 



200 ALGEBRA. 

But if the quantities have no common surd, they can only 
be subtracted by means of the sign — -. 

EXAMPLES. 

1. To find the difference between v/320 and ^/80. 
First, v/320=V(64 X 5) = 8 ✓ 5;and /S0= % /{ 16X5) =4^5. 
Then, 8^/5 — 4^/5 = 4^/5 the difference sought. 

2. To find the difference between yi'ZS and ^/54. 
Fint,l/128=i/(64*2)=4y2; andy54 = V(27x2)=3y2- 
Then, 4J/2 - 3^/2 = the difference required. 

3. Required the difference of y/75 and ^/48. Ans. ^/3. 

4. Required the difference of $/256 and ]/32. Ans 2^4. 

5. Required the difference of y/% and Ans. Jv'S. 

6. Find the difference of t/| and yf. Ans. ^^/^ 

7. Required the difference of y f and (/V* Ans.fj^/75. 

8. Find the difference of v/24a*o a and ^/M'* 4 . 

Ans. ^/(36 3 — 206)^/6. 

PROBLEM VI. 

To multiply Surd Quantities together. 

Reduce the surds to the same index, if necessary ; next 
multiply the rational quantities together, and the surds to- 
gether ; then annex the one product to the other for the 
whole product required ; which may be reduced to more 
simple terms if necessary. 

EXAMPLES. 

1. Required to find the product of 4y/\2 and 3^2. 
Here, 4*3X y/ 12x^2 = 12 v /(12v2) = 12 v /24=12 v /(4x6) 

= 12X2X^/6=24^/6, the product required. 

2. Required to multiply \%/} by . 

Here lXtf/f xyt^iV* V*=rV Xl/«=A X * X ^ 18 
=J r ^/18, the product required. 

3. Required the product of SyZ and 2^/8. Ans. 24. 

4. Required the product of 1^/4 and jyl2. Ans. 

5. To find the product of f and rWi* Ans. jtV15. 

6. Required the product of 2^/14 and 3y4. Ans. 12^/7. 

7. Required the product of 2a* and a 3 . Ans. 2n f . 

8. Required the product of (a+A)^ and (a+&)^. 



Ml 

9. Required the product of 2x+^b and 2x - y/b. 

10. Required the product of (af 2y/6)*, and 

11. Required the product of 2x* and 3x'»- 

12. Required the product of 4x* and2y". 

PSOBLEH VII. 

7b divide one Surd Quantity by another. 

Reduce the surds to the same index, if necessary ; then 
take the quotient of the rational quantities, and annex it to 
the quotient of the surds, and it will give the whole quotient 
required ; which may be reduced to more simple terms if 
requisite. 

EXAMPLES. 

1. Required to divide 6 1/ 90 by 3^/8. 

Here6-r- 3.^(96 + 8) = 2^12 =2 /(4X3) =2 X2v/3 
= 4v/3, the quotient required. 

2. Required to divide 12^/280 by 3$/5. 

Here 12 -5- 3 = 4, and 280 + 5 = 56 = 8 X 7 = 2 3 . 7 ; 
Therefore 4x2 XV 7 = 18 tne quotient required. 

3. Let 4y/50 be divided by 2^/5. Ann. 2^10. 

4. Let 6 \/100 be divided 3^5. Ans. 2^/20. 
5 Let fv/jV be divided by j v'f. Ans. fv/5. 

6. Let f be divided by | \/%. Ans. 

7. Let | y/a, or |a^, be divided by fa*. Ans. f a^. 

8. Let be divided by <A. 

9. To divide 3a" by 4a". 



PROBLEM VIII. 



To involve or raise Surd Quantities to any Power. 

Raise both the rational part and the surd part Or multi- 
tiply the index of the quantity by the index of the power to 
which it is to be raised, and to the result annex the power of 
die rational parts, which will give the power Tec\uvc«A» 

Vol. I. 27 



202 



ALGEBRA. 



EXAMPLES. 

1. Required to find the square of Jo^. 

Firs', (f ) a = } X {- = A> a « d («*)■ = a 4 X2 = a i =r^ 
Therefore, Jafy = ^a, is the square required. 

2. Required to find the square of J<A. 
First, i X i = i, and (a*) f = a* = aj/a ; 
Therefore = Ja ^/a is the square required* 

3. Required to find the cube ef }^/6 or f X 6^. 

First, (*y = § X § X § - and (6*) 3 = 6* = 6^6 ; 
Theref. (|^tt) 3 = ft X6 V 6= V the cube required. 

4. Required the square of 2^/2. Ans. 4{/4» 

5. Required the cube of 3^, or </3. Ans. 3 

6. Required the 3d power of £ %/3. Ans. J v& 

7. Required to find the 4th power of Ans. £. 

l 

8. Required to find the with power of a n . 

9. Required to find the square of 2 + v^3. 

PROBLEM IX. 

To evolve or extract ihe Roots of Surd Quantities* . 

Extract both the rational part and the surd part. Or 
divide the index of the given quantity by the index of the 



* The square roet of a binomial or residual surd, « -|- 6, or a — K 
may he found thus : Take Va* — b 2 = c ; 

then Vo+T= + V*-^ ; 

and V« — o = V— g V ~2~' 

Thus, the square root of 4 + 2v3 = 1 + v3 ; 

and the square root of 6 — 2v5 = V5 — 1. 

But for the cnbe, or any higher root, no general role is known. 

For more on the subject of Surds, see BonrycastU's Algebra, the 8rev 
edition, and the EUmentary Treatise ofAbgcbra, by Mr. J. R. Ynmg. 



ARI T HME TICAL PROGRESSION. 203 

root to be extracted ; then to the result annex the root of 
the rational part, which will give the root required. 

RXAHPLE8. 

1* Required to find the square root of 16^/6. 
First, v/16 = 4, and (6^)* = 6* * = 6*; 
theref. (16 y 6)* = 4 . 6* = 4(/G, » the sq. root required. 

2. Required to find the cube root of ^ 
Tint, ^ = J, and (i/3)* = s' " 5 * 3 = 3* ; 

theref. ^/3^ = | . 3^ = it/3, is the cube root required. 

3. Required the square root of 6 s . Ans. 6^/6. 

4. Required the cube root of |a 3 6. Ans. l atyb. 

5. Required the 4th root of 16a* . Ans. 2y/a. 

6. Required to find tie mth root of aA 

7. Required the square root of a 9 — 6a y/b + 96. 



ARITHMETICAL PROPORTION AND PRO- 
GRESSION. 

Arithmetical Proportion is the relation whic'i two 
quantities, of the same kind, bear to each other, in respect to 
their difference. 

Four quantities are said to be in Arithmetical Proportion, 
when the difference between the first and seconJ is i vju... io 
the difference between the third and fourth. 

Thus, 3, 7, 12, 16, and a 9 a + 6, c, c + 6, are arith- 
metically proportional. 

Arithmetical Progression is when a series of quantifies 
V either increase or decrease by the same common difference. 

Thus, 1, 8, 5, 7, 9, 11, dec. and a, a -f- 6, a + 26, a +36, 
«+ 46, a + 56, &c. are series in arithmetical progression, 
whose common differences are 2 and 6. 

The most useful part of arithmetical proportion and pro. 
fwssion has been exhibited in the Arithmetic. The same 
may be given algebraically, thus : 



204 ALGEBRA. 

Let a denote the least term, 

z the greatest term, 
d the common difference, 
n the number of the terms, 
and s the sum of the series ; 
then the princip U properties are expressed by these equa- 
tions, viz. 

1. x = a + d. (n — 1) 

2. a = z — €*.(»— 1) 

3. s = (a 

4. a = (z — \d . n — l )n, 

5. « = (a + \d . n — l)n. 

Moreover, when the first term a is or nothing, the 
theorems become z = d (n — 1) 
and $ = Jam. 

EXAMPLES FOR PRACTICE. 

1. The first term of an increasing arithmetical series is 1, 
the common difference 2, and the number of terms 21 ; re- 
quired the sum of the series ? 

First, 1 + 2 X 20 = 1 + 40 = 41, is the last term. 
1 4- 41 

Then —~— x 20 = 21 X 20 = 420, the sum required. 

A 

2. The first term of a decreasing arithmetical series is 199, 
the common difference 3, and the number of terms 67 ; re- 
quired the sum of the series ? 

First, 199 — 3 . 66 = 199 — 198 = 1, is the last term. 
199 4- 1 

Then X 67 = 100 X 67 = 6700, the sum re- 

quired. 

3. To find the sum of 100 terms of the natural numbers 
1, 2, 3, 4, 5, 6, dec. And. 505a 

4. * Required the sum of 99 terms of the odd numbers 
1, 3, 5, 7, 9, dec. Ans. 9801. 



• The tarn of any number (w) of terms of the Arithmetical aerie* of 
odd numbers 1, 3, 5, 7, 9, &c. is equal to the square (*a ) of that nnfi> 
ber. That is, ' 
If 1, 3, ft. 7, 9 t &c. be the numbers, 1hpn will 
1 2 s , 3 f , 4\ 6», be the sums of 1, 2 3, fcc. terms, 
Thus»0-4-l= 1 or 1M be sum of 1 term, 
14-3= 4 or 2', the sum of 2 terms, 
4 -j- 5 = -9 or 3-', the sum of 3 terms, 
9 + 7 = 16 or 4» , the sum oC 4 tenai, &c 



ARITHMETICAL PBOGBESflXON. 



90S 



5. The first term of a decreasing arithmetical series is 10, 
the common difference 1, and the number of terms 21 ; re- 
quired the sum of the series ? Ans. 140. 

6. One hundred stones being placed on the ground, in a 
straight line, at the distance of 2 yards from each other ; 
how far will a person travel, who shall bring them one by 
one to a basket, which is placed 2 yards from the first stone ? 

Ans. 11 miles and 840 yards. 



APPLICATION OF ARITHMETICAL PROGRES- 
SION. 

Qu. i. A Tin angular Battalion * consists of thirty ranks, 
in which the first rank is formed of one man only, the second 
of 3 ; the 3d of 5 ; and so on : What is the strength of such 
* triangular battalion ? Answer, 900 men. 

Qu. ii. A detachment having 12 successive days to march, 
with orders to advance the first day only 2 leagues, the 
second 3£, and so on, increasing l£ league each day's march : 
What is the length of the whole march, and what is the last 
day's march ? 

Answer, the last day's march is 18J leagues, and 123 
leagues is the length of the whole march. 

Qu. in. A brigade of sappers*)*, having carried on 15 
yards of sap the first night, the second only 13 yards, and 



For, by the 1st theorem, 1 + 2 (n - 1) - 1 -f2n - 2 = 2* - 1 » 
the last term, when the number of term* is n ; to this last term 2* — 1, 
■dd the first term 1, gives 2n the sum of the extremes, or n half the sum 
of the extremes ; then, by the 3d theorem, nXn = n" is the sum of all 
the terms. Hence it appears, in general, that half the sum of the extremes 
b always the same as the number of the terms, n ; and that the sum of 
alt the terms is the same rs the square of the same number, as. 

See more on Arithmetical Proportion in the Arithmetic. 

* By triangular battalion, is to be understood, a body of troops ranged 
In the form of a triangle, in which the ranks exceed each other by an 
equal number of men: if the first rank consist of one man only, and 
the difference between the ranks be al«o 1, then it? form is that of an 
equilateral triangle; and when the difference between the ranks is 
mora than 1, its form may then be an isosceles or scalene triangle. 
The practice of forming troops in this order, which is now laid aside, 
Was formerly held in greater esteem than forming them in a solid square, 
at admitting of a greater front, especially when the troops were to make 
simply a stand on all sides. 

t A brigade of sappers consists generally of 8 men, divided tc\\ut\V] 
feto two parties. While one of these parties is advancing fat wp, \\v<* 
•(kerb faro \3hwg the gabions, fa* cines, and other neceuory \m*o\ a tm«D\A * 



206 



ALGEBRA. 



so on, decreasing 2 yards every night, till at last they car* r 
ried on in one night only 3 yards : What is the number of 
nights they were employed ; and what is the whole length of 
the sap. ||r 

Answer, they were employed 7 nights, and the length of 
the whole sap was 03 yards. 

Qu. iv. A number of gabions* being given to be placed 
in six ranks, one above the other, in such a manner as that 
each rank exceeding one another equally, the first may con- 
sist of 4 gabions, and the last of 9 : What is the number of 
gabions in the six ranks; and what is the difference between 
each rank ? 

Answer, the difference between the ranks will be 1, and 
the number of gabions in the six ranks will be 39. 

Qu. v. Two detachments, distant from each other 37 
leagues, and both designing to occupy nn advantageous post 
equi-distant from each other's camp, set out at different 
times ; the first detachment increasing every day's march 1 
league and a half, and the second detachment increasing each 
day's march 2 leagues : both the detachments arrive at the 
same time ; the first after 5 days' march, and the second 
after 4 days' march : What is the number of leagues marched 
by each detachment each day ? 

The progression r 7 T , 2fc, 5^, 6^, answers the con- 
ditions of the first detachment : and the progression If-, 3}, 
5f , 7|, answers the condition of the second detachment. 



and when the first party is tired, (he second takes its place, and so on, 
till each man in turn has been at (he head of the nap. A sap is a >ma11 
ditch, between 3 and 4 feet in breadth and depth ; and is distinguished 
from the trench by its breadth only, the trench having between 10 and 
15 feet breadth. As an encouragement to tappers, the pay for all the 
work carried on by the whole brigade is given to the survivors. 

• Gabions are baskets, open at both ends, made of ozier twigs, and 
of a cylindrical form ; those made use of at the trenches are 2 fret 
wide, and about 3 feet high ; which, being filled with earth, serve aia 
shelter from the enemy's fire: and those made u«e of to construct bat- 
teries, are generally higherand broader. There is another sort of gabion, 
made usa of to raise a low parapet : its height is from 1 to 2 feel, and 1 
foot wide at top, hut somewhat less at bottom, to give room for placing 
the muszle of a firelock between them : these gabions serve instead 0? 
sand bags. A sand bag is generally made to contain about a cubic foot 
of earth. 



PILIXO OF BALLS. 



207 



OF COMPUTING SHOT OR SHELLS IN A FINISHED PILE. 

Shot and Shells are generally piled in three different 
forms, called triangular, square, or oblong piles, according 
as their base is cither a triangle, a square, or u rectangle. 

Fig. 1. C G Fig. 2 




abcdef, fig. 3, is an oblong pile. 



A. triangular pile is formed by the continual laying of tri- 
angular horizontal courses of shot one above another, in 
such a manner, as that the sides of these courses, called rows, 
decrease by unity from the bottom row to the top row, which 
ends always in 1 shot. 

A square pile is formed by the continual laying of square 
horizontal courses of shot one above another, in such a man. 
ner, as that the sides of these courses decrease by unity from 
the bottom to the top row, which ends also in 1 shot. 

In the triangular and the square piles, the sides or faces 
being equilateral triangft&, the shot contained in those faces 
form an arithmetical progression, having for first term unity, 
and for last term and number of terms, the shot CAmV&va&& 

•7 



208 



AXGKBRA. 



in the bottom row ; for the number of horizontal rows, or 
the number counted on one of the angles from the bottom to 
the top, is always equal to those counted on one side in the 
bottom : the sides or faces in either the triangular or square 
piles, are called arithmetical triangles ; and the numbers 
contained in these, arc called triangular numbers : abc, fig. 
1, kfg, fig. 2, are arithmetical triangles. 

The oblong pile may be conceived as formed from the 
square pile abcd ; to one side or face of which, as ad, a 
number of arithmetical triangles equal to the face have been 
added : and the number of arithmetical triangles added to 
the square pile, by means of which the oblong pile is formed, 
is always one less than the shot in the top row ; or which is the 
same, equal to the difference between the bottom row of the 
greater side and that of the lesser. 

Qu. vi. To find the shot in the triangular pile jlbod, fig. 
1, the bottom row ab consisting of 8 shot. 

Solution. The proposed pile consisting of 8 horizontal 
courses, each of which forms an equilateral triangle ; that is, 
the shot contained in these being in an arithmetical progres- 
sion, of which the first and last term, as also the number of 
terms, ore known ; it follows, that the sum of these particu- 
lar courses, or of the 8 progressions, will be the shot con- 
tained in the proposed pile ; then 

The shot of the first or lower ) 

triangular course will be $ (8 + l)X4— 36 

the second - - - - (7 + 1) X 3J- = 28 

the third - - - - (6 + 1) X 3 = 21 

the fourth - - - - (5 + 1) X 2i = 15 

the fifth - - - - (4 + 1) X 2 = 10 

the sixth - - - - (3 + 1) X 1 J = 6 

the seventh - - - . (2 + 1) X 1 = 3 

the eighth - - - - (1 + 1) X J = 1 

Total 120 shot 

in the pile proposed. 

Qu. vn. To find the shot of the square pile efgh, fig. 2, 
the bottom row sf consisting of 8 shot. 

Solution. The bottom row containing 8 shot, and the 
second only 7 ; that is, the rows farming the progression, 
8, 7, 6, 5, 4, 3, 2, 1, in which each of the terms being the 
square root of the shot contained in each separate square 



FILING OF BALLS. 



209 



; course employed in forming the square pile ; it follows, that 
the sum of the squares of these roots will be the shot requir- 
ed ; and the sum of the squares divided by 8, 7, 6, 5, 4, 3, 
2, 1, being 204, expresses the shot in the proposed pile. 

Qu. viii. To find the shot of the oblong pile abcdep, fig. 
3 ; in which bf = 16, and bc = 7. 

Solution. The oblong pile proposed, consisting of the 
square pile abcd, whose bottom row is 7 shot ; besides 9 
arithmetical triangles or progressions, in which the first and 
last term, as also the number of terms, are known ; it follows, 
that, 

if to the contents of the square pile - - 140 
we add the sum of the 9th progression - 252 

&\, r their total gives the contents required - - 392 shot. 

REMARK I. 

The shot in the triangular and the square piles, as also 
the shot in each horizontal course, may at once bo ascer- 
tained by the following table : tho vertical column a con- 
tains the shot in the bottom row, from 1 to 40 inclusive ; 
the column b contains the triangular numbers, or number 
of each course ; the column c contains the sum of the tri- 
angular numbers, that is, the shot contained in a triangular 
pile, commonly called pyramidal numbers ; the column n 
contains the square of tho numbers of the column a, that is, 
the shot contained in each square horizontal course ; and 
the column £ contains the sum of these squares or shot in a 
square pile. 



Vol. I 



28 



210 ALQMBEAm 



c 


B 


A 


D 


E 


Pyramidal 


Triangular 


Natural 


Square of 
the natural 


Sum of (beta 
square 
numbers. 


number* 


d umbers. 


numbers* 


numbers. 


1 


1 


1 


1 


1 


4 


3 


2 


A 

4 


c 
O 


10 


6 


3 




1 A 

14 


20 


10 


4 


lo 


OA 

4U 


35 


15 


5 


2d 


00 


56 


21 


6 


Qfi 

OD 




84 


28 


7 


a n 

49 


1 Afk 

140 


120 


36 




64 


204 


165 


45 


9 


81 


AO C 

285 


220 


55 


10 


100 


385 


286 


66 


11 


121 


506 


364 


78 


12 


144 


690 


455 


91 


13 


169 


819 


560 


105 


14 


196 


1015 


680 


120 


15 


225 


1240 


816 


136 


16 


256 


1496 


969 


153 


17 


289 


1785 


1140 


171 


18 


324 


2109 


1330 


190 • 


19 


361 


2470 


1540 


210 


20 


400 


2870 


1771 


231 


21 


441 


3311 


2024 


253 


22 


484 


3795 


9 MOO 


976 


23 


529 


4324 


2600 


300 


24 


576 


4900 


2925 


325 


25 


625 


5525 


3276 


351 


26 


676 


6201 


3654 


378 


27 


729 


6930 


4060 


406 


28 


784 


7714 


4495 


435 


29 


841 


8555 


4960 


465 


30 


900 


9455 


5456 


496 


31 


961 


10416 


5984 


528 


32 


1024 


11440 


6545 


561 


33 


1089 


12529 


7140 


595 ! 


34 


1156 


13685 


7770 


630 


35 


1225 


14910 


8436 


666 


36 


1296 


16206 


9139 


703 


37 


1369 


17575 


9880 


741 


38 


1444 


19019 


10660 


780 


39 


1521 


20540 


11480 


820 


40 


1600 


22140 



Thus, tho bottom row in a triangular pile, consisting of 
shot, the contents will be 1330 ; and when of 19 in the squfl»-i* 



FILING 09 BALLS. 



211 



t m . pile, 3470.— In die same manner, the contents either of a 
71 square or triangular pile being given, the shot in the bottom 
row may be easily ascertained. 

The contents of any oblong pile by the preceding table 
may be also with little trouble ascertained, the less side not 
exceeding 40 shot, nor the difference between the less and 
the greater side 40. Thus, to find the shot in an oblong pile, 
the less side being 15, and the greater 35, we are first to 
find the contents of the square pile, by means of which the 
oblong pile may be conceived to be formed ; that is, we arc 
to find the contents of a square pile, whose bottom row is 
15 shot : which being 0B4O, we are, secondly, to add these 
1240 to the product 2400 of the triangular number 120, 
answering to 15, the number expressing the bottom row of 
the arithmetical triangle, multiplied by 20, the number of 
jtito. triangles* and their sum, being 3640, expresses the 
"^ppUber of shot V-ihn proposed oblong pile. 

• SEHARK II. 

The following algebraical expressions, deduced from the 
investigations of the sums of the powers of numbers in 
arithmetical progression, which are men upon many gunners' 
callipers' 11 , serve to compute with ease and expedition the shot 
or shells in any pile. 

That serving to compute any triangular > (n+2)X(n+l)Xn 
pile, is represented by $ 6 

That serving to compute any square ) (n+ l)X( 2n+l)xn 
pile, is represented by $ (5 

In each of these, the letter n represents the number in the 
bottom row : hence, in a triangular pile, the number in the 
bottom row being 30 ; then this pile will be (30+2) X (30+1) 
X V = 4960 shot or shells. In a square pile, the number 
in the bottom row being also 30 ; then this pile will be 
(30 + 1) X (60 + 1) X V = W5 5 shot or shells. 



• Callipers are large compasses, with bowed shanks, serving to tnke 
the diameters of convex and concave bodies. The gunners' callipers 
consist of two thin rules or plates, which are moveable quite round a 
Joint, by the plates folding one over the other : the length of each rule 
or plate is 6 inches, the breadth about 1 inch. It is usual to represent, 
-*>tt the plates, ■ variety of scales, tables, proportions, &c. such as are 
esteemed useful to be known by persons employed about artillery ; hut, 
«xeept the measuring of the caliber of shot and cannon, and the measur- 
ing of salient and re-enterine angles, none of the articles, with YiVfa>i 
the callipers are usually filled, an essential to thai taitroxneal 



212 



ALGEBRA. 



That serving to compute any oblong pile, is represented by 

(2n + 1 + 3m) X (n + 1) X n . «» « ■ « . , . 

^ ! ' — - — ' — - , in which the letter n denotes 

o 

the number of courses, and the letter m the number of shot, 
less one, in the top row ; hence, in an oblong pile the num- 
ber of courses being 30, and the top row 31 ; this pile will 
be COTT+90 X 30 + 1 X y = 23405 shot or shells. 

REMARK III. 

One practical rule, of easy reelection, will include the 
three cases of the triangular, square^ and rectangular, com- 
plete piles. 

Thus, recurring to the diagrams 1, 2, and 3, we shall 
have, balls in 

(bd + a + c) X £bdo = triar* liar pile. 

(ef + ef + g) X £gfh = aquare pile. 

(bf + bf +ae) X £abc = rectangular pita 
Hence, for a general rule : add to the number of balls or 
shells in one side of the base, the numbers in its two paral- 
lels at bottom and top (whether row or ball), the sum being 
multiplied by a third of the slant end or face, gives the 
number in the pile. % 



GEOMETRICAL PROPORTION, AND PRO- 
GRESSION. 

Geometrical Proportion contemplates the relation of 
quantities considered as to what part or what multiple ono 
is of another, or how often one contains, or is contained 
in, another. — Of two quantities compared together, the first 
is called the Antecedent, and the second the Consequent, 
Their ratio is tho quotient which arises from dividing the 
one by the other. 

Four Quantities are proportional, 'when the two couplets 
have equal ratios, or when the first is the same part or mul- 
tiple of the second, as the third is of the fourth. Hius, 
3, 6, 4, 8, and a, ar, &, fer, are geometrical proportional*. 
or hr 

For f = f =2, and -r = -y = r. And they are staled 

thus, 3 : 6 : : 4 : 8, &c. See the Arithmetic. 
Geometrical Progression is one in which the terms have 



GEOMETRICAL PROGRESSION. 213 

all successively the same ratio ; as 1, 2, 4, 8, 16, dec. where 
the common ratio is 2. 

The general and common property of a geometrical pro- 
gression is, that the product of any two terms, or the square 
of any one single term, is equal to the product of every other 
two terms that are taken at an equal distance on both sides 
from the former. So of these terras, 

1, 2, 4, 8, 16, 82, 64, czc. 

1X04 = 2X 32 = 4X 16 = 8X 8=64. 

In any geometrical progression, if 
a denote the least term, 
x the greatest term, 
r the common ratio, 
n the number of the terms, 
8 the sum of the series, or all the terms ; 



any of these quantities may be found from the others, 
by means of these general values or equations, viz. 



2. z = a X r*~ l . 

1 — 

4. n = a = l°g» r + log, g — log * a 

log. r log.r 

5. * = X a = r- X— r = -. 

r — 1 r — 1 r*~ l r - 1 

When the series is infinite, then the least term a is nothing, 

and the sum s = 

r— 1 

In any increasing geometrical progression, or series be- 
ginning with 1, the 3d, 5th, 7th, czc. terms will be squares ; 
the 4th, 7th, 10th, czc. cubes ; and the 7th will be both a 
square and a cube. Thus, in the series 1, r, r", r 3 , r* 9 r 5 , 
r*, r 7 , r 1 , r*, &c. r 2 , r 4 , r 1 , r*, are squares ; r*, r", r 9 , cubes ; 
and r° both a square and a cube. 

In a decreasing geometrical progression, the ratio, r, is a 
1— r» 

fraction, and then s = - -a. If n be infinite, this becomes 



* = 1 — r ' a ^ >e * n 8 ^ **** term r t 



214 ALGSUU* 

When four quantities, a, or, b, br, or 2, 6, 4, 12, are pro- 
portional ; then any of the following forms of those quantities 
are also proportional, viz. 

1. Directly, a : or : : b : br ; or 2 : 6 : : 4 : 12. 

2. Inversely, «r : a : : fcr : b ; or 6 : 2 : : 12 : 4. 

3. Alternately, a : b ; : or : br ; or 2 : 4 : : 6 : 12. 

4. Compoundedly, a : a+ar ::b: b+br ; or 2 : 8 : : 4 : 10. 

5. Dividedly, a : or— a : : b : ftr—ft ; or 2 : 4 : : 4 : 8. 

6. Mixed, ar-fa: or — a : : br+b : ftr— 6 ; or 8 : 4 : : 16 : 8. 

7. Multiplication, ae : arc : : be : ftrc ; or 2.3 : 6.3 : : 4 : 12, 

8. Division, — : — : : b : br ; or 1 : 3 : : 4 : 12. 

c c 

9. The numbers a, 6, c, J, are in harmonical proportion, 
when a : d : : a b : c d ; or when their reciprocals 

-j-, -i, are in arithmetical proportion. 



EXAMPLES. 



1. Given the first term of a geometrical series 1, the ratio 
2, and the number of terms 12 ; to find the sum of the series 7 
First, 1 X 2 11 = 1 X §048, is the last term. 

2048 X 2-1 4096 — 1 Ant% - A . . , 

Then : = = 4095, the sum required. 

<£ — 1 1 

„ 2. Given the first term of a geometric series |, the ratio 
and the number of terms 8 ; to find the sum of the series 1 
First, | X (i) 7 = }X T } T = sly , is the last term. 
Then (i - *U X i) + (W) = (*- T * T ) -7- i = f ff Xf 
= a{ { , the sum required. 

3. Required the sum of 12 terms of the series 1, 3, 9,27, 
81, &c. Ans. 265720. 

4. Required the sum of 12 terms of the series, 1, £, -fr. 
A. &c. Ans. fflfff 

5. Required the sum of 100 terms of the series, 1, 2, 4, 8, 
16, 32, &c. Ans. 1267650600228220401496703205375. 

See more of Geometrical Proportion in the Arithmetic. 



INFINITE SERIES. 

An Infinite Series is formed either from division, dividing 
by a compound divisor, or by extracting the root of a com* 
pound surd quantity, or by other general processes ; and is 



- ntrnuTB series. 215 

such at, being continued, would run on infinitely, in the 
manner of a continued decimal fraction*. 

But, by obtaining a few of the first terms, the law of the 
progression will be manifest ; so that the series may thence be 
continued, without actually performing the whole operation. 

problem I. 

To reduce Fractional Quantities into Infinite Series by 
Division. 

Divide the numerator b)- the denominator, as in common 
division ; then the operation, continued as far as may be 
thought necessary, will give the infinite series required. 

EXAMPLE. 

2ab 



1. To change — p-r into an infinite scries. 
j7 ^ a + b 

2oft..(2&- 
2ab + 2b 2 



2b 2 2&' 2b* 
* + b)2ab..{2b- — + -^ — ^ +&c. 

a or or 



— 2b 2 

2b* 

— 2b 2 — — 

a 



2b 3 



a 

2P W 
a + a 2 



2b* 
a 2 

2b* 2b 9 
a 2 a 3 



* The doctrine of infinite series was commenced by Dr. Wallis ; 
who, fn his arithmetical works published in 1657, first reduced the frac- 
tal |-~ by a perpetual division into the infinite series a + ox + at* + 
, if s + ar4 + &c. 



316 AL0BBJU. 

2. Let c ^ an 8 e ^ * nto 811 infinite series. 

1 — a) l....(l + a + a , + o» + a«+ &c. 
a 

a — a» 



a* -a' 



3. Expand — into an infinite series. 



Ans.^X(l-f + ^-| + &c.) 

o a or or 



4. Expand ^3-^ into an infinite series. 



a a a + a 

1 x 

5. Expand ^ ^ into an infinite series. 

Ans. 1 — 2x + 2* 1 — 2x 3 + 2x\ &c. 

a a 

6. Expand - — r-rr, into an infinite series. 

r (* + 6) a 

Ans. 1 f- -3- 3, &c 

a <r a 3 

7. Expand ^ ^ ^ = J, into an infinite series. 

PROBLEM II. 

2b reduce a Compound Surd into an Infinite Series. 

Extract the root as in common arithmetic ; then the 
operation, continued as far as may be thought necessary, will 
give the series required. But this method is chiefly of use 
in extracting the square root, the operation being too tedious 
for the higher powers. 



ZHFUTITB SERIES. 



217 



EXAMPLES. 

1. Extract the root of a 3 — x 2 in an infinite series. 
„._*»(«____ _ — - _ T &c. 



2« — — i 

2a) 



X 4 

— x a + — 
^ 4a* 



X s X 4 x 4 



X 4 X 9 X* 

4^ §0*" + 64a« 



X« ** . V 



8a 4 04a° 

x 8 X s 

■ -4 &c. 

8a 4 T 16a° 



dec. 



04a« 



2. Expand ^/(l + 1) = \/2, into on infinite series. 

Ans. l + £-J- + -,V-r5F &c. 

3. Expand ^/(l — 1) into an infinite scries. 

Ans. i-yV— t{? &c. 

4. Expand </{c? + x) into an infinite series. 

5. Expand ^/(o 3 — 2bx — x 3 ) to an infinite series. 

PROBLEM III. 

To extract any Root of a Binomial: or to reduce a Binomial 
St . d into an infinite Series. 

This will be done by substituting the particular letters of 
the binomial, with their proper signs, in the following gene- 
ral theorem or formula, viz. 

, _ v ^ ^ _ m in — 7i , m — 2ji 

(* + Pft) w = P n + — aq + ~2^" " Ba "* f*n~~ Ctt + &c- 

Vol. I. 29 



218 



ALCKBSA. 



and it will give the root required : observing that p denotes 

m 

the first term, a the second term divided by the first, « the 
index of the power or root ; and a, b, c, d, dec. denote the 
several foregoing terms with their proper signs. 



EXAMPLES. 

. 1. To extract the sq. root of a 1 + V 9 in an infinite series* 
Here p « a 1 , a = and — = g 2 therefore 

m m 

p w = (a a )*= (a*) = a = a, the 1st term of the series. 

— Aa = i XaX^=^ = b, the 2d term. 

m-n 1—2 v i» v 6» ft* _ . . 

-s — sa = —j— x — X — = — ^r— 1 ■= c, the 3d term. . 
2n 4 2a a* 2.4<r 

w _2n 1-4 i« V 36* t . ... 

Hence a + ^ - g — + j-g- - &c. or 
a + 2«""8^ + 16^ j2^ 7 «fec.u.the senw required. 

2. To find the value of ; — ^—7^, or its equal (a — x)~~* in an 
(a — x)* n N ' 

infinite series*. 

Here p=o, q=— =-o- , «, and — = ^? = -2 ; theref. 
a n 1 



* ATofe. To facilitate the application of the rule to fractional eiem- 
pies, it is proper to observe, that any surd may be taken from the de- 
nominator of a fraction and placed in the numerator, and vice versa, by 
only changing the sign of its indei. Thus, 

L = 1 x ar-a or only x-*; and = 1 X (« + or (a + *)-»; 

(o» + x») J x («* - **H ; &c. 

The theorem above given is only the Binomial Theorem so expreav 
ed as to facilitate its application to roots and series. 



INFIlCtTB SXStES. 219 

= [a)- 9 = a~* = ^- ■= a, the first term of the series. 

— aq = -2X~X— = ~=2a-^r = b, the 2d term. 
* tr a or 

jr — ft 2jf —x 3x* 

— Tjia =-| x^X-— X = ^~- = 3«-V = c, the 3d. 
2» ■ a 3 a a* 

«— 2» 4 ^Si 1 ^ — x 4x» . g . 

aft 1 a 4 a a 5 

Hence or* + 2a- 3 * + 3a- 4 *" + 4a-** 3 + dec. or 

1 , 2x , 3*» 4r* 5x* . . . . , 

— r + — H — — + — — + — - dec is the series required. 

Qi Or ft* ft* fl J 

a 9 

8. To find the value of , in an infinite series. 

a— a? 

X* X 9 X 4 

Ans.a+*+- + -+-&e. 

4. To eqpand ✓ „ — in a series. 

. 1 at* , 3x« 5x* . 

5. To expand t r-ri in an infinite series. 

(«-*)" 

a i . 26 , 3i» 4i» 56 4 . 

Ans. 1 H h — r + — ^- H — - <fcc. 

o a* a*. a 4 

6. To expand ^/o 1 — x a or (a 3 — x*)^ in a series. 

x* x 4 x* 5x f 
Ans. «__-__ I6^~i28?^. f . 

7. To find the value of ?/ (a 3 - 6 s ) or (tf 3 — 6 3 )* in a series. 

6 3 5b 9 m 

8. To find the value of V (<* + «*) or (ci'+x 5 )* in a series. 

X s 2c 10 6x l 8 

o — 6 

9. To find the square root of ^ in an infinite series. 



220 ALGEBSA. 



a* 

10. Find the cube root of -tttt i° a series. 

o» . 2&« 

3a 3 9a 8 81a» 



Ans.l__+--.- — &c 



INFINITE SERIES : PART THE SECOND. 



PROBLEM I *• 



A series being given, to find the several orders of dif- 
ferences of the successive terms. 

Rule i. Subtract the first term from the second, the 
second from the third, the third From the fourth, and so on ; 
the several remainders, will constitute a new series, called the 
first order of differences. 

ii. In this new series, take the first terra from the second, 
the second from the third, dec. as before, and the remainders 
will form another new series, called the second order of dif. 
ferences. 

in. Proceed in the same manner for the third, fourth, 
fifth, <3fc. orders, until either the differences become 0, or the 
work will be carried as far as is thought necessaryf . 



EXAMPLES. 



1. Given the series 1, 4, 8, 13, 19, 26, &c. to find the 
several orders of differences. 



* The study of this second part of Infinite Series may be conve- 
niently postponed till Simple and Quadratic Equations have been 
learnt. 

t Let a, 6, e, d, e, &c. be the terms of a given series, then if d = the 
first term of the nth order of differences, the following theorem will 

exhibit the value of d : vis. ;fc a =P *6 + n . n "7* . c + n . *T"^ - . 

n— 2 , . 11— 1 n— 2 is— 3 . . , . 

—3— • d ± * • — y- • —3— • —5— • « Hhi fcc- (to n + 1 terms) 

= d, where the upper signs must be taken when n is an even number, 
and the lower signs when it is odd. 

If the differences be very great, the logarithms of the quantities may 
be used, the differences of which will be much smaller than those of ^ 
the cjuantities themselves ; and at the close of the operation the natural 
number answering to the logarithmical result will be the answer. See 
Emsnan's ZHfnimtal Mcihod; prop. 1. 



INFIN I TE SSBIE8. 391 

Thus 1, 4, 8, 13, 19, 26, dec. the given series. 
Then, 3, 4, 5, 6, 7, dec. the first differences. 
And 1, 1, 1, 1, dec. the second differences. 
Also 0, 0, 0, dec. the third differences, 

where the work evidently must terminate. 

2. Given the series 1, 4, 8, 16, 32, 64, 128, dec. to find the 
several orders of differences. 

Here 1, 4, 8, 16, 32, 64, 128, dec. given series. 

And 3, 4, 8, 16, 32, 64, dec. 1st diff. 

1, 4, 8, 16, .32, &c 2nd diff. 

3, 4, 8, 16, &c. 3rd diff. 

1, 4, 8, dec. 4th diff. 

3, 4, fcc. 5th diff. 

1, &c. 6th diff. 

3. Find the several orders of differences in the series 
1, 2, 3, 4, &c. 

Ans. First diff. 1, 1, 1, 1, &c. Second diff. 0, 0, 0, &c. 

4. To find the several orders of differences in the series 
1, 4, 9, 16, 25, &c. of squares. 

Ans. First differences 3, 5, 7, 9, dec. Second, 2, 2, 2, 
&c. Third, 0, 0, dec. 

5. Required the orders of differences in the series 1, 8, 
27, 64, 125, &c. being cubes. 

6. Given 1, 6, 20, 50, 105, dec. to find the several orders 
of differences* 



PROBLEM, n. 



To Find any term of a given series. 

Rule i. Let a, ft, c, d, e, dec. be the given series ; d r , d", 
d m , <Z* V , dec. respectively, the first term of the first, second, 
third, fourth, dec. order of differences, as found by the ore- 
ceding article; n = the number denoting the place or the 
term required. 

n. Then will a + ^1. d' + ^=1 . ^ . d> + 

1 1 A 1 

w— 2 n— 3 n— 1 n-2 n-3 n — i 

■ to the nth term required. 



388 ALGEBRA* 



XX AMPLE 8* 



1. To find the 10th term of the series 2, 5, 9, 14, 20, dec 
Here 2, 5, 9, 14, 20, dec. series. 
3, 4, 5, 6, dec. 1st diff. 
1, 1, 1, dec. 2nd diff. 
0, 0, dec 3rd diff. 
Where d' = 3, d' = 1, d'" = 0, also a = 2, n = 10 ; 

wherefore a H j— . d' H — ---g— . d f = (2 -| j — 

X3 + 12pIxi^?Xl=)2+27 + 86 = e6 = the 

10th term required. 

2. To find the 20th term of the series 2, 6, 12, 20, 30, dee. 
Here a = 2, n = 20 ; and Art. 12. 

2, 6, 12, 20, 30, dec. series. 
4, 6, 8, 10, dec. 1st diff. 
2, 2, 2, dec. 2nd diff. or d' = 4, d' = 2 ; 

whence a -\ j— . a' + — p- . — ^— . d" = (2 + — X 4+ 

^ X ^X2 = )2 + 76 + 342 = 420 = the 20th term 
required. 

3. Required the 5th term of the series, 1, 3, 6, 10, dec. 

Ans. 16. 

4. To find the 10th term of the series, 1, 4, 8, 13, 19, dee. 

Ans. 64. 

5. Required the 20th term of the series, 1, 8, 27, 64, 125, 
dec. Ans. 8000. 



PROBLEM HI. 



If the succeeding terms of a given series be at an unit's 
distance from each other, to find any intermediate term by 
interpolation. 

Rule 1. Let y be the term to be interpolated, x its 
distance from the beginning of the series, d', d% d h \ d ir , dec* 
the tint terms of the several orders of differences. 



INFINITE SEBIES. 223 

2. Then willa + *d' + * . ^ . d' +*. ^ . <T"+ 
* . . ^-jp • ^-j^ . <# v + &c. = y, the term required. 

EXAMPLES. 

1. Given the logarithmic sines of 3° 4', 3° 5', 3° 6', 8° 7', 
and 3° 8', to find the sine of 3° 6' 16*. 

Series. Logarithm*. Istdiff. 2nddiff. 3rddif. 

8*4' 8-7283366 23516 1QA 

3 5 8-7306882 23390 ~ifl 1 

3 6 8-7330272 23263 ~\ZL — * 

3 7 8-7353535 23140 ~ KM 

3 8 8-7876675 

Here x "= (3*6' 15'- 3° 4' = 2' 15' = ) f = the distance 
of the term y, to be interpolated; a = 8-7283366, d' = 
23516, d" = - 126, d'" = 1, and y = a + <rd' + *. 

2=1' * + x ?=1. . = (a + K + Hd' + 

M*" = ) 8-7283366 + 0052911 — -00001771875 + 
•0000000117 = 8-73360999296, the log. sine required. 

2. Given the series -fa, fa, iV iV> t0 find the tenn 
which stands in the middle, between fa and fa. Ans. y| T . 

3. Given the logarithmic sines of 1° 0', 1° \', 1 Q 2', and 
1° 3', to find the logarithmic sine of 1° 1' 40". Ans. 8-2537533. 



PROBLEM IV. 

2b find any intermediate Term by Interpolation, when the 
first Differences of a Series of equidifferent Terms are 
small. 

Rule 1. Let a, b, c, d, e, <kc. represent the given series, 
and n = the number of terms given. 

2. Then will a — nb + n . n ~ 1 . c — n*. . n ~ 2 

. d + n . n ^-^ . . . e + dec. = 0, from whence 
55 o 4 

by transportation, dec. any required term may be obtained*. 



• For the investigation of these roles, see Emenon't Diffemdiel 
Msthod. 



▲L0JBBJU. 



EXAJCPLBS. 

1. Given the square root of 10, 11, 12, 13, and 15, to find 
the square root of 14. 

Here n = 5, and e is the term required. 
a = (-/10 = ) 31622776 
b = (^11 = ) 3-3166248 
c = (v/12=) 8-4641016 
d = L/12 = ) 3-6055512 
/ = (^15=) 8-8729833 
And since n = 5, the series must be continued to 6 terms. 

Therefore a — nft + n.— . c — n . — . .d + 
n — 1 n — 2 n — 3 n — 1 n — 2 n — 3 n — 4 

./=0. 

Whence, by transposition, in order to find c we shall have 

n — 1 n — 2 n— 8 , . n — 1 , 

n. g . 3 . —j— .e = — a + nft — n. — .c + n 

n— 1 9i — 2 , . n -1 n — 2 n — 3 n — 4 r 

.- 2 -.- 3 ~,<*+».-2-.-3 j-.-g-./; *" 

in numbers becomes 5c = — 3-1622776 + (5 X 3-3166248) 
— (10 x 3-4641016) + (10 X 3-6055512) + 3-8729833 =* 

56-5116193— 37-8032936=18-7088257, and e= 18 7083257 

5 

ss 3-74166514 = the root, nearly. 

2. Given the square roots of 37, 38, 39, 41, and 42, to 
find the square root of 40. Ans. 6-32455532. 

3. Given the cube roots of 45, 46, 47, 48, and 49, to find 
the cube root of 50. Ans. 3-684033. 



PROBLEM V. 



7b revert a given Series. 

When the powers of an unknown quantity are contained 
in the terms of a aeries, the finding the value of the unknown 
quantity in another series, which involves the powers of the 



nornriTE series. 



225 



quantity to which the given series is equal, and known quan- 
tities only, is called reverting the series*. 

Rule 1. Assume a series for tho value of the unknown 
quantity, of the same form with the series which is required 
to be reverted. 

2. Substitute this series and its powers, for the unknown 
quantity and its powers, in the given series. 

3. Make the resulting terms equal to the corresponding 
terms of the given series, whence the values of the assumed 
co-efficients will be obtained. 



examples. 



1. Let ax + bx* + cr* + dx* + &c. = z be given, to find 
the value of x in terms of z and known quantities. 

Let z*= x, then it is plain that if z n and its powers be sub- 
stituted in the given series for x and its powers, the indices 
of z will be n, 2n, 3n, 4n, &c. and 1 ; whence n= 1, and 
the differences of these indices are 0, 1, 2, 3, 4, &c. Where- 
fore the indices of the series to be assumed, must have the 
same differences ; let therefore this series be az + bz 3 + cz 3 
+ dz 4 + ©zc. = x. And if this series be involved, and sub- 
stituted for the several powers of x, in the given series, it will 
become 

axz + obz 2 + acz 3 + avz* + dec. 

* + 6aV + 26ab* 3 + 2b ac z K + dec. 

* * * + £hV + dec. 

* * + CA 3 Z 3 + 3cA a HZ l +&C 

* * * + dA 4 Z 4 . + &C 

Whence, by equating the terms which contain like powers 
of *, we obtain (oaz = z, or) a = -i- ; {aitz 2 + 6a V = 0, 

whence) b = ( — = ) - ~, (acz*+ 2 W+ caV=0, 

t 2oab+ca 3 K W-ac 
whence) c = ( = ) - — 5 - — ; J> = ( — - 

&bAC+bi?+3cA a B+dA* . 5abc-5b 3 -a>d _ . 

1 x= ) , &c. and conse- 



* Other methods of reversion are given by different mathemalic;axv%» 
Tltie above is selected for it* simplicity. 
Vol. I. 30 



326 ALGEBRA. 

quently x=(\z+Bz > +cz i + &c.=)-^ ^ + — s*— 

= . 2* + sc. the senes required. 

a 

This conclusion forms a general theorem for every similar 
series, involving the like powers of the unknown quantity. 

2. Let the series x — x 2 + x 3 — x* + dec. = *, be propos- 
ed l'>r reversion. 

Here a = 1, b = — 1, c = 1, d = — 1, &c. these values 
being substituted in the theorem derived from the preceding 
example, we thence obtain x = z + z 2 + z 3 +z 4 + &c. the 
answer required. 

^2 x 3 jgi 

3. Let x — + - — — + = y, be given for rever- 

sion. 

Substituting as before, we have a = l,6= — |, c =4, 
and d =— £, &c. These values being substituted, we shall 

y2 y3 yA 

have x = y + + ~r + &c. from which if y be given, 
2 24 

and sufficiently small for the series to approximate, the value 
of x will be known. 



PROBLEM VI. 

To find the Sum of n Terms of an Infinite Series. 

Rule 1. Let a, b, c, d, c, &c. be the given scries, s = the 
sum of n terms, and d', d% d"\ d w , &c. respectively the first 
terms of the several orders of differences, found by prob. 1. 

2. Then will na + n . ^ . d' + n . 5=1 . ^ . d*+ 

n-l n-2 n-3 n— 1 n— 2 n— 3 n— 4 

. d lv + &c. = s, the sum of n terms of the series, as was re- 
quired. 

Case 1. To find the sum of n terms of the series 1, 2, 
3, 4, 5, dec. 

First, 1, 2, 3, 4, 5, &c. the given series. 
1, 1, 1, 1, &c, first differences. 
0, 0, 0, &c. second differences. 

Here o=l, d'=l, <T=0 ; then will na + n . ^ . cT 



INFINITE SERIES. 22ff 

^ ^""^ ' — , which, (since a and d' each = 1) = 

2n+«* — * v » . a + 1 A , . , 

= ) ~ = 8, the sum required. 

EXAMPLES. 

1. Let the sum of 20 terms of the above series be re- 
quired. 

„ on j n^n+T 20X21 01A . 

Here » = 20, and * = — ^— = — - — = 210, the ans. 

2. Let the sum of 1000 terms be required. Ans. 500500. 

3. Let the sum of 12345 terms be required. 

Case 2. To find the sura of n terms of the series, 1, 3, 
5,7,9, <&c. 

Here 1, 3, 5, 7, 9, &c. the given series. 

2, 2, 2, 2, &c. . . . first difference. 
0, 0, 0, &c. . . . second difference. 

Wherefore a = 1, d r = 2, <T = 0, and na + n . <*' 

ft 3 — ft 

== (na H — . d' = (since a = 1 and a* = 2) n + n 3 — 

n =)** = *, the sum required. 

EXAMPLE. 

To find the sum of 10 terms of the above series. 

Here » = 10, and s =- (ft 3 =) 100, the answer. 
Case 3. To find the sum of it terms of the series of 
squares 1, 4, 9, 16, 25, &c. 

Here 1, 4, 9, 16, 25, ©zc. the series. 

3, 5, 7, 9, &c 1st difF. 

2,2,2, <fcc 2nddi(F. 

0, 0, &c 3rd diff. 

Whence a =* 1, d' = 3, d" = 2, d'" = 0, and na + n . 

_.^ + ._._.^ = ( w+ .3^.. — . + 2n. 



n— 2 _ 3a* -n n*-3n ? -f-2n n .Ai T l.-'n 1 1 
~' "3 T~ + 3 " ' ft * 

4hs sum required. 



838 ALGEBRA. 



EXAMPLE. 



Let the sum of 30 terms of the above series be required. 
Here „= 30; wherefore^l^±i)=?^«- 
9455, the answer. See the table, pa. 210, 



PROBLEM VII. 

To find the Sums of Series, by the Method of Subtraction. 

This method will be rendered evident by two or three 
simple examples. 

EXAMPLE 1. 

Let 1 + i + ± + } + ozc. in inf. = s 
then \ + J + } + I + ozc. in inf. = * — 1. 

by sub. ig + i- 4 + i- + & c. mm/. = 1. 

EXAMPLE 2. 

Let 1 + | + * + &c. = * 
then i + i + i + i + <&c. = s = |. 

2 2 2 2 
* 8ub - O + 274. + 375 + 4T6 + &C ' = * 

°'- 5 - 2 'i!3 + ^4 + 3 1 5 + i + &C - = *- 

EXAMPLE 3. 



1 

1.2 


+ 


1 

2.3 


+ 


1 

3.4 


1 

2.3 


+ 


1 

3.4 


+ 


1 

4.5 



b y 8Ub -r|-3 + 2li + ro + &c - e= * 



INFINITE SBBXE8. 



EXAMPLE 4. 

Find the s»m of the series ^ + ji-g + — + &c. 

Take away the last factor out of each denominator, and 
1,1 , 1 , - 
" rane 23 + + 678 + * C - S!! * 

* >ub OS + SXft + £§X0 + * C ' - * 



KXAKFLK 4. 

Find the sum of the infinite series 

_1 , 1 1 , I 

8.4.6.8 4.6.8.10 6.8.10.12 8.10.12.14 

Ans. ^t- 

EXAMPLE 5. 

Find the sum of the infinite series 

1 , 1 , 1_ , . 

*" * q ll i^ q n ia l* ■ n ia inr oa ' flcc « 



3.5.8.11 4 5.8.11.14 4 8.11.14.17 r 11.14.17.20 

Ans. 

PROBLEM vni. 

To sum an infinite series by supposing it to arise from the 
expansion of some fractional expression. 

Rule. Assume the series equal to a fraction, whose de- 
nominator is such, that when the series is multiplied by it, 
the product may be finite ; this product being equal to the 
numerator of the assumed fraction, determines its ^jue. 

EXAMPLES. 

1. Required the sum of the infinite series x a? * 



280 ALGEBRA. 

Assume the series = -r^— 
1 — x 

then x + x 2 + x 3 + d&c. 
into 1 — x 

x + x 3 + x 3 + d&c. 
— x* — x 3 — dec. 

2 = X 

.\ x + x* + x 3 + dec* = j-^-. 

Thus, if x = J, then i + * + i + &c = | -^ j = 1 ; 
if x = J, then i + i + ft + &c. = $ -r | = J. 

2. Required the sum of the infinite series x + 2x a + 8r* 
+ dec. 



Assume the series = 



(l—x) a 1— 2x+x a ' 
then x + 2x* + 3X 3 + dec. 
into 1 — 2x x a 



x + 2r a + 3r» + dtc. 
_2x 3 — 4i 3 — dec. 
+ V + d:c. 



V x + 2i a + 3x 3 + dec. = - a . 

(1— x) 

If x = J, then ^ + f + | + T V + dec. = i -T- -J = 2. 
If x = J, then J + J + ft + ? * r + dtc. = | -j- J = f. 
And so on, in other cases *. 

3. Find the sum of the infinite series x + 4x 2 + 9x* + 
16x 4 + dcc. . *(l+x) 



* TbvPpfrfeceding is only a sketch of an inexhaustible subject. For 
the algebraical investigation of infinite series, consult Dodson's Math* 
malical Repository, and Mr. J. R. Younu's Mnebra. The subject, how- 
ever, is much more extensively treated by means of the flaiional 
•BjJjsift 



231 



SIMPLE EQUATIONS. 

An Equation is the expression of two equal quantities 
with the sign of equality (=) placed between them. Thus 
10 — 4 = 6 is an equation, denoting the equality of the 
quantities 10 — 4 and 6. 

Equations are either simple or compound. A Simple 
Equation, is that which contains only one power of the un- 
known quantity, without including different powers. Thus, 
x — a = b + c, or ax a = b 9 is a simple equation, containing 
only one power of the unknown quantity x. But x 2 — 2ax 
■= b 2 is a compound one. 

GENERAL RULE. 

Reduction of Equations, is the finding the value of the 
unknown quantity. And this consists in disengaging that 
quantity from the known ones; or in ordering the equa- 
tion so, that the unknown letter or quantity may stand 
alone on one side of the equation, or of the mark of equality, 
without a co-efficient ; and all the rest, or the known quan- 
ties, on the other side. — In general, the unknown quantity 
is disengaged from the known ones, by performing always 
the reverse operations. So, if the known quantities arc con- 
nected with it by + or addition, they must be subtracted ; if 
by minus ( — ), or subtraction, they must be added ; if by 
multiplication, we must divide by them ; if by division, we 
must multiply ; when it is in any power, wo must extract 
the root ; and when in any radical, we must raise it to the 
power. As in the following particular rules ; which are 
founded on the general principle, that when equal operations 
are performed on equal quantities, the results must still be 
equal ; whether by equal additions, or subtractions, or mul- 
tiplications, or divisions, or roots, or powers. 

PAB^ftcULAR BULK T. 

When known quantities arc connected with the unknown 
hy + or — ; transpose them to the other side of the equa- 
tion, and change their signs. Which is only adding or sab- 
tracting the same quantities on both sides, vu ottat to 



232 



all the unknown terms on one side of the question, and all 
the known ones on the other side*. 

Thus, if x + 5 = 8 ; then transposing 5 gives x=8 — 5=3. 

And if x— 3 + 7 = 9; then transposing the 3 and 7, gives 
x=9 + 3 — 7 = 5. 

Also, if* — a + b^cd 9 then by transpc sing a and ft, it 

is x =» a — b + cd. 
In like manner, if 5x — 6 = 4x + 10, then by transposing 
6 and 4x, it is 5x — 4x — 10 + 6, or x = 1G. 

RULE II. 

When the unknown term is multiplied by any quantity ; 
divide all the terms of the equation by it. 

Thus, if ax=ab — 4a; then dividing by a, gives x=&— 4. 

And, if 3x + 5 = 20 ; then first transposing 5 gives 
3x = 15 ; and then by dividing by 3, it is x = 5. 

In like manner, if ax+3a6=4c a ; then by dividing by a, it 

4e* 4c 2 
is x+36= — ; and then transposing 36, gives x = 36. 



RULE III. 

WiiEif the unknown term is divided by any quantity ; 
we must then multiply all the terms of the equation by that 
divisor ; which takes it away. 

Thus, if | = 3 + 2: then mult, by 4, gives x = 12 + 8=20. 

And, if- = 3b + 2c — d: 
a 

then mult, by a, it gives x = Sab + 2ac — ad. 



* Here it is earnestly recommended that the pupil be accustomed, 
at every line or step in the reduction of the equations, to name the par- 
ticular operation to be performed in the equation in the last line, in or- 
der to produce the next form or state of the equation, in applying each 
of these rules, according as the particular form of the equation may 
require ; applying them according to the order in which they are here 
placed : and beginning every line with the words Then by, as in the 
following specimens of Examples; which two words will always bring 
to his recollection, that he is to pronounce what particular operation 
he is to perform on the last line, in order to give the next ; allotting 
always a single line for each operation, and ranennq the equations neat- 
ly just under each other, in the several line*, as they are successively 
prodaced. 



SIMPLE EQUATIONS. 



238 




Then by transposing 3, it is * r 
And multiplying by 5, it is 3r 
Lastly, dividing by 3, gives x 



10. 

50. 



RULE IV. 



When the unknown quantity is included in any root or 
surd : transpose the rest of the terms, if there he any, by 
Rule 1 ; then raise each side to such n power as is denoted 
by the index of the surd ; viz. square each hide when it is 
the square root ; cube each side when it is the cube loot ; 
&c. which clears that radical. 

Thus, if y/x — 3=4; then transposing 3, gives y/x = 7 ; 

And squaring both sides gives x = 40. 

And, if y/(2r+ 10) = 8: 

Then by squaring it, it becomes 2r + 10 = 04 ; 

And by transposing 10, it is 2r = 54 ; 

Lastly, dividing bv 2. gives x 27. 

Also, if 3/(3r + 4/+ 3 = ; 
Then by transposing 3, it is y(3r + 4) = 3 ; 
And by cubing, it is 3x -f- 4 — 'SI : 
Also, by transp> ; :ig4, it h !>r = 23 ; 
Lastly, dividing by 3, gives x = 7J. 



When that side of the equation which contains the un- 
known o/jantity is a complete power, or can easily be reduced 
to one, by rule 1, 2, or 3 ; then extract the runt of the said 
power on both sides of the equation ; lhat is, extract the 
square root when it is a square power, or the cube root when 
it is a cube, &c. 

Thus, if x 2 + 8t + 1G = 30, or (x 4) a = 30 ; 
Then by extracting the root, it is x + 4 = ; 
And by transposing 4, it is x = — 4 = 2. 

And if 3x : — 10 = 21 + 35. 
Then, by transposing If), it is 3< 2 = 75 ; 
And dividing by 3, gives x 1 ~ 25 ; 
And extracting the root, gtv«:s i = 5. 

Also, if Jx 3 — fl = 24. 
Then transposing 0, jives J.r* = 30 ; 



KT.LK v. 



Vol. L 



31 



284 



ALGEBRA. 



And multiplying by 4, gives 3s 3 = 120 ; 

Then dividing by 3, gives x* = 40 ; 

Lastly, extracting the root, gives x = ^40 = 6*824556* 



RULE VI. 

Whe.v there is any analogy or proportion, it is to be 
changed into an equation, by multiplying the two extreme 
terms together, and the two means together, and making the 
one product equal to the other. 

Thus, if 2x : 9 : : 3 : 5. 
Then mult, the extremes and means, gives 10* = 27 ; 
And dividing by 10, gives x = 2 T 7 T . 

And if Jx : a : : 56 : 2c. 
Then mult. c.\!rcmes and means, gives jcx = bob ; 
And multiplying by 2, gives 3cx = 10ad; 

t i r r i o l0ab 
Lastly, dividing by 3c, gives x = 

Also, if 10— x : lx : : 3 : 1. 
Then mult, extremes and means, gives 10 — x = 2x ; 
And transposing x, gives 10 =■= 3x ; 
Lustly, dividing by 3, gives 3£ = x. 



RULE VII. 

When the same quantity is found on both sides of an equa- 
tion, with the same sign, cither plus or minus, it may be left 
out of both : and when every term in an equation is either 
multiplied or divided by the same quantity, it may be struck 
out of them all. 

Thus, if 3x + 2a = 2a + b : 
Then, by taking away 2a, it is 3x = b. 
And, dividing by 8, it is x = ±b. 

Also, if there be 4ax + 6ab = 7ac. 
Then striking out or dividing by a, gives 4x + Gb = 7c. 
Then by transposing 66, it becomes 4x = 7c — 6b; 
And then dividing by 4, gives x = jc — $6. 

Again, if fx — f = V — f - 
Then, taking away the £, it becomes fx = f s ° ; 
And taking away the 3's, it is 2x = 10 ; 
Lastly, dividing by 2, gives x = 5. 



SIMPLE EQUATIONS. 



286 



MISCELLANEOUS EXAMPLES. 

1. Given 7* — 18 = 4x + 6 ; to find the value of x. 
Firsts transposing 18 and 4x gives 3x = 24 ; 
Then dividing by 3, gives x = 8. 

2. Given 20 — 4* — 12 = 02 — 10* ; to find x. 
First, transposing 20 and 12 and lOx, gives 6x = 84 ; 
Then dividing by 6, gives x = 14. 

3. Let 4ax — 5b = Sdx + 2c be given : to find x. 
First, by trans. 56 and 9dx 9 it is 4ax — 3cfx = 55 + 2c : 

Then dividing by 4a — 3d, gives x = |~~~^» 

4. Let 5x" — 12x = 9x + 2X 3 be given ; to find x. 
First, by dividing by x, it is 5x — 12 = 9 -f- 2x ; 
Then transposing 12 and 2x, gives 3x=21 ; 
Lastly, dividing by 3, gives x =■• 7. 

5. Given 9ax 3 — 15abx* = Gar* + 12ax 3 ; to find x. 
First, dividing by Sax 3 , gives 3x — 56 = 2x + 4 ; 
Then transposing 56 and 2x, gives x=56 -fi. 

— XXX 

6* Let - — - + ^ =: 2 be given, to find x. 

First, multiplying by 3, gives x — f r-f- 3 ! x - 6 ; 
Then, multiplying by 4, gives x + V*x =24. 
Also multiplying by 5, gives 17x = 120 : 
Lastly, dividing by 17, gives x = 7y T . 

~ ^- x — 5 , x , _ x — 10 „ , 

7. Given — — f- - = 12 — ; to find x. 

First, mult, by 3, gives x — 5 + £r=3G — x + 10 ; 
Then transposing 5 and x, gives *2» , 4-- , ;r=-51 ; 
And multiplying by 2, gives 7x = 102 ; 
Lastly, dividing by 7, gives x=l 1$. 

3x 

8. Let v^-j- + 7 = 10, be given ; to find x. 

First, transposing 7, give3 ^^=3 ; 
Then squaring the equation, gives J x —9 



236 ALGEBRA* 

Then dividing by 3, gives \x = 3 ; 
Lastly, multiplying by 4, gives x = 12. 

6a 3 

9. Let 2x + 2v/(a 2 + x 3 ) = , be given, to find*. 

First, mult, by ^/(a 2 +x t ) 9 gives 2s v /(a 2 +x a )+2a , +2x> 
= 5a 3 . 

Thentransp. 2a 3 and 2X 3 , gives 2x v ^(a a +r , )=3a 1 — 2x»; 
Then by squaring, it is 4x* X (a* + x 3 ) = (3a 1 — 2*") 1 ; 
That is, 4aV -f- 4x 4 = 9a 4 — 12aV + 4x 4 ; 
By taking 4x 4 from both sides, it is 4a 3 x 3 =9a 4 — 12aV; 
Then transposing 12a a x 3 , gives 16a 3 x 3 =9a 4 ; 
Dividing by a 3 gives lCx 3 = 9a 3 ; 
And dividing by 16, gives x 2 =f 9 a 9 ; 
Lastly, extracting the root, gives x=Ja. 

EXAMPLES FOR PIIACTICK. 

1. Given 2r — 5 + 16 = 21 ; to find x. Ans. x=5. 

2. Given Ox — 15 = x + 6 ; to find x. Ans. *=4i. 
5. Given 8— 3x r 12 = 30-5x+4 ; to find x. Ans. x=7. 

4. Given x + Jx — Jx=rl3 ; to find x. Ans. x— 12. 

5. Given 3x+ Jx+2=5x— 4; to find x. Ans, *=4. 

6. Given 4ax+Ja — 2=ax — bx ; to find x. 

AnS ' X=: 9a+3S- 

7. Given -}x — -f \x = J ; to find x. Ans. x — ff. 

8. Given v /(4-fx)=4— v/x ; to find x. Ans. x = 2J. 

X" 

9. Given 4a + x = ; to deter, x. Ans. x = — 2a. 

4(7 -f- x 

10. Given x /{la l + x 3 )= «/(4& 4 + x 4 ) ; to find x. 

Ans. x = \/— sT' 
2a" 

11. Given y/x + y/(2a+x) = — ~— , ; to find x. 

</{Za-\-x) 

Ans. x = |a. 

12. Given —V + ----- = 2b ; to find x. 

l+2x 1 — 2x 

Ans. x=£ \/"^" # 

13. Given a+x= v/(a 3 -fx ^/^P+a*)) ; to find x. 

Ans. x = — — <*• 
a 



SIMPLE EQUATIONS. 



287 



OF REDUCING DOUBLE, TRIPLE, &C. EQUATIONS, CONTAINING 
TWO, THREE, OR MORE UNKNOWN QUANTITIES. 

PROBLEM I. 

lb exterminate two Unknown Quantities; Or, to reduce the 
two Simple Equations containing them, to a Single one. 



RULE I. 

Find the value of one of the unknown letters, in terms of 
the other quantities, in each of the equations, by the methods 
already explained. Then put those two values equal to 
each other for a new equation, with only one unknown quan- 
tity in it, whose value is to be found as before. 

Note. It is evident that we must first begin to find the 
values of that letter which is easiest to be found in the two 
proposed equations. 

EXAXPLE8. 

1. Given ^ 5^ 2y = 14 | ; t0 find x and y * 
In the 1st equat. trans. 3y and div. by 2, gives x = ^ ^ ; 

14+2y 

In the 2d trans. 2y and div. by 5, gives x = — — £ ; 

o 

. 1. . 1 • 14 + 2 y l?-3y 

Putting these two values equal, gives — — ~ = — ^ i 

Then mult, by 2 and 5, gives 28 + Ay = 85 — 15y ; 
Transposing 28 and 15y, gives 1% = 57 ; 
And dividing by 19, gives y = 3. 
And hence x = 4. 

Or, effect the same by finding two values of y, thus 

17-2x 

In the 1st equat. tr. 2x and div. by 3, gives y = — g — ; 

5x— 14 

In the 2d tr. 2y and 14, and div. by 2, gives y = — — ; 

5a; -14 17— 2x 
Putting these two values equal, gives — — — = — g — \ 



288 ALGEBRA. 

Mult, by 2 and by 3, gives 15x — 42 = 34 — 4x ; 
Transp. 42 and 4x 9 gives 19x = 76 ; 
Dividing by 19, gives x = 4. 
, Henoe y = 8, as before. 

2. Given { jjlSfa* j 5 tofindxandy. 

Ans. x = a + b 9 and y = — Ji. 

3. Given 3* + y = 22, and 3y + x = 18; tofindxandy. 

Ans. x =s 6, and y « 4. 

4. Given + jy = 3J | ? to ^ n< * x y# 

Ans. x = 6, and y = 3. 

, 2a: . 3y 22 _ 3a? . 2y . 67 „ _ 

5. Given T + ^= y and- + ^=-; to find. 

and y. Ans. a? =3, and y = 4. 

6. Given x + 2y = s y and x* — 4y* = <P ; to find x and y. 

An»- * = — ^— , and jf = — ^— . 

7. Given x — 2y = d> and x : y : : a : b ; to find x and y. 

^•^^^^^ 

RULE II. 

Find the value of one of the unknown letters, in only one 
of the equations, as in the former rule; and substitute this 
value instead of that unknown quantity in the other equation, 
and there will arise a new equation, with only one unknown 
quantity, whose value is to be found as before. 

Note. It is evident that it is best to begin first with that 
letter whose value is easiest found in the given equations. 

EXAMF E8. 

1. Given |*J + 1* | ; to find x and y. 

This will admit of four ways of solution ; thus ; First, 

17 3 tf 

in the 1st eq. trans. 3y and div. by 2, gives x = — — - . 

This val. subs, for x in the 2d, gives 85 — 2y = 14; 
Hub. by 2, this becomes 86 — 15y — 4y = 28; 



snmx equations. 289 

Transp. 15y and 4y and 28, gives 57 = 19y ; 
And dividing by 19, gives 3 = y. 

l*en, = ^ = 4. 

14 + 2v 

Hly, in the 3d trans. 2y and div. by 5, gives x = — - — - ; 

* This subst. for * in the 1st, gives ^jtl? + 3y =s 17 ; 

Mult, by 5, gives 28 + 4y + 15y = 85 ; 
Transpo. 2*5, gives 19y = 57 ; 
And dividing by 19, gives y = 3. 
14 + 2y 

Then * = — - s= 4, as before. 

o 



8d1y, in the 1st trans. 2x and div. by 3, gives y = ? 

34 4 X 

This subst for y in the 2d, gives Ex ^ = 14 ; 

Multiplying by 3, gives 15x — 34 + 4x = 42 ; 
Transposing 34, gives 19x = 76 ; 
And dividing by 19, gives x = 4. 
17 2x 

Hence y = — = 3, as before. 

3 



5x — 14 

4thly, in the 2d tr. 2y and 14 and div. by 2, gives y = — - — . 

15x 42 

This substituted in the 1st, gives 2x H - = 17 ; 

Multiplying by 2, gives 19x — 42 = 34 ; 
Transposing 42, gives 19x = 70 ; 
And dividing by 19, gives x = 4 ; 
5 X 14 

Hence y = — = 3, as before. 



2. Given 2x + 3y = 29, and 3x — 2y = 11 : to find x 
and y. Ans. x = 7, and y = 5. 

,3. Given | * * ^Z. l \ ^ ; to find x and y. 

Ans. x atv^^^^ 



MO ALGEBRA. 

4. Given | ^^20 | ;*<> find x and y. 

Ans. x = 6, and y = 4. 

5. Given g + 3y = 21, and | + 3x = 29 ; to find r 
and y. Ans. x = 9, and y = 6L 

6. Given 10 - | = | + 4, and + f - 2 = ' 
3y — x 

-~r 1 ; to find x and y. Ans. x = 8, and y =■ 6. 

7. Given x : y : : 4 : 3, and x 3 — y 1 = 37 ; to find % 
and y. Ans. x = 4, and y = 8.^ 



RULE III. 



Lbt the given equations be so multiplied, or divided, &c. 
and by such numbers or quantities, as will make the terms 
which contain one of the unknown quantities the same in 
both equations ; if they are not the same when fir3t pro- 
posed. 

Then by adding or subtracting the equations, according 
as the signs may require, there will result a now equation, 
with only one unknown quantity, as before. That is, add 
the two equations when the signs are unlike, but subtract 
them when the signs are alike, to cancel that common 
term. 

Note. To make two unequal terms become equal, as 
above, multiply each term by the co-efficient of the other. 



EXAMPLES. 



r g,. __. 9 ^ 

Given i 2x + 5y = 16 \ ' t0 find x and y ' 

Here we may either make the two first terms, containing rr , 
equal, or the two 2d terms, containing y, equal. To mak.^ 
the two first terms equal, we must multiply the 1st equation 
by 2, and the 2d by 5 ; but to make the two 2d terms equaT, 
we must multiply the 1st equation by 5, and the 2d by 3 ; 
Ma follows. 



SIMPLE EQUATIONS. 



241 



1. By making the two first terms equal : 

Mult the 1st equ. by 2, gives lOx — 6y = 18 

And mult, the 2d by 5, gives lOx + 25y = 80 

Subtr. the upper from the under, gives Sly = 62 
And dividing by 31, gives y = 2. 

Hence, from the 1st given equ. * = — -— ? = 3. 

2. By making the two 2d terms equal : 

Mult, the 1st equat. by 5, gives 25x — 15y = 45 ; 
And mult, the 2d by 3, gives Ox + 15y = 48 ; 
Adding these two, gives 31x = 93 ; 

And dividing by 31, gives x = 3. 

c y 9 

3 Hence, from the 1st equ. y = — - — = 2.' 



MISCELLANEOUS EXAMPLES. 



1. Given + 6y = 21, and ^ + 5x = 23 ; to find 
x and y. Ans. x = 4, and y = 3. 

2. Given + 10 = 13, and + 5= 12 ; to find 
x and y. Ans. a: = 5, and y = 3. 

3 . Give „?^ + *=10,and^ + ? = 14 ; tofind 

5 4 o o 

x and y. Ans. x = 8, and y = 4. 

4. Given 3x+4y=38, and 4r — 3y ~ 9 ; to find x and y. 

Ans. x = 6, and y = 5. 



problem: hi. 

y 

To exterminate three or more Unknotcn Quantities ; Or, to 
reduce the simple Equations, containing them, to a Single 
one* 

RULE. 

This may be done by any of the three methods in the last 
problem : viz. 

1. After the manner of the first rule in the last problem, 
find the value of one of the unknown letters in each of the 
given equations ; next put two of these values equal to each 
other, and then one of these and a third value equal, and so 
on for all the values of it ; which gives a new set of c^ftvcycA, 

Vol. 1. 32 



243 



AXGEBSJU 



with which the same process is to be repeated, and so on till 
there is only one equation, to be reduced by the rules for a 
single equation. 

2. Or, as in the 2d rule of the same problem, find the value 
of one of the unknown quantities in oue of the equations only; 
then substitute this value instead of it in the other equations ; 
which gives a new set of equations to be resolved as before, 
by repeating the operation. 

3. Or, as in the 3d rule, reduce the equations, by multi- 
plying or dividing them, so as to make some of the terms to 
agree : then, by adding or subtracting them, as the signs 
may require, one of the letters may be exterminated, dec. as 
before. 



EXAMPLES. 



Cx + y + z = 91 
I. Given < x 4- 2y +3z = 16 > ; to find x 9 y, and z. 

(x + 3y +4* = 21) 
1. By the 1st method : 
Transp. the terms containing y and z, in each cqua. gives 
x = 9 — y — 2, 
*= 16 — 2y — 3z, 
x = 21 — 3y — 4z; 

Then putting the 1st and 2d values equal, and the 2d and 3d 
values equal, give 

9— y— z = 16— 2?/ — 32, 
16 _2y —32 =21 - 3y — 4z ; 
In the 1st trans. 9, 2, and 2y, gives y = 7 — 2z ; 
In the 2d trans. 16, 3z, and 3y, gives y = 5 — z ; 
Putting these two equal, gives 5 — z = 7 — 22. 
Trans. 5 and 2z, gives 2 = 2. 
Hence y = 5 — 2 = 3, and x = 9 — y — 2 — 4- 
2dly. By the 2d method : 

From the 1st equa. x =• 9 — y — 2 ; 
This value of x substit. in the 2d and 3d, gives 
9 + y + 22 = 16, 
9 + 2y + 32 = 21 ; 
lii the 1st trans. 9 and 22, gives y = 7 — 2z ; 
This substit. in the last, gives 23 —2 = 21; 
Trans, z and 21, gives 2 = 2. 
Hence again y = 7 — 22 = 3, and a: = 9 — y — * — 



SIMPLE EQUATIONS. 248 

8d1y. By the 3d method : subtracting the 1st equ. from 
the 2d, and the 2d from the 3d, gives 
y + 2z = 7, 
y+ * == 5 ; 
Subtr. the latter from the former, gives z = 2. 
Hence y = 5 — z = 3, and x = 9 — y — : = 4. 



(*+ y+ * = 18) 
2. Given ^x + 3y + 2* = 38V; 

+ + i* = i<0 

Ans. x = 
+ 4y + i* = 27) 
+ Jy + i* = 20 } ; to 

+ *y + = 16 J 



to find x, y, and z. 



Ads. x = 4, y = 6, z = 8. 

> + 4y + i* = 27, 

3. Given < * + |y + |z = 20 J. ; to find x, y, and z, 

Ans. x = 1, y = 12, z = 60. 

4. Given * — y = 2, x — z = 3, and y + z = 9 • to 
find x, y, and z. Ans. x = 7, y = 5, z = 4. 



5. Given < 3x + 4y + 5z = 46 ) ; to find x, y, and z. 

( 2x + 6y + 8z = 68 ) 
ix(x + y + z)= 4jL^ 

6. Given J y (x + y + z) = ; to find x, y, and z. 

+ * + = 10?J 

J 

A COLLECTION OF QUESTIONS PRODUCING SIMPLE 
EQUATIONS. 

Quest. 1. To find two numbers, such, that their sum 
shall be 10, and their difference 6. 

Let x denote the.greater number, and y the less"" 
Then, by the 1st condition x + y = 10, 
And by the 2d - - . x — y = 6, 
Transp. y in each, gives x =■= 10 — y, 
and x = 6 + y ; 
Put these two values equal, gives 6 + y= 10 — y; 
Transpos. 6 and — y, gives - 2y = 4 ; 
Dividing by 2, gives - - y = 2. 
And hence - - • - x = 6 -f- y = 8. 



* la these solutions, as many unknown letters are always used as 
there are' unknown numbers to be found, purposely for exercise in the 
nodes of reducing the equations : avoiding the short ways of notation, 
which, though they may give neater solutions, afford less eiertV6«\u 
practising the several rules in reducing equations, 



244 



AL€SBBA. 



Quest. 2. Divide 100Z among a, b, c, so that ▲ may hate 
20Z more than b, and b 10/ more than c. 

Let x = a's share, y = b's, and z = c's. 
Then x + y + % — 100, 
x = y + 20, 
y = z + 10. 

In the 1st suhstit. y + 20 for x, gives 2y + z + 20 = 100 ; 
' In this substituting z + 10 for y, gives 3* + 40 = 100 ; 
By transposing 40, gives - - 3z = 60 ; 
And dividing by 3, gives - - z = 20. 
Hence y = z + 10 = 30, and x = y + 20 = 50. 

Quest. 3. A prize of 500Z is to be divided between two 
persons, so as their shares may be in proportion as 7 to 8 ; 
required the share of each. 

Put x and y for the two shares ; then by the question, 

7 : 8 : : x : y, or mult, the extremes, 
and the means, 7y = 8x, 

and x + y = 500 ; 
Transposing y, gives x = 500 — y ; 
This substituted in the St, gives 7y = 4000 — Sy ; 
By transposing 8y, it is*f 5y = 4000 ; 
By dividing by 15, it gives y = 266f ; 
And hence x = 500^- y = 233£. 

Quest. 4. What fraction is that, to the numerator of 
which if 1 be added, the value will be \ ; but if 1 be added 
to the denominator, its value will be £ ? 

x 

Let — denote the fraction. 

y 

Then by the quest. * ^ 1 = 4, and — = 4. 

The 1st mult, by 2 and y, gives 2x + 2 = y ; 
The 2d mult, by 3 and y + 1, is 3x = y + 1 ; 
The upper taken from the under leaves x — 2 = 1; 
By transpos. 2, it gives x = 3. 
And hence y = 2x + 2 = 8 ; and the fraction is f . 

Quest. 5. A labourer engaged to serve for 30 days oflK 
these conditions : that for every day he worked, he was 
receive 20d, but for every day he played, or was absent, he» 
was to forfeit lOd. Now at the end of the time he had t» 
receive just 20 shillings, or 240 pence. It is required ts» 



SIMPLE EQUATIONS. 



2*5 



find how many days he worked, and how many he was 
idle? 

Let x he the days worked, and y the days idled. 
Then 20x is the pence earned, and lOy the forfeits ; 
Hence, by the question - x + y = 30, 

and 20x - lOy = 240 ; 
The 1st mult, by 10, gives lOx + lOy = 300 ; 
These two added, give - 30x =■ 540 ; 
This div. by 30, gives . x = 18, the days worked ; 
Hence - y = 30 — x = 12, the days idled. 

Quest. 6. Out of a cask of wine which had leaked away 
30 gallons were drawn ; and then, being guaged, it appear- 
ed to be half full ; how much did it hold? 

Let it he supposed to have held x gallons, 

Then it would have leaked }x gallons, 

Conseq. there had been taken away }x + 30 gallons. 

Hence Jx = \x + 30 by the question. 

Then mult, by 4, gives 2x = x + 120 ; 

And transposing x, gives x = 120 the gallons it held. 

Quest. 7. To divide 20 into rVo such parts, that 3 times 
tlie one part added to 5 times the other may make 76. 

Let x and y denote the two parts. 
Then by the question - - x + y = 20, 

and 3x + by = 76. 
Mult, the 1st by 3, gives - 3x + 3y = 60 ; 
Subtr. the latter from the former, gives 2y = 16 ; 
And dividing by 2, gives - - y = 8. 
Hence, from the 1st, - x = 20 — y = 12. 

Quest. 8. A market woman bought in a certain number 
of eggs at 2 a penny, and as many more at 3 a penny, and 
•old them all out again at the rate of 5 for two-pence, and 
so doing, contrary to expectation, found she lost 3d ; what 
number of eggs had she ? 

Let x = number of eggs of each sort, 
Then will -£x = cost of the first sort, 
And \x = cost of the second sort ; 
But 5:2 : : 2x (the whole number of eggs) : |* ; 
Hence jx = price of both sorts, at 5 for 2 pence ; 
Then by the question |r + J* — f x = 3 ; 
JVfult. by 2, gives - x + \x — -fx = 6 ; 
And mult, by 3, gives 5x — *£x — IB; 
Also mult, by 5, gives x = 90, the number of tfjgt *t 
each sort. 



246 



ALGEBRA* 



Quest. 9. Two persons, a and b, engage at play. Be- 
fore they begin, a has 80 guineas, and b has 60. After a 
Certain number of games won and lost between them, a rises 
with three times as many guineas as b. Query, how many 
guineas did a win of b ? 

Let x denote the number of guineas a won. 
Then a rises with 80 + x> 
And n rises with 60 — x ; 
Theref. by the quest. 80 + x = 180 — Sx ; 
Transp. 80 and 3x, gives Ix = 100 ; 
And dividing by 1, gives x = 25, the guineas won. 



QUESTIONS FOB PRACTICE. 

1. To determine two numbers such, that their difference 
may be 4, and the difference of their squares 64. 

Ans. 6 and 10. 

2. To find two numbers with these conditions, viz. that 
half the first with a third part of the second may make 9, 
and that a 4lh part of the first with a 5th part of the second 
may make 5. Ans. 8 and 15. 

3. To divide the number 20 into two such parts, that a 
3d of the one part added to a 5th of the other, may make 6. 

Ans. 15 and 5. 

4. To find three numbers such, that the sum of the 1st 
and 2d shall be ?, the sum of the 1st and 3d 8, and the sum 
of the 2tl and 3d 0. Ans. 3, 4, 5. 

5. A father, dying, bequeathed his fortune, which was 
2800/, to his son and daughter, in this manner ; that for eve- 
ry half crown the son might have, the daughter was to have 
a shilling. What then were their two shares ? 

Ans. The son 2000/ and the daughter 800L 

6. Three persons, a, b, c, make a joint contribution, which 
in the whole amounts to 400/ : of which sum b contributes 
twice as much as a and 20/ more ; and c as much as a and 
b together. What sum did each contribute ? 

Ans. a 60/, b 140/, and c 2001. 

7. A person paid a bill of 100/ with half guineas and 
crowns, using in all 202 pieces ; how many pieces were there 
of each sort ? Ans. 180 half guineas, and 22 crowns. 



SIMPLE EQUATIONS. 



247 



8. Says a to u, if you give me 10 guineas of your money, 
, I , Bhall then have twice ?.s much us you will have left : but 

says n to a, give me 10 of your guineas, and then I shall 
have 3 times as many as you. How many had each ? 

A us. a 22, n 2G. 

9. A person goes to a tavern with a curtain quantity of 
money in his poc!:et, where ho spends 2 .shillings ; he then 
borrows as much money as h: j had left, and going to another 
tavern, he there spends '2 shillings also ; then borrowing 
again as much money as was left, he went to a third tavern, 
where likewise he spent 2 shillings ; and thus repeating tho 
same at a fourth tavern, he then had nothing remaining. 
What sum had he at first ! A us. 3*. Orf. 

10. A man with his wife and child dine together at an 
inn. The landlord charged 1 shilling for the child ; and for 
the woman he charged as much as for the child and \ as 
much as for the man ; and for the man he charged as much 
as for the woman and child together. How much was that 
for each ? Ans. The woman 20c/ and the man 32</. 

11. A cask, which held (50 gallons, was tilled with a 
mixture of brandy, wine, and cyder, in this manner, viz. 
the cyder was gallons more than tho brandy, and the 
wine was as much as the cyder and \ of the brandy. How 
much was there of each ? 

Ans. Brandy 15, cyder 21, wine 24. 

12. A general, disposing his armv into a square form, 
finds that he has 2 SI men more than a perfect square : but 
increasing the side by 1 niau, he men wants 25 men to be a 
complete square. How iininv men had lie under his com- 
mand ? " Ans. '24000. 



13. What number is that, to which if 3. 5, and 8, bo 
severally added, the three Minis shall be in geometrical pro- 
gression ? Ans. 1. 

14. The stock of three traders amounted to 7*50/ : the 
shares of the first and second exceeded that of the third 
by 240 : and the sum of the 2d and 3d exceeded the first 
by 360. What was the share of each .' 

Ans. The 1st 200, the 2d 300, the 3d 200. 

15. What two numbers are those, which, !;cing in the 
ratio of 3 tu 1, their product is equal to 12 times their sum 



248 



ALGEBRA. 



16. A certain company at a tavern, when they came to 
settle their reckoning, found that had there been 4 more in 
company, they might have paid a shilling each less than 
they did ; but that if there had been 3 fewer in company, 
they must have paid a shilling each more than they did. 
What then was the number of persons in company, what 
each paid, and what was the whole reckoning ? 

Ans. 24 persons, each paid 7s, and the whole 
reckoning 8 guineas. 

17. A jockey has two horses : and also two saddles, the 
one valued at 18/. the other at 3/. Now when he sets the - 
better saddle on the 1st horse, and the worse on the 2d, it 
makes the first horse worth double the 2d ; but when he 
places the better saddle on the 2d horse, and the worse on 
the first, it make* the 2d horse worth three times the 1st. 
What then were the values of the two horses ? 

Ans. The 1st 0/, and the 2d 9/. 

18. What two numbers arc as 2 to 3, to each of which if 
6 be added, the sums will be as 4 to 5 ? Ans. and O. 

19. What arc those two numbers, of which the greater is 
to the less as their sum is to 20, and as their difference is to 
10? Ans. 15 and 45. 

20. What two numbers are those, uhosc difference, sum, 
and product, are to each other, as the three numbers 2, 
3, 5 ? Ans. 2 and 10. 

21. To find three numbers in arithmetical progression, of 
which the first is to the third as 5 to 0, and the sum of all 
three is 03. Ann. 15, 21, 27. 

22. It is required to divide the number 2i into two such 
parts, that the quotient of the greater part divided by the 
less, may be to the quotient of the less part divided by the 
greater, as 4 to 1. Ans. 10 and 8. 

23. A gentleman being asked the age of his two sons, 
answered, that if to the sum of their ages 18 be added, the 
result will be double the age of the elder ; but if be taken 
from the difference of their ages, the remainder will be equal 
to the age of the younger. What then were their ages ? 

Ans. 30 and 12. 

24. To find four numbers such, that the sum of the 1st, 
2d, and 3d shall be 13 ; the sum of the 1st, 2d, and 4th, 
15 ; the sura of the 1st, 3d, and 4th, 18 ; and lastly, the sum 
of the 2d, 3d, and 4th, 20. Ans. 2, 4, 7, 9. 



4/ / 



r*6 

i^tk M 6x> y » r 
f& i** to ic~/r J 



tiro 



*-¥ 'ft* 



-11 






i+t.s.i+tjfo ° isL^ri* -fa 



2-* It* 



•V v «*r iiZ.C-?> r**y~i u-*y> 

v . 1 



v --vr 



lor>fr|i 



T 



■»-*-/f 



xaej. 



QUADRATIC EQUATIONS. 



249 



• 

25. To divide 48 into 4 such parts, that the first increased 
by 3, the second diminished by 3, the third multiplied by 3, 
and the 4th divided by 3, may be all equal to each other. 

Ans. 6, 12, 3, 27. 



QUADRATIC EQUATIONS. 

Quadratic Equations are either simple or compound. 

A simple quadratic equation, is that which involves the 
square only of the unknown quantity. As ax 3 = b. The 
solution of such quadratics has been already given in simple 
equations. 

A compound quadratic equation, is that which contains 
the square of the unknown quantity in one term, and the 
first power in another term. As or 1 + bx = c. 

All compound quadratic equations, after being properly 
reduced, fall under the three following forms, to which they 
must always be reduced by preparing them for solution. 

1. x 3 + ar = 6 

2. x a — ax = b 

3. x* — ax = —b 

The general method of solving quadratic equations, is by 
what is called completing the square, which is as follows : 

1. Reduce the proposed equation to a proper simple form, 
as usual, such as the forms above ; namely, by transposing 
all the terms which contain the unknown quantity to one 
aide of the equation, and the known terms to the other ; 
placing the square term firs., and the single power scr md ; 
dividing the equation by the co-efficient of the square or 
first term, if it has one, and changing the signs of all the 
terms, when that term happens to be negative, as that 
term must always be made positive before the solution. 
Then the proper solution is by completing the square as 
follows, viz. 

2. Complete the unknown side to a square, in this man. 
ner, viz. Take half the co-efficient of the socond term, «r»d 
square it ; which square add to both sides of the equation, 
then that side which contains the unknown quantity will be 
a complete square. 

Vol. L 33 



250 



ALGEBRA. 



3. Then extract the square root on both sides of the 
equation*, and the value of the unknown quantity will be 
determined, making the root of the known side either + or 
— , which will give two roots of the equation, or two values of 
the unknown quantity. 



* As the square root of any quantity may be cither + or—; there- 
fore all quadratic equations admit of two solutions. Thus, the square 
root of -j- n * 's either -j- n or — n; for -j- n X + « and — n X — • 
are each equal to -|- n a . But the square root of — n 8 , or V — is 
imaginary or impossible, as neither -j- n nor — n, when squared, gives 

— i* 

So. in the first form, x*~±-az - ft, where x -j- fa is found = V(b + 
the root may be either -J- V(b + fa 2 ), or — V(b -j- fa 2 \ since either 
of them being multiplied by itself produces b -\- fa\ And this ambi- 
guity is expressed by writing the uncertain or double sign + before 
V(b 4- fa*) ; thus x = ± V(b + fa 2 ) — fa. 

In this form, where i - Hh V(b -\- fa 2 ) — fa, the first value of a:, via. 
x = -j- V(6 + d« v ) — 4» is always affirmative ; for since J« y -f b is 
greater than fa-, the greater square must necessarily have the greater 
root; therefore v(b fa 7 ), will always be greater than Vjir 3 , or its 
equal fa ; and consequently -j- V(b-\- fa*) — fa will always be affirm- 
ative. 

The second value, viz. x — — \/(b -)- fa 2 ) — fa will always \te nega- 
tive, because it is composed of two negative terms. Therefore when 
x 9 -}- ax = b, we shall have x — -j- \'{b -f- ja 3 ) — for the affirmative 
value of x % and x = — -f- Ja 2 ) — for the negative value of x. 

In the second form, where x = ± V(6 -f- -f- fa tne fi^t value, 
viz. i — + H" ~r ?l a is always affirmative, since it is composed 
of two affirmative terms. But the second value, viz. x = — V(6-f fa 7 ) 
•4- £a, will always be negative; for since b-\- fa 2 is greater than 4«* t 
therefore v(6 -{- fa") will be greater than vfa 2 , or its equal ^cr ; and 
consequently — V(n -|- fa' : ) \- fa is always a negative quantity. 

Therefore, when x 2 — ar — 0, we shall havez — -j- \ '(b -j- fa 2 ) fa 
for the affirmative value of x ; and x = — - v(& -)- fa'<) -j- j« for toe 
negative value of x; so that in both the first arid second forms, the un- 
known quantity has always two value>, one of which is positive, and 
the other negative. 

But, in the third form, whero .t - -± \ '(fa 2 — b) -' r fa, both the 
values of x will be positive, when fa 2 is greater than b. For the first 
value, viz. x =z + V(\a l — b) -f- fa will then be affirmative, being com- 
posed of two affirmative terms. 

The second value, viz. s - — V(fa- — b) -i- fa is affirmative also ; 
for since fa 2 is greater than fa 2 — b, therefore Vfa- or fa is greater than 
V(fa 2 — b) ; and consequently — v'(i«- — 6) -f- A<i will always be an 
affirmative quantity. So that, when x* — ax — — b, wc shall have z 

— + v(fa l — M 4a. and also x — \ {fa 2 — b) -f- fa, for the 
values of x, both positive. 

But in this third form, if b be greater than fa 1 , the solution of the pro- 
posed question will be i n possible. For bince the square of any quan- 
tity (whether that quantity be affirmative or negative) is always affirma- 
tive, the square root of a negative quantity is impossible, and cannot 
be assigned. But when b is greater than fa\ then fa* — b is a nega- 
tive quuntity ; and therefore its root V(fa J — b) is impossible, or ima- 
ginary ; consequently, in that case, z = fa + V(fa~ — b), or the two 
rood or values of x, are both impo^'ible^ or ioiaginary quantities. 



QUADRATIC KQUATIONS. 



251 



Note, 1. The root of the first side of the equation, is 
always equal to the root of the first term, with half the co- 
efficient of the second term joined to it, wi;ii its sign, whether 
+ or — . 

2. All equations, in which there are two terms including 
the unknown quantity, and which have the index of the one 
just double that of the other, are resolved like quadratics, by 
completing the square, as above. 

Thus, x* + ax 2 6, or x** + ax n = b, or x + ax* = 6, 
or (x 2 ± ax) 2 ±. m{x 2 :£ ax) = 6, are analsgous to quadra- 
tics, and the value of the unknown quantity may be deter, 
mined accordingly. 

3. For the construction of Quadratics, see vol. ii. 

EXAMPLES. 

1. Given x* + 4x = 60 ; to find x. 

First, by completing the square, x' 2 + 4r + 4 = 64 ; 

Then, by extracting the root, x + 2 = ±. 8 ; 

Then, transpos. 2, gives x = 6 or — 10, the two roots* 

9 2. Given x a — 6x + 10 = 65 ; to find x. 
First, trans. 10, gives x 2 — 6x = 55 ; 
Then by complet. the sq. it is x 2 — 6x + 9 = 64 ; 
And by extr. the root, gives x — 3 = ± 8 ; 
Then trans. 3, gives x = 1 1 or — 5. 

3. Given 3x 2 — 3* + 9 *= 8| ; to find x. 
First div. by 3, gives x 2 — x + 3 = 2 J ; 
Then transpos. 3, gives x 2 — x = — J ; 
And compl. the sq. gives x 2 — x + J = ^ ; 
Then extr. the root gives x — ^ = i ; 
And transp. J, gives x = | or 

4. Given i* 2 — £r + 30J = 52£ ; to find x. 
First by transpos. 30J, it is \x 2 — \x = 22$ ; 
Then mult, by 2 gives x 2 — §x=44£ ; 

And by compl. the sq. it is x 2 — §x4-J=44J. 
Then extr. the root gives x — £ = tt 6} ; 
And trans. £, gives x = 7 or <— 6£ ; 

5. Given ax 2 — bx = c ; to find x. 
First by div. by a, it is x 1 — - s = — ; 



9B& AXGBBRA. 

Then compl. the sq. gives a*—- a: +^ f =- 

b 4ac+ b* 

And extrnc. the root, gives * — ^ = :£ — ^ — ; 

Then transp. gives * = ± + 

6. Given a* — 20**= 6 ; to find a?. 

First by compl. the sq. gives a?— 2ox+o a ==a a 4-o ; 
And extract, the root, gives a*— a = ± ^/(a 2 — b) ; 
Then transpos. a, gives a* = ± ^/(a 1 + 6) + a ; 
And extract, the root, gives a? = + ^/[a + ^/(a'+i)]. 



SXAMFLB8 FOR PRACTICE*. 

J. Given a?— 62-7 = 33 ; to find 2. Ans. 2 = 10 or-4. 



* 1. Cubic equations when occurring in pairs, may usually be reduced 
to quadratic*, by extermination. Thus, 

Suppose 4z 4- 3x* -r- 6z = 150 > 
and 3z*-f 2x a + 2l = 106 5 
Then molt. 1st equa. by 3, and 2d by 4, 

1223 4- 9xa 4- 15z = 450 
12x 3 4-8** + gz =420 

By aobtr. z* 4- 7x = 30 

Compl. the sq. z» 4- 7x + = 30 + ±f. = xfi. 

Extr. the root z + f = ± V 

z = £=3 or — 10. 

2. Sometines, when the unknown square has a co-efficient, the fol- 
lowing method may be advantageously adopted : viz. 

Having transposed the known terms to one side and the unknown 
terms to the other, multiply each side by 4 times the co-efficient of the 
unknown square. 

Add the square of the co-efficient of the simple power of the un- 
known quantity, to both sides; the first side will then be a complete 
square. 

Extract the root, and the value of the unknown quantity will be ob- 
tained. 

Thus, if 5x a + 4z = 28. 

Then mult, by 4 X 5 t lOOx* 4- 80x = 660 
Add 4 2 . . 100x2 4- 80x4-16* = 576 
Eitr. the root, lOz -f 4 = ± 24 
Transposing . lOz = 20 or — 28 
Dividing by 10, x = 2, or — 2 8. 
The principal advantage of this method, which is due to the Indians, 
Is that it dots not introduce fractions into the operation. It will have 
the same advantage in cases where the square has 00 co-efficient, if that 
of the simple fwwer be an odd number. 



QUADRATIC EQUATIONS. 



208 



2. Given X 1 — 5x— 10 = 14 ; to find x. Ans. x = 8 or — 3. 

3. Given 5x* + Ax — 90 = 114 ; to find x. 

Ans. a? = 6 or — ftf • 

4. Given Jx 1 — $x + 2 =» 9 ; to find x. Ana. x = 4 or - 3J. 

5. Given 3x* — 2x" = 40 ; to find a?. Ans. x = 2 or —2. 

6. Given Jx— J^/x = 1J ; to find x. An*, x =9 or 2}. 

7. Given ix* + fx = J ; to find x. * 

Ans. x = — | ± | */70 — -7277668 or —2-0611000. 

8. Given x 8 + 4x* = 12 ; to find x. 

Ans. x = = 1-259921, or y — 6 = - 1-817121. 

9. Given X 2 + 4x = a* + 2 ; to find x. 

Ans. x — v'(« 2 +6)^2. 

questions pboduoing quadratic equations. 

1. To find two numbers whose difference is 2, and pro- 

duct 80. 

Let x and y denote the two required numbers. 
Then the first condition gives x — y = 2, 
And the second gives xy •= 80. 
The n transp. y in the 1st gives x = y + 2 ; 
This value of x suhstitut. in the 2d, is y 3 + 2y = 80 ; 
Then comp. the square gives y 3 + 2y + 1 = 81 ; 
And extrac. the root gives y + 1 = 9 ; 
And transpos. 1 gives y = 8 ; 
And therefore x = y + 2 = 10. 

2. To divide the number 14 into two such parts, that their 

product may be 48. 
Let x and y donote the two parts. 
Then the 1st condition gives x + y *= 14, 
And the 2d gives xy = 48. 
Then transp. y in the first gives x =-* 14 — y ; 
This value subst. for x in the 2d, is 14y — j/ 5 = 48 ; 
Changing all the signs, to make the square positive, 
gives y 2 — 14y = — 48 ; 
Then com pi. the square gives y 3 — 14y + 49 = 1 ; 
And extrac. the root gives y — 7 = 4-1; 
Then transpos. 7, gives y = 8 or 6, the two parts. 

3. What two numbers are those, whose sum, product, and 
difference of their squares, are all equal to each other ? 

Let x and y denote the two numbers. 

Then the 1st and 2d expression give x + y =. xy* 

And the 1st and 3d give x + y = a?— . 



254 



ALGEBRA. 



Then the last equa. div. by x + y, gives 1 = x — y ; 
And transpos. y, gives y + 1 = x ; 
This val. substit. in the 1st gives 2y + 1 = y' + y ; 
And transpos. 2y, gives 1 = y 2 — y ; 
Then complet. the sq. gives f = y* — y + j ; 
And extracting the root gives J ^5 = y — £ ; ■ 
And transposing \ gives i v/5 + J = y ; 
And therefcre x = y + 1 = J + 
And if these expressions be turned into numbers, by ex- 
tracting the root of 5, &c. they give x = 2-6180 +, and 
y= 1-6180 +. 

4. There are four numbers in arithmetical progression, of 
wtth the product of the two extremes is 22, and that of the 
means 40 ; what are the numbers ? 

Let x = the less extreme, 

and y = the common difference ; 
Then r, x+y, x-f 2y, x-fc3y, will be the four numbers. 
Hence, by the 1st condition .t 2 + 3.ry = 22, 
And by the 2d x 2 + Sry + 2y a = 40. 
Then subtracting the first from the 2d gives 2y* = 18 ; 
And dividing by 2 gives y 1 = 9 ; 
And extracting the root gives y = 3. 
Then substit. 3 for y in the 1st, gives x 2 r|n 9x = 22 ; 
And completing the square gives x 3 "+ 9x + V = "f 9 ; 
Then extracting the root gives x -f- J- = y ; 
And transposing g- gives a: = 2 the least number. 
Hence the four numbers are 2, 5, 8, 11. 

5. To find 3 numbers in geometrical progression, whose 
sum shall be 7, and the sum of their squares 21. 

Let x, y, and z denote the Ihree numbers sought. * 
Then by the 1st condition xz = y 2 , 
And by the 2d x + y + z = 7, 
And by the 3d x a + + z* = 21. 
Transposing y in the 2d gives x + z = 7 — y ; 
Sq. this equa. gives x 2 ■+ 2xx + 2 a = 49 — 14y + y"; 
Substi. 2y 2 for 2xs, gives x 3 + 2y= + z 2 = 49 — 14y f ; 
Subtr. y 2 from each side, leaves x 3 + y 2 + z 2 = 49 — 14y ; 
Pulling the two values of x 2 + y 3 + z 3 > 21 = 49 _ - m 
equal to each other, gives $ y * 
Then transposing 21 and 14y, gives 14 v = 28 ; 
And dividing by 14, gives y = 2. 
Then substit. 2 for // in the 1st equa. gives xz = 4, 
And in the 4th, it gives x + z = 5 ; 
Transposing z in the last, gives x = 5 — z ; 
This subst. in the next above, gives bz — z* = 4 ; 



QUADRATIC EQUATIONS. 



255 



Changing all the signs, gives r 1 — hz = — 4 ; 

Then completing the square, gives z 2 — 5z + V = { ? 

And extracting the root gives z — \ = ±: \\ 

Then transposing £, gives * and i = 4 and 1, the two 

other numbers* ; 
So that the three numbers are 1, 2, 1. 

QUESTIONS KOR FRACTICK. 

1. What number is that which added to its square makes 
42? Ans. C, or — 7. 

2. To find two numbers such, that the less may be to \jfie 
greater as the greater is to 12, and that the sum of Jhf ir 
squares may be 45. j\ns. 3 anR>. 

3. What two numbers are those, whose difference is 2, 
and the difference oi' their cubes 98? Ans. 3 and 5. 

4. W T hat two numbers arc those, whose sum is 0, and the 
sum of their cubes 72 ? * Ans. 2 and 4. 

5. What two numbcis are those, whose product is 20, and 
the difference of their cubes (51 Ans. 4 and 5. 

6. To divide the number 11 into two such parts, that the 
product of their squares may be 784. Ans. 4 and 7. 

7. To divide the number 5 into two such parts, that the 
sum of their alternate quotients may b«; that is of the 
two quotients of each part divided by the other. 

Ans. i r s nd 4. 

8. To divide 12 into two such parts, Ihsit their product 
may be equal to 8 limes their dilfei'Micc. Ans. 1 and 8. 

1). To divide the number 10 into two such parts, that the 
square of 4 times the less part, may be 112 more than the 
square of 2 times the greater. Ans. 4 and 6. 

10. To find two numbers such, that the sum of their 
squares may be 80, and their sum multiplied by the greater 
may produce 104. J^ns. 5 and 8. 

11. What number is that, which bcin<* divided by the 
product of its two dibits, the quotient is V ; ; but when 9 is 
subtracted from it, there remains a uurnher Laving the same 
digits inverted ? A ns. 32. 

12. To divide 20 into three !>;• its suc : i. the continual 
product of all throe may b* 2"0, ;nnl th.it th« difference of 
the first and second may b^ 2 h <..* than thi* diflc re nee of the 
second and third. Ans. 5, 6, 9. 

13. To find three numbers in arithmetical progression^ 
such that the sum of their squares may he wuixYifc tori 



256 ALGEBRA. 

arising by adding together 3 times the first and 2 times the 
second and 3 times the third, may amount to 32. 

Ans. 2, 4, 6. 

14. To divide the number 13 into three such parts, that 
their squares may have equal differences, and that the sum 
of those squares may be 75. Ans. 1, 5, 7. 

15. To find three numbers having equal differences, so 
mat their sum may be 12, and the sum of their fourth powers 
062. Ans. 3, 4, 5.. 

16. To find three numbers having equal differences, and 
such that the square of the least added to the product of the 
two greater may make 28, but the square of the greatest 
adflsd to the product of the two less may make 44. 

W Ans. 2, 4, 0. 

17. Three merchants, a, r, c, on comparing their gains 
find, that among them all they have gained 1444/ ; and that 
b's gained added to the square root of a's made 920/ ; but if 
added to the square root of c*s it made 912/. What were 
their several gains ? Ans. a 400, b 900, c 144. 

.18. To find three numbers in arithmetical progression, so 
that the sum of their squares shall be 93 ; also if il e first 
be multiplied by 3, the second by 4, and the third by" 5, 
the sum of the products may be 66." Ans. 2, 5, 8. 

19. To find two numbers such, that their product added 
to their sum may make 47, and their sum taken from the 
sum of their squares mr.y leave 02. Ans. 5 and 7* 



RESOLUTION OF CUBIC AND HIGHER 
EQUATIONS. 

A Cubic Equation, or Equation of the 3d degree or 
power, is one that contains the third power, of the unknown 
quantity. As i 3 - ar- + bx = c. 

A Biquadratic, or Double Quadratic, is an equation that 
contains the 4th power of the unknown quantity : 
As x* — ax 3 -f bx J — cx = d. 
An Equation of the 5th Power or Decree, is one that 
contains the 5th power of the unknown quantity. 

As x 3 — ux A + W — cx 2 + clx =-= e. 
And so on, for all other higher powers, \fhere it is 
to be noted, however, that all the powers, or terms, in the 




=4 y=J" 



ne < — **- 







\ . «*- •••• 



CUBIC, X4UAT10X8. 



equation, are supposed to be freed from surds or fractional 
exponents. 

There are many particular and prolix rules usually given 
for the solution of some of the above-mentioned powers 
or equations. But they may be all readily solved by the 
following easy rule of Double Position, sometimes called 

Trial -and-E rror *. 

i 

RULE. 

1. Find, by trial, two numbers, as near the true root as 
you can, and substitute them separately in the given equa- 
tion, instead of the unknown quantity ; and find how much 
the terms collected together, according to their signs 7- or 
— , differ from the absolute known term of the equation, 
marking whether these errors are in excess or defect. 

2. Multiply the difference of the two numbers, found or 
taken by trial, by either of the errors, and divide the pro- 
duct by the difference of tho errors, when they are alike, 
but by their sum when they arc unlike. Or say, As the 
difference or sum of the errors, is to the difference of the 
two numbers, so is either error to the correction of its sup- 
posed number. 

3. Add the quotiont, last found, to the number belonging 
to that error, when its supposed number is too little, but 
subtract it when too great, and the result will give tho true 
root nearly. 

4. Take this root and the nearest of the two former, or 
any other that may be found nearer : and, by proceeding in 
like manner as above, a root will be had still nearer than 
before. And so on, to any degree of exactness required. 

Note 1. It is best to employ always two assumed num- 
bers that shall differ from each other only by unity in the 
last figure on the right hand ; because then the difference, 
or multiplier, is only 1. It is also best to use always the 
least error in the above operation. 

NoU 2. It will be convenient also to begin with a single 



* See, farther, that portion of vol. ii. which relates to equations, their 
construction, be. 

A new and ingenious general method of solving equations has been 
recently discovered by Messrs. //. Atkinson, Holared, and Horner, inde- 
pendently of each other. For the best pratical view of this new <Ki«\ta&. 
and its applications, consult the Elementary Treatise of Algebra, TAx, 
J. R, Young; a work which deserves our cordial reuimoiefedtttoft* 
Vol. L 34 



258 



-"ALGEBRA. 



figure at first, trying several single figures till thero be found 
the two nearest the truth, the one too little, and the other 
too great ; and in working witli them, find only one more 
figure. Then substitute this corrected result in the equation, 
for the unknown letter, and if the result prove too little, 
substitute also the number next greater for the second sup. 
position ; but contrarywise, if the former prove too great, 
then take the next less number for the second supposition ; 
and in working with the second pair of errors, continue the 
quotient only so far as to have the corrected number to four 
places of figures. Then repeat the same process again with 
this last corrected number, and the next greater or less, as 
the case may require, carrying the third corrected number 
to tight figures ; because each new operation commonly 
doubles the number of true figures. And thus proceed to 
any extent that may be wanted. 



EXAMPLES. 



Ex. 1. To find the root of the cubic equation ar 1 + jr + 
x= 100, or the value of x in it. 



Here it is soon found that 
x lies between i and 5. As- 
sume therefore these two num- 
bers, and the operation will be 
as follows : 
k 1st Sup. 2d Sup. 

4 - x 5 
Hi - X s - 25 
61 - - 125 



84 
100 

—10 



sums - 
but should be 

- errors - 



+55 



the sum of which is 71. 
Then as 71 : 1 : : 1G : -2 
Hence x = 4 '2 nearly. 



Again, suppose 4*2 and 4-3, 
and repeat the work aa fol- 
lows : 



155 
100 



1st Sup. 
42 
1704 
74 088 

05-928 
100 



x 

x 2 

X 3 

sums 



2d Sup. 
4-3 
. 18-49 
. 79-507 

102-297 
100 



072 errors +2-297 



the sum of which is &l 
! As 0-309 :-l : : 2-297: 0-036 
I This taken from - 4-300 



loaves x nearly 



= 4-264 



CUBIC, Ac. EQUATIONS. 



250 



Again, suppose 4*264, and 4*265, and work as follows: 



4*264 


X 


4*265 


18181696 


X* 


18*100225 


77-526752 


x 1 


77*581310 


99-972448 


sums 


100036535 


100 




100 


-0027552 


errors - 


+0 036535 



tbe sum of which is -064087. 
Then as -064087 : -001 : : 027552 : 0004299 



To this adding - 4*264 
gives x very nearly = 4-2644209 



The work of the example above might have been much 
shortened, by the use of the Table of Powers in the Arith- 
metic, which would have given two or three figures by in- 
spection. But the example has been worked out so particu- 
larly as it is, the better to show the method. 

Ex. 2. To find the root of the equation z 3 — 15x 2 + 63x 
= 50, or the value of x in it. 

Here it soon appears that x is very little above 1. 



Suppose therefore 1 *0and 1*1, 
and work as follows : 



10 



1*1 



63 - 63x - 69*3 
—15 — 15* 1 —1815 
1 . X s - 1-331 



49 

50 



~ sums • 52*481 
50 



—l - errors +2*481 
3*481 sum of the errors. 
As 8*481 : 1 : : *1 : -03correct. 
1*00 

Hence xa 1-03 nearly. 



Again, suppose the two num. 
bers 1*03 and 1-02, &c. as 
follows : 
1*03 . i - 1-02 

64*89 - 63* 64*26 
—15*9135- IS* 3 — 15*6060 
1-092727 x 3 1 -061208 



50 069227 sums 49-715208 
50 50 



+ •0(S9227er rors — -284792 
•284792* 



As -354019 : -01 : : -069227: 
•0019555 
This taken from 1 -03 



leaves x nearly = 1*02804 



960 



Note 3. Every equation has as many roots as it contains 
dimensions, or as there are units in the index of its highest 
power. That is, a simple equation has only one value of 
the root ; but a quadratic equation has two values or roots, 
a cubic equation has three roots, a biquadratic equation has 
four roots, and so on. 

When one of the roots of an equation has been found by 
approximation, as above, the rest may be found as follows. 
Take, for a dividend, the given equation, with the known 
term transposed, with its sign changed, to the unknown side 
of the equation ; and, for a divisor, take x minus the root 
just found. Divide the said dividend by the divisor, and 
the quotient will be the equation depressed a degree lower 
than the given one. 

Find a root of this new equation by approximation, as be- 
fore, or otherwise, and it will be a second root of the origin- 
al equation. Then, by means of this root, depress the se- 
cond equation one degree lower, and from thence find a third 
root ; and so on, till the equation be reduced to a quadratic ; 
then the two roots of this being found, by the method of com- 
pleting the square, they will make up the remainder of the 
roots. Thus, in the foregoing equation, having found one 
root to be 1 '02804, connect it by minus with x for a divisor, 
and the equation for a dividend, &c. as follows : 

x — 1-02804 ) s 3 - 15r» + 63* — 50 ( x* - 13-97196* + 

48-63627 = 0. 

Then the two roots of this quadratic equation, or - - - 
X s — 13-97196* = — 48-63627, by completing the square, 
are 6-57653 and 7*39543, which are also the other two roots 
of the given cubic equation. So that all the three roots of 
that equation, viz. r 1 - Ibx 2 + 63j? = 50, 

and the sum of all the roots is found to be 
15, being equal to the co-efficient, of the 2d 
term of the equation, which the sum of the 
roots always ought to be, when they are 
right. 

Note 4. It is also a particular advantage of the foregoing 
rule, that it is not necessary to prepare the equation, as for 
other rules, by reducing it to the usual final form and state 
of equations. Because the rule may be applied at once to an 
unreduced equation, though it be ever so much embarrassed 



arc 1 02804 
and 6-57653 
and 7-39543 



sum 15-00000 



CUBIC. 



mvATiont. 



9tl 



by mid and compound quantities. As in the following ex- 
ample : 

Ex. 3. Let k be required to find the root x of the equation 
y(144** — (jr» + 20)») + + 24) 1 ) » 114, or 

the value of x in it. 

By a few trials it is soon found that the value of x is but 
little above 7. Suppose therefore first that x is=7, and then 
x = 8. 

First, when x = 7, Second, when * = 8. 

47-906 . -/[144* 1 — («» + 20)n - 46-476 
65-384 . v/flOO* 1 — (x a + 24)«] 



113- 290 

114- 000 

—0-710 
+1759 



- the sums of these 

- the true number 

• the two errors 



115-750 
114-000 

+1-750 



As 2-469 : 1 



0-710 : 0-2 nearly. 
70 



Therefore x = 7-2 nearly. 



Suppose again x = 7-2, and then, because it turns out too 
great, suppose x also = 7-1, dec. as follows : 



Supp. x = 7-2, 

47-990 . v /[144x a — (s 8 + 20) 2 ] 
66-402 . ^[196**— (x* + 24) 2 ] 

414*392 - the sums of these 
114-000 - the true number 



+0-392 
0123 



the two errors 



Supp. * = 7-l, 
- 47-973 
. 65-904 



113- 877 

114- 000 

—0128 



As -515 : 1 



*123 : -024 the correction, 
7-100 add 



Therefore x = 7*124 nearly the root required. 

Nate 5. The same rule also, among other more difficult 
forms of equations, succeeds very well in what are caiUd 
exponential ones, or those which have an *^a*a&« 



an 



ALGEBRA. 



ty in the exponent of the power ; as in the following ex- 
ample : 

Ex. 4. To find the value of x in the exponential (equation 

For more easily resolving such kinds of equations, it is 
convenient to take the logarithms of them, and then com- 
pute the terms by means of a table of logarithms. Thus, 
the logarithms of the two sides of the present equation are 
x X log. of x = 2, the log. of 100. Then, by a few trials, 
it is -soon perceived that the value of x is somewhere be- 
tween the two numbers 3 and 4, and indeed nearly in the 
middle between them, but rather nearer the latter than the 
former. Taking therefore first x = 3 5, and then =■ 3*6, 
and working with the logarithms, the operation will be as 
follows : 



First Supp. x = 3-5. . 
Log. of 3-5 = 0-544068 
then 3-5 Xlog.3 5=1 -904238 



Second Supp. x = 3*6. 
Log. of 3 6 = 0*556303 
then 3-6 X log. 3 6=2-002689 



the true number 2-000000 I the true number 2*000000 



error, too little, — -095762 
•002689 



error, too great, + -002689 



•098451 sum of the errors. Then. 



As -098451 : -1 : : -002689 : 00273 the correction 
taken from 3-60000 



leaves - 3*59727 = x nearly. 



By repeating the operation with a larger table of loga- 
rithms, a nearer value of x may be found 3*597285. 

This method, indeed, may be a little improved in practice : 
for since x* = a, we have by logarithms x X log. x = log. a ; 
and again, log. x + log. log. x = log. log. a. We have 
therefore only to find a number, which, added to its log. will 
will be equal to the log. of the log. of the given number ; 
and the natural number answering to this number, is the va- 
lue of x required. 

In illustration of the above, take the 12th example : — 
«• = 123456789. First, log. 123456789 = 8 -0915143, and 
log. 8*0915148 = -9080298. Searching in a table of loga- 
liSaoB, we find the nearest number -93651 ; which added to 



CUBIC, AiCrn EQUATIONS. 263 

its logarithm — 1-0715124 = -9080224. The next higher 
number '93652 + its log. = -9080371. Hence 

-9080371 -9080298 

•9080224 -9080224 74 — 147 = -503 



147 74 



Therefore, the number sought is '93651503, the natural 
number answering to which is 8-640026 the value of x, 
which is true to the last figure, the value given by Dr. Hut* 
ton being 8-6400268. 

The common logarithmic solution fails when a is less than 
unity, its log. being then negative. In this case, assume 
x = 1 -f- y, and a =- 1 e, which transforms the given equa. 
x* = a f to & = y. Taking the logs, twice, we get y log. 
t = log. y, and log. y + log. of log. e = log. of log. y ; or, 
putting log. y = », and log. of log. e = *, we have v + * = 
log. v, an equation easy to solve. 

Ex. 5. To find the value of x in the equation x 3 + lOx 3 
+ 5* = 260. Ans. x = 4-1 179857. 

Ex. 6. To find the value of x in the equation x 3 — 2x=50. 

Ans. 3-8648854. 

Ex. 7. To find the value of x in the equation x 3 + 2x 2 — 
23x = 70. Ans. x = 5-13457. 

Ex. 8. To find the value of x in the equation x 3 — 17x 2 
+ 54x = 350. Ans. x = 14-95407. 

Ex. 9. To find the value of x in the equation x 4 — 3x* — 
75x = 10000. Ans. x = 10-2609. 

Ex. 10. To find the value of x in the equation 2x 4 — 16x 3 
+ 40X 3 — 30x = — 1. Ans. x = 1-284724. 

Ex. 11. To find the value of x in the equation X s + 2x 4 
+ Sx 3 + 4X 8 + 5x = 54321 . Ans. x = 8*414455. 

Ex. 12. To find the value of x in the equation x* = 
123456789. Ans. x = 8 6400268. 

Ex. 13. Given 2x 4 — 7x 3 + llx 3 — 3x = 11, to find *. 

Ex. 14. To find the value of x in the equation. 

(3x* — 2y/x + 1)? — (x 2 — 4* v /x + 3 v 'x)* = 56. 

Ans. x = 18-360877. 



90* 



To resolve Cubic Equations bp Cardan's Rule. 

Though the foregoing general method, by the application 
of Double Position, be the readiest way, in reul practice, of 
finding the roots in numbera of cubic equations, as well as 
of all the higher equations universally, we may here add the 
particular method commonly called Cardan's Rule, for re- 
solving cubic equations, in case any person should choose 
occasionally to employ that method ; although it is only ap- 
plicable when two of the roots are impossible. 

The form that a cubic equation ihust necessarily have, to 
be resolved by this rule, is this, viz. z 1 + az = 6, that is, 
wanting the second term, or the term of the 2d power z*. 
Therefore, after any cubic equation hns been reduced down 
to its final usual form, x 3 + px* + qx = r, freed from the 
co-efficient of its first term, it will then be necessary to take 
away the 2d term px 2 ; which is to be done in this manner ; 
Take Jp, or £ of the co-efficient of the second term, and 
annex it, with the contrary sign, to another unknown letter 
z, thus z — Jp ; then substitute this for x, the unknown 
letter in the original equation x 3 + px 2 + qx = r, and there 
will result this reduced equation z' J ±az b, of the form 
proper for applying the following, or Cardan's rule. Or 
take c = J a, and d = J6, by which the reduced equation 
takes this form, z 3 + 3cz = 2d. 

Then substitute the values of c and d in this 

form, % = V[d + + J)] + V[d - y/{* + O], > 

or , = „[* + + O] - w ^ w+ ^ y \ 

and the value of the root z, of the reduced equation z 3 + 
az = b, will be obtained. Lastly, take x = z — |p, which 
will give the value of r, the required root of the original 
equation x 3 + px 3 + qx = r, first proposed. 

One root of this equation being thus obtained, then de- 
pressing the original equation one degree lower, after the 
manner described, p. 260, the other two roots of that equa- 
tion will be obtained by means of the resulting quadratic 
equation. 

Note. When the co-efficient a, or c, is negative, and c 3 is 
greater than this is called the irreducible case, because 
then the solution cannot be generally obtained by this rule*. 



* Suppose a root to consist of the two parts z and y, so that (x + y) 
as jr; which sum substituted for z, in the given equation *3 + a* = 



cubic, 4ce» MVAirom. 985 

Ex. To find the roots of the equations 3 — 6ac* +10* = 8. 
First to take away the 2d term, its co-efficient being — 6, 
its 3d part is — 2 ; put therefore x = * + 2 ; then 

= z 3 + 6s 9 + 12z + 8 
- 6**= -6* 1 — 24* — 24 
+ 10* = + 10* + 20 



theref. the sum z 3 # - 2* + 4 = 8 
or z 3 * — 2* = 4 

Here then a = — 2, 6 = 4, c = — |, d = 2. 

Theref. V[rf+^(W)]-l/[2+^(4-A)]«l/(2+^ W)^ 

and Vt^-^+c 3 )] ^/[2-v/(4-A)]-l/(2-^ W)» 
3/(2- y v /3)=0-42265 ^ 

then the sum of these two is the value of * = 2. Hence 
x = * + 2 = 4, one root of * in the eq. a 3 — Ox* + lOx = 8. 

To find the two other roots, perform the division, Sec. as 
in p. 261, thus: 

x _ 4 ) x 3 — 6x a + lOx — 8 ( X s — 2x + 2 =s 
x 3 — 4x" 



— 2**+ lOx 

— 2r» + 8x 



2r — 8 
2x — 8 



it becomes x 3 + y 3 + 3xy (x+ v) +a (x + y) = 6. Again, suppose 
3sjf = — a ; which substituted, the last equation becomes x 3 -|- = 6. 
Now, from the square of this equation subtract four times the equation 
xy = — Ja, and there results x 8 — 2xy + 3^ = * 9 + aS* 3 * tbe square 
root of which is x 3 — y 3 = V (6* + A fli )- This being added to and 
taken from the equation x 3 + y 3 = 6, gives 

• C 2xs = 6 + V (6 a + A *) = 6 4- 2 V [(J6) 3 4- (J*) 3 !, 
i 2y* = 6 - V (6* — aS « ) = * — 2 V + (J*) 3 ] ; or 

{ S! = 5S ± » V [S J ^ } • Hence ' di * d '"* * * — 
extracting the cube roots, we have x = l/d -f- v (d 2 -f c 1 ), and y =s 
%/d — V(d 2 + c 3 ) ; the sum of these two gives the first form of the root 
s above stated. And that the 2d form is equal to the first will be evident 
by reducing the two 2d quantities to the same denominator. 

When c is negative, and c 3 greater than o* f the root *pp*axiY*w& 
imaginary form. 

Vox. I. 35 



906 of BiKPix umnunr. 

Hence x 3 — 2* = — 2, or ** — 2*+ 1 «= — 1, and *— 1 
= rhv' — 1 = lor=l — — l,thetwo 

other roots sought. 

Ex. 2. Given X s — 6a* + 36x = 44, to find *. 

Ans. * = 2*32748. 
Ex. 3. To find the roots of x> — 7x* + 14x*= 20. 

Ans. x = 5, or = 1 + ^/ — 3, or = 1 — ^/ — 3. 

Ex. 4. Find the three roots of x 3 + 6x = 20. 



OF SIMPLE INTEREST. 

As the interest of any sum, for any time, is directly pro- 
portional to the principal sum, and to the time ; therefore 
the interest of 1 pound, for 1 year, being multiplied by any 
given principal sum, and by the time of its forbearance, in 
years and parts, will give its interest for that time. That is, 
if there be put 

r = the rate of interest of 1 pound per annum, 
p = any principal sum lent, 
t = the time it is lent for, and 

a = the amount or sum of principal and interest ; then 
is pfi = the interest of the sum p, for the time t, and conseq. 
p + prt or p X (1 + rt) = dy the amount for that time. 

From this expression, other theorems can easily be de- 
duced, for finding any of the quantities above mentioned : 
which theorems, collected together, will be as follows : 

1st, a = p + prt the amount ; 2d, p = ■ J* ■ the principal ; 

3d, r = the rate ; 4th, t = the time. 

pt pr 

For Example. Required to find in what time any princi- 
pal sum will double itself, at any rate of simple interest. 

In this case, we must use the first theorem, o =-p + prt, 
in which the amount a must be made = 2p, or double the 
principal, that is, p + prt = 2p, or prt = p, or rt = 1 ; 

and hence t = -. 

r 

Hence r being the interest of 11 for 1 year, it follows, that 
the doubling at simple intern^ ia to the quotient of 



COMPOUND INTEREST. 907' 

any sum divided by its interest for 1 year. • So, if the rate of 
interest be 5 per cent, then 100 -s- 5 = 20, is the time of 
doubling at that rate. Or the 4th theorem gives at once 

# a—p 2p—p 2-1 1 . ' r 

t = = — — — = = the same as before. 

pr pr r r 



COMPOUND INTEREST. 

Besides the quantities concerned in Simple Interest, 
namely, 

p = the principal sum, 

r = the rate of interest of 11 for 1 year, 

a = the whole amount of the principal and interest, 

t = the time* 

there is another quantity employed in Compound Interest, 
viz. the ratio of the rate of interest, which is the amount of 
II for 1 time of payment, and which here let be denoted by 
B, viz. 

K = 1 + r, the amount of 11 for I time. 

Then the particular amounts for the several times may 
be thus computed, viz. As II is to its amount for any time, 
so is any proposed principal sum, to its amount for the same 
time ; that is, as 

}l : R : : p : pa, the 1st year's amount, 
1Z : R : : pn : pR a , the 2d year's amount, 
11 : R : : pR 2 : J>R 3 , the 3d year's amount, 
and so on. 

Therefore, in general, pn 1 = a is the amount for the 
t year, or t time of payment. Whence the following genera! 
theorems are deduced : 

1st, a = j>r« the amount ; 2d, p = ~ the principal ; 

Jd,K = y-theratio; 4tM = ^ ^4°---^^ 
' v p loe. of r * - ■ 



OOXPOTCTD 

From which, any one of the quantities may bo found, 
when die rest are given. 

As to the whole interest, it is found by barely subtracting 
die principal/) from the amount a. 

Example. Suppose it be required to find, in how many 
years any principal sum will double itself, at any proposed 
rate of compound interest. 

In this case the 4th theorem must be employed, making 
m a* 2p ; and then it is 

I log* a~log.jp log, 2 p — log. p _ log. 2 

leg. n log. n log. r" 

So, if the rate of interest be 5 per cent per annum ; then 
B = 1 + -06 -» 1*05 ; and henee 



log. 2 '301030 
log. 1-05 * -021189 



14-2067 nearly ; 



that is, any sum doubles itself in 14£ years nearly, at the 
rate of 5 per cent, per annum compound interest. 

Hence, and from the like question in simple interest, above 
given, are deduced the times in which any sum doubles itself, 
at several rates of interest, both simple and compound ; viz. 



At" 
3 

f 

4 

4* 

5 

6 

7 

8 

9 

10 J 


per cent. -per annum 
interest, 1/. or any 

other sum, will 
double itself in the 

following years. 


AtSimp.Int. 


At Comp.Int 


in 50 
40 
33i 
28| 
25 

22J Kj 
20 2 
16| ? 
14| 
12* 

Hi 

I 10 


in 35 0028 
28-0701 
23-4498 
201488 
17-6730 
15-7473 h- 
14-2067 1 
11-8957? 
10-2448 
9 0065 
8 0432 
7-2725 



The following Table will very much facilitate calculations 
of compound interest on any sum, for any number of years, 
at various rates of interest. 



cokpouhd iivtibxst. 968 
The Amounts of 12 in any Number of Tears. 



Yrs. 


3 


3} 


4 




5 


6 


1 


1*0300 


1-0350 


1-0400 


1-0450 


1-0500 


1-0600 


2 


1-0609 


10712 


1-0816 


1-0920 


1-1025 


11236 


3 


1-0927 


11087 


1-1249 


11412 


1-1576 


11910 


4 


1-1255 


1-1475 


1-1699 


1-1925 


1-2155 


1-2625 


* 


1-1593 


1-1877 


1-2167 


1-2462 


1-2763 


1-3382 


6 


1-1948 


1-2293 


1-2653 


1-3023 


1-3401 


1-4185 


7 


1-2299 


1-2723 


1-3159 


1-3609 


1-4071 


1-5036 


8 


1-2668 


1-3168 


1-3686 


1«4221 


1-4775 


1-5939 


9 


1-3048 


1-3629 


1-4233 


1-4861 


1-5513 


1-6895 


10 


1-3439 


1-4106 


1-4802 


1-5530 


1-6289 


1-7909 


11 


1-3842 


1-4600 


1-5895 


1-6229 


1-7103 


1-8983 


12 


1-4258 


1-5111 


1-6010 


1-6959 


1.7959 


2-0122 


13 


1-4685 


1-5640 


1-6651 


1-7722 


1-8856 


21329 


14 


1-5126 


1-6187 


1-7317 


1-8519 


1-9799 


2-2609 


15 


1-5580 


1-6753 


1-8009 


1-9353 


20789 


2-3966 


16 


1-6047 


1-7340 


1-8730 


2 0224 


21829 


2-5404 


17 


1-6528 


1-7947 


1-9479 


21134 


2-2920 


2-6928 


18 


1-7024 


1-8575 


2 0258 


2-2085 


2-4066 


2-8543 


19 


1-7535 


1-9225 


2-1068 


2-3079 


2-5270 


30256 


20 


1-8061 


1-9828 


21911 


2-4117 


2-6533 


3-2071 



The use of this Table, which contains all the powers, r', 
to the 20th power, or the amounts of 1Z, is chiefly to calcu- 
late the interest, or the amount of any principal sum, for any 
time, not more than 20 years. 

For example, let it be required to find, to how much 5231 
will amount in 15 years, at the rate of 5 per cent, per annum 
compound interest. 

In the table, on the line 15, and in the column 5 per cent 
is the amount of 1/, viz. . - 2 0789 
this multiplied by the principal - 523 

gives the amount - - 1087-2647 
or .... 1087Z5*3J<*. 
and therefore the interest 564Z 5* 3\d. 

Note 1. When the rate of interest is to be determined to 
any other time than a year ; as suppose to £ a year, or J a 
year, dec. : the rules are still the same ; but then t will ex- 
press that time, and k must be taken the amount foe tihsX 
time also. 



270 



ANNUITIES. 



Nole 2, When the compound interest, or amount, of any 
sum, is required for the parts of a year ; it may be determin- 
ed in the following manner : 

1st, For any time which is some aliquot part of a year : — 
Find the amount of 11 for 1 year, as before ; then that root 
of it which is denoted by the aliquot part, will be the amount 
of 11. This amount being multiplied by the principal sum, 
will produce the amount of the given sum as required. 

2d, When the time is not an aliquot part of a year : — 
Reduce the time into days, and take the 365th root of the 
amount of 11 for 1 year, which will give the amount of the 
same for 1 day. Then raise this amount to that power whose 
index is equal to the number of days, and it will be the 
amount for that time. Which amount being multiplied by 
the principal sum, will produce the amount of that sum as 
before. — And in these calculations, the operation by loga- 
rithms will be very useful. 



OF ANNUITIES. 



Annuity is a term used for any periodical income, arising 
from money lent, or from houses, lands, salaries, pensions, 
&c. payable from time to time, but mostly by annual pay- 
ments. 

Annuities are divided into those that are in Possession, 
and those in Reversion : the former meaning such as have 
commenced ; and the latter such as will not begin till some 
particular event has happened, or till after some certain time 
has elapsed. 

When an annuity is forborn for some years, or the pay- 
ments not made for that time, the annuity is said to be in 
Arrears. 

An annuity may also be for a certain number of years ; 
or it may be without any limit, and then it is called a Per- 
petuity. 

The Amount of an annuity, forborn for any number of 
years, is the sum arising from the addition of all the annui- 
ties for that number of years, together with the interest due 
upon each after it becomes due. 



ANNUITIES. 271 

The Present Worth or Value of an annuity, is the price 
or sum which ought to be given for it, supposing it to be 
bought off, or paid all at once. 

Let a = the annuity, pension, or yearly rent ; 
n = the number of years forborn, or lent for ; 
R = the amount of 11 for 1 year ; 
m =s the amount of the annuity ; 
v = its value, or its present worth. 

Now, 1 being the present value of the sum b, by propor- 
tion the present value of any other sum a, is thus found : 

as r : 1 : : a : ~ the present value of a due 1 year hence. 
In like manner ~ is the present value of a due 2 years 
hence ; for r : 1 : : - : So also ~, — , —, &c. will 

R R a R 3 R* R 4 

be the present values of a, due at the end of 3, 4, 5, &c. 
years respectively. Consequently the sum of all these, or 

a i a i a i a ic ,1.1.1.1. X ^ 

a continued to n terms, will be the present value of all the n 
years' annuities. And the value of the perpetuity, is the sum 
of the series to infinity. 

But this series, it is evident, is a geometrical progression, 

having ~ but for its first term and common ratio, and the 

number of its terms n ; therefore the sum v of all the terms, 
or the present value of all the annual payments, will be 

„ = * * R " X a, or — J *—^-^- X 

^ 1 R 1 R n 

R 

When the annuity is a perpetuity ; n being infinite, r* 
is also infinite, and therefore the quantity ~ becomes = 0, 

therefore — ^-j- X ~ also = ; consequently the expression 

becomes barely v = — — _ ; that is, any annuity divided by 

R "™ ■ ~ •* 

the interest of 11 for 1 year, gives the value of tto \*%T^etoai- 
ty. So, if the rate of Interest be 5 per cent. 



372 ANNUITIES. 

Then 100a -f- 5 = 20a is the value of the perpetuity at 
5 per cent : Also 100a -f- 4 = 25a is the value of the per- 
petuity at 4 per cent. : And 100a -r- 3 = 33£a is the value 
of the perpetuity at 3 per cent. : and so on. 

Again, because the amount of 11 in n years, is R n , its 
increase in that time will be R n — 1 ; but its interest for one 
single year, or the annuity answering to that increase, is 
r — 1 ; therefore, as r — 1 is to r a — 1, so is a to m ; that 
n n — 1 

is, m = — X a. Hence, the several cases relating to 

r — 1 

Annuities in Arrear, will be resolved by the following 
equations : 



m = 



v = 



n = 



Rn— 1 
R— 1 


X a = 


CR« ; 


R» — 1 


X- = 


m 


R— 1 


X R" 


? ? 


R — 1 
R n 1 


X m = 


R— 1 
R n — 1 


log. m - 


-log. V 


'log. 



a = — ^ X m = — — ^ x vr* ; 

77iR — m + a 



log. R log. R 

Log & — * ^* m — v 



Rp R n ' R 1 

In this last theorem, r denotes the present value of an 
annuity in reversion, after p years, or not commencing till 
after the first p years, being found by taking the difference 

between the two values ——4 X — and = — , for n 

R— 1 R n R— 1 RP 

years and p years. 

But the amount and present value of any annuity for any 
number of years, up to 21, will be most readily found by the 
two following tables. 



ANNUITIES. 



273 



TABLE I. 

The Amount of an Annuity of 1/ at Compound Interest. 



YrB. 


at 3 per c. 


31 perc. 


4 per c. 


4 J perc. 


5 per c. 


6 per c. 


1 


10000 


10000 


1 0000 


1 0000 


10000 


10000 


2 


20300 


2 0350 


20400 


2 0450 


20500 


20600 


3 


3 0909 


3*1062 


3*1216 


3- 1370 


3 1525 


31836 


4 


41836 


4-2149 


42465 


42782 


43101 


43746 


5 


5 3091 


53625 


5 4163 


5-4707 


55256 


56371 


6 


64684 


65502 


66330 


67169 


68019 


69753 


7 


76625 


77794 


78983 


80192 


8 1420 


83938 


8 


88923 


90517 


92142 


9 3800 


95491 


98975 


9 


10 1591 


103685 


10-5828 


10-8021 


110266 


11 4913 


10 


114639 


117314 


120061 


122882 


12 5779 


131808 


11 


12-8078 


131420 


13 4864 


138412 


14-2068 


149716 


12 


14 1920 


146020 


150258 


15-4640 


159171 


16-8699 


13 


156178 


16 1130 


16 6268 


171599 


17-7130 


18*8821 


14 


170863 


17 6770 


182919 


189321 


19 5986 


21 0151 


15 


18-5989 


19 2957 


203236 


20 7841 


21 5786 


232760 


16 


20 1569 


209710 


21 8245 


22 7193 


236575 


25-6725 


17 


21 7616 


22-7050 


236975 


24-7417i 25 8404 


282129 


IS 


234144 


244997 


25 6454 


268551 


28 1324 


309057 


19 


25 1169 


26 3572 


276712 


290636 


305390 


3376C0 


20 


268704 


282797 


29*7781 


31 3714 


33 0660 


367856 


21 


286765 


302695 


31 9692 


33 7831 


35 7193 


399927 



table ii. The Present Value of un Annuit}' of 1/. 



Yrs. 


at 3 perc. 


3£ per c. 


4 per c. 


1 


09709 


09662 


09615 


2 


1 9135 


1-8997 


1*8861 


3 


28286 


2*8016 


27751 


4 


37171 


36731 


36299 


5 


4-5797 


4*5151 


44518 


6 


5-4172 


5*3286 


52421 


7 


62303 


61145 


60020 


8 


70197 


6*8740 


77327 


9 


77861 


7-6077 


74353 


10 


8*5302 


8*3166 


8 1109 


11 


9-5256 


90016 


8-7605 


12 


9-9540 


96633 


93851 


13 


10-6350 


10 3027 


9-9857 


14 


11 2961 


10*9205 


105631 


15 


11 9379 


11 5174 


11 1181! 


16 


12 5611! 120941 


11-65231 


17 


131661 


126513 


12 1657 


18 


13-7535 


13 1S97 


12-659?! 


19 


143238 


13*7098 


13 1339 


20 


148775 


142124 


135903 


21 


154150/ 


14 69801 


14 0292 



4} por c. I 5 per c. 



09569 
1 8727 
27490 
3*5875 

4- 3^C0 
51579 

5- 8927 
65959 
7-2688i 
7*9127| 
8*5289' 
91 186! 
9*6S29| 

10 2228: 
10-7396| 

11 2340. 

11 7072: 

12 16001 
1259331 

13 0079 
13 4047' 



09524 
1-8594 
27233 
35460 
4 :)23b 
50757 
57864 
64632 
7* 1078 
7 7217 
83G54 
88633 
9-3S3S 
989S6 
10-3797 



6 per c. 



95. 4 

2 (>7oC 
34651 
42124 
49173 
55824 
62095- 
68017 
73601 
78869 
8-3838 
88527 
92950 
971*3 
10 8378: 10 1069 



11-2741 
11 6S96 
120853 



10-4773 
10*8276 
11 158V 



Vol. I. 



36 



274 



ANNUITIES. 



To find the Amount of any Annuity forlorn a certain number 
of years. 

Take out the amount of 12 from the first table, for the 
proposed rate and time ; then multiply it by the given 
annuity ; and the product will be the amount, for the same 
number of years, and rate of interest. And the converse to 
find the rate of time. 

Exam. To find how much an annuity of 507 will amount 
to in 20 years, at 3£ per cent, compound interest. 

On the line of 20 years, and in the column of 3 J per cent, 
stands 28*2797, which is the amount of an annuity of 12 for 
the 20 years. Then 28-2797 X 50, gives 1413-9851 = 
1413/ 19* 8d for the answer required. 

To find the Present Value of any Annuity for any number 
of years. — Proceed here by the 2d table, in the same manner 
us above for the 1st table, and the present worth required 
will be found. 

Exam. 1. To find the present value of an annuity of 502, 
which is to continue 20 years, at 3J per cent. — By the table, 
the present value of 11 for the given rate and time, is 
14-2124; therefore 14-2124 X 50 = 710-022 or 7102 12* 4d 
is the present value required. 

Exam 2. To find the present value of an annuity of 202, 
to commence 10 years hence, and then to continue for 11 
years longer, or to terminate 21 years hence, at 4 per cent, 
interest. — In such cases as this, we have to find the difference 
between the present values of two equal annuities, for the 
two given times ; which therefore will be done by subtracting 
the tabular value of the one period from that of the other, 
and then multiplying by the given annuity. Thus, 
tabular value for 21 years 14*0292 
ditto for 10 years 8*1109 



the difference 5-9183 
multiplied by 20 



gives - 118*3602 
or - - 1 1 8Z 7* 3 £c2 the answer. 



END OF THE ALGEBRA. 



275 



GEOMETRY, 



DEFINITIONS. 

1. A Point is that which has position, bat 
no magnitude, nor dimensions ; neither length, 
breadth, nor thickness. 

2. A Line is length, without breadth or 
thickness. 

S. A Surface or Superficies, is an extension 
or a figure of two dimensions, length and 
breadth ; but without thickness. 

4. A Body or Solid, is a figure of three di- 
mensions, namely, length, breadth, and depth, 
or thickness. 

5. Lines are either Right, or Curved, or 
Mixed of these two. 

6. A Right Line, or Straight Line, lies all iu 
the same direction, between its extremities ; 
and is the shortest distance between two points. 

When a Line is mentioned simply, it means 
a Right Line. 

7. A Curve continually changes its direction 
between its extreme points. 

8. Lines are either Parallel, Oblique, Per- 
pendicular, or Tangential. 

. 9. Parallel Lines are always at the same 
perpendicular distance ; and they never meet, 
though ever so far produced. 

10. Oblique lines change their distance, and 
would meet, if produced on the side of the least 
distance. 

11. One line is Perpendicular to another, 
when it inclines not more on the one side 



276 



GEOMETRY. 



than the other, or when the angles on both 
sides of it are equal. 

12. A line or circle is Tangential, or is a 
Tangent to a circle, or other curve, when it 
touches it, without cutting, when bath are 
produced. 

13. An Angle is the inclination or open- 
ing of two lines, having different directions, 
and meeting in a point. 

14. Angles are Right or Oblique, Acute 
or Obtuse. 

15. A Right Angle is that which is made 
by one line perpendicular to another. Or 
when the angles on each side are equal to 
one another, they are right angles. 

16. An Oblique Angle is that which is 
made by two oblique lines ; and is either less 
or greater than a right angle. 

17. An Acute Angle is less than a right 
angle. 

18. An Obtuse Anglo is greater than a 
right angle. 

19. Superfices are either Plane or Curved. 

20. A Plane Superficies, or a Plane, is that with which 
a right line may, every way, coincide. Or, if the line touch 
the plane in two points, it will touch it in every point. But, 
if not, it is curved. 

21. Plane Figures are bounded either by right lines or 
curves. 

22. Plane figures that are bounded by right lines have 
names according to the number of their sides, or of their 
angles ; for they have as many sides as angles ; the least 
number being three. 

83. A figure of three sides and angles is called a Triangle. 
And it receives particular denominations from the relations 
of its sides and angles. 

24. An Equilateral Triangle is that whose 
three sides are all equal. 

' 

25. An Isosceles Triangle is that which has 
two sides equal. 






DEFINITIONS. 



277 



26. A Scalene Triangle is that whose three 
sides are all unequal. 

27. A Right-angled Triangle is that which 
has one right angle. ^ 

28. Other triangles are Oblique-angled, and 
are either obtuse or acute. 

20. An Obtuse-angled Triangle has one ob- 
tuse angle. 

30. An Acute-angled Triangle has all its 
three angles acute. 

31. A figure of Four sides and angles is call- 
ed a Quadrangfe, or a Quadrilateral. 

32. A Parallelogram is a quadrilateral which 
has both its pairs of opposite sides parallel. 
And it takes the following particular names, 
viz. Rectangle, Square, Rhombus, Rhomboid. 

33. A Rectangle is a parallelogram, having 
a right angle. 

34. A Square is an equilateral rectangle ; 
having its length and breadth equal. 

35. A Rhomboid is an oblique-angled paral- 
lelogram. 

36. A Rhombus is an equilateral rhomboid ; 
having all its sides equal, but its angles ob- 
lique. 

37. A Trapezium is a quadrilateral which 
hath not its opposite sides parallel. 

88. A Trapezoid has only one pair of oppo- 
site sides parallel. 

39. A Diagonal is a line joining any two op* 
poaite angles of a quadrilateral. 




□ 

/J 

40. Plane figures that have more than four sides are, in 
general, called Polygons : and they receive other particular 
names, according to the number of their sides or angles. 
Thus, 

41. A Pentagon is a polygon of five sides ; a Hexagon, of 
six sides; a Heptagon, seven; an Octagon, eight; %ttocu 
agon, nine ; a Decagon, ten ; an Undecagon, tawa \ 
vodeoagon, twelve sides. 



278 



GEOMETRY. 



42. A Regular Polygon has all its sides and all its angles 
equal. — If they are not both equal, the polygon is Irregular. 

43. An Equilateral Triangle is also a Regular Figure of 
three sides, and the Square is one of four : the former being 
also called a Trigon, and the latter ^tetragon. 

44. Any figure is equilateral, when all its sides are equal : 
and it is equiangular when all its angles arc equal. When 
both these arc equal, it is a regular figure. 

45. A Circle is a plane figure bounded by 
a curve line, called the Circumference, which 
is every whore equidistant from a certain point 
within, called its Centre. 

The circumference itself is often called a 
circle, and also the Periphery. 

46. The Radius of a circle is a line drawn 
from the centre to the circumference. 

47. The Diameter of a cirle is a line drawn 
through the centre, and terminating at the 
cireumfcrencc on both sides. 



48. An Arc of a circle is any part of the 
circumference. 





49. A Chord is a right line joining the ex. 
tremitics of an arc. 



50. A Segment is any part of a circle 
bounded by an arc and its chord. 

51. A Semicircle is half the circle, or a 
segment cut off by a diameter. 

The half circumference is sometimes called 
the Semicircle. 

52. A Sector is any part of a circle which 
is bounded by an arc, and two radii drawn to 
its extremities. 

• 53. A Quadrant, or Quarter of a circle is 
a sector having a quarter of the circumference 
for its arc, and its two radii are perpendicular 
to each other. A quarter of the circumference 
is sometimes called a Quadrant. 




DEFINITIONS. 



279 



B A d 



54. Tho Height or Altitude of a figure is 
a perpendicular let fall from an angle, or its 
vertex, to the opposite side, called the base. 

55. In a right-angled triangle, the side op- 
posite the right angle ailed the Hypothc- 
nuse ; and the other two sides are called the 
Legs, and sometimes the Base and Perpen- 
dicular. 

56. When an angle is denoted by three 
letters, of which one stands at the angular 
point, and the other two on the two sides, 
that which stands at the angular point is read 
in the middle. 

57. The circumference of every circle is 
supposed to be divided into 360 equal parts 
called degrees ; and each degree into 60 Mi- 
nutes, each Minute into 60 Seconds, and so 
on. Hence a semicircle contains 160 degrees, 
and a quadrant 90 degrees. 

58. The Measure of an angle, is an arc of 
any circle contained between the two lines 
which form that angle, the angular point 
being the centre ; and it is estimated by the 
number of degrees contained in that arc. 

59. Lines, or chords, are said lo be Equi- 
distant from the centre of a circle, when per- 
pendiculars drawn Lo them from the centre 
are equal. 

69. And the right line on which the Great- 
er Perpendicular falls, is said to be farther 
from the centre. 

61. An Angle In a Segment is that which 
is contained by two lines, drawn from any 
point in the arc of the segment, to the two 
extremities of that arc. 

62. An Angle On a segment, or an are, is that which is 
contained by two lines, drawn from anv point in the opposite 
or supplemental part of the circumference, to tho extremities 
of the arc, and containing the arc between (hem. 

63. An Angle at the circumference, is that S^/V^ 
whose angular point or summit is any where ( f\ 
in the circumference. And an angle at the ' / ^ v 
centre, is that whose angular point is at the 
centre. 





280 



GEOMETRY. 



64. A right-lined figure is Inscribed in a 
circle, or the circle Circumscribes it, when 
all the angular points of the figure are in the 
circumference of the circle. 

65. A right-lined figure Circumscribes a 
circle, or the circle is Inscribed in it, when all 
the sides of the figure touch the circumference 
of the circle. 

66* One right-lined figure is Inscribed in 
another, or the latter circumscribes the former, 
when all the angular points of the former are 
placed in the sides of the latter. 



67. A Secant is a line that cuts a circle, 
lying partly within, and partly without it. 

66. Two triangles, or other right-lined figures, are said to 
be mutually equilateral, when all the sides of the one are 
equal to the corresponding sides of the other, each to each ; 
and they are said to be mutually equiangular, when the 
angles of the one are respectively equal to those of the other. 

68. Identical figures are such as are both mutually equi- 
lateral and equiangular ; or that have all the sides and all the 
angles of tho one, respectively equal to all the sides and all 
the angles of the other, each to each ; so that if the one figure 
were applied to, or laid upon the other, all the sides of the one 
would exactly fall upon and cover all the sides of the other ; 
the two becoming as it were but one and the same figure. 

70. Similar figures, are those that have all the angles of 
the one equal to all the angles of the other, each to each, and 
the sides about the equal angles proportional. 

71. The Perimeter of a figure, is the sum of all its sides 
taken together. 

72. A Proposition, is something which is either proposed 
to be done, or to be demonstrated, and is either a problem or 
a theorem. 

73. A Problem, is something proposed to be done. 

74. A Theorem, is something proposed to be demonstrated. 

75. A Lemma, is something which is premised, or demon- 
trated, in order to render what follows more easy. 

76. A Corollary, is a consequent truth, gained immediate- 
ly from some preceding truth, or demonstration. 

77. A Scholium, is a remark or observation made upon 
something going before it. 




281 



, AXIOMS. 

1. Things which are equal to the aame thing arc equal to 
each other. 

2. When equals are added to equals, the wholes are 
equal. 

3. When equals are taken from equals, the remainders 
are equal. 

4. When equals are added to uncquals, the wholes are un- 
equal. 

5. When equals are taken from unequal*, the remainders 
are unequal, 

6. Things which are double of the same thing, or equal 
things, are equal to each other. ^ 

7. Things which are halves of the same thing, are equal. 
. .8. Every whole is equal to all its parts taken together. 

0. Things which coincide, or fill the same space, are iden- 
tical, or mutually equal in all their parts. 

20. All right angles are equal to ono another. 

21. Angles that have equal measures, or arcs, are equal. 



TIIEOREX T. 



in 
all 



If two triangles have two sides and the included angle 
in the one, equal to two sides and the included angle 
the other, the triangles will be identical, or equal in 
respects. 

In the two triangles jlbc, def, if 
the side ac be equal to the side dp, 
and the side bc equal to the side ef, 
and the angle c equal to the angle f ; 
then will the two triangles be identical, 
or equal in all respects. ^ 

For conceive the triangle abc to be applied to, ot ^Yas&& 
on, the triangle def, in such a manner that the pov&l c m*$ 

Vol 1 37 




\ 



482 •EOXETftT. 

coincide with the point r, and the side ac with the side sv 9 
which is equal to it. 

Then, since the angle f is oqunl to the angle c (by hyp.), 
the side bc w«ill fall on the side r.r. Also, because ac is 
equal to or, and nc equal to kf (by hypO, the point a will 
co»»cido witn the point i>, and ihe jx inr a with the point k ; 
consequently the side An will coincide with the side dk. 
Therefore the two triangles are ilt-nlical, and have all their 
other corresponding parts equal (ax. ?)), namely, the side ab 
equal to the side de, the angle a to the angle d, and the angle 
b to the angle e. q. e. d. 

THEOREM II. 

When two triangles have two angles and the included 
side in the one, equal to two angles and the included side in 
the other, the triangles are identical, or have their other sides 
and angle equal. 

Let the two triangles abc, def, q -g 

have the angle^i equal to the anglo 

D, the angle b equal to t L - 1 — 

and the side ab equal to 1 
then these two triangles 1 
tical. 

For, conceive the triangle abc to be placed on the triangle 
def, in such manner that the side ab muy lull exactly on the 
equal side dk. Then, since the angle a is equal to the angle 
i> (by hyp.) } the side ac must fall on the side df ; and, in 
like manner, because the angle h is equal to the angle >:, the 
side bc must fall on the side kf. Thus the three sides of the 
triangle a»c will be exactly placed on the, three sides of the 
triangle dkf : consequently the two triangles are identical 
(ax. 9), having the other two sides ac, bu, equal to the two 
df, rf, and the remaining angle c equal to the remaining 
angle f. q. e. d. 

theorem III. 

In an isosceles triangle, the angles at the base are equal. . 
Or, if a triangle have two sides equal, their opposite angles 
will also bc equal. 

If the triangle abc have the side ac equal 
to the side bc : then will the angle b be equal 
to the angle a. 

For, conceive the angle c to be bisected, 
or divided into two equal parts, by the line 
cd, making the angle acd equal to the angle — jj- 

BCD. 



I to the anglo A 

the angle e, / I 

> the side df. ; / I 

\ will be iden- * t, 





THEOREM. 



Then, the two triangles, acd, bci>, have two sides and 
the contained angle of tho one, equal to two sides and the 
contained angle of tho other, viz. the side ac equal to Br, 
the angle acd equal to bcd, and the side cd common ; there- 
fore these two triangles arc identical, or equal in all respocts 
(th. 1) ; and consequently tho angle a equal to the angle b. 
q. E. 1). 

Carol. 1. Hence the lino which hisects the vertical angle 
of an isosceles triangle, bisects the base, and is also perpen- 
dicular to it. 

Carol . 2. Hence too it appears, that every equilateral tri- 
angle, is also equiangular, or has all its angles equal. 



THEOREM IT, 

When a triangle has two of its angles equal, the sides 
opposite to them are also equal. 



If the triangle abc, have the angle cab 
equal to the angle cba, it will also have the 
side ca equal to the side ch. 

For, if ca and cb he not equal, let ca be 
the greater of tho two, and let da be equal 




t3 en, and join db. Then, because da, ab, 
arc equal to cb, ba, each to each, and the angle dab to 
cba (hyp.), the triangles dab, cba, are equal in all respects 
(th. 1), a part to the whole, which is absurJ ; therefore 
ca is not greater than cb. fn the same way it may be 
proved, that cb is not greater than ca. They are therefore 
equal. f. d. 

Cord. Hence every equiangular triangle is also cquu 
lateral. 

THEOREM V. 

Whex two triangles have all the three sides in the one, 
equal to all the three sides in the other, the triangles are 
identical, or have also their three triangles equal, each to each* 

Let the two triangles abc, abd, 
have their three sides respectively, 
equal, viz. the side ab equal to ab, 
ac to ad, and nc to bd ; then shall 
the two triangles be identical, or have 
their angles equal, viz. those anglos 




•K0ML1KV. 



that are opposite to the equal sides ; g 

namely, the angle bac to the angle 

bad, the angle abc to the angle abd, 

and the angle c to the angle d. A 
For, conceive the two triangles to 

be joined together by their longest j) 

equal sides, and draw the line cd. 

Then, in the triangle acd, because the side ac is equal 

to ad (by hyp.), the angle acd is equal to the angle adc 

(th. 3). In like manner, in the triangle rcd, the angle BCD 

is equal to the angle bdc, because the side bc is equal to bd. 
Hence then, the angle acd being equal to the angle adc, 
and the angle bcd to the angle bdc, by equal additions the 
sum of the two angles acd, bcd, is equal to the sum of the 
two adc, bdc, (ax. 2), that is, the whole angle acb equal to 
the whole angle adb. 

Since then, the two sides ac, cb, are equal to the two 
sides ad, db, each to each, (by hyp.), and their contained 
angles acb, adb, also equal, the two triangles abc, abd, 
are identical (th. 1), and have the other angles equal, viz. 
the angle bac to the angle bad, and the angle abc to the 
angle akd. u. l. d. 

TilKOKJ m VI. 

Win:* one line meets another, the angles which it makes 
on the same side of the otli^r, are together equal to two right 
angles. 

Let the line ah meet the line cd : then 
will the two angles abc, abd, taken to- 
gether, ho equal to two right angles. 

For, first, when the two angles abc, abd, 
are equal to each other. tlic\ are both of 
them right angles (def. 15.) 

But when the angles are unequal, suppose bb drawn per- 
pendicular to cd. Then, since the two angles ebc, fbd, are 
right angles (def. 15), and the an^le ebd is equal to the two 
angles eba, add, together (ax. 8), the three angles, ebc, kba, 
and abd, arc equal to two right angles. 

But the two angles ebc, eba, arc together equal to the 
angle abc (ax. 8). Consequently the two angles abc, abd, 
are also equal to two right angles, q. e. d. 

Corol. 1. Hence also, conversely, if the two angles abc, 
abd, on both sides of the line ab, make up together two 
right angles, then cb and bd form one continued right 
line cd. 





THEOREMS. 



385 



Corol. 2. Hence, alt the angles which can be made, at 

any point b, by any number of linos, on the same side of 
the right line cd, are, when taken all together, equal to two 
right angles. 

Cord. 3. And, ns all the angles that can be made on the 
other side of the line cn are also equal to two right angles ; 
therefore all the angles that can be made quite round a point 
b, by any number of lines, are equal to four right angles. 

Coral. 4. Hence also the whole circumfer- 
ence of a circle, being the sum of the mea- 
sures of all the angles that can be made about 
the centre f (def. 57), is the measure of four 
right angles. Consequently, a semicircle, or 
180 degrees, is the measure of two right an- 
gles ; and a quadrant, or 90 degrees, the measure of one 
right angle. 

THEOREM VII. 

When two lines intersect each other, the opposite angles are 
equal. 

Let the two lines ad, vu, intersect in 
the point e ; then will the angle aec be /C 
equal to the angle bed, and the angle > i 
aed equal to the angle ceb. a } /TQ 

For, since the lino cc meets the line / 
ab, the two angles aec, bec, taken to- D 
gethcr, arc equal to two ricjht angles (lli. 6). 

In like manner, the line he, meeting the line cd, makes 
the two angles dkc, red, equal to two right angles. 

Therefore the sum of the two angles aec, bec, is equal to 
the sum of the two bec, ki d (ax. 1 ;■. 

And if the angle nr.c, which is common, be taken away 
from both these, the remaining angle aec will be equal to 
the remaining angle bed (ax. 3). 

And in like manner it may be shown, that the angle aid 
is equal to the opposite angle bec. 

theorem viii. 

When one side of a triangle is produced, the outward 
angle is greater than either of the two vkw&x& c^c*\\» 
angles. 




980 



ozomnr. 



Let abo be a triangle, having the 
aide ab produced to i> ; then will the 
outward angle cno be greater than 
cither of the inward opposite angles a 
or c. 

For, conceive the side bc to be bi- 
sected in the point e, and draw the 




line ak, producing it till ef bc equal 
to ak ; and join»BF. 

Then, since the two triangles aec, hf.f, have the side 
ae — the side kf, and the side ce = the side be (by suppos.) 
and the included or opposite angles at e also equal (th. 7}, 
therefore those two triangles arc equal in all respects 
(th. 1), and have the angle c = the corresponding angle 
ebf. But the angle crd is greater than the angle ebf ; 
consequently the said outward angle cbd is afso greater than 
the angle <:. 

In like manner, if cb bo produced to g, and ab b? 1 i- 
sected, it may bc shown that the outward angle abc, or it* 
equal cbd, is greater than the other angle a. 

THEOREM IX. 

TnE greater side, of cvrry triangle, is oiprs'tc to the 
greater ungle ; and the greater angle opposite to the greater 
aide. 

Lot abc be a triangle, having the side q 



ab greater than the side ac ; then will the 
angle acb, opposite the greater side ab, be 
greater than the angle b, opposite the less 
aide ac. 




For, on the greater side ab, take the 
part ad equal to the less side ac, and join cd. Then, since 
bcd is a triangle, the outward angle Anc if. greater than the 
inward opposite angle b (th. 8). But the angle acd is equal 
to the said outward angle abc, because ad is eqal to ac 
(th. 3). Consequently the angle acd also is greater than the 
angle b. And since the angle acd is only a part of acb, 
much more must the whole angle acb be greater than the 
angle n. a. k. d. 

Again, conversely, if the angle c he greater than the angle 
b, then will the side ab, opposite the former, be greater than 
the side ac, opposite the latter. 

For, if ab bc not greater than ac, it must bo cither 
eguai to it, or leas than k. But it cannot be equal, for 



TRSOBEKB- 



then the angle c would be equal to the angle b (th- 3), which 
it is not, by the supposition. Neither can it bo less, for then 
the angle c would be less than the angle n, by the former 
part of this ; which is also contrary to the supposition. Tho 
aide ab, then, being neither equal to ac, nor less than it, 
must necessarily be greater, a. u. d. 



tiieobem x. 




The sum of any two sidtes of a triangle is greater than the 
third side. 

Let abc be a triangle ; then will the 
sum of any two of its sides be greater 
than the third side, as for instance, 
ac + cb greater than ab. 

For, produce ac till cn be equal to 
CB, or ad equal to the sum of the two 
AC + cb; and join bi> Then, because 
CD is equal to en (by constr.), the angle i> is equal to the 
angle cbd(iIi. 3). But the angle ahd is greater than the 
angle cbd, consequently it must also be greater than the 
angle in And, since the greater side of any triangle is op. 
posite to the greater nngle (th. 9), the side ai> (of the tri- 
angle ahd) is greater tbun the side au. But ad is equal to 
ac and cd, or ac and cb, taken together (by constr.) ; there- 
fore ac + cb is also greater than ab. q. e. d. 

Carol. The shortest distance between two points, is a 
single right line drawn from the one point to the other. 



THE0BE3I XI. 



The difference of any two sides of a triangle, is less than 
the third side. 

Let abc be a triangle ; then will the D 
difference of any two sides, as ab — ac, 
be less than the third side sc. 

For, produce the less side ac to d, 

till ad be equal to the greater side ab, . 

so that <"D may be the difference of tl.o -A. B 
two sides ah — ac ; and join bd. Then, 
because ad is equal to ab (by constr.), the opposite angels D 
and abd are equal (th. 3). J kit the angle cbd is less than 
the angle abd, and consequently also less than the eaual 
angle d. And since the greater side of any \xv»&$& 




OEOatCTKT. 



opposite to the greater angle (th. 9), the side cd (of the tri- 
anglo bcd) is less than the side bc. a. e. d. 

Otherwise* Set off upon ah a distance ai equal to ao. 
Then (th. |0) ac + cb is greater than ab, that is, greater 
than ai + ib. From these, take away the equal parts AC, 
Ai, respectively ; and there remains cb greater than ic. Con* 
sequently, ic is less than cb. a. e>- d. 

THEOREM XII. 

When a lino intersects two parallel lines, it makes the 
alternate angles equal to each other. 

Let the line ef cut the two parallel 
line ab, cd ; then will the angle aef be 
equal to the alternate angle efd. 

For if they are not equal, one of them 
must be greater than the other; let it be 
efd for instance, which is the greater, if 
possible ; and conceive the line fh to bc 
drawn, cutting off the part or angle efb equal to the angle 
AEF, and meeting the line ab in the point n. 

Then, since the outward angle aef, of the triangle bef, 
is greater than the inward opposite angle efb (th. 8) ; and 
since these two angles also are equal (by the constr.) it fol- 
lows, that those angles are both equal and unequal at the 
same time : which is impossible. Therefore the angle efd 
is not unequal to the alternate angle aef, that is, they are 
equal to each other, q. e. d. 

Corol. Right lines which are perpendicular to one, of two 
parallel lines, are also perpendicular to the other. 

THEOREM XIII. 

When a lino, cutting two othrr lines, makes the alter- 
nate angles equal to each oilier, those two lines are paral- 
lel. 

Let the lino F.r, cutting the two lines 
ab, cn, make the alternate angles aef, 
dfe, equal to each other ; then will ab 
be parallel to cd. 

For if they be not parallel, let some 
other line, as fg, be parallel to ab. 
Then, because of these parallels, the 
angle aef is equal to the alternate angle efg (th. 12). But 
the angle aef is equal to the angle efd (by hyp.) There- 
fore the anglo efd is equal to the angle efg "(ax. 1) ; that is, 
a part is equal to the whole, which is impossible. Therefore 
oo line but cd can be parallel to ab. q. e. d. 





THEOREMS. 



280 



Cord. Those lines which are perpendicular to the same 
lines, are parallel to each other. 




THEOREM XIV. 

When a line cuts two parallel lines, the outward angle is 
equal to the inward opposite one, on the same side ; and 
the two inward angles, on the sume side, equal to two right 
angles. 

Let the line ef cut the two parallel 
lines ab, cd ; then will the outward an- 
gle eob be equal to the inward oppo- 
site angle ghd, on the same side of the 
line ef ; and the two inward angles 
bgh, ghd, taken together, will be equal 
to two right angles. 

For since the two lines ab, cd, are 
parallel, the angle agh is equal to the alternate angle cud, 
(th. 12.) But the angle agh is equal to the opposite angle 
egb (th. 7). Therefore the angle egb is also equal to the 
angle ghd (ax. 1). q. e. d. 

Again, because the two adjacent angles egb, bgh, are to- 
gether equal to two right angles (th. (5; ; of which the angle 
xgb has been shown to be equal to the angle ghd ; therefore 
the two angles bgh, gud, taken together, are also equal to 
two right angles. 

Carol. 1. And, conversely, if one line meeting two other 
lines, make the angles on the same side of it equal, those 
two lines are parallels. 

Corol. 2. If a line, cutting two other lines, make the sum 
of the two inward angles on the same side, less than two 
right angles, those two lines will not be parallel, but will 
meet each other when produced. 

THEOREM XV. 

Those lines which arc parallel to the same line, are 
parallel to each other. 

Let the lines ab, cd, be each of them G - p 

parallel to the line ef ; then shall the lines A iS 



ab, cd, be parallel to each other. C 35 

For, let the line Gibe perpendicular jr """I p 
to ef. Then will this line be also per- I 
pendicular to both the lines ab, cn (corol. th. 12), and con- 
sequently the two lines ab, cd, are parallels (corol. th. 13\. 

a. i>. 

Vol. I. 38 



390 



OKOKETAY. 



THEOREM XVI. 



Whex one sido of a triangle is produced, the outward 
angle is equal to both the inward opposite angles 
together. 

Let the side ab, of the triangle 
ahc, be produced to d ; then will the 
outward angle cbd be equal to the sum 
of the two inward opposite angles a 
and c. 

For, conceive be to be drawn pa- 
rallel to the side ac of the triangle. 
Then iic, meeting the two parallels ac, be, makes the alter- 
nate angles c and cbe equal (th. 12). And ab, cutting the 
same two parallels ac, be, makes the inward and outward 
angles on the same side, a and ebd, equal to each other 
(th. 14). Therefore, by equal additions, the sum of the two 
angles a and c, is equal to the sum of the two cbe and : 
that is, to the whole angle cbd (by ax. 2), <i» E. D. 




THEOREM XVII. 




In any triangle, the sum of all the three angles is equal to 
two right angles. 

Let abc be any plane triangle ; then 
the sum of the three angles a + b + c 
is enuul to two right angles. 

For, let the side ab he produced to D. 
Then the outward angle cbd is equal 
to the sum of the two inward opposite 
angles a + c (th. 16). To each of these equals add the in- 
ward angle b, then will the sum of the three inward angles 
a + b + c be equal to the sum of the two adjacent angles 
abc + cbd (ax. 2). But the sum of these two last adjacent 
angles is equal to two right angles (th. 6). Therefore aba 
the sum of the three angles of the triangle a + b + c ii 
equal to two right angles (ax. 1). q. e. d. 

Carol. 1. If two angles in one triangle, be equal to two 
angles in another triangle, the third angles will also be equal 
(ax* 3), and the two triangles equiangular. 

Cord. 2. If one angle in one triangle, be equal to mm 
angle in another, the sums of the remaining angles will afcj» 
be equal (ax. 3). 



THEOREMS. 



291 



CoroZ. 3. If one angle of n triangle be right, the sum of 
the other two will also be equal to a right angle, and each of 
them singly will be acute, or less thun a right angle. 

Cmxi. 4. The two least angles of every triangle are acute, 
or each less thaa a right angle. 



THEOREM XVIII. 




Iff any quadrangle, the sum of all the four inward angles, is 
equal to four right angles. 
Let a bcd be a quadrangle ; then the 
sum of the four inward angles, a + b + 
c + d is equal to four right angles. 

Let the diagonal ac be drawn, dividing 
the quadrangle into two triangles, abc, adc. 
Then, because the sum of the three angles 
of each of these triangles is equal to two 
right angles (th. 17) ; it follows, that the sum of all (he 
angles of both triangles, which make; up the lour angles of 
the quadrangle, must be equal to four ri*» In angles (ax. 2). 

ci. K. D. 

Cord. 1. Hence, if three of the angles be right ones, tl.e 
fourth will also be a right angle. 

Carol. 2. And if the sum of two of the four angles be 
equal to two right angles, the sum of the remaining two will 
also be equal to two right angles. 



THEOREM XIX. 

In any figure whatever, the sum of all the inward angles, 
taken together, is equul to twice as many right angles, 
wanting four, as the figure has sides. 

Let abcde be any figure ; then the 
eum of all its inward angles, a + h + 
c + D + E, is equal to twice as many 
right angles, wanting four, as the figure 
lias sides. 

For, from any point r, within it, draw 
lines, pa, i»b, re, &c. to all the. angles, 
dividing the polygon into as m iny tri- 
angles as it has sides. Now the sum of the three angles of 
each of these triangles, is equal to two riijht angles (h. 17) ; 
therefore the sum of the angles of alt the triangles is equal 
to twice as many right angles as the figure has sides. But 
the sum of all the angles about the point nYikh m w> 




292 



GEOMETRY. 




many of f lie angles of the triangles, but no part or the to- 
ward angles of the polygon, is equal to four right angles 
(corol. 3, th. 6), and must he deducted out of the former 
sum. Hence it follows that the sum of all the inward angles 
of the polygon alone, a + b + c + i> + e» is equal to twice 
as many right angles as the figure has sides, wanting the 
said four right angles, u. e. d. 

TI1EOREX XX. 

When every side of any figure is produced out, the sum 
of all the outward angles thereby made, is equal to four right 
angles. 

Let a, b, c, dtc. be the outward 
angles of any polygon, made by pro- 
ducing all the sides ; then will the sum 
a + b + c + i) + e, of all those outward 
angles, be equal to four right angles. 

For every one of these outward an- 
gles, together with its adjacent inward 
angle, make up two right angles, as 
A+a equul to two right angles, being 
the two angles made by one line meeting another (th. 6). 
And there being as many outward, or inward angles, as the 
figure has sides ; therefore the sum of all the inward and 
outward angles, is equal to twice as many right angles as 
the figure has sides. But the sum of all the inward angles 
with four right angles, is equal to twice as many right angles 
as the figure has sides (th. 19). Therefore the sum of all 
the inward and all the outward angles, is equal to the sum 
of all the inward angles and four right angles (by ax. 1). 
From each of these take away all the inward angles, and 
there remains all the outward angles equal to four right angles 
(by ax. 3j. 

THEOREM XXI. 

A perpendicular is the shortest line that can be drawn 
from a given point to an indefinite line. And, of any other 
linos drawn from the same point, those that are nearest the 
perpendicular are less than those more remote. 

If ab, ac, ad, &c. be lines drawn from 
the given point a, to the indefinite line de, 
of which ab is perpendicular ; then shall 
the perpendicular ab be less than ac, and 
ac less than ad, &c. 

For, the angle b being a t\gV\t quo, the 




TREOBEKI. 



298 



angle c is acute (by cor. 3, th. 17), and therefore less than 
the angle b. But the less angle of a triangle is subtended 
by the leas side (th. 9). Therefore the side ab is less than 
the side ac. 

Again, the angle acb being acute, as before, the adjacent 
angle acd will be obtuse (by th. tf) ; consequently the angle 
d is acute (corol. 3, th. 17), and therefore is less than the 
angle c. And since the less side is opposite to the less angle, 
therefore the side ac is less than the side ad* q. s. d. 



Carol. A perpendicular is the least distance of a given 
point from a line. 



THEOREM XXII. 

Thb opposite sides and angles of any parallelogram are 
equal to each other ; and the diagonal divides it into two 
equal triangles. 

Let abcd be a parallelogram, of which 
the diagonal is bd ; then will its opposite 
sides and angles be equal to each other, 
and the diagonal bd will divide it into two 
equal parts, or triangles. 

For, since the sides ab and dc are pa- 
rallel, as also the sides ad and bc (defin. 



32), and the line bd meets them ; therefore the alternate 
angles are equal (th. 12), namely, the angle abd to the angle 
cdb, and the angle adb to the angle cbd. Hence the two 
triangles, having two angles in the one equal to two angles 
in the other, have also their third angles equal (cor. 1, th. 17), 
namely, the angle a equal to the angle c, which are two of 
the opposite angles of the parallelogram. 

Also, if to the equal angles abd, cdb, be added the equal 
angles cbd, abd, the wholes will be equal (ax. 2), namely, 
the whole angle abc to the whole ado, which are the other 
two opposite angles of the parellelogram. ft. e. d. 

Again, since the two triangles are mutually equiangular 
and have a side in each equal, viz. the common side bd ; 
therefore the two triangles are identical (th. 2), or equal in 
all respects, namely, the side ab equal to the opposite side 
dc, and ad equal to the opposite side bc, andlta^fata 
triangle abd equal to the whole triangle bcd. <fc- fe« t>% 



294 



GEOMETRY. 



Corol. 1. Hence, if one angle of a parallelogram be a right 
angle, ail the other three will also be right angles, and the 
parallelogram a rectangle. 

Corol. 2. Henc; also, the sum of any two adjacent angles 
of a parallelogram is equal to two right angles. 

THEOREM XXIII. 

Every quadrilateral, whose opposite sides are equal, is a 
parallelogram, or has its opposite sides parallel. 



Let a bcd be a quadrangle, having the 
opposite sides equal, namely, the side ab 
equal to nc, and ad equal to hc ; then 
shall these equal sides be also parallel, 
and the figure a parallelogram. 




For, let the diagonal bd he drawn. 
Then, the triangles, abd, cud, being 
mutually equilateral (by hyp.), they are 
also mutually equiangular (th. 5), or have their corresponding 
angles equal ; consequently the opposite sides are parallel 
(th. 13) ; viz. the side ah parallel to nc, and ad parallel to 
bc, and the figure is a parallelogram, a. k. d. 

THEORKM XXIV. 

Those lines which join the corresponding extremes of 
two equal and parallel lines, are themselves equal and 
parallel. 

Let ab, nc, be two equal and parallel lines ; then will 
the lines ad, bc, which join their extremes, be also equal 
and parallel. [Sec the fig. above.] 

For, draw the diagonal bd. Then, because ab and dc are 
parallel (by hyp.), the angle abd is equal to the alternate 
angle bdc (th. 12). Hence then, the two triangles having 
two sides and the contained alines equal, viz. the side ab 
equal to the aide dc, and the side bo common, and the con- 
tained angle abd equal to the contained angle bdc, they 
have the remaining sides and angles also respectively equal 
(th. 1) ; consequently ad is equal to bc, and also parallel to 
it (th. 12). u. £. d. 

THEORKM XXV. 

Parallelograms, as also triangles, standing on the 
same base, and between the samo parallels, are equal to 
each other. 



THEOREMS. 



295 



Let abcd, abep, be two parallelograms, J> C T E 
and abc, abf, two triangles, standing on V A A 7 
the same base ab, and between the same \ / V ; / 
parallels ab, de ; then will the parallelo- \ / 
gram abcd be equal to the parallelogram y/ V/ 
abbp, and the triangle abc equal to the Js^ JJ 
triangle abf* 

For, since the line de cuts the two parallels af, be, and 
the two ad, bc, it makes the angle e equal to the angle afd, 
and the angle d equal to the angle bce (th. 14) ; the two 
triangles adf, bce, are therefore equiangular (cor. 1, th. 17) ; 
and having the two corresponding sides ad, bc, equal 
(th. 22), being opposite sides of a parallelogram, these two 
triangles are identical, or equal in all respects (th. 2). If 
each of these equal triangles then be taken from the whole 
space abed, there will remain the parallelogram abef in 
the one case, equal to the parallelograms abcd in the other 
(by ax. 3). 

Also the triangles abc, abf, on the same base ah, and 
between the same parallels, are equal, being the halves of 
the said equal parallelograms (th. 22). q. r. d. 

CoroL 1. Parallelograms, or triangles, having the same 
base and altitude, are equal. For tho altitude is the same as 
the perpendicular or distance between the two parallels, which 
is every where equal, by the definition of parallels. 

Corel. 2. Parallelograms, or triangles, having equal bases 
and altitudes, are equal. For, if the one figure be applied 
with its base on the other, the bases will coincide or be the 
same, because they are equal : and so the two figures, having 
the same base and altitude, are equal. 

theorem xxvi. 

If a parallelogram and a triangle, stand on the same 
base, and between the same parallels, the parallelogram 
will be double the triangle, or the triangle half the paral- 
lelogram. 

Let abcd be the parallelogram, and abe a 
triangle, on the same base ab, and between 
the same parallels ab, de ; then will the 
parallelogram abcd he double the triangle 
abe, or the triangle half the parallelo- 
gram. 

For, draw the diagonal ac of the parallelogram, divid\\i% 
it into two equal parts (th. 22). Then bec&uafe lY\& XxwoqgtK* 




296 



GEOMETRY. 



abc, abe, on the same base, and between the same parallels, 
are equal (th. 25) ; and because the one triangle abc is half 
the parallelogram abcd (th. 22), the other equal triangle 
abe is also equal to half the same parallelogram abcd. 
q. e. d. 

Card. 1. A triangle is equal to half a parallelogram of the 
same base and altitude, because the altitude is the perpendi- 
cular distance between the parallels, which is every where 
equal, by the definition of parallels. 

Carol. 2. If the base of a parallelogram be half that of a 
triangle, of the same altitude, or the base of the triangle be 
double that of the parallelogram, the two figures will be 
equal to each other. 



THEOREM XXVn. 







Rectangles that are contained by equal lines, are equal 
to each other. 

Let bd, fh, be two rectangles, having jj C H G 
the sides ab, bc, equal to the sides ef, "7 
fo, each to each ; then will the rectangle / 
bd be equal to the rectangle fh. / 

For, draw the two diagonals ac, eg, A. B £ F 
dividing the two parallelograms each into 
two equal parts. Then the two triangles abc, efg, are 
equal to each other (th. 1), because they have the two sides 
ab, bc, and the contained angle b, equal to the two sides 
ef, fg, and the contained angle f (by hyp.}. But these 
equal triangles are the halves of the respective rectangles. 
And because the halves, or the triangles, are equal, the 
wholes, or the rectangles db, hf, are also equal (by ax. 6). 
a. e. d. 

Carol. The squares on equal lines are also equal ; for 
every square is a species of rectangle. 



THEOREM XXVIII. 



The complements of the parallelograms, which are about 
the diagonal of any parallelogram, are equal to each other. 

Let ac be a parallelogram, bd a dia- 
gonal, eif parallel to abof dc, and <;iu J) G 
parallel to "ad or bc, making ai, ic, com- K j 
plements to the parallelograms eg, iif, E 



which are about the diagonal db : then / ~j 

will the complement ai be equal to the h 
complement xc 




i 



THEOREMS. 207 

For, since the diagonal db bisects the three parallelograms 
ac, eg, hp (th. 22) ; therefore, the whole triangle dab being 
equal to the whole triangle dcb, and the parts dei, ihb, re* 
sportively equal to the parts dgi,ifb, the remaining parts Ar, 
ic, must also be equal (by ax. 3). q. e. d. 



THEOREM XXIX. 



A trapezoid, or trapezium haying two sides parallel, is 
equal to half a parallelogram, whose base is the sum of those 
two sides, and its altitude the perpendicular distance between 
them. 

Let abcd be the trapezoid, having its 
two sides ab, dc, parallel ; and in ah 
produced take be equal to dc, so that 
ae may be the sum of the two parallel 
sides ; produce dc also, and let kf, gc, G B E 

bh, be all three parallel to ad. Then is 
af a parallelogram of the same altitude with the trapezoid 
abcd, having its base ae equal to the sum of the parallel 
sides of the trapezoid ; and it is to be proved that the trape- 
zoid abcd is equal to half the parallelogram af. 

Now, since triangles, or parallelograms, of equal bases and 
altitude, are equal (corol. 2, th. 25), the parallelogram no is 
equal to the parallelogram he, and the triangle cob equal to 
the triangle chb ; consequently the line bc bisects, or equal- 
ly divides, the parallelogram af, and abcd is the half of it. 

Q. E. D. 




THEOREM XXX. 



J3 G H C 



The sum of all the rectangles contained under one whole 
line, and the several parts of another line, any way divid- 
ed, is equal to the rectangle contained under the two whole 
lines. 

Let ad be the one line, and ab the 
other, divided into the parts ae, ef, 
fb ; then will the rectangle contained 
by ad and ab, be equal to tlio sum of 
the rectangles ot ad and ae, and ad and 
ef, and ad and fb : thus expressed, 

AD • AB = AD . AE + AD . EF + AD . FB. 

For, make the rectangle ac of the two whole lines ad, 
ab ; and draw eg, fh, perpendicular to ab, or parallel to 
ad, to which they are equal (th. 22). Then \ta ^iWAa 
rectangle ac is made up of all the other TectaaeW 

Vol. I. 39 









A. E f B 



208 



GEOMETBY. 



G H C 



fc. Hut these rectangles arc contain- 
ed by ad and ae, eg and ef, i n and fb ; 
which aro equal to the rectangles of ad 
and ae, ad and ef, ad and fb, because 
ad is equal to each of the two eg, fh. 
Therefore the rectangle \d. ab is equal 
to the sum of all l he other rectangles ad • 

AE, AD . EF, AD . FB. Q. E. D. 

Corol. If a right line be divided into any two parts, the 
square on the whole line, is equal to both the rectangles of 
the whole line and each of the parts. 



X FB 



THEOREM XXXI. 

The square of the sum of two lines, is greater ti.an the 
sum of their squares, by twice the rectangle of the said 
lines. Or, the square of a whole line, is equal to the 
squares of its two parts, together with twice the rectangle of 
those parts. 

Let the line ab be the sum of any two 
lines ac, ch ; then will the square of ab js- H ij 

be equal to the squares of .u?, cb, together 
with twice the rectangle of ac . cb. That 
1S, AB a = AC 2 + cb 3 + 2ac . cn. 



I 



For, lot \ni)E be the square on the sum * C B 

or whole line ab, and acfg the square 
on the part ac. Produce cf and ar to the other sides at H 
and i. 

From the lines ch, ci. which are equal, being each equal 
to the biiVs of the square ab or bd (th. 22), take the parts 
CF, gf, which arc also equal, being the sides of the square 
Ai'. am 1 there remains *u equal to fx, which are also equal 
to mi. or, being the opposite sides of the parallelogram, 
fleucc the figure hi is equilateral : and it has all its angles 
riiiht onus teorol. 1, th. 22) : it is therefore a square on the 
line fi, or the square of its equal cn. Also the figures ef, 
fb, arc equal to two rectangles under ac and cb, because 
c;r is equal to ac. and fh or fi equal to cb. But the 
whole square ad is made up of the four figures, viz. the two 
tiquarch af, id, and the two equal rectangles ef, fb. That 
i? ? the .-square of aji is equal to the squares of ac, cb, toge- 
ihcr with twice th" rectangle of ac, cb. a. u. d. 

0>r. /. Hence, if a line be divided into two equal parts ; 
the .square of the whole line will be equal to four times the 
square of half the hue. 



I 

THEOREMS. 



299 



E 



ID 



THEOREM XXXII. 

The square of the difference of two lines, is less than the 
sum of their squares, by twice the rectangle of the said 
lines. 

Let ac, bc, be any two lines, and ab 

their difference : then will the square of ab & 

be less than the squares of ac, bc, by 
twice the rectangle of ac and bc. Or, 

AB* s* AC 1 + BC 1 — 2AC . BC. 

For, let abde be the square on the dif- 
ference ab, and acfo the square on the A. B 
line ac. Produce ed to h ; also produce K I 

db and hc, and draw ki, making bi the square of the other 
line bc. 

Now it is visible that the square ad is less than the two 
squares af, bi, by the two rectangles ef, di. But gf is 
equal to the one line ac, and ue or fii is equal to .the other 
line bc ; consequently the rectangle kf, contained under i:« 
and of, is equal to the rectangle of ac and bc. 

Again, fh being equal to ci or bc or Dir, by adding the 
common part hc, the whole hi will bo equal to the whole rc, 
or equal to ac ; and consequently the figure di is equal to 
the rectangle contained by ac and bc. 

Hence the two figures ef, di, arc two rectangles of the 
two lines ac, bc ; and consequently the square of vn is 
less than the squares of ac, bc, by twice the rectangle 

AC . BC. Q. E. B. 

THEOREM XXXUI. 

The rectangle under the sum and difference of two lines, is 
equal to the difference of the squares of those lines*. 
Let ab, ac, be any two unequal lines ; 

then will the difference of the squares of 

ab, ac, be equal to a rectangle under 

their sum and difference That is, 

AB 1 — AC 3 = AB -f- AC . AB — AC. 

For, let abde be the square of ab, and 
acfo the square of ac. Produce db 
till bh be equal to ac ; draw hi parallel 
to ab or ed, and produce fc both ways 
to i and k. 



E 
G 



C 



F 



* This and the two preceding theorems, arc evinced algebraically, 
by the three expression! 

( a + 6 ) J = a ' + %ib + 6» a* + 6» -f- 2ab 
(« — 6 )3 = a> — 2a6 4 fta = a> + 63 — 2a6 
(a + 6)(ft-6):=«*— 6*. 



800 



GEOMETRY. 



Then the difference of the two squares ad, af, is evi- 
dently the two rectangles ef, kb. But the rectangles ef, 
bi are equal, being contained under equal lines ; for ex and 
bh are each equal to ac, and ge is equal to cb, being each 
equal to the difference between ab and ac, or their equals 
ae and ag. Therefore the two ef, kb, are equal to the two 
xb 9 bi, or to the whole kh ; and consequently kh is equal 
to the difference of the squares ad, af. But kh is a rect- 
angle contained by dh, or the sum of ab and ac, and by kd, 
or the difference of ab and ac. Therefore the difference of 
the squares of ab, ac, is equal to the rectangle under their 
sum and difference. Q. e. d. 

theorem xxxiv. 

In any right angled triangle, the square of the hypo- 
thenuse, is equal to the sum of the squares of the other two 
sides* 

Let abc be a right-angled triangle, Q 
having the right angle c ; then will the 
square of the hypothenuse ab, be equal 
to the sum of the squares of the other -° 
two sides ac, cb. Or ab 3 = ac* 

+ BC a . 

For, on ab describe the square ae, 
and on ac, cb, the squares ag, bh ; 
then draw ck parallel to ad or be ; 
and join ai, bf, cd, ck. 

Now, because the line ac meets the two cg, cb, so as to 
make two right angles, these two form one straight line gb 
(corol. 1, th. 6). And because the angle fac is equal to the 
angle dab, being each a right angle, or the angle of a square ; 
to each of these equals add the common angle bac, so will 
the whole angle or sum fab, be equal to the whole angle or 
sum cad. But the line fa is equal to the line ac, and the 
line ab to the line ad, being sides of the same square ; so 
that the two sides fa, ab, and their included angle fab. are 
equal to the two sides ca, ad, and the contained angle cad, 
each to each : therefore the whole triangle afb is equal to 
the whole triangle acd (th. 1). 

But the square ag is double the triangle afb, on the 
same base fa, and between the- same parallels fa, ob 
(th. 26) ; in like manner the parallelogram ak is double the 
triangle acd, on the same base ad, and between the same 
parallels ad, ck. And since the doubles of equal things, 
are equal (by ax. 6) ; therefore the square ag is equal to the 
parallelogram ak. 




THEOREMS. 



301 



la like manner, the other square bh is proved equal to 
the other parallelogram bk. Consequently the two squares 
ag and bh together, are equal to the two parallelograms ak 
and bk together, or to the whole square ak. That is, the 
sum of the two squares on the two less sides, is equal to the 
square on the greatest side. a. e. d. 

Cord. 1. Hence, the square of either of the two less sides, 
is equal to the difference of the squares of the hypothenuse 
and the other side (ax. 3) ; or, equal to the rectangle con- 
tained by the sum and difference of the said hypothenuse 
and other side (th. 33). 

Carol. 2. Hence also, if two right-angled triangles have 
two sides of the one equal to two corresponding sides of the 
other ; their third sides will also be equal, and the triangles 
identical. 



THEOREM XXXV. 




In any triangle, the difference of the squares of the 
two sides, is equal to the difference of the squares of the 
segments of the base, or of the two lines, or distances, 
included between the extremes of the base and the perpen- 
dicular. 

Let abc be any triangle, having 
cd perpendicular to ab ; then will 
the difference of the squares of ac, 
bc, be equal to the difference of 

the squares of ad, bd ; that is, _ _ A ^ ^ 

AC* BC 3 = AD 2 BD 2 . A BD A DB 

For, since ac 2 is equal to ad 2 + cd 2 > , „. N 

and bc 2 is equal to bd 2 + c d 2 \ W ihm M > ; 
Theref. the difference between ac 2 and bc 2 , 
is equal to the difference between ad 2 + cd 2 

and bd 2 + cd 2 , 

or equal to the difference between ad 2 and bd 3 , 

by taking away the common square cd 3 . q. e. d. 

Carol. The rectangle of the sum and difference of the two 
sides of any triangle, is equal to the rectangle of the sum 
and difference of the distances between the perpendicular 
and the two extremes of the base, or equal to the rectangle 
of the base and the difference or sum of the segments, 
according as the perpendicular falls within or without the 
triangle. 



302 



GEOMETRY. 



That is, (ac+bc) . (ac — bc) = (ad+bd) . (ad — bd) 
Or, (ac+bc) . (ac — bc) = ab . (ad — bd) in the 2d fig. 
And (ac+bc) . (ac— bc) = ab . (ad+bd) in the lrt fig. 



THEOREM XXXVI. 



In any obtuse-angled triangle, the square of the side sub- 
tending the obtUBe angle, is greater than the sum of the 
squares of the other two sides, by twice the rectangle of 
the base and the distance of the perpendicular from the ob- 
tuse angle. 

Let abc be a triangle, obtuse angled at b, and cd perpen- 
dicular to ab ; then will the square of ac be greater than the 
squares of ab, bc, by twice the rectangle of ab, bd. That 
is, ac 3 = ab 3 + bc 8 + 2a n . bd. See the 1st fig. above, or 
below. 

For, ad 2 = ab 2 + bd 3 + 2ab . bd (th. 31). 

And ad 1 + en 8 = ab 3 + bd* + cd 2 + 2ab . bd (ax. 2). 

But ad 2 + cd 2 = ac 3 , and bd 2 + cd* = bc 2 (th. 34). 
Therefore ac 2 = ab 2 + bc 2 + 2a b . bd. q. e. d. 



THEOREM XX XVI I. 



In any triangle, the square of the side subtending an acute 
angle, is less than the squares of the base and the other aide, 
by twice the rectangle of the base and the distance of the 
perpendicular from the acute angle. 

Let abc be a triangle, having ^ „ 

the angle a acute, and cd perpen- ^ 
dicular to ab ; then will the square ''■ 
of bc, be less than the squares 

of ab, ac, by twice the rectangle 

of ab, ad. That is, bc = == ab 2 + W .B D A. J)B 
ac* — 2ad . AB. 

For BD 3 = AD 3 + AB 2 — 2ad . AB (th. 32). 

And BD 3 + DC 2 = AD 2 + DC 2 + AB 2 — 2 AD . AB (aX. 2), 

Therefore bc 3 = ac 2 + ab 2 — 2ad . ab (th. 34). q. s. d. 




THEOREM XXXVIII. 



Ik any triangle, the double of the square of a line drawn 
from the vertex to the middle of the base, together with 



THEOREMS. 



303 



double the square of the half base, is equal to the sum of the 
squares of the other two sides. 

Let abc be a triangle, and <;d the line 
drawn from the vertex to the middle of 9 
the base ab, bisecting it into the two equal 
parts ad, db ; then will the sum of the 
squares of ac, ch, be equal to twice the 
sum of the squares of <;«, ad ; or ac 3 + 
cb 3 = 2cd 3 + 2\D a . 



/ 
' j 



A DEB 



For AC 8 = CD 3 + ad 3 + 2ad • de (th. 36). 
And bc 3 = CD* + bd 2 — 2ad . db (th. 37). 
Therefore ac 3 + bc 3 = 2cd 3 + ad 3 + bd 3 

= 2cd 3 -f- 2ad 3 (ax. 2). <j. e. d. 



tiikork;i xxxix. 

In an isosceles triangle, the square of a line drawn from 
the vertex to any point in the base, together with the rect- 
angle of the segments of the base, is equal to the square of 
one of the equal sides of the trim.gle. 

Let abc be the isosceles triangle, and cd c 
a line drawn from the vertex to any point 
D in the base : then will the square of ac, 
be equal to the square of ci>, together 
with the rectangle of ad and on. That is, 

AC 3 = CD 3 + AD . DB. 

For AC 3 — CD 3 = AE 3 — DE 3 (th. 35). 

= AD . DB (th. 33). 

Therefore, ac 2 = cd 2 -f- ad . db (ax. 2). a. k. d. 




A 13 



THEOREM XL. 

In any parallelogram, the two diagonals bisect each other ; 
and the sum of their squares is equal to the sum of the 
squares of all the four sides of the parallelogram. 

Let abcd be a parallelogram, whost; "JJ C 
diagonals intersect each other in e : then 
will ae be equal to ec, and be to i:i> : and 
the sum of the squares of ,u\ m>, will be 
equal to the sum of the square s of ab, ik\ 
cd, da. That is, 

ae = ec, and bj: — ed, 
and ac 3 + bd' = ab s + bc 3 -t ci> r ua • 




804 



GEOMETRY. 



For, the triangles aeb, dec, are equiangular, because 
they have the opposite angles at e equal (th. 7), and the two 
lines ac 9 bd, meeting the parallels ab, dc, make the angle 
bae equal to the angle dce, and the angle abe, equal to the 
angle cde, and the side ab equal to the side dc (th. 22) ; 
therefore these two triangles are identical, and have their 
corresponding sides equal (th. 2), viz. ae = ec, and be a ed. 

Again, since ac is bisected in e, the sum of the squares 
ad 3 + DC 3 = 2AE 3 + 2de s (th. 38). 

In like manner, ab' + bc* =■ 2ae s + 2be 3 or 2de 3 . 

Theref. ab s + bc 3 + CD* + da 1 = 4 a e 3 + 4de 3 (ax. 2). 

But, because the square of a whole line is equal to 4 
times the square of half the line (cor. th. 31), that is, ac* = 
4ae\ and bd 3 = 4de 3 : 

Theref. ab 3 + bc 3 + cd 3 -4 da 3 = ac 2 f bd 3 (ax. 1). 

Q. E. D. 

Cor. 1. If ad = dc, or the parallelogram be a rhombus; 
then ad 3 = ae 3 + ED 3 , CD 3 =■ de* + ce 1 , dsc. 

Cor. 2. Hence, and by th. 34, the diagonals of a rhom- 
bus intersect at right angles. 



THEOREM XLI. 



If a line, drawn through or from the centre of a circlet 
bisect a chord, it will bc perpendicular to it; or, if it be 
perpendicular to the chord, it will bisect both the chord and 
the arc of the chord. 

Let ab be any chord in a circle, and ct> 
a line drawn from the centre <- to the 
chord. Then, if the chord be bisected in 
the point d, cd will bc perpendicular to 
ab. 

Draw the two radii ca, cb. Then the 
two triangles acd, bcd, having ca equal to 
cb (def. 44), and cd common, also ad equal 
to db (by hyp.) ; they have all the three sides of the one, 
equal to all the three sides of the other, and so have their 
angles also equal (th. 5). Hence then, the angle ado being 
equal to the angle bdc, these angles are right angles, and the 
line cd is perpendicular to ab (def. 11). 

Again, if cd bo perpendicular to ah* then will the chord 




THEOUM*. 



Mft 



ab be bisected at the point d, or have ad equal to db ; and 
the arc abb bisected in the point e, or have ae equal bb. 

For, having drawn ca, cb, as before : Then, in the tri- 
angle abc, because the side ca is equal to the side cb, their 
opposite angles a and b are also equal (th. 3). Hence then, 
in the two triangles acd, bcd, the angle a is equal to the 
angle b, and the angles at d are equal (def. 11) ; therefore 
the third angles are also equal (corol. 1. th. 17). And 
having the side cd common, they have also the side ad equal 
to the side db (th. 2). 

Also, since the angle ace is equal to the angle bob, the 
arc ae, which measures the former (def. 57), is equal to the 
arc be, which measures the latter, since equal angles must 
have equal measures. 

Corol. Hence a line bisecting any chord at right angles, 
passes through the centre of the circle. 



THEOREM XLII. 



If more than two equal lines oan be drawn from any 
point within a circle to the circumference, that point will bo 
the centre. 

Let abc be a circle, and d a point 
within it : then if any three lines, da, 
db, dc, drawn from the point d to the 
circumference, be equal to each other, 
the point d will be the centre. 

Draw the chords ab, bc, which let 
be bisected in the points e, f, and join 
de, df. 

Then, the two triangles, dak, dbe, 
have the side da equal to the side db by supposition, and 
the side ae equal to the side eb by hypothesis, also the side 
db common : therefore these two triangles are identical, and 
have the angles at e equal to each other (th. 5) ; conse. 
quently de is perpendicular to the middle of the chord ab . 
(def. 11), and therefore passes through the centre of the 
circle (corol. th. 41). 

In like manner, it may be shown that df passes through 
the centre. Consequently the point d is the centre of the 
circle, and the three equal lines da, db, dc, are radii. 

0» 

Vol. 1. 40 




306 



•KOMETRY. 



THEOREM XLIII. 



If two circles placed one within another, touch, the centre* 
of the circles and the point of contact will be all in the same 
right line. 

Let the two circles abc, ade, touch one A. 
anottier internally in the point a ; then 
will the point a and the centres of those 
circles be all in the same right line. 

Let f be the centre of the circle ahc, 
through which draw the diameter afc. 
Then, if the centre of the other circle g 
can be out of this line ac, let it be sup- 
posed in some other point as o ; through which draw the line 
fis, cutting the two circles in b and d. 

Now in the triangle afg, the sum of the two sides fg t 
ca, is greater than the third side af (th. 10), or greater than 
its equal radius fb. From each of these take away the 
common part fg, and the remainder ga will be greater 
than the remainder gb. But the point g being supposed 
the centre of the inner circle, its two radii, ga, gd, are equal 
to each other ; consequently od will also be greater than gb. 
But ade being the inner circle, gd is necessarily less than 
gb. So that up is both greater and less than gb ; which is 
absurd. Consequently the centre o cannot be out of the 
lino afc. q. e. n. 




THEOREM XLIV. 



If two circles touch one another externally, the centres of 
the circles and the point of contact will be all in the same 
right line. 

Let the two circles ybc, ape, touch one 
another externally at the point \ : then will / ^ 
the point of contact a and the centres of the I i 
two circles be all in the same right line. \w /I 

Let f be the centre of the circle aim:. 
through which draw the diameter afc, and / 
produce it to the other circle at e. Then, if \ 



F 



* - - — ..v, ... . 

the centre of the other circle ade can be out \C 
ofyhc line fk, let. i(, if possible, be supposed 
in some other point as g ; and draw the lines 
a«:, rune, cutting the two circles in b and d. 



./ 



THEOREMS. 



507 



Then, in the triangle afg, the sum of the two sides af, 
ao, is greater than the third side fg (th. 10). But, f and o 
being the centres of the two circles, the two radii ga, gd, 
•are equal, as are also the two radii af, fb. Hence the sura 
of ga, af, is equal to the sum of ou, bf ; and therefore this 
latter sum also, od, bf, is greater than gf, which is absurd. 
Consequently the centre o cannot be out of the line ef. 

4.B.D. 



THEOREM XLV. 

Asnc chords in a circle, which are equally distant from 
the centre, are equal to each other ; or if they be equal to 
each other, they will be equally distant from the centre. 

Let ab, cd, be any two chords at equal C 
distances from the centre g ; then will /7\A\ 
these two chords ab, cd, be equal to each fc l \ / Vp V 
other. I / qT \ J 

Draw the two radii ga, gc, and the V V/ 
two perpendiculars ge, of, which are the B >s*_^>T3 
equal distances from the centre g. Then, 
the two right-angled triangles, gae, gcf, having the side ga 
equal the side gc, and the side ge equal the side gf, and 
the angle at e equal to the angle at f, therefore those two 
triangles are identical (cor. 2, th. 34), and have the line 
ae equal to the line cf. But ab is the double of ae, and 
cd is the double of cf (th. 41) ; therefore ab is equal to cd 
(by ax. 6). u. e. d. 

Again, if the chord ab be equal to the chord cd ; then 
will their distances from the centre, ge, gf, also be equal 
to each other. 

For, since ab is [equal cd by supposition, the half ai: is 
equal the half cf. Also the radii ga, gc, being equal, as 
well as the right angles e and f, therefore the third sides are 
equal (cor. 2, th. 34), or the distance ge equal the distance 
of. a. fi. d. 



theorem xlvi. 

A line perpendicular to the extremity of a radius^ va 
tangent to the circAe. 



906 



•BOMBTBY. 



Let the line adb be perpendicular to the 
radius cd of a circle ; then shall ab touch 
the circle in the point d only. 

From any other point s in the line ab 
draw cfb to the centre, cutting the circle 
in f. 

Then, because the angle d, of the trian- 

51e cde, is a right angle, the angle at e is acute (cor. 3, th. 
7), and consequently less than the angle d. But the greater 
side is always opposite to the greater angle (th. 9) ; there* 
fore the side ce is greater than the Bide cd, or greater than 
its equal cf. Hence the point e is without the circle ; and 
the same for every other point in the line ab. Consequently 
the whole line is without the circle, and meets it in the point 
d only. 



J THEOREM XLVIZ. 

Whew a line is a tangent to a circle, a radius drawn to 
the point of a contact is perpendicular to the tangent. 

Let the line ab touch the circumference of a circle at the 
point d ; then will the radius cd be the perpendicular to the 
tangent ab. [See the last figure.] 

For the line ab being wholly without the circumference 
except at the point d, every other line, as ce, drawn from 
the centre c to the line ab, must pass out of the circle to 
arrive at this line. The line cd is therefore the shortest that 
can be drawn from the point c to the line ab, and conse- 
quently (th. 21) it is perpendicular to that line. 

Carol. Hence, conversely, a line drawn perpendicular to 
a tangent, at the point of contact, passes through the centre 
of the circle. 

THEOREM XLVIII. 

The angle formed by a tangent and chord is measured by 
half the arc of that chord. 

Let ab be a tangent to a circle, and cd a chord drawn 
from the point of contact c ; then is the angle bcd measured 
by half the arc cfd, and the angle acd measured by half the 
arc con. 

Draw the radius ec to the point of contact, and the radios 
bf perpendicular to the chord at h. 



i 




THEOREMS. 809 

Then the radius kf, being perpendicular 
to the chord cd, bisects the arc cfd (th. 
41 ). Therefore cf is half the arc cfd. 

In the triangle ceh , the angle h being a 
right one, the sum of the two remaining 
angles e and c is equal to a right angle (cor. 
3, th. 17), which is equal to the angle bce, 
because the radius ce is perpendicular to 
the tangent. From each of these equals take the common 
part or angle c, and there remains the angle e equal to the 
angle bcd. But the angle b is measured by the arc cf (def. 
57), which is the half of cfd ; therefore the equal angle 
bcd must also have the same measure, namely, half the 
arc cfd of the chord cd. 

Again, the line gef, being perpendicular to the chord cd, 
bisects the arc cod (th. 41). Therefore co is half the arc 
cod. Now, since the line ce, meeting fo, makes the sum 
of the two angles at s equal to two right angles (th. 6), and 
the line cd makes with ab the sum of the two angles at c 
equal to two right angles ; if from these two equal sums 
there be taken away the parts or angles ceh and bch, 
which have been proved equal, there remains the angle 
ceo equal to the angle ach. Rut the former of these, 
cfo, being an angle at the centre, is measured by the are 
cg (def. 57) ; consequently the equal angle acd must also 
have the same measure co, which is half the arc cod of the 
chord CD. Q. E. D. 

Carol. 1. The sum of two right angles is measured by 
half the circumference. For the two angles bcd, acd, 
which make up. two right angles, are measured by the arcs 
cf, co, which make up half the circumference, fo being a 
diameter. 

Carol. 2. Hence also one right angle must have for its 
measure a quarter of the circumference, or 90 degrees. 



THE OBEX XLIX. 

Ah angle at the circumference of a circle is measured by half 
the arc that subtends it. 



Let bac be an angle at the circumference ; 3) A 18 
it has for its measure, half the arc bc which 
subtends it. 

For, suppose the tangent db passing 
through the point of contact a ; then, the 



3) A. IS 



GEOMETRY. 



angle dac being measured by half the arc arc, and the angle 
dab by half the arc ab (th. 48) ; it follows, by equal tub. 
traction, that the difference, or angle bag, must be measured 
by half the arc bc, which it stands upon. a. e. d. 

THEOREM L. 

All angles in the same segment of a circle, or standing on 
the same arc, are equal to each other. 

Let c and d be two angles in the game 
segment acdb, or, which is the same thing, 
standing on the supplemental arc aeb ; then 
will the angle c be equal to the angle d. 

For each of these angles is measured by 
half the arc aeb ; and thus, having equal B 
measures, they are equal to each other (ax. 11). 




THEOREM LI. 



An angle at the centre of a circle is double the angle at the 
circumference, when both stand on the same arc. 

Let c be an angle at the centre c, and d 
an angle at the circumference, both standing 
on the same arc or same chord ab : then will 
the angle c bo double of the angle i>, or the 
angle d equal to half the angle c. 

For, the angle at the centre c is measured 
by the whole arc aeb (def. 57), and the angle at the circum- 
ference d is measured by half the same arc aeb (th. 49) ; 
therefore the angle d is only half the angle c, or the angle c 
doubles the angle d. 




THEOREM LII. 



An angle in a semicircle, is a right angle. 

If abc or adc be a semicircle ; then any D 
angle d in that semicircle, is a right angle. 

For, the angle d, at the circumference, 
is measured by half the arc abc (th. 49), 
that is, by a quadrant of the circumference. 
But a quadrant is the measure of a right 
angle (cor. 4, th. 6; or cor. 2, th. 48). 
Therefore the angle d is a rigtit angle. 




THE0REHS. 



311 



THEOREM LIU. 

The angle formed by a tangent to a circle, and a chord 
drawn from the point of contact, is equal to the angle in the 
alternate segment. 

If ab be a tangent, and ac a chord, and 
D any angle in the alternate segment adc ; 
then will the angle d be equal to the angle 
bac made by the tangent and chord of the 
arc a ec. 

For the angle d, at the circumference, 
is measured by half the arc aec (th. 49) ; 
and the angle bac, made by the tangent and chord, is also 
nfeasured by the same 1 half arc aec (th. 48) ; therefore these 
two angles are equal (ax. 11). 

THEOREM LIV. 

The sum of any two opposite angles of a Quadrangle in- 
scribed in a circle, is equal to two right angles. 

Let abcd be any quadrilateral inscribed 
in a circle ; then shall the sum of the two 
opposite angles a and c, or u and d, be equal 
to two right angles. 

For the angle a is measured by half the 
arc dcb, which it stands on, and the angle 
c by half the arc dab (th. 49) ; therefore 
the sum of the two angles a and c is measured by half the 
sum of these two arcr., that is, by half the circumference. 
But half the circumference is the measure of two right angles 
(cor* 4, th. G) ; therefore the sum of the two opposite angles 
a and c is equal to two right angles. In like manner it is 
dhown, that the sum of the other two opposite angles, d and 
B, is equal to two right angles, u. k. d. 

THEOREM LV. 

Ir any side of a quadrangle, inscribed in a circle, be pro- 
duced out, the outward angle will be equal to the inward 
opposite angle. 

If the side An, of the quadrilateral 
abcd, inscribed in a circle, be produced 
to e ; the outward angle dak will bo equal 
to the inward opposite angle c. 






ns 



GEOMETRY. 



For, the mim of the two adjacent angles dab and dab is 
equal to two right angles (th. 6) ; and the sum of the two 
opposite angles c and dab iB also equal to two right angles 
(th. 54) ; therefore the former sum, of the two angles dab 
and dab, is equal to the latter Bum, of the two c and dab (ax. 
1). From each of these equate taking away the common 
angle dab, there remains the angle dab equal the angle c. 

Q. E. D. 



THEOREM LVI. 



Awv two parallel chords intercept equal arcs. 

Let the two chords ab, cd, be parallel : 
then will the arcs ac, bd, be equal ; or 

AC = BD. 

Draw the line bc. Then, because the 
lines ae, cd, are parallel, the alternate an- 
gleB b and c are equal (th. 12}. But the 
angle at the circumference b, is measured by half the arc 
ac (th. 49) ; and the other equal angle at the circumference 
c is measured by half the arc bd : therefore the halves of the 
arcs ac, bd, and consequently the arcs themselves, are also 
equal, q. e. d. 



THEOREM LVII. 



When a tangent and chord are parallel to each other, they 
intercept equal arcs. 

Let the tangent abc be parallel to the 
chord df ; then are the arcs bd, bf, equal ; 
that is, bd = bf. 

Draw the chord bd. Then, because the 
lines ab, df, are parallel, the alternate 
angles d and b are equal (th. 12). But 
the angle b, formed by a tangent and chord, is measured by 
half the arc bd (th. 48) ; and the other angle at the circum- 
ference d is measured by half the arc bf (th. 49) ; therefore 
the arcs bd, bf, are equal, a. e. d. 





THBOBJUC8. 



313 



THEOREM LVin. 




Tn angle formed, within a circle, by the intersection of 
(wo chords, is measured by half the sum of the two inter- 
cepted arcs. * 

Let the two chords ab, cd, intersect at 
the point e : then the angle aec, or deb, is 
measured by half the sum of the two arcs 

AC, DB. 

Draw the chord ap parallel to cd. Then 
because the lines ap, cd, are parallel, and ab 
cuts them, the angles on the same side a 
and deb are equal (th. 14). But the angle at the circumfer- 
ence a is measured by half the arc bf, or of the sum of fd 
and db (th. 49) ; therefore the angle £ is also measured by 
half the sum of fd and db* 

Again, because the chords af, cd, are parallel, the arcs ac v 
fd, are equal (th. 56) ; therefore the sum of the two arcs ac, 
db, is equal to the sum of the two fd, db ; and consequently 
the angle e, which is measured by half the latter sum, is also 
measured by half the former, q. e. d. 



theorem: lix. 



The angle formed, out of a circle, by two secants, is mea- 
sured by half the difference of the intercepted arcs. 

Let the angle e be formed by two se- 
cants eab and ecd ; this angle is measur- 
ed by half the difference of the two arcs 
Ac, db, intercepted by the two secants. 

Draw the chord af parallel to cd. Then, 
because the lines af, cd, are parallel, and 
ab cuts them, the angles on the same side a 
«nd bed are equal (th. 14). But the angle a, at the circum- 
ference, is measured by half the arc bf (th. 49), or of the 
difference of df and db : therefore the equal angle b is also 
measured by half the difference of df, db. 

Again, because the chords, af, cd, are parallel, the arcs 
ac, fd, are equal (th. 56) ; therefore the diffettfic* ot ^* 

Vol. I. 41 




314 



GEOMETRY. 



two arcs AC) db, is equal to the difference of the two »r, db. 
Consequently the angle e, which is measured by half the 
latter difference, is also measured by half the former. 

q. e. d. 



THEOREM LX. 



The angle formed by two tangents, is measured by half the 
difference of the two intercepted arcs. 

Let eb, ed, be two tangents to a circle 
at the points a, c; then the angle e is 
measured by half the difference of the two 

arcs CFA, CGA. 

Draw the chord af parallel to ed. 
Then, because the lines, af, ed, are pa- 
rallel, and eb meets them, the angles on 
the same side a and e are equal (th. 14). 
But the angle a, formed by the chord af and tangent ab, 
is measured by half the arc af (th. 48) ; therefore the equal 
angle e is also measured by half the same arc af, or half the 
difference of the arcs cfa and cf, or cga (th. 57). 




Carol. In like manner it is proved, that 
the angle e, formed by a tangent ecd, 
and a secant eab, is measured by half 
the difference of the two intercepted arcs 
ca and cfb. 




theorem lxi. 



When two lines, meeting a circle each in two points, cut 
one another, either within it or without it; the rectangle 
of the parts of the one, is equal to the rectangle of the 
parts of the other ; the parts of each being measured from 
the point of meeting to the two intersections with the cir- 
cumference. 



THEOREMS. 



815 



Let the two lines ab, cd, meet each 
trther in e ; then the rectangle of ae, eb, 
will be equal to the rectangle of ce, ed. 

Or, AE . EB = CE . ED. 

For, through the point e draw the dia- 
meter fo ; also, from the centre h draw 
the radius dh, and draw hi perpendicular 
to CD. 

Then, since dbh is a triangle, and the 
perp. hi bisects the chord cd (th. 41), the 
line cb is equal to the difference of the 
segments di, ei, the sum of them being 
be. Also, because h is the centre of the 
circle, and the radii dh, fh, oh, are all equal, the line eg 
is equal to the sum of the sides dh, he ; and ef is equal to 
their difference. 

But the rectangle of the sum and difference of the two 
sides of a triangle is equal to the rectangle of the sum and 
difference of the segments of the base (th. 35) ; therefore 
the rectangle of fe, eo, is equal to the rectangle of ce, ed. 
In like manner it is proved, that the same rectangle of fe, 
eg, is equal to the rectangle of ae, eb. Consequently the 
rectangle of ae, eb, is also equal to the rectangle of ce, ed 
(ax. 1). Q. E. D. 

Coral. 1. When one of the lines) in the 
second case, as de, by revolving about the 
point e, comes into the position of the tan- 
gent ec or ed, the two points c and d run- 
ning into one ; then the rectangle of ce, ed, 
becomes the square of ce, because ce and de 
are then equal. Consequently the rectangle 
of the parts of the secant, ae . eb, is equal 
to the square of the tangent, ce 9 . 

Carol. 2. Hence both the tangents ec, ef, drawn from 
the same point e, are equal ; since the square of each is equal 
tolhe same rectangle or quantity ae • eb. 



THEOREM LXII. 

In equiangular triangles, the rectangles of the corresponding 
or like sides, taken alternately, are equal. 





816 



GKOVETMT. 




Let abc, dbf, be two equiangular 
triangles, having the angle a = the 
angle d, the angle b = the angle e, 
and the angle c = (he angle f ; also 
the like si-les ab, de, and ac, df, be- 
ing those opposite the equal angles : 
then will the rectangle of ab, df, be 
equal to the rectangle of ao, de. 

In ba produced take ag equal to or ; and through the 
three points b, c, g, conceive a circle bcoh to be described, 
meeting ca produced at h, and join oh. 

Then the angle o is equal to the angle c on the same arc 
bh, and the angle h equal to the angle b on the same arc 
co (th. 50) ; also the opposite angles at a are equal (th. 7) : 
therefore the triangle agh is equiangular to the triangle 
acb, and consequently to the triangle dfb also. But the 
two like sides ao, df, are also equal by supposition ; conse- 
quently the two triangles aoh, dfk, are identical (th. 2), 
having the two sides ao, ah, equal to the two df, de, each 
to each. 

But the rectangle ga . ab is equal to the rectangle ha • ac 
(th. 61) : consequently the rectangle df . ab is equal to the 
rectangle de . ac. a. e. d. 



THEOREM LXI1I. 



The rectangle of the two sides of any triangle, is equal 
to the rectangle of the perpendicular on the third side and 
the diameter of the circumscribing circle. 

Let cd be the perpendicular, and ce 
the diameter of the circle about the triangle 
abc ; then the rectangle ca . cb is = the 
rectangle cd • ce. 

For, join be : then in the two triangles 
acd, ecb, the angles a and e are equal, 
standing on the same arc bc (th. 50) ; also 
the right angle d is equal the angle b, which is also a right 
angle, being in a semicircle (th. 52) : therefore these two 
triangles have also their third angles equal, and are equian- 
gular. Hence, ac, ce, and cd, cb, being like sides, nib- 
tending the equal angles, the rectangle ac . cb. of the first 
and last of them, is equal to the rectangle ce . cd, of the 
other two (th. 62). 




THEOREMS. 



317 



THEOREM LXIV. 

The square of a line bisecting any angle of a triangle, 
together with the rectangle of the two segments of the oppo- 
site side, is equal to the rectangle of the two other sides in- 
cluding the bisected angle. 



Let cd bisect the angle c of the triangle 
abc ; then the square cd 9 + the rectangle 
ad . db is = the rectangle ac . cb. 

For, let cd be produced to meet the cir- 
cumscribing circle at e, and join ae. 




Then the two triangles ace, bcd, are 
equiangular : for the angles at c are equal 
by supposition, and the angles b and e are equal, standing 
on the same arc ac (th. 50) ; consequently the third angles 
at a and Dare equal (cor. 1, th. 17) : also ac, cd, and ce, 
cb, are like or corresponding sides, being opposite to equal 
angles : therefore the rectangle ac . cb is = the rectangle 
cd . ce (th. 62). But the latter rectangle cd . ce is = cd* + 
the rectangle cd . de (th. 30) ; therefore the former rect- 
angle ac . cb is also = cd 3 + cd . de, or equal to cd 2 + 
ad . db, since cd . de is = ad • db (th. 61). a. e. d. 



theorem lxv. 

The rectangle of the two diagonals of any quadrangle 
inscribed in a circle, is equal to the sum of the two rect- 
angles of the opposite sides. 

Let abcd be any quadrilateral inscribed 
in a circle, and ac, bd, its two diagonals : 
then the rectangle ac . bd is = the rect- 
angle ab . dc + the rectangle ad . sc. 

For, let ce be drawn, making the angle 
bce equal to the angle dca. Then the two 
triangles acd, bce, are equiangular ; for 
the angles a and b are equal, standing on the same arc dc ; 
and the angles dca, bce, are equal by supposition ; conse- 
quently the third angles adc, bec, are also equal : also, ac, 
bc, and ad, .be, are like or corresponding sides, being oppo- 
site to the equal angles : therefore the rectangle ac . be it 

the rectangle ad . bc (th. 62). 




318 



GEOMETRY* 



Again, the two triangles abc, dec, are equiangular : for 
the angles bac, bdc, are equal, standing on the same arc bc ; 
and the angle dce is equal to the anglo bca, by adding the 
common angle ace to the two equal angles dca, bce ; there- 
fore the third angles e and abc are also equal : but ac, dc, 
and ab, de, are the like sides : therefore the rectangle AC . 
de is = the rectangle ab . dc (th. 62). 

Hence, by equal additions, the sum of the rectangles 
ac . be + ac . de is = ad . bo + ab . dc. Hut the for- 
mer sura of the rectangles ac . be + ac . de is = the rect- 
angle ac . bd (th. 30) : therefore the same rectangle ac • 
bd is equal to the latter sum, the rect. ad . bc + the rect. 
ab . dc (ax. 1). Q. E. D. 

CoroL Hence, if abd be an equilateral triangle, and c 
any point in the arc bcd of the circumscribing circle, we have 
ac = bc + dc. For ac . bd being = ad . bc + ab - dc ; 
dividing by bd == ab = ad, there results ac = bc + dc* 



OF RATIOS AND PROPORTIONS. 



DEFINITIONS. 

Dep. 76. Ratio is the proportion or relation which one 
magnitude bears to another magnitude of the same kind, 
with respect to quantity. 

Note* The measure, or quantity, of a ratio, is conceived, 
by considering what part or parts the leading quantity, called 
the Antecedent, is of the other, called the Consequent ; or 
what part or parts the number expressing the quantity of the 
former, is of the number denoting in like manner the latter. 
So, the ratio of a quantity expressed by the number 2, to a 
like quantity expressed by the number 6, is denoted by 2 
divided by 6, or | or £ : the number 2 being 3 times con- 
tained in 6, or the third part of it. In like manner, the ratio 
of the quantity 3 to 0, is measured by £ or £ ; the ratio of 
4 to 6 is } or |; that of 6 to 4 is $ or £ ; &c. 

77. Proportion is an equality of ratios. Thus, 

78. Three quantities are said to be proportional, when the 
ratio of the first to the second is equal to the ratio of the 



THEOREMS. 



319 



second to the third. As of the three quantities a (2), b (4), 
c (8), where $==1 = 1, hoth the same ratio. 

79. Four quantities are said to be proportional, when the 
ratio of the first to the second, is the same as the ratio of the 
third to the fourth. As of the four, a (4), n (2), c (10), o (5), 
where a = y> = 2, both the same ratio. 

Note* To denote that four quantities, a, b, c, d, are pro* 
portional, they are usually stated or placed thus, a : b : : c : d; 
and read thus, a is to b as c is to d. But when three quan- 
tities are proportional, the middle one is repeated, and they 
are written thus, a : b : : b : c. 

The proportionality of quantities may also be expressed 
very generally by the equality of fractions, as at pa. 118. 

Thus, if - = then a : b : : c : d, also b : a : r c : p, and 

B D 

A : o : : b : d, and c : a : : b : d. 

80. Of three proportional quantities, the middle one is 
said to be a Mean Proportional between the other two ; and 
the last, a Third Proportional to the first and second. 

81. Of four proportional quantities, the last is said to be 
a Fourth Proportional to the other three, taken in order. 

82. Quantities are said to be Continually Proportional, or 
in Continued Proportion, when the ratio is the same between 
every two adjacent terms, viz. when the first is to the second, 
as the second to the third, as the third to the fourth, as the 
fourth to the fifth, and so on, all in the same common ratio. 

As in the quantities 1, 2, 4, 8, 16, &c. ; where the com- 
mon ratio is equal to 2. 

83. Of any number of quantities, a, b, c, d, the ratio of 
the first a, to the last d, is said to be Compounded of the 
ratios of the first to the second, of the second to the third, 
and so on to the last. 

84. Inverse ratio is, when the antecedent is made the 
consequent, and the consequent the antecedent. — Thus, if 
1 : 2 : : 3 : 6 ; then inversely, 2 : 1 : : 6 : 3. 

85. Alternate proportion is, when antecedent is compared 
with antecedent, and consequent with consequent. — As, if 
1 : 2 : : 3 : 6 ; then, by alternation, or permutation, it will be 
1 : 3 : : 2 : 6. 

86. Compound ratio is, when the sum of the antecedent 
and consequent is compared, either with the to\u&qp£feV> « 



820 



GEOMETRY* 



with the antecedent.— Thus, if 1 : 2 : : 3 : 6, then by com- 
position, l+2:l::3 + 6:3, an( ] 1 -f 2 : 2 : : 3 + 
6 : 6. 

87. Divided ratio, is when the difference of the antecedent 
and consequent is compared, either with the antecedent or 
with the consequent. — Thus, if 1 : 2 : : 3 : 6, then, by di- 
vision, 2 — 1:1 : : 6— 3 : 3, and 2 — 1 : 2 : : 6 — 3:6. 

Nate. The term-Divided, or Division, here means subtract, 
ing, or parting ; being used in the sense opposed to com- 
pounding, or adding, in def. 86. 

THEOREM LXVI. 

Equimultiples of any two quantities have the same ratio as 
the quantities themselves. 

Let a and b be any two quantities, and m\, j»b, any equi- 
multiples of them, m being any number whatever : then will 
wia and mB have the same ratio as a and b, or a : b : : jra : 

fllB. 

mB b . 
For — = -, the same ratio. 

HIA A 

Cord. Hence, like parts of quantities have the same ratio 
as the wholes ; because the wholes are equimultiples of the 
like parts, or a and r are like parts of ma and ms. 

THEOREM LXVII. 

If four quantities, of the same kind, be proportionals ; 
they will be in proportion by alternation or permutation, 
or the antecedents will have the same ratio as the conse- 
quents*. 



* The author's object in these propositions was to simplify the doc- 
trine of ratios and proportions, by imagining that the antecedents and 
consequences may always be divided into parts that are commensura- 
ble. But it is known to mathematicians that there are certain quantities 
or magnitudes, such as the side and the diagonal of a square, which 
cannot possibly be divided in that manner by means of a common mf* 
sure. The theorems themselves are true, nevertheless, when applied to 
these incommensurabUs ; since no two quantities of the same kind can 
possibly be assigned, whose ratio cannot be expressed by that of two 
numbers, so near, that the difference shall be less than the least number 
that can be named. From the greater of two unequal magnitudes we 
may take, or suppose taken, its Hoif y from the remaining half, its half, 



THEOREMS. 821 

Let a : b : : uiA : »iB ; then will a : wia : : b : jrb. 

For — = —i and — - = ~> both the same ratio. 
a 1 b 1 



and so on, by continual bisections, until there shall at length be left a 
magnitude less than the least of two magnitudes ; or. indeed, less than 
the least magnitude that can be assigued ; and this principle furnishes a 
ground of reasoning. 

Or, somewhat differently, let a and b be two constant quantities, a 
and b two variable quantities, which we can render as small as we 
please, if we have an equality between i-j-fl, and b -}- 6, or, in other 
words, if the equation a+o = b-|-6 holds good whatever are the va- 
lues of a and 6, it may be divided into two others, a = b, between the 
constant quantities, and, a = b t between the variable quantities, and 
which latter must obtain for all their states of magnitude. For if, on 
the contrary, we suppose a = b ^ q, we shall have I — b = 6 — a = 
q, an absurd result ; since the quantities a and 6 being susceptible of 
diminishing indefinitely* their difference cannot always be = q. This 
is the principle which constitutes the method of limits. In general, one 
magnitude is called a limit of another, tchen we can make this latter ap- 
proach so near to the former, that their difference shall be less than any given 
magnitude* and yet so that the two magnitudes shall never become strictly 
equal. 

Let us here apply the principle to the demonstration of this proposi- 
tion, that the ratio of two angles acb, nop, is equal to that of the arcs, 
o6, np, comprised between their sides, and drawn from their respective 
summits as centres with equal radii. 

If the 'arcs pn f ba, are 
commensurable, their 
common measure 6m 
will be contained n 
times in iro, r times in 
ba ; so that we shall - 
have the equal ratios 

¥L = -. Through each 
6a r 

point of division, m, n', <fcc. draw the lines mc, n'c, &c. to the summits 
c, and o, the angles proposed will be divided into n, and r, equal angles, 

6cm, men', poq, qor, be. We shall, therefore, have = *. Hence 
' r ' boa r 

is =?—, since each of tbem is equal to the ratio -. 
bca 6a r 

If the arcs are incommensurable, divide one of them, 6a, into a num- 
ber r of equal parts, 6m, mn', be. and set off equal parts pq, qr } &c. upon 
the other arc pn; and let s be the point of division that falls nearest to n. 
Draw oss. Then, by the preceding, 6a, ps, being commensurable, we 

shall have ? — = the angle pos = pon + iros, arc ps = pn + ns. 
bca ba 

Therefore, 

pon ifos pn_ ns 
bca bca ~ ba ba' 
Here nos and ns are susceptible of indefinite variation, according as 
we change the common measure, 6m, of 6a ; they may, therefore, ba 
Vol. I. 42 




832 GEOMETRY. 

Otherwise. Let a : b : : c : d ; then shall b r a : 

A C 

For, let - = - = r ; then a = bt, and c = dt : there- 

B D 



fore b 



A, C TT bI.d 1 

-, and d =■ -. Hence - = and - = — , 



B D 

' Whence it is evident that — = — (ax. 1), or b : a : : d : c» 

A C ' 

In a similar manner may most of the other theorems he 
demonstrated. 

THEOREM LXTUI. 

If four quantities be proportional ; they will be in proportion 
by inversion, or inversely. 

Let a : b : : ira : ms ; then will b : a : : mB : ma. 

For — = — , both the same ratio. 
mB b 

THEOREM LXIX. 

If four quantities be proportional ; they will be in proportion 
by composition and division. 

Let a : b : : mx : mB ; 1 
Then will b ± a : a : : ms ± mx : mx, 
and b ± a : b : : kb dfc mx : mB. 
_ mx A , f»B B 

For — =* — — ; and — = — — . 

«iB rb m\ b±a mB±.m\ b rb a 

Carol. It appears from hence, that the sum of the greatest 
and least of four proportional quantities, of the same kind, 
exceeds the sum of the other two. For, since . - - - 
a : a + b : : m\ : mx + ms, where a is the least, and 
fflA + «»b the greatest ; then m + 1 . a + «b, the sum of 
the greatest and least, exceeds m + 1 . a + b, the sum of 
the two other quantities. 

THEOREM JLXX. 

If, of four proportional quantities, there be taken any 
equimultiples whatever of the two antecedents, and any equi- 

rendered as small as we please, while the other quantities remain the 
same. Consequently, by the nature of limits, as above explained, we 

have the equal ratios -~ = or for : bac ::pn: ba. 

bca oa r 



THEOREMS. 323 

multiples whatever of the two consequents ; the quantities 
resulting will still be proportional. 

Let a : b : : «a : wib ; also, let pA and pmA be any 
equimultiples of the two antecedents, and qs and qmB any 
equimultiples of the two consequents ; then will .... 
px : qB : : pmA : qmB. 

For VOIL = 2? both the same ratio. 
pmA pA 

■ t 

THEOREM LXXI. 

If there be four proportional quantities, and the two con- 
sequents be either augmented or diminished by quantities 
that have the same ratio as the respective antecedents ; the 
results and the antecedents will still be proportionals. 

Let a : b : : wa : mn, and n\ and nmA any two quan- 
tities having the same ratio as the two antecedents ; then will 
a : b ± nA mA : ms ± nmA. 

„ mi* ± nmA b ± tia , 

For = , -both the same ratio. 

IRA A 

THEOREM LXXII. 

If any number of quantities be proportional, then any 
one of the antecedents will be to its consequent, as the 
sum of all the antecedents, is to the sum of all the conse- 
quents. ^ * 

Let a : b : : mA : wib : : tiA : ub, &c. ; then will - - . 
a : b : : a + wa + iia : b + mB + iib, &c. ' 

B+ma+itB (l+ro+n)B b , 

For : = 77-; ; — r~ = — , the same ratio. 

A+roA+ftA (l+m+n)A A 



THEOREM LXXHI. 



If a whole magnitude be to a whole, as a part taken from 
the first, is to a part taken from the other ; then the re- 
mainder will be to the remainder, as the whole to the 
whole. 



Let a : b : : 




then will a b ; ; a a : b b. 

n % 



324 



GEOMETRY. 



m 

B— — B 

For as — , both the same ratio. 

m a 

A — A 

theorem: lxxiv. 

If any quantities be proportional ; their squares, or cubes, 
or any like powers, or roots, of them, will also be propor- 
tional. 

Let a : b : : i?ia : i?ib ; then will a* : b*' : : m n A n : m*B". 

_ m n B n b» . . . 
r or — — - = — , both the same ratio. 
ro*A w a" 

See also, th. vni. pa, 118. 

THEOREM LXXV. 

If there be two sets of proportionals ; then the products 
or rectangles of the corresponding terms will also be pro- 
portional. 

Let a : b : : wa : mB, 

and c : d : : nc : hd ; 

then will ac : bd : : mnAC : uitibd. 

mnBD bd . , . 
ror = — , both the same ratio. 

fflttAC AC 

THEOREM LXXVI. 

If four quantities be proportional ; the rectangle or pro- 
duct of the two extremes, will be equal to the rectangle or 
product of the two means. And the converse. 

Let a : b : : ota : ms ; 
then is a X wb = b Xj»a = hiae, as is evident. 

THEOREM LXXVII. 

If three quantities be continued proportionals ; the rect- 
angle or product of the two extremes, will be equal to the 
square of the mean. And the converse. 

Let a, «a, m 9 A be three proportionals, 
or a : tfiA : : mA ; m a A ; 
then is a X vp?k =* mV, a» \* owtaciSi. 



THEOREMS. 



325 



THEOREM LXXVI11. 

If any number of quantities be continued proportionals 
the ratio of the first to the third, will be duplicate or the 
square of the ratio of the first and second ; and the ratio of 
the first and fourth will be triplicate or the cube of that of 
the first and second ; and so on. 
' Let a, mA, jti'a, jti 3 a, dec. be proportionals ; 

then is — = — ; but -4- = —7 ; and -4- = \ ; dtc. 
mA m m a A mr wta mr 

THEOREM LXXIX. 

Triangles, and also parallelograms, having equal altitudes, 
are to each other as their bases. 

Let the two triangles adc, def, have I CK. 
the same altitude, or be between the same 
parallels ae, ce ; then is the surface of 
the triangle adc, to the surface of the 
triangle def, as the base ad is to the AB BOH i s 
base de. Or, ad : de : : the triangle 
adc : the triangle def. 

For, let the base ad be to the base de, as any one num- 
ber m (2), to any other number n (3) ; and divide the 
respective bases into those parts, ab, bd, do, gm, he, all 
equal to one another ; and from the points of division draw 
the lines bc, fg, fh, to the vertices c and f. Then will 
these lines divide the triangles adc, dkf, into the same 
number of parts as their bases, each equal to the triangle 
abc, because those triangular parts have equal bases and 
altitude (cor. 2, th. 25) ; namely, the triangle abc equal to 
each of the triangles bdc, dfg, gfh, hfe. So that the tri- 
angle adc, is to the triangle dfe, as the number of parts m 
(2) of the former, to the number n (3) of the latter, that is, 
as the base ad to the base de (def. 79)*. 

In like manner, the parallelogram adki is to tho parallelo- 
gram defk, as the base ad is to the base de ; each of these 
having the same ratio as the number of their parts, m to ft* 

Q. E. D. 




* If the bases ad, de, of two triangles that have a common ^%t\VL%, 
aire incommensurable to each other, the ratio of tto trvi&tjto Ha* wjM^ 
gUnd'wg, equal to that of their bases. 



926 



GEOMETRY* 



THEOREM LXXX. 

Triangles, and also parallelograms having equal bases, are 
to each other as their altitudes. 



Let abc, bep, be two triangles 
having the equal bases ab, be, and 
whose altitudes are the perpendicu- 
lars co, fh ; then will the triangle 
abc : the triangle bef : : co : fh. 

For, let bk be perpendicular to 




ab, and equal to cg ; in which let 
there be taken bl = fh ; drawing ak and al. 

Then triangles of equal bases and heights being equal 
(cor. 2, th. 25), the triangle abk is = abc, and the triangle 
abl = bef. But, considering now abk, abl, as two tri- 
angles on the bases bk, bl, and having the same altitude ab, 
these will be as their bases (th. 79), namely, the triangle 
abk : the triangle abl : : bk : : bl. 

But the triangle abk = *abo, and the triangle abl = bef, 

also bk = cg, and bl = fh. 
Theref. the triangle abc : triangle bef : : cg : fh. 

And since parallelograms are the doubles of these triangles, 
having the same bases and altitudes, they will likewise have 
to each other the same ratio as their altitudes, q. e. d. 

Corel. Since, by this theorem, triangles and parallelo- 
grams, when their bases are equal, arc to each other as their 
altitudes; and by the foregoing one, when their altitudes are 
equal, they are to each other as their bases ; therefore uni- 



For, first, if possible, let the triangle kcd 
be to the triangle acd, not as ed to ad, but 
as some other line bd greater than ed, is to 

AD. 

Let ah be a part, or measure of ad. less 
than be, and let di be that multiple frf ah, 
which least exceeds de, and which by the note B I E .D MA 
to th. 67, may be made as small as we please. 

Let cb, ci, be drawn, i evidently falls between e and b, because (by 
hyp.) ei is less than am. But icd : acd : : id : ad, by th. 79. Also, 
by hyp. ecd : acd : : bd : ad, greater than the. ratio of id : ad, or of 
Ico : acd ; and consequently, ecd is greater than icd : which is impoui- - 
bU, By a like reasoning it may be shown, that ecd cannot be to acd, 
.as a line feu than ed, is to ad. Consequently, it must be ecd : acd : : *D 
I ad. 

Similar reasoning, founded upon vYte \rc*wdvcv^ tvoU, &$oUes alto te 
the case of parallelograms. 




THEOREM!. 



327 



versally, when neither are equal, they are to each other in 
the compound ratio, or as the rectangle or pibduct of their 
bases and altitudes. 



c 

~~A 


Q 
B 


r d 





THEOREM LXXXI. 

If four lines be proportional ; the rectangle of the ex- 
tremes will be equal to the rectangle of the means. And, 
conversely, if the rectangle of the extremes, of four lines, 
be equal to the rectangle of the means, the four lirttes, taken 
alternately, will be proportional. 

Let the four lines a, b, c, d, be A 

proportionals, or a : b : : c : d ; B 
then will the rectangle of a and d be ^ 
equal to the rectangle of b and c ; 
or the rectangle a . d = b . c. 

For, let the four lines be placed 
with their four extremities meeting 
in a common point, forming at that 

point four right angles ; and draw lines parallel to them to 
complete the rectangles p, q, r, where p is the rectangle of 
a and d, a the rectangle of b and c, and r the rectangle of 
b and d. " 

Then the rectangles r and r, being between the same 
parallels are to each other as their bases a and b (th. 79) ; 
and the rectangles q and r, being between the same parallels, 
are to each other as their bases c and d. But the ratio of 
a to b, is the same as the ratio of c to n, by hypothesis : 
therefore the ratio of p to a, is the same as the ratio of q to 
r ; and consequently the rectangles f and q are equal. 

Q. E. D# 

Again, if the rectangle of a and d, be equal to the 
rectangle of b and c ; these lines will be proportional, or 
a : b : : c : D.j 

For, the rectangles being placed the same as before : then, 
because parallelograms between the same parallels, are to one 
another as their bases, the rectangle r : r : : a : b, and 
q : r : : c : d. But as p and a are equal, by supposition, 
they have the same ratio to n, that is, the ratio of a to b is 
equal to the ratio of c to d, or a : b : : c : d. q. e. d. 

Carol. 1. When the two means, namely, the second and 
third terms, are equal, their rectangle becomes a square of 
the second term, which supplies the place of both the second 
and third, And hence it follows, that when three lines an 



828 



GEOMETRY. 



proportionals, the rectangle of the two extremes is equal to 
the square of the mean ; and, conversely, if the rectangle of 
the extremes be equal to the square of the mean, the three 
lines are proportionals. 

Corel. 2. Since it appears, by the rules of proportion in 
Arithmetic and Algebra, that when four quantities are pro* 
portional, the product of the extremes is equal to the product 
of the two means ; and, by this theorem, the rectangle of the 
extremes is equal to the rectangle of the two means ; it fol- 
lows, that the area or space of a rectangle is represented or 
expressed by the product' of its length and breadth multiplied 
together. And, in general, a rectangle in geometry is similar 
to the product of the measures of its two dimensions of length 
and breadth, or base and height. Also, a square is similar 
to, or represented by, the measure of its side multiplied by 
itself. So that, what is shown of such products, is to be un- 
derstood of the squares and rectangles. 

CoroL 3. Since the same reasoning, as in this theorem, 
holds for any parallelograms whatever, as well as for the 
rectangles, the same property belongs to all kinds of paral- 
lelograms, having equal angles, and also to triangles, which 
are the halves of parallelograms ; namely, that if the sides 
about the equal angles of parallelograms, or triangles, be 
reciprocally proportional, the parallelograms or triangles will 
be equal ; and, conversely, if the parallelograms or triangles 
be equal, their sides about the equal angles will be recipro- 
cally proportional. 

Corol. 4. Parallelograms, or triangles, having an angle in 
each equal, are in proportion to each other as the rectangles 
of the sides which arc about these equal angles. 



THEOREM LXXXII. 



If a line be drawn in a triangle parallel to one of its sides, 
it will cut the other two sides proportionally. £ / 

Let de be parallel to the side bc of the A 
triangle abc ; then will ad : db : : ae : ec. /\ 

For, draw be and cd. Then the tri- X)/ \j£ 

angles dbe, dce, are equal to each other, /x^\A 
because they have the same base de, and ^\ 
are between the same parallels de, i:c B C 
(th. 25). But the two triangles, ade, bde, 
on the bases ad, db, have the same altitude ; and the two 
triangles ade, cde, on the bases ae, ec, have also the same 



THEOREMS. 329 

altitude ; and because triangles of the same altitude are to 
each other as their bases, therefore 

the triangle ade : bde : : ad : db, 
and triangle ade : cde : : ae : ec. 

But bde is ~ cde ; and equals roust have to equals the 
same ratio ; therefore ad : db : : ae : ec. a* e. d. 

Carol. Hence, also, the whole lines as, ao, are propor- 
tional to their corresponding proportional segments (corol. 
th. 60), 

viz. ab : ac : : ad : ae, 
and ab : ac : : bd : ce. 



theorem lxxxiii. 




A Line which bisects any angle of a triangle, divides the 
opposite side into two segments, which are proportional to 
the two other adjacent sides. 

Let the angle acb, of the triangle abc, 
be bisected by the line cd, making the 
angle r equal to the angle s : then will the 
segment ad be to the segment db, as the 
side ac is to the side cb. Or, - - - - 
ad : db : : ac : cb. 

For, let be be parallel to cd, meeting 
AC produced at e. Then, because the line bc cuts the two 
parallels cd, be, it makes the angle cbe equal to the alter- 
nate angle s (th. 12), and therefore also equal to the angle 
r, which is equal to * by the supposition. Again, because 
the line ae cuts the two parallels uc, be, it makes the angle 
E equal to the angle r on the same side of it (th. 14). Hence, 
in the triangle bck, thfjangles b and e, being each equal to 
the angle r, are equal to each other, and consequently their 
opposite sides cb, ce, are also equal (th. 3). 

But now, in the triangle abe, the line cd, being drawn 
parallel to the side be, cuts the two other sides ab, ae, pro- 
portionally (th. 82), making ad to db, as is ac to ce or to 
its equal cb. q. e. d. 

Vol. L 43 



830 



GEOXETXT. 



THEOREM LXXXIV. 



Euuiangular triangles are similar, or have their like sides 
proportional- 
Let abc, def, be two equiangular tri- 
angles, having the angle a equal to the 
angle d, the angle b to the angle e, and 
consequently the angle c to the angle f ; 
then will ab : ac : : de : df. A. B 

For, make dg = ab, and dh = ac, and F 
join gii. Then the two triangles abc, 
dgh, having the two sides ab, ac, equal 
to the two dg, dii, and the contained an- 
gles a and d also equal, are identical, or 
equal in all respects (th. 1), namely, the 
angles b and c are equal to the angles g and n. But the 
angles b and c are equal to the angles e and f by the hypo- 
thesis ; therefore also the angles g and 11 are equal to the 
angles e and f (ax. 1), and consequently the line qh ia paral- 
lel to the sido ef (cor. 1, th. 14). 

Hence then, in the triangle def, the line gh, being parallel 
to the side ef, divides the two other sides proportionally, 
making do : dii : : de : df (cor. th. 82). But dg and 
dh are equal to ab and ac ; therefore also - - - - • 
ab : ac : : de : df. a. e. d. 





THEOREM LXXXV. 



Triangles which have their sides proportional, arc equi- 
angular. 

In the two triangles abc, def, if 
ab : de : : ac : df : : bc : ef ; the two 
triangles will have their corresponding /\ 
angles equal. 

For, if the triangle abc bc not equian- A B 
gular with the triangle dkf, suppose some <*• P 

other triangle, as deg, to be equiangular 
with abc. But this is impossible : for if 
the two triangles abc, deg, were equian- 
gular, their sides would be proportional 
(th. 84). So that, ab being to de as ac 
to dg, and ab to de as bc to eg, it follows that do and eg, 
being fourth proportionals to the same three quantities, as 
well as the two df, ef, the Conner, dg, eg, would be equal 



THEOREMS. 



881 



to the latter, df, bp* Thus, then, the two triangles def, 
deg, having their three sides equal, would be identical 
(th. 5) ; which is absurd, since their angles are unequal. 

THEOREM LXXXVI. 

Triangles, which have an angle in the one equal to an 
angle in the other, and the sides about these angles pro- 
portional, are equiangular. 

Let ABO, def, be two triangles, having the angle a = the 
angle d, and the sides ab, ac, proportional to the sides 
db, dp : then will the triangle abc be equiangular with the 
triangle dep. 

For, make do = ab, and dh = ac, and join on. 

Then, the two triangles abc, doh, having two sides equal, 
and the contained angles a and d equal, are identical and 
equiangular (th. 1), having the angles o and h equal to the 
angles b and c. But, since the sides dg, du, are proportional 
to the sides dk, dp, tho line air is parallel to ef (th. 82) ; 
hence the angles e and f are equal to the angles o and h 
(th. 14), and consequently to their equals b and c. Q. e. d. 
[See fig. th. lxxxiv.] 

THEOREM LXXXV1I. 

In a right-angled triangle, a perpendicular from the 
right angle, is a mean proportional between the segments of 
the hypothenuse; and each of the sides, about the right 
angle, is a mean proportional between the hypothenuse and 
the adjacent segment. 

C 

Let abc be a right-angled triangle, and 
cd a perpendicular from the right angle c 
to the hypothenuse ab ; then will A D B 

cd be a mean proportional between ad and db ; 
ac a mean proportional between ab and ad ; 
bc a mean proportional between ab and bd. 

Or, ad : cd : : cd : db ; and ab : bc : : bc : bd ; and 
ab : ac : : ac : ad. 

For, the two triangles, abc, adc, having the right angles 
at c and d equal, and the angle a common, have their third 
angles equal, and are equiangular, (cor. 1, th. 17). In, tikfe 
manner, the two triangles abc, bdc, having xY&TUjgoX wct^e* 




382 



GEOMETRY. 



at c and d equal, and the angle b common, hare their third 
angles equal, and are equiangular. 

Hence then, all the three triangles, abc, adc, bdc, being 
equiangular, will have their like sides proportional (th. 84) ; 

viz. ad : CD : : cd : db; 
and xr: ac : : ac : ad; 
and ab : bc : : bc : bd. q. e .d. 

Carol. 1. Because the angle in a semicircle is a right 
angle (th. 52) ; it follows, that if, from any point c in the 
periphery of the semicircle, a perpendicular be drawn to the 
diameter ab ; and the two chords ca, cb, be drawn to the 
extremities of the diameter : then are ac, bc, cd, the mean 
proportionals as in this theorem, or (by th. 77), .... 

CD* = AD . DB ; AC 3 = AB . AD ; and BC 1 = AB . BD. 

Corol. 2. Hence ac 3 : bc 9 : : ad : bd. 

Corol. 3. Hence we have another demonstration of 
th. 34. 

For since ac 3 = ab . ad, and bc 3 = ab . bd ; 
By addition ac 3 + bc 9 = ab (ad + bd) = ab 3 . 



THEOREM LXXXVIII. 



Equiangular or similar triangles, are to each other as the 
squares of their like sides. 

Let abc, def, be two equiangular 
triangles, ab and de being two like 
sides : then will the triangle abc be to 
the triangle def, as the square of ab 
is to the square of dk, or as ah 2 to de 2 . 

For, the triangles being similar, they 
have theirlikesidesproportional (th.84), 
and are to each other as the rectangles 
of the like pairs of their sides (cor. 4, 
th. 81) ; 

theref. ab : de : : ac : df (th. 84), 

and ab : de : : ab : de of equality : 
theref ab 3 : de 9 : : ab • ac : de . df (th. 75). 
But A abc : A def : : ab . ac : de . df (cor. 4, th. 81), 
theref. A abc : A def : : ab 3 : de 9 . a. s. d. 




a £ 



THEOREMS • 



383 



THEOREM LXXX1X. 



All similar figures are to each other, as the squares of their 
like sides. 



Let abcde, fohik, be 
any two similar figures, the 
like sides being ab, fg, and 
bc, oh, and so on in the 
same order: then will the 
figure abcde be to the figure 
fohik, as the square of ab 
to the square oft or as ab 3 to fo 9 . 

For, draw be, bd, ok, 01, dividing the figures into an 
equal number of triangles,* by lines from two equal angles 
b and g. 

The two figures being similar (by suppos.), they are equi- 
angular, and have their like sides proportional (def. 67). 

Then, since the angle a is = the angle f, and the sides 
ab, ae, proportional to the sides fo, fk, the triangles 
abe, fok, are equiangular (th. 88). In like manner, the 
two triangles bcd ghi, having the angle c = the angle h, 
and the sides bc, cd, proportional to the sides gh, hi, are 
also equiangular. Also, if from the equal angles aed, fkj, 
there be taken the equal angles aeb, fko, there will remain 
the equals bed, gki ; and if from the equal angles cde, hik, 
be taken away the equals cdb, hio, there will remain the 
equals bde, oik ; so that the two triangles bde, gik, having 
two angles equal, are also equiangular. Hence each trian- 
gle of the one figure, is equiangular with each corresponding 
triangle of the other. 

But equiangular triangles arc similar, and are proportional 
to the squares of their like sides (th. 88). 

Therefore the A abb : A fgk : : ab 3 : fg 3 , 
and A bcd : A ohi : : nc 3 : gii 2 , 
and A bde : A gik : : de 3 : iK a . 

But as the two polygons are similar, their like sides are 
proportional, and consequently their squares also propor- 
tional ; so that all the ratios ab 2 to fg 3 , and bc 9 to oh 9 , and 
dm 9 to ik 9 , are equal among themselves, antlxonsequently 
the corresponding triangles also, abe to fob* and bcd to 
ohi, and bde to gik, have all the same ratio, viz. that of 
ab 9 to fg 3 : and hence all the antecedents, or the figure 
abcde, have to all the consequents, or the figure fghik, 
the same ratio, viz. that of ab 3 to fo 9 QCh. T&y o.. i>» 



384 



GEOMETRY* 



THEOREM XC. 




Similar figures inscribed in circles, have their like sides, 
and also their whole perimeters, in the same ratio as the 
diameters of the circles in which they are inscribed. 

. Let ABCDE, FGHIK, T> 1 

be two similar figures, ^^>J* 
inscribed in the circles 
whose diameters are al 
andFM; then will each 
side ab, bc, die. of the 
one figure be to the like 
side gf, oh, &c. of the 

other figure, or the whole perimeter ab + bc + die. of the 
one figure, to the whole perimeter fg + oh + dec. of the 
other figure, as the diameter al to the diameter fm. 

For, draw the two corresponding diagonals ac, fh, as 
also the lines bl, gm. Then, since the polygons are similar, 
they are equiangular, and their like sides have the same ratio 
(def. 67) ; therefore the two triangles abc, fgh, have the 
angle b = the angle g, and the sides ab, bc, proportional 
to the two sides fg, Gir, consequently these two triangles 
are equiangular (th. 80), and have the angle acb = fhg. 
But the angle acb = alb, standing on the same arc ab ; 
and the angle fhg = fmg, standing on the same arc fg ; 
therefore the angle alb = fmg (ax. 1). And since the 
angle abl = fgm, being both right angles, because in a 
semicircle ; therefore the two triangles abl, fgm, having 
two angles equal, are equiangular ; and consequently their 
like sides are proportional (th. 84) ; hence ab : fg : : the 
diameter al : the diameter fm. 

In like manner, each side bc, cd, dec. has to each side 
oh, in, dec. the same ratio of al to fm ; and consequently 
the sums of them are still in the same ratio, viz. ab + bc + 
cd, &c. : fg + gh + hi, dec. : : the diam. al : the diam. 
fm (th. 72). Q. E. D. 



THEOREM XCI. 

Similar figures inscribed in circles, are to each other a» the 
squares of the diameters of those circles. 

Let abcde, foiiis, be two similar figures, inscribed in 
the circles whose diameters are al and fm ; then the surface 
of the polygon abode will be to the surface of the polygon 

FGHIK, as AL 3 to FM 3 . 



THEOREMS. 



385 



For, the figures being similar, are to each other as the 
squares of their like sides, ab 2 to Ft; 3 (th. 86). But, by the 
last theorem, the sides ab, fg, are as the diameters al, fm ; 
and therefore the squares of the sides ab 3 to fg 3 , as the 
squares of the diameters al* to fx 1 (th. 74). Consequently 
the polygons abcde, fohik, are also to each other as the 
squares of the diameters al 3 to fm 3 (ax. 1). u. e. d. 
[See fig. th. xc] 

THEOREM XCII. 

The circumferences of all circles are to each other as their 
diameters 1 '. 



* The truth of theorems 92, 93, and 94, may be established more 
satisfactorily than in the text, upon principles analogous to those of the 
two last notes. 

Theorem. The area of any circle abd is equal to the rectangle con- 
tained by the radius, and a straight line equal to half the circumference. 

If not, let the rectangle be lest than the circle 
aid, or equal to the circle fhh : and imagine ed 
drawn to touch the interior circle in f, and meet 
the circumference abd in e and d. Join cd, 
cutting the arc of the interior circle in k. Let 
fii be a quadrantal arc of the inner circle, and 
from it take its half, from the remainder Us half, 
and so on, until an arc fi is obtained, less than 
fit. Join ci, produce it to cut ed in l, and make 
fo = fl : so snail lo be the side of a regular polygon circumscribing the 
circle frh. It is manifest that this polygon is less than the circle abd, 
because it is contai.ud within it Because the trinngle gcl is half the 
rectangle of base gl and altitude cf, the whole polygon of which gcl 
is a constituent triangle, is equal to half the rectangle whose base is the 
perimeter of that polygon and altitude cf- But that perimeter is less 
than the circumference abd, because each portion of it, such as gl, Is 
less than the corresponding arch of circle having radius cl, and there- 
fore, * fortiori, less than the corresponding arch of circle with radius 
ca. Also ce is less than ca. Therefore the polygon of which one side 
is ol, is less than the rectangle whose base is half the circumference abd 
and altitude ca; that is, (by hyp.) less than the circle fwh, which it 
conta *s: which is absurd. Therefore, the rectangle under the radius 
and half the circumference is not less than the circle abd. And by a 
similar process it may be shown that it is not greater. Consequently, 
H is equal to that rectangle, q. e. d. 

Theorem. The circumferences of two circles abd, abd, are as their 
radii. 

If possible, let the radius ac, 
be to the radius ac, as the cir- 
cumference abd to a circum- 
ference ihk less than abd. Draw 
the radius cic, and the straight 
line/tg- a chord to the circle 
abd, and a tangent to the circle 
ihk in i. From eb, a quarter of 
the circumference ofa&t, take 





336 



GEOMETRY. 



Let d, d y denote the diameters of two circles, and c, e 9 
their circumferences ; 

then will d : d : : c : c, or d : c : : d : c. 

For (by theor. 90), similar polygons inscribed in circles 
have their perimeters in the same ratio as the diameters of 
those circles. 

Now as this property belongs to all polygons, whatever 
the number of the sides may be ; conceive the number of the 
sides to be indefinitely great, and the length of each inde- 
finitely small, till they coincide with the circumference of 
the circle, and be equal to it, indefinitely near. Then the 
perimeter of the polygon of no infinite number of sides, is 
the same thing as the circumference of the circle. Hence it 
appears that the circumferences of the circles, being the same 
as the perimeters of such po /gone, are to each other in the 
same ratio as the diameters of the circles, a- e. d. 

THEOREM XCIII. 

The areas or spaces of circles, arc to each other as the 
squares of their diameters, or of their radii. 

Let a, a, denote the areas or spaces of two circles, and 
d, d, their diameters ; then a : a : : d 3 : d 2 . 

For (by theorem 91) similar polygons inscribed in circles 
are to each other as the squares of the diameters of the 
circles. 



away its half, and then the half of the remainder, and so on, until there 
be obtained an arc ed less than eg; and from d draw ad parallel to//, 
it will be the side of a regular polygon inscribed in the circle abd, yet 
evidently greater than the circle ihk, because each of its constituent tri- 
angles, as acd contains the corresponding circular sector eno. Let ad 
be the side of a similar polygon inscribed in the circle adb, and join ac, 
cd, similarly to ac, ed. The similar triangles acd, acd t give ac : ac : : 
ad : ad, and : : perim. of polygon in abd : perim. of polygon in abd. 
But, by the preceding theorem, ac : ac : : circumf. abd : circumf. abd. 
The perimeters of the polygons are, therefore, as the circumferences of 
the circles. But, this is impossible ; because, (by hyp.) the perim. of 
polygon in abd is less than the circumf. ; while, on the contrary, the 
perim. of polygon in adb is greater than the circumf. ihk. Conse- 
quently, ac is not to ac, as circumf. adb, to a circumference Ust than 
adb. And by a similar process it may be shown, that ac is not to ac, as 
the circumf. dbd, to a circumference less than abd. Therefore ac : ac 
:: circumf. abd : circumf. abd. e. d. 

Corol. Since by this theorem, we have c : c : : r : r, or, if c = »R, 
c = *r ; and, by the former, area (a) : area (a) : : £rc : |rc : we btvo 
a : a : : fan? : far* : : R* : t* •. d* c» : A 



THEOREMS. 



837 



Hence, conceiving the number of the sides of the polygons 
to be increased more and more, or the length of the sides to 
become less and less, the polygon approaches nearer and 
nearer to the circle, till at length, by an infinite approach, 
they coincide, and become in effect equal ; and then it fol- 
lows, that the spaces of the circles, which are the same as of 
the polygons, will be to each other as the squares of the 
diameters of the circles, q. e. d. 

Carol. The spaces of circles are also to each other as the 
squares of the circumferences ; since the circumferences are 
in the same ratio as the diameters (by theorem 92). 

THEOREM XCIV. 

% 

Thb area of any circle, is equal to the rectangle of half its 
circumference and half its diameter. 

Conceive a regular polygon to be in- 
scribed in the circle ; and radii drawn to 
all the angular points, dividing it into as 
many equal triangles as the polygon has 
sides, one of which is abc, of which the 
altitude is the perpendicular cd from the 
centre to the base ab. 

Then the triangle abc, being equal to a rectangle of half 
the base and equal altitude (th. 26, cor. 2), is equal to the 
rectangle of the half base ad and the altitude cd ; con- 
sequently the whole polygon, or all the triangles added to- 
gether which compose it, is equal to the rectangle of the 
common altitude cd, and the halves of all the sides, or the 
half perimeter of the polygon. 

Now conceive the number of sides of the polygon to be 
indefinitely increased ; then will its perimeter coincide with 
the circumference of the circle, and consequently the altitude 
cd will become equal to the radius, and the whole polygon 
equal to the circle. Consequently the space of the circle, or 
of the polygon in that state, is equal to the rectangle of the 
radius and half the circumference, a. e. d. 




Vol. I. 



44 



338 



OF PLANES AND SOLIDS. 



DEFINITIONS. 



Def. 88. The Common Section of two Planes, is the 
line in which they meet, or cut each other. 

80. A Line is Perpendicular to a Plane, when it is per* 
pendicular to every line in that plane which mmka it* 

90. One Plane is Perpendicular to Another, when eray 
line of the one, which is perpendicular to the line of their 
common section, is perpendicular to the other. 

91. The Inclination of one Plane to another, or the angle 
they form between them, is the angle contained by two lines, 
drawn from any point in the common section, and at right 
angles to the same, one of these lines in each plane. 

92. Parallel Planes, are such as being produced ever so 
far both ways, will never meet, or which are every where at 
an equal perpendicular distance. 

93. A Solid Angle, is that which is made by three or 
more plane angles, meeting each other in the same point. 

94. Similar Solids, contained by plane figures, are such as 
have all their solid angles equal, each to each, and are bounded 
by the same number of simitar planes, alike placed. 

95. A Prism, is a solid whose ends are parallel, equal, and 
like plane figures, and its sides, connecting those ends, are 
parallelograms. 

96. A Prism takes particular names according to the figure 
of its base or ends, whether triangular, square, rectangular, 
pentagonal, hexagonal, &c. 

97. A Right or Upright Prism, is that which .has the 
planes of the sides perpendicular to the planes of the erifa 
or base. 

98. A Parallelopiped, or Parallelopipedon, is 
a prism bounded by six parallelograms, every 
opposite two of which are equal, alike, and pa- 
rallel. 



DEFINITIONS. 



889 



A 



DO. A Rectangular Parallelopidedon, is that whose bound- 
ins planes are all rectangles, which are perpendicular to each 
other. 

100. A Cube, is a square prism, being bound- 
ed by six equal square sides or faces, and are 
perpendicular to each other. 

101. A Cylinder is a round prism, having 
circles for its ends ; and is conceived to be form- 
ed by the rotation of a right line about the cir- 
cumferences of two equal and parallel circles, 
always parallel to the axis. 

102. The Axis of a Cylinder, is the right 
line joining the centres of the two parallel circles, about 
which the figure is described. 

108. A Pyramid, is a solid, whose base is any 
right-lined plane figure, and its sides triangles, 
having all their vertices meeting' together in a 
point above the base, Called the vertex of the 
pyramid. 

104. A pyramid, like the prism, takes particular names 
from the figure of the base. 

105. A Cone, is a round pyramid, having a 
circular base, and is conceived to be generated 
by the rotation of a right line about the circum- 
ference of a circle, one end of which is fixed at 
a point above the plane of that circle. 

106. The Axis of a cone, is the right line, joining the 
vertex, or fixed point, and the centre of the circle about 
which the figure is described. 

107. Similar Cones and Cylinders, are such as have their 
altitudes and the diameters of their bases proportional. 

108. A Sphere, is a solid bounded by one curve surface, 
which is every where equally distant from a certain point 
within, called the Centre. It is conceived to be generated 
by the rotation of a semicircle about its diameter, which re- 
mains fixed. 

109. The Axis of a Sphere, is the right line about which 
the semicircle revolves ; and the centre j is the same as that 
of the revolving semicircle. 

110. The Diameter of a Sphere, is any right line passing 
through the centre, and terminated both ways by the surface. 

111. The Altitude of a solid, is the perpendicular drawn 
from the vertex to the opposite side or base. 



840 



GEOMETRY. 



THEOREM XCV. 

A perpendicular is the shortest line which can be drawn 
from any point to a plane. 

Let ab be perpendicular to the plane 
db ; then any other line, as ac, drawn 
from the same point a to the plane, will 
be longer than the line ab. 

In the plane draw the line bo, joining d 
thepoints m . 

Then, because the line ab is perpendi- 
cular to the plane de, the angle b is a right angle (de£ 90), 
and consequently greater than the angle c ; therefore the 
line ab, opposite to the less angle, is less than any other line 
ac, opposite the greater angle (th. 21). q. e. d. 

THEOREM ZCVI. 

A perpendicular measures the distance of any point from a 
plane. 

The distance of one point from another is measured by a 
right line joining them, because this is the shortest line which 
can be drawn from one point to another. So, also, the dis- 
tance from a point to a line, is measured by a perpendicu- 
lar, because this line is the shortest which can be drawn 
from the point to the line. In like manner, the distance 
from a point to a plane, must be measured by a perpendicu- 
lar drawn from that point to the plane, because this is the 
shortest line which can be drawn from the point to the 
plane. 

THEOREM XCVII. 

The common section of two planes, is a right line. 

Let acbda, aebfa, be two planes cut- 
ting each other, and a, b, two points in 
which the two planes meet ; drawing the 
line ab, this line will be the common in- 
tersection of the two planes. 

For, because the right line ab touches 
the two planes in the points a ajd b, it 





THEOREMS. 



841 



touches them in all other points (def. 20) ; this line is there- 
fore common to the two planes. That is, the common in- 
tersection of the two planes is a right line. a. b. d. 

Cord. From the same point in a plane, there cannot be 
drawn two perpendiculars to the plane on the same side of 
it. For, if it were possible, each of these lines would be 
perpendicular to the straight line which is the common inter- 
section of the plane and another plane passing through the 
two perpendiculars, which is impossible. 



THEOREM XCVIII. 

If a line be perpendicular to two other lines, at their com- 
mon point of meeting ; it will be perpendicular to the plane 
of those lines. 

Let the line ab make right angles with -q 
the lines ac, ad ; then will it be per- 
pendicular to the plane cde which passes 
through these lines. y^^Ty^ 

If the line ab were not perpendicular to E^^sT/) 
the plane cde, another plane might pass 
through the point a, to which the line ab 
would be perpendicular. But this is im- 
possible ; for, since the angles bac, bad, are right angles, 
this other plane must pass through the points c, d. Hence, 
this plane passing through the two points a, c, of the line 
ac, and through the two points a, d, of the line ad, it will 
pass through both these two lines, and therefore be the same 
plane with the former, a. e. d. 



THEOREM XCIX. 



If two planes cut each other at right angles, and a line 
be drawn in one of the planes perpendicular to their 
common intersection, it will be perpendicular to the other 
plane. 

Let the two planes acbd, aebf, cut 
each other at right angles ; and the line 
co be perpendicular to their common sec- 
tion ab ; then will co be also perpendicu- 
lar to the other plane aebf. 

For, draw eg perpendicular to ab. 
Then, because the two lines, ac, ox, are 
perpendicular to the common intersection 



53 



GEOXETftY. 



ab, the angle cge is the angle of inclination of the two 
planes (def. 92). But since the two planes cut each other 
perpendicularly, the angle of inclination oge is a right 
angle. And since the line cg is perpendicular to the two 
Jines oa, oe, in the plane aebf, it is therefore perpendicular 
to that plane (th. 98). a. e. d. 

Carol. 1. Every plane, ACB,'passing through a perpendicular 
<jg to another plane aebf, wiH be perpendicular to that other 
plane. For, if acb be not perpendicular to the plane aebf, 
some other plane on the same side of aebf, and passing 
through ab, will be perpendicular to it. Then, if from the point 
g a straight line be drawn in this other plane perpendicular to 
the common intersection, it will be perpendicular to the plane 
abbf. But (hyp.) co is perpendicular to that plane. There- 
fore, there will be, from the same point g, two perpendicu- 
lars to the same plane on the same side of it, which is im- 
possible (cor. 97). 

Carol. 2. If from any point g in the common intersection 
of the two planes acb and aebf perpendicular to each other, 
a line be drawn perpendicular to either plane, that line will 
be in the other plane. 



THEOBEM C. 



If two lines be perpendicular to the same plane, they will be 
parallel to each other. 

Let the two lines ab, cd, be both per- 
pendicular to the same plane ebdf ; then 
will ab be parallel to cd. . ■ ■■ 

For, join b, d, by the line bd in the El B ' h > B 
plane. The plane abd is perpendicular to I ■■ • J 
the plane ef (cor. 1, th. 99) ; and therefore 
the line cd, drawn from a point in the common intersection 
of the two planes, perpendicular to ef, will be in the plane 
abd (cor. 2, th. 99). But, because the lines ab, cd, are 
perpendicular to the plane ef, they are both perpendicular 
to the line bd in that plane, and they have been proved to 
be in the same plane abd ; consequently, they are paral- 
lel to each other (cor. th. 13). a. e. d. 

Carol. If two lines be parallel, and if one of them be per- 
pendicular to any plane, the other will also be perpendicular 
to the same plane. 



THE0RBMS. 



348 



THEOREM CI. 

If one plane meet another plane, it will make angles 
with that other plane, which are together equal to two right 
angles. 

Let the plane acbd meet the plane aebf ; these planes 
make with each other two angles whose sum is equal to two 
right angles. 

For, through any point g, in the common section ab, 
draw cd, ef, perpendicular to ab. Then, the line cg 
makes with ef two angles together equal to two right angles. 
But these two angles are (by def. 92) the angles of inclina* 
tion of the two planes. Therefore the two planes make an- 
gles with each other, which are together equal to two right 
angles. 

Corel. In like manner, it may be demonstrated, that planes 
which intersect have their vertical or opposite angles equal ; 
also, that parallel planes have their alternate angles equal ; 
and so on, as in parallel lines. 



• % THEOREM CII. 

If two planes be parallel to each other ; a line which is 
perpendicular to one of the planes, will also be perpendicular 
to the other. 



Let the two planes cd, ef, be parallel, 
and let the line ab be perpendicular to the 
plane cd ; then shall it also be perpendi- 
cular to the other plane ef. 

For, from any point g, in the plane ef, 
draw gh perpendicular to the plane cd, and 
draw ah, bg. 




Then, because ba, gh, are both perpendicular to the 
plane cd, the angles a and h are both right angles. And 
because the planes cd, ef, are parallel, the perpendiculars 
ba, gh, are equal (def. 93). Hence it follows that the lines 
bg, ah, are parallel (def. 9). And the line ab being per- 
pendicular to the line ah, is also perpendicular to the parallel 
fine bg (cor. th. 12). 

In like manner it is proved, that the line ab is perpendicu- 
lar to all other lines which can be drawn from Ita^wsiX Vto. 



844 



OEOXBTET. 



the plane ef. Therefore the line ab is perpendicular to the 
whole plane ef (def. 00). q. b. d. 



theobek cm. 

If two lines be parallel to a third line, though not in the 
same plane with it ; they will be parallel to each other. 



Let the lines ab, cd, be each of them 
parallel to the third line ef, though not in 
the same plane with it ; then will ab be pa- J) 
rallel to cd. 

For, from any point o in the line ef, let 
oh, 01, be each perpendicular to ef, in the H 
planes eb, id, of the proposed parallels. 

Then, since the line ef is perpendicular 
to the two lines gh, 01, it is perpendicular 
to the plane ghi of those lines (th. 98). And because ef 
is perpendicular, to the plane cm, its parallel ab is also per- 
pendicular to that plane (cor. th. 99). For the same reason, 
the line cd is perpendicular to the same plane ghi. Hence, 
because the two lines ab, cd, are perpendicular to the same 
plane, these two lines are parallel (th. 99). q. e. d. 



THEOREM CIV. 



If two lines, that meet each other, be parallel to two 
other lines that meet each other, though not in the same 
plane with them ; the angles contained by those lines will be 
equal. 

^*et the two lines ab, bc, be parallel to B 
the two lines, de, ef ; then will the angle ^ 
abc beequal to the angle def. ^ 

For, make . the lins ab, bc, de, ef, all 
equal to each other, and join ac, df, ad, 
be, cf. 

Then, the lines ad, be, joining the equal £> 
and parallel lines ab, de, are equal and 
parallel (th. 24). For the same reason, cf, be, are equal 
and parallel. Therefore ad, cf, are equal and parallel 
(th. 15) ; and consequently also ac, df (th. 24). Hence, 
the two triangles abc, def, having all their sides equal, 
each to each, have their angles also equal, and consequently 
the angle abc == the angle def. a. e. d. 



THEOREMS. 



845 



THEOREM CV. 




Tax sections made by a plane cutting two other parallel 
planes, are also parallel to each other. 

Let the two parallel planes ab, cd, be 
cut by the third plane efhg, in the lines 
sr, oh : these two sections kf, oh, will 
be parallel. 

Suppose eo, fh, be drawn parallel to 
each other in the plane efhg ; also let 
ei, FK, be perpendicular to the plane 
cd ; and let ig, kh, be joined. 

Then eg, fh, being parallels, and ei, fk, being both 
perpendicular to the plane cd, are also parallel to each other 
(th. 99) ; consequently the angle hfk is .equal to the angle 
gei (th. 104). But the angle fkh is also equal to the angle 
big, being both right angles ; therefore the two triangles are 
equiangular (cor. 1, th. 17) ; and the sides fk, ei, being the 
equal distances between the parallel planes (def. 93), it fol- 
lows that the sides fh, eg, are also equal (th. 2). But these 
two lines are parallel (by suppos.), as well as equal; con" 
sequently the two lines ef, gh, joining those two equal 
parallels, are also parallel (th. 24). a. e. d. 



theorem cvi. 



If any prism be cut by a plane parallel to its base, the sec- 
tion will be equal and like to the base. 

Let ag be any prism, and il a plane 
parallel to the base, ac ; then will the plane 
il be equal and like to the base ac, or the 
two planes will have all their sides and all 
their ancles equal. 

For, the two planes ac, il, being parallel 
by hypothesis ; and two parallel planes, cut 
by a third plane, having parallel sections 
(th. 105) ; therefore nt is parallel to ab, and 
xjl to bc, and lm to cd, and im to ad. But ai and bk are 
parallels (by def. 95) ; consequently ak is a parallelogram ; 
and the opposite sides ab, ik, are equal (th. 22). In like 
manner, it is shown that kx is = bc, and lm = cd, and im 
s= ad, or the two planes ac, il, are mutually equilateral. But 
these two planes having their corresponding side* ^an&&, 

Vol. I. 45 




846 



GEOMETRY. 



have the angles contained by them also equal (th. 104), 
namely, the angle a = the angle i, the angle b « the 
angle k, the angle c = the angle l, and the angle » = the 
angle m. So that the two planes ac, il, have all their 
corresponding sides and angles equal, or they are equal mwA 
like. a. b. d. 



THBOBKM CVII* 



If a cylinder be cut by a plane parallel to its base, the 
section will be a circle, equal to the base. 

Let af be a cylinder, and ghi any sec- 
tion parallel to the base abc ; then will 0111 
he a circle, equal to abc. 

Forf let the planes kb, kf, pass through 
the .axis of the cylinder mk, and meet the 
section ghi in the three points h, i, l ; and 
join the points as in the figure. 

Then, since kl, ci, are parallel (by def. 
102) ; and the plane ki, meeting the two 
parallel planes abc, ghi, makes the two sections kg, li, pa- 
rallel (th. 105) ; the figure klic is therefore a parallelogram, 
and consequently has the opposite sides li, kc, equal, where 
kc is a radius of the circular base. 

In like manner it is shown that lh is equal to the radius 
kb; and that any other lines, drawn from the point L to 
the circumference of the section ghi, are all equal to radii 
of the base ; consequently ghi is a circle, and equal to abc. 

B. D. 




THEOREM CV1II. 



All prisms and cylinders, of equal bases and altitudes, ate 
equal to each other. 

Let ac, df, be two 
prisms, and a cylinder, 



on equal bases, ab, de, 
and having equal alti- 
tudes bc, ef ; then will P 
the solids ac, df, be 
equal*. 

For, let pq, bs, be 



7 



©J 



S It 



E 



* This, and some other demonstrations relative to solids, are upon the 
defective principle of Indivisibles, introduced by CaodUriuM in tie year 
1635. Unfortunately, demonstrations npon sounder principles would 
sot accord with the brevity of this Course. 



847 



may two sections parallel to the bases, and equidistant from 
them. Then, by the last two theorems, the section pq is 
equal to. the base, ab, and the section rs equal to the base 
bb. But the bases, ab, db, are equal, by the hypothesis ; 
therefore the sections pq, rs, are equal also. In like manner, 
it may be shown, that any other corresponding sections are 
equal to one another. 

Since then every section in the prism ac is equal to its 
corresponding seetion in the prism or cylinder df, the prisms 
and cylinder themselves, which are composed of an equal 
number of all those equal sections, must also be equal. 

Q. K. D. 

Corol. Every prism, or cylinder, is equal to a rectangular 
parallelopipedon, of an equal base and altitude. 



THEOREM CIZ. 



Rectangular parallelopipedons, of equal altitudes, are to 
each other as their bases 3 ". 

Let ac, eo, be two rectan- 
gular parallelopipedons, having 
the equal altitudes ad, eh ; 
then will the solid ac be to the 
solid bo, as the base ab is to the 
base bp. 



Q R 5 C 

HO 



A LIS' 




For, let the proportion of the 
base ab to the base ef, be that 
of any one number m (3) to 

any other number n (2). And conceive ab to be divided 
into m equal parts, or rectangles, ai, lk, mb (by dividing an 
into that number of equal parts, and drawing il, km, paral- 
lel to bit). And let ef be divided, in like manner, into n 
equal parts, or rectangles, eo, pf : all of these parts, of both 
bases, being mutually equal among themselves. And through 
the lines of division Jet the plane sections lb, ms, pv, pass 
parallel to aq, et. 

Then the parallelopipedons ar, ls, mc, ev, pg, are aU 
equal, having equal bases and altitudes. Therefore the solid 
ac ia to the solid eg, as the number of parts in the former, 
to the number of equal parts in the latter ; or as the number 



* Here, also, the principle of former notes may readily be, 
the case of ineommeasorablef. 



848 GBOBBTRT. 

of parts in ab to the number of equal parts in Br, that is, as 
the base ab to the base bf. q. b. d. 

Carol. From this theorem, and the corollary to the last, it 
appears that all prisms and cylinders of equ Jl altitudes, are 
to each other as their bases ; every prism and cylinder being 
equal to a rectangular parallelopipeaon of an equal base and 
altitude. 



THEOREM CZ. 



V 



Rectangular parallelopipedons, of equal bases, are to each 
other as their altitudes. 

Let ab, cd, be two rectan- n 
gular parallelopipedons, stand, 
ing on the equal bases ae, cf ; 
then will the solid ab be to the 
solid cd, as the altitude eb is to 
the altitude fd. 

For, let ag be a rectangular 
parallelopipedon on the base ** ^ 

ab, and its altitude eg equal to the altitude fd of the solid 

CD. 

Then ag and cd are equal, being prisms of equal bases and 
altitudes. But if hb, hg, be considered as bases, the solids 
ab, ag, of equal altitude ah, will be to each other as those 
bases hb, hg. But these bases hb, hg, being parallelograms 
of equal altitude he, are to each other as their bases eb, 
bo ; therefore the two prisms, ab, ag, are to each other as 
the lines eb, eg. But ag is equal to cd, and eg equal to fd ; 
consequently the prisms ab, cd, are*to each other as their al<* 
titudes, eb, fd ; that is, ab : cd : : eb : fd. q. e. d. 



Corol. 1. From this theorem, and the corollary to theorem 
108, it appears, that all prisms and cylinders, of equal bases, 
are to one another as their altitudes. * 

Corol. 2. Because, by corollary 1, prisms and cylinders 
are as their altitudes, when their bases are equal. And by 
the corollary to the last theorem, they are as their bases, 
when their altitudes are equal. Therefore, universally, 
when neither are equal, they are to one another as the pro- 
duct of their bases and altitudes. And hence also these 
products are the proper numeral measures of their quantities 
or magnitudes. 



THEOREMS. 



849 



THEOREM CXI. 



Sdolaji prisma and cylinders are to each other, at the 
cubes of their altitudes, or of any other like linear dimensions. 

Let abcd, efgh, be two similar 
prisms ; then will the prism cd be 
to the prism oh, as ab 3 to ef 3 or 
ad 3 to EH 3 . 

For the solids are to each other 
as the product of their bases and 
altitudes (th. 110, cor. 2), that is, 
as ac • ad to so • ih. But the 
bases, being similar planes, are to each other as the squares 
of their like sides, that is, ac to eg as ab 3 to ef 1 ; therefore 
the solid cd is to the solid oh, as ab 3 . ad to ef 3 . eh. But 
bd and fh, being similar planes, hare their like sides pro. 

portional, that is, ab : ef : : ad : eh, - or ab 3 : 

ef 3 : : ad 3 : eh 3 : therefore ab 3 . ad : ef 3 . eh : : ab 3 : ef 3 , 
or : : ad 3 : eh 3 ; conseq. the solid cd : solid oh : : ab 3 : ef 3 : : 
ad 3 : eh 3 , a. E. D. 



B 



THEOREM CXII. 

In any pyramid, a section parallel to the base is similar to 
the base ; and these two planes are to each other as the 
squares of their distances from the vertex. 

Let abod be a pyramid, and efo a sec. 
tion parallel to the base bcd, also aih a 
line perpendicular to the two planes at h 
and i : then will bd, eg, be two similar 
planes, and the plane bd will be to the 
plane eg, as ah 1 to ai 3 . 

For, join ch, fi. Then because a plane 
cutting two parallel planes, makes parallel 
sections (th. 105), therefore the plane abc, 
meeting the two parallel planes bd, eg, makes the sections 
bc, ef, parallel : In like manner, the plane acd makes the 
sections cd, fg parallel. Again, because two pair of parallel 
lines make equal angles (tb. 104), the two ef, fg, which 
are parallel to bc, cd, make the angle efo equal the angle 
bcd. And in like manner it is shown, that each angle in 
the plane eg is equal to each angle in the plane bd, and con- 
sequently those two planes are equiangular. 




OBOXBTBY. 



Again, the three lines as, ac, ad, making with the 
parallels bc, bf, and cd, fo, equal angles (th. 14), and 
* we angles at a being eommon, the two triangles abc, ad, 
are equiangular, as also the two triangles acd, afg, and 
.%fm therefore their like sides proportional, namely t * - - - 
40 f ajt ; : bo : bf : . 2 cd : re. And in like manner it 
way be shown, that all the lines in the plane fo, are pro- 
portional to all the corresponding lines in the base bd. 
Hence these two planes, haying their angles equal, and their 
sides proportional, are similar, by def. 68. 

But, similar planes being to each other as the squares of 
their like sides, the plane bd : bo : : bc 9 : bf?, or : { ac 9 : 
AT*, by what is shown above. Also, the two IfAPttfAf* 
.abc, aif, having the angles h and 1 right ones ph. m\ 
wad the angle a common, are equiangular, and have there- 
fore their luce sides proportional, namely, ac : af : : ah : Afc 
or AC* ; af* : : Ad 9 : ai 9 . Consequently the two planes bd, 
30, which are as the former squares ac 9 , af 8 , will be also as 
the latter squares ah 3 , ai 9 , that is bd : bo : : 

AH 9 : AI 9 . Q. E. D. 

THEOREM CX1II. 

In a cone, any section parallel to the base is a circle ; and 
this section is to the base, as the squares of their distances 
from the vertex. 

Let abod be a cone, and ohi a secti on A. 
parallel to the base bcd ; then will ghi 
be a circle, and bcd, obi, will be to each 
other, as the squares of their distances 
from the vertex. 

For, draw alf perpendicular to the two 
parallel planes ; and let the planes ace, 
adb, pass through the axis of the cone 
akb, meeting the section in the three points 

, H, I, K. 

9 Then, since the section ohi is parallel to the base bcd, and 

the planes ck, dk, meet tbem, hk is parallel to cb, and 
IK to ps (th. 105). And because the triangles formed by 
these lines are equiangular, kh : bc : : ax : ab : : ki : bd. 
But bc is equal to bd, being radii of the same circle ; there- 
fore xi is also equal to kh. And the same may be shown of 
driy other lines drawn from the point k to the perimeter of 
the section ohi, which is therefore a circle (def. 44). 

Again, bv similar triangles, al : af : : ax : ab, or 
; ; XI : bd, hence al 9 : af 3 ; : ki 9 : bd 9 ; but ki 9 : ed 1 : 2 




THEOREMS. 



circle ohi : circle bcd (tb. OS) ; therefore al 9 : af 3 : : 
circle obi : circle bcd. q. e. d. 



THEOREM CXIV. 




All pyramids, and cones, of equal bases and altitudes, are 
equal to one another. 

Let abc, def, be 
any pyramids and 
cone, of equal bases 
bc, ef, and equal 
altitudes ao, dh : 
then will the pyra- 
mids and cone abc 
and def, be equal. 

For, parallel to the bases and at equal distances \x, do, 
from the vertices, suppose the planes ik, lm, to be drawn. 

Then, by the two preceding theorems, - 

do* : dh 8 : : lm : ef, and 
an 3 : ao 2 : : ik : bc. 

But since an-, ao 2 , are equal to do 3 , dh 3 , respectively, 
therefore ik : bc : : lm : ef. But bc is equal to ef, 
by hypothesis : therefore ik is also equal to lm. 

In like manner it is shown, that any other sections, at 
equal distance from the vertex, are equal to each other. 

Since then, every section in the cone, is equal to the cor- 
responding section in the pyramids, and the heights are equal, 
the solids abc, def, composed of all those sections, must be 
equal also. q. e. d. 



THEOREM CXV. 

Every pyramid is the third part of a prism of the same base 
and altitude. 

Let abcdef be a prism, and bdef a 
pyramid, on the same triangular base def : 
then will the pyramid bdef be a third part 
of the prism abcdef. 

For, in the planes of the three sides of 
the prism, draw the diagonals bf, bd," cd. 
Then the two planes bdf, bcd, divide the 
whole prism into the three pyramids bdef, 
dabc, dbcf, which are proved to be all equal to one another, 
as follows. 

Since the opposite ends of the prism are equal to eu&Vk^fabT, 



V 



Mt OBOJCXTBY. 

the pyramid whoM base is abc and vertex d, if equal to the 
pyramid whose base is dbf and vertex b (th. 114), being 
pyramids of equal base and altitude. 
' But the latter pyramid, whose base is dbf and vertex B, 
is the wne solid as the pyramid whose base is bxf and 
vertex n, and this is equal to the third pyramid whose base 
is bcf and vertex d, being pyramids of the same altitude and 
equal bases bef, bcf. 

Consequently all the three pyramids, which compose the 
prism, are equal to each other, and each pyramid* is the 
third part of the prism, or the prism is triple of the pyramid. 

4. X. D. 

Hence also, every pyramid, whatever its figure may be, is 
the third part of a prism of the same base and attitudes 
since the base of the prism, whatever be its figure, my be 
divided into triangles, and the whole solid into triangular 
prisms and pyramids. 

Cord. Any cone is the third part of a cylinder, or of a 
prism, of equal base and altitude ; since it has been proved 
that a cylinder is equal to a prism, and a cone equal to a 
pyramid, of equal base and altitude. 

Scholium. Whatever has been demonstrated of the pro- 
portionality of prisms, or cylinders, holds equally true of 
pyramids, or cones ; the former being always triple the 
latter ; viz} that similar pyramids or cones are as the cubes 
of their like linear sides, or diameters, or altitudes, fcc. 
And the same for all similar solids whatever, viz. that they 
are in proportion to each other, as the cubes of their like 
linear dimensions, since they are composed of pyramids every 
way similar. 

THEOREM CXVI. 

If a sphere be cut by a plane, the section will be a circlet. 

Let the sphere abbf be cut by the plane B 
aub ;.then will the section adb be a circle* 

If the section pass through the centre of 
the sphere, then will the distance from the 
centre to every point in the periphery of 
that section be equal to the radius of the 
sphere, and consequently such section is a 

carafe. Such, in truth, is the circle safb ¥ 

in the figure. 

Draw the chord ab, or diameter of the section adb ; per* 
psndicnhr to which, or to the said section, draw the axis of 




# 



THKOSXM8. 358 

the sphere ecgf, through the centre c, which will bisect the 
chord ab in the point « (th. 41). Also, join ca, cb ; and 
draw cd, od, to any point d in the perimeter of the sec- 
tion ADB. 

Then, because co is perpendicular to the plane adb, it is 
perpendicular both to ga and od (def. 90). So that coa, 
cod are two right-angled triangles, having the perpendi- 
cular co. common, and the two hypothenuses ca, cd, equal, 
being both radii of the sphere ; therefore the third sides ga, 
gd, are also equal (cor. 2, th. 34). In like manner it is 
shown, that any other line, drawn from the centre o to the 
circumference of the section adb, is equal to ga or gb ; con- 
sequently that section is a circle. 

Scholium. The section through the centre, having the 
same centre and diameter as the sphere, is called a great 
circle of the sphere ; the other plane sections being little 
circles. 

THEOREM. CXVn. 

Evert sphere is two.thirds of its circumscribing cylinder. 

Let abcd be a cylinder, circumscribing \ J g 
the sphere efgh ; then will the sphere 
sfgh be two-thirds of the cylinder abcd. 

For, let the plane ac be a section of the 
sphere and cylinder through the centre i. 
Join ai, bi. Also, let fih be parallel to 
ad or bc, and eig and kl parallel to ab 
or dc, the base of the cylinder ; the latter ■ 
line kl meeting bi in m, and the circular section of the 
sphere in w. « 
- Then, if the whole plane hfbc be conceived to revolve 
about the line rf as an axis, the square fg will describe 
a cylinder ag, and the quadrant ifg will describe a hemi- 
sphere efg, and the triangle ifb will describe a cone lab. 
Also, in the rotation, the three lines or parts kl, kit, km, as 
radii, will describe corresponding circular sections of those 
solids, namely, kl a section of the cylinder, kn a section of 
the sphere, and km a section of the cone. 

Now, fb being equal to n or ig, and kl parallel to fb, 
then by similar triangles ik is equal to km (th. 82^~> And 
since, in the right-angled triangle ikn, in* is equal to ik 1 
+ kn 3 (th. 34) ; and because kl is equal to the radius io 
or in, and km = ik, therefore kl 2 is equal to km 3 + xir* 9 
or the square of the longest radius, of tto 8t&~V\t&\^ 

Vol. I • 46 




864 GEOMETRY. 

sections, is equal to the sum of t^e squares of the two others* 
And because circles are to each other as the squares of their 
diameters, or of their radii, therefore the circle described by 
kl is equal to both the circles described by km and kn ; or 
the section of the cylinder, is equal to both the corresponding 
sections of the sphere and cone. And as this is always the. 
case in every parallel position of kl, it follows, that the cy- 
linder eb, which is composed of all the former sections, is 
equal to the hemisphere efg and cone iab, which are com- 
posed of all the latter sections. 

But the cone iab is a third part of the cylinder eb (cor. 2, 
th. 115) ; consequently the hemisphere efg is equal to the 
remaining two-thirds ; or the whole sphere efgh equal to 
two-thirds of the whole cylinder abcd. q. e. d. 

Coral. 1. A cone, hemisphere, and cylinder of the same 
base and altitude, are to each other as the numbers 1, 2, 3. 

CoroL 2. All spheres are to each other as the cubes of 
their diameters ; all these being like parts of their circum- 
scribing cylinders. 

CoroL 3. From tho foregoing demonstration it also ap- 
pears, that the spherical zone or frustum egxf, is equal to 
the difference between the cylinder eglo and the cone in a, 
all of the same common height ik. And that the spherical 
segment pfn, is equal to the difference between the cylinder 
ablo and the conic frustum aqmb, all of the same common 
altitude fk. 



355 



PROBLEMS. 



PROBLEM I. 

To bisect a line ab ; that is, to divide it into two equal parts. 

From the two centres a and b, with any c 
equal radii, describe arcs of circles, inter, 
secting each other in c and d ; and draw 
the line cd, which will bisect the given line A 
ab in the point e. 

For, draw the radii ac, bc, ad, bd. 
Then, because all these four radii are equal, 
and the side cd common, the two triangles 
acd, bcd, are mutually equilateral : consequently they are 
also mutually equiangular (th. 5), and have the angle ace 
equahto the angle bce. 

Hence, the two triangles ace, bce, having the two sides 
ac, ce, equal to the two sides bc, ce, and their contained 
angles equal, are identical (th. 1), and therefore have the 
side ae equal to eb. q. e. d. 



problem n. 

To bisect an angle bac*. 

From the centre a, with any radius, de- 
scribe an arc, cutting off the equal lines 
ad, ae ; and from the two centres d, e, 
with the same radius, describe arcs inter, 
secting in f ; then draw af, which will 
bisect the angle a as required. 





* A very ingenious instrument for trisecting an angle, is described in 
the Mechanic's Magasioe, No. 22, p. 344. 



A 



For, join df, if. Then \ the two triangles isr, a»f. 
baring the two aide* ad 9 df, equal to the two ae, bf (beiag 
equal radii), and the side af common, they are mutuafiy 
equilateral ; consequently they are also mutually equiangular 
(th. 5), and have the angle bat equal to the angle oaf. 

Scholium. In the same manner is an arc of a circle bi- 
sected. 

fboblex m. 

At a given point o, in a line ab, to erect a perpendicular. 

From the given point o, with any radius, 
cut off any equal parts cd, cb, of the given 
line ; and, from the two centres d and s, 
with any one radius, describe arcs intersect- _ 
ing in f ; then join of, which will be per- -4tT"0"TtB 
pendicular as required* 

For, draw the two equal radii df, ef. Then the two 
triangles cdf, cef, having the two sides cd, df, equal to 
the two ok, bf, and cf common, are mutually equilateral ; 
consequently they are also mutually equiangular (th. 5), and 
have the two adjacent angles at c equal to each other ; there- 
fore the line cf is perpendicular to ab (def. 11). 

Otherwise* 

Whew the given point c is near the end of the line* 

From any point d assumed above the 
line, as a centre, through the given point 
c describe a circle, cutting the given line 
at e ; and through e and the centre d, 
draw the diameter edf; then join cf, 
which will be the perpendicular required. 

For the angle at c, being an angle in a semicircle, is a 
right angle, and therefore the line cf is a perpendicular (by 
def. 15). 

PROBLEM IV. 

From a given point a, to let fall a perpendicular on a given 
line bc. 

From the given point a as a centre, with 
any convenient radius, describe an arc, cut- 
ting the given line at the two points d and 
b ; and from (he two centres d, e, with any 
radius, describe two arcs, intersecting at f ; 
then draw aof, which will be perpendicular 
to bc as required. 





S07 



For, draw the equal radii ad, as, and df, ep. Then the 
two triangles adf, aef, having the two sides ad df, equal 
to the two ab, bf, and af common, are mutually equilateral ; 
consequently they are also mutually equiangular (th. 5), and 
have the angle dag equal the angle bag. Hence then, the 
two triangles ado, abg, having the two sides ad, ao, equal 
to the two ab, ao, and their included angles equal, are there- 
fore equiangular (th. 1), and have the angles at o equal ; 
consequently ao is perpendicular to bc (def. 11). 

Otherwise. 

WftEN the given point is nearly opposite the end of the line. 

From any point d, in the given line 
bc, as a centre, describe the arc of a 
circle through the given point a, cutting 
bc in e ; and from the centre e, with the 
radius ba, describe another arc, cutting 
the former in f ; then draw agf, which 
will be perpendicular to bc as required. 3f 

For, draw the equal radii da, df, and ea, bf. Then the 
two triangles dak, dfe, will be mutually equilateral ; conse- 
quently they are also mutually equiangular (th. 5), aqd have 
the angles at d equal. Hence, the two triangles dag, dfo, 
having the two sides da, dg, equal to the two df, do, and 
the included angles at d equal, have also the angles at o 
equal (th. 1) ; consequently those angles at o are right 
angles, and the line ag is perpendicular to dg. 




PROBLEM V. 



At a given point a, in a line ab, to make an angle equal to 
a given angle c. 

From the centres a and c, with any one • jg^ 

radius, describe the arcs de, fg. Then, 
with radios de, and centre f, describe an 



arc, cutting fg in g. Through g draw C D 
the line ag, and it will form the angle re* ~ 
quired. \ 

For, conceive the equal lines or radii, j± jtjj 
db, fg, to be drawn. Then the two trian- 
gles CDs, afg, being mutually equilateral, are mutually equi- 
angular (th. 5), and have the angle at a eqiud to ta* vu^a fe» 



868 



GEOMETRY* 



PROBLEM VI. 



Through a given point a, to draw a line parallel to a given 
line bo. 

From the given point a draw a line ad EA 
to an j point in the given line bc. Then 
draw the line eaf making the angle at a 
equal to the angle at d (by prob. 5) ; bo j5~C 
shall ef be parallel to bc as required. 

For, the angle d being equal to the alternate angle a, the 
lines bc, ef, are parallel, by th. 13. 



problem vii. 



To divide a line ab into any proposed number of equal 
parts. 

Draw any other line ac, forming any 
angle with the given line ab ; on which EVA 
set off as many of any equal parts ad, de, \ 
ef, fc, as tho line ab is to be divided into. 3\V \ \ 

Join bc ; parallel to which draw the other ^IHCKB 
lines fg, eh, di : then these will divide ab 
in the manner as required. — For those parallel lines divide 
both the sides ab, ac, proportionally, by th. 82. 



PROBLEM Vin. 

To find a third proportional to two given lines ab, ac. 

Place the two given lines ab, ac, 

forming any angle at a ; and in ab take A B 

also ad equal to ac. Join bc, and A- ^ 
draw de parallel to it ; so will ae be 
the third proportional sought. 

For, because of the parallels, bc, de, A 2) '0 

the two lines ab, ac, are cut propor- 
tionally (th. 82) ; so that ab : ac : : ad or ac : ae ; there- 
fore ae is the third proportional to ab, ac. 

PROBLEM IX. 

To find a fourth proportional to three lines ab, ac, ad. 

Place two of the given lines ab, ac, making any angle at 
A ; also place ad on ab. Join bc ; and parallel to it draw 




PROBLEMS. 



850 



de : so shall ae be the fourth propor- 
tional as required. 

For, because of the parallels bc, de, 
the two sides ab, ac, are cut propor- 
tionally (th. 82) ; so that .... 
ab : ac : : ad : ae. 

PROBLEM X. 

To find a mean proportional between two lines ab, bc. 

Place ab, bc, joined in one straight 
line ac : on which, as a diameter, describe 
the semicircle adc ; to meet which erect 
the perpendicular bd ; and it will be the 
mean proportional sought, between ab and 
bc (by cor. th. 87). . 

problem XI. 

To find the centre of a cii 

Draw any chord ab ; and bisect it p 
pendicularly with the line cd, which will 
a diameter (th. 41, cor.). Therefore 
bisected in o, will give the centre, as reqi 
ed. 

PROBLEM XII. 

To describe the circumference of a circle through three given 
points a, b, c. 

From the middle point b draw chords 
ba, bc, to the two other points, and bi- •"-Hs^JL-JK--* 
sect these chords perpendicularly by lines ^^T^Si 
meeting in o, which will be the centre. A<^^y[^5\C 
Then from the centre o, at the distance I /q\ J 
of any one of the points, as oa, describe \. J 
a circle, and it will pass through the two 
other points b, c, as required. 

For the two right-angled triangles oad, obd, having the 
sides ad, db, equal (by constr.), and od common, with the 
included right angles at d equal, have their third sides oa, 
ob, also equal (th. 1). And in like manner it is shown that 
oc is equal to ob or oa. So that all the three oa, ob, go* tid- 
ing equal, will be radii of the same circta. 





OROMSTRY. 




PROBLEM XIII. 

To draw a tangent to a circle, through a given point 

When the given point a is in the circum- 
ference of the circle : Join a and the centre 
o ; perpendicular to which draw bac, and it 
will be the tangent, by th. 46. 

But when the given point a is out of the 
circle : Draw ao to the centre o ; on which 
as a diameter describe a semicircle, cutting 
the given circumference in d ; through R 
which draw badc, which will be the tangent 1 
cs required. 

For, join do. Then the angle ado, in a 
semicircle, is a right angle, and consequent- 
ly ad is perpendicular to the radius do, or 
is a tangent to the circle (th. 46). 

PROBLEM xiv. 

On a given line b to describe a segment of a circle, to contain 
a given angle c. 

At the ends of the given line make 
angles dab, dba, each equal to the 
given angle c. Then draw ae, be, 
perpendicular to ad, bd ; and with the 
centre e, and radius ea or eb, describe 
a circle ; so shall afb be the segment 
required, as any angle f made in it will 
be equal to the given angle c. 

For, the two lines ad, bd, being 
perpendicular to the radii ea, eb (by constr.), are tangents 
to the circle (th. 46) ; and the angle a or b, which is equal to 
the given angle c by construction, is equal to the angle f in 
the alternate segment aeb (th. 53). 

problem xv. 

To cut off a segment from a circle, that shall contain a given 
angle c. 

Draw any tangent ab to the given 
circle ; and a chord ad to make the E 
angle dab equal to the given angle c ; 
then dka will be the segment required, 
any angle s made in it being equal to 
the given angle c. 




PROBLEMS. 



861 



For the angle a, made by the tangent and chord, whioh 
ia equal to the given angle c by construction, is also equal to 
any angle a in the alternate segment (th. 63). 



problem zvi. 

To make an equilateral triangle on a given line ab. 

From the centres a and b, with the 
distance ab, describe arcs, intersecting in 
c. Draw ac, bo, and abc will be the 
equilateral triangle. 

For the equal radii, ac, bc, are, each of 
them, equal to ab. 



problem xvn. 

To make a triangle with three given lines ab, ac, bc 

With the centre a, and distance ac, 
describe an arc. With the centre b, and 
distance bc, describe another arc, cutting 
the former in c. Draw ab, bc, and abc 
will be the triangle required. 

For the radii, or sides of the triangle, 
ac, bc, are equal to the given lines ac, 
ac, by construction. 




A 



PROBLEM XVIII. 

To make a square on a given line ab. 

Raise ad, bc, each perpendicular and j>_ 
equal to ab ; and join dc ; so shall abcd be 
the square sought. 

For all the three sides ab, ad, bc, are 
equal, by the construction, and dc is equal 
and parallel to ab (by th. 24) ; so that all the 
four sides are equal, and the opposite ones are parallel. 
Again, the angle a or b, of the parallelogram, being a right 
angle, the angles are all right ones (cor. 1, th. 22). Hence, 
then, the figure, having all its sides equal, and alL'towGq&a* 
right, is a square (def. 34). 

Vol. I 47 




GEOMETRY. 



problem XIX. 



To make a rectangle, or a parallelogram, of a given length 
and breadth, ab, bc. 

Erect ad, bc, perpendicular to ab, and 
each equal to bc ; then join dc, and it is 
done. 

The demonstration is the same as the _ 

last problem. B— — C 

And in the same manner is described any oblique paral- 
lelogram, only drawing ai> and bc to make the given oblique 
angle with ab, instead of perpendicular to it. 




PROBLEM xx* 

To inscribe a circle in a given triangle abc. 

Bisect any two angles a and b, with 
the two lines ad, bd. From the inter- 
section d, which will be the centre of 
the circle, draw the perpendiculars db, 
df, DU, and they will be the radii of the 
circle required. 

For, since the angle dae is equal to 
the angle dag, and the angles at e, o, 
right angles (by const r.), the two triangles, adb, ado, are 
equiangular ; and, having also the side ad common, they are 
identical, and have the sides de, do, equal (th. 2). In like 
manner it is shown, that df is equal to de or do. 

Therefore, if with the centre d, and distance de, a circle 
be described, it will pass through all the three points, e, f, g, 
in which points also it will touch the three sides of the tri- 
angle (th. 46), because the radii de, df, do, are perpendicu- 
lar to them. 

PROBLEM XXI. 

To describe a circle about a given triangle abc. 

Bisect any two sides with two of the 
perpendiculars de, df, do, and d will 
be the centre. 

For, join da, db, dc Then the two 
right-angled triangles dae, dbe, have 
the two sides, de, ea, equal to the two 
de, eb, and the included angles at e 
equal : those two triangles are therefore 




P10BLEHS. 



383 



identical (th. 1% and have the side da. equal to db. In like 
manner it it shown, that dc is also equal to da or db. So 
that all the three da, db, dc, being equal, they are radii of 
a circle passing through a, b, and c. 



fgOBLBH XXII. 



To inscribe an equilateral triangle in a given circle. 

Through the centre c draw any diameter a 
ab. From the point b as a centre, with the y^/fV^N, 
radius bc of the given circle, describe an f / |\ \ 
arc dcb. Join ad, as, db, and ade is the f X-JsA ) 
equilateral triangle sought x^^l^^ v/ 

For, join db, dc, bb, bc. Then dcb ^ X^TT^^E 
is an equilateral triangle, having each side ^jj^ 
equal to the radius of the given circle. In 
like manner, bcb is an equilateral triangle. But the angle 
ade is equal to the angle abb or cbe, standing on the same 
arc ae ; also the angle akd is equal to the angle cbd, on the 
same arc ad ; hence the triangle dab has two of its angles, 
adb, aed, equal to the angles of an equilateral triangle, and 
therefore the third angle at a is also equal to the same ; so 
that the triangle is equiangular, and therefore equilateral. 



PROBLEM XXIII. 



To inscribe a square in a given circle* 



Draw two diameters ac, rd, crossing 
at right angles in the centre e. Then 
join the four extremities a, b, c, d, with 
right lines, and these will form the in- 
scribed square abcd. 

For the four right-angled triangles 
aeb, bp.c, ced, dea, are identical be- 
cause they have the sides ea, eb, ec, ed, 
all equal, being radii of the circle, and 
the four included angles at e all equal, 
being right angles, by the construction. Therefore all their 
third sides ab, bc, cd, da, are equal to one another, and the 
figure abcd is equilateral. Also, all its four angles, a, b, c, d, 
are right ones, being angles in a semicircle. Cooaec^t&Vj 
the figure is a square. 




GEOMETRY. 



PROBLEM XXIV. 



To describe a square about a given circle. 

Draw two diameters ac, bd, crossing 
at right angles in the centre s. Then 
through their four extremities draw fg, 
ib, parallel to ac, and fi, oh, parallel 
to bd, and they will form the square 

FGHI. 

For, the opposite sides of parallelo- 
grams being equal, fg and in are each 
equal to the diameter ac, and fi and on each equal to the 
diameter bd ; so that the figure is equilateral. Again, be- 
cause the opposite angles of parallelograms are equal, all the 
four angles f, c, h, i, are right angles, being equal to the 
opposite angles at e. So that the figure fgmi, having its 
sides equal, and its angles right ones, is a square, and its sides 
touch the circle at the four points a, b, c, d, being perpen- 
dicular to the radii drawn to those points. 




PROBLEM XXV. 



To inscribe a circle in a given square. 

Bisect the two sides fg, fi, in the points a and b (last fig.). 
Then through these two points draw ac parallel to fc or ih, 
and bd parallel to fi or gh. Then the point of intersection 
x will be the centre, and the four lines ea, eb, ec, ed, radii 
of the inscribed circle. 

For, because the four parallelograms ef, eg, kh, ei, have 
their opposite sides and angles equal, therefore all the four 
lines ea, eb, ec, ed, are equal, being each equal to half a 
side of the square. So that a circle described from the centre 
E, with the distance ea, will pass through all the points 
a, b, c, d, and will be inscribed in the square, or will touch 
its four sides in those points, because the angles there are 
right ones. 

PROBLEM XXVI. 

To describe a circle about a given square. 
(See fig. Prob. xxiii.) 
Draw the diagonals ac, bd, and their intersection s will 
be the centre. 

For the diagonals of a square bisect each other (th. 40), 
making ea, eb, ec, ed, all equal, and consequently these 
are radii of a circle naasin^ vUtou^U the four points a, b, c, p. 



FftOBUMS. 



805 



FKOBLEM XXVII. 

To cut a given line in extreme and mean 

Let ab be the given line to be divided 
in ex I re me and mean ratio, that is, so as 
that the whole line may be to the greater 
part, as the greater part is to the less part. 

Draw bc perpendicular to AR/and equal 
to half ab. Join ac ; and with centre c 
and distance cb, describe the circle bd ; 
then with centre a and distance ad, de- 
scribe the arc dk ; so shall ab be divided 
in e in extreme and mean ratio, or so that 
ab : ae : : ab : eb. 

Produce ac to the circumference at f. Then, auf being 
a secant, and ab a tangent, because b is a right angle : there- 
fore the rectangle af . ad is equal to ab 9 (cor. 1, th. 61) ; con* 
sequently the means and extremes of these are proportional 
(th. 77), viz. ab : af or ad + df : : ad : ab. But ab 
is equal to ad by construction, and ab = 2bc = df ; 
therefore, ab : ae + ab : : ae : ab ; 

and by division, ab : ae : : ae : eb. 

PROBLEM XXVIII. 

To inscribe an isosceles triangle in a given circle, that 
shall have each of the angles at the base double the angle at 
the vertex. ^ 



Draw any diameter ab of the given 
circle ; and divide the radius cb, in the 
point d, in extreme and mean ratio, by the 
last problem. From the point b apply the 
chords be, bf, each equal to the greater 
part cd. Then join ae, af, ef ; and akf 
will be the triangle required. 




For, the chords be, bf, being equal, 
their arcs are equal ; therefore the supplemental arcs and 
chords ae, af, are also equal ; consequently the triangle aef 
is isosceles, and has the angle e equal to the angle f ; alto 
the angles at u are right angles. 

Draw cf and df. Then, bc : cd : : cd : bd, or 
bc : bf : : bf : bd by constr. And ba : bf : : bf : bo 
(by th. 87). But bc = £ba ; therefore bo = £bo = go \ 
therefore the two triangles gbf, ©df, ara VtouvicaX VJ3fc%\V 



ratio. 



A 




t 



c 

806 . C GEOMETRY. 

£ 

and each equiangular to arf and agf (th. 87). Therefore 
their doubles, bfd, afe, are isosceles and equiangular, as 
well as the triangle hcf ; having the two sides bc, cf, equal, 
and the angle b common with the triangle bfd. But cd 
is = df or bf ; therefore the angle c = the angle dpc 
(th. 4) ; consequently the angle bdf, which is equal to the 
sum of these two equal angles (th. 16), is double of one of 
them c ; or the equal angle b or ceb double the angle c. 
So that cbf is an isosceles triangle, having each of its two 
equal angles double of the third angle c. Consequently the 
triangle arf (which it has been shown is equiangular to the 
triangle c f) has al*o each of its angles at the base double 
the angle a at the vertex. 

PROBLEM XXIX. 

To inscribe a regular pentagon in a given circle. 

Inscribe the isosceles triangle abc, 
having each of the angles abc, acr, 
double the angle bac (prob. 28). Then 
bisect the two arcs adr, arc, in the 
points d, e ; and draw the chords ad, dr, 
ar, ec, so shall adrce be the inscribed 
equilateral pentagon required. 

For, because equal angles stand on 
equal arcs, and double angles on double arcs, also the angles 
abc, acb, being each double the angle bac, therefore the 
arcs adr, arc, subtending the two former angles, are each 
double the arc bc subtending the latter. And since the two 
former arcs are bisected in d and r it follows that all the 
five arcs ad, dr, rc, ck, ea, are equal to each other, and con- 
sequently the chords also which subtend them, or the live 
sides of the pentagon, are a 1 I equal. 

Note. In (he constniction, the points d and e are most 
easily found, by applying bd and ce each equal to bc. 

problem xxx. 

To inscribe a regular hexagon in a circle. 

Apply the radius ao of the given circle 
as a chord, ar, rc, cd, &c. quite round 
the circumference, and it will complete 
the regular hexagon abcdef. 

Draw the radii ao, ho, co, do, eo, fo, 
completing six equal triangles ; of which 
any one, as abo, being equilateral (by 
const r.) its three angles are all equal (cor. 
2, tb. 3), and any one of them, as aob, is one-third of the' 





FB0BLEMS. 



987 




whole, or of two right angles (th. 17), or 
one-sixth of four right angles. But the 
whole circumference is the measure of four 
right angles (cor. 4, th. 6). Therefore the 
arc ar is one.sixth of the circumference of 
the circle, and consequently its chord ab one 
aide of an equilateral hexagon inscribed in the 
circle. And the same of the other chords. 

Corel. The side of a regular hexagon is equal to the 
radius of the circumscribing circle, or to the chord of one- 
sixth part of the circumference*. 

PROBLEM XXXI. 

To describe a regular pentagon or hexagon about a circle. 

Izv the given circle inscribe a regular 
polygon of the same name or number 
of sides, as abcdk, by one of the fore- 
going problems. Then to all its angular 
points draw tangents (by prob. 13), and 
these will form the circumscribing poly- 
gon required. 

For all the chords, or sides of the 
inscribed figure, ab, bc, dec. being equal, and all the radii 
oa, ob, dec. being equal, all the vertical angles about the 
point o are equal. But the angles oef, oaf, oao, obg, 
made by the tangents and radii, are right angles ; therefore 
oef + oaf = two right angles, and oao + obg = two right 
angles ; consequently, also, aoe + afe = two right an- 
gles, and oab + a«b « two right angles (cor. 2, th. 18). 
Hence, then, the angles aoe + afe being = aob + aob, of 
which aob is = aoe ; consequently the remaining angles r 
and o are also equal. In the same manner it is shown, that all 
Hthe angles f, g, h, i, k, are equal. 

Again, the tangents from the same point fe, fa, are equal, 
as also the tangents ao, gb, (cor. 2, th. 61) ; and the angles 
f and g of the isosceles triangles afe, agb, are equal, as well 
as their opposite sides ae, ab ; consequently those two tri- 
angles are identical (th. 1), and have their other sides ef, fa, 
ag, gb, all equal, and fg equal to the double of any one of 
them. In like manner it is shown, that all the other sides 
gh, hi, ik, ki, are equal to fg, or double of the tangents gb, 
bh, &c. 




* The bent way to describe n polygon of any number of sides, the 
length of one side being given, is to find the radius of the cATOraMafa» 
log circle by means of the table, at pa. 413, and \ta t*\» «t \*. 



368 



GKOKETBY. 



Hence, then, the circumscribed figure is both equilateral 
and equiangular, which was to be shown. 

Cord* The inscribed circle touches the middle of the sides 
of the polygon. 

PSOBLEX XXXII. 



To inscribe a circle in a regular polygon. 

Bisect any two sides of the polygon 
by the perpendiculars go, fo, and their 
intersection o will be the centre of the 
inscribed circle, and oo or of will be the 
radius. 

For the perpendiculars to the tangents 
af, ao, pass through the centre (cor. r 
th. 47) ; and the inscribed circle touches C 
the middle points f, o, by the last corollary. Also, the two 
sides, ao, ao, of the right-angled triangle aog, being equal 
to the two sides af, ao, of the right-angled triangle aof, the 
third sides of, oo, will also be equal (cor. th. 45). There- 
fore the circle described with the centre o and radius og, will 
pass through f, and will touch the sides in the points g and f. 
And the same for all the other sides of the figure. 




PROBLEM XXXIII. 



To describe a circle about a regular polygon. 

Bisect any two of the angles, c and d, 
with the lines co, do ; then their inter, 
section o will be the centre of the circum- 
scribing circle ; and or, or od, will be the 
radius. 

For, draw ob, oa, of, &c. to the an- 
gular points of the given polygon. Then 
the triangle ocd is isosceles, having the angles at c and D 
equal, being the halves of the equal angles of the polygon 
BCD, cdb ; therefore their opposite sides co, do, are equal, 
(th. 4). But the two triangles oci>, ocb, having the two sides 
oc, cd, equal to the two oc, cb, and the included angles ocd, 
Ocb, also equal, will be identical (th. 1), and have their third 
•ides bo, od, equal. In like manner it is shown, that all the 
lines oa, ob, oc, od, or, aro equal. Consequently a circle 
described with the centre o fjnd radius oa, will pass through 
all the other angular points, b, c, d, dec. and will circum- 
msribe the polygon. 




PROBLEMS. 



PROBLEM XXXIV. 



To make a square equal to the sum of two or more given 
squares. 

Let ib and ac be the sides of two 
given squares. Draw two indefinite 
lines ap, aq, at right angles to each 
other ; in which place the sides ab, ac, 
of the given squares ; join bc ; then a 
square described on bc will be equal to 
the sum of the two squares described 
on ab and ac (th. 34). 

In the same manner, a square may be made equal to the 
sum of three or more given squares. For, if ab, ac, ad, be 
taken as the sides of the ffiven squares, then, making ae=bc, 
ad = ad, and drawing he, it is evident that the square on db 
will be equal to the sum of the three squares on ab, ac, ad. 
And so on for more squares. 




PROBLEM XXXV. 



To make a square equal to the difference of two given 
squares. 

Let ab and ac, taken in the same 
straight line, be equal to the sides of the 
two given squares. — From the centre a, 
with the distance ab, describe a circle ; 
and make cd perpendicular to ab, meet- A C *B 

ing the circumference in d : so shall a square described on 
cd be equal to ad' j — ac\ or ab 3 — ac 9 , as required (cor. th. 34). 



PROBLEM XXXVI. 



To make a triangle equal to a given quadrangle abcd. 

Draw the diagonal ac, and parallel j) q 

to it de, meeting ba produced at e, and /P J9\ 

join ce ; then will the triangle ceb be st* 
equal to the given quadrilateral abcd. • / 



XL 




For, the two triangles ace, acd, E A B 
being on the same base ac, and between 
the same parallels ac, de, are equal (th. 25) ; therefore, if 
abc be added to each, it will make bce eqaaUo kwct> Ujl^v 
•Vol. I. 48 



870 OBOMBTBY. 



PROBLEM XXXVII. 




To make o triangle equal to a given pentagon abcds. 

Draw da and db, and also ef, co, 
parallel to them, meeting ab produced 
at v and o ; then draw dp and do ; so 
shall the triangle dfo be equal to the 
given pentagon abcde. 

For the triangle dfa = dka, and 
the triangle dob = dcb (th. 25) ; 
therefore, by adding dab to the equals, 
the sumw are equal (ax. 2), that is, dab + dap + dbg» dab 
+ dae + dbc, or the triangle dfo = to the pentagon . 



problem xxxviii. 



To make a rectangle equal to a given triangle abc. 

Bisect the base ab in d : then raise 
de and bf perpendicular to ab, and 
meeting cf parallel to ab, at e and f : 
so shall dp be the rectangle equal to the 
given triangle abc (by cor. 2, th. 26). 



problem xxxix. 

To make a square equal to a given rectangle abcd. 

Produce one side ab, till be be 

equal to the other side bc. On ae as O . V 

a diameter describe a circle, meeting 3 ) r J j ^,... 

bc produced at f : then will bp be the \ \ I 

side of the square bfoh, equal to the f__ j ! j 

given rectangle bd, as required ; as AH B E 
appears by cor. th. 87, and th. 77. 




871 



APPLICATION OF ALGEBRA 



TO 



GEOMETRY. 



Whbx it is proposed to resolve a geometrical problem 
algebraically, or by algebra, it is proper, in the first place, 
to draw a figure that shall represent the several parts or con- 
ditions of the problem, and to suppose that figure to be the 
true one. Then having considered attentively the nature 
of the problem, the figure is next to be prepared for a solu- 
tion, if necessary, by producing or drawing such lines in it as 
appear most conducive to that end. This done, the usual 
symbols or letters, for known and unknown quantities, are 
employed to denote the several parts of the figure, both the 
known and unknown parts, or as many of them as necessary, 
as also such unknown line or lines as may be easiest found, 
whether required or not. Then proceed to the operation, 
by observing the relations that the several parts of the figure 
have to each other ; from which, and the proper theorems 
in the foregoing elements of geometry, make out as many 
equations independent of each other, as there are unknown 
quantities employed in them : the resolution of which equa- 
tions, in the same manner as in arithmetical problems, will 
determine the unknown quantities, and resolve the problem 
proposed. 

As no general rule can be given for drawing the lines, and 
selecting the fittest quantities to substitute for, so as always 
to bring out the most simple conclusions, because different 
problems require different modes of solution ; the best way 
to gain experience, is to try the solution of the same problem 
in different ways, and then apply that which succeeds best, 
to other cases of the same kind, when they afterwards occur. 
The following particular directions, however, may be of 
some use. 

l«t, In preparing the figure, by drawing lines, let them be 
either parallel or perpendicular to other lines in the figure, 
or so as to form similar triangles. And if an angle be given, 
it will be proper to let the perpendicular be opposite to that 
-angle, and to fall from one end of a given line, if ^awfttat* 



372 



APPLICATION or ALGEBRA 



2d, In selecting the quantities proper to substitute for, 
those are to be chosen, whether required or not, which lie 
nearest the known or given parts of the figure, and by means 
of which the next adjacent parts may be expressed by addi- 
tion and subtraction only, without using surds. 

3d, When two lines or quantities are alike related to other 
parts of the figure or problem, the best way is, not to make 
use of either of them separately, but to substitute for their 
sum, or difference, or rectangle, or the sum of their alternate 
quotients, or for some line .or lines, in the figure, to which 
they have both the same relation. 

4th, When the area, or the perimeter, of a figure is given, 
or such parts of it as have only a remote relation to the parts 
required : it is sometimes of use to assume another figure 
similar to the proposed one, having one side equal .to unity, 
or some other known quantity. For, hence the other parts 
of the figure may be round, by the known proportions of the 
like sides, or parts, and so an equation be obtained. For 
examples, take the following problems. 

PROBLEM I. 

In a right-angled triangle, having given the base (8), and 
the sum of the hypotenuse and perpendicular (9) ; to find 
both these two sides. 

Let abc represent the proposed triangle 
right-angled at b. Put the base ab = 3 = b, 
and the sum ac + bc of the hypothenuse 
and perpendicular = 9 = s ; also, let x de- 
note the hypothenuse ac, and y the perpen- 
dicular BC. 

Then by the question - . . x + y = s, 
and by theorem 34, - - . . x* = y % + b*, 
By tr»:nspos. y in the 1st equ. gives x = s — y t 
This value of x substi. in the 2d, 

gives .... s^-Qsy + y 1 = f + b* 9 
Taking away y*on both sides leaves j*... 2sy = b*, 
By transpos. Usy and 6 j , gives - s^b' = Usy, 

* 3 — b 2 

And dividing by 2s, gives - - — — = y = 4 = bc. 
Hence x = * — y = 5 = ac. 

^ N. B. In this solution, and the following ones, the nota. 
tkm is made by using as many unknown letters, x and jr, as 




TO OEOMKBT. 



378 



there are unknown tides of the triangle, a separate letter for 
each ; in preference to using only one unknown letter for 
one side, and expressing the other unknown side in terms 
of that letter and the given sum or difference of the sides ; 
though this latter way would render this solution shorter and 
sooner ; because the former way gives occasion for more and 
better practice in reducing equations ; which is the very end 
and reason for which these problems are given at all. 

PROBLsk'lI. 

In a right-angled triangle, having given the hypothenuse (5) ; 
and the sum of the base and perpendicular (7) ; to find both 
these two sides. 

Let abc represent the proposed triangle, right-angled at 
B. Put the given hypothenuse ac = 5 = a, and the sum 
ab + bc of the base and perpendicular = 7 = s ; also let x 
denote the base ab, and y the perpendicular bc. 

Then by the question - - - x + y = s t 
and by theorem 34 ... x*+ y*= a 2 , 
By transpos. y in the 1st, gives x = s — y, 
By substitu. this value for x, gives **— 2sy 4- 2y a = a\ 
By transposing s*, gives - - 2y* — 2sy — a* — s 7 , 
By dividing by 2, gives - - y 2 — sy = \a* — ±s* 9 
By completing the square, gives y* — sy + Js* = }a* — \s* f 
By extracting the root, gives - y — \s = %/(i a3 ~" i* 9 ) 
By transposing is, gives - - y = J* ± y/\\a 2 — \s*) = 

4 and 3, the values of x and y. 

problem in. 

In a rectangle, having given the diagonal (10\ and the peri- 
meter, or sum of all the four sides (28) ; to find each of the 
sides severally* 

Let abcd be the proposed rectangle ; 
and put the diagonal ac = 10 = d, and 
half the perimeter ab + bc or ad + 
dc — 14 =. a : also put one side ab = ar, 
and the other side bc = y. Hence, by 
right-nngled triangles, - - - . - x* + y 7 = d\ 
And by the question - - - . - x+y — a, 
Then by transposing y in the 2d, gives x — a — y, 
This value substituted in the 1st, gives a? — < SUu)V*^=^ 




874 



APPLICATION OF ALGEBRA 



Transposing a*, gives . - 2y' — 2ay = d 1 — «*, 
And dividing by 2, gives - y 1 — ay = \d l — ±a\ 
By completing the square, it is y* — ay + \a* = \<P — \a\ 
And extracting the root, gives y — ±a = y/(±d l — Ja 1 ), 
And transposing \a, gives - y = £a 2: v^(i^ J — l* 1 ) 3 ^ 
or 6, the values of x and y. 



problem IV. 

Having given the base anfoerpendicular of any triangle ; to 
find the side of a square inscribed in the same. 

Let abc represent the given triangle, q 
and EFGit its inscribed square. Put the 
base ab =■ by the perpendicular cd = a, 
and the side of the square gf or on = 
Di x ; then will ci = cd — di = 
a — x. 

Then, because the like lines in the 
similar triangles abc, gpc, are propor- 
tional (by theor. 84, Geom.), ab : cd : : ge : ci, that 
is, 6 : a : : x : a — x. Hence, by multiplying extremes and 
means, qb — bx = ax, and transposing bx, gives ab = ax 

ab 

+ bx ; then dividing by a + b, gives x = a ^ ^ = gf otch 

the side of the inscribed square : which therefore is of the 
same magnitude, whatever the species or the angles of the 
triangles may be. 




problem v. 



In an equilateral triangle, having given the lengths of the 
three perpendiculars, drawn from a certain point within, on 
the tlirce sides ; to determine the sides. 

Let abc represent the equilateral tri- 
angle, and oe, df, dg, the given per- 
pendiculars from the point d. Draw the 
lines da, dh, dc, to the three angular 
points ; and let fall the perpendicular cu 
on the base ab. Put the three given per- 
pendiculars, de = a, df = b, d<; = c, 
and put x = a 11 or bii, half the side of 
the equilateral triangle. Then is ac or bc = 2.r, and by 
right-angled triangles the perpendicular cu = */(ac 2 — ah 1 ) 




•S3T 




c 


) - 














TO GEOMETRY. 



375 



Now, since the area or space of a rectangle, is expressed 
by the product of the buse arid height (cor. 2, th 81, Geom.), 
and that a triangle is equal to half a rectangle of equal baso 
and height (cor. 1, th. ^6), it follows that, 
the whole triangle abc is = Jab X cn = x X x y/3 = x 2 ^3, 
the tnungle abd = Jab X do = x X c = cr, 
the triangle bcd =■ Jbc X dk = x X a = ax, 
the triangle acd = \ac X df = * X b = bx. 
But the three last triangles make up, or are equal to, the 
whole former, or great tnungle ; 

that is, x* = ax + br + cx ; hence, dividing by *, gives 
r y/3 = a +6 + c, and dividing by ^/3, gives 
a-f. & + r 

x = 6 — , half the side of the triangle sought. 

Also, since the whole perpendicular cn is =- x y/3, it is 
therefore = a + b + c. That is, the whole perpendicular 
cn, is just equal to the sum of all the three smaller perpen- 
diculars db + nr + do taken together, wherever the point 
d ii situated. 

PROBLEM VI. 

In a right-angled triangle, having given the base (3\ and 
the difference between the hypoihenuse and perpendicular 
(1) ; to find both these two sides. 

PROBLEM VII. 

In a right-angled triangle, having given the hypothenuse 
5), and the difference between the base and perpendicular 
1) ; to determine both these two sides. 

problem viu. 

Having given the area, or measure of the space, of a rect. 
angle, inscribed in a given triangle ; to determine the sides 
of the rectangle. 

problem u. 

In a triangle, having given the ratio of the two sides, 
together with both the segments of the base, made by a per- 
pendicular from the vertical angle ; to determine the sides of - 
the triangle. 

problem x. 

In a triangle, having given the base, the sum of the other 
two sides, and the length of a line drawn ftom\Yi« <<i«*ta*\ 



376 APPLICATION OF ALGEBRA 

angle to the middle of the base ; to find the sides of the 
triangle. 

PROBLEM XI. 

In a triangle, having given the two sides about the verti- 
cal angle, with the line bisecting that angle, and terminating 
in the base ; to find the base. 

PROBLEM XII. 

To determine a right-angled triangle ; having given the 
lengths of two lines druwn from the acute angles, to the 
middle of the opposite sides. 

PROBLEM XIII. 

To determine a right-angled triangle ; having given the 
perimeter, and the radius of its inscribed circle. 

PHOBLEM XIV. 

To determine a triangle ; having given the base, the per- 
pendicular, and the ratio of the two sides. 

PROBLEM xv. 

To determine a right-angled triangle ; having given the 
hypothenuse, and the side of the inscribed square. 

PROBLEM XVI. 

To determine the radii of three equal circles, described in 
a given circle, to touch each other and also the circum- 
ference of the given circle. 

PROBLEM XVII. 

In a right-angled triangle, having given the perimeter, or 
sum of all the sides, and the perpendicular let fall from the 
right angle on the hypothenuse ; to determine the triangle, 
that is, its sides. 

PROBLEM XVIII. 

To determine a right-angled triangle ; having given the 
hypothenuse, and the difference of two lines drawn from the 
two acute angles to the centre of the inscribed circle. 




-$&~ i-$A t&^h. w-a+yi- t t 



I "IT 



If- 






hZe^dr^ 




1 




to 



TO GEOMETRY. 377 
PROBLEM XIX. 

To determine a triangle ; having given the base, the per- 
pendicular, and the difference of the two other sides. 

PROBLEM XX. 

To determine a triangle ; having given the base, the per- 
pendicular, and the rectangle or product of the two sides* 

PROBLEM XXL 

To determine a triangle ; having given the lengths of three 
lines drawn from the three angles, io the middle of the oppo* 
site sides. 

PROBLEM XXXI. 

In a triangle, having given all the three sides ; to find the 
radius of the inscribed circle. 

PROBLEM XXIII. 

To determine a right-angled triangle ; having given the 
aide of the inscribed square, and the radius of the inscribed 
circle. 

PROBLEM XXIV. 

To determine a triangle, and the radius of the inscribed 
circle ; having given the lengths of three lines drawn from 
the three angles, to the centre of that circle. 

PROBLEM XXV. 

To determine a right-angled triangle ; having given the 
hypothenuse, and the radius of the inscribed circle. 

PROBLEM XXVI. 

To determine a triangle ; having given the base, the line 
bisecting the vertical angle, and the diameter of the circum. 
scribing circle. 



Vol. I. 



49 



97 



PLANE TRIGONOMETRY. 



DEFIMTIOX8. 

1. Plank Trigonometry treats of the relations and cal- 
culations of the sides and angles of plane triangles. 

2. The circumference of every circle (as before observed 
in Geom. Def. 56) is supposed to be divided into 360 equal 
parts, called Degrees ; also each degree into 60 Minutes, 
and each minute into 60 Seconds, and so on. Hence a se- 
micircle contains 180 degrees, and a quadrant 90 degrees. 

3. The Measure of an angle (Def. 57, Geom.) is an arc 
-of any circle contained between the two lines which form 
that angle, the angular point being the centre ; and it is esti- 
mated by the number of degrees contained in that arc. 

Hence, a right angle, being measured by a quadrant, or 
quarter of the circle, is an angle of 90 degrees ; and the 
sum of the three angles of every triangle, or two right an* 
gles, is equal to 180 degrees. Therefore, in a right-angled 
triangle, taking one of the acute angles from 90 degrees, 
leaves the other acute angle ; and the sum of the two angles, 
in any triangle, taken from 180 degrees, leaves the third 
angle ; or one angle being taken from 180 degrees, leaves 
the sum of the other two angles. 

4. Degrees arc marked at the top of the figure with a 
small °, minutes with ', seconds with *, and so on. Thus, 
57° 30' 12", denote 57 degrees 30 minutes and 12 seconds. 

5. The Complement of an arc, is 
what it wants of a quadrant or 90°. 
Thus, if ad be a quadrant, then bd is 
the complement of the arc ab ; and, 
reciprocally, ab is the complement of 
bd. So that, if ab be an arc of 50°, & 
then its complement bd will be 40°. 

6. The Supplement of an arc, is 
what it wants of a semicircle, or 180°. T 
Thus, if ade be a semicircle, then 
bde is the supplement of the arc ab ; and, reciprocally, ab 




DJEFiirrnoifs. 



87V 



is the supplement of the arc bde. So that, if ab be an arc 
of 50°, then its supplement bob will be 130°. 

7. The Sine, or Right Sine, of an arc, is the line drawn 
from one extremity of the arc, perpendicular to the diameter 
which passes through the other extremity. Thus, bf is the 
sine of the arc ab, or of the supplemental arc bde. Hence 
the sine (bf) is half the chord (bo) of the double arc 
(bao). 

8. The Versed Sine of an arc, is the part of the diameter 
intercepted between the arc and its sine. So, af is the versed 
sine of the arc ab, and ef the versed sine of the arc edb. 

9. The Tangent of an arc, is a line touching the circle in 
one extremity of that arc, continued from thence to meet a 
line drawn from the centre through the other extremity ; 
which last line is called the Secant of the same arc. Thus, 
ah is the tangent, and en the secant, of the arc ab. Also, 
ei is the tangent, and ci the secant, of the supplemental arc 
bob. And this latter tangent and secant are equal to the 
former, but are accounted negative, as being drawn in an 
opposite or contrary direction to the former. 

10. The Cosine, Cotangent, and Cosecant, of an arc, 
are the sine, tangent, and secant of the complement of that 
arc, the Co being only a contraction of the word comple- 
ment Thus, the arcs ab, bi>, bein * the complements of 
each other, the sine, tangent, or secant of the one of these, 
is the cosine, cotangent, or cosecant of the other. So, bf, 
the sine of ab, is the cosine of bd ; and bk, the sine of bd, 
is the cosine of ab : in like manner, ah, the tangent of ab, 
is the cotangent of bd ; and dl, the tangent of db, is the 
cotangent of ab ; also, ch, the secant of ab, is the cosecant 
of bd ; and cl, the secant of bd, is the cosecant of ab. 

Cord. Hence several important properties easily follow 
from these definitions ; as, 

1st, That an arc and its supplement have the same sine, 
tangent, and secant ? but the two latter, the tangent und 
secant, are accounted negative when the arc is greater than 
a quadrant or 90 degrees. 

2d, When the arc is 0, or nothing, the sine and tangent 
are nothing, but the secant is then the radius oa, the least it 
can be. As the arc increases from 0, the sines, tangents, 
and secants, all proceed increasing, till the arc becomes a 
whole quadrant ad, and then the sine is tkc greatest it can 
be, being the radius en of the circle ; and both the tangent 
and secant are infinite. 

3d, Of any arc ab, the versed sine af, and cosine bk, 
or cf, together make up the radius ca of the evecta.— 



•80 



FLAHK TUOONOVXTBT. 



adius ca, the tangent ah, and the^ secant ch, form a right» 
angled triangle c % h. So also do the radius, sine, and conn** 
form another right-angled triangle chf or cbk. As also the 
radius, cotangent, and cosecant, another right-angled triangle 
cdl. And all these right-angled triangles are similar to each 
other. 

11. The sine, tangent, or 
secant of aa angle, is the sine, 
tangent, or secant of the arc 
by which the angle is mea- 
sured, or of tho degrees, dic- 
ta the same arc or angle. 

12. The method of con- 
structing the scales of chords, 
sines, tangents, and secants, 
usually engraven on instru- 
ments, for practice, is exhi- 
bited in the annexed figure. 

13. A Trigonometrical 
Canon, is a table showing 
the length of the sine, tan- 

Sent, and secant, to everyjg 
egree and m minute of the 
quadrant, with respect to the 
radius, which is expressed by 
unity or 1, with any number 
of ciphers. The logarithms 
ef these sines, tangents, and 
secants, are also ranged in the 
tables ; and these are most commonly used, as they perform 
the calculations by only addition and subtraction, instead of 
the multiplication and division by the natural sines, etc. ac- 
cording to the nature of logarithms. Such tables of log. 
sines and tangents, as well as the logs of common numbers, 
greatly facilitate trigonometrical computations, and are now 
very common. Among the most correct are those published 
by the author of this Course. 




PROBLEM I. 



To compute the Natural Sine and Cosine of a Given Arc. 

Tni9 problem is resolved after various ways. One of these 
is as follows, viz. by means of the ratio between the diameter 



PBOBLSK*. 881 

and circumference of a circle, together with the known series 
for the sine and cosine, hereafter demonstrated. Thus, the 
aemicircumference of the circle, whose radius is 1, being 
3-14169^653589703 &c, the proportion will therefore be, 
as the number of degrees or minutes in the semicircle, 
is to the degrees or minutes in the proposed arc, 
so is 3 14169265 dec, to the length of the said arc. 
This length of the arc being denoted by the letter a ; and 
its sine and cosine by « and c ; then will these two be ex- 
pressed by the two following series, viz. 

_ _ jr r , _a*_ a* _ 

9 a 2.3 + 2.3.4.5 2.3.4.5.0.7 

= a 3 , a* a 7 , . 

° 6 120 5040 

2 ^2.3.4 2.3.4.5.6 T C 
a 9 o* a 8 
= l — 2 + 34-720 + &C - 

Exam. 1. If it be required to find the sine and cosine of 
1 minute. Then, the number of minutes in 180° being 
10800, it will be first, as 10600 : 1 : : 3 14150205 dec. : 
•000290888208665 = the length of an arc of one minute. 
Therefore, in this case, 

a= -0002908882 
and^a 3 = -000000000004 dec. 
the diflT. \ss= 0002908882 the sine of 1 minute. 
Also, from 1- 

take ia* = 0000000423079 dec. 

leaves c = -9999999577 the cosine of 1 minute. 

Exam. 2. For the sine end cosine of 5 degrees. 
Here as 1H)° : 5« : : 3 141 59205 &c. : -08726646 = a the 
length of 5 degrees. Hence a == -08726K46 
— Ja* * - -00011076 
+ ^ a s = -00000004 



these collected give $ = '08715574 the sine of 5\ 

And, for the cosine, 1 = 1* 

— ±a 2 = — -00380771 
+ = -00000241 



these collected give c = 



•99619470 the cosine of 5°. 



383 



PL ARB TBIGOXOXXTBY. 



After the same manner, the sine and cosine of any other 
arc may be computed. Rut the greater the arc is, the slower 
the series will converge, in which case a greater number of 
terms must be taken, to bring out the conclusion to the same 
degree of exactness. 

Or, having found the sine, the cosine will be found from 
it, by the property of the right-angled triangle cbf, vis. the 
cosine cp = y/{cu* — bp 11 ), or c =• y/{l — ?). 

There are also other methods of constructing the canon 
of sines and cosines, which, for brevity's sake, are here 
omitted : some of them, however, are explained under 
the analytical trigonometry in the second volume of this 
Course. 



problem n. 

To compute the Tangents and Secants. 

Tub sines and cosines being known, or found by the 
foregoing problem ; the tangents and secants will be easily 
found, from the principle of similar triangles, in the follow, 
ing manner : 

In the first figure, where, of the arc ab, bf is the sine, 
cf or bk the cosine, ah the tangent, ch the secant, dl the 
cotangent, and cl the cosecant, the radius being ca or cb or 
cd ; the three similar triangles cfb, cah, cdl, give the fol- 
lowing proportions : 

1**, cf : fb : : ca : ah ; whence the tangent is known, 
being a fourth proportional to the cosine, sine, and radius. 

2d, cp : cb : : ca : ch; whence the secant is known, 
being a third proportional to the cosine and radius : or, 
being, indeed, the reciprocal of the cosine when the radius 
is unity. 

3d, bf : fc : : cd : dl ; whence the cotangent is known, 
being a fourth proportional to the sine, cosine, and radius. 

Or, aii ; ac i i cd : dl ; whence it appears that the co- 
tangent is a third proportional to the tangent and radius ; 
or the reciprocal of the tangent to radius 1. 

4th bf : bc : : cd : cl ; whence the cosecant is known, 
being a third proportional to the sine and radius ; or the re- 
ciprocal of the sine to radius 1. 

As for the log. sines, tangents, and secants, in the tables, 
they are only the logarithms of the natural sines, tangents, 
and secants, calculated as above. 

Having given an idea of the calculation and use of sines, 
tangents, and secant*, wt> may uow proceed to resolve the 



PROBLEMS. 



889 



several cases of Trigonometry ; previous to which, however, 
it may be proper to add a few preparatory notes and ob- 
servations, as below. 

Note 1. There are three methods of resolving triangles, 
or the cases of trigonometry ; namely, Geometrical Con- 
struction, Arithmetical Computation, and Instrumental Opera- 
tion ; of which the first two will here be treated. 

In the First Method, The triangle is constructed, by 
making the parts of the given magnitudes, namely, the sides 
from a scale of equal parts, and the angles from a scale of 
chords, or by some other instrument. Then measuring the 
unknown parts by the same scales or instruments, the solu- 
tion will be obtained near the truth. 

In the Second Method, Having stated the terms of the 
proportion according to the proper rule or theorem, resolve 
it like any other proportion, in which a fourth term is to fie 
found from three given terms, by multiplying the second 
and third together, and dividing the product by the first, 
in working with the naturul numbers ; or, in working with 
the logarithms, add the logs, of the second and third terms 
together, and from the sum take the log. of the first term ; 
then the natural number answering to the remainder is the 
fourth term sought. 

Note 2. Every triangle has six parts, viz. three sides and 
three angles. And in every triangle proposed, there must 
be given three of these parts, to find the other three. Also, 
of the. three parts that are given, one of them at least must 
be a side ; because, with thf same angles, the sides may be 
greater or less in any proportion. 

Note 3. All the cases in trigonometry, may be comprised 
in three varieties only ; viz. 

1**, When a side and its opposite angle are given. ' 
■ 2d, When two sides and the contained angle are given. 

3d, When the three sides are given. 

For there cannot possibly be more than these three varie- 
ties of cases ; for each of which it will therefore be proper 
to give a separate theorem, as follows : 



THEOREM I. 

When a Side and Us Opposite Angle are two of the Give* 
Parts. 

Then the unknown parts will be found by this theorem : 
viz. The sides of the triangle have the same proportion, to 
each other, as the sines of their opposite ua^tatYm*. 



884 



PLANE TRIGONOMETRY. 



That is. As any one aide, 

la to the sine of its opposite angle ; 

So is any other side, 

To the sine of its opposite angle. 
Demonstr. For, let abc be the pro- C 
posed triangle, having ab the greatest Y^f\ 
aide, and bc the least. Take ad = sSv* \ 

bc, considering it as a radius ; and let |\i \ 

fall the perpendiculars dk, cf, which £ 3£ Jf B 
will evidently be the sines of the an- 

§lea a and a, to the radius ad or bc. 
Tow the triangles ade, acf, are equiangular ; they therefore 
have their like sides proportional, namely, ac : cf : : ad or 
Be : de ; that is, the side ac is to the sine of its opposite an- 
gle b, as the aide bc is to the sine of its opposite angle a. 

Note 1. In practice, to find an angle, begin the proportion 
with a side opposite to a given angle. And to find a aide, 
begin with an angle opposite to a given side. 

Note 2. An angle found by this rule is ambiguous, or un- 
certain whether it be acute or obtuse, unless it be a right 
angle, or unless its magnitude be such as to prevent the 
ambiguity ; because the sine answers to two angles, which 
are supplements to each other ; and accordingly the geome- 
trical construction forms two triangles with the same parts 
that are given, as in the example below ; and when there is 
no restriction or limitation included in the question,- either 
of them may be taken. The number of degrees in the table, 
answering to the sine, measurc%he acute angle ; but if the 
angle be obtuse, subtract those degrees from 180°, and the 
remainder will be the obtuse angle. When a given angle if 
obtuse, or a right one, there can be no ambiguity ; for then 
neither of the other angles can be obtuse, and the geometri- 
cal construction will form only one triangle. 



EXAMPLE I. 

In the plane triangle abc, 

I ab 345 yards 
Given { bc 232 yards 
a 37" 20' 
Required the other parts. 

1. Geometrically. 

Draw an indefinite line ; on which set off ab = 345, 
from some convenient scale of equal parts. — Make the angle 




THEOREM I . 



885 



a = 37°i. — With a radius of 232, taken from the same 
scale of equal parts, and centre b, cross ac in the two 
points, c, c. — Jjastly, join bc, bo, and the figure is construct, 
ed, which gives two triangles, and shows that the case is am. 
biguous. 

Then, the sides ac measured by the scale of equal parts, 
and the angles b and c measured by the line of chords, or 
other instrument, will be found to be nearly as below ; viz. 

ac 174 z.b27« Z.cll5 ft |. 

or 3741 or 78} or 64 J. 

2. Arithmetically. 

First, to find the angles at c. 

As side bc 232 . log. 2-3654880 

To sin. op. £ a 37° 20' . . 9*78^7958 

So side ab 345 - 25378191 

To sin. op. c 1 1 5' 36* or 61° 24' 9 9551269 

add Z.A 37 20 37 20 

the sum 152 56 or 101 44 

taken from 180 00 180 00 

leaves b 27 04 or 78 16 

Then, to find the side ac. 

As sine Z a 37° 20' 

To op. side bc 2S2 

„ . . i 270 0* 
So sin. Z b J 78 16 

To op. side ac 174 07 
or 374*56 



- log. 9-7827956 

- ' 2-3654*80 

9*6580371 
9*9908291 
'84407*93 
2-5735218 



EXAMPLE II. 



In the plane triangle abc, 

C ab 365 poles 
Given < £a 57° 12' 
24 45 

Required the other parts. 

EXAMPLE III. 

In the plane triangle abc, 

( ac 120 feet 
Given < bc 112 feet 
( Za 57" 27' 

Required the other parts. 
Vol. I. 50 



Ans. < a 



c 

AC 
BC 



98° 3' 
154 



f £b 64*34' 21' 
or 115 25 39 
£c57 58 39 
or 7 7 21 
ab 112-65 feet 

lor 16-47 fa*. 



386 



PLANE TBIOOH OMETXY. 



THEOREM II. 



When too Sides and their Contained Angle are given* 

F1S8T find the sum and the difference of the given sides. 
Next subtract the given angle from 180°, and the remainder 
will be the sum of the two other angles ; then divide that 
by 2, which will give the half sum of the said unknown an- 
gles. Then say, 

As the sum of the two given sides, 

Is to the difference of the same sides ; 

So is the tang, of half the sum of their op. angles. 

To the tang, of half the diff. of the same angles. 

Add the half difference of the angles, so found, to their 
half sum, and it will give the greater angle, and subtracting 
the same will leave the less angle : because the half sum of 
any two quantities, increased by their half difference, gives 
the greater, and diminished by it gives the less. 

All the angles being thus known, the unknown side will 
be found by the former theorem. 

Note. Instead of the tangent of the half sum of the un. 
known angles, in the third term of the proportion, may be 
used the cotangent of half the given angle, which is the 
same thing. 

Demon. Let abc be a plane triangle of which ac, ci, 
and the included angle c are given : c being acute in the 
first figure, obtuse in the second. 

On ac, the longer side, set off cd = cb the shorter ; join 



bd, and bisect it in e ; also, bisect ad in o, and join », cs, 
producing the latter to r. 

Now J(ac + cb) = £(2gd + 2dc) = co 
and ^(ac — cb) = |(2ag) = ag 

also {(a + b) = £(cdb + CBD ) x cm 
and |(b — a) == abc — J sum = abd: 



C 




THEOREM II. 397 

also, because cs bisects the base of the isosceles triangle cbd, 
it is perpendicular to it : 

Therefore ec = tangent of cbd ) . M 
kp = tangent of abd 5 t0radMI8BB - 
Lastly, because in the triangle acf, oe is parallel to af 
(Geom. th. 82) we have 

co : oa : : ce : ef ; that is, 
J(ac + cb) : |(ac — cb) : : tan. £(b + a) : tan. J(b — a) ; 
or, siuce doubling both the antecedent and consequent of 
the first ratio does not change the mutual relatiou of its 
terms, we have 

ac + cb : ac — cb : : tan. J(b + a) : tan. J(b — a), q. e. d. 



EXAMPLE I. 



In the plane triangle abc, 
C ab 345 yards 
Given < ac 174*07 yards 
( £ a 37° 20' 
Required the other parts. 



1. Geometrically. 

Draw ab = 345 from a scale of equal parts. Make the 
angle a == 37° 20'. Set off ac = 174 by the scale of equal 
parts. Join bc, and it is done. 

Then the other parts being measured, they are found to 
be nearly as follow ; viz. the side bc 232 yards, the angle 
b 27°, and the angle c 115°{. 



2. Arithmetically. 

The side ab 345 From 180° 00' 

the side ac 174*07 take Lk 37 20 

their sum 519*07 sum of c and b 142 40 

their differ. 170*93 half sum of do. 71 20 

As sum of sides ab, ac, - - 519 07 log. 2*7152259 

Todiff.ofsides ab.ac, - - 170*93 - 2*2328183 

So tang, half num^flc and b 71- 2ff - 10 4712979 

To tang, half diff. L s c and b 44 10 - 9-9888903 

these added give 115 36 

and subtr. give z. b 27 4 



888 



PLANK TKJOOMOXKTBY. 



Then, by the former theorem. 

As sin Z c 1 15" 3ff or 64 24' - log. 0-95518S0 

To its ou. side ar 345 - - - 2-53781M 

Sown. ofZ a37 2ff • - - 07827U58 

To its op. side bc 232 - • - 2-3054880 

EXAMPLE n. 

In the plane triangle abc, 

( ab 365 poles ( bc 800-86 

Given I ac 154-33 Ana. { L b 24« 45' 

a 57" 12* file 08 3 
Required the other parts. 

EXAMPLE III* 

In the plane triangle abc, 

( ac 120 yards i ab 112-65 
Given I bc 112 yards Ans. 1 Z± 57 e 2T 0" 

( Zc 57" 58' 39" ( ^b 65 34 21 
Required the other parts. 



TIIEOREM III. 

When the Three Sides of a Triangle a r give a. 

First, let full a perpendicular from the greatest angle on 
the opposite side, or base, dividing it into two segments, and 
the whole triangle into two right-angled triangles : then the 
proportion will be, 

As the base, or sum of the segments, 
Is to the sum of the other two sides ; 
So is the difference of those sides, 
To the diff. of the segments of the base. 

Then take half this difference of the segments, and add 
it to the half sum, or the half base, for the greater segment, 
and subtract the same for the less segment. 

Ht nee, in each of the two right-angled triangles, there 
will be known two sides, and the right angle opposite to one 
of them ; consequently the other angles will be found by the 
first theorem. 

Demonsfr. By theor. 35, Geom. the rectangle of the sum 
and difference of the two sides, is equal to the rectangle of 
the sum and difference of the two segments. Therefore, by 
forming the sides of these rectangles into a proportion by 



THfOREV Iff. 



389 



tbeor. 76, Geometry, it will appear that the sums and dif- 
ferences are proportional as in this theorem. 

N. B. Before you commence a solution of an example to 
this case, ascertain whether the triangle be rigta-angled or 
not, by determining whether the square of the rongest side 
be equal or unequal to the sums of the squares of the other 
two. If equal, the exircple may be referred to the notes 
to theorem nr. 



EXAMPLE i. 

In the plane triangle abc, 
Given 4 345 yards 

*• ~ de8 J bc 174-07 
To find the angles. 

1. Geometrically. 



Draw the base ab = 345 by a scale of equal parts. With 
radius 232, and centre a, describe an arc ; and with radius, 
174, and centre b, describe another arc, cutting the former 
in c. Join ac, bc, and it is done. 

Then, by measuring the angles, , they will be found to be 
nearly as follows, vie. 

Z a 27% L b 37°i, and Lc 11 5° J. 



2. Arithmetically* 

Having let fall the perpendicular cp, it will be, 
As the base ab : ac + bc : : ac — bc : ap — bp, 
that is, as 345 : 406 07 : : 57 03 : 68-18 = ap — bp, 
its half is - 34 09 
the half base is 172-50 
the sum of these is 206-59 = ap. 
and their diff. is 138*41 = bp. 



Then, in the triangle apc, right-angled at p, 

As the side ac - - 232 - log. 2-3054880 

To sin. op. iiP . . 90 s . - 10*0000000 

So is the side ap . - 208-59 . 2-3151093 

To sin. op. L acp . - 62° 56' . 9-94t6213 

which taken from - 90 00 

leaves the Lh. 27 04 



990 



plaice naooHoxmr. 



Again, in the triangle bpc, right-angled at p 9 

As the aide bc . . 174-07 - log. 2-2407230 

Toain. op. £p - 00" . . 10-0000000 

8oiaaide bp - - 188-41 . . 2-1411675 

To ein op. L bcp - - 52° 40' . . 9-0004486 

which taken from - 90 00 

leaves the Lb 87 20 

Also the /acp 62° 66' 

added to Zbcp 52 40 

gives the whole £acb 115 36 

So that all the three angles are as follow, viz. 
the iLA27'4'; the Z.B3T20'; the Lc 115* 36'. 

The angles a and b may also easily be found by the ex- 

» AC BO 

pressions sec. a = — , sec. b = — , or the equivalent logs. 



EXAMPLE II. 



In the plane triangle abc, 



r ( ab 365 poles { L a 57* 12' 

ss-issa Has - 

To find the angles 



exampw in. 



r; w «n (ab120 (^a 57*27' 0' 

\ AC 112-65 Ans. < /1b 57 58 39 
the sides J BcU2 (^c 64 34 21 

To find the angles. 



The three foregoing theorems include all the eases of 
plane triangles, both right-angled and oblique. But there 
are other theorems suited to some particular forms of tri- 
angles (see vol. ii.), which are sometimes more expeditious 
in their use than the general ones ; one of which, as the case 
for which it serves so frequently occurs, may be here ex- 
plained. 



TRSOBBM IV. 



4 

an 



THEOREM IV. 

When a Triangle is Right-angled ; any of the unknown part* 
may be found by the following proportions : viz. 

As radius 

Is to either leg of the triangle ; 
So is tang, of its adjacent angle, 
To its opposite leg ; 
And so is secant of the same angle, 
To the hypothenuse. 

Demonstr. ab being the given leg, in the * 
right-angled triangle abc : with the centre 
a, and any assumed radius ad, describe an 
arc de, and draw dp perpendicular to ab, 
or parallel to bc. Then it is evident, from 
the definitions, that df is the tangent, and 
af the secant of the arc de, or of the 
angle a which is measured by that arc, to the radius ad. 

Then, because of the parallels bc, df, it will be - 

as ad : ab : : df : bc and : : af : ac, which is the same as 
the theorem is in words. 

Note. The radius is equal, either to the sine of 90°, or the 
tangent of 45 r ; and is expressed by 1, in a table of natural 
sines, or by 10 in the log. sines. 

EXAMPLE I. 




In the right-angled triangle abc, 

Given \ TAf 48' \ To A0 and B& 

1. Geometrically. 

Make ab = 162 equal parts, and the angle a=53* 7' 48' ; 
then raise the perpendicular bc, meeting ac in c. So shall 
ac measure 270, and bc 216. 



2. Arithmetically. 

As radius log. 10*0000000 

To leg ab 162 2-2005150 

So tang. 53° 7 48" 10-1240871 

TolegBc 216 - S-S&M&fllY 



S03 



PLANE raieONOMBTRY. 



So secant Lk - 53° T 48 7 - 10 2218477 
To hyp. ac 270 . 2-4318627 



EXAMPLE H. 




In the right-angled triangle abc, 

n . 5 the leg ab 180 . (ac 302-0146 

UlTcn ) the Z a 62 40' An8, I bc 348-2464 

To find the other two side*. 

Note. There is sometimes given another method for right- 
angled triangles, which is this : 

abc being such a triangle, make one 
leg ab radius ; that is, with centre a, 
and distance ab, describe an arc bf. 
Then it is evident that the other leg bc 
represents the tangent, and the hypo- 
thenuse ac the secant, of the arc bf, or 
of the angle a. 

In like manner, if the leg bc be made 
radius ; then the other leg ab will re- 
present the tangent, and the hypothenuse ac the secant, of 
the arc bo or angle c. 

But if the hypothenuse be made radius ; then each leg 
will represent the sine of its opposite angle ; namely, the leg 
ab the sine of the arc ae or angle c, and the leg bc the sine 
of the arc cd or angle a. 

Then the general rule for all these cases is this, namely, 
that the sides of the triangle bear to each other the same 
proportion as the parts which they represent. 

And this is called, Making every side radius. 

Note 2. When there are given two sides of a right-angled 
triangle, to find the third side ; this is to be found by the 
property of the squares of the sides, in theorem 34, Geom. 
vis* that the square of the hypothenuse, or longest side, is 
equal to both the squares of the two other sides together. 
Therefore, to find the longest side, add the squares of the 
two shorter sides together, and extract the square root of 
that sum ; but to find one of the shorter sides, subtract the 
one square from the other, and extract the root of the re- 
mainder. Or, when the hypothenuse, h, and either the base, 
b, or the perpendicular, p, are given : then half the sum of 
log. (a + p) and log. (u — p) — log. b ; and half the sum 
ofJog. (h -f b) and log. (h — b) = log. p. 



QflSFUL JPOftMGUB* 



When b and p are given, the following logarithmic ope- 
ration may sometimes be advantageously employed ; viz. 
Find n the number answering to the log. diff., 2 log. p — log. 
b ; and make b + * = x : then, £ (log. m + log. b) = log. h, 
the hypothenuse. 

The truth of this rule is evident : for, from the nature 



of logarithms. — = if; 



whence b + n = b + -t- = 



b 9 +p 9 

— - — =* m; and £ (log. n + log.fl) = J log. mb = | log. 
( B * + p>) = log. i/(B a + p*) = log. H. 

Or, stall more simply, find 10 + the diff. (log. p — log. b) 
in the log. tangents. The corresponding log. secant added 
to log. b == log. H. 

Note, also, as many right-angled triangles in integer num- 
bers as we please may be found by making 
»* + a 9 5= hypothenuse 
m 9 — n 9 = perpendicular 
2mn ■= base 
m and n being taken at pleasure, m greater than ft. 

Before we proceed to the subject of Heights and Distances 
" we shall give, 

A CONCISE INVESTIGATION OF SOME OF THE MOST USEFUL 
TRIGONOMETRICAL FORMULAE. 

Let ab, ac, ad, be three arches, such that bc = c*, aud 
o the centre. Join ao, oc, bo. Draw deq and oi per- 
pendicular, and bdc || to oa. Join Ma and bisect it by the 
radius on ; and draw ah || to bp. 
Then is ah = sin. ac 
oh = cos. ac; 

alSO DE = EQ = 8in. AD 

EK = oi = sin. ab 
ojk = sin. ad + sin. ab 
dk = sin. ad — sin. ab 

BI = IM = COS. AB 
OR => KI = COS. AD 
MK = COS. AB + COS. AD 
BK = COS. AB — COS. AD 

Because the angles at k are right angles : 
arc bd + arc.Mli = 180*, and arc dc + arc mn = Ifr 

.% MP = VH = OG = COS. DC = COS. \ ' V 

Vol, I 51 % V 




394 PLANE TRIGONOMETRY. 

also, because ao = 4(ab + ad) = £baq = angle aoc (at 
centre) = bdq (at circumf.) = bmq (on same arc) 
.% triangles aoh, bdk, qmk, are equiangular. 

Hence — 

I. oa : ah : : mq : qk ; 
that is, rad. : sin. ac : : 2 cos. bc : sin. ad + sin. ab 

II. ao : oh : : bd : dk ; 

or, rad. : cos. ac : : 2 sin. bc : sin. ad — siii. ab 

III. ao : oh : : on : kk ; 

or, rad. : cos. ac : : 2 cos. bc : cos. ab + cos. ad 

IV. ao : ah : : db : dk ; 

or, rad. : sin. ac : : 2 sin. bc : cos. ab — cos ad ; 

also, 

V. BK . KM = DK . KQ, that is (COS. AB COS. AD) 

(cos. ab+cos. ad) = (sin. ad— sin. ab) (sin. AD+sin. ab). 

By reducing the above four proportions into equations, 
making rad. = 1, we obtain two distinct classes of formula*, 
thus :— 

First Class, ac = a, cb = b ; then ad = a + 6, ab = a— b l 

1. sin. (a + b) + sin. (a — I) = 2 sin. a cos. b 

2. sin. (a + b) — sin. (a — b) = 2 cos. a sin. 6 

3. cos. (a — 6) + cos. (a + 6) = 2 cos. a cos. 6 

4. cos. (a — b) — cos. (a + b) = 2 sin. a sin. b 

Second Class, ad = a, ab = b ; then ac = £(a + 6), 
bc = £(a — 6). 

5. sin. a + sin. 6=2 sin. i(a + b) cos. £(a — b) 

6. sin. a — sin. 6 = 2 cos.£(a + b) sin. £(a — b) 

7. cos. 6 + cos. a ss2 cos.£(a + b) cos. \\a — 6) 

8. cos. b — cos. a = 2 sin. i(a + &) sin, £(a — 6) 

The first class is useful in transforming the products of 
sines into simple sines, and the contrary. 

The second facilitates the substitution of sums or. differ- 
ences of sines for the products, and the contrary. 

Taking the sum and the difference of equations 1 and 2, 
also of 3 and 4, remembering that sin. = cos. tan. we obtain 
the following : 

Third Class. 

9. sin. (a + b) = sin. a cos. b + sin. b cos. a 
= cos. a co*. h (tan. a + tan. b) 



USEFUL FOBKiflLfi. 305 

10. sin. (a — 6) = sin. a cos. b — sin. b cos. a 

= cos. £i cos. 6 (tan. a — tan. b) 

11. cos. (a + b) = cos. a cos. & — - sin. a sin. b 

■s cos. a cos. 6 (1 — tan. a tan. b) 

12. cos. (ci — b) = cos. a cos. 6 + sin. a sin. b. 

= cos. a cos. b (1 +tan. a tan. b). 

From these, making a = 6, we readily obtain the ex- 
pressions for sines and cosines of double arcs ; also dividing 
equation 9 by 11, and equation 10 by 12, we obtain ex- 
pressions for the tangents of a + b and a — b. Thus we 
have : — 

Fourth Class. 

13. sin. 2a = 2 sin. a cos. a = 2 ooe.'a tan. a 

14. cos. 2a = cos. 2 a — sin. 2 a = cos. 2 a (1 — tan. 2 a) 

sin. , . . v / • t\ tan. a + tan. 6 

15. (a + 6) = tan. (a + 6) = - — — z 

cos. v ' 1— tan. a tan. 6 

sin. , , v /ix tan. o — tan. b 

(a — b) = tan. (a — ft) = r . 

cos. v ' ' 1 + tan. a tan. 6 . 



16 



_ 4 2 tan. a 
17. tan. 2a = 



18. cot. 2a = 



1— tan. 2 a 
1— tan. 2 a 
2 tan. a 



Substituting in the second class, 

for sin. i(a+b), cos. i(a + b) tan. $(a+b), 
.and for sin. i(a— b), cos. £(a — b) tan. we have : — 

Fifth Class. 

19. cos. &+cos. a=2 cos. £(a+&)cos. 4(a— 6). — Seeequa. 7. 

20. cos. 6 — cos.a = tan. $(a+b) tan. £(a— 5) 2 cos. |(a+&) 
cos. £(a-&) = ton- tan. i(a— &)(cos.&+cos. a) 

21. sin. a+sin. 6 = tan. ±(a+&} 2 cos. i( a +&) cos. i(a-5) 

= tan. £(a+6) (cos. a+cos. b) 

22. sin. a -sin. b = tan. ^(a-6) 2 cos. J( a +&) cos. }(*—*) 

= tan. |(a— 6) (cos. a+cos. 6) 

^ sin. a+sin. 5 tan. i{a+b) 

23. r — r = - — 77 — 7x - from 21 and 22. 

sin. a— sm. o tan. J(a— 6) 

^ sin. a+sin. 6 w . r rt « 

24. r = tan. Ua + 6) : from 21. 

cos.a+cos. o 

^ sin. a— sin. i A , , r M 

25. ; r = tan. Ua —6) : from 22. 

cos.a+cos. b ,v 



300 o* nktolm 



Excunplts for Exercise. 

1. Demonstrate that in any right-angled plane triangle tiur 
following properties obtain : viz. 

oerp. base 

(1.) £_£.=£tan* ang. at base. (2.) =tan. ang. at vertex* 

* 'base * v 'perp. ^ 

(3.) j^EL'sssin. ang. at base. (4.) — sin. ang. at vertex* 

(5. ) « sec. ang. at base. (6.) =sec. ang. at vertex* 

2. Demonstrate that tan< a + sec. a = tan. (45* + i^)* 

l*4"tan * a 

3. Demonstrate that sec. 2a = - — - — Vi> tnat 

1 — tanv * 

l+tan. 8 A sec. 9 a 



cosec, 



2a = 



2 tan. a 2 tan. a 

4. Given Axy = ay 8 + ns* ; to find x and y the sine and 
cosine of an arc. 

5. Demonstrate that of any arc, tan. a — sin. 2 = tan. 2 sin. 2 - 

6. Demonstrate that if the tan. of an arc be = ^/n, the 

sine of the same arc is = y/ n 



n+1" 



OF HEIGHTS AND DISTANCES, &c. 

6y the mensuration and protraction of lines and angles, 
are determined the lengths, heights, depths, and distances of 
bodies or objects. 

Accessible lines are measured by applying to them some 
certain measure a number of times, as an inch, or a foot, or 
yard. But inaccessible lines must be measured by taking 
angles, or by such-like method, drawn from the principles of 
geometry. 

When instruments are used for taking the magnitude of the 
angles in degrees, the lines are then calculated by trigonome- 
try : in the other methods, the lines are calculated fr#m the 
principle of similar triangles, or some other geometrical 
property, without regard to the measure of the angles* 



AlfD MBTANCEf . 



397 



Angles of elevation, or of depression, are usually taken 
either with a theodolite, or with a quadrant, divided into de- 
grees and minutes, and furnished with a plummet suspended 
from the centre, and two open sights fixed on one of the radii, 
or else with telescopic sights. 



To lake an Angle of Altitude and Depression with tike 
Quadrant. 




Let a be any object, as the sun, 
inoon, or a star, or the top of a 
tower, or hill, or other eminence : 
and let it be required to find the 
measure of the angle abc, which a 
line drawn from the object makes 
above the horizontal line bc. 

Place the centre of the quadrant 
in the angular point, and move it 
round there as a oentre, till with one eye at n, the other 
being shut, you perceive the object a through the sights ; 
then will the arc gh of the quadrant, cut off by the plumb- 
line, bii, be the measure of the angle abc as required. 

The angle abc of depression of 
any object a, below the horizontal 
line bc, is taken in the same manner; 
except that here the eye is applied to 
the centre, and the measure of the 
angle is the arc gh, on the other 
side of the plumb-liner 

The following examples are to be constructed and calcu- 
lated by the rules of Trigonometry. 



B 




EXAMPLE t. 



Having measured a distance of 200 feet, in a direct hori- 
zontal line, from the bottom of a steeple, the angle of eleva- 
tion of its top, taken at that distance, was found to be 47° 30' ; 
hence it is required to find the height of the steeple. 



Construction. 

Draw an indefinite line ; on which set off ac a 200 equal 
parts, for the measured distance. Erect the indefinite per- 
pendicular ab ; and draw cb so as to make the an&la,$ «a 



OF HEIGHTS 



47° SO 7 , tho angle of elevation ; and it is done. Then ab, 
measured on the scale of equal parts, is nearly 218J. 

Calculation. 



As radius - 10-0000000 

To ac 200 - - 2-3010300 

So tang. L c 47° 80' 10-0379475 

To ab 218-26 required 2-3380775 




Or, by the nat. tangents, we have ac X tan. bca = 
200 X 1-091308 = 218-2616 = ab. 

EXAMPLE It. 

What was the perpendicular height of a cloud, or of a 
balloon, when its angles of elevation were 35° and 64°, as 
taken by two observers, at the same time, both on the same 
side of it, and in the same vertical plane ; the distance be* 
tween them being half a mile or 880 yards ? And what was 
its distance from the said two observers ? 

Construction. 

Draw an indefinite ground line, on which set off the 
given distance ab = 880 ; then a and b are the places of 
the observers. Make the angle a = 35°, and the* angle 
B = 64° ; then the intersection of the lines at c will be the 
place of the balloon : whence the perpendicular cd, being let 
fall, will be its perpendicular height. Then, by measure- 
ment are found the distances and height nearly as follow , 
viz. ac 1631, bc 1041, dc 936. 



C 



Calculation. 

First, from L b 64' 
take L a 35 
leaves L. acb 29 




— ' — / ; 

A 33 D 
Then in the triangle abc, 
As sin. Zacb 29° ... 0-6855712 

To op. side ab 880 - 2-9444827 

So sin. Lk 35° - 9-7585913 

To op. side bc 1041-125 3-0175028 



AND DISTANCES. 



399 



As sin. ^acb 29° - . - 9-6855712 

To op. side ab 880 ... 2-9444827 

So sin. Z.B 1 16° or 64° - 9-9536602 

To op. side ac 1631-442 - - 3-2125717 



And in the triangle bcd, ( 

As sin. L d 90° - - - 10-0000000 

To op. side bc 1041-125 - . 3-0175028 

So sin. L b 64° - - - 0-9536602 

To op. side cd 935-757 - - 2-9711630 



EXAMPLE in. 



Having to find the height of an obelisk standing on the 
top of a declivity, I first measured from its bottom a distance 
of 40 feet, and there found the angle, formed by the oblique 
plane and a line imagined to go to the top of the obelisk, 
41 ° ; but after measuring on in the same direction 60 feet 
farther, the like angle was only 23° 45'. What then was 
the height of the obelisk ? » 



Construction. 

Draw an indefinite line for the sloping plane or declivity, 
in which assume any point a for the bottom of the obelisk, 
from which set off the distance ac = 40, and again cd = 60 
equal parts. Then make the angle c = 41°, and the angle 
d = 23° 45' ; and the point b where the two lines meet will 
be the top of the obelisk. Therefore ab joined, wilr be its 
height. — Draw also the horizontal line de perp. to ab. 

Calculation. 

From the L c 41° 00' 
take the /d 23 45 
loaves the £ dbc 17 15 



Then in the triangle dbc, 

As sin. /.dbc 17° 15' 
To op. side dc 60 
So sin. L d 23 45 
To op. side cb 8M88 




9-4720856 

1-7781513. 

9-6050320 

i-9\vwn 



400 



OF 1IBEGHTS 



And in the triangle abc, 



As sum of sides cb, ca 
To diff. of sides cb, ca 
So tang. }(a + b) 
To tang. ] (a — b) 



121*488 - 20845333 

41-488 . 1-6179225 

60* 3C - 10-4272623 

42 24} - 9*9606516 



the diff. of these is £cba 27 5} 
the sum is L cab 111 54} 



Lastly, as sin. £cba 27° 5'} . - 9*0582842 

To op. side ca 40 . 1*6020600 

So sin. Lc - 41° 0' - - 9*8169429 

To op. side ab 57-623 - - 1*7607187 



Also the L ade ^= bac — 90 r = 21° 54'}. 



EXAMPLE IV. 



Wanting to know the distance between two inaccessible 
trees, or other objects, from tho top of a tower 120 feet high, 
which lay in the same right line with the two objects, I took 
the angles formed by the perpendicular wall and lines con* 
ceived to be drawn from the top of the tower to the bottom 
of each tree, and found them to be 33° and 64°}. What is 
the distance between the two objects ? 

Construcliou. 

Draw the indefinite ground line 
bd, and perpendicular to it ba = 
120 equal parts. Then draw the 
two lines ac, ad, making the two 
angles bac, bad, equal to the 
given measures 33' and 64°£. So 
shall c and d be the places of the 
two objects. 

Calculation. ^ St - 

First, in the right-angled triangle abi , 

As radius - 10-0000000 

To ab . 190 ... 4i 2 0791812 

Sotaug. /I bac 33 l » - - . 9-8125174 

Tob<; . 77-«» - - 1-8916980 




ASD DISTANCES. 



401 



Then in the right-angled triangl* abd, 

As radius 10 'OOOOOOO 

To ab • - 120 . 2 0701812 

So tang. L bad - 64° 3a - - 10*3215089 

To bd • 251-586 - . 2*4006851 
From which take bc 77*929 
leaves the dist. cd 173*056, as required. 

Or thus, by the natural tangents, 

From nat. tan. dab - 64° 30' = 2*0905436 

Take nat. tan. cab - . 33 =0 6494076 



Difference - • - 1*4471360 

If drawu into ab - 120 



The result gives cd - = 173-05632 



EXAMPLE V. 

Being on the side of a river, and wanting to know the 
distance to a house which was seen on the other side, I mea- 
sured 200 yards in a straight line by the side of the river ; 
and then, at each end of this line of distance, took the hori- 
zontal angle formed between the house and the other end of 
the line ; which angles were, the one of them 68° 2 , and the 
other 73° 15'. What were the distances from each end to 
the house ? 



Construction. 

Draw the line ab = 200 equal parts. 
Then draw ac so as to make the angle 
a = 68 2', and bc to make the angle 
b=73' 15 . So shall the point c be the 
place of the house required. 




The calculation, which i* left for the student's exercise, 
gives ac = 30619, bc = 296 54. 

Exam. vi. From the edge of a ditch, of 36 feet wide, 
surrounding a fort, having taken the angle of elevation of 
the top of the wall, it was found to be 62*40' : required the 
height of the wall, and the length of a ladder to reach from 
my station to the'top of it ? At t height of wall 69-64, 

Ans " {ladder, 7%Afe*\. 

Vol. I. 52 



Or KXZOHTt 



Exam, vii. Required the length of a shoar, which being 
to strut 11 feet from the upright of a building, will support 
a jamb 23 feet 10 inches trooMhe ground 7. 

Ans. 26 feet 3 inches. 

Exam. viii. A ladder, 40 feet long, can be so placed, that 
it shall reach a window 33 feet from the ground, on one side 
of the street ; and by turning it over, without moving the 
foot out of its place, it will do the same by a window 21 feet 
high, on the other side : required the breadth of the street T 

Ans. 56-649 feet. 

Exam. ix. A maypole, whose top was broken off by a 
blast of wind, struck the ground at 15 feet distance from the 
foot of the pole : what was the height of the whole maypole, 
supposing the broken piece to measure 39 feet in length ? 

Ans. 75 feet 

Ex ax. x. At 170 feet distance from the bottom of a tower, 
the angle of its elevation was found to be 52° 30* : required 
the altitude of the tower ? Ans. 221-55 feet 

Exam. xi. From the top of a toner, by the sea-side, of 
143 feet high, it was observed that the angle of depression 
of a ship's bottom, then at anchor, measured 35° ; what was 
the ship's distance from the bottom of the wall ? 

Ans. 204-22 feet. 

Exam. xii. What is the perpendicular height of a hifl ; 
its angle of elevation, taken at the bottom of it, being 46°, 
and 200 yards farther off, on a level with the bottom, the 
angle was 31 ° ? Ans. 286-28 yards. 

Exam. xiii. Wanting to know the height of an inacces- 
sible tower ; at the least distance from it, on the same hori- 
zontal plane, I took its angle of elevation equal to 58° ; then 
going 300 feet directly from it, found the angle there to be 
only 32° : required its height, and my distance from it at the 
first station ? k i height 307-53 

An8# } distance 198-15 

Exam. xiv. Being on a horizontal plane, and wanting to 
know the height of a tower placed on the top of an inacces- 
sible hill ; I took the angle of elevation of the top of the hill 
40", and of the top of tha tower 51° ; the measuring in a 
line directly from it to the distance of 200 feet farther, I 
found the angle to the top of the tower to be 33° 45'. What 
is the height of the tower ? Ans. 93-33148 feet 

Exam. xv. From a window near the bottom of a house, 
which seemed to be on a level with the bottom of a steeple, 



AND MSTANCM. 



I took the angle of elevation of the top of the steeple equal 
40° ; then from another window, 18 feet directly above the 
former, the like angle was 37* 10' : required the height and 
distance of the steeple. A t height 21(1 44 

An } distance 250-79 

Exam. xvi. Ranting to know the height of, and my 
distance frofn, an object on the other side of a river, which 
appeared to be on a level with the place where 1 stood, 
close by the side of the river ; and not having room to 
measure backward, in the same line, because of the im. 
mediate rise of the bank, I placed a mark where I stood, 
and measured in a direction from the object, up the ascend, 
ing ground, to the distance of 264 feet, where it was evi. 
dent that I wan above the level of the top of the object ; 
there the angles of depression were found to be, viz. of the 
mark left at the river's side 42°, of the bottom of the object 
27°, and of its top 19*. Required the height of the object, 
and the distance of the mark from its bottom ? 

a \ height 57-26 
Ans " I distance 150-56 

Exam. xvii. If the height of the mountain called the 
Peak of Teneriffe be 2£ miles, as it is very nearly, and the 
angle taken at the top of it, as formed between a plumb-line 
and a line conceived to touch the earth in the horizon, or 
farthest visible point, be 88° 2 ; it is required from these 
measures to determine the magnitude of the whole earth, 
and the utmost distance that cpn be seen on its surface from 
the top of the mountain, supposing the form of the earth to 
be perfectly globular ? 

. (dist. 185-943 1 ji 
in, -Jdiam. 7918! mLCS - 

Exam. xvui. Two ships of war, intending to cannonade 
a fort, are, by the shallowness of the water, kept so fur from 
it, that they suspect their guns cannot reach it with effect. 
In order therefore to measure the distance, they separate 
from each other a quarter of a mile, or 440 yards ; then each 
ship observes and measuses the angle which the other ship 
and the fort subtends, which angles are 83" 45' and 85° 15'- 
What is the distance between each ship and the fort ? 

. $2292 26 yards. 
An *' {2296-05 

Exam. xix. Wanting to know the breadth of a river, I 
measured a base of 500 yards in a straight line close by one 
aide of it ; and at each end of this line I found the anglea 
Mbtended by||be other end and a tree, close to the haok** 



404 



OF HEIGHT! AMD DISTAXCBS. 



the other side of the river, to be 53° and 79* VH* What was 
the perpendicular breadth of the river t 

Ana. 520-48 yards. 

Exam. xx. Wanting to know the extent of a piece of water, 
or distance between two headlands ; I measured from each 
of them to a certain point inland, and found^the two distances 
to be 786 yards and 840 yards ; also the horizontal angle 
subtended between these two lines was 55° 40. What was 
the distance required ? Ans. 741 -2 yards. 

Exam. xxi. A point of land was observed, by a ship at 
sea, to bear east-by .south ; and after sailing north-east 12 
miles, it was found to bear south.east-by-east. It is required 
to determine the place of that headland, and the ship's dis- 
tance from it at the last observation ? Ans. 26-0728 miles. 

Exam. xxii. Wanting to know the distance between a 
house and a mill, which were seen at a distance on the other 
side of a river, I measured a base line along the side where 
I was, of 000 yards, and at each end of it took the angles 
subtended by the other end and the house and mill, which 
were as follow, viz. at one end the angles were 58° 2tf and 
95° 20', and at the other end the like angles were 53° 3a and 
98° 45'. What then was the distance between die house and 
mill ? Ans. 959-5866 yards. 

Exam, xxiii. Wanting to know my distance from an in. 
accessible object o, on the other side of a river ; and having 
no instrument for taking angles, but only a chain or cord for 
measuring distances ; from each of two stations, a and b, 
which were taken at 500 yards asunder, I measured in a di- 
rect like from the object o 100 yards, viz. ac and bd each 
equal to 100 yards ; also the diagonal ad measured 550 
yards, and the diagonal bc 560. \\ hat was the distance of 
the object o from each station a and b ? 

ao 536-81 



n ' \ bo 500-47 

Exam. xxiv. In a garrison besieged are jjjiree remarkable 
objects, a, b, c, the distances of which frrtm each other are 
discovered by means of a map of the place, and are as fol- 
low, viz. ab 266], ac 530, bc 327 £ yards. Now, having to 
erect a battery against it, at a certain spot without th(? place, 
and being desirous to know whether my distances from the 
three objects be such, as that they may from thence be bat- 
tered with effect, I took, with an instrument, the horizontal 
angles subtended by these objects from the station s, and 
found them to be as follow, viz. the angle asb 13° 30, and 
the angle B8c28°5tf. Refuted the three distances, a a, 



MX2ISUBATIOX OF ttAlfBS. 



406 



ib, sc ; the object b being situated nearest me, and between 
the two others a and c. i sa 757*14 

Ans. < sb 587*10 
(sc 655-80 

Exam. xxv. Required the same as in the last example, 
when the object a is the farthest from my station, but still seen 
between the two others as to angular position, and those an* 
gles being thus, the angle asb 88 45', and bsc 2*4 80 , also 
the three distances, ab 600, ac 800, bc 400 vaids ? 

(•a 710-3 
Ans. {sb 1041*85 
(sc 93414 

Exam. xxvi. If db in the figure at pa. 378 represent a por- 
tion of the earth's surface, and n the point where the level* 
ling instrument is placed, then lb will be the difference 
between the true and the apparent level ; and you nre re- 
quired to demonstrate that, for distances not exceeding 5 or 
6 miles measured on the earth's surface, bl, estimated in 
feet, is equal to } of the square of bd, taken in miles. 



MENSURATION OF PLANES. 



The Area of any plane figure, is the measure of the space 
contained within its extremes or bounds ; without any re- 
gard to thickness. 

This area, or the content of the plane figure, is estimated 
by the number of little square* that may be contained in it ; 
the side of those little measuring squares being an inch, or a 
foot, or a yard, or any other fixed quantity. And hence tl e 
area or content is said to be so many square inches, or square 
feet, or square yards, &c. 

Thus, if the figure to be measured be 
the rectangle abcd, and the little square 
b, whose side is one inch, be the mea- 
suring unit proposed : then as often as 
the said little square is contained in the 
rectangle, so many square inches the 
rectangle is said to contain, which in 
the present case is 12. 



D 


4 




c 


J 
















A. 


B 

m 



406 



PROBLEM I. 

7b find the Area of any Parallelogram ; whether it be a Square, 
a Rectangle, a Rhombus, or a Rhomboid. 

Multiply the length by the perpendicular breadth, or 
height, and the product will be the area*. 

EXAMPLES. 

Ex* 1 • To find Ihe area of a parallelogram, the length being 
12*25, and breadth or height 8*5. 

12*25 length 
8*5 breadth 



C125 
0800 



104*125 area. 



Ex. 2. To find the area of a squire, whose side is 35*25 
chains. Ans. 124 acres, 1 rood, 1 perch. 

Ex. 3. To find the area of a rectangular board, whose 
length is 12} feet, and breadth 9 inches. Ans. 9f feet 

Ex. 4. To find the content of a piece of land, in form of a 
rhombus, its length being 6-20 chains, and perpendicular 
breadth 5*45. Ans. 3 acres, 1 rood, 20 perches. 

Ex. 5. To find the number of square yards of painting in 
a rhomboid, whose length is 37 feel, and height 5 feet S 
inches. Ans. 21^ square yards. 



* The truth of this rule is proved in the Geom. theor. 81. cor. 9. 

The same is otherw.e proved ihu*: Let thts foregoing met Angle he 
the figure piopnsed ; and let the length end lirenrith he dvided inlmeft* 
nil purls, each equal to lite linear measuring imit t being here 4 forth* 
length, and H for th»» hreadtli ; and let the opposite points of division bt 
connected hy right line*.— Then it is evident that three Hues divide tbt 
rectangle into a n um her of little -squares, each eniial to the sqoait 
measuring unit i ; and further, that the n urn her of these little squares, 
or the area of the figure, is eq-ial to the tuimher of linear measuring spin 
in the length, related at olten as there are linear measuring units in the 
hrearlth, or height; thai h, equal to the length drawn into the height; 
whirh here is 4 X 3 or I '2. 

And it is proved (Geom theor. 25, cor. 2), that any oblique parallelo- 
gram if equal to a rectangle, of equal length and perjtendirular breadth. 
Therefore the rale U general for all |*railelograiofl whatever. 



0F NAHM. 



PBOB1 EM II. 



To find the Area of a Triangle. 

Rulk im Multiply the base by the perpendicular height, 
and take half the product for the area*. Or, multiply the 
one of these dimensions by half the other. 



EXAMPLES. 

Ex. 1. To find the area of a triangle, whose J>ase is 625* 
and perpendicular height 520 links ? 

Here 6^5 X 260 = 162500 square links, 

or equal 1 acre, 2 roods, 20 perches, the answer. 
Ex. 2. How many square yards contains the triangle, 
whose base is 40, and perpendicular 30 feet ? 

A ns. 66} square yards. 
Ex. 3. To find the number of square yards in a triangle, 
whose base is 49 feet, and height 25J feet. 

Ans. 684}, or 68*7361. 
Ex. 4. To find the area of a triangle, whose base is 18 
feet 4 inches, and height 11 feet 10 inches ? 

Ans. 108 feet, 5} inches. 
Rule n. When two sides and their contained angle are 

S'ven : Multiply the two given sides together, and take half 
eir product : Then say, as radius is to the sine of the given 
angle, so is that half product, to the area of the triangle. 

Or, multiply that half product by the natural sine of the 
■aid angle, for the areaf. 



* The truth of this rule is evident, because any triangle it the half of 
a parallelogram of equal base and altitude, by Geom. theor. 26, 

t For. let ab, ac, be the two given sides, in- Q 
eluding the given angle a. Now £ ab X cp is the 
area, by the first rule, cr being the perpendicular. 
But by trigonometry, as sin. JL r, or radius : 
ac : : sin. ^ a : cr, which is therefore = ac X 
era. Z. a, taking radius = 1. Therefore the area 
£a* X cp is = Iab X ac X tin. £ a, lo radius I ; 
•r, as radios : sis. 4 a x : Jab x ac : the area. 




408 



X I2UU RATI05 



Ex. 1. What hi the area of a triangle, whose two sides are 
80 and 40, and their contained angle 28* 57' ? 

By Natural Number*. By Lcgariikm. 
First, J X 40 X 30 = 600, 

then, 1 : 600 : : -484046 sin. 28- 57' log. 0-684887 

600 2 778151 



Anawer 290-4276 the area answ. to 2 468088 

Ex. 3* How many square yards contains the triangle of 
which one angle is 45% and its containing sides 25 and 21} 
feet ? Ans. 20-86047. 

Ruue in. When the three sides are given : Add all the 
three sides together, and take half that sum. Next, subtract 
each aide severally from the said half sum, obtaining three 
remainders. Then multiply the said half sum and those three 
remainders all together, and extract the square root of the 
last product, for the area of the triangle f. 



t For, let 6 denote the base a a of the triangle abc (see the last fig.), 
also a the side ac, and e the side ac. Then, by Ibeor. 3, 

Trigon. as 6 : « + e : : « — ei **~ - = ap — pb the diff. of the seg> 
ments; 

- , , oa — ec bb 4- aa — cc 4l _ 
theref. J6-f — — = — = the segment ap ; 

hence V^ao 1 — ap*) = the perp. cr, that is, 



CP. 



66 + «« — «» 
V(as-(— ^ ))= - 

Wb->~a< +2b€i — »i-f g«y — f i _ 
V 466 " 

Bot Jab X cp is the area, that is, 

16 X cp = V- J — jg 1 

_ vi ma-bb-cc + 2bc^-na + bb + cc + V>c ) (A) 

. ( « +J±f y ii+ *±! x — *±- e v tt* r •% 

— v\ 2 2 2 x 2 

= V { t X (f — «) X (« — A) X (• — c) J, which is toe role, 
denotes half the sum of the three sides. 

The espressioa marked (A), if we put s = 6 + e, and d for 6 — f, to 
equivalent to J V \ — d — di) j ; which, in most cases, furnishes 
a satire commodious rale for practice than rule in. here gfren ; espe- 
Ssatyv t£ the computet have a table of squares at band, 



OF PLANES. 



409 



If the aides of the triangle be large, then add the logs, of 
the half sum, and of the three remainders together, and half 
their sum will be the log. of the area. 

Ex. 1. To find the area of the triangle whose three sides 
are 20, 30, 40. 

20 45 45 45 

30 20 30 40 

40 — — — 

25 1st rem. 15 2d rem. 5 3d rem. 

2) 00 — — — 
45 half sum 

Then 45 X 25 X 15 X 5 = 84375, 
The root of which is 290-4737, the area* 

Ex. 2. How many square yards of plastering are in a 
triangle, whose sides, are 30, 40, 50 feet ? Ans. 66}. 

Ex. 3. How many acres, &c. contains the triangle, whose 
aides are 2569, 4900, 5025 links ? 

Acs. 61 acres, 1 rood, 39 perches. 



PROBLEM III. 

To find the Area of a Trapezoid. 

Add together the two parallel sides ; then multiply their 
sum by the perpendicular breadth, or the distance between 
them ; and take half the product for the area. By Geora. 
theor. 29. 

Ex. 1. In a trapezoid, the parallel sides are 750 and 1225, 
and the perpendicular distance between them 1540 links ; to 
find the area. 
1225 
750 

1975 X 770 = 152075 square links —-15 arc. 33 perc. 

Ex. 2. How many square feet are contained in the plank, 
whose length is 12 feet 6 inches, the breadth at the greater 
end 15 inches, and at the less end 11 inches ? 

Ans. 13}} feet. 

Ex. 3. In measuring along one side ab of a quadrangular 
field, that side, and the two perpendiculars let fall on it from 
the two opposite corners, measured as foUqw \ 
content. 

Vol. I 53 



410 



WCWITftATTOXf 



ap = 110 links 



Ads. 4 acres, 1 rood, 5*792 perches. 



au= 745 

AB = 1110 

cp = 352 
pa- 595 




PROBLEM IT* 



To find the Area of any Trapezium. 



Divide the trapezium into two triangles by a diagonal ; 
then find the areas of these triangles, and add them together. 

Or thus, let fall two perpendiculars on the diagonal from 
the other two opposite angles ; then add these two perpen- 
diculars together, and multiply that sum by the diagonal, 
taking half the product for the area of the trapezium. 

Ex. 1. To find the area of the trapezium, whose diagonal 
is 42, and the two perpendiculars on it 16 and 18. 

Here 16+18= 34, its half is 17. 
Then 42 X 17 = 714 the area. 

Ex. 2. How many square yards of paving are in the tra* 
pezium, whose diagonal is 65 fe<;t, and the two perpendicu- 



Ex. 3. In the quadrangular field abcd, on account of ob- 
structions there could only be taken the following measures, 
viz. the two sides bc 265 and ad 220 yards, the diagonal 
ac 378, and the two distances of the perpendiculars from the 
ends of the diagonal, namely, ae 160, and cf 70 yards. Re- 
quired the construction of the figure, and the area in acres, 
when 4840 square yards make an acre ? 



To find the Area of an Irregular Polygon. 

Draw diagonals dividing the proposed polygon into tra» 
peziums and triangles. Then find the areas of all these se- 
parately, and add them together for the content of the whole 
polygon. 



lars let fall on it 28 and 33£ feet ? 



Ans. 222y 9 yards. 



Ans. 17 acres, 2 roods, 21 perches. 



problem v. 




or PLAinct. 



abcdefga, in which are given the following diagonals and 
perpendiculars: namely, 



To find the Area of a Regular Polygon. 

Rule i. Multiply the perimeter of the polygon, or sum 
of its sides, by the perpendicular drawn from its centre on 
one of its sides, and take half the product for the area 3 ". 

Ex. 1. To find the area of a regular pentagon, ench side 
being 25 feet, and the perpendicular from the centre on each 
side 17-2047737. 

Here 25 X 5 = 125 is the perimeter. 
And 17-2047737 X 125 = 2150 5967125. 
Its half 1075*208356 is the area sought^ 

Rule n. Square the side . of the polygon ; then multiply 
that square by the tabular area, or multiplier set against its 
name in the following table, and the product will be the 
areaf. 



• This is only in effect resolving the polygon into as ninny equal tri- 
angles as it has sides, hy drawing lines from ihe cent re to all the angles; 
then finding their areas, and adding them all together. 

t This rule is founded on the property, thitt like polvgon«, being simi- 
lar figures, are to one another as the Mpiares of their Ifke sides : w»»icb 
is proved in the Geora. theor. 89. Now, the multipliers iu the table, 
«re the areas of the respective polygons to the side 1. Whence tLe 
rule is manifest. 



ac 55 



Ads. 1878}. 



fd 52 
€0 44 
em 13 

BR 18 

oo 12 
sp 8 
d?23 




PROBLEM VI. 



413 



XXRtCBATIOIf 



No. of 
Side*. 


Names. 


3~ 


Trigon or triangle 


4 


Tetragon or square 


5 


Pentagon 


6 


Hexagon 


7 


Heptagon 


8 


Octagon 


9 


Nonagon 


10 


Decagon 


11 


Undccagon 


12 


Dodecagon 



Areas, or 


Radius ot cir- 


Multipliers. 


cum. cin !e. 


0-4330127 


05778503 


1-0000000 


0-7071068 


1-7204774 


0-8506508 


2-5980762 


1-0000000 


: -6339124 


1 1523824 


4-828*271 


1-3065628 


6-1818242 


1-4619022 


7-6942088 


1-6180340 


9-3656399 


1-7747324 


11 1961524 


1-9318517 



Exam. Taking here the same example as before, namely, 
a pentagon, whose side is 25 feet. 9 
Then 25 2 being = 625, 
And the tabular area 1 7204774 ; 
Theref. 1-7204774 X 625 = 1075-298375, as before. 
Ex. 2. To find the area of the trigon or equilateral tri- 
angle, whose side is 20. Ans. 173-20508. 
Ex. &• To find the area of the hexagon whose side is 20. 

Ans. 1039-23048. 
Ex. 4. To find the area of an octagon whose side is 20. 

Ans. 1931-37084. 
Ex. 5. To find the area of a decagon whose side is 20. 

Ans. 3077-68352. 

Note. If ab * 1, and n the number of sides of the poly- 
gon, then area of polygon — n times area of the triangle 
abc = ft ad . dc — n ad tan. c?ad (to rad. ad) = in tan. cad 

180° 

= in cot. ai d = Jn cot. . The ra- 
ft 

dius of the circumscribing circle, to side 1, 
is evidently equal fo J si*c. cad. Mulri-' 
plying, therefore, the radius of the table by 
the numeral value of any proposed side, the 
product is the radius of a circle in which 
that polygon may be inscribed ; and from which it may 
readily be constructed. 




OF PLANES. 



41S 



PROBLEM VII. 

To find the Diameter and Circumference of any Circle, the 
one from the other. 

This may be done nearly, by either of the four following 
proportion*, 

viz. As 7 is to 22, so is the diameter to the circumference. 
Or, As 1 is to 3-1410, so is the diam. to the circumf. 
Or, As 113 to 355, so is the diam. to the circumf.* 
And, as 1 : '318309 : : the circumf. : the diameter. 




* For let abcd benny circle, whose centre is 
z, and lei ab, bc, he any two equal arcs. Draw 
the several chords as in the figure, and join be; 
also draw the diameter da. which produce to r, 
till bp be equal to the chord bd. 

Then the two isosceles triangles df.r, dbf, are 
equiangular, because they have the angle at d 
common; consequently de : db : : dr : df. But 
the two triangles afb, dcb, are identical, or equal 
in all re«pects, because they have the angle f — 
the an^le biic, being each equal to the angle 
adb, these being subtended by the equal arcs ab, 
bc; also the exterior angle fab of the quadrnn- . 
gle abcd, is equal to the opposite interior angle 
at c ; and the two triangles have also the side bf = the side bd ; there- 
fore the side af is also equal in the side dc. Hence the proportion above, 
viz. de : db : : db : df = da -J- af, becomes dk : db : : db : 2de -|- dc. 
Then, by taking the rectangles of the extremes and means, it is db-> = 
2db : -|-dk . DC. 

Now, if the radius de be taken = 1, this exprestUafv becomes db 3 = 
2-|- i>c. and hence the root db =. V(2-f dc). That is, if the measure 
of the supplemental chord of any arc be increased by the number 2, the. 
sq tare root of the sum will be the supplemental chord of half that arc 
N »w, to apply this to the calculation of the circumference of the cir- 
cle, let the arc ac be taken equal to \ of the circumference, and be suc- 
cessively bisected by the above theorem : thus the chord ac of -J- of the 
1 circumference, is the side of the inscribed regular hexagon, and is there- 
, fore equ.il to the radius ae or I : hence, in the right-angled triangle a CD, 
it « ill be dc ^ V(ad - ac ) = V( ,i — 1) = V3 = 1-7320508U76, the 
supplemental chord of X of the periphery. 

Then, by the foregoing theorem, by . always bisecting the aces, end 
adding 2 to the last square root, there will be found the supplemental 
chords of the l2th, the 24th, the 48th, the 96th, &c, parts of the peri- 
phery ; thus. 



V 3 -7320 108076 : 
^/ 3-93 1 85 165*5 : 
^3 9828897227 : 
^3-9957178165 : 
v/3-9989291743 : 
v/3-9997322757 : 
v/3-9999330678 : 
V3-9999832669 - 



1-9318516525 
1-9828097227 
1-9957178465 

: 1-9939291743 
1*9997322757 

; 1-9999330G78 
1-9999832669 



° 1 


r ^ i 
















US 
— o 


T¥ 


J8 

c 


3 JX ' 




*u 













414 



MBIfSITBATIOlT 



1 . . ' V 

Ex. 1. To.findtfie circumference of the circle whose dia- 
meter is 20. 

By the first rale, as 7 : 22 : : 20 : 62j, the answer. 

Ex. 2. If the circumference of the earth be 24877-4 miles, 
what is its diameter ? 

By the 2d rule, as 3-1416 : 1 : : 24877-4 : 7918-7 nearly 
the diameter. 

By the 3d rule, as 355 : 113 : : 24877-4 : 7918-7 nearly. 
i>; u.e4lhrule,asl : -318309 : : 24877 4 : 7918-7 nearly. 



PROBLEM VIII. 

To find the Length of any Arc of a Circle. 

Multiply the decimal -017453 by the degrees in the 
given arc, and that product by the radius of the circle, for the 
length of the arc*. 

Since then it is found that 3*9999832669 is the square of fhe supple- 
mental chord of the 1536th part of the periphery, let this number be 
Inker- from 4, which is the square of the diameter, nnd the remainder 
0-0000167331 will tie the square of the chord of the said 1 546th part of 
the periphery, and consequently the root \/O 00O0 167331 = 0-OO40906 U2 
is the length of that chord; this number then being multiplied by 1536 
gives 6 2S3I7BS for the perimeter of a regular polygon of 1536 sides in* 
scribed in the circle; which, as the sides of the polygon nearly coin- 
cide with the circumference of the circle, must also express the length 
of the circumference itself, very nearly. 

But now, to show how near this determination is to 
the truth, let a^p = 0-0040906 1 12 represent one side 
of such a regular polygon of 1536 side*, and «rt a side 
of another similar polygon described about the circle; 
and from the centre e let the perpendicular zqR be 
drawn, bisecting a p and st in q nud r. Then >ince 
a<i is = 4 ap - 00i» »4Vi05rt. and ka = I. therefore 
»q'i — ka'— ai^ 1 = -9991)958167, and consequently its 
root gives ei = -99991)791)8-1 ; then because of the pa- 
rallels ap, st, it is p.q : kr : : ap : st : : as the whole in- 
scribed perimeter: to the circumscribed one. that is, 
as -9999979084 : 1 : : 6 283176* : 6 28*1920 the perime- 
ter of the circumscribed polygon. Now, the circum- 
ference of the circle being greater than the perimeter of the inner poly- 
gon, but less limn that of the outer, it must consequently b« greater 
than 6-283178& 

but less than 6-2831940, 

and must therefore be nearly equal to £ their sum, <>r 6-2831854, which 
% in fact is true to the last figure, which fhould be a 3, instead of the 4. 
Hence the circumference heing 6-2831854 w hen the diameter is 2, it 
will be the half of that, or 3 1415^27. when the diameter is I, to which 
the ratio in the rule, viz. 1 to 3- 14 16. is very near. Also the first ratio 
in the rule, 7 to 22 or 1 to 3f 3- 1428 &c. is another near a pproii na- 
tion. But the third ratio, 1 13 to 355, — 1 to 3- 14 15929, is the nearest. 

* It having been found, in the demonstration of the foregoing problem, 
that when the radius of a cArc\e\t \,\V& ItugLh of the whole circumie- 




OF FLANKS. 



415 



Ex. 1. To find the length of an arc of 30 degrees, the 
radius being feet. Ans. 4-71231. 

Ex. 2. To find the length of an arc of 12 10', or 12 J, the 
radius being 10 feet. Ans. 2*1234. 

PROBLEM IX. 

To find the Area of a Circle*. 

Rule i. Multiply half the circumference by half the 
diameter. Or multiply the whole circumference by the whole 
diameter, and take \ of the product. 

Rule ft. Square the diameter, and multiply that square 
by the decimal a 7854, for the area. 

Rule in. Square the circumference, and multiply that 
square by the decimal -07958. 

Ex. 1. To find the area of a circle whose diameter is 1Q> 
and its circumference 31 '416. 

By Rule 1. By Rule 2. By Rule 3. 

31-416 -7854 31-416 

10 10 3 = 100 31-416 



4)314 16 ~* 986-965 

78-54 70 34 -07958 



78-54 



So that the area is 78*54 by all the .three rules. 



rence is 6 2831854, which consists of 360 degrees ; therefore as 3*0" : 
6-2831854: : 1 : 017453, kc. the length of the arc of I degree. Hence the 
decimal -017453 multiplied hy any number of degrees, will give the length 
of the arc of those degrees. And because the circumferences and arcs are 
in proportion as the diameters, or as the radii of the circles, therefore as 
the radius 1 is to any other radius r, so is the length of the arc above 
mentioned, to '017453 X degrees in the arc X r> which is the length of 
that arc, as in the rule. 

* The first rule is proved in the Geom. theor. 94. 

And the 9d and 3 1 rules are deduced from the first rule, in this man- 
ner.— By that rule, de -i- 4 is the area, when d denote* the diameter, and 
t the circumference. But, by prob. 7, e is = 3- 14 IrW ; therefore ihe said 
area de -r 4, becomes (tX*14lfi<*-r>4 = -78544', which gives the 2d 
rale. — Also, by tho snme prob. 7, d is = c -f- 3 1416; therefore again 
the same first area dr. -f- 4, becomes (c 3*1416) X (c 4) = e- 
12-5664, which is = c* X 07958, by taking the reciprocal of 12 5664, or 
changing ibat divisor into the inuitipler -07958 ; which gives the 3d rule. 

CoroL Hence the areas of different circles are in proportion to one 
another, as the square of their diameters or as the square oC Umax ^maNtr 
fereoces- as before proved in the Geom. Iheor. 



416 



MEKST7IUTI0N 



Ex. 2. To find the area of a circle, whose diameter is 7, 
and circumference 22. Ans. 38$. 

Ex. 3. How many square yards are in a circle, whose 
diameter is 3 J feet ? Ans. 1 -069. 

Ex. 4. To find the area of a circle, whose circumference 
is 12 feet. Ans. 11-4595. 



PROBLEM X. 



To find the Area of a Circular Ring, or of the Space included 
between the Circumferences of two Circles ; tlie "bite being 
contained within the other. 

Take the difference between the areas of the two circles, 
as found by the last problem, for the area of the ring. — Or, 
which is the same thing, subtract the square of the less dia- 
meter from the square of the greater, and multiply their dif- 
ference by -7854. — Or, lastly, multiply the sum of the dia- 
meters by the difference of the same, and that product by 
•7854 ; which is still the same thing, because the product of 
the sum and difference of any two quantities, is equal to the 
difference of their squares. 

Ex. 1. The diameters of two concentric circles being 10 
and 6, required the area of the ring contained between their 
circumferences. 

Here 10 + 6 = 16 the sum, and 10 — 6 = 4 the diff. 
Therefore -7854 X 10 X 4 ^= -7854 X 64 - 50 2656, 
the area. 

Ex 2. What is the area of the ring, the diameters of whose 
bounding circles are 10 and 20 ? Ans. 235 62. 

PROBLEM XI. 

To find the Area of the Sector of a Circle. 

Rule i. Multiply the radius, or half the diameter, by half 
the arc of the sector, for the area. Or, multiply the whole 
diumeter by the whole arc of the sector, and take £ of the 
product. The reason of which is the same as for the first 
rule to problem 9, for the whole circle. 

Rule ii. Compute the area of the whole circle : then say, 
as 360 is to the degrees in the «rc of the sector, so is the area 
of the whole circle, to X\ve <rt toa «&&&t. 



OP PLAICES. 



41? 



This is evident, because the sector is proportional to the 
length of the arc, or to the degrees contained in it. 

Ex. 1. To find the area of a circular sector, whose arc 
contains 18 degrees ; the diameter being 3 feet. 
1. By the first Rule. 
First, 3-1416 X 3 = 0-4248, the circumference. 
And 360 : 18 : : 0-4848 : -47124, the length of the arc. 
Then -47124 X 3 ^4= 1-41372 + 4 « -35343, the area. 

2; By the 2d Rule. 
First, -7854 X 3" = 7-0686, the area pf the whole circle. 
Then, as 360 : 18 : : 7 0686 : -35343, the area of the 
sector. 

Ex. 2. To find the area of a sector, whose radius is 10, 
and arc 20. Ans. 100. 

Ex. 3. Required the area of a sector, whose radius is 25, 
and its arc containing 147° 20'. Ans. 804-3086. 



PROBLEM XII. 



To find the Area of a Segment of a Circle. 

Rule I. Find the area of the sector having the same arc 
with the segment, by the last problem. 

Find also the area of the triangle, formed by the chord of 
the segment and the two radii of the sector. 

Then add these two together for the answer, when the 
segment is greater than a semicircle : or subtract them when 
it is less than a semicircle. — As is evident by inspection. 

Ex. 1. To find the area of the segment acbda, its chord 
▲b being 12, and the radius ae or ce 10. 

First, As af : sin. L. d 00° : : ad : sin. 
86° 52' J = 36*87 degrees, the degrees in the 
aec or arc ac. Their double, 73*74, are 
the degrees in the whole arc acb. \ J 

Now -7854 X 400 = 314-16, the area of VX-/ 

the whole circle. . / ' 

Therefore 360° : 73-74 : : 31416 : ^3504, area of the 
sector acre. 

Again, ^/(ab» — ad 2 ) = ^(100 — 36) = ^/64 = 8 = de. 
Theref. ad X de =* 6 X 8 = 48, the area of the triangle 

AEB. 

Hence sector acbe — triangle akb = 16-3504, area of 
seg. acbda. 

Rule n. Multiply the square of the radius of the circle 
by either half the difference of the arc acb and its sine (Jrtdrt 
Vol. I. 54 



418 



MHIUBJkUOM 



to the radius 1), or half the sum of the; arc and its sine, Ac* 
cording as the segment is less or greater than a semicircle ; 
the product will be the area. 

The reason of this rule, also* is evident from an mspcetioo 
of the diagram. 

Exam, the same as before, in which am « IS, ax a* 10 ; 
and from the former computation are acb = 78* 44'f- 

Then, by Button's Mathematical Tables, pp. 840, ccc 
arc 73° 44'f, to radius 1 =* 1-2870059 
sin. 78° 44'}, to radius 1 = -9600010 



2) -3270009 



•1685084 



whence, -1635034 X 10* = 16-35034, the area of thcseg- 

ment ; very nearly as before. 

Ex. 2. What is the area of the segment, whose height is 
18, and diameter of the circle 50 ? Ans. 636-375. 

Ex. 3. Required the area of the segment whose chord is 
16, the diameter being* 20 ? Ans. 44-728, or 269-432. 

PROBLEM XIII. 

To measure long Irregular Figures. 

Take or measure the breadth at both ends, and at several 
places, at equal distances. Then add together all these ia* 
termediate breadths and half the two extremes, which sum 
multiply by the length, and divide by the number of parts, 
for the area*. 



5 {jlj^t? 



* This rule is made out as follows: 
— Let abcd be the irregular piece ; 
having the several breadths ad, bf, oh, 
iK y bc, at the equal distances ak, eg, 
oi, u. Let the several breadths in or- . 
der be denoted by the corresponding A E & I B 
letters a, b t c, d. c, and the whole length 

ab by l\ then compute the areas of the parti into which the Store » 
divided by the perpendiculars, as so many trapezoids, by prob. 3, aid 
add them all together. Thus, the sum of the parts is, 

= x it + tjf x y+ - x ii+Sf • x * 

= «• + & + c + 4 + ft X V = <JR -V » + « + <*) V. 



or rums. 



JVbfe. If the perpendiculars or breadths be not at equal 
distances, compute all the parts separately, as so many trape- 
zoids, and add them all together for the whole area. 

Or else, add all the perpendicular breadths together, and 
divide their sum by the number of them for the mean 
breadth, to multiply by the length ; which will give the whole 
area, not fiur from the truth. 

Ex. 1. The breadths of an irregular figure, at five equi. 
distant places, being 8-2, 7-4, 9-2, 10-2, 8-6 ; and the whole 
length 39 ; required the area. 

8*2 35-2 sum. 

8-6 39 

2 ) 16-8 sum of the extremes. 8168 

8-4 mean of the extremes. 1066 



7-4 4 ) 1372-8 

9-2 343-2 An*. 
10-2 

36-2 sum. 

Ex. 2. The length of an irregular figure being 84, and the 
breadths at six equidistant places 17*4,20-6, 14-2, 16-5, 20*1, 
24-4 ; what is the area? Ana. 1550-64. 



FBOBLBM XIV. 

To find Ae Area of an Ellipsis or Oval 

Multiply the longest diameter, or axis, by the shortest ; 
then multiply the product by the decimal -7854, for the area. 
As appears from cor. 2, theor. 3, of the Ellipse, in the Conic 
Sections. 

Ex. 1. Required the area of an ellipse whose two axes are 
70 and 50. Ans. 2748-9. 

Ex. 2. To find the area of the oval whose two axes are 24 
and 18. Ans. 339-2928. 



which if the whole area, agreeing with the rule : m being the arithmeti- 
cal memo between the extremes, or half the sum of them both, and 4 
the number of the parti. And the same for any other aombet «C ?ute 
whatever. 



480 XBMlUBAnOIC 



NKOBLXH XV. 

To find ike Area of an Elliptic Segment. 

Find the area of a corresponding circular segment, having 
the same height and the same vertical axis or diameter. Then 
say, as the said vertical axis is to the other axis, parallel to 
the segment's base, so is the area of the circular segment 
before found, to the area of the elliptic segment sought. 
This rule also comes from cor. 2, theor. 3, of the Ellipse. 

Ex. 1. To find the area of the elliptic segment, whose 
height is 20, the vertical axis being 70, and the parallel axis 
60. ins. 648 13. 

Ex. 2. Required the area of an elliptic segment, cut off 
parallel to the shorter axis ; the height being 10, and the two 
axes 25 and 35. Ans. 162-03. 

Ex. 3. To find the area of the elliptic segment, cut off pa- 
rallel to the longer axis ; the height being 5, and the axes 25 
and 35. Ans. 97-8425. 

PROBLEM XVI. 

To find the Area of a Parabola, or iU Segment. 

MultiAy the base by the perpendicular height; then 
take two-thirds of the product for the area. As is proved in 
theorem 17 of the Parabola, in the Conic Sections. 

Ex. 1. To find the area of a parabola ; the height being 2, 
and the base 12. 

Here 2 X 12 = 24. Then % of 24 = 16, is the area. 

Ex. 2. Required the area of the parabola, whose height is 
10, and its base 16. Ans. 106f. 



MENSURATION OF SOLIDS. 

By the Mensuration of Solids are determined tho spaces 
included by contiguous surfaces ; and the sum of the met* 
sures of these including surfaces is the whole surface or su. 
perficies of the body. 

The measure of a solid, is called its solidity, capacity, or 
content. 



o» solids, 431 

Solids are measured by cubes, whose sides are inches, or 
feet, or yards, dec. And hence the solidity of a body is said 
to be so many cubic inches, feet, yards, dec. as will fill its 
capacity or space, or another of an equal magnitude. 

The least solid measure is the cubic inch, other cubes 
being taken from it according to the proportion in the fol- 
lowing table, which is formed by cubing the linear pro* 
portions. 

Table of Cubic or Solid Measures. 

1728 cubic inches make ' 1 cubic foot 

27 cubic feet - 1 cubic yard 

166| cubic yards - 1 cubic pole 

04000 cubic poles . 1 cubic furlong 

512 cubic furlongs 1 cubic mile. 

PROBLEM 1. 

To find the Superficies of a Prism or Cylinder. 

Multiply the perimeter of one end of the prism by the 
length or height of the solid, and the product will be the sur- 
face of all its sides. To which add also the area of the two 
ends of the prism, when required*. 

Or, compute the areas of all the sides and ends separately, 
and add them all together. 

Ex. 1. To find the surface of a cube, the length of each 
side be ng 20 feet. Ans. 2400 feet. 

Ex. 2. To find the whole surface of a triangular prism, 
whose length is , 20 feet, and each side of its end or base 18 
inches. Ans. 91*048 feet. 

Ex. 3. To find the convex surface of a round prism, or 
cylinder, whose length is 20 feet, and the diameter of its base 
is 2 feet. Ans. 125-664. 

Ex. 4. What must be paid for lining a rectangular cistern 
with lead, at 3d. a pound weight, the thickness of the lead 
being such as to weigh 71b. for each square foot of surface ; 
the inside dimensions of the cistern being as follow, viz. the 
length 3 feet 2 inches, the breadth 2 feet 8 inches, and depth 
2 feet 6 inches? Ans. 32. 5s. 9jd. 



* The truth of this will easily Appear, by considering that the sides of 
any prism are parallelograms, whose common length is the same as Ibe 
length of the solid, and their breadths taken all together make op the 
perimeter of the ends of the same. 

And the rale is evidently the tame for the tatta* oAiqjYk&ki. 



MJUOTJRATSOIf 



PBOBLBMII. 



To find the Surface of a Pyramid or Cone. 

Multiply the perimeter of the bate by the riant height, 
or length of the side, and half the product will evidently be 
the surface of the sides, or the sum of the areas of all the tri- 
angles which form it. To which add the area of the end or 
base, if requisite. 

Ex. 1. What is the upright surface of a triangular pyra- 
mid, the slant height being 20 feet, and each side of the base 
3 feet? Ana. 90 feet. 

Ex. 2. Required the convex surface of a cone, or circular 
pyramid, the slant height being 50 feet, and the diameter of 



To find the Surface of the Frustum of a Pyramid or Cone, 
being the lower pari, when the top is cut off by a plane pa- 
rallel to the base. 

Add together the perimeters of the two ends, and multiply 
their sum by the slant height, taking half the product for the 
answer. — As is evident, because the sides of the solid are 
trapezoids, having the opposite sides parallel. 

Ex. 1. How many square feet are in the surface of the 
frustum of a square pyramid, whose slant height is 10 feet ; 
also each side of the base or greater end being 3 feet 4 inches, 
and each side of the less end 2 feet 2 inches ? Ans. 110 feet. 

Ex. 2. To find the convex surface of the frustum of a 
cone, the slant height of the frustum being 12± feet, and 
the circumferences of the two ends 6 and 8*4 feet. 



To find the Solid Cohtent of any Prism or Cylinder. 

Find the area of the base, or end, whatever the figure of 
it may be ; and multiply it by the length of the prism or cylin- 
der, for the solid content*. 




Ans. 607-511. 



PROBLEM m. 



Ans. 90 feet. 



PROBLEM IV. 



• This rale appears from the Geom. tbeor. 110, cor. 2. The fMBe » 
more particularly shown at follows: Let the annexed rectangular parti 



OF MUM. 



Note. For a cube, take the cube of its aide by multiplying 
this twice by itaelf ; and for a parallelopipedon, multiply the 
length, breadth, and depth all together, for the content. 

Ex. 1. To find the solid content of a cube, whose side is 
24 inches. Ans. 18834. 

Ex. 2. How many cubic feet are in a block of marble, its 
length being 3 feet 2 inches, breadth 2 feet 8 inches, and 
thickness 2 feet 6 inches ? Ans. 21}. 

Ex. 3. How many gallons of water will the cistern con- 
tain, whose dimensions are the same as in the btst example, 
when 277} cubic inches are contained in one gallon ? 

Ans. 131-53. 

Ex. 4. Required the solidity of a triangular prism, whose 
length is 10 feet, and the three sides of its triangular end or 
base are 3, 4, 5 feet. Ans. 60. 

Ex. 5. Required the content of a round pillar, or cylinder, 
whose length is 20 feet, and circumference 5 feet 6 inches. 

Ans. 48*1450 feet. 



PROBLEM V. 

To find the Content of any Pyramid or Cone. 

Find the area of the base, and multiply that area by the 
perpendicular height ; then take £ of the product for the 
content*. 



lelopipedon be the solid to be measured, 
and the cube r the solid measuring unit, its 
side being 1 inch, or 1 foot, Jtc. ; also, let 
the length and breadth of the base abcd, 
and also the height ah, be each divided 
into spaces equal to the length of the base 
of the cube p, namely, here 3 in the length 
and 2 in the breadth, matins; 3 times 2 or 6 
squares in the base ac, each equal to the 
base of the cube p. Hence it is manifest 
that the parallelopipedon will contain the 
cube p, as many times as the base ac con- 
tains the base of the cube, repeated as often 
as the height ah contains the height of the cube. That is, the contest 
of any parallelopipedon is found, by multiplying the- are a of the base by 
the altitude of that solid. 

And because all prisms and cylinders are equal to parallelopfoedom 
of equal bases and altitudes, by Geom. theor. 108, it follows that tne rale 
b general for all such solids, whatever the figure of the base may be. 

* This rule follows from that of the prism, because any pyramid if -l 
of a prism of equal base and altitude ; by Geom. theot AvVout. 




XBSrsVBATIOZf 



Ex. 1. Required the solidity of a square pyramid, each 
aide of its base being 30, and its perpendicular height 25. 

Ads. 7500. 

Ex. 2. To find the content of a triangular pyramid, whose 
perpendicular height is 30, and each side of the base 8. 

Ans. 38-971143. 

Ex. 3. To find the content of a triangular pyramid, its 
height being 14 feet 6 inches, and the three sides of its base 
5, 6, 7 feet. Aim. 71*0352. 

Ex. 4. What is the content of a pentagonal pyramid, its 
height being 12 feet, and each side of its base 2 feet ? 

Ans. 27-5276. 

Ex. 5. What is the content of the hexagonal pyramid, 
whose height is 6*4 feet, and each side of its base 6 inches 1 

Ans. 1-38564 feet. 

Ex. 6. Required the content of a cone, its height being 
lty feet, and the circumference of its base 9 feet, 

Ans. 22-56093. 



PROBLEM VI. 



To find the Solidity of a Frustum of a Cone or Pyramid. 

Add into one sum, the areas of the two ends, and the 
mean proportional between them : and take \ of that sum 
for a mean area ; which being multiplied by tne perpen- 
dicular height, or length of the frustum, will give its con- 
tent*. 



* Let abcd be any pyramid, of which bcdofe is 
a frustum. And put a 2 for the area of I he base rcd, 
b 9 the area of the top, rfo, A the height ih of the 
frustum, and c the height ai of the top part above 
it. Then c -\- A = ah is the height of the whole 
pyramid. 

Hence, by the last prob. (c-fM is the content 
of the whole pyramid abcd. and {b'c the content 
of the top part akfo ; therefore the difference - - - 
|a*(c + A) — J6 ? c is the content of the frustum 
bcdgfjc. fiat the quantity c being no dimension of 
the frustum, it mutt be expelled from this formula, by substituting iti 
value, found in the following manner. By Geom. theor. 1 12, « J : £» : : 
(c + A)« : c\ or « : b : : e + A : c, hence (ficom. th. 69) a — b : b : : A : c, 

6A cA 

and« — b : a : : A : c 4- A ; hence therefore c = —j- and c -f- A = — . ; 




OP SOLIDS. 



435 



Note. This general rule may be otherwise expressed, as 
follows, when the ends of the frustum arc circles or regular 
polygons. In this latter case, square one side of each poly, 
gon, and also multiply the one side by the other ; add all 
these three products together ; then multiply their, sum by 
the tabular area proper to the polygon, and take one-third of 
the product for the mean area, to be multiplied by the length, 
to give the solid content. And in the case of the frustum- 
of a cone, the ends being circles, square the diameter or the 
circumference of each end, and also multiply the same two 
dimensions together ; then take the sum of the three pro- 
ducts, and multiply it by the proper tabular number, viz. by 
•7854 when the diameters are used, or by '07958 in using 
the circumferences ; then taking one-third of the product, to 
multiply by the length, for the content. 

Ex. 1. To find the number of solid feet in a piece of tim- 
ber, whose bases are squares, each side of the greater end 
being 15 inches, and each side of the less end 6 inches ; also, 
the length or the perpendicular altitude 24 feet. Ans. 19 J. 

Ex. 2. Required the content of a pentagonal frustum, 
whose height is 5 feet, each side of the base 18 inches ; and 
each side of the top or less end 6 inches. Ans. 9-31925 feet. 

Ex. 3. To find the content of a conic frustum, the altitude 
being 18, the greatest diameter 6, and the least diameter 4. 

Ans. 527-788P. 

Ex. 4. What is the solidity of the frustum of a cone, the 
' altitude being 25, also the circumference at the greater end 
being 20, and at the less end 10 ? Ans. 464-216. 

Ex. 5. If a cask, which is two equal conic frustums joined 
together at the bases, have its bung diameter 28 inches, the 
bead diameter 20 inches, and length 40 inches ; how many 
gallons of wine will it hold ? Ans. 79-0613. 



then these values of c and e +h being substituted for them in the ex- 
pression for the content of the frustum gives that content 

= *a> - X fl -^j = V* X^=p = lhx(* + ab + b*); which 

to the rale above given ; ak being the mean between a 9 and a*. 

Note. If d, d be the corresponding linear dimensions of ihe ends, 6 
their difference, m the appropriate multiplier, h the height of the frus- 
tum, then is the content = JmA (3ixf + 6) ; which is a coavenieot prac- 
tical expression. 



Vol- I. 55 ' 



426 



MKIfttJBATIOIf 



PROBLEM VII. 

To find the Surface of a Sphere, or any Segment. 

Rule i. Multiply the circumference of the sphere by 
its diameter, and the product will be the whole surface of 
it*. 

Rulk u. Square the diameter and multiply that square by 
3*1416, for the surface. 

Rulk hi. Square the circumference ; then either multi- 
ply that square by the decimal '31 83, or divide it by 8-1416, 
for the surface. 

Note. For the surface of a segment or frustum, multiply 



* These rules come from the following theorems for the surface of a 
sphere, viz. That the said surface is equal to the curve surface of ^cir- 
cumscribing cylinder; or Hint it is equal to 4 great circles of the same 
sphere, nr of l he ?ame diameter; which are thus proved. 

Let abcd be a cylinder, circumscribing the 
Sphere eioh ; the former genertitr d by the J± 
rotation of the rectangle fbch about the axis 
or dimeter fii ; and the. latter by the rota- 
tion of the semicircle fgh about the same di- 
ameter fh. Draw two lines kl, mi», perpen- 
dicnlar to the axis, intercepting the paits i.n, 
op, of the cylinder and sphere ; then will the 
ring or cylindric surface generated by the ro- 
tation of i n, be equal to the ring or fpherical 
surface generated by the arc op. For, first, ™ 
suppose the parallels ki. and ms to be indefi- 
nitely near together; drawing 10, and also oo. parallel to iji. Then the 
t\\ o triangles iko. oqp, being equiangular, it is, as op : oq or LR : : so or 
ki- : ko :: circumference described by kl : circiimf. described by so; 
therefore the rectangle op.xcirciimf. of ko is equal to the rectangle l*X 
circumf. of ki.; that is, the ring described by op on the sphere, iseqoal 
to the ring described by l.n on the rylindt-r. 

And as this is every where the case, therefore the sums of any corres- 
ponding number of tln-se are also equal ; that is. the whole surface of 
the sphere, described by the whole semicircle fgh, is equal to the whole 
curve surface of the cylinder, described by the height bc ; as well as the 
surface of any > cement described by fo, equal to the surface of the cor- 
responding segment described by el. 

Corol. I. Hence 'he smface of the sphere is equal to 4 of its great cir- 
cles or equal to the circumference efgh, or of dc, multiplied by tat 
height bc. or by the diameter fh. 

Corol. 2. Hence also, the surface of any such part, as a segment or 
frustum, or zone, is equal to the same circumference of the sphere, mal- 
tiplied by the height of the said part. And consequently such spheri- 
cal curve surfaces are to one auotber in the same proportion as their 
altitudes. 





OP SOLIDS. 



437 



the whole circumference of the sphere by the height of the 
part required. 

Ex. 1. Required the convex superficies of a sphere, whose 
diameter is 7, and circumference 22. Ans. 154. 

Ex. 2. Required the superficies of a globe, whose diameter 
is 24 inches. Ans. 1809-5616. 

Ex. 3. Required the area of the whole surface of the 
earth y its diameter being 71)57 J miles, and its circumferenco 
25000 miles. Ans. 198943750 sq. miles. 

Ex. 4. The axis of a sphere being 42 inches, what is the 
convex superficies of the segment whose height is 9 inches ? 

Ans. 1187-5248 inches. 

Ex. 5. Required the convex surface of a spherical zone, 
whose breadth or height is 2 feet, and cut from n sphere of 
12» feet diameter. Ans. 78-54 feet. 



PROBLEM Vin. 

7b find the Solidity of a Sphere or Globe. 

Rule i. Multiply the surface by the diameter, and take 
| of the product for the content*. Or, which is the same 
thing, multiply the square of the diameter by the circum. 
ference, and take } of the product. 

Rulr H. Take the cube of the diameter, and multiply it 
by the decimal '5236, for the content. 

Rule in. Cube the circumference, and multiply by 
•01688. 

Ex. 1. To find the solid content of the globe of the earth, 
supposing its circumference to be 25000 miles. 

Ans. 263750000000 miles. 

Ex. 2. Supposing that a cubic inch of cast iron weighs 
•269 of a lb. avoird. what is the weight of an iron bull of 
5-04 inches diameter ? 



* For. nnt d = the diameter, c = the circumference, end t — the 
surface of the sphere, or of its circumscribing cylinder ; also, a — the 
number 3-1416. 

Then, \t is =. the base of the cylinder, or one gre«t circle of the 
sphere ; and d is the height of the cylinder : then-fore \ds is lite cniifent 
of the cylinder. But { of the cylinder is the sphere, liy th. 117, Gcom. 
that it, § of $d$ % or^ds is the sphere ; which is the first rule. 

Again, because the surface « is = ad*; iherefore ±*h _ \nd* — 5MM\ 
U the content, as in the 2d rule. Also, d being = c a, therefore 
t*d* = a* = •01088, the 3d rule for the content* 



488 



MJBH S UKATfON 



PROBLEM K. 



7b find the Solid Content of a Spherical Segment. 

* Rule i. From 3 tiroes the diameter of the sphere 
take double the height of the segment; then multiply the re* 
mainder by the square of the height, and the product by the 
decimal -5230, for the content. 

Rule ii. To 3 tiroes the square of the radius of the 
segment's base, add the square of its height ; then multiply 
the sum by the height, and the product by '5236, for the 
content. 

Ex. 1 . To find the content of a spherical segment, of 2 
feet in height, cut from a sphere of 8 feet diameter. 

Ans. 41-888. 

















• By corol. 3, of theor. 117, Geom. it ap- 
pears thnt the spheiic segment prit, is equal 
to the difference between the cylinder ablo, 
and the conic frustum abmq. 

But, putting d = ab or fh the diameter of £ 
the sphere or cylinder, /* = fk the height of 
the segment, r = pk the radius of its lm«c, 
and a = 3*1416; ihen the content of the 
tone abi is = \ad* X Jfi — : and by 

the similar cones abi, qmi, as ri 1 : si 1 : : 

^ad x : ^ad* X (r-^j—) 1 = the cone <*m ; therefore the cone abi — 

the cone qsti = &ad* — J^i X (*-""- V = iad* — b*dk* + J** 1 

is = the conic frustum of abm<*. 
And \atl h is — the cylinder ablo. 

Then the difference of these tw o is jmfti — \ah^ = \ah* X (3rf —21), 
for the spheric segment pfn ; whirh is ihe first rule. 
Again, because pk* = fk X kh (cor. to theor. 87, Geom.) or r 1 = k 

(d- A), therefore = /*, and 2d - <J> = ^+«=^L+i; 

3ri -4- A* 

which being substituted in the former rule, it becomes \ah* X — ^ — 

= +*A X ( *r'-M ), which is (he 2d rule. 

Note. By subtracting a segn ent from n half sphere, or from another 
tegmeiit, the content o( my faiiitam or tone may be found. 



or MUM. 429 

*Ex. 2. What is the solidity of the segment of a sphera, its 
height being 9, and the diameter of its base 20 ? 

Ana. 1795*4344. 



Arte. The general rules for measuring the most useful 
figu-es having been now delivered, we may proceed to apply 
them to the several practical uses in life, as follows. 



[ 430 ] 



LAND SURVEYING. 



SECTION I. 

DESCRIPTION AND USE OF THE INSTRUMENTS. 

1. OF THE CHAIN. 

Land is measured with a chain, called Gunter's Chain, 
from its inventor, the length of which is 4 poles, or 22 yards, 
or 66 feet. It consists of 100 equal links ; and the length 
of each link is therefore j'fo of a yard, or of a foot, or 
7 -92 inches. 

Land is estimated in acres, roods, and perches. An acre 
is equal to 10 square chains, or as much as 10 chains in length 
and 1 chain in hreadth. Or, in yards, it is 220 X 22 = 4840 
square vards. Or, in poh»s, it is 40 X 4 = 160 square poles. 
Or, in 'links, it is 1000 X 100 — 100000 square links: 
these being all the same quantity. 

Also, an acre is divided into 4 parts called roods, and a 
rood into 40 parts called perches, which are square poles, or 
the square of a pole of yards long, or the square of £ of a 
chain, or of 25 links, which is 625 square links. So that the 
divisions of land measure will be thus : 

625 sq. links — 1 pole or perch 
40 perches = 1 rood 
4 roods = 1 acre. 

The lengths of lines measured with a chain, are best set 
down in links as integers, every chain in length being 100 
links ; and not in chains and decimals. Therefore, after the 
content is found, it will he in square links; then cut off five 
of the figures on the right h;ind for decimals, and the rest will 
be acres. These decimals are then multiplied by 4 for roods, 
and the decimals of these again by 40 for perches. 



LAUD SUB YE TING. 



481 



Exam. Suppose tho length of a rectangular piece of ground 
be ?£2 Lnks, and its breadth 385 ; to fi.ul the area in acres, 
roods, aud perches. 

792 3 04020 

385 4 



•K680 

6336 40 
2376 



3 (49*0 



7-8*200 



Ans. 3 aores, roods, 7 perches. 

2. OF THE PLAIN TABLE. 



This instrument consists of a plain rectangular board, of 
any convenient size : the centre of which, when used, is fixed 
by means of screws to a three-legged stand, having a hall 
and socket, or other joint, at the top, by means of which, 
when the legs are fixed on the ground, the table is inclined 
in any direction. 

To the table belong various parts, as follow. 

1. A frame of wood, made to fit round its edges, and to 
be taken off, for the convenience of putting a sheet of paper 
on the table. One side of this frame is usually divided into 
equal parts, for drawing lines across the table, parallel or 
perpendicular to the sides ; and the other side of the frame 
is divided into 360 degrees, to a centre in the middle of the 
table ; by means of which the table may be used as a theo- 
dolite, dec. 

2. A magnetic needle arid compass, either screwed into the 
side of the table, or fixed beneath its centre, to point out the 
directions, and to be a check on the sights. 

3. An index, which is a brass two-foot scale, with either 
a small telescope, or open sights set perpendicularly on the 
ends. These sights and one edge of the index are in the same 
plane, and that is called the fiducial edge of the index. 

To use this instrument, take a sheet of paper which will 
cover it, and wet it to make it expand ; then spread it flat on 
the table, pressing down the frame on the edges, to stretch 
it and keep it fixed there ; and when the paper is become 
dry, it will, by contracting again, stretch itself smooth and 
flat from any cramps and unevenness. On this paper is to 
be drawn the plan or form of the thing measured. 

Thus, begin at any proper part of the ground, and make 
a point on a convenient part of the paper or tabic, to reore. 
sent that place on the ground ; then fix. vu \tax ^oveft. 



482 



LAND 



leg of (he compasses, or a fine steel pin, and apply to it the 
fiducial edge of the index, moving it round till through the 
aights you perceive some remarkable object, as the corner of 
a field, dtc. ; and from the station-point draw a line with the 
point of the compasses along the fiducial edge of the index, 
which is called setting or taking the object : than set another 
object or corner, and draw its line ; do the same by another ; 
and so on, till as many objects are taken as may be thought 
fit* Then measure from the station towards as many of toe 
objects as may be necessary, but not more, taking the requi- 
site offsets to corners or crooks in the hedges, laying the 
measures down on their respective lines on the table. Then 
at any convenient place measured to, fix the table in the 
same position, and set the objects which appear from that 
place ; and so on, as before. And thus continue till the 
work is finished, measuring such lines only as are necessary, 
and determining as many as may be by intersecting lines of 
direction drawn from different stations. 

Of shifting the Paper on the Plain Table. 

When one paper is full, and there is occasion for more, 
draw a line in any manner through the farthest point of the 
last station line, to which the work can be conveniently laid 
down ; then take the sheet off the table, and fix another 
on, drawing a line over it, in a part the most convenient for 
the rest of the work ; then fold or cut the old sheet by the 
line drawn on it, applying the edge to the line on the new 
sheet, and, as they lie in that position, continue the last sta- 
tion line on the new paper, placing on it the rest of the 
measure, beginning at where thu old sheet left off. And so 
on from sheet to sheet. 

When the work is done, and you would fasten all the 
sheets together into one piece, or rough plan, the aforesaid 
lines are to be accurately joined together, in the same man* 
ncr a8 when the lines were transferred from the old sheets 
to the new ones. But it is to be noted, that if the said join, 
ing lines, on the old and new sheets, have not the same in- 
clination to the side of the table, the needle will not point to 
the original degree when the table is rectified ; and if the 
needle be required to respect still the same degree of the 
compass, the easiest way of drawing the line in the same 
position, is to draw them both parallel to the same sides of 
the table, by means of the equal divisions marked on the 
other two sides. 



SUItVBYlNQ. 



m 



3. OF THE THEODOLITE. 

The theodolite is a brazen circular ring, divided into 360 
degrees, &c. and having an index with sights, or a telescope, 
placed on the centre, about which the index is moveable ; 
also a compass fixed to the centre, to point out courses and 
check the sights ; the whole being fixed by the centre on a 
stand of a convenient height for use. 

In using this instrument, an exact account, or field-book, 
of all measures and things necessary to be remarked in the 
plan, ntust be kept, from which to make out the plan on re- 
turning home from the ground. 

Ilegin at such part of the ground, and measure in such 
directions as are judged most convenient ; taking angles or 
directions to objects, and measuring such distances as appear 
necessary, under the same restrictions as in the use of the 
plain table. And it is safest to fix the theodolite in the 
original position at every station, by means of fore and back 
objects, and the compass, exactly as in using the plain table ; 
registering the number of degrees cut off by the index when 
directed to each object ; and, at any station, placing the 
index at the same degree as when the direction towards that 
station was taken from the last preceding one, to fix the 
theodolite there in the original position. 

The best method of laying down the aforesaid lines of 
direction, is to describe n pretty large circle ; then quarter it, 
and lay on it the several numbers of degrees cut off by the 
index in each direction, and drawing lines from the centre to 
all these marked points in the circle. Then, by means of a 
parallel ruler, draw from station to station, lines parallel to 
the aforesaid lines drawn from the centre to the respective 
points in the circumference. 

4. OF THE CROSS. 

The cross consists of two pair of sights set at right angles 
to each other, on a staff having a sharp point at the bottom, 
to fix in the ground. 

Tho cross is very useful to measure small and crooked 
pieces of ground. The method is, to measure a base or chief 
line, usually in the longest direction of the piece* from corner 
to corner ; and while measuring it, finding the places where 
perpendiculars would fall on this line, from the several cor- 
nere and bends in the boundary of the piece, with the cross, 
by fixing it, by trials, on such parts of the line, as that 
through one pair of the sights both ends of the line may 
appear, and through the other pair the rom«^uKv&%taro&i 

Vol. I. 66 



m 



urn 



or corners : and then measuring die lengths of the eni Tp« r 

pendiculors. 

BEXAMS* 

Besides the fore*mentioned instruments, whic't ere mcst 
commonly used, there are some others ; as, 

The perambulator, used for measuring roads, and other 
great distances, level ground, and by the sides of rivers. It 
has a wheel of 8} feet, or half a pole, in circumference* by 
the turning of which the machine goes forward ; and the 
. distance measured is pointed out by an index, which is moved 
round by clock-work. 

Levels, with telescopic or other sights, are used to find 
the' level between place and place, or how much one place 
is higher or lower than another. And in measuring any 
sloping or oblique line, either ascending or descending, a small 
pocket level is useful for showing how many links lor each 
chain are to be deducted, to reduce the tine to the hori- 
zontal length. 

An offset. staff is a very useful instrument, for measuring 
the offsets and other short distances. It is 10 links in length, 
being divided and marked at each of the 10 links. 

'J en small arrows, or rods of iron or wood, are used to 
mark the end of every chain length, in measuring lines. 
And sometimes pickets, or staves with flags, are set up as 
marks or objects of direction. 

Various scales are also used in protracting and measuring 
on the plan or paper ; such as plane scales, line of chords, 
protractor, compasses, reducing scale, parallel and pc rpen- 
dicular rules, dec. Of plane scales, there should be several 
sizes, as a chain in 1 inch, a chain in } of an inch, a chain 
in £ an inch, &c. And of these, the best for use are those 
that are laid on the very edges of the ivory scale, to mark off 
distances, without compasses. 



SECTION IL 

THE PRACTICE OF SURVEYING. 

This part contains the several works proper to be dc n*» in 
the field, or the ways of measuring by all the instaiaeots, 
aud in all situations. 



•UBVBViire, 



PBO^EH I. 

7b measure a Line or Distance. 

To measure a line on the ground with the chain : Having 
provided a chain, with 10 small arrows, or rods, to fix one 
into the ground, as a mark, at the end of every chain ; two 
persons take hold of the chain, one at each end of it ; nnd 
all the 10 arrows are taken by one of them, who goes fore- 
most, and is called the leader ; the other being called the 
follower, for distinction's sake. 

A picket, or station-staff, being set up in the direction of 
the line to be measured, if there do not appear some murks 
naturally in that direction, they measure straight towards it, 
4he leader fixing down an arrow at the end of every chain, 
which the follower always takes up, as he comes at it, till 
all the ten arrows are used. 'They are then all returned to 
the leader, to use over again. And thus the arrows are 
changed from the one to the other at every 10 chains length, 
till the whole line is finished ; then the number of changes 
of the arrows shows the number of tens, to which the fol. 
lower adds the arrows he holds in his hand, and the number 
of links of another chain over to the mark or end of the 
line. So, if there have been 3 changes of the arrows, and 
the follower hold 6 arrows, and the end of the line cut off 
45 links more, the whole length of the line is set down in 
links thus, 3645. 

When the ground is not level, but either ascending or de- 
scending ; at every chain length, lay the offset-staff, or link- 
staff, down in the slope of the chain, on which lay the small 
pocket level, to show now many links or parts the slope line 
is longer than the true level one ; then draw the chain for- 
ward so many links or parts, which reduces the line to the 
horizontal direction. 



problem n. 
To lake Angles and Bearings. 

Let b and c be two objects, or two 
pickets set up perpendicular ; and let 
it be required to take their bearings, 
or the angles formed between them 
at any station a. 




1. With the Pimm Table. 



The table being covered with * paper, and fixed on "ts 
ataod ; plant it at the station a, and fix a fine pin, or a foot 
of the compasses, in a proper point of the paper, to repre- 
sent the place a ; Close by ihe side of this pin lay the fiducial 
edge of the index, and turn it about, still touching the pin, 
till one object b can be seen through the sights : then by the 
fiducial edge of the index draw a lute. In the same manner 
draw another line in the direction of the other object o» 
And it is done. 

2. With the Theodolite, 4*c 

Direct the fixed sights along one of tho lines, as ab, by 
turning the instrument about till the mark b is seen through 
these sights ; and there screw the instrument fast. Then 
turn the moveable index round, till through its sights the 
other mark c is seen. Then the degrees cut by the index, 
on the graduated limb or ring of the instrument, show the 
quantity of the angle. 

3. With the Magnetic Needle and Compass. 

Turn the instrument or compass so, that the north end 
of the needle point to the flower-de-luce. Then direct the 
sights to one mark ns b, and note the degrees cut by the 
needle. Next direct the. sights to the other mark c, and 
note ngnin the degrees cut by the needle. Then their sum 
or difference, as the case may be, will give the quantity of 
the angle bac. 

4. By Measurement with the Chain, <fc. 

Measure one chain length, or any other length, along 
both directions, as to b and c. Then measure the distance 
6c, and it is done.— This is easily transferred to paper, by 
making a triangle Abe with these three lengths, and then 
measuring the angle a. 



0UBV12T1XO. 



417 



pboblbh in. 



To survey a Triangular Field abc. 
1. By the Chain. 



ap 794 

AB 1321 

fc 826 



C 




Having set up marks at the come , which is to be done 
in all cases where there are not marks naturally ; measure 
with the chain from a to p, where a perpendicular would 
fall from the angle c, and set up a mark at p, noting down 
the distance a p. Then complete the distance ab, by mea- 
suring from p to b. Having set down this measure, return 
to p, and measure the perpendicular pc. And thus, having 
the base and perpendicular, the area from them is easily 
found. Or having the place r of the perpendicular, the 
triangle is easily constructed. 

Or, measure all the three sides with the chain, and note 
them down. From which the content is easily found, or the 
figure is constructed. 



Measure two sides ab, ac, and the angle a between them. 
Or measure one side ab, and the two adjacent angles a and 
b. From either of these ways the figure is easily planned ; 
then by measuring the perpendicular cp on the plan, and 
multiplying it by half ab, the content is found. 

problem IV. 

To Measure a Four-sided Field. 
1. By the Chuin. 




Measure along one of the diagonals, as ao ; and eitfce 
the two perpendiculars be, bf, as in the last prohWA * 



2. By taking some of the Angles. 



4*8 



LAHP 



else the tides ab, bo, cd, da. From either of which the 
figure may be planned and computed as before directed* 

Otherwise, h C\aU. 

Measure, on the longest side, the distances at, aq, ab ; and 
the perpendiculars rc, <u>. 

2. J9y tatoig mne ©ffAe itagfat. 



ap 110 I 852 rc 
aq 745 505 qd 

AB 1110 I 



Measure the diagonal ac (see the last fig. but one), and 
the angles cab, cad, acb, acd. — Or measure the four sides, 
and any one of the angles, as bad. 



Thus. 
ac 501 
cab 37° W 

CAD 41 15 

ub 72 25 
acd 54 40 



Or thus. 
ab 486 
bc 304 
cd 410 
da 462 
bad 78" 35'. 



PROBLEM. V, 



To survey any Field by the Chain only. 

Having set up marks at the corners, where necessary, of 
the proposed field abcdefg, walk over the ground, and con* 
aider how it can best be divided into triangles and trapeziums ; 
and measure them separately, ns in the last two problems. 
Thus, the following figure is divided into the two trapeziums 
abco, gdef, and the triangle ocn. Then, in the first tra- 
pezium, beginning at a, measure the diagonal ac, and the 
two perpendiculars cm, iro. Then the base gc and the 
perpendicular vq. Lastly, the diagonal df, and the two 
perpendiculars pE, og. All wbich measures write against 
the corresponding parts of a rough figure drawn to resemble 
the figure surveyed, or set them down in any other form you 
choose. 



«0 



Thus. 



Am 


135 


130 


mo 


Aft 


410 


180 


TIB 


AC 


550 






cq 


152 


230 


q» 


CG 


440 






FO 


237 


120 


oo 


FP 


288 


80 




FD 


520 








Or thus. 

Measure all the sides ab, bc, cd, dk, ef, fg, ga ; and the 
diagonals ac, cg, gd, df. 

Otherwise. 

Many pieces of land may be very well surveyed, by mea- 
suring any base line, either within or without them, with the 
perpendiculars let fall on it from every corner. For they 
are by those means divided into several triangles and trape- 
zoids, all whose parallel sides are perpendicular to the base 
line ; and the sum of these triangles and trapeziums will be 
equal to the figure proposed if the base line fall within it ; if 
not, the sum of the parts which are without being taken from 
the sum of the whole which are both within and without, will 
leave the area of the figure proposed. 

In pieces that are not very large, it will be sufficiently 
exact to find the points, in the base line, where the several 
perpendiculars will fall, by means of the crass, or even by 
judging by the eye only, and from thence measuring to the 
corners for the lengths of the perpendiculars. — And it will be 
most convenient to draw the line so as that all the perpen- 
diculars may fall within the figure. 

Thus, in the following figure, beginning at a, and mea- 
suring along the line ao, the distances and perpendiculars on 
the right and left are as below. 



Ab 
AC 

xd 

AC 

A/ 
AO 



315 
440 
585 
610 
990 
1020 



350 6b 
70 cc 

320 dv 
50 

470 /f 





4t» 



LA*B 



PBOBUm TI. 



To 



ike Of***. 



jjtitiwm being a crooked hedge, or brook, dec. Front 
a measure in a straight direction along the side of it to b. 
And in measuring along this line ab, observe when you m 
directly opposite any bends or corners of the boundary, as si 
c, 4, e, ic.; and from these measure the perpendicular 
offsets dfc, di, fcc. with the ofiset.staff, if they are not very 
large, otherwise , with the chain itself ; and the work is done* 
'fhe register, or field-book, may be as follows : 



Oft. left. 


Baaa line am 










A 


c* 


62 


45 


AC 


di 


84 


220 


Ad 


ek 


70 


840 


Kt 


fl 


96 


510 


*f 


gm 


57 


034 


*g 


m 


91 


785 


AB 



Ac d e J 9 3 



FROBLEV VH. 

To survey any PieUjeUh the Plain Table. 
1. Prom one Station. 




PtAirr the table at any angle as 
c, from which all the other angles, 
or marks set up, can be seen ; turn 
the table about till the needle point 
to the flower-de-luce ; and there 
screw it fast. Make a point for c 
on the paper on the table, and lay 

the edge of the index to c, turning 

it about c till through the sights ' A. JB 
you see the mark d : and by the edge of the index draw a 
dry or obscure line : then measure the distance cd, and lsy 
that distance down on the line cd. Then turn the index 
about the point c, till the mark £ be seen through the sights, 
by which draw a line and measure the distance to b, layiag 
it on the line from c to s. In like manner determine the 
positions of ca and cb, V*y \wnxva£the sights auccossircly to 



•UEVEYIXG. 



441 



a and b ; and lay the length of those lines down. Then 
connect the points, hy drawing the black lines cd, de, ka, 
ab, bo, for the boundaries of the field. 

From a Station within the Field. 




When all the other parts cannot 
be seen from one angle, choose some 
place O within, or even without, if v 
more convenient, from which the other 
part 8 can be seen. Plant the table 
at O, then fix it with the needle 
north, and mark the point O on it. 
Apply the index successively to (), 
turning it round with the sights to 
each angle, a, b, c, d, e, drawing dry lines to them hy the 
edge of the index ; then measuring the distances oa, or, 
and laying them down on those lines. Lastly, draw the 
boundaries ab, bc, cd, de, ea. 

3. By going round the Figure. 

When the figure is a wood, or water, or wlien from some 
other obstruction you cannot measure lines across it ; begin 
at any point a, and measure around it, either within or 
without the figure, and draw the directions of all the sides, 
thus: Plant the table at a ; turn it with the needle to the 
north or flower-de-luce ; fix it, and mark the point a. Apply 
the index to a, turning it till you can see the point k, and 
there draw a line : then the point b, and there draw a line : 
then measure these lines, and lay them down from a to Kund 
b. Next move the table to b, lay the index along the line 
ab, and turn the table about till you can see the mark a, and 
screw fast the table ; in which position also the needle will 
again point to the flower-de-luce, as it will do indeed at every 
station when the table is in the right position. Here turn 
the index about b till through the sights you seek the mark c ; 
there draw a line, measure bc, and lay the distance on that 
line after you have set down the table at c. Turn it then 
again into its proper position, and in like manner find the 
next line cd. And so on quite around by e, to a again. 
Then the proof of the work will be the joining at a : for if 
the work be all right, the last direction ka on the ground, 
will pass exactly through the point a on the paper ; and the 
measured distance will also reach exactly to a. If these do 
not coincide, or nearly so, some error has been committed, 
and the work must be examined over again. 

Vol. I. 57 



442 



LAND 



PllOBLEX VIII. 

To surrey a Field xci'h the Theodolite, «fc. 

I. From One Point or Station* 

When all the angles can be seen from one point, as the 
angle c first fig. to lust prob.), place the instrument at c, and 
turn it about, till through the fixed sights you see the mark 
a, and there fix it. Then turn the moveable index about 
till the mark a be seen through the sights, and note the de- 
grees cut on the instrument. Next turn the index suc- 
cessively to k and n, noting the degrees cut off* at each ; which 
gives all the angles h<;a, bck, bod. Lastly, measure the 
lines cb, ca, ce, cd ; and enter the measures in a field-book, 
or rather, against the corresponding parts of a rough figure 
drawn by guess to resemble the field. 

2. From a Point within or without. 

Plant the instrument at o (last fig.), and turn it about till 
the fixed sights point to any object, as a ; and there screw it 
fast. Then turn the moveable index round till the sights 
point successively to the other points n, c, b, no'ing the 
degrees cut off at each of them ; which gives all the angles 
round the point o. Lastly, measure the distances oa, ob, oc, 
od, oe, noting them down as before, and the work is done. 

3. By going round the Field. 

By measuring round, either % EL-- 
within or without the field, pro. 
ceed thus. Having set up marks 
at b, u, &c. near the corners as 
usual, plant the instrument at 
any point a, and turn it till the 
fixed index be in the direction 
ab, and there screw it fast : then 
turn the moveable index to the / 
direction ac ; and the degrees cut off will be the angle a. 
Measure the line ab, and plant the instrument at b, and! 
there in the same manner observe the angle a. Then mea- 
sure bc, and observe the angle c. Then measure the dis- 
tance cd, and take the angle d. Then measure de, and 
take the angle e. Then measure ef, and tak^ the angle r. 
And lastly, measure the distance fa. 

To prove the work ; add all the inward angles, a, b, c, 
Ac. together ; for when the work is right, their sum will be 
equal to twice as many right angles as the figure has sides, 




SUtVIYJNG. 



443 



wanting 4 right angles. But when there is an angle, as f, 
that bends inwards, and you measure the external angle, 
which is less than two right angles, subtract it from 4 right 
angles, or 360 degrees, to give the internal angle greater than ■ sfc-; 
a semicircle or 180 degrees. 

Otherwise. < 

Instead of observing the internal angles, we may take the 
external angles, formed without the figure by procuring the 
sides farther out. And in this case, when the work is right, 
their sum altogether will be equal to 360 degrees. But when 
one of them, as f, runs inwards, subtract it from the sum of 
the rest, to leave 360 degrees. 

problem rx. 

To survey a Field with crooked Hedges, 4*c. 

With any of the instruments, measure the lengths and 
positions of imaginary lines running as near the sides of the 
field as you can ; and. in going along them, men sure the 
offsets in the manner before taught ; then you will have the 
plan on the paper in using the plain table, drawing the 
crooked hedges through the ends of the offsets ; but in sur- 
veying with the theodolite, or other instrument, set down 
the measures properly in a field-book, or memorandum* 
book, and plan them after returning from the field, by lay. 
ing down all the lines and angles. 




So in surveying the piece arcde, set up marks, a, b. r, #2, 
dividing it so as to have as few sides as may be. Then begin 
at any station, a, and measure the lines a//, 6c, cd, d>i> taking 
their positions, or the angles, a, 6, c, d ; and, in going along 
the lines, measure all the offsets, as at m, n, o, p, &c. along 
every station. line. 

And this is done either within the field, or without, as 
may be most convenient. When there wc* oV^rt^Mrai 



LAUD 



within, as wood, water, hills, ozc. then measure without, t» 
in the next following figure. 




problem x. 

7b Survey a Field, or any other Thing, by ttco Stations. 

This is performed by choosing two stations from which 
all the marks and objects can be seen ; then measuring the 
distance between the stations, and at each station, taking 
the angles formed by every object fr< m the station line or 
distance. 

The two stations may be taken either within the hounds, 
or in one of the sides, or in the direction of two of the ob- 
jects, or quite at a distance and without the bounds of the 
objects or part to be surveyed. 

In this manner, not only grounds may be surveyed, with- 
out even entering them, but a map may be taken of t'.ie 
principal parts of a county, or the chief places nf a town, 
or any part of a river or coast surveyed, or any other inac- 
cessible objects ; by taking two stations, on two towers, or 
two hills, or such-like. 

B 



8T7HVEY1JG. 



445 



PRO B LEX XI. 

To survey a large Estate. 

If the estate be very large, and contain a great number of 
field*, it cannot welt be done by surveying all the fields 
singly, and then putting them together ; nor can it be done 
by taking all the angles and boundaries that enclose it. For 
in these cases, any small errors will be so much increased, as 
to render it very much distorted. But proceed a* below. 

1. W ilk over the estate two or three times, in order to 
get a perfect idea of it, or till you can keep the figure of it 
pretty we I in mind. And to help your memory, draw an 
eye-draught of it on paper, at least of the principal parts 
of it, to guide you ; setting the names within the fields in 
that draught. 

2. Choose two or more eminent places in the estate, for 
station?*, from which all the princip.il parts of it can be seen : 
selecting th' se stations as far distant from one another as 
convenient. 

3. Take such nng'es, between the station*, as you think 
necessary, and measure the distances from station to station, 
always in a ri^ht line : these things must be done, till you 
get an many angles and lines as are sufficient for determining 
all the points of station. And in measuring any of these 
station distance*, mark accurately where these lines meet 
with any hedges, ditches, roads, lanes, paths, rivulets, dec. ; 
and where any remarkable object is placed, by measuring 
its distance from the station- line ; and where a perpendicular 
from it cuts that line. And thus as you go along unv mnin 
station. line, take offsets to the ends of all hedges, am/ to any 
pond, house, mill, bridge, 6zc. noting every thing down that 
is remarkable. 

4 As to the inner parts of the estate, they must be deter- 
mined, in like manner, by new station lines ; for, after the 
main stations are determined, and every thing adjoining to 
them, then the estate must be subdivided into two or three 
parts by new station lines ; taking inner stations at proper 
places, where you can have the best view. Measure these 
Stat ion- lines as you d d tl e first, and all their intersections 
with hedges, and offsets to such objects as appear. Then 
proceed to survey the adjoining fields, by taking the angles 
that the sides make with the station. 'ine, at the intersections, 
and measuring the distances to each corner, from the inter, 
sections. For the station. lines will be the bases to \\W \V» 
future operations ; the situation of aft pun* Ym&% «D&t*Vi 



448 



LAKD 



dependent on them ; nnd therefore they should be taken of 
as great length a* possible ; and it is best for them to run 
along Home of the hedges or boundaries of one or more fields, 
or to pass through some of their angles. All things being 
determined fur these stations, you must take more inner 
stations, and continue to divide and subdivide till at last you 
come to single fields : repeating the same work for the inner 
stations as for the outer ones, till all is done ; and close the 
work as often as you can, and in as few lines as possible. 

5. An estate nmy be so situated that the whole cannot be 
surveyed together ; because one part of the estate cannot be 
seen from another. In this case, you may divido it into 
three or four parts, and survey the parts separately, as if 
they were lands belonging to different persons ; and at last 
join them together. 

6. As it is necessary to protract or lay down the work as 
you proceed in it, you must have a scale of a due length to 
do it by. To get such n scale, measure the whole length of 
the estate in chains ; then consider how many inches long 
the map is to he ; and from these will be known how many 
chains you must have in an inch ; then make the scale ac- 
cordingly, or choose one already made. 



PROBLEM XII. 

To survey a County, or large Trad of Land. 

1. Choosk two, three, or four eminent places, for stations; 
such as the tops of high hills or mountains, towers, or church 
steeples, which may he seen from one another ; from which 
most of the towns and other places of note may also be seen ; 
and so as to be as far dis ant from one another as possible. 
On these places raise beacons, or long poles, with flags of 
different colours flying at them, so as to be visible from all 
the other stations. 

2. At all the places which you would set down in the map, 
plant long poles, with Hags at them of several colours, to 
distinguish the places from one another ; fixing them on the 
tops of church steeples, or the tops of houses ; or in the 
centres of smaller towns and villages. 

These marks then being set up at a convenient number of 
places, and such as may be seen from both stations ; go to 
one of these stations, and, with an instrument to take angles, 
standing at that station, take ail the angles between the other 
station nnd each of these marks. Then go to the other 
station, and take all the angles between the first station and 



SURVEY:*©. 



447 



each of the former marks, setting them down with the 
others, each against its fellow with the same colour. You 
may, if convenient, also take the angles at some third station, 
which may serve to prove the work, if the three lines inter* 
sect in that point where any mark stands. The marks must 
stand till the observations are finished al both stations ; and 
then they may be taken down, and set up at new places. 
The same operations must be performed at both stations, for 
these new places ; and the like for others. The instrument 
for taking angles must ba an exceeding good one, made on 
purpose with telescopic sights, and of a good length of ra- 
dius. 

3. And though it be not absolutely necessary to measure 
any distance, because, a stationary line being laid down from 
any scale, all the other lines will be proportional to it ; yet 
it is better to measure some of the lines, to ascertain the 

' distances of places in miles, and to know how many geome- 
trical miles there are in any length ; as also from thence to 
make a scale to measure any distance in miles. In measuring 
any distance, it will not be exact enough to go along the 
high roads ; which, by reason of their turnings and windings, 
hardly ever lie in a right line between the stations ; which 
must cause endless reductions, and require great trouble to 
make it a right line ; for which reason it can never be exact. 
But a better way is to measure in a straight line with a chain, 
between station and station, over hills and dales, or level 
fields and all obstacles. Only in case of water, woods, 
towns, rocks, banks, <kc. where we cannot pass, such parts 
of the line must be measured by the methods of inaccessible 
distances ; and besides, allowing for ascents and descents, 
when they are met with. A good compass, that shows the 
bearing of the two stations, will always direct us to. go 
straight, when the two stations cannot be seen ; and in the 
progress, if we can go straight, offsets may be taken to any 
remarkable places, likewise noting the intersection of the 
station. line with all roads, rivers, dtc. 

4. From all the stations, and in the whole progress, we 
must be very particular in observing sea-coasts, river-mouths, 
towns, castles, houses, churches, mills, trees, nicks, sands, 
roads, bridges, fords, ferries, woods, hills, mountains, rills, 
brooks, parks, beacons, sluices, floodgates, locks, &c, and in 
general every thing that is remarkable. 

5. After we have done with the first and main station* 
lines, wbich command the whole county : we must then 
take inner stations, at some places already determined ; which 
will divide the whole into several partitions : and from these 
•tations we must determine the places of as traiw) ^ga 



land 



remaining towns as we can. And if any remain in that 
part y we must take nnre stations, at Home places already 
determined ; from which we may determine the rent. And 
thus go through all the parts of *lho county, taking atalkm 
after station, till we have determined the whole. And in 
general the station distances must always pass through such 
remarkable points as have been determined before, by the 
former stations. 



rKOBLEM XIII. 

To survey a Town or City. 

This may be done with any of the instruments for taking ^ 
angles, but best of all with the plain table, where every 
minute part is drawn while in sight. Instead of the common 
surveying or Gunter's chain, it will he best, for this purpose, 
to have a chain 50 feet long, divided into 50 links of one 
foot each, and an offset- staff «;f 10 feet long. 

Begin at the meeting of two or more of the principal 
streets, through which you enn have the longest prospect?, 
to get the longest station lines : there having fixed the in* 
strument, draw lines jf direction along those streets, using 
two men as marks, or poles set in wooden pedestals, or per- 
haps some remarkable places in the houses at the fan her 
ends, as windows, doors, corners, dtc. Measure those lines 
with the chain, taking onsets with the staff, at all corners of 
streets, bendings, or windings, and to all remurkable thing*, 
as churches, markets, halls, colleges, eminent houses, &c. 
Then remove the instrument to another station, nlon"r one of 
these lines ; and there repeat (he same process as before. 
And so on till the whole is finished. 




SURVEYING* 



449 



Thus, fix the instrument at a, and draw lines in the 
direction of all the streets meeting there ; then measure ab, 
noting the street on the left at m. At the second station b, 
draw the directions of the streets meeting there ; and mea- 
sure from b to c, noting the places of the streets at n and o 
as you pass by them. At the third station c y take the di- 
rection of all the streets meeting there, and measure cd. At 
d do the same, and measure de, noting the place of the 
cross streets at p. And in this manner go through all the 
principal streets. This clone, proceed to the smaller and 
intermediate streets ; and lastly to the lanes, alleys, courts, 
yards, and every part that it may be thought proper to re- 
present in the plan. 

THEOREM XIV. 

To lay down the Plan of any Survey. 

Ir the survey was taken with the plain table, we have a 
rough plan of it already on the paper which covered the 
table. But if the survey was with any other instrument, a 
plan of it is to be drawn from the measures that were taken 
in the survey ; and first of all a rough plan on paper. 

To do this, you must have a net of proper instruments, 
for laying down both lines and angles, &c. ; as scales of va- 
rious sizes, the more of them, and the more accurate, the 
better, scales of chords, protractors, perpendicular and pa- 
rallel rulers, <kc. Diagonal scales are best for the lines, 
because they extend to three figures, or chains, and links, 
which are 100 parts of chains. But in using the diagonal 
scale, a pair of compasses must be employed, to take on the 
lengths of t!io principal lines very accurately. But a scale 
with a thin edge divided, is much readier for laying down 
the perpendicular offsets to crooked hedges, and for marking 
the places of those offsets on the station- line ; which is done 
at only one application of the cd«;e the scale to that line, 
and then pricking off all at once the distances along it. 
Angles are to be laid down, either with a good scale of 
chords, which is perhaps the most accurate way, or with a 
large protractor, which is much readier when many angles 
are to be laid down at one point, as they arc pricked off all 
at once round the edge of the protractor. 

In general, all ILies and angles must be laid down on the 
plan ii> the same order in which they were measured in the 
field, and in which they are written in the field-book ; lay- 
ing down first the angles for the position of \vma, wraX. 

Vol. I. 58 



LAJCD 



lengths of the lines, with the places of the offsets, and then 
the lengths of the offsets themselve*, all with dry or obscure 
lines; then a black line drawn through the extremities of 
nil the offset*, will be the edge or bounding line of the field, 
ccc. After the principal bounds and lines are laid down, 
and made to fit or close properly, proceed next to the smaller 
objects, till you have entered every thing that ought to ap- 
pear in the plan, as houses, brooks, trees, hills, gates, stiles, 
roads, lanes, mill*, bridges, woodlands, Ac. dec. 

The north side of a map or plan is commonly placed 
uppermost, and a meridian is somewhere drawn, with the 
compass or flower-de-luce pointing north. Also, in a vacant 
part, a scale of equal parts or chains is drawn, with the title 
of the map in conspicuous characters, and embellished with 
a compartment. Hills are shadowed, to distinguish them in 
the map. Colour the hedges with different colours ; repre- 
sent hilly grounds by broken hills and valleys ; draw single 
dotted lines for foot-paths, and double ones for horse or car- 
riage roads. Write the name of each field and remarkable 
place within it, and, if you choose, its contents in acres, 
roods, and perches. 

In a very large estate, or a county, draw vertical and ho- 
rizontal lines through the map, denoting the spaces between 
them by letters placed at the top, and bottom, and sides, for 
readily finding any field or other object mentioned in a 
table. 

In mapping counties, and estates that have uneven grounds 
of hills and valleys, reduce all oblique lines, measured up* 
hill and down. hill, to horizontal straight lines, if that was 
not done during the survey, before they were entered in the 
field-book, by making a proper allowance to shorten them. 
For which purpose there is commonly a small table engraven 
on some of the instruments for surveying. 

THE NEW METHOD OF SURVEYING. 

PROBLEM XV. 

To survey and plan bp the new Method. 

Is the former method of measuring a large estate, theac 
curacy of it depends both on the correctness of the instru- 
ments, and on the care in taking the angles. To avoid the 
errors incident to such a multitude of angles, other methods 
hu\c of late years been used by some few skilful surveyors : 
the most practical, expeditious, and correct, seems to be the 



SURVEYING. 



451 



following, which is performed, without taking angles, by 
measuring with the chain only. 

Choose two or more eminences, as grand stations, and 
measure a principal base line from one station to another ; 
noting every hedge, brook, or other remarkable object, as you 
pass by it ; measuring also such short perpendicular lines to 
the bends of hedges as may be near at hand. From the ex- 
tremities of this base line, or from any convenient parts of 
the same, go off with other lines to some remarkable object 
situated towards the sides of the estate, without regarding 
the angles they make with the base line or with one another ; 
still remembering to note every hedge, brook, or other ob- 
ject, that you pass by. These lines, when laid down by in- 
tersections, will, with the base line, form n grand triangle on 
the estate ; several of which, if need he, being thus mea- 
sured and laid down, you may proceed to form other smaller 
triangles and trapezoids on the sides of the former : and so 
on till you finish with the enclosures individually. By which 
means a kind of skeleton of the estate may first be obtained, 
and the chief lines serve as the bases of such triangles and 
trapezoids as are necessary to fill up all the interior parts. 

The field-book is ruled into three columns, ns usual. In 
the middle one are set down the distances on the chain. line, 
at which any mark, offset, or other observation, is made ; 
and in the right and left hand columns are entered the off- 
sets and observations made on the right and left hand re- 
spectively of the chain-line ; sketching on the sides the shape 
or resemblance of the fences or boundaries. 

It is of groat advantage, both for brevity and perspicuity, 
to begin at the bottom of the leaf, and write upwards ; de- 
noting the crossing of fences, by lines drawn across the mid. 
die column, or only a part of such a line on the right and 
left opposite the figures, to avoid confusion ; and the corners 
of fields, and other remarkable turns in the fences where off- 
sets are taken to, by lines joining in the manner the fences 
do ; as will be best seen by comparing the book with the 
plan annexed to the field-book following, p. 454. 

The letter in the left-hand corner at the beginning of every 
line, is the mark or place measured from; and that at the 
right-hand corner at the end, is the mark measured to : but 
when it is not convenient to go exactly from a mark, the 
place measured from is described such a distance from one 
mark towards another ; and where a former mark is not mea- 
sured to, the exact place is ascertained by saying, turn to the 
right or left hand, such a distance to such a mark, it being 
always understood that those distances are tuken in the 
chain line* 



4» 



LAND 



The characters used are, f for turn to the right handf 
} for turn to the left hand, and placed over an offset, 
to show that it is not taken at right angles with the chain- 
line, but in the direction of some straight fence ; being 
chiefly used when crossing their directions ; which is a better 
way of obtaining their true places than by offsets at right 
angles. 

When a line is measured whose position is determined, 
either by former work (as in the case of producing a given 
line, or measuring from one known place or mark to another) 
or by itself (as in the third side of the triangle), it is called 
a fast line, and a double tine across the book is drawn at the 
conclusion of it ; but if its position is not determined 'as in 
the second side of the triangle), it is called a loose line, and a 
single line is drawn across the book. When a line becomes 
determined in position, and is afterwards continued farther, 
a double line half through the hook is drawn. 

When a loose line is measured, it becomes absolutely ne- 
cessary to measure some other line that will determine its 
position. Thus, the first line ah or bh, being the base of a 
triangle, is always determined ; but the position of the second 
side hj does not become determined, till the third side jb is 
measured ; then the position of both is determined, and the 
triangle may be constructed. 

At the beginning of a line, to fix a loose line to the mark 
or place measured from, the sign of turning to the right or 
left hand must be added, as at h in the second, and j in the 
third line ; otherwise a stranger, when laving down the 
work, may as easily construct the triangle hjb on the wrong 
side of the lino ah, us on the right one : hut this error can- 
not be fallen into, if the sign above named be carefully ob- 
served. 

In choosing a line to fix n loose one, care must be taken 
that it does not make a very acute or obtuse angle ; as in the 
^ triangle pur, by the an«rle at b being very obtuse, a small 
deviation from truth, even the breadth of a point at p or r, 
would make the error at b, when constructed, very consi. 
derable ; but by constructing the triangle pnq, such a devia. 
tion is of no consequence. 

Where the words leave off are written in the field* book, it 
signifies that the taking of offsets is from thence discominu. 
ed ; and of course something is wanting between that and 
the next offset, to be afterwards determined by measuring 
some other line. 

The field. book for this method, and the plan drawn from 
it, arc contained in the four following pages, engraven on 
copper plates ; answerable to which the pupil is to draw a 



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SO - 



t 



f. .a , 



■UBVJBVIXG* 



458 



plan from the measures in the field-book, of a larger size, 
viz. to a scale of a double size will be convenient, such a 
scale being also found on most instruments. In doing this, 
begin at the commencement of the field-book, or bottom of 
the first page, and draw the first line ah in any direction at 
pleasure, and then the next two sides of the first triangle bhj 
by sweeping intersected arcs ; and so all the triangles in 
the same manner, after each other in their order ; and after- 
wards setting the perpendicular and other offsets at their 
proper places, and through the ends of them drawing the 
bounding fences. 

Note. That the field-book begins at the bottom of the first 
page, and reads up to the top ; hence it goes to the bottom 
of the next page, and to the top ; and thence it passes from 
the bottom of the third page to the top, which is the end of 
the field-book. The several marks measured to or from, 
are here denoted by the letters of the alphabet, first the small 
ones, a, b, c, d y &c, and after them the capitals A, B, C, D 9 
dec. But instead of these letters, some surveyors use the 
numbers in order, 1, 2, 3, 4, dtc. 

OF THE OLD KIND OF FIELD-BOOK. 

In surveying with the plain table, a field-book is not used, 
as every thing is drawn on the table immediately when it is 
measured. But in surveying with the theodolite, or any 
other instrument, some kind of a field book must be used to 
write down in it a register or account of all that is done and 
occurs relative to the survey in hand. 

This book every one contrives and rules as he thinks fittest 
for himself. The following is a specimen of a form which 
has been formerly used. . It is ruled in three columns, as in 
the next page. 

Here Q 1 is the first station, where the angle or bearing 
is 105° 25'. On the left, at 73 links in the distance or prin- 
cipal line, is an offset of 92 ; and at 610 an offset of 24 to a 
cross hedge. On the right, at 0, or the beginning, an offset 
25 to the corner of the field ; at 248 Brown's boundary 
hedge commences ; at 610 an offset 35 ; and at 954, the end 
of the first line, the denotes its terminating in the hedge. 
And so on for the other stations. 

A line is drawn under the work, at the end of every sta- 
tion line, to prevent confusion. 



454 



LAKD 



Form ofthU FiM.Book. 



Offset* and Remarks 
on the left. 


Stations, 
Bearings, 

and 
Distances. 


Ofiveta and Remarks 
on the right. 


80 
OS 

a cross hedge 34 


1 
105° 25' 
00 
73 
248 
610 
054 


25 corner 

Brown's hedge 

35 

00 


house corner 51 
34 


2 
53* 10* 
25 
120 
764 


21 

20 a tree 
40 a stile 


a brook 30 

foot-path 16 
cross hedge 18 


3 
67* 20 
61 
248 
635) 
810 
973 


35 

16 a spring 
20 a pond 



Then the plan, on a small scale drawn from the above 
field- book, will be a* in the following figure. But thn pupil 
may draw a plan of 3 or 4 time* the size on his paper book. 
The dotted lines denote the 3 chain or measured linea, and 
the bluck lines the boundaries on the right and left. 




But some skilful surveyors now make use of a different 
method for the field-book, namely, beginning at the bottom 



455 



of the page and writing upwards ; sketching also a neat 
boundary on either hand, resembling the parts near tho 
measured lines as they pas* along ; an example of which was 
given in the new method of surveying, in the preceding 
pages. 

In smaller surveys and measurements, a good way of set- 
ting down the work, is to draw by the eye, on a piece of 
paper, a figure resembling that which id to be measured ; 
and so writing the dimensions, as they are found, against 
the corresponding parts of the figure. And this method 
may be practised to a considerable extent, even in the larger 
surveys. 

SECTION III. 



OF COMPUTING AND DIVIDING. 

PROBLEM XVI. 

To compute the Contents of Fields. 

1. Compute the contents of the figures as divided into 
triangles, or trapeziums, by the proper rules for these figures 
laid down in measuring ; multiplying the perpendiculars by 
the diagonals or bases, both in links, and divide by 2 , the 
quotient is acres, after having cut off five figures on the right 
for decimals. Then bring there decimals to roods and 
perches, by multiplying first by 4, and then by 40. An 
example of which is ghen in the description of the chain, 
pag. 480. 

2. In small and separate pieces, it is usual to compute their 
contents from the measures of the lines taken in surveying 
them, without making a correct plan of them. 

3. In pieces bounded by very crooked and winding hedges, 
measured by offsets, all the parts between the offsets are most 
accurately measured separately as small trapezoids. 

4. Sometimes such pieces as that last mentioned are com* 
puted by finding a mean breadth, by adding all the offsets 
together, and dividing the sum by the number of them, ac- 
counting that for one of them where the boundary meets 
the station* line (which increases the number of them by I, 
for the divisor, though it does not increase the sum or quan- 
tity t > be divided) ; then multiply the length by that mean 
breadth. 

5. But in larger pieces and whole estate** wmm£&%<& 



456 



LARD 



many fields, it is the common practice to make a rough plan 
of the whole, and from it compute the contents, quite inde. 
pendent of the measures of the lines and angles that wort 
taken in surveying. For then new lines are drawn in the 
fields on the plans,' so as to divide them into trapeziums and 
triangles, the bases and perpendiculars of which are measured 
on the plan by means of the scale from which it was drawn, 
and so multiplied together for the contents. In this way, 
the work is very expeditiously done, and sufficiently correct; 
for such dimensions are taken as afford the most easy method 
of calculation ; and among a number of parts, thus taken 
and applied to a scale, though it be likely that some of the 
parts will be taken a small matter too little, and others too 
great, yet they will, on the whole, in all probability, very 
nearly balance one another, and give a sufficiently accurate 
result. After all the fields and particular parts are thus 
computed separately, and added all together into one sum ; 
calculate the whole estate independent of the fields, by di- 
viding it into large and arbitrary triangles and trapeziums, 
and add these also together. Then if this sum be equal to 
the former, or nearly so. the work is right ; but if the sums 
have any considerable difference, it is wrong, and they must 
be examined, and re-computed, till they nearly agree. 

6. But the chief art in computing, consists in finding the 
contents of pieces bounded by curved or very irregular lines, 
or in reducing such crooked sides of fields or boundaries to 
straight lines, that shall enclose the same or equal area with 
those crooked sides, and so obtain the area of the curved 
figure by means of the right-lined one, which will commonly 
be a trapezium. Now this reducing the crooked sides to 
straight ones, is very easily and accurately performed in this 
manner: — Apply the straight edge of a thin, clear piece of 
lantern. horn to the crooked line, which is to be reduced, 
in such a manner, that the small parts cut off from the 
crooked figure by it, may be equal to those which are taken 
in : which equality of the parts included and excluded you 
will presently be able to judge of very nicely by a little prac- 
tice : then with a pencil, or point of a tracer, draw a line by 
the straight edge of the horn. Do the same by the other 
sides of the field or figure. So shall you have a straight- 
sided figure equal to the curved one ; the content of which, 
being computed as before directed, will be the content of the 
crooked figure proposed. 

Or, instead of the straight edge of the horn, a horse-hair, 
or fine thread, may be applied across the crooked sides in 
the same manner ; and the easiest way of using the thread, it 
to string a amaVV ateufat how with it, either of wire, or cane. 



IUKVJCYM6. 



457 



or whale-bone, or such-like slender elastic matter ; for the 
bow keeping it always stretched, it can be easily and neatly 
applied with one hand, while the other is at liberty to make 
two marks by the side of it, to draw the straight line by. 

EXAMPLE. 

Thus, let it be required to find thejeontents of the same 
figure as in Prob. ix, page 443, to a scale of 4 chains to an 
inch. 




Draw the 4 dotted straight lines ab, bc, cd, da, cutting 
off equal quantities on both sides of them, which they do as 
near as the eye can judge : so is the crooked figure reduced 
to an equivalent right-lined one of 4 sides, abcd. Then 
draw the diagonal b», which, by applying a proper scale 
to it, measures suppose 1256; Also the perpendicular, or 
nearest distance from a to this diagonal, measures 456 ; and 
the distance of c from it, is 428. 

Then, half the sum of 456 and 428, multiplied by the 
diagonal 1256, gives 555152 square links, or 5 acres, 2 roods, 
8 perches, the content of the trapezium, or of the irregular 
crooked piece. 

As a general example of this practice, let the contents be 
computed of all the fields separately in the foregoing plan 
facing page 453, and, by adding the contents altogether, the 
whole sum or content of the estate will be found nearly equal 
to 103} acres. Then, to prove the work, divide the whole 
plan into two parts, by a pencil line drawn across it any way 
near the middle, as from the corner I on the right, to the 
corner near s on the loft ; then, by computing these two 
large parts separately, their sum must be nearly equal to the 
former sum, when the work is all right. 

Vol. I. 59 



458 



LAND SURVEYING. 



PROBLEM XVII. 

To Transfer a Plan to Another Paper, $c. 

After the rough plan is completed, and a fair one is 
wanted ; this may be done by any of the following methods* 

First Method.— Lay the rough plan on the clean paper, 
keeping them always pressed flat and close together, by 
weights laid on them? Then with the point of a fine pin 
or pricker, prick through all the corners of the plan to be 
copied. Take them asunder, and connect the pricked points 
on the clean paper, with lines ; and it is done. This method 
is only to he practised in plans of such figures as are small 
and tolerably regular, or bounded by right lines. 

Second Method* — Rub the back of the rough plan over 
win black-lead powder ; and lay this blacked part on the 
clean paper on which the plan is to be copied, and in the 
proper position. Then, with the blunt point of some hard 
substance, as brass, or such-like, trace over the lines of the 
whole plan ; pressing the tracer so much, as that the bJack 
lead under the lines may be transferred to the clean paper : 
after w hich, take off the rough plan, and trace over the leaden 
marks with common ink, or with Indian ink. — Or, instead of 
blacking the rough plan, we may keep constantly a blacked 
paper to lay between the plans. 

Third Method. — Another method of copying plans, is by 
means of squares. This is performed by dividing both ends 
and sides of the plan which is to be copied into any conve- 
nient number of equal parts, and connecting the correspond- 
ing points of division with lines ; which will divide the plan 
into a number of small squares. Then divide the paper, on 
which the plan is to be copied, into the same number of 
squares, each equal to the former when the plan is to be 
copied of the same size, but greater or less than the others, 
in the proportion in which the plan is to be increased or 
diminished, when of a different size. Lastly, copy into the 
clean squares the parts contained in the corresponding squares 
of the old plan ; and you will have the copy, either of the 
same size, or greater or less in any proportion. 

Fourth Method. — A fourth method is by the instrument 
called a pentagraph, which also copies the plan in any size 
required : for this purpose, also, Professor Wallace's eido* 
graph may be advantageously employed. 

Fifth method. — A very neat method, at least in copying 
from u fair plan, is this. Procure a copying frame or glass, 
made in this manner : namely, a large square of the best 



AlTOFfCBRS 9 WORK. 



window glass, set in a broad frame of wood, which can be 
raised up to any angle, when the lower side of it rests on a 
table. Set this frame up to any angle before you, facing a 
strong light ; fix the old plan and clean paper together, with 
several pins quite around, to keep them together, the clean 
paper being laid uppermost, and over the face of the plan to 
be copied. Lay them, with the back of the old plan, on the 
glass ; namely, that part which you intend to begin at to 
copy first ; and by means of the light shining through the 
papers, you will very distinctly perceive every -line of the plan 
through the clean paper. In this state then trace all the 
lines on the paper with a pencil. Having drawn that part 
which covers the glass, slide another part over the glass, and 
copy it in the same manner. Then another part : and so 
on, till the whole is copied. Then take them asunder, and 
trace all the pencil lines over with a fine pen and Indian ink, 
or with common ink. And thus you may copy the finest 
plan, without injuring it in the least. 



OF ARTIFICERS' WORKS, 



AND 

TIMBER MEASURING. 



1. OF THE CARPENTER'S OR SLIDING RULE. 

The Carpenter's or Sliding Rule, is an instrument much 
used in measuring of timber and artificers' works, both for 
taking the dimensions, and computing the contents. 

The instrument consists of two equal pieces, each a foot 
in length, which are connected together by a folding joint. 

One side or face of the rule is divided into inches, and 
eighths, or half-quartoM On the same fuce also are several 
plane scales divided mm twelfth parts by diagonal lines; 
which are used in plannutg dimensions that are taken in feet 
and inches. The edge of the rule is commonly divided 
decimally, or into tenths ; namely, each foot into 



400 



ARTIFICERS* WORK. 



parts, and each of these into ten parts again ; so that by 
means of this last scale, dimensions are taken in feet, tenths, 
and hundredths, and multiplied as common decimal numbers, 
which is the best way. 

On the one part of the other face are four lines, marked 

b, c, d ; the two middle ones b and c being on a slider, 
which runs in a groove made in the stock. The same num. 
berg serve for both these two middle lines, the one being 
above the numbers, and the other below. 

These four lines are logarithmic ones, and the three a, s, 
c, which are all equal to one another, are double lines, as 
they proceed twice over from 1 to 10. The other or lowest 
line, d, is a single one, proceeding from 4 to 40. It is also 
called the girt line, from its use in computing the contents 
of trees and timber ; and on it are marked wo at 17*15, and 
AO at 18-05, the wine and ale gage points, to make this in- 
strument serve the purpose of a gaging rule. 

On the other part of this face, there is a table of the value 
of a load, or 50 cubic feet of timber, at all prices, from 6 
pence to 2 shillings a foot. 

When 1 at the beginning of any line is accounted 1, then 
the 1 in the middle will be 10, and the 10 at the end 100 ; 
but when 1 at the beginning is counted 10, then the 1 in the 
middle is 100, and the 10 at the end 1000 ; and so on. And 
all the smaller divisions are altered proportionally. 



n. ARTIFICERS' WORK. 

Artificers compute the contents of their works by several 
different measures. As, 

Glazing and masonry, by the foot ; Painting, plastering, 
paving, 6zc. by the yard, of 9 square feet : Flooring, 
partitioning, roofing, tiling, 6zc. by the square of 100 
square feet : 

And brickwork, either by the yard of 9 square feet, or by 
the perch, or square rod or pole, containing 272} square 
feet, or 30} square yards, being the square of the rod or 
pole of 16} feet or 5} yards long. 
As this number 272} is troublesome to divide by, the } is 
often omitted in practice, and the content in feet divided only 
by the 272. 

All works, whether superficial or solid, are computed by 
the rules proper to the figure of them, whether it be a tri- 
angle, ot rectang\e, & ^axa\taV)^«d, or any other figure. 



mUCKLAYSBft' WORK* 401 



III- BRICKLAYERS' WORK. 

Brickwork is estimated at the rate of a brick and a half 
thick. So that if a wall be more or less than this standard 
thickness, it must be reduced to it, as follows : 

Multiply the superficial content of the wall by the number 
of half bricks in the thickness, and divide the product by 3. 

The dimensions of a building may be taken by measuring 
half round on the outside and half round on tho inside ; the 
sum of these two gives the compass of the wall, to be multi- 
plied by the height, for the content of the materials. 

Chimneys are commonly measured as if they were solid, 
deducting only the vacuity from the hearth to the mantle, on 
account of the trouble of them. All windows, doors, dec. are 
to be deducted out of the contents of the walls in which they 
are placed. 

The dimensions of a common bare brick are, 8J inches 
long, 4 inches broad, and 2\ thick ; but including the half 
inch joint of mortar, when laid in brickwork, every dimen- 
sion is to be counted half an inch more, making its length 
9 inches, its breadth 4), and thickness 3 inches. So that 
every 4 courses of proper brickwork measures just 1 foot or 
12 inches in height. 

EXAXPLE8. 

Exam. 1. How many yards and rods of standard brick* 
work are in a wall whose length or compass is 57 feet 3 
inches, and height 24 feet 6 inches ; the wall being 2\ bricks 
or 5 half bricks thick ? Ans. 8 rods, 17} yards. 

Exam. 2. Required the content of a wall 62 feet 6 inches 
long, and 14 feet 8 inches high, and 2| bricks thick ? 

Ans. 169-753 yards. 

Exam. 3. A triangular gable is raised 17| feet high, on 
an end wall whose length is 24 feet 9 inches, the thickness 
being 2 bricks : required the reduced content ? 

Ans. 32-08J yards. 

Ex ax. 4. The end wall of a house is 28 feet 10 inches 
long, and 55 feet 8 inches high, to the eaves ; 20 feet high 
is 2£ bricks thick, other 20 feet high is 2 bricks thick, and 
the remaining 15 feet 8 inches is l£ brick thick ; above which 
is a triangular gable, of 1 brick thick, which rises 42 courses 
of bricks, of which every 4 courses make a foot. What is the - 
whole content in standard measure ? 

&n&. Y*x&&« 



402 



carpenters' and joiners' work. 



IV. MASONS' WORK. 

To Masonry belong all sorts of stone work ; and the mea- 
sure made use of is a foot, either superficial or solid. 

Walls, columns, blocks of stone or marble, &c. are mea- 
sured by the cubic foot ; and pavements, slabs, chimney- 
pieces, &c. by the superficial or square foot. 

Cubic or solid measure is used for the materials, and square 
measure for the workmanship. 

In (he solid measure, flic true length, breadth, and thick, 
ness arc taken and multiplied continually together. In the 
superficial, there must be taken the length and breadth of 
every part of the projection which is seen without the general 
upright face of the building. 

EXAMPLES* 

Exam. 1. Required the solid content of a wall, 53 feet 
6 inches long, 12 feet 3 inches high, and 2 feel thick ? 

Ans. 1310J feet. 

Exam. 2. What is the solid content of a wall, the length 
being 21 feet 3 inches, height 10 feet 9 inches, and 2 feet 
thick ? Ans. 521-375 feet. 

Exam. 3. Required the value of a marble slab, at 8*. per 
foot ; the length being 5 feet 7 inches, and breadth 1 foot 
10 inches ? Ans. 1Z. 1*. lOJd. 

Exam. 4. In a chimney-piece, suppose the 
length of the mantle and slab, each 4 feet 6 inches 
breadth of both together - 3 2 
length of each jamb -.44 
breadth of both together. 1 9 

Required the superficial content ? Ans. 21 feet 10 inches. 



V. CARPENTERS' AND JOINERS' WORK. 

To this branch belongs all the wood-work of a house, 
such as flooring, partitioning, roofing, &c. 

Large and plain articles are usually measured by the 
square foot or yard, &c. ; but enriched mouldings, and some 
other articles, are often estimated by running or lineal mea. 
sure ; and some things are rated by the piece. 

In measuring of Joists, take the dimensions of one joist, 



carpenters' and joiners' work. 



463 



and multiply its content by the number of them ; consider- 
ing that each end is let into the wall about & of the thick- 
ness, as it ought to be. 

Partitions are measured from wall to wall for one dimen- 
sion, and from floor to floor, as far as they extend, for the 
other. 

The measure of Centering for Cellars is found by making 
a string pass over the surface of the arch for the breadth, 
and taking the length of the cellar for the length : but in 
groin centering, it is usual to allow double measure, on ac- 
count of their extraordinary trouble. 

In Roofing, the dimensions, as to length, breadth, and 
depth, are taken as in flooring joists, and the contents com- 
puted the same way. 

In Floor -boarding, take the length of the room for one di- 
mension, and the breadth for the other, to multiply together 
for the content. 

For Stair-cases, take the breadtii of all the steps, by mak- 
ing a line ply close over ihem, from the top to the bottom, and 
multiply the length of this line by the length of a step, for 
the whole area. — By the length of a step is meant the length 
of the front and the returns nt the two ends ; and by the 
breadth is to be understood the girts of its two oater sur- 
faces, or the tread and riser. 

For the Balustrade, take the whole length of the upper 
part of the hand-rail, and girt over its end till it meet the 
top of the newel-post, for the one dimension ; and twice the 
length of the baluster on the landing, with the girt of the 
hand-rail, for the other dimension. 

For Wainscoting, take the compass of the room for the 
one dimension ; and the height from the floor to the ceiling, 
making the string ply close into all the mouldings, for the 
other. 

For Doors, take the height and the breadth, to multiply 
them together for the area. — If the door be panneled on 
both sides, take double its measure for the workmanship ; 
but if one side only be panneled, take the area and its half 
for the workmanship. For the Surrounding Architrave, girt 
it about the uppermost part for its length ; and measure over 
it, as far as it can be seen when the door is open, for the 
breadth. 

Window-shutters, Bases, &c. are measured in like manner. 
In measuring of Joiners' work, the sttin^ \a roa&fe v& ^ 



464 



SLATERS AND TILERS' WOBK. 



close into all mouldings, and to every part of the work over 
which it passes. 

EXAMPLES. 

Ex ax. 1. Required the content of a floor, 48 feet 6 inches 
long, and 24 feet 3 inches broad? An*. 11 sq. 76} feet. 

Exam. 2. A floor being 36 feet 3 inches long, and 16 feet 
6 inches broad, how many squares are in it ? 

Ans. 5 sq. 98} feet. 

Exam. 3. How many squares are there in 173 feet 10 
inches in length, and 10 feet 7 inches height, of partitioning ? 

Ans. 18*3973 squares. 

Exam. 4. What cost the roofing of a house at 10*. 6d. 
a square ; the length within the walls being 52 feet 8 inches, 
and the breadth 30 feet 6 inches ; reckoning the roof £ of 
the flat? Ans. 121. 12s. ll}d. 

Exam. 5. To how much, at 0*. per square yard, amounts 
the wainscoting of a room ; the height, taking in the cornice 
and mouldings, being 12 feet 6 inches, and the whole com- 
pass 83 feet 8 inches ; also the three window-shutters are 
each 7 feet by 8 inches by 3 feet G inches, and the door 7 feet 
by 3 feet inches ; the door and shutters, being worked on 
both sides, arc reckoned work and half work ? 

Ans. 36/. 12*. 2*d. 



VI. SLATERS' AND TILERS' WORK. 

In these articles, the content of a roof is found by mul- 
tiplying the length of the ridge by the girt over from eaves 
to caves ; making allowance in this girt for the double row 
of slates at the bottom, or for how much one row of slates or 
tiles is laid over another. 

When the roof is of a true pitch, that is, forming a right 
angle at top; then the breadth of the building, with its half 
added, is the girt added over both sides nearly. 

Iu angles formed in a roof, running from the ridge to the 
eaves, when the angle bends inwnnN. it is called a valley ; 
but when outwards, it is onllcd n hip. 

Deductions are made ibr chimney shafts or window holes. 



PLASTERERS* won. 



EXAMPLES. 

Exam. 1. Required the content of a slated roof, the 
length being 45 feet 9 inches, and the whole girt 34 feet 3 
inches? Ans. 174^ yank. 

Exam. 2. To how much amounts the tiling of a house, 
at 25*. 6d. per square ; the length being 43 feet 10 inches, 
and the breadth on the fiat 27 feet 5 inches ; also the caves 
projecting 16 inches on each side, and the roof of a true 
pitch ? Ans. 241. 9s. fyd. 



VH. PLASTERERS' WORK. 

Plasterers' work is of two kinds ; namely, ceiling, which 
is plastering on laths ; and rendering, which is plastering on 
walls : which are measured separately. 

The contents are estimated either by the foot or the yard, 
or the square, of 100 feet. Enriched mouldings, dec. are 
rated by running or lineal measure. 

Deductions are made for chimneys, doors, windows, &c. 

examples. 

Exam. 1. How many yards contains the ceiling which is 
43 feet 3 inches long, and 25 feet 6 inches broad ? 

Ans. 122}. 

Exam. 2. To how much amounts the ceiling of a room, 
at 10J. per yard : the length being 21 feet 8 inches, and the 
breadth 14 feet 10 inches ? Ans. 1/. 9s. 8Jrf. 

Exam. 3. The length of a room is 18 feet 6 inches, the 
breadth 12 feet 3 inches, and height 10 feet 6 inches ; to 
how much amounts the ceiling and rendering, the former at 
Sd. and the latter at 3d per yard : allowing for the door of 
7 feet by 3 feet 8, and a fire-place of 3 feet square ? 

Ans. 1Z. 13*. 3jd. 

Exam. 4. Required the quantity of plastering in a room, 
the length being 14 feet 5 inches, breadth 13 feet 2 inches, 
and height 9 feet 3 inches to the under side of the cornice, 
which girts 8} inches, and projects 5 inches from the wall 
on the upper part next the ceiling ; deducting only for a door 
7 feet by 4? 

Ans. 53 yards 5 feet 3} inches of rendering 
18 5 6 of ceiling 

39 Of} ofwrarcfe. 
Vol. I. 60 



406 



CLAftflUtl' WOfcX. 



VIII. PAINTERS 9 WORK. 

Painters' work is computed in square yards. Every part 
is measured where the colour lies ; and the measuring line is 
forced into all the mouldings and corners. 

Windows are done at so much a piece. And it is usual to 
allow double measure for carved mouldings, dec. 

EXAMPLES. 

Exam. 1. How many yards of painting contains the room 
which is 65 feet 6 inches in compass, and 12 feet 4 inches 
high ? Ans. 89}£ yards. 

Exam. 2. The length of a room being 20 feet, its breadth 
14 feet 6 inches, and height 10 feet 4 niches ; how many 
varus of painting are in it, deducting a fire-place of 4 feet 
by 4 feet 4 inches, and two windows each 6 feet by S feet 

2 inches ? Ans. 73^ yards. 
Exam. 3. What cost the painting of a room, at 6d. per 

yard ; its length being 24 feet 6 inches, its breadth 16 feet 

3 inches, and height 12 feet 9 inches ; also the door is 7 feet 
by 3 feet 6, and the window-shutters to two windows each 
7 feet 9 by 3 feet 8 ; but the breaks of the windows them- 
selves are 8 feet 6 inches high, and 1 foot 3 inches deep ; in- 
cluding also the window cills or seats, and the soffits above, 
the dimensions of which are known from the other dimen. 
sions : but deducting the fire-place of 5 feet by 5 feet 6 ? 

Ans. 31 3*. 10}d. 



IX. GLAZIERS' WORK. 

Glaziers take their dimensions, either in feet, inches, and 
parts, or feet, tenths, and hundredths. And they compute 
their work in square feet. 

In taking the length and breadth of a window, the cross 
bars between the squares are included. Also windows of 
round or oval forms are measured as square, measuring them 
to their greatest length and breadth, on account of the waste 
in cutting the glass. 

examples. 

Exam. 1. How many square feet contains the window 
which is 4*25 feet long, and 2-75 feet broad ? Ans. llf 



PAVEIt'* WORK. 



407 



Exam. 2. What will the glazing a triangular sky-light 
tome to, at lOd. per foot ; the base being 12 feet 6 inches, 
and the perpendicular height 6 feet 9 inches ? 

Ans. II. 15s. \\d. 

Exam. 3. There is a house with three tiers of windows, 
three windows in each tier, their common breadth 3 feet 1 1 
inches : 

now the height of the first tier, is 7 feet 10 inches 
of the second 6 8 
of the third 5 4 
Required the expense of glazing at I4d per foot ? 

Ans. 13*. lis. lO^d. 

Exam. 4. Required the expense of glazing the windows 
of a house at 13i. a foot ; there being three stories, and three 
windows in each story : 

the height of the lower tier is 7 feet 9 inches 
of the middle 6 6 
of the upper 5 3| 
and of an oval window over the door 1 10} 
the common breadth of all the windows being 3 feet 9 
inches ? Ans. 121. 5s. 6d. 



X. PAVERS' WORK. 

Pavers' work is done by the square yard. And the con* 
tent is found by multiplying the length by the breadth. 



EXAMPLS8. 

Exam. 1. What cost the paving a foot-path, at Ss. 4d. a 
yard ; the length being 35 feet 4 inches, and breadth 8 feet 
3 inches ? Ans. 51. 7s. ll$d. 

Exam. 2. What cost the paving a court, at 3*. 2d. per 
yard ; the length being 27 feet 10 inches, and the breadth 
14 feet 9 inches ? Ans. 71. 4s. 5}d. 

Exam. 3. What will be the expense of paving a rectan. 
gular court-yard, whose length is 63 feet, and breadth 45 
feet ; in which there is laid a foot-path of 5 feet 3 inches 
broad, running the whole length, with broad stones, at 8*. 
a yard ; the rest being paved with pebbles at 2s. 6d. a yard ; 

Ans. 401 5s. 10^1. 



XI. PLUMBERS* WORK. 



Pmnsfts' work is rated at so much a pound, or else fey 
the hundred weight of 113 pounds. 

J Sheet lead, used in roofing, guttering, Ac is from II t» 
101b. to the square foot. And a pipe of an inch bore ia cess* 
feoely l&or 141b, to tbe yard in length, 

jEgAJt* 1. How much weighs the lead which is 80 feet 
C inches long, and 3 feet 3 inches broad, at 8|lb. to the 
a^iarefeot? P Ana. 1091 ^ lb. 

Exam. 2. What cdst the covering and guttering* reef 
With lead, at 18s. the cwU ; the length of the roof being 43 
feet, and breadth or girt over it 32 feet ; the guttering 57 
feet long, and 2 feet wide ; the former 9*831 lb. and the latter 
7<37ftlb. to the square foot ? Ana, 1151. 9s. ljd- 



XII. TIMBER MEASURING. 

PROBLEM I. 

7b find the Area, or Superficial Content of a Board or 
Plank. 

Multiply the length by the mean breadth. 

Note. When the board is tapering, add the breadths at 
the two ends together, and take half the sum for the mean 
breadth. 0{ else take the mean breadth in the middle. 

By the Sliding Rule. 

Set 12 on b to the breadth in inches on a ; then against the 
length in feet on b, is the content on a, in feet and fractional 
parts. 

bxaxflbs. 

£xax. 1. What is the value of a plank, at lfd. per feet, 
whose length is 12 feet 6 inches, and mean breadth 11 
iftfbes? Ans. ls.5d. 



^ TIMBU MSA1UBIK9. 469 

Exam. 2. Requred the content of a board, whose length is 
11 feet 2 inches, and breadth 1 foot 10 inches ? 

Ans. 20 feet 5 inches 8*. 
Exam. 3. What is the value of a plank, which is 12 feet 
9 inches long, and 1 foot 3 inches broad, at 2±d. a foot? 

Ans. 3* . 3| A 

Exam. 4. Required the value of 5 oaken planks at 3d. 
per foot, each of them being 17 1 feet long ; and their several 
breadths as follows, namely, two of 13£ inches in the middle, 
one of 14 J inches in the middle, and the two remaining 
ones, each 18 inches at the broader end, and 11 J at the nar. 
rower ? Ans. II. 5s. 9±d. 

PROBLEM II. 

To find the Solid Content of Squared or Four-tided Timber. 

Multiply the mean breadth by the mean thickness, and 
the product again by the length, for the content nearly. 

By the Sliding Rule. 

C D D C 

As length : 12 or 10 : : quarter girt : solidity. 
That is, as the length in feet on c, is to 12 on d, when 
the quarter girt is in inches, or to 10 on d, when it is in 
tenths of feet ; so is the quarter girt on d, to the content 
on c. 

Note 1. If the tree taper regularly from the one end to 
the other ; either take the mean breadth and thickness in 
the middle, or take the dimensions at the two ends, and half 
their sum will be the mean dimensions : which multiplied as 
above, will give the content nearly. 

2. If the piece do not taper regularly, but be unequally 
thick in some parts and small in others ; take several different 
dimensions, add them all together, and divide their sum by 
the number of them, for the mean dimensions. 

EXAMPLES. 

Exam. 1. The length of a piece of timber is 18 feet 
6 inches, the breadths at the greater and less end 1 foot 
6 inches and 1 foot 3 inches, and the thickness at the greater 
and less end 1 foot 3 inches and 1 foot ; required the solid 
content? Ans. 28 feet 7 inches. 




Exam. 2. What is the content of the piece of MfhBef, 
whoee length is 24£ feet, and the mean breadth and thick* 
ness each 1-04 feet ? Ana. 26} feet. 

Exam. 3. Required the content of a piece of timber, 
whose length is 20-38 feet, and its ends unequal squares, the 
side of the greater being 19 J inches, and the side of the less 
9} inches t Ana. 29*7562 feet 

Exam. 4. Required the content of the piece of timber, 
whose length is 27*36 feet ; at the greater end the breadth 
Is 1*78, and thickness 1*23 ; and at the less end the breadth 
is 1-04, and thickness 0*91 feet ? Ans. 41 -278 feet. 

PROBLEM m. 

To find the Solidity of Round or Unfavored Timber. 

Multiply the square of the quarter girt, or of J of the 
toean circumference, by the length, for the content. 

By the Sliding Rule. 

Ae the length upon c : 12 or 10 upon d :: 
quarter girt, in 12ths, or lOths, on d : content on c. 

Note 1. When the tree is tapering take the mean dimen- 
sions as in the former problems, either by girting it in the 
fciddle, for the mean girt, or at the two ends, and taking half 
the sum of the two ; or by girting it in several places, then 
adding all the girts together, and dividing the sum by the 
number of them, for the mean girt. But when the tree is 
very irregular, divide it into several lengths, and find the 
content of each part separately. 

2. This rule, which is commonly used, gives the answer 
about i less than the true quantity in the tree, or nearly 
what the quantity would be, after the tree is hewed square 
in the usual way : so that it seems intended to make an 
allowance for the squaring of the tree. 

On this subject, however, Hutton's Mensuration, part v. 
sect. 4, may be advantageously consulted. 



EXAMPLES. 



Exam. 1. A piece of round timber being 9 feet 6 inches 
long, and its mean quarter girt 42 inches ; what is the 
content ? Ans. 116$ feet 

Exam. 2. The length of a tree is 24 feet, its girt at the 
thicker end 14 feet, and at the smaller end 2 feet ; required 
the content 7 Ans. 96 feet 



T1XBXK UASUBHIO. 471 

Exam. 3. What is the content of a tree whose mean 
girt is 3*15 feet, and length 14 feet 6 inches ? 

Ans. 8-9922 feet. 

Exam. 4. Required the content of a tree, whose length 
is IH feet, which girts in five different places as follows, 
namely, in the first plaoe 9*43 feet, in the second 7*92, in 
the third 6-15, in the fourth 4*74, and in the fifth 3*16 ? 

Ans. 42-519525. 



[ 472 J 



1 

• 



CONIC SECTIONS. 



DEFINITIONS. 



1. Conic Sections are the figures made by a plane cut- 
ting a cone. 

2. According to the different positions of the cutting 
plane there arise five different figures objections, namely, a 
triangle, a circle, an ellipsis, an hyperboK, and a parabola : 
the three last of which only are peculiarly called Conic Sec- 
tions. 



3. If the caning plane pass through 
the vertex of the, cone, and any part of 
the base, the section will evidently be a 
triangle ; as vab. 



4. If the plane cut the cone parallel to 
the base, or make no angle with it, the 
section will be a circle ; as abd. 




5. The section dab is an ellipse 
when the cone is cut obliquely through 
both sides, or when the plane is inclin- 
ed to the base in a less angle than the 
side of the cone is. 



6. The section is a parabola, when 
the cone is cut by a plane parallel to 
the side, or when the cutting plane and 
the side of the cone make equal angles 
with the base. 




DHflNITlOm. 



473 



7. The section is an hyperbola, when 
the cutting plane makes a greater angle 
with the base than the side of the cone 
makes. 



8. And if all the sides of the cone be 
continued through the vertex, forming 
an opposite equal cone, and the plane 
be also continued to cut the opposite 
cone, this latter section will be the op- 
posite hyperbola to the former ; as dac. 




9. The Vertices of any section, are the points where the 
cutting plane meets the sides of that vertical triangular sec- 
tion which is perpendicular to it ; as a and b. 

Hence the ellipse and the opposite hyperbolas, have each 
two vertices ; but the parabola only one ; unless we consider 
the other as at an infinite distance. 

10. The Axis, or Transverse Diameter, of a conic section, 
is the line or distance ab between the vertices. 

Hence the axis of a parabola is infinite in length, \b being 
only a part of it. 



Ellipse. Hyperbolas. Parabola. 




1 1 . The centre c is the middle of the axis. 

Hence the centre of a parabola is infinitely distant from 
the vertex. And of an ellipse, the axis and centre lie within 
the curve ; but of an hyperbola, without. 

12. A Diameter is any right line, as ab or db, drawn 
through the centre, and terminated on each side by the curve ; 
and the extremities of the diameter, or its intersections with 
the curve, are its vertices. 

Hence all the diameters of a parabola are parallel to the 
axis, and infinite in length. Hence also evei^ ftassfttax <& 
Vol. I. 61 



474 



CONIC SECTIONS, 



the ellipse and hyperbola has two vertices ; but of the pan. 
bola, only one ; unless we consider the other as at an infinite 
distance. 

13. The Conjugate to any diameter, is the line drawn 
through the centre, and parallel to the tangent of the curve 
at the vertex of the diameter. So, fo, parallel to the tangent 
at i>, is the conjugate to de ; and 111, parallel to the tangent 
at a," is the conjugate to ab. 

Hence the conjugate hi, of the axis ab, is perpendicular 
to it. .. 

14. An Ordinate to any diameter, is a line parallel to its 
conjugate, or to the tangent at its vertex, and terminated by 
the diameter and curve. So dk, el, are ordinates to the axis 
ab ; and mn, xo, ordinates to the diameter dk. 

Hence the ordinates of the axis are perpendicular to it. 

15. An Absciss is a part of auy diameter contained between 
either of its vertices and an ordinate to it ; as ak or bk, or 
dn or en. 

Hence, in the ellipse and hyperbola, ever}' ordinate has 
two determinate abscisses ; but in the parabola only one ; the 
other vertex of the diameter being infinitely distant. 

1G. The Parameter of any diameter, is a third proportional 
to that diameter and its conjugate, in the ellipse and hyper- 
bola, and to one absciss and its ordinate in the parabola. 

17. The Focus is the point in the axis where the ordinate 
is equal to half the parameter. As k and l, where dk or el 
is equal to the semi -parameter. The name focus being given 
to this point from the peculiar property of it mentioned in the 
corol. to thcor. 5* in the Ellipse and Hyperbola following, and 
to theor. in the Parabola. 

Hence, the ellipse and hyperbola have each two foci ; but 
the parabola only one. 




18. If t)al, fhg, be two opposite hyperbolas, having as 
for their first or transverse axis, and ab for their second or 
conjugate axis. And if dm, fbg, be two other opposite hy. 
pcrbolns having the same axes, hut in the contrary order, 
namely, ab their fatal txxvs, *iA » tat\t n*»wA\ vWwvlvfiee 



DEFINXTIOXS* 



475 



two latter curves d*e, fbg, are called the conjugate hyper- 
bolas to the two former dae, pro ; and each pair of opposite 
curves mutually conjugate to the other ; being all for con- 
venience of investigation referred to one plane, though they 
are only posited two and two in one plane ; as will appear 
more evidently from the demonstration of th. 2. Hyperbola. 

19. And if tangents be drawn to the four vertices of the 
^curves, or extremities of the axes, forming the inscribed 
rectangle iiikl ; the diagonals hck, icl, of this rectangle, 
are called the asymptotes of the curves. And if these lfamp. 
totes intersect at right angles, or the inscribed rectangle be 
a square, or the two axes ab and ab be equal, then the hy- 
berbolas are said to be right-angled, or equilateral. 



SCHOLIUM. 



The rectangle inscribed between the four conjugate hy- 
perbolas, is similar to a rectangle circumscribed about an 
ellipse, by drawing tangents in like manner, to the four ex. 
tremities of the two axes ; and the asymptotes or diagonals 
in the hyperbola, arc analogous to those in the ellipse, cut- 
ting this curve in similar points, and making that pair of 
conjugate diameters which arc equal to each other. Also, 
the whole figure formed by the four hyperbolas, is as it 
were, an ellipse turned inside out, cut open at the extre. 
mi tics, d, e, f, g, of the said equal conjugate diameters, and 
those four points drawn out to an infinite distance ; the cur. 
vature being turned the contrary way, but the axes, and the 
rectangle passing through their extremities, continuing fixed. 

And further, if there be four cones 
cscrf, cop, cmp, cno, having all the 
same vertex c, and all their axes in the 
same plane, and their sides touching or 
coinciding in the common intersecting 
lines mco, ncv ; then if these four 
cones be all cut by one plane, parallel 
to the common plane of their axes, there 
will be formed the four hyperbolas, gqr, 
fst, vkl, win, of which each two op- 
po8itesare equal ; and each pair resembles 
the conjugates to the other two, as here 
in the annexed figure ; but they are not 
accurately the conjugates, except only 
when the four cones are all equal, and 
then the four hyperbolic sections are all equal also. 




[ 476 ] 



OF THE EIJJPSE. 

THEOREM X. 

The Squares of the Ordinntes of the Axis are to each other 
as the Rectangles of their Abscisses. 

Let avb be a plane passing through 
the apju of the cone ; agiii another 
section of the cone perpendicular to 
the plane of the former ; ab the axis 
of this elliptic section ; and fg, hi, or- 
dimftes perpendicular to it. Then it 
will be, as fg 3 : hi 3 : : af . eb : ah . iib. 

For, through the ordinates fg, hi, 
draw the circular sections xgl, min, 
parallel^ to the base of the cone, having kl, mn, for their 
diameters, to which fg, hi, are ordinate, as will as to the 
axis of the ellipse. 

Now, by the similar trangles afl, ahn, and bfk, bhh, 

it is AF l AH * FL : UN, 

and fb : iib : ; kf : mu ; 

hence, taking the rectangles of the corresponding terms, 
it is, the rect. af . fb : ah . hb : : kf . fl : mu . iin. 

But, by the circle, kf . fl = f.; 8 , and mh . hn = hi 9 ; 
Therefore the rect. af . fb : ah . hb : : fg 8 : hi*, q. e. d. 




^ THEOREM II. 

As the Square of the Transverse Axis 
Is to the Square of the Conjugate : 
So is the Rectangle of the Abscisses 
To the Square of their Ordinate. 




ft 

For, by theor. 1, ac . cb : ad . db : : c« 9 : de* ; 
But, if c be the centre, then ac.cbs ac 1 , and ca is the 
semi-conjugate. 



OF TBS SLUMS. 477 

Therefore ac 9 : ad • db : : ac* : db 9 ; 

or, by permutation, ac 9 : ac 9 : : ad • db : de 9 ; 
or, by doubling, ab s : ab* : : ad . db : db 1 . a. s. d. 

Cord. Or. by div. ab : — : : ad . db or ca 9 — cd 9 : db 1 . 

' ab 

that is, ab : p : : ad . db or ca 9 — CD 9 : db 9 ; 

where p is the parameter^-, by the definition of it. 

That is, As the transverse, 9 
Is to its parameter, 
So is the rectangle of the abscisses, 
To the square of their ordinate. 

THEOREM III. 

As the Square of the Conjugate Axis 

Is to the Square of the Transverse Axis, 

So is the Rectangle of the Abscisses wf the Conjugate, or 
the difference of the Squares of the Semi-conjugate and 
Distance of the centre from any Ordinate of that Axis, 

To the Square of the Ordinate. 



That is, 
cb 9 : : ad . db or ca 9 — cd 1 : 



For, draw the ordinate ed to the transversa ab. 
Then, by theor. 1, ca 9 : ca 1 : : de 1 : ad .*bb or ca 1 — cd*, 

or ca 9 : ca 9 : : cd* : ca 9 — cJe 9 , 

But ca 9 : ca 9 : : ca 9 : ca 9 , 

theref. by subtr. ca 9 : ca 9 : : ca 9 — cd 1 or ad .db : da 9 . 

Q. E. D. 

Carol. 1. If two circles be described on the two axes as 
diameters, the one inscribed within the ellipse, and the other 
circumscribed about it ; (hen an ordinate in the circle will 
be to the corresponding ordinate in the ellipse, as the axis of 
this ordinate, is to the other axis. 

That is, ca : ca : : dg : de, 
and ca : ca dg : dz. 
For, by the nature of the circle, ad • db = dg* ; theref. 
by the nature of the ellipse, ca 9 : ca 9 : : ad • db or do* : de 9 , 
or ca : ca ; : iw \ iv* 




conic SBCTIOKt. 



In like manner . ca : ca : : dg : o*e. 
Also, by equality - dg : de or <:d : : o*e or tc 2 dg. 
Therefore ego is a continued straight line. 

Corel. 2. Hence also, as the ellipse and circle are made up 
of the same number of corresponding ordinates, which are 
all in the same proportion of the two axes, it follows that 
the areas of the whole circle and ellipse, as also of any like 
parts of them, are' in the same proportion of the two axes, 
or as tfcfe square of the diameter to the rectangle of the two 
axes ; that is, the areas of the two circles, and of the ellipse, 
are as the square of each axis and the rectangle of the two ; 
and therefore the ellipse is a mean proportional between the 
two circles. 

THEOREM IV. 

The Square of the Distance of the Focus from the Centre, 
is equal to the Difference of the Squares of the Semi- 
axes. 

Or, the square of the Distance between the Foci, is equal to 
the Difference of the Squares of the two Axes. 



That is, cf 9 = ca 8 - ca* 
or f/ j = ab 1 — a6 J 

b 

For, to the focus f draw the ordinate fe ; which, by the 
definition, will be the semi parameter. Then, by the nature 
of the curve - - ca 8 : ca 1 : : ca 2 - of* : Ft* ; 
and by the def. of the para, ca 8 : ca 2 : : ca 8 : fk 2 ; 
therefore - . c/i a = ca 2 - cr 2 ; ♦ 

and by addit. nnd subtr. cf 2 = ca 2 — ca'; 
or, by doubling, • rf 2 = ab 3 — ab*. q. e. p. 

Carol. 1. The two semi-axes, nnd the focal distance from 
the centre, are the sides of a right. angled triangle era ; and 
the distance Fa from the focus to the extremity of the con- 
jugate axis, is = ac the semi-transverse. 

Carol. 2. The conjugate semi-axis ca is a mean proper, 
ttonal between af, fb, or between a/*, /b, the distances of 
either focus from the two vertices. 

For co" = ca? — cf* =■ Vsfc- "V ^ • — w \ = af . fb. 




Or THE ELLIPSE. 



THEOREM V. 



x The Sum of two lines drawn from the two Foci to meet , 
at any Point in the Curve, is equal to the Transverse 
Axis. 




For, draw ag parallel and equal to ca the semi-conjugate ; 
and join cg meeting the ordinate de in h ; also take ci a 
4th proportional to ca, cf, cd. 
Then by theor. 2, ca 8 : ag 2 : : ca 3 — cd 3 : de 3 ; 
and, by sim. tri. ca 2 : ag 3 : : ca 3 — cd 3 : ag 8 — dh* ; 
consequently de 3 = ag 3 — du 2 = ca 3 — dh 3 . 
Also, fd = cf ^ cd, and fd 2 = cf 2 — 2of . cd + cd 3 ; 
And, by right-angled triangles, fe 3 = fd 3 + de 2 ; 
therefore fe 2 = cf 2 + ca 2 — 2cf . cd + cd 2 — dh 3 ; 
But by theor. 4, cf 3 + ca 3 = ca 2 , % 
and by supposition, 2cf . cd = 2ca • ci ; 
theref. fe 3 = ca 3 — 2ca . ci + cd 3 — dh 3 . 
Again, by supp. ca 3 : cd 3 : : cf 3 or ca 3 — ag 3 : cr ; 
and, by sim. tri. ca 3 : cd 8 : : ca 3 — ag 3 : cd 3 — dh 3 ; 
therefore • ci 3 = cd 3 — dh 8 ; 
consequently fe 2 =■= ca 3 — 2ca . ci + cr*. 
And the root or side of this square is fe = ca — ci = ai. 
In the same manner it is found that fz = ca + ci = bi. 
Conseq. by addit. fe + ft = ai + bi = ab. q. k. d. 

Cord. 1. Hence ci or ca — fe is a 4th proportional to 
ca, cr, CD. 

Coral. 2. And /e — fe = 2ci ; that is, the difference be* 
tween two lines drawn from the foci, to any point in the 
curve, is dquble the 4th proportional to ca, cf, cd. 

CoroL 3. Hence is derived the common method of de- 
scribing this curve mechanically by points, or with a thread, 
thus: 



480 



come accnoxi. 



In the transverse take the foci f,/, 
mod any point i. Then with the radii 
ai, bi, and centres r,/, describe arcs 
intersecting in b, which will be a 
point in the curve. In like manner, 
assuming other points i, as many 
other points will be found in the 
curve. Then with a steady hand, 
the curve line may be drawn through all the points of inter- 
section E. 

Or, take a thread of the length ab of the transverse axis, 
and fix its two ends in the foci f, /, by two pins. Then 
carry a pen or pencil round by the thread, keeping it always 
stretched, and its point will trace out the curve line. 



THEOREM VI. 

If from any Point i in the Axis produced, a Line il be 
drawn touching the Curve in one Point l ; and the Or- 
dinate lh be drawn ; and if c be the Centre or Middle 
of ab : Then shall cm be to ci as the Square of am to the 
Square of ai. 



cm : 




For, from the point i draw any other line if.h to cut the 
curve in two points is and h ; from which let fall the perpen- 
diculars ed and hg ; and bisect do in k. 

Then, by theor. 1, ad . db : ag . gb : : de 2 : gh 9 , 
and by sim. triangles, id* : ig 3 : : de 9 : gh 9 ; 
there f. by equality, ad . db : ag . gb : : id* : ig*. 

But DB = CB + CD = AC + I'D — AG + DC CG = 2cE + AG, 

and GB = CB CG = AC CG = AD+ DC — CG = 2cK + AD 5 

theref. ad . 2ck + ad . ag : ag . 2c k + ad . ag : : id* : 
and, by div. dg . 2ck : ig 8 — id 2 or dg . 2ik : : ad . 2ck + 

AD . AG : ID a , 

or 2ck : 2ik : : ad . 2ck + ad . ag : id', 
or ad . 2ck : ad . 2ik : : ad . 2ck + ad . ag : id 9 ; 
theref. by div. ck : ik : : ad • ag : id' — ad . 2ik, 
and, by corop. ck : ic : : ad . ag : id 9 — ad . id + ia 9 
or - ck : ci : : ad . ag : ai 9 . 




OF THE ELLIPSE. 



481 




But, when the line ih, by revolving about the point i, 
comes into the position of the tangent il, then the points e 
and ii meet in the point l, and the points d, k, g, coincide 
with the point m ; and then the last proportion becomes 
cm : ci : : am 3 : ai s . q. e. d. 

THEOREM VII. 

If a Tangent and Ordinate be drawn from any Point in the 
Curve, meeting the Transverse Axis ; the Semi -transverse 
will be a Mean Proportional between the Distances of the 
said two Intersections from the Centre. 

That is, 

ca is a mean proportional 
between cd and or ; 

or cd, ca, cr, are con- 
tinued proportionals. 



For. by tlieor. G, cd : ct : : ad 9 : at 2 
that is, cd : ct : : (ca — co) a : (ct — ca) 3 , 
or - cd : ct : : cd 3 + CA a : ca 9 + ct 1 , 
and - cd : dt : : cd 3 + ca 9 : ct 9 — cd 9 , 
or - cd : dt : : cd ! + ca 2 : (ct + cd)dt, 
or - cd 9 : cd . dt : : cd* + ca 2 : (cd . dt) -f- (ct . dt), 
hence cd 2 : ca 9 : : cd . dt : ct . dt, 
and - cd': ca j : : cd : ct. 

therefore (t^. 78, Geora.) cd : ca : : ca : ct. q. e. d. 

Cord. 1. Since cr is always a third proportional to cd, 
ca ; if the points d, a, remain constant, then will the. point 
T be constant also ; nnd therefore all the tangents will meet 
in this point t, which are drawn from the point e, of every 
ellipse described on the same axis ar, where they are cut by 
the common ordinate def drawn from the point d. 

Cord. 2. When the outer ellipse, by enlarging, becomes 
a circle, as at the upper figure at e, then by drawing et 
perp. to ce, and joining t to the lower e, the tangent to the 
point e at the ellipse is obtained. 

theorem viii. 

If there be any Tangent meeting four Perpendiculars to the 
Axis drawn from these four Points, namely, the Centre, the 
two Extremities of the Axis, and the Point of Contact ; 
those four Perpendiculars will be Proportions. 
Vol. I. 62 



488 



COKIC SKCTIOM* 




That is, 
AO : de : : ch : bi. 



For, by theor. 7, tc : ac : : ac : dc, 
theref. by div. ta : ad : : tc : ac or cb, 
and by comp. ta : td : : tc : th, 
and by sim. tri. ao : de : : cn : hi. q. E. D. 

Cord. 1. Hence ta, td, tc, tb ) ^ . , 

and TG, TB, TH, TI \ *"> P~pOrtM»^ 

Fur these are as ao, dr, ch, bi, by similar triangles. 

Corel. 2. Draw ai to bisect de in p ; then since 
ta : te : : tc : ti, the triangles tab, tci are similar, as well 
as the triangles aed, cbi, and adp, abi. 
Hence - ad : dk : : cb : bi 
and - ad : dp : : ab : bi 

.\ de : dp : : ab : cb : : 2 : 1 ; which sug- 
gests another simple practical method of drawing a tangent 
to an ellipse. 



theorem ix. 



If there be any Tangent, and two Lines drawn from the 
Foci to the Point of Contact ; these two lines will make 
equal Angles with the Tangent. 



That is, 
the JL vet = L fve. 




For, draw the ordinate dk and /a parallel to fe. 
By cor. 1 , theor. 5, ca : cd : : cf : ca — fk, 
and by theor. 7, ca : cd : : ct : ca ; 
therefore ct : cf : : ca : ca — fe ; 

and by add. and sub.TF : rf : : fe : 2ca— fe or /e by th. 5. 
But by sim. tri. tf : rf : : re :fe ; 
therefore ft: — fe> and conseq. = /.foe. 
But because fe is parallel to fe, the k ^fet ; 
therefore ^.fet = Lfot. q. e. d. 



Or TUB ILLIP8X 



488 



Cord. As opticians find that the angle of incidence is equal 
to the angle of reflection, it appears from this theorem, that 
rays of Tight issuing from the one focus, and meeting the 
curve in every point, will be reflected into lines drawn from 
those points to the other focus. So the ray fn is reflected 
into fe. And this is the reuson why the points f,/, ure call- 
ed the foci, or burning points. 

theorex x. 

All the Parallelograms circumscribed about an Ellipse are 
equal to one another, and each equal to the Reclaugle of 
the two Axes. 



That is, 
the parallelogram pqrs 
the rectangle ab • ab. 




S 

Let' kg, eg y be two conjugate diameters parallel to the 
aides of the parallelogram, and dividing it into four less and 
equal parallelograms. Also, draw the ordinate* i>k, de, and 
ck perpendicular to pq; and let the axis c.\ produced meet 
the sides of the parallelogram, produced if uecessury, in t 
and t. 

Then, by theor. 7, ct : ca : : ca : cd, 

and - - c/ : ca : : ca : erf ; 

theref. by equality, err : ct : : cd : cd ; 

but, by sim. triangles, ct : ct : : td : cd, 

theref. by equality, td : cd : : cd : cd, 

and the rectangle td . dc = is the square cd\ 

Again, by theor. 7, cd : ca : : ca : ct, 

or, by division, cd : ca : : da : at, 

and by composition, cd : db : : ad : dt ; 

conseq. the rectangle cd . dt = cd 9 =* ad . dh*. 

But, by theor. 1, ca*: ca*:: (ad . db or) cd* : de?, 

therefore - ca : ca . : cd : dk ; 

or - - - ca i dk : : ca : cd ; 

By th. 7, - ct : ca : : ca : ca\ 



* Cord. Bet-nil <« cd 1 = ad . ns — ca^ — «d", 

therefore ca* = cd 1 ~f- cH* . 
In like manner, c« J = ss'-f *V. 



484 



oohic tactions 



by equality 
by sim. tri. 
theref. by equality, 
But, by trim. tri. 
theref. by equality, 
and the rectangle 
But the reet. 
theref. the rect. 
conseq. the rect. 



ct : ca : : ca : db, 

ct : ct : : de : de, 

ct : ca : ; ca : de. 

ct : ck : : ce : de; 

cr : ca : : ca : ce, 

ck • ce ~ ca . ca. 

ck . ce = the parallelogram cbp*, 

ca • c<i — the parnllelograra cbic, 

ab • ab ~ the parallelogram pqrb- q.e*d. 



THEOREM XX. 



The Sum of the Squares of every Pair of Conjugate Diame- 
ters, is equal to the same constant Quantity, namely, the 
Sum of the Squares of the two Axes. 



That is, 
ab* + a&* = eo 8 + eg 9 ; 
where eo, eg, are any pair of con- 
jugate diameters. 




For, draw the ordinates ed, ed y 
Then, by cor. to Theor. 10, ca 3 = cir + cd\ 
and - - - co 9 = de' 4- de 1 ; 
therefore the sum ca 8 + va 1 = cd 2 + dk* + cd? + de*. 
But, by right-angled as, ce 51 = cn a 4- de', 
and ce 2 = cd 3 + de 1 ; 

therefore the sum t e 2 4- cc 2 = cn 2 + de 2 + cd 1 + de 1 . 
consequently - ca 2 + ca 2 ce 2 + ce* ; 
or, by doubling, ab 2 + ab 2 = kg 5 + eg 9 . q. b. d. 



THEOREM XII, 



The difference between the semi -transverse and a line drawn 
from the. focus to any point in the curve, is equal to a 
fourth proportional to the semi -transverse, the dixtnnce 
from the centre to the focus, and the distance from the 
centre to the ordinate belonging to that point of the 
curve. 



OF trs uunii 



485 




That is, 
ac — fe =■ ci, or fe = ai ; 
and/is — ac ci, or /k = bi. 
Where ca : cf : : cd : ci the 4th 
proportional to ca, cf, cd. 



For, draw ao parallel and equal to ca the semi-conjugate ; 
and join co meeting the ordinate de in h. 
Then, by theor. 2, ca 2 : ah 2 : : ca" - cd 2 : dr 2 : 
and, by sim. tri. ca 9 : ao* : : ca* — cd 2 : ^o a — dh" ; 
consequently dk 2 »ao j -dh*=oi*— dh*. 
Also - fd=cf^-cd, end fd^cf 2 — 2cf . cd+cd 2 ; 
but by right-angled triangles, fd'+dk^fk 2 ; 
therefore - FK^ci^+ca 2 — 2cf . cd+cd 2 — du 1 . 
But by theor. 4, ca , +CF a =CA*; 
and, by supposition, 2cf . cd=2ca . ci ; 
therefore - - fb 2 =ca 2 — 2c a . ci+cd 2 — do 1 . 
But by supposition, ca 2 : cd' : : cf 8 or ca 2 — ag 2 : ct a . 
and, by sim. tri. ca 2 : cn a : 
therefore - - ci^cd 2 — nn a ; 
consequently - fe 9 «»ca 2 — 2ca . ci + cr 2 . 
And the root or side of this square is fe =-= ca — ci = ax. 
In the same manner is found /e= ca+ci=bi. q. v. d. 

Carol. 1. Hence ci or ca — fe is a 4th proportional to 

CA, CF, CD. 

Carol. 2. And fp. — fe =■ 2ci ; that is, the diffi rence 
between two lines draw from the foci, to any point in the 
curve, is double the 4th proportional to ca, cf, cd. 



: ca 2 — ao 2 : cd 2 — dh 2 ; 



THEOREM XIII. 



If a line be drawn from either focus, perpendicular to a tan- 
gent to any point of the curve ; the distance of their inter- 
sections from the centre will be equal to the semi-transverse 
axis. 



That is, if fp, 
fp, be perpendi- 
cular to the tan- 
gent T17?, then 
shall cp and cp 
be each equal to 
ca or CB. 




486 



come SECTIONS. 



For through the point of contact e draw rs, and /* 
meeting fp produced in o. Then the L gkp = ^ pep, 
being each equal to the L f&pt and the angles at p being 
right, and the side pk being common, the two triangles gep, 
fep are equal in all respects, and woe = fe, and gp = fp. 
Therefore, since fp = £fg, and fc = \rf, and the angle at 
f common, the side cp will be -= £/b or J ab, that is cp = ca 
*©r cb. And in the same manner cp = cx or cb. o> e. v. 

Card. 1. A circle described on the transverse axis, as a 
diameter, will pasc through the points p, p ; because all the 
lines ca, cp, cp 9 cb, being equal, will be radii of the circle. 

Cord. 2. cp is parallel to /e, and cp parallel to fb. 

Cord. 8. If at the intersections of any tangent, with the 
circumscribed circle, perpendiculars to the tangent be drawn, 
they will meet the transverse axis in the two foci. That is, 
the perpendiculars pf, pf give the foci r y f, 

THEOREU XIV. 

The equal ordi nates, or the ordi nates at equal distances 
from the centre, on the opposite sides and ends of aa 
ellipse, have their extremities connected by one right line 
passing through the centre, and thut line is bisected by 
the centre. 

That is, if cd = r«, or the ordinate de = gii ; 
then shall cr = ch, and k<;ii will be a right line. 




For when cd = cg, then also is de = on by cor. 2, th. 1. 
But the L d =r ^ «, being both right angles ; 
therefore the third side ce = en, and the / dce = ^.gcb, 
and consequently ech is a right line. 

Coroh 1. And, conversely, if ecu be a right line passing 
through the centre ; then shall it he bisected by the centre, 
or have ce — ch ; also de will be = oh, and cd = co. 



OF TU SLUMS. 487 

Carol. 2. Hence also, if two tangents be drawn to the two 
ends b, h of any diameter eh ; they will be parallel to each > 
other, and will cut the axis at equal angles, and at equal 
distances from the centre. For, the two cd, ca being equal 
to the two co, cb, the third proportionals <t, cs will be 
equal also ; then the two sides ce, ct being equal to the two 
ch, cs, and the included angle ect equal to the included 
angle bcs, all the other corresponding parts are equal : and^ 
so the L t = L s, and te parallel to ns. 

Carol. 3. And hence the four tangents, at the four extre- 
mities of any two conjugate diameters, form a parallelogram 
circumscribing the ellipse, and the pairs of opposite sides are 
each equal to the corresponding parallel conjugate diameters. 
For, if the diameter eh be drawn parallel to the tangent ts 
or hs, it will be the conjugate to eh by the definition ; and 
the tangents to e, k will be parallel to each other, and to the 
diameter eh for the same reason. 



THEOREX XV. 

If two ordinates ed, ed be drawn from the extremities e, e, 
of two conjugate diameters, and tangents be drawn to 
the same extremities, and meeting the axis produced in 
t and r; 

Then shall cd be a mean proportional between cd, da, and 
cd a mean proportional between cd, dt. 




For, by theor. 7, cd : ca : 

and by the same, cd : ca i 

theref. by equality, cd : cd : 

But, by sim. tri. dt : cd 

theref. by equality, cd : cd 

In like manner, cd : cd : 

Corel. 1. Hence cd : cd : : cs : ct. 



: ca : ct, 
: <:a : cn ; 
: cr : ct, 
: ct : cb; 
: cd : dt. 
: cd : da. a. 



s. p. 



488 



come lECTiom. 



« Carol. 2. Hence also cd : erf : : de : de. 

And the rectangle cd.de = erf . rff , or A cdb = a ede. 

Corof. 8. Also W J = cd . dt, 
and cd* ■= cd . rfa, 

Or erf a mean proportional between cd, dt ; 
and cd a mean proportional between cd y da. 

P THEOREM XVI. 

TTie name figure being constructed as in the last theorem, 
each ordinate will divide the axis, and the semi-axis added 
to the external part, in the same ratio. 

[See the last fig.] 

That if», da : dt : : dc : db, 
and dx : rfa : : dc : rfn. 
For, by Theor. 7, cd : ca : : ca : ct, 
and by div. en : ca : : ad : at, 
and by com p. cd : db : : ad : dt, 
or, - - - da : dt : : dc : db. 
In like manner, o*a : rfa : : rfc : ds. 



a. e. D. 



Corol. \. Hence, and from cor. 3 to the last, it is, 



cd 1 = CD . 
cd 3 = cd 



DT = AD 
rf*R = Ad , 



DB = CA a — CD*. 



ds = CA 8 



crf\ 



Corol. 2. Hence also, ca 2 = cd 2 + cd 2 , 
and ca 3 = de* + de*. 

Corol. 3. Further, because ca 2 : ca 1 : : ad . db or erf 2 : de*, 
therefore ca : ca : : erf : de, 
likewise ca : ca : : cd : rfe. 

THEOREM XVII. 



If from any point in the curve there be drawn an ordinate, 
and a perpendicular to the curve, or to the tangent at that 
point : then, the 

Dist. on the trans, between the centre and ordinate, cd, 
Will be to the dist. pd, ™ "> 

As sq. of the trans, axis 
To sq. of the conjugate. 



EL 



ca" : ca' 



That is, 
: : dc : dp. 



D P C 



For, by theor. % ca? ; ccr *. kt> * to 



• 



OF THE ELLIPSE. 



489 



But, by rt. angled As, the rect. td . dp-=pe s ; V 
and, by cor. 1, theor. 16, cd • dt=ad . db ; 

therefore - - - ca 2 : ca' : : td . dc : td . dp, 
or - - - - - ac 2 : ca 3 : : dc : dp. q. e. d. 



THEOREM. XVIII. 



If there be two tangents drawn, the one to the extremity**- 

of the transverse, and the other to the extremity of any^^ 
other diameter, each meeting the other's diameter pro- 
duced ; the two tangential triangles so formed, will be 
equal. 



For, draw the ordinate de. Then # 
By sim. triangles, cd : ca : : ce : cn ; 
but, by theor. 7, cd : ca : : ca : ct ; 
the re f. by equal, ca : ct : : ce : ex. 

The two triangles crt, can, have then the angle c common, 
and the sides about that angle reciprocally proportional ; those 
triangles are therefore equal, namely, the a cet = a can. 
Cord. 1. From each of the equal tri. cet, can, 
take the common space cape, 
and there remains the external A pat = apne. 
Corel. 2. Also from the equal triangles cet, can, 
take the common triangle ced, 
and there remains the A ted = trapez. aned. 



The same being supposed as in the last proposition ; then any 
lines kq, qg, drawn parallel to the two tangents, shall 
also cut off equal spaces. That is, 



That is, 
the triangle cET=the 
triangle can. 




theorem xix. 



Vol. I. 



68 




For, draw the ordinate pb. Then 
the three aim. triangles can, cde, cgh, 
are to each other as ca*, cd 1 , cg s ; 
th. by div. the trap, aned : trap, anhg : : ca 2 — cd* : ca j — ce*. 
But, by theor. 1, de 8 : oq 9 : : ca s — cd* : ca 2 — cc 1 . 

theref. by equ. trap, aned : trap, anhg : : de* : oq*. 
But, by8im. As, tri. ted : tri. kqo : : de* : gq\ 
theref by equality, ankp : ted : : anho : kqc. 
But, by cor. 2, theor. 18, the trap, aned = A ted; 
and therefore the trap, anho =" A kqg. 
In like manner the trap. anA#= AKgg. o. e. d. 
Carol. 1. The three spaces anhg, tehg, kqg, are aW equal. 
Carol. 2. From the equals anhg, kqg, 
take the equals anA^, Kqg, 
and there remains ghuo =» gqoa. 
Corol. 3. And from the equals #Ahg, gqao 9 
take the common space £?lkg, 
and there remains the Alqh = At? A. 
Corol. 4. Again, from the equals kqo, ttshg, 
take the common space klhg, 
and there remains tclk = -A Left. 

Corol. 5. And when 
by the lines kq, gh, 
moving with a parallel 
motion, kq comes into 
the position ie, where ^ 
cr is the conjugate to T 
ca; then 



the triangle kqg becomes the triangle ntc, 
and the space anhg becomes the triangle anc ; 
and therefore the Airc = A anc = tec. 
Corol. 6. Also when the lines kq and hq, by moviag 
with a parallel motion, come into the position ce f me, 




OP THE ELLIPSE. 



tke triangle lqh becomes the triangle cent, 
and the space telk becomes the triangle tec ; 
and theref. the Ac cm = A tec = Aanc = &uic. 



theorem xx. 



Any diameter bisects all its double ordinotes, or the lines 
drawn parallel to the tangent at its vertex, or to its con- 
jugate diameter. 



That is, if oq be pa- 
rallel to the tangent te, 
or to ce, thenehall La = 

Mr- 



For, draw qh, qh perpendicular to the transverse. 
Then by cor. 3, theor. 19, the A lqh ="= vqh ; 
but these triangles are also equiangular ; 
consequently their like sides are equal, or La = 14. 

Cord* Any diameter divides the ellipse into two equal 
parts. 

For, the ordinates on each aide being equal to each other, 
aud equal in number ; all the ordinates, or the area, on one 
side of the diameter, is equal to all the ordinates, or the areu, 
on the other side of iU 




TBEOKEX XXI. 



As the square of any diameter 
Is to the square of its conjugate, • 
So is the rectangle of any two abscisses 
To the square of their ordinate. 

That is, ce 8 : ce 9 :: el • lo or ce 3 — cl 2 : lq*. 



For, draw the tangent 
te, and produce the or- 
dinate ul to the trans- 
verse at k. Also draw 
<w» ex perpendicular to 
the transverse, and meet. 
ingEG in n and x. 

Then, similar triangles 




r 



\ 



403 come iBcnoKt. 

being as the squares of their like sides, it is, 

by sim. triangles, Acet : A cut : : ce* : cl 8 ; 

or, by division, Ackt : trap, tklk : : ck*: ce 1 — cx a . 

Again, by sim. tri. A : A t<m : : cc* : l<*'. 
But, by cor. 5thcor. 19, the Ac*m = acet, 
and, by cor. 4 theor. 10, the Ai<w = trap, tslk ; 
theref. by equality, ce : ce 1 : : en* — cl* : m*, 
* or - • ce 3 : ce* : : el . lg : i.q s . q. b. d. 

Cord. 1. The squares of the ordi nates to any diameter, 
are to one another as the rectangles of their respective 
abscisses, or as the difference of the squares of the semi- 
diameter and of the distance between the ordinate and centre. 
For they are all in the same ratio of ce 2 to ce\ 

Corel. 2. The above being a similar property to that be* 
longing to the two axes, all the other properties before laid 
down, for the axes, may be understood of any two conjugate 
diameters whatever, using only the oblique ordinates of these 
diameters, instead of the perpendicular ordinates of (he axes ; 
namely, all the properties in theorens 6, 7, 8, 14, 15, 16, 18, 
and )9. 



THEOREM XXII. 



If any two lines, that any where intersect each other, meet 
the curve each in two points ; then 
the rectangle of the segments of the one 
is to the rectangle of the segments of the other, 
as the square of the diam. parallel to the former 
to the square of the diam. parallel to the latter. 



That is, if cr and cr be y' 
parallel to any two lines ^ 



phq, pnq 



then shall 
ph . hq : pn 




For, d raw the diameter che, and the tangent te, and its 
parallels pk, hi, mii, meeting the conjugate of the diameter 
cr in the points t, k, i, m. Then, because similar triangles 
are as the squares of their like sides, it is, 

by sim. triangles cu a : up 2 : : acri : £gpk, 

and . - - . v:u A \ uiv 1 % \ kvaiw ^hk ; 

t!*eref. by division, en? *. o# — ^v? *. *. \ *x-vnu 



OF THE ELLIPSE. 



Again, by aim. tri. ce 3 : cn 9 : : acts : Ackh; 

and by division, cr 1 : cr* — cu 9 : : Acte : tehm. 
But, by cor. 5 theor. 19, the Acte = A cm, 
and by cor. 1 theor. 19, tf.hg kphg, or tehm = kpbm ; 
the re 1*. by equ. ce 9 : ce 3 — ch 9 : : cr* : gp 9 —- gh 9 or ph . HQ. 
In like manner ck* : ce* — ch 9 : : cr 9 : pu . uq. 
Theref. by equ. cr 3 : cr 3 : : ph . na : pH. Hq. q. e. d. 

Carol. 1. In like manner, if any other line p'n'q, parallel 
to cr or to pq, meet phq ; since the rectangles ph'q, p'n'q' 
are al*o in the same ratio of cr 9 to cr 9 ; therefore reel. 
phq, : pnq : : pii'q : pu'q'. 

Also, if another line p ho? be drawn parallel to pq or cr ; * 
because the rectangles, p Aq', p hq are still in the same ratio, 
therefore, in general, the rect. phq : puq : : p ha : p hq'. 

That is, the rectangles of the pans of two parallel lines, 
are to one another, as the rectangles of the parts of two other 
parallel linen, any where intersecting the former. 

Cord. 2. And when any of the lines only touch the curte, 
instead of cutting it, the rectangles of such become squares, 
and the general property still attends them. 




Corel. 3. And hence « : re : : If : /e. 



[484] 
OF THE HYPERBOLA. 



THEOREM I. 



The Squares of the Ordinate* of the Axis are to each other 
as the Rectangles of their Abscisses. 

Let avb be a plane passing 
through the vertex and axis of 
the opposite cones; agih an- 
other section of them perpendi- 
cular to the plane of the former ; 
ab the axis of the hyperbolic 
sections; and fo, hi. ordinatcs 
perpendicular to it. Then it will 
be, as fg 2 : hi 3 : : af . fb : ah . hb. 

For, through the ordinates 
fg, hi, draw the circular sections 
kgl, min, parallel to the base of 
the cone, having ki., m\, for their diameters, to which fg, 
hi, are ordinates, as well as to the cxis of the hyperbola. 

Now, by the similar triangles afl, ahn, and bfk, bhm, 
it is af : ah : : fl : hn, 
and fh : hr : : kf : mh ; 

hence, taking the rectangles of the corresponding terms, 

it is, the rect. af . fb : ah . iiu : : kf . fl : mh . hn. 
But, by the circle, kf . fl ■» fg 3 , and mh . hn = hi 3 ; 
Therefore the rect. af . fb : ah . hb : : kg 9 : hi 2 . 

q. e. D. 




THEOREM II. 



As the Square of the Transverse Axis 
Is to the Square of the Conjugate : 
So is the Rectangle of the Abscisses 
To the Square of their Ordinate. 



That is, ab 3 

ac 3 : ac? : ; - 




OF TflS RTPSRBOLA. 



For, by theor. 1 , ca . cb : ad . db : : ca* : dr* ; 
But, if c be the centre, then ac . cb = ac 1 , and ca is the 

setni.conj. 

Therefore . ac 1 : ad . db : : cc* : de* ; 
or, by permutation, ac 9 : oc 9 : : ad . db : dk* ; 
or, by doubling, ab' : a& 9 : : ad . db : db 1 . <l. e. d. 

ab 9 

Cord. Or, by div. ab : — : : ad • db or cd*— ca 1 : de*, 

J AB 

that is, ab i p 1 1 ad • db or cd* — ca* : de* ; 
ab 3 

where p is the parameter — by the definition of it. 

That is, As the transverse, 
Is to its parameter, 
So is the rectangle of the abscisses, 
To the square of their ordinate. 



Otherwise, thus ; 



Let a continued plane, cut 
from the two opposite cones, the 
two mutually connected oppo- 
site hyperbolas hag, hag, whose 
vertices arc a, a, and bases u<», 
hg y parallel to each other, fall, 
ing in the planes of the two pa- 
rallel circlet* lgk, Jgk. Through 
c, the middle point of ah, let a 
plane be drawn parallel to that 
of lok, it will cut in the cone 
lvk a circular section whose di- 
ameter is mn ; to which circu- 
lar section, let d be a tangent at t. 

Then, by sim. tri. 
Acm, AFL 
and, by sim. tri. 
acn, an 




ac : cm 



AF : I*L ! 



ac : cn : : of : fk. 



/, ac • ca i cm • eft 1 1 af • Fa : lf 
or, ac* : c£* : : af . ra : fg*. 



FK, 



In like manner, for the opposite hyperbola 
ac* \cP n hf .fa :fg*. 

Hens ct is what.ic usually denominated the semMSonjvgsje 
to the opposite hyperbolas hak, hak : but it is evidently «sf 
in the same plane whh them. 



406 



OOXIC SKCTIOXS. 



THEOREM III. 

As the Square of the Conjugate Axis 

In to the Square of the Transverse Axis, 

So is the Sum of the Squares of the Semi •conjugate, and 

distance of the Centre from any Ordinate of the Axis, 
To the Square of the Ordinate. 

m 



ca 9 



That i*, 
ca 1 : : ca 1 + cd 9 



dm 9 . 




For, draw the ordinate ed to the transverse, ab. 



ca 9 

CA 9 

ca 9 

CA 9 



Then, by theor. 1, ca 9 
or - - ca* 
B^ut - ca 3 
tfieref. by compos, ca 9 
In like manner, ca 9 : ca 9 : : 
Corol. By the last theor. ca 9 
and by this theor. ca 9 : 
therefore - de* 
In like manner, 



de 9 
ca 2 

CA 9 

ca 9 
ca 9 

I* 9 
Df 9 . dE* 



AD . DB Or CD 1 — CA 9 , 
C*E 9 — CA 9 . 
. CA 9 . 

+ erf 9 <*e 9 . 

+ CD' : DC 9 . Q.. E. D. 

: : cd 9 — ca 9 : de 9 9 



:cd 9 +ca 9 

: CD 9 — CA 9 

: cd 9 — ca 1 



:d* 3 , 
: cd 9 



4- CA 9 

cd 9 + ca 9 



THEOREM IV. 



The Square of the Distance of the Focus from the Centre, is 
equal to the Sum of the Squares of the Semi-axes. 

Or, the Square of the Distance between the Foci, is equal to 
the Sum of the Squares of the two Axes. 

That is, f . 

cf 9 = ca 9 + ca 9 , or Hrt 
pfi ab 9 + ab*. 



For, to the focus f draw the ordinate fe ; which, by the 
definition, will be the semi- parameter. Then, by the nature 
of the curve . ca 9 : ca 9 : : cf 9 — ca 9 : fe 1 ; 

and by the def. of the para, ca 9 : ca 9 ; : ca 9 : fe 9 ; 
therefore - - ca 2 = cf 9 — ca 9 ; 
and by addition • cf 2 = ca 9 + ca 9 ; 
or, by doubling, — -V afc* * 4. *• D. 




Or THE HYPERBOLA. 



497 



Corol. 1 . The two semi-axes, and the fncal-distance from 
the centre, are the sides of a right-angled triangle c\a ; and 
the distance \a is = cf the focal distance. 

Corol. 2. The conjugate semi-axis ca is a mean propor- 
tional between af, fb, or between a/*,/b, the distances of 
either focus from the two vertices. 

For ca* = cf 3 — ca 2 = (cf + ca) . (cf — ca) = af . fb, 

THEOREM V. 

The Difference of two Lines drawn from the two Foci to 
meet at any Point in the Curve, is equal to the Transverse 
Axis. 




For, draw ag parallel and equal to ca the semi. conjugate ; 
and join cg, meeting the ordinate de produced in h ; also 
take ci a 4th proportional toe a, cf, cd. 

Then, by th. 2, ca : ag : : cd 3 — ca 3 : de 1 ; 
and, by sin. As, ca 2 : ag 3 : : CD i — ca 3 : dh 3 — ag' ; 
consequently de 3 = dh 3 — ag 3 = dh 3 — ci ? . 
Also, fd = cf cd, and fd j = cf 3 — 2cf . cd -f • cd 3 ; 
and, by right. angled triangles, fe 3 = fd 2 + de 3 . 
therefore fe 3 = cf 3 — ca J — 2cf . cd + cd 2 + dh 
But, by theor. 4, cf 3 -r~ ca 3 = ca s , 
and, by supposition, 2c f . cd = 2c a . ci ; 
theref. fe 3 = ca 3 — 2ca . ci + cd 3 + dh" ; 
Again, by suppos. ca 2 : cd 2 : : cf* or ca 2 + ag 3 : cr ; 
and, by sim. tri. ca 3 : cd 3 : ; ca 3 + ag 3 : cd 3 + dh 1 ; 
therefore - ci 3 = cd 3 + dh 3 ■= en 3 ; 
consequently fe 3 = ca 3 — 2ca . cr +ci 3 . 

And the root or side of this square is fe = ci — ca = ai. 
In the same manner, it is found that/k = ci + ca = bi. 
Conseq. by subtract. f% — fe = bi — ai = ab. q. b. d, 

Corol. 1. Hence ch = ci is a 4th proportional to ca, cf, 
cd. 

Corol. 2. And /e + fe = 2ch or 2ci ; or fe, ch, fk 9 are 
in continued arithmetical progression, the common difference 
being ca the semi-transverse. 

Vol. I. 64 



4fe8 COKIC SECTIONS. 

\ CoroL 8. Hence is derived the common method of deecrib- 
*irtg this curve mechanically by points, thus: 

In the transverse ab, produced, take the foci f,/, nnd anj 
point i. Then with the radii ai, bi, and centres f,/„ describe 
arcs intersecting in e, which will be a point in the curve. Id 
like manner, assuming other points i, as many other points 
will he found in the curve. 

Then, with a siendy hand, the curve line may be drawn 
through all the points of intersection r. 

In the same manner are constructed the other two or con- 
jugate hyperbolas, using the axis ab instead of ab. 



THEOREM VI. 

If from any Point i in the Axis, a Line il be drawn touching 
the Curve in one Point l ; and the Ordinate lm be drawn ; 

: and if c be the Centre or the Middle of ab : Then shall cm 
be to ci as the Square of am to the Square of ai. 

**>- 

That is, 
cm ; ci : : am 2 : Ai a . 



For, from the point i draw any line irh to cut the curve 
in two points e and 11 : from which let fall the perps. ed, hg ; 
and bisect no in k. 

Then, by theor. 1, \d . db : ac . gb : : de 3 : gh 9 , 
and by sim. triangles, ur* : iG a : : de 3 : gh*; 
theref. by equality, ad . db : ag . gb : : id 9 : ig*. 

But DB = CH + CD = CA + CD = CG + CD — AG = 2cK AG, 

and tut = cn + ct: = c\ + cc = cg + cd — ad = 2ck— ad ; 
thcrcf. a i) . 2c k — ad . ag : ag . 2ck — ad . ag : : id 3 : ig 3 , 
and. by div. »g . 2ck : ig 3 — id 2 or dg . 2ik : : ad . 2c B 

— ad . ag : id 2 ; 
or - 2c k : 2ik : : ad . 2ck — ad . ag : id 3 ; 
or ad . 2c k : ad . 2ik : : ad . 2c k — ad . ag : id 3 ; 
theref. by div. ck : ik : : ad . ag : ad . 2ik — id 8 , 
and, by div. ck : ci : : ad . ag : id 2 — • ad . (id + ia), 
or - ck : ci : : ad . ag : ai 2 . 

But, when the line in, by revolving about the point i, comes 
into the position of the tangent il, then the points e and h meet 
in the point l, nnd the points d, k, g, coincide with the point 
>i ; and then the last proportion becomes cm : ci ; : am 3 : ai 3 . 

E. D. 




.» / 

OF THE HTPBfiBOLA. 



THEOREM VII. 

If a Tangent and Ordinate be drawn from any Point in the 
Curve, meeting the Iransverse Axis ; the Semi- 1 runs vewc 
will he a Mean Proportional between the Distances of the 
said Two Intersections from the Centre. 

That is, 

ca is a mean proportional between 
cd and ct ; or cd, ca, it, are con- 
tinued proportionals. 

For, by th. 6, cd : ct : : ad* : at 3 , 
that is, - cd : ct : : (cn — ca) s : (ca — ct) 1 , 
or - - cd : ct : : cd' + ua 3 : ca 2 + cr 3 , 
and - - cd : or : : cd 3 4- ca 1 : cd 2 — ct 3 , 
or - - cd : dt ; : cd* 1 + ca 9 : (cd + ct) dt, & 
or cd* : cd . dt : : cd 8 + ca 3 : cd . dt + ct . td ; 
hence cd 2 : ca 3 : : cn . dt : ct . td, 
and co a : ca 3 ; : cd : ct, 

theref. (th. 78, Geom.) cd : ca : . ca : ct. q. e. d. 

Carol. Since ct is always a third proportional to cn, ca ; 
if the points d, a, remain constant, then will the point r be 
constant also ; and therefore all the tangents will meet in this 
point t, which are drawn from the point e, of every hyperbo- 
la described on the same axis ar, where they are cut by the 
common ordinate deb drawn from the point d. 

THEOREM VIII. 

If there be any Tnngent meeting four Perpendiculars to the 
Axis drawn from these four Points, namely, the Centre, 
the two Extremities of the Axis, and the Point of Contact ; 
those four Perpendiculars will be Proportionals. 



That is, 
ao : de : : en : bi. 



ac : nc, 

tc : ac or cd, 

TC ' TB, 

cu : HI. 



499 

■ I 





For, by thcor. 7, tc : ac : . 
theref. by div. ta : ai> : : 
and by com p. ta : td : : 
mad by sim. tri. ao : de : : 



600 



CONIC 1X011091. 



Cord. Hence ta, td, tc, tb ) | 

and TO, TE, Til, ti \ 

For these are as ag, de, cii, bi, by similar triangles. 



ore also proportionals. 



THEOREM. IX. 

If thore be any Tangent, and two Lines drawn from the Foci 
to the Point of Contact ; these two Lines will make equal 
Angles with the Tungent. 

That is, 1 l^n^^FD 

the JL fbt = L foe. 




For, draw the ordinate pe, and fe parallel to fe. 
By, cor. 1, theor. 5, ca : cn : : cf : ca + fe, 

cd : : cr : ca ; 
cf : : ca : ca + fe ; 
rf : : fe : 2ca + fe or /e by th. 5. 

if :: fe :/k; 

-fe y and conseq. Le = &f$e. 
fe is parallel to fe, the Le = Z.fkt ; 

Z. FKT = £fEC. Q. K. D. 



and by th. 7, ca 
therefore - ct 
and by add. and sub. tf 
But by sini. tri. i f 
therefore - fv. 
But, because 
therefore the 

Corol. As opticians find that the angle of incidence is equal 
to the angle of reflection, it appears, from this proposition, 
that rays of light issuing from the one focus, and meeting the 
curve in every point, will be reflected into lines drawn from 
the other focus. So the ray ft: is reflected into fe. And 
this is the reason why the points f, /, are called foci, or 
burning points. 



THEOREM X. 



All the Parallelograms inscribed between the four Conjugate 
Hyperbolas are equal to one another, and each equal to 
the Rectangle of the two Axes. 



That is, 
the parallelogram pqrs 
the rectangle ab . ab. 




4 



OF THE HYPEBBOLA. 



Ml 



Let eg, eg, be two conjugate diameters parallel to the i 
of the parallelogram, and dividing it into four less and equal 
parallelograms. Also, draw the ordi nates dk, de, andec 
perpendicular to pq ; and let the axis produced meet the sides 
of the parallelograms, produced, if necessary, in T and t. 

Then, by theor. 7, ct : ca : : ca : cd, 
and - • cl : ca : : ca : cd ; 
theref. hy equality, ct : cl : : cd : cd ; 
but, by sim. triangles, ct : ct : : td : cd, 
theref. by equality, td : cd : : cd : cd, 
and the rectangle td • dc is '= the square cd?. 
Again, hy theor. 7, cd : ca : : ca : ct, 
or, by division, cd : ca : : da : at, 

and, by composition, cd : db : : da : dt ; 
conseq. the rectangle cd . dt : : cd 2 = ad • db*. 



But, by theor. X*' okr 

therefore - ca 

or . ca 

By theor. 7, . ca 

By equality - ct 

By sim. tri. - ct 

theref. by equality, ct 

But, by sim. tri. ct 

theref. by equality, ck 

and the rectangle ck , 

But the rect. ck 

theref. the rect. ca 

conseq. the rect. ab 



co 8 : : (ad . db or) aP ; 



ca 

de : 
ct 

CA 

CT ; 
ca : 
ck : 
ca : 



cd 

: ca 
cd 

: ca 
de 

: ca 
; ce : 
ca : 



Ct = CA 



DE, 
id. 
CA. 
DE. 

DE ; 

de. 
de ; 
ce. 
ca. 



ce = the parallelogram cepc, 
ce = the parallelogram cere, 
ab = the paral. pqrs. q. e. d. 



THEOREM XI. 



The Difference of the Squares of every Pair of Conjugate 
Diameters, is equal to the same constant Quantity, namely, 
the Difference of the Squares of the two Axes. 



That is. 
ab" — ab 2 — kg* — eg ; 
where eg, eg are any conjugate 
diameters. 




• Corol. Becnu*e erf 1 = ai> . dk = cd* — ca 9 . 

therefore ca 9 = cd 1 — cd 2 . 
la Lice manner ejff = «V* — d* 9 . 



Mt 



come SECTIOXS. 



For, draw the ordirmtcs kd, rd. 
Then, by cor. to theor. 10, ca 3 = cd 3 — cd* 9 
and - - - - ca 1 = de 2 — dk 3 ; 



theref. the difference ca 3 — 


ca 3 


= t'i) s 


+ DK 3 — 


cd*—de*. 


Bui, by right-angled As, 


CE 7 


s= CD* 


+ DK 3 , 




and - 


ce 1 


= cd* 


+ de 2 ; 




theref. the difference ce 3 — 


ce 2 


— CD 3 


+ DK 3 — 


cd* —de 9 , 


consequently . ca 3 — 


Ctl 2 


= CK 3 


— ck 3 ; 




or, by doubling, ab 3 — 


ab 2 


= EG 3 


-eg 2 . 


a, e. d. 



THEOREM XII. 

All the Parallelograms are equal which are formed between 
the Asymptotes and Curve, by Lines drawn Parallel to 
the Asymptotes. 



That is the lines gk, f.k, ap, aq, 
being parallel to the asymptotes <;ii, <7 ; 
then the para I. igek = pural. cpaq. 



For, let a be the vortex of the curve, or extremity of the 
semi-transverse axis a<\ perp. to which draw al or a/, which 
will be equal to the semi-conjugate, by definition 19. Also, 
draw hkdi/i parallel to l/, 

Then, by theor. k 2, v\ s : al= : : co 3 — ca 3 : de 3 , 
and, by parallels, ia 2 : al 2 : : ( n 3 : mi - ; 
theref. by subtract. ca 2 : al 3 : : <\\ 3 : dii* — dk 3 or 

rect. ii k . v.h ; 
conseq. the square al 2 = the rect. in: . f7i. 

But, by sim. tri. i a : al : : gk ; eh, 
and, by the same, a a : a/ : : kk : v.h ; 
theref. by comp. pa . \q . al 3 . : gk . f.k : iik . v.h ; 
and because al 3 = iik . k/i, theref. pa . aq = <;e . ek. 

But the parallelograms cgk.k, cpaq, being equiangular, 
are as the rectangles gk . kk and pa . aq. 

Therefore the parallelogram gk ~ the paral. pq. 

That is, all the inscribed parallelograms are equal to one 
another. q. e. d. 

CoroL 1. Because the rectangle gkk or cgk is constant, 
therefore ok is reciprocally as cg, or cg : cp : : pa : gk. 
And hence the asvmnUiU'. continually approaches towards 
the curve, bul i\g\*:t vw^^X^ \\ . vvt w^ywaUy 




C K Q £ h 



OF THE OYPEBBOLA. 



as co increases ; and it is always of some magnitude, except 
when co is supposed to be infinitely great, for then gk is 
infinitely small, or nothing. So that the asymptote co may 
be considered us a tangent to the curve at a point infinitely 
distant from c. 

Carol. 2. If the abscisses cd, 
cv, cg, &c. taken on the one 
asymptote, be in geometrical pro- 
gression increasing ; then shall the 
ordi nates uh, ki, ok, &c. parallel 
to the other asymptote, be a de- 
creasing geometrical progression, £r 
having the same ratio. For, all 
the rectangles cdh, cei, cor, &c. being equal, the ordinates 
dh, ki, gk, dec, are reciprocally as the abscisses cd, ck, cg, 
die. which are geometricals. And the reciprocals of geome- 
trical are also geometricals, and in the same ratio, but de- 
creasing, or in converse order. 




THEOREM XIII. 

The three following Spaces between the Asymptotes and the 
Curve, arc equal ; namely, the Sector or Trilinear Space 
contained by an Arc of the Curve and two Radii, or Lines 
drawn from its Extremities to the Centre ; and each of 
the two Quadrilaterals, contained by the said Arc, and 
two Lines drawn from its Extremities parallel to one 
Asymptote, and the intercepted Part of the other Asymp- 
tote. 



That is, 

The sector cae = paeg = qaek, 
all standing on the same arc ae. 



For, bytheor. 12, cpaq = cgkk ; 
subtract the common space cgiq, 
there remains the paral. pi = the par. ik ; 
To each add the trilineal iae, then 
the sum is the quadr. pa kg = qakk. 

Again, from the quadrilateral caek 
take the equal triangles, caq, cek, 
and there remains the sector cak = qaek. 
Therefore cae = qaek = faeg. q. e. d. 




MM 



cosiic lEcnoifi. 



SCHOLIUM. 



In the figure to theorem 12, cor. 2, if c d = 1, and ce, ca, 
&c. be any numbers, the hyperbolic Bpaces hdei, iegk, &c. 
are analogous to the logarithms of those numbers. For, 
whilst the numbers cd, ce, cg, 6zc. proceed in geometries! 
progression, the correspondent spaces proceed in arithmetical 
progression ; and therefore, from the nature of logarithms 
are respectively proportional to the logarithms of those num. 
hers. If the angle c were a right angle, and cn = dh =1 ; 
then if ce were "= 10, the space deih would be 2*30258509, 
&c. ; if co were = 100, then the space dgkh would be 
4-60517018 : these being the Napierean logarithms to 10 and 
100 respectively. Intermediate a re are corresponding to in- 
termediate abscissae would be the appropriate logarithms. 
These are usually called Hyperbolic logarithms ; but I he 
term is improper : for by drawing other hyperbolic curves 
between hik and its asymptotes, other systems of logarithms 
would be obtained. Or, by changing the angle between the 
asymptotes, the same thing may he effected. Thus, when 
the angle c is a right angle, or has it:, sine = 1. the hyperbo- 
lic spaces indicate the Napierean logarithms; hut when the 
angle is 25 44' 27}", whose sine is = -43420148, A:c. the 
modulus to the common, or ttriggs's, logaiithms, the spaces 
deih, &c. measure those logarithms. I*) both eases, if spaces 
to the right of dh are regarded as posit ice, those Jo the left 
will be negative; whence it follows that the logarithms of 
numbers less than 1 arc negative also. 



The sum or difference of the semi -transverse and a line drawn 
from the focus to any point in the curve W equal to a fourth 
proportional to the semi. transverse, the distance from the 
centre to the focus, and the distance from the centre to the 
ordinate belonging to that point of the curve. 



THZOREM XIV. 



fe+ac=xi, or FK=AI ; 
and fit — ac=ci, or fK—m. 
•Where c\ : cf : : cn : ci the 
4th propor. to ca, cf, cd. 



That is, 




OF TUB HYPERBOLA. fiM 

For, draw ao parallel and equal to ea the semi. conjugate { 
and join co meeting the ordinate de produced in h. 

Then, by theor. 2, CA a : ao* : : cd 2 — ca 2 : de 2 ; 
and, by aim. As, ca 2 : ao 2 : : co a — ca 8 : dh 3 — AG a ; 
consequently db 2s5s dh 2 — ag 2 =dh 2 — ca*. 

Also fd=cf^cd, and fd 3 =cf 3 — 2cf . cd+cd 2 ; 
but, by right-angled triangles, fd 2 +de 2 =-fr 2 ; 
therefore fe^cf 3 — ca 7 — 2cf • cD+cD a +DH i . 

But by theor. 4, cf 2 — ca'=x'A a , 
and, by supposition, 2cf . cd=2ca . ci ; 
tlieref. fe 2 =ca 2 — 2ca . ci+cd 2 +dh 2 . 

But, by supposition, ca 2 : cd 2 : : of 2 or ca 2 +ag 2 : ci 1 ? 
and, by sim. as, ca 2 : cd 2 : : ca 2 +ao 2 : cd 2 +dh 2 ; c c- 
therefore - ci^cdM-dh^ch 2 ; 
consequently - fb^ca 2 — 2ca . ci+ci 3 . 

And the root or aide of this square is fe=ci — ca=ai. 

In the same manner is found /e— ci+ca=bi. a. e. b. 

Coral. 1. Hence ch=ci is a 4th propor. to ca, cf, cd. 

Cord. 2. And /e+fe=2ch or 2ci ; or fk, ch, /e are in 
continued arithmetical progression, the commr* difference 
being ca the semi- transverse. \ 

Corol. 3. From the demonstration it appears, that de 2 = 
dh 2 — ao 2 = dm 2 — ca\ Consequently dh is every where 
greater than de ; and so the asymptote cgh never meets the 
curve, though they be ever so far produced : but dh and de 
approach nearer and nearer to a ratio of equality as tlfey 
recede farther from the vertex, tity at an infinite distance they 
become equal, and the asymptote is a tangent to the curve at 
an infinite distance from the vertex. 



theorem xv. 

If a line be drawn from either focus, perpendicular to a tan- 
gent to any point of the curve; the distance of their in- 
tersection from the centre will be equal to the semi-trans- 
verse axis. 



Vol. I. 



65 




• That is, if ff, fp be per- 
pendicular to the tangent 
Trp, then ahall cp ami cp be 
each equal to ca or ca. 



For, through the point of contact ■ draw n, aad /a, nenV 
ing fp produced in e. Then, the £obp« £*bp, being each 
equal to the l/ap, and the anglee at f being n^at, and the 
aide pb being common, the two trangtes, obf, i» emeaaal 
in all respects, and ae «■ » ra, and* or ■* fp. Tho tufa e, 
aince re^po, and re=^rf p and the eagle, at f cxNaaaocvthe 
aide cr will be =4/e or 4ab, that ia cp«*ca sccb. 

And in the same manner op«CA or ca. e> sfeBt 

Corn/. 1. A circle deecribed on the transverse axis, as' a' 

diameter, will pass through the points r, p; because all the 
lines ca, cp, en, being equal, will be radii of the circle. 

Carol. 2. cr is parallel to /a, and cp parallel to Fa. 

Coral. 3. If at the intersections of any tangent with the 
circumscribed circle perpendiculars to the tangent be drawn, 
-they will meet the transverse axis in the two loci. That is, 
the perpendiculars ff, pf give the foci f, /. 



THSOBBM XVI. 



The equal ordinates, or the ordinatea at equal 

from the centre, on the opposite sides and ends of aa 
hyperbola, have their extremities connected by one right 
line passing through the centre, and that line is 
by the centre. 



That is, if en £3 co, or the 
ordinate dk = oh ; then shall 
cb = cr, and ech will be a 
right line. 




For, when cd = co, then also is ra = on by cor. 9 theor. 1. 
But the L d = L o, being both right anglee ; 
therefore the third aide ca =■= cw, and the L MB « L oca, 
and conaequetrtty ttt^ottafe* 



Carol. 1. And, conversely, if ech be a right line pairing 
through the centre ; then shall it be bisected by the centre; 
or have cr « ch, also de will be = on, and-cn = c«. 

Carol, 2. Hence also, if two tangents be drawn to the two 
ends e, H of any diumeier eh ; they will be parallel to each 
other, and will cut the axis at equal tingles, and at equal dis- 
tances from the centre. For, the two cd, ca being equal to 
the two co, cr, the third proportionals or, <:» will be equal 
also ; then the two sides ce, ct being equal to the two 1.11, 
< a, and the included angle ect equal to the included angle 
Res, all the other corresponding parts are equal : and so the 
L t = JL s, and te parallel to ns. 

Carol. 3. And hence the four tangents, at the lour ex- 
tremities of any two conjugate diameters, form a parallelogram 
inscribed between the hyperbolas, and the pairs of opposite 
sides are each equal to the corresponding parallel conjugate 
diameters. — For, if the diameter eh be drawn parallel to the 
tangent te or hs, it will be the conjugate to eh by the defini- 
tion ; and the tangents to e, A will be parallel to each other, 
add to the diameter eh, for the same reason. 

theobem xvir. 

If two ordinates ed v ed be drawn from the extremities f, e, 
of two conjugate diameters, and tangents be drawn to 
the same extremities, and meeting the axb produced in 
t and r ; 

Then shall cd be a mean proportional between cd, da, 
and cd a mean proportional between cd, dt. 




For, by theor. 7, cd : ca : : ca : -ct, , 

and by the same, cd 2 ca : ; ca : cr ; 

thtref. by equality, cd : cd : 2 ca : ct. 

But by sim. tri. # nr : cd : : ct : cr ; 

theref. by equality, (So : cd : : cd : dt. 

In like manner, cd 2 cd : : cd : d*. a. e. d. 
Coral. 1. Hence cd : cd : : cr : cr. % 
Carol. 2. Hence also cn : cd : vdt : de. 
Aad the rect. cd. djs = cd.de, or A cde ~ &c<fe. 



JjffiM-S. Akocf~oD.»r,telcaP«e«.dta. 
7 Or cd a mean proportional between ca» vr % 
and cd a mean proportional between cd, dsu 

The mm igvro befog constructed aetata* lael awea»oeitj»B> 
pmch ordinate will divide tbe axis, and the ■■— a m added 

to the external partem the tame ratio. 

[See the last fig.] 
That is, da : dt : : dc ; m, 
and dx :da : :4c: da. 
For, by theor. 7, cd : ca 8 : ca : cr, 
and by dir. - cd : oa : : ad t at£ 
and by eomp. ■ cd : db t : ad : dt, 
or • • da : or : : dc : db. 
In like manner, 4a : da : : dc : dm. q. b» d. 

Carol. 1. Hence, and from cor. 9 to the last prop* it is 

Cd 2 — CD • DT = AD • DB = CD 5 CA*, 

and cd* = cd . da = Ad . da = ca" = cd* . 
Carol. 2. Hence also ca s =cd s — cd', and ca*=de*— db*. 
Corel. 3. Farther, because ca 9 : ca 2 : : ad . dbotcJ' : db*« 
therefore ca : ca : : cd : db. 
likewise ca : ca : : cd : de. 

THEOBEM XIX. 

If from any point in the curve there be drawn an ordinate, 
and a perpendicular to the curve, or to the tangent at that 
point : then the 

Dirt, on the trans, between the centre and ordinate, cd, 
Will be to the dist. pd, 
As square of trans, axis 
To square of the conjugate. 



That is, 
ca s : co 2 : : dc : dp. 



For, by theor. 2, ca* : cd* : : ad . db : db 1 , 
But, by it. angled as, the reel, td . dp = de 1 , 
and, by cor. 1 theor. 16, cd . dt = ad . db; 
therefore • - ca* : co 9 : : td . dc : td • dp, 

or % • • • CA? \ OQ> % \ TO 4»S*B> 




OF TBS BYMMOLA* 



an 



THBOftJBK XX. 



If there be two tangents drawn, the one to the extremity of 
the transverse, and the other to the extremity of any other 
diameter, each meeting the other's diameter produced: the 
two tangential triangles so formed, will be equal. 



For, draw the ordinate ox. Then 
By aim. triangles, cd : ca : : cb : cir ; 
but, by theor. 7, cd : ca : : ca : ct ; 
theref. by equal, ca : cb : : cb : ext. 

The two triangles cet, can have then the angle c com- 
mon, and the sides about that angle reciprocally proportional ; 
those triangles are therefore equal, viz. the Acet = A can. 

q. B. D. 

CoroL 1. Take each of the equal triangles cbt, can, 
from the common space cape, 
and there remains the external A pat = A pub. 

Cord. 8. Also take the equal triangles cbt, can, 
from the common triangle cbd, 
and there remains the A ted = trapez. anbd. 



The same being supposed as in the last proposition ; then any 
lines kq, go, drawn parallel to the two tangents, shall also 
cut off equal spaces. 



That is, 
the triangle cbt = 
the triangle can* 




THEOREM XXI. 



That is, * 
the akqg = trapez. anho. 
and AKf£ = trapez* ana£. 




F$ draw the ordinate Djk Then 
The three mm. trinngl^VAW, cdb, oob 9 
are to each other as ca% gd% oe* ; 
th. by di? . the trap, anbd : trap, akho : : cd*~ca* : co*— CA*. 
Bor, by thear. 1, ob* t «Q f : : cb*— ca*,: og&— 

theref. by equ. trap. axbb : trap, akhq : : db" : oq\ 
Bat, by aim. As, tri. tkd : tri* kqg i : bb* : ao? ; 
theref. by equal* aned tbd : : a»bg : boo. 

Bot| by cor. 3 theor. 20, the trap, aitkd .= A tbd ; 

jind therefore the trap, anho = A kqo. 
Ill like manner the trap. A*hg — Aiff. q. b. b. 

Carat. 1. The threo apacea abhg, tkbg, kqo are aD 
equal. 

ConL 3. From the equals akbg, kqg, 
* take the equ*la anA^, Kqg, 

' and there remaina gkua =» gfQG. 

Carol. 8. And from the equals £Ahg, £?qg, 
take the common space ^i.ho, 
and there remains the Alqh = A 14*. 

Cord. 4. Again, from the equals kqg, tbhg, « 
take the common space klho, 
and there remaina tklk = alqh. • 



CoroZ. 5. And when by 
the lines kq, gh, moving 
with a parallel motion, kq 
comes into the position ir, 
where ca is the conjugato 
to Ca ; then 




the triangle kqg becomes the triangle ikc, 
and the space ax kg becomes the triangle axc ; 
and therefore the Ainc - aanc~ Atcc. 

Corol. 6. Also when the lines kq nnd no, by moving with 
a parallel motion, come into the position re, i*e f 
the triangle lqh becomes the triangle cm, 
i nd th ' spac/; tklk l>ec< mes thi triangle tkc ; 
and theref. the acix a atkc = A axc » Aibc. 

THKOKEX XXn. 

Any diameter bisects nil its double ordinate*, or the lines 
drawn parallel to the tangent at its vertex, or to its 'coojo- 
gate diameter. 



or TBI HYtSUOLA. 



611 



That is, if oq be paral- 
lel to the tangent te, or 
to c* , then shall lq = Lq* 



For, draw qii. qh perpendicular to the transverse. 
Then by cor. 3 theor. 21, the a lqh =?= &Lqh ; 
but these triangles are also equimigiilar; 
conseq. their like sides are oqual, or ui = Lq. 
Corel.* 1. Any diameter divides the hyperbola into two 
equaj^parts. 

For, the ordinates on each side being equal to each other, 
and equal in number; all the ordinates, or the urea, oa one 
side of the diameter, is equal to all the ordinates, or the area, 
on the other side of it. 

Cord. 2. In like manner, if the ordinate he produced to 
the conjugate hyperbolas at q', q\ it may be proved that 
lq' = tq . Or if the tangent te be produced, then kv =ew. 
Also the diameter gckh bisects all lines 'drawn parallel to tk 
or oq, and limited either by one hyperbola, or by its two con* 
jugate hyperbolas. 

THBOREX XXin. 

As the square of any diameter 
Is to the square of its conjugate, 
So is the rectangle of any two abscisses 
To the square of their ordinate. 
That is, ce* : cc* : : el . lg or cx a — cb* : lq*. 
For, draw the tangent 

te, and produce the ordi- 
nate ql to the transverse 

at k. Also draw qh, ex 

perpendicular to the trans* . 

verse, and meeting eg in 

h and m. Then, similar 

triangles being as the 

squares of their like sides, 

it is, 

by sim. triangles, £cet : A cut : : ce* : cl* ; 

i 





aftay division, AflfT : tap. tub : : on* : €»• — .«■'. 

Again, by sim. tri. |fli : alqb : : ce a ;LQ 9 . ^ 

Bat, by oor. 5 theor. 81, the Aoex «* a err, 

and, by cor. 4 theor. 31, the Alqji trap, raw ; 

theref. by equality, ©a 8 : or* : : cl 8 en* : aa*, 

or • • • os* : oe* : : bl • lo : 14 1 . . o» a* ft. 

Gorrf. 1. The squares of the ordinatee to any diaiajntar, 
are to one another aa the rectangles of their respective ab» 
actssee, or aa the difference of the squares of the eeau- dia- 
meter and of the distance between the ordinate and centre. 
For they ate all in the same ratio of ca f to ce". 

Carol. 8. The above being a similar property lo that Be- 
longing to the two axes, all the other properties before laid 
down, for the axes, may be understood of any two ronisyfe 
diameters whatever! using only the oblique onlinatee af these 
diameters instead of the perpendicular ordinatee of the exes ; 
namely, all the properties in theorems 6, 7, 8, 16, 17, SO, SI. 

Carol. 8. Likewise, when the ordinates are continued to 
the conjugate hyperbolas at a', g\ the same properties Ml 
obtain, substituting only the sum for the difference of the 
squares of ce and cl, ../ 
That is, cB f : ce* : : cl 1 + cb* : W a . 1 
And so lq* : LQ* : : cl*— cb* : cl 1 -r 

Carol. 4. When/ by the motion of La' parallel to haal( 
that line coincides with ev, the last corollary becomes 
cb 1 .: ce* : : 2c K* 2 Bv* f 
or ce* : bv* : : 1 : 2, 
or ce : bv 2 1 
or as the side of a square to its diagonal. 
That is, in all conjugate hyperbolas, and all their dia- 
meters, any diameter is to its parallel tangent, in the constant 
ratio of the aide of a square to ita diagonal. 

THBOBEM XXIV. 

If any two lines, that any where intersect each other, mast 
the curve each in two points ; then 

The rectangle of the segments of the one ^ 
Is to the rectangle of the segments of the other, 
As the square of the diam. parallel to the former 
To the square of the diam. parallel to the latter. 



Or THX HYPRRB3LA. 



51S 



That is, if cr and 
cr be parallel to nny 
two lines phq, pnq ; 
then shall cr 8 : cr* : : 
fh • hq : pu • H£. 




For, draw the diameter ens, and the tangent tr, and its 
parallels pk, ri, hii, meeting the conjugate of the diameter 
cr in the points t, k, i, m. Then, because similar triangles 
are as the squares of their like side*, it is, 

by sim. triangles, cr 3 : of* : : Acai : Agpk, 

and - - cr* : oh 9 : : Acri : A cum ; 

tlicref. by division, cr* : op 3 — on' : : cki : kpuh. 

Again, by sim. tri. ck* : cii a : : Actk : A cam ; 

and by division, cr 1 : c» 3 — ck j : : Acre : teiix. 
But, by cor. 5 theor. 21, the Acte = Acir, 
an J by cor. 1 theor. 21, tkiio = kph j, or tehm = kphm ; 
tlieref. by equ. C* 1 : ch'-ck* : : rR J : gp 1 — oh 1 or ph . hq. 
In like m inner ce ! : ch j -ck 1 : : cr 1 : pa . Hq. 
Tberef. byeq'i. cu J : cr 1 : : ph . hh : pH . Hfl. q. r. d. 

Coral. 1. In like minner, if any other line p'fTf', parallel 
to cr or to pq % meet phq ; since th > rectangles ph q. p %iq' 
are also in the same ratio of cr j to cr J ; therefore the rect. 
phq : pnq : : pii'q : pn'q. 

Also, if another line p Aq' be drawn parallel to pq or cr ; 
because the rectangles p Aq , phq are still in the same ratio, 
therefore, in general, the rectangle phq : pnq : : p'Aq' : phq'. 
That id, the rectangles of the p irts of two parallel lines, are 
to one another, as the rectangles of the parts of two other 
parallel lines, any where intersecting the former. 

Cord. 2. And when any of the lines only touch the curve, 
instead of cutting it, the rectangles of audi become squares, 
and the general property still attends them. 



Vol. L 



66 



914 




Carol. 3. And hence tb : re : : Is : ft. 



THEOREM XXV. 

If a line be drawn through any point of the curves, parallel 
to either of the axes, and terminated at the asymptotes ; 
the rectangle of its segments, measured from that point, 
will 1>e equal to the square of the serai-axis to which it is 

parallel. 



That is, 
the rect. iiek or n*K = ca\ 
and reel, hsk or hek ca 8 . 




e 



For, draw al parallel to ca, and «x to ca. Then 
hy the parallels. ca 8 : ca'or al* : : cd 8 : dh* ; 
and, by tlicor. 2, ca 8 : c«* : : cd"— ca* : v*. 9 ; 
theref. by subtr. ca 2 : ca* : : c%" : dh 8 — de* or HEX* 
But the antecedents ca 3 , ca 8 are equal, 
theref, the consequents ca 2 , hex must also be equal* 

In like manner it is again, 
by the parallels, ca 3 : ca 2 or al 3 : : cd j : dh> ; 
and by theor. 3, ca 3 : ca 3 : : cd 3 + ca s : ne* ; 
there!, by subtr. ca 3 : ca 3 : : ca 3 : ne 3 — dh 3 or hjk. 

But the antecedents ca 3 , ca 3 are the same, 

theref the conseq. ca 3 , hck must be equal. 

In like manner, by changing the axes, is hsk or hek = ca** 

Corol. 1. Because the rect. hek = the rect. h«k. 
therefore eh : en : : en : ek. 
Aud cxma&^iftaxYy \&*ta*Y*J greater than He. 



OF THE ltVF*XBOLA. 



MS 



CSdro/. ft. the rectangle Ark (he rect. hb*, 
for, by aim. tri. feA : 6k : : e& : EX. 

8CHOLIUM. 

ft is evident that thii proposition is general for any line 
oblique to the axis aha, namely, that the rectangle (if the 
seg n mta of any line, cut by ttoe curve, and t originated by the 
asymptotes, is e.p*l to tfhe squ ire of the semi-diamHter to 
which the line is parallel. Since the demonstration is drawn 
from properties that are common to art diameters. 

THEOREM zxvi. 

All the rectangles are eqiml which are made of the seg. 
mints of any parallel lines cut by the curve, and limited 
by the asymptotes. 




For, each of the rectangles hex or hsk is equal to the 
squire of the parallel semi-diamater c? ; and each of the rect- 
angles huh or hek is oqu il to the sqi ire of the p.inllel semi* 
diameter oi. and therefore the rectangles of the segments 
of all parallel 4ines are equal to one another. q. k. d. 

C)rol. 1. The rectangle hrk being constantly the same, 
whether the point R is taken on the one side or the other of 
the point of contact i of the tangent parallel to iik, it follows 
that the parts uk, kk, of any such line hk, are equal. 

And because the rectangle HtK is constant, whether the 
point e is taken in the one or the other of the opposite hy- 
perbolas, it follows, that the parte He, k«, are also equal. 

Ctr&L 2. And when hx crimes into the position of the 
tangent mi, the last corollary becotftds il = id, arid lit — in, 
and lm = dn. 

Hence also the diameter cm bisect* all the parallels to dl 
which are terminate J by the asymprto, a&outj yav — 



Carol. 8. From the proposition, and the bet corollary, it 
follows that the constant rectangle hkk or khk hi *= il*. And 
the equal constant rect. ueK or me = mut or ix 9 — uP. 

Coral. 4. And hence il = the parallel setni-diameter cs- 
For, the rect.KHE = il*, 
and the equal rect. ees = in" — il*, 
theref. il* =«" — iL* f or n> = 2iL a ; 
but, by cor. 4 tbeor. 23, is* ■= 2cs> 9 
and therefore • il = c*. 

And an the asymptotes pass through the opposite angles of 
all the inscribed parallelograms. 

THEOBB3I XXVn. 

The rectangle of any two lines drawn from any point ia 
the curve, parallel to two given lines, and limned by 

the asymptotes, is a constant quuntity. 

That i«, if ap, eg, pi be parallels, 
as also AQi kk, dm p;irallelf, 
then shall the rect. pap. = rect. gek = red. ira. 




For, produce ke, md to the other asymptote at u», l. 

Then, by the parallels, me : <;e : : ld : id ; 

but - - - ek : ek : : dm : dm ; 

theref. the fectangle hkk : gek : :ldm : idm. 

But, by the last thwr. the rect. hkk = ldm ; 

and therefore the rect. gek = idm = paq. q. a. a. 

THEOREM XXVIII. 

Every inscribed triangle, formed by any tangent and the 
two inter* epted parts of the asymptotes is equal lo a 
constant quantity ; namely, double the inscribed paral- 
lelogram. 

Thai \s, lYte \x\*Ng& ct% — % 



or «m nraaou. 



•17 




For, since the tangent ts is 
bisected by the point of contact 
e, (th. 26, cor. 2), nnd kk is 
parallel to tc, and ge to ck ; 
therefore ck, hs, ge, are all 

equal, as are also co, gt, kk. ^ — , 

Consequently the triangle urB C JC g 

= the triangle kes, and each equal to half the constant in- 
scribed parallelogram gk. And therefore the whole triangle 
CT8, which is composed of the two smaller triangles and the 
parallelogram, is equal to double the constant inscribed paral. 
lelogram gk. q. e. d. 

TnEOREJf XXIX. 

If from tho point of contact of any tangent, and the two in- 
tersections of the curve with a line parallel to the tangent, 
three parallel linen be drawn m any direction, and ter- 
minated by either asymptote ; those three lines shall be 
in continued proportion. 

That is* if hkm and the 
tangent 1 1* be parallel, then 
are the parallels dh, ei, 
ok in continued propor- 
tion. 

C D E L 
For, by the parallels, ei : il : : dh : nx ; 
and, by the same, ei : il : : gk : m ; 
thoref. by compos, ei 9 : il 9 : : dh . gk : hxx ; 
but, by theor. 26, the rect. hxr = il* ; 
and theref. the rect. dh . ok s ei 1 , 
or - • dh : ei : : ei : gk. q. e. d. 

theorem xxx. 

Dr.iw the semi-diameters ch, cix, ck ; 
Tneu shall the sector chi ^ the sector cik. 





For, because hk and all its parallels are bisected by cis, 
therefore the triangle c.\h = tri. cxk % 



■If 



and the segment wm » aeg, nm f 
consequently the sector ao = sec. cik. 

Carol. If the geometrical proportional* hh, mm 9 w be 
parallel to the other asymptote, the spaces drib, biko wiB 
be equal ; for they are equal to the equal sectors chi, cik. 

So that by taking any geometrical proportionuU cd, cm, 
go* 4tc. and drawing dm, bi, «k, &c. parallel to the other 
asymptote, as also the radii uh. ci, ck ; 

then the sectors chi, cut, dec 

or the space* dhik, eiku, dec. 1 

will be all equal among themselves. 

Or the sectors chi, chx, &c. 

or the spaces dhir, dhko, dtc. 

will be in nrithmeticaj progression. 
And therefore these sectors, or space*, wilt be analogous to 
the logarithms of the lines or bases cd, ck, cu, dsc* } namely, 
cm or dbib the log. of the ratio of 

co to ck, or of cs to cg, dec. ; or of ei to dii, or of gk to si, Ac ; 
and chk or phk«; the log. of the ratio of cd to co, die. 
or of ok to dh, Sic. 



OF THE PARABOLA. 



THEOREM I. 

The Abscisses are proportional to the Squares of their 
Ordinates. 

Let avm be a section through 
the axis of the cone, and agih a 
parabolic section by n plane per. 
pendicular to the former, and 
parallel to the side vm of the 
cone ; also let afh be the com. 
mon intersection of the two 
planes, or the axis of the para- 
bola, and fg, hi ordinates per- 
pendicular to it. 

Then it will be, as ap : ah : : fg* : Hi f . 
For, through the ordinates fg, hi, draw the circular sec- 
tions, kgl, min, parallel to the base of the cone, having kl, 
MR for their diamctera, \o yg^ hi ace ordinates, as well 
as to the axis of \ta ^itoVhA*. 




OF THJB PAJUBALA. 619 

Then, by similar triangles, af : ah : : fl : hn ; 
but, because of the parallels, kf = mh ; 

therefore - - - af : ah : : kf . fl : mh . hn. 
But, by the circle, kf . fl = FG 2 /and mh . hn — hi 1 ; 
Therefore - - - af : ah : : fg 9 : hi 9 . o,. E. d. 

PqS hi' 

Carol. Hence the third proportional — or — is a con* 

r r AF AH 

stant quantity, and is equal to the parameter of the axis, by 
defin. 16. 
Or af : fg : : fo : f the parameter. 
Or tho rectangle p . af = fg 9 . 



THEOREM II. 



As the Parameter of the Axis : 
Is to the Sum of any Two Ordinates : : 
So is the Difference of those Ordinates : 
To the Difference of their Abscisses. 

That is, 
f : gh + db : : gii — de : dg, 
Or, p : ki : : ih : ie. 



For, by cor. theor. 1, p . ag gh 9 , 
and - • - p . ad -= de 9 ; 
theref. by subtraction, p . dg = gh 9 — db 9 . 
Or, - - • p . dg = ki . ih, 
therefore - - p : Kt : : ih : dg or ei. q. e. d. 

Cord. Hence, because p . ei = ki . ih* 
and, by cor. theor. 1, p . ag — gh 9 , 
therefore - - ag : ei : : gh 9 : ki • ih. 

So that any diameter ei is as the rectangle of the segments 
ki, ih of the double ordinate kh. 




theorem hi. 



The Distance from the Vertex to the Focus is equal to J of 
the Parameter, or to Half the Ordinate at the Focus. 



That is, 
af - Jfe = {r, 
where f is the focus. 




590 O0XIC tKCTIOXfl* 

For, the genera! property is af : fb : : rmi f. 

But, by definition 17, - fe * \r ; 

therefore also - at = Jfe = Jr. q. b* d» 



TBKORBM IV. 



A Line drawn from the Focus to any Point in the Curve, is 
equal to the Sum of the Focal Distance and the Absciss 
of the Ordinate to that Point. . 

6 

That is, 

FE = FA + AD = GD, 

taking ao ~ af. 



For, since fd = ad ^ af, 




theref. by squaring, 
But, by cor. theor. 1, 
theref. by addition, 
But, by right-ang. tri. 
therefore 

and the root or side is 
or - 



FD' •=» AF 8 - 2AF . Ad + AD*, 
DE a = P . AD - 4AF . AD ; 
FD 5 + DK 3 = AF 1 + 2AP . AD + AD*. 
FD fl + DE a = FE* J 
FE 2 = AF a + 2\F . AD + AD*, 
FE = AF + Al>, 

fk =t gd, by taking ag = af. 

u. E. D. 

Carol 1. If, through the point g, the HHH G» HHH 
line on be drawn perpendicular to the 
axis, it in called the directrix of the 
parabola.* The property of which, 
from this theorem, it appears, in this : 
That drawing any lines he parallel to 
the axis, he is always equal to fe the 
distance of the forus from the point e. 

Corol. 2. Hence also the curve is easily described by peinls. 
Namely, in the axis produced take ag = af the focal dis- 
tance, and draw a number of lines ee perpendicular to the 
axis ad; then with the distances gd, gd, gd, dec. as. radii, 
and the centre f, draw arcs crossing the parallel ordi nates 
in e, e, f., ©zc. Then draw the curve through all the points 
E, s, K. 



l 5 — ! 



* Rach of the other conic sections has n directrix ; hut the conside- 
ration of it dotttnoA nccwT \\\ v\w tftttfatat«<tm^taY?«dof tavetUgstinf 
the general properties ©A \\\« cwv«*. 



Of TBI PAJUIOLA. 



«1 



THEOREM V. 



If a Tangent be drawn to any Point of the Parabola, meet- 
ing the Axis produced; and if an Ordinate to the, Axis 
be drawn from the point of Contact; then the Absciss of 
that Ordinate will he equal to the external Part of the 
Axis, measured from the Vertex. 




That is, 
if tc touch the curve 
at the point c, 
then is at = ah. 



Let cc, an indefinitely small portion of a parabolic curve, 
be produced to meet the prolongation of the axis in t ; and 
let cm be drawn parallel to ex, and cs parallel to ag the 
axis. Let, also, p = parameter of the parabola. 

Then, by aim. tri. cs : sc : : cm : ma -f- at = mt, 

MT • CS 

.\ cs = . 

CM 

Also, th. 1. cor. p . Am = mc* = ms* + 2ms . sc + sc*, 

= mc* -f 2mc . sc + sc 1 , 
and p . am = mc'. 

Consequently, omitting sc 1 as indefinitely small, and sub* 
trading the latter equa. from the former, we have 
p . (aju — am) =" p . cs = 2cs . mc : 
or, substituting for cs its value above, 



mt . cs 



= 2cs . mc ; 



or p . MT = 2mc* : 
Consequently, mt • 



CM 

:2p . AM (th. 1.) 

= 2am, and ma = at. 



Q. E. D. 



THEOREM VI. 



If a Tangent to the Curve meet the Axis produced ; then 
the Line drawn from the Focus to the Point of Contact, 
will be equal to the Distance of the Focus from the Inter- 
section of the Tangent and Axis. 
Vol. I. 07 



east 



coanc sBcnsjs** 



That i*, 
tv ss. rr. 



K <* 

For, draw the ordinate vc to the point of contact c 

Then, hy theor. 5, .\r = ad ; 
therefore • ft ~ af + ad. 
Kut, by theor. 4. fc = af + ad ; 

thcref. by equality, fc — ft. Q- b. d. 

Corol. 1. If ro be drawn perpendicular to th* curve, or to 
the tangent, at c ; then Khali kg = fc = FT. 

For, draw rii perpendicular t ) tc, which will also bisect 
tc, because ft = fc ; and therefore, hy the nature of the 
pantile!*, fii ulso bisects tg in f. And consequently fg = 
ft — fc. 

So tli it f is the centre of a cire'e passing through t, c, g. 

Corol. 2. The subnormal dg is a constant quantity, and 
equal to half the parameter, or to 2af, double the focal 
distance. For, since ice; is a right angle, 
therefore 1 1> or <Jad : dc : : dc : dg ; 
but by the def ad : dc : : dc : parameter ; 
therefore do = half the parameter = 2 aw 

Corel. 3. The tangent at the vertex an, is a mean propor- 
tional between AFand ad. 
For, because fiit is a right angle, 
therefore - ah is a mean between af, at, 
or I etween - af, ad, because ad = at. 
Likewise, - fii is a mean between fa, ft, 
or between fa, tc. 

Corol. 4. The tangent tc makes equal angles with fc and 
the axis ft ; as well as with fc and ci. 
For, because ft = fc, 
Therefore the L fct = L ftc. 
Also, the angle c.cf -= the angle gck, 
drawing ick parallel to the axis ag. 

Corol. 5. And because the angle of incidence gck is =* 
the angle of reflection gcf ; therefore a ray of light falling 
on the curve in the direction kc, will be reflected to the focus 
f. That is, all rays parallel to the axis, are reflected totbs 
focus, or burning uuuvx* 




Or THE PAS ABOLA. 



62* 



THEOREM VII. 

If there be any Tangent, and a Do'ihle Ordinate c'riwn 
from the Point of Contact, and also any Line parallel to 
the Axis, limited by the Tangent and Double Ordinate : 
Then ahull the Curve divide that Line in the sum; Ratio 
as the Line divides the Double Ordinate. 



That is. 
ie : ek : ; ck : kl. 



For, by sim. triangles, cx : ki : : cd : dt or 2da ; 
but, by the def. the paratn. p : cl : : < o : 2da ; 
therefore, by equality, p : ck : : «x : ki. 
• But, by theor. 2, v : ck ::kl:rr; 

therefore, by equality, cl : kl : : ki . kb ; 
and, by division, • ck : kl : ; ie : ek. q. e. d. 

THEOREM VIII. 

The same being supposed as in theor. 7 ; then shall the 
External Part of the Line between the Curve and Tun- 
gen*, be proportioual to the Square of the intercepted Part 
of the Tangent, or to the Square of the intercepted Part 
of the Double Ordinate. 



That is, ie is as ci* or as ck', 

and IK, TA, ON, PL, C2C. 

are as ci 3 , ct', co*,'cf 3 , die. 
or as ck 3 , cd 9 , cm 3 , cl*, &c. 



For, by theor. 7, ie : ek : : ck : kl, 
or, by equality, ie : kk : : ck 2 : ck . kl. 
But, by cor. th. 2, ek isai the rect. ck . kl, 
therefore - ik id as ck 3 , or as ci 3 . Q. e. d. 

Cord. As this property is common to every position of 
the tangent, if the lines ie, ta, o\, due. be appended on the 
points i, r, o, dtc. and moveable about ihetn, and of such 
lengths as that their extremities k, a, n, &c. be in the curve 
of a pa rub da in some one position of the tangent ; Ineu 
making the tangent revolve about the point c, u ui» v «u.i* 





694 



CONIC SBCTKKCS* 



that the extremities e, a, h, dee. will always fern the cum 
of some parabola, in every position of the tangent. 



THXORXH u. 

, The Abscisses of any Diameter, are as the Squares of their 
Ordinates. 



That is, cq, cr, csg&c. 
are as qr 1 , ra*, sn*, &c. 
Or cq, : cr : : qjb 1 : ra s , 
&c. 




For, draw the tangent ct, and the externals, si, at, it o, 
die. parallel to the axis, or to the diameter, cs. 

Then, because the ordinates, qe, ka, sn, 6lc. are parallel 
to the tangent ct, by the definition of them, therefore all 
the figures iq, tk, os, dtc. are parallelograms, whose op- 
posite sides arc equal ; 

namely, - • ^k, ta, on, dec. 

are equal to - cq, cr, cs, dtc. 

Therefore, by theor. 8, cq, cr, cs, &c. 

are as - - \ ci 3 , ct 2 , co a , die. 

or as their equals - qk 2 , ha*, sn% &c. q. k. d* 

Cord, Here, like as in theor. 2, life difference of the ah* 
scisses is as the difference of the squares of their ordinates, 
or as the rectangles under the sum and difference of the 
ordinates, the rectangle of the sum and difference of the 
ord mutes being equal to the rectangle under the difference 
of the abscisses and the parameter of that diameter, or a# 
third proportional to any absciss and its ordinate. 



THEOREM X. 

If a Line be drawn parallel to any Tangent, and cut the 
Curve in two Points ; then, if two Ordinates be drawn to 
the Intersections, and a third to the Point of Contact, 
those three Ordinates will be in Arithmetical Progression, 
or the Sum of the Extremes will be equal to Double the 
Mean. 



Or Till FABABA&A. 888 



That is, 

BO + HI =• 2CD. 




For, draw bk parallel to the axis, and produce hi to &• 
Then, by aim. triangles, kk : hk : : tb or 2ap : cd ; 
but, bu theor. 2, - bk : hk : : kl : r the param. 
thereu by equality, 2ad : kl : : cd : p. 
But, by the defin. 2ad f 2cd : : cd : t» ; 
theref. the 2d terms are equal, kl = 2cd, 
that is, . . bo + hi = 2cd. q. b. d. 

Carol. When the point b is on the other side of ai ; then 
hi — ob = 2cd. 



THEOREM XI. 

Any diameter bisects all its Double Ordinates, or lines 
parallel to the Tangent at its Vertex. 



• 

That is, 





T 

H X 
jj X 











For, to the axis ai draw the ordinates eg, cd, hi, and M5 
parallel to them, which is equal to cd. ♦ 
Then, by theor. 10, 2mn or 2cd = eg + hi, 
therefore m is the middle of eh. 

* 

And, for the same reason, all its parallels are bisected. 

Q. b. D. 

Schol. Hence, ns the abscisses of any diameter and their 
ordinates huve the same relations as those of the axis, namely, 
that the ordinates nre bisected by the diameter, and their 
squares proportional to the abscisses ; ho all the other pro- 
perties of the axis and ||s ordinates and abscisses, before da* 
monst rated, will likewise hold good for any diameter and its 
ordinates and abscisses. And also those of the parameters^ 



cone ascnoss. 



understanding the parameter of any diameter, as a third pio- 
portional to uny absciss and its ordinate. Some of the most 
materia] of which are demonstrated in the tour following 
theorems. 

TUEORXTl III. 



The Parameter of any Diameter is equal to four Times the 
Line drawn from the » ocus to the Vertex of that Diane* 
ter. 




M N 



For. draw the ordinate m i parallel to the tangent ct : also 
cd, perpendicular to the axis a.n, and fh perpendicular to 
the tangent ct. 

Then the abscisses ad, cm or at, being equal, by theor. 5, 
the parameters will be as the. squares of the ordinate;? co, xa 
or ct, by the definition ; 

that is, . - !•:/).. cd s . ct*, 

But by sim. tri. - fii : ft : : co : ct ; 

therofore - . v : p : : fii* . ft*. 

But, by cor. 3, th. 6, fii 3 =- fa . ft ; 

therefore - • i» : p : : fa . ft : ft* ; 

or, by equality, - v : p : : i a : ft or fc 

Bui. by theor. 3, p = 4 fa, 

and therefore - p ~ 4ft or 4fc. q. e. d. 

Cord. Hence the parameter p of the diameter cm is equal 
to 4fa + 4ad, or to p + 4ad, that is, the parameter of the 
axis added to 4a o. 



THEOREM XIII. 

If an Ordinate to any Di:i meter pass through the Focun, it 
will be equal to Half Us Parameter ; and its Absciss equal 
to One Fourth ot vYv& torm* Vaxvuc&vwu 



OF TUX PARABOLA. 



an 



That is, cm = lp, 
and me = \p. 



For, join fc, and draw the tangent ct 
By (he parallels, cm = ft ; 
and, hy theor. 6, fc = ft; 
also, by theor. 12, fc = }p ; 
therefore - - cm = Jp. 
Again, by the defin. cm or jp 




MF. 



MR 



£. D. 



and consequently me = jp = 2cm. 

Corel. 1. Hence, of any diameter, the double ordinate 
which passes through the focus, is equal to the parameter, or 
to quadruple its absciss. 

C rol. 2. flence, and from cor 1, 
to theor. 4, and theor. 6 and 12, it 
appears, that if the directrix on be 
drawn, and any lines he, he, paral- 
lel to the axis ; then every parallel 
he will he equal to ef, or \ of the pa- 
rameter of the diameter to the point e. 




THEOBKM XIV. 



If there be a Tangent, and any Line drawn from the Point 
of Contact and meeting the Curve in some other Point, as 
also another Line parallel to the Axis, and limited by the 
First Line and the Tangent : then shall the Curve divide 
this Second Line in the same Ratio as the Second Line 
divides the First Lino. 



That is, 
ie : ek : : ck : KL. 




For, draw lp parallel to ik, or to the axis. 

Then by theor. 8, ne : fl : : ci* : cf% 

or, by sini. tri. - ib : fl : : ck ( : cl\ 

Also, by sim. tri* ik : fl : : ck : cl, 

or - - - ik ; fl : : ck* : ck • cl ; 



9 



638 como octioni. * 

therefore by equality, ir : ik : : ck . cl : cl" ; 
or - - ie : ik : : ck : cl ; 

and, by division, ie : bk : : ck : kl. a. r. d % 
Cord. When ck = kl, then ie = bk = Jik. 

TneoBBX xv. 

If from any Point of the Curve there be drawn a Tangent, 
and also Two Right Line* to cut the Curve ; and Dia- 
meters be drawn through the Points of Intersection b and 
i. t meeting those Two Right Lines in two other Prints o 
and k : then will the Line kg joining these last Twe 
Points be parallel to the Tangent. 




For, by theor. 14, ck : kl :: ei : kk ; 
and by composition, ck : ci. :: ki : ki ; 
and by the paiallels ck : cl :: oh : lh. 
Hut, by aim. tri. - ck : cl : ; ki : lh ; 
theref. by equal. - ki : lh : : gu : lh : 
consequently - ki = oh, 

and therefore - kg is parallel and equal to t h. q* b. b. 

TIIKOREM xvi. 

If an ordinate be drawn to the point of contact of any tan- 
gent, and another ordinate produced to cut the tangent ; 
it will be, as the difference of the ordinates 

Is to the difference added to the external part, 

So is double the first ordinate 

To the sum of the ordinates. 

That is, kh : ki : : kl : KG. 




For, by cor. \, faew. \, * \ xvc -» *. dc : da. 



OF THE PARABOLA. 



529 



and - . - p : 2dc : : dc : dt or 2da. 

Hut, by aim. triangles, ki : kc : : dc : dt ; 

therefore, by equality, j» : 2dc : : ki : kc, 

or, - . - p : ki : : kl : kc. 

Again, by theor. 2, p : kh : : kg : kc ; 

1 he re fore by equality, kh : ki : : kl : kg. q. e. d. 

Carol. 1. Hence, by composition and division, 
it is, kh : ki : : gk : gi, 
and hi : hk : : hk : kl, 
also ih : ik : : ik : ig ; 

that is, ik is a mean proportional between ig and ih. 

Carol. 2. And from this last property a tangent cun easily 
be drawn to the curve from any given point i. Namely, 
draw ihg perpendicular to the axis, and take ik a mean pro- 
portional between ih, ig ; then dmw kc parallel to the axis, 
and c will be tbe point of contact, through which and the 
given point i the tangent ic is to be drawn. 

THEOREM XVII. 

If a tangent cut any diameter produced, and if an ordinate 
to that diameter be drawn from the point of contact ; 
then the distance in the diameter produced, between the 
vertex and the intersection of the tangent, will be equal 
to the absciss of that ordinate. 

That is, ie = ek. 
For, by the last th. is : ek : : ck : kl. 
But, by theor. 11, ck = kl, 
and therefore ie — ek. 



Coral. 1. The two tangents ci, li, at the extremities of 
any double ordinate cl, meet in the same point of the dia- 
meter of that double ordinate produced. And the diameter 
drawn through the intersection of two tangents, bisects the 
line connecting the points of contact. 

CoroL 2. Hence we have another method of drawing a 
tangent from any given point i without the curve. Namely, 
from i draw the diameter ik, in which take ek = ei, and 
through k draw cl parallel to the tangent at e ; then c and L 
are the points to which the tangents must be drawn from i. 

THEOREM XVIII. 

If a line be drawn from the vertex of any diameter, to cut 
the curve in some other point, and an ot$voaX« >taoX 
Vox. /. 68 




580 



conic tKcnoirt. 



diameter be drawn to that point, as also another ordinate 
any where cutting the line, both produced if necessary : 
'i he three will be continual proportionals, namely, the two- 
ord. nates and the part of the latter limited by the said line 
drawn from the vertex. 

That is, de, 6H, 01 are 
crntinual proportionals, or 
de : gh : : gh : gi. 



For, by thcor. 9, - - . de 8 : gh* : : ad : ao ; 
i ml, by sim. tri. - - - de : oi : : ad : ag ; 
theref. by equality, - - de : gi : : hk* : gh 9 , 
that is, of the three de, gh, ci, 1st : 3d : : 1st* : 2d*, 

therefore 1st : 2d : : 2d : 3d, \ 7 y 

that is, de : Grr : : gh : gi. q. e. d. 

Cbrol. 1. Or their equals gk, gh, gi, are proportionals ; 
where ek is parallel to the diameter ad. 

CoroL 2. Hence it is de : ag : : p : gi, where p is 
the parameter, or ag : gi : : de : p. 
For, by the dcfiu. ag : gh : : gh : p. 
Cord. 3. Hence also the three mn, mi, mo, are propor- 
tionals, where mo is parallel to the diameter, and am parallel 
to il.e ordinatcs. 

For, by theor. 9, - mn, mi, mo, 
or their equals - ap, ag, ad, 
are as the squares of pn, gh, de, 
or of their equals gi, gh, gk, 
which are proportionals by cor. 1. 

theorem xix. 

If a diameter cut any parallel lines terminated by the curve ; 
the segments of the diameter will be as the rectangle of 
the segments of those lines. 

That is, ek : em : : ck . kl : nm . mo. 
Or, ek is as the rectangle ck . kl. 

For, draw the diameter 
rs to which the parallels 
cl, no are ordinatcs, and 
the ordinate eq parallel to 
them. 

Then ck i» \V\e toffet- 
cnce, and kx \Y\e sum of 
the ordinatcs cb \ oVso 





OF TBS PARABOLA. 



531 



ttx the difference, and no the sum of the ordinate* eq, ns. 
And the differences, of the abscisses, are or, as, or kk, km. 

Then by cor. theor. 9, an : qs : : ck • kl : nm . mo, 
that in • • kk : km : : ck . kl : nm . mo. 

Carol 1. The rect. cl . kl = rect, ek and the pa mm. of i>s. 
For the rect. ck . kl = rect. u« and the param of rs. 

Carol. 2. If any line cl be cut by two di « meters, kk, csii ; 
the rectangles of the parts of the line, are as the segments 
of the diameters. 

For kk is as the rectangle cz • kl, 
and ch is as the rectangle cii.iil; 
therefore kk : oh : : ck . kl : cu . .hl. 

Carol. 3. If two parallels, cl, no, he cut by two diame- 
ters, km, gi ; the rectangles of the parts of the parallel* will 
be as the segments of the respective diameters. 

For - - - ek : km : : ck . kl : nm . no, „ 
and . . . ek .: gh : : ck • kl : ch . hl, 
theref. by equal, km : on : : nm . mo : ch . hl. 

Carol. 4. When the parallels come info the position of 
the tangent at p, their two extremities, or points in the curve, 
unite in rhe point of contact p ; and the rectangle of the parts 
becomes the square of the tangent, and the same properties 
still follow them. 

So that, ev : pv : : pv : 
• ow : pw : : pw : 



ev 

ev ; 



gw : 
gh : 



: pv 
pv a 



: p the param. 

: PW», 

; cn. ii l. 



theorem xx. 'I 

If two parallels intersect any other two parallels ; the rect. 
angles of the segments will be respectively proportional. 
That is, ck . kl : " m - 

E 




. 10. 



For, by cor. 3 theor. $3, pk : at : : ck . kl : gi • in 
And by the same, ' pk : ut : : dk • KB : ni • io ; 
therefc byoquaL<iK.KL: ox. « ax . w ;si.io. 



CoroL When one of the pain of intersecting lines comes 
k. into the position of their parallel tangents, meeting and limit* 
ing oach other, the rectangles of their segments become ihe 
squares of their respective tangents. So that the constant 
ratio of the rectangles, is that of the square of their parallel 
tangents, namely, 

ce • kl: dk • sis : : tang 9 , parallel to cl : tang 9 , parallel to ac 

THEOBJE* XXI. 

If there be three tangents intersecting each other ; seek 
segments will be in the same proportion. 
That is, gi : in : cu : gd : : dh 
For, through the points 
q, i, d, h, draw the diame- 
ters ok, il, dm, iin ; ns 
also the lines ci, ei, which 
are double ordinates to the 
diameters ok, iin, by cor. 1 
theor. 16 ; therefore 
the diameters gk, dm, iin, 
bisect the lines cl, ce, le ; 
hence km = cm — ck = Jce — {cl = )us *e L x r ne, 
and mn * me — ne = ! ce — £le = Jcl « ck or KL, 
But, by parallels, gi : in : : kl : ln, 
and - . cg : cd : : ck : km, 
also - - dh : he : : mn : ne. 
But the 3d terms kl, ck, mn are all equal ; 
as also the 4th terms ln, km ne. 
Therefore the first and second terms, in all the lines, are 
proportional, namely, gi : m : : cc : gd : . dii : he. q. e. d» 




THEOREM XXII. 

The Area or Space of a Parabola, is equal to Two-Thirds of 
its Circumscribing Parallelogram. 

*Lct acb he a semi.pnrubola, cb ihe axis, f the focus, ed 
the directrix ; then if the line af 
be supposed to revolve about f 
as a centre, while the line ae 
moves along the directrix per- 
pendicularly to it, the nrea gene- 
rated by the motion of ae, will 
always be equal to double the 
area genera ted hy fa ; and con- 
sequently the whole external area aegd = double the area 




or raft PARABOLA- 



Ml 



For draw a k parallel, and indefinitely near, to ab ; and 
draw the diagonals ak' and a'k ; then by th. 6, cor. 4, the 
angles b'a'a and ka'a are equal, aa' being considered as part 
of the tangent at a' ; and in the same manner, the angles 
baa' and faa are also equal to each other ; and since ba = 
af, and b'a' — a'f ; the triangles ea a' and b'a'a are each equal 
to the triangle a a'f ; hut the triangle baa' =■ the triangle 
bb'a, being on the same base and between the same parallels ; 
therefore the sum of the two triangles bb'a and baa, or the 
quadrilateral space eaa'b' is double the trilateral space aa'f ; 
and as this is the case in every position of fa', b'a', it fol- 
lows that the whole external area baco = double the inter- 
nal area afc. 

Hence, Take do = fb, and complete the parallelogram 
do he, which is double the triangle abf ; therefore the area 
abc = J the area haco, or 7 of the rectangle aegh, or | of 
the rectangle abci, because bc = |bo ; that is, the area of 
a parabola = Jof the circumscribing rectangle. q. b. d.* 



THBOREM XXIII. 

The Solid Content of a Paraboloid (or Solid generated by 
the Rotation of a Parabola about its Axis), is equal to Half 
its Circumscribing Cylinder. 

Let ghdd be a cylinder, 
in which two equal para bo. 
loids are inscribed ; one bad 
having its base bcd equal to 
the lower extremity of the 
cylinder ; the other gch in- 
verted with respect to the 
former, but of equal base 
and altitude. Let the plane 
lr parallel to each end of the cylinder, cut all the three so. 
lids, while a vertical plane may be supposed to cut them so 
as to define the parabolas shown in the ngure. 
Then, in the semi-parabola acb, p . ap = pm 1 , 
also, in the semi-parabola aco, p . cv = pit*, 
consequently, by addition, p . (ap + cp)=p . ac « pm* + fb*. 
But, p . ac = cb* = PL*. 
Therefore pl 3 = pm 3 + ph* : 

That is, since circles are as the squares of their radii, the 













p y 






" e 



* This Jem »n it ration whs given by Lieut. Drummond of the Roynl 
Engineers, * hen be wai a gentleman Cadet at the Royal Military Acs- 




lo a Ctinder »^#«e Hei»j- i« pr. ud ft* ft**e li*lf ike 
to of the two CirtoUr Kmc* ea. sc. 



Let * - 3 1410 : 




B D C 



Th*n, by th* l**t theor. % *pe Y Air = the ». I id arc, 

and, by the wifnc \jc X Ar 1 = the solid aec, 

tl§#?r^f. the difT. £pc X 'ai* 1 — *f ;=the fiust. begc. 

Rut kit 1 — ai" = ur X ' hi* -r ir% 

Iheref. Jf* X di X 'ad 4- kt) = the fro* begc. 

But, by th. I, p X sit — ih. j , and/> X ir = fg ? : 

the re f. X dj X (in/ 1 + fg*, = the frost, begc. 

q. e. d. 

fhoblems, &c for exercise in comc sections. 

1. Demonstrate flint if a cylinder be cut obliquely the sec- 
tion will In; an ellipse. 

2. Show how to draw a tangent to an ellipse whose foci are 
r, f } from a given point i*. 

3. Show how to draw a tangent to a given parabola from 
ii given point v. 

4. The diameters of an ellipse are 16 and 12. Required 
the parameter and the area. 

5. The bane and altitude of a parabola are 12 and 0. Re* 
quired the parameter, and the semi-ordinates corresponding 
to the abscissa* 2, .'1, and 4. 

(I. In the actual formation of arches, the voussoirs or arch, 
gtoauft arc *j cuV a& \o >»tai«3% ^on^adiaUar 



CONIC SBCTI09CS. 



5S5 



to the respective points of the curve upon which they stand. 
By what const ruction* may this be effected for the parabola 
and the ellipse ? 

7. Construct accurately on paper, a parabol.i whose base 
shall be 12 and altitude 9. 

8. A cone, the diameter of whose base is 10 inches, and 
*hose altitude is 12, is cut obliquely by a plane, which enters 
at 3 inches from the vertex on one slant slide, and comes out 
at 3 inches from the base on the opposite slant side* Requir- 
ed the dimensions of the section ? 

9. Suppose the same cone to be cut by a plane parallel to 
one of the slant sides, entering the other slant side at 4 
inches from the vertex, what will be the dimensions of the 
section ? 

10. Let any straight line efb be drawn through f, one 
of the foci, of an ellipse, and terminated by the curve in b 
and r ; then it is to be demonstrated that kf . fr = eb.{ pa- 
rameter. 

11. Demonstrate that, in any conic section, a straight line 
drawn from a focus to the intersection of two tangents makes 
equal angles with straight lines drawn from the same focus to 
the points of contact. 

12. In every conic section the radius of curvature at any- 
point is to half the parameter, in the triplicate ratio of the 
distance of the focus from that point to its distance from the 
tangent. 

Also, in every conic section the radius of curvature is pro- 
portional to the cube of the normal. 

Also; let fc be the radius of curvature at any point, p, in> 
an ellipse or hyperbola whose tranverse axis is ab, conju- 

fpF . rf)i 

gate ab, and foci f and /: then is pc = ~ -^-t-. 

° ' Jab . ab 

Required demonstrations of these properties* 



OH Til COMIC NLCTBOH! *S W MW ** AMMftMl *0*A- 
TIOHC CALLED THS BQ0ATIO]* if Tl| W*TB. • 




* 1. Fortke gltipm. 

Let I denote ab, the transverse, or any diameter; 

* sb|h its conjugate ; 

* = ak, any absciss, from the extremity of the diam. 

y = ok. the correspondent ordinate : the two being jointly 
denominated co-ordinates. 
Then, theor. 2, ab* : hi* : : ak . kb : dk 3 , 
that is, 1* : e\ : : x(f — x) : y*, hence iy « c"(<Jr — ar*), 

or y = -y- y/(tx — x 9 ), the equation of the curve. 

And from these equations, any one of the four letters or 
quantities, f, c, x, y, may easily be found, by the reduction of 
equations, when the other three are given. 

Or, if p denote the parameter, = c 1 1 by its definition ; 

then, by cor. th. 2, f.: p : : *' v *-x) : y\ or y*= -y- (fx— x*), 
which is another form of the equation of the curve. 
Otherwise. 

If f = ac the semiaxis ; e «= cr the semiconjugate ; then 
p =; c* -r t the semi parameter ; x = ck the absciss counted 
from the centre ; and y = dk the ordinate as before. 
Then is ak = f— x, and kb = t + x, and ak . kb = {t — x) X 
(H-*)*** 8 — X s . 

Then, by th. 2, : c* : : fi— x* : y', and <y = c*(f* — x»), 

or y = -j- ^/(<* — X s ), the equation of the curve. 

Or, tip:: <■— x 1 : y\ and y« = (C — x*), another form 

of the equation to \\ie cwrc*\ Vrom ^\v\tVv <me of the 
quantities may be found, HiYawi vVv^ rax 



ZQUATIONS OF THJB CVBVS. 



697 



2. For tJie Hyperbola. 

Because the general property of the opposite hyperbolas, 
with respect to their abscisses and ordinatcs, is the same as 
that of the ellipse, therefore the process here is the very same 
as in the former cade for the ellipse ; and the equation to the 
curve must come out the same also, with sometimes only the 
change of the sign of a letter or term, from + to — , or from 
— to +, because here the abscisses lie beyond or without 
the transverse diameter, whereas they lie between or upon 
them in the ellipse. Thus, making the same notation for the 
whole diameter, conjugate, absciss, and ordinate, as at first in 
the ellipse ; then, the one absciss ak being x, the olher bk 
will be t + x, which in the ellipse was t — r ; so the sign of 
x must be changed in the general property and equation, 
by which it becomes f 3 : c* : : x[t + x) : y 8 ; hence ftp = 

c* (tx + x*) and y = -j- y/ (tx + jc 2 ), the equation of the 

curve. 

Or using p the parameter, as before, it \s,t : p : : x(t+x) : 
y* or y* = -j- {tx + x* ), another form of the equation to the 
curve. 

Otherwise, by using the same letters f, c, p, for the halves 
of the diameters and parameter, and x for the absciss ck 
counted from the centre ; then is ak = x — t, and bk = x+t, 
and the property P : c* : : (< — t) X (x+ 1) : y\ gives *V=r 

€* (x* — f) 9 or y = T ✓(x*— / a ), where the signs of t* and x % 

are changed from what they were in the ellipse. 

Or again, using the semi parameter, tip:: x* — £ : y 3 , and 

y* = -j- (x 3 — f) the equation of the curve. 

But for the conjugate hyperbola, as in the figure to theo- 
rem 3, the signs of both x' J and t % will be positive ; for the 
property in that theorem being ca 3 : ca 2 : : cd* + ca" : ne 3 , 
it is f 3 : c* : : x* + f 3 : y > = D« a , or < a y s = c 3 (x a + < > ), and y =* 

\ y/(*? + 0> tne equation to the conjugate hyperbola. 

Or, as t : p : : x* + c* : y 8 , and y 3 = (x 3 + f 3 ) also the 
equation to the same curve. 



Vol. I. 



69 



a* x * y*rrvML <r*. jdt imn. vtraa; ». wmi 
as, >i a9, sa 4r?inacei pan£ei to 
aansmou*. a/ « ir =«,cisz. 
Mil a* = «- T"i*&- sr tM«r. 2&l at . iv 
as t« . ». *>p = rj, :a* to tie 

*/p*rV*a, n«a *.v* 1 Triton sjs4 csm- 
aac** take* parauet to the asvwaassBflL 
If lite a;>*r»*4a fc* arx rectangular ar . sr. am. r vfi be 
tquai to & vfiare. 

3. Fir U* P mrmhoU- 

if x tenrtoi any ataciss beginning mt the vertex, and y 

ordinate: alvi /> tne :*ararnei*r. Then, oy oor. theorem 1, 
ak Mi* k f# p. <jt 1 y y p : hence yz = jf 'is the equa- 
tion f/i t.v; \,htwa'+. Or, if 6i = abscissa and 6 the corre*- 

fforidiri^ ^rriiordiriat'.-, ilien ~ j = y% is the equation. 

1. For the Circle. 

H':raui«: the circle is only a species of the ellipse, in which 
two fixoH an; equal to each other ; therefore, m«bmg the 
(wo fli'irnctorH / and r equal each to d in the foregoing equa* 
Iioiih to thf! ellip.se, they become y a = cir — -r 1 , when the 
nhnciMH t h<;giriH at the vertex of the diameter : and y a = 
|d a x*, when the absciss begins at the centre. Or y = 
(iJr 4 - .■!■*), and y= ^/(r 2 — s 2 ), respectively, when r is the 
ruiliiiM. 

Scholium. 

lu vvcry onu of theso equations, we perceive that they rise 
to tlm *Jd or quadratic degree, or to two dimensions ; which 
in hImu the number of points in which any one of these curves 
nut) l»<i nit by a right lino. Hence also it is that these four 
eurvoM are niihI to he linos of the 2d order. And these four 
urn nil the. lines that aro of that order, every other curve hav- 
ing muim higher equation, or may be cut in more points by a 
right lino. 

Wo nun heir add an important observation with regard 
to all eurves o\y rwwed by equations : viz. that the origin of 




iukbxts of noraiMBTKY. 58t 

when aU the terms ef its equation are affected by one of the 
variable quantities x or y; and when, on the contrary, 
there is in the equation one terra entirely known, then the 
origin of the co-ordinates cannot be on a point of the curve* 
In proof of this, let the general equation of a curve be asf* 
+ bx*yi =0 ; then, it is evident that if we take »= 0, 
we shall likewise haveiy* = 0, ory =0; and consequently 
the origin of the co-ordinates is a point in the curve. So 
again, if, in the same equation, we take y = 0, it will result 
thai ax = 0, and * =0, which brings us to the same thing as 
before. Bat, if the equation of the curve include one known 
term, as, for example, aw* + fc'y* + cy* — = ; then 

taking x = a, we shall have cy — =*= 0, or y = '/ 

which proves that the corresponding point p, of die curve, 

is distant from the origin of the z'a by the quantity V 

A simitar troth will flow from making y = 0, when the same 

equation will give * « 



ELEMENTS OF ISOPERIMETRY. 

Dtf. 1. When a variable quantity has it* rimtations re- 
jpdated by a certain law, or confined within certain limits, it 
is called a maximum when it has reached the greatest mag- 
nitude it can possibly attain ; and, on the contrary, when k 
has arrived at the least possible magnitude, it is called a mi- 

Isepcrimetors> or f mpeiimeti ic ai Figures, me those 
which have equal perimeters. 

Def. 3. The Locus of any point, or intersection, dec. is 
the right line or curve in which these are always situated. 

The problem in which it is required to find, among figures 
of the same or of different kinds, those which, within equal 
perimeters, shall comprehend the greatest surfaces, has long 
engaged the attention of mathematicians. Since the admir- 
able invention of the method of Fluxions, this problem has 
been elegantly treated by some of the writers on that branch 
of analysis ; especially by Haclaqrin and Simpson* k wud*> 



640 



ELEXKm Or fSOPBUMXTKT. 



more extensive problem was investigated at the time of 
" the war of problems," between the two brothers John and 
James Bernoulli : namely, " To find, among all the isoperi- 
metrical curves between given limits, such a curve, that, con- 
structing a second curve, the ordinates of which shall be 
functions of the ordinates or arcs of the former, the area of 
the second curve shall be a maximum or a minimum." While, 
however, the attention of mathematicians was drawn to the 
most abstruse inquiries connected with isoperiraetry, the cfe- 
menls of the subject were lost sight of. Simpson was the first 
who called them back to this interesting branch of research, 
by giving in his neat little book of Geometry a chapter on the 
maxima and minima of geometrical quantities, and some of w 
the simplest problems concerning isoperi meters. The next 
who treated this subject in an elementary manner was Simon 
Lhuillier, of Geneva, who, in 1782, published his treatise 
De Relatione mulua Capacitatis el Terminorvm Figurarum, 
die. His principal object in the composition of that work 
was to supply the deficiency in this respect which he found in 
most of the Elementary Courses ; and to determine, with re- 
garb? to both the most usual surfaces and solids, those which 
possessed the minimum of contour with the same capacity ; 
and, reciprocally, the maximum of capacity with the same 
boundary. M. Legend re has also considered the same sub- 
ject, in a manner somewhat different from either Simpson or 
Lhuillier, in his EUments dc Geomttrie. An elegant geo- 
metrical tract, on the same subject, was also given by Dr. 
Horsley, in the Philos. Trans, vol. 75, for 1775 ; contained 
also in the New Abridgement, vol. 13, page 653*. The chief 
propositions deduced by these four geometers, together with 
a few additional propositions, are reduced into one system in 
the following theorems. 



" Another work on the same general subject, containing mnny valua- 
ble theorems, has been published since the first edition of this volume, 
by Dr. CrtsucU of Trinny College, Cambridge. 



{541 ] 



SECTION I. 



SURFACES. 



THEORXM I. 

Of all triangles of the same base, and whose vertices fall 
in a right Tine given in position, the one whose perimeter 
is a minimum is that whose sides are equally inclined to 
that line. 

Let ab be the common base of a series of triangles abc\ 
abc, &c. whose vertices c', c, fall in the right line lm, jpven 
in position, then is the triangle of least 
perimeter that whose sides ac, bc, are 
inclined to the line lm in equal angles. 

For, let bm be drawn from b, per. 
pendicularly to lm, and produced till 
dm = bm : join ad, and from the point 
c where ad cuts lm draw bc : also, from 
any other point c', assumed in lm, draw c'a, c'b, cd. Then 
the triangles dmc, bmc, having the angle dcm = angle acl 
(th. 7 Geom.) = mcb (by hyp.), dmc =■ bmc, and dm = bm, 
and mc common to both, have also dc = bc (th. 1 Geom.). 

So also, we have c'i> = c'b. Hence ac + cb = ac + cd 
~ ad, is less than ac' + c'p (theor. 10 Geom.), or than its 
equal ac' + c'b. And consequently, ab + bc + ac is less 
than ab + bc' + ac'. q. e. d. 

Cor. 1. Of all triangles of the same base and the same al- 
titude, or of all equal triungles of the same base, the isosceles 
triangle has the smallest perimeter. 

For, the locus of the vertices of all triangles of the same 
altitude will be a right line lm parallel to the base ; and 
when lm in the above figure becomes parallel to ab, since 
mcb = acl, mcb =? cba (th. 12 Geom.), acl = cab ; it 
follows that cab = cba, and consequently ac = cb (th. 4 
Geom.) 

Cor. 2. Of all triangles of the same surface, that which 
has the minimum perimeter is equilateral. 

For the triangle of the smallest perimeter, with the same 
surface, must be isosceles, whichever of the sides be con- 
sidered as base : therefore, the triaogU of ra^i^^\TMtfust 




542 



ELEMENTS OF ISOFEllIMETBT. 



has each two or each pair of its sides equal, and consequently 
it is equilateral. 

Cor. 3. Of all rectilinear figures, with a given magnitude 
and a given number of sides, that which has the smallest 
perimeter is equilateral. 

For so long as any two adjacent sides are not equal, we 
may draw a diagonal to become a base to those two sides, and 
then draw an isosceles triangle equal lo the triangle so cut 
off, but of less perimeter : whence the corollary is manifest 

Sc?iolium. 

?■■ 

To illustrate the second corollary above, the student may 
proceed thus : assuming an isosceles triangle whoso base is 
not equal to either of the two sides, and then, taking for anew 
base one of those sides of that triangle, he may construct an- 
other isosceles triangle equal to it, but of a smaller perimeter. 
Afterwards, if the base and sides of this second isosceles tri- 
angle are not respectively equal, he may construct a third 
isosceles triangle equal to it, but of a still smaller perimeter ; 
and so on. In performing these successive operations, he 
will find that the new triangles will approach nearer and 
nearer to an equilateral triangle. 

THEOREM II. 

Of all triangles of the same base, and of equal perimeters, 
the isosceles triangle has the greatest surface. 

Let abc, abo, be two triangles of the same 
base ab and with equal perimeters, of which 
the one auc is isosceles, the other is not : 
then the triangle aim- has a surface (or an 
altitude) greater than the surface (or than 
the altitude) of the triangle \bd. 

Draw c'd through o, parallel to ab, to 
cut ve (drawn perpendicular to ab) in c : then it is to be 
demonstrated that ce is greater than c'k. 

The triangles ac'b, adb, are equal both in base and alti- 
tude ; but the triangle Ac'n is isosceles, while adb is scalene : 
therefore the triangle ac'b has a smaller perimeter than the 
triangle adb (th. 1 cor. 1), or than acb (by hyp.). Conse- 
quently ac < ac ; and in the right-angled triangles aec, 
aec, having ae common, we have c'e < ce *. q. e. d. 




♦ When two mathtma\\tB\ qpwfl&>»% <&wx»kXm\ <^ 



SURFACES. 



548 



Cor. Of all isoperimetrical figures, of which the number 
of sides is given, that which is the greatest has all its sides 
equal. And in particular, of all isoperimetrical triangles, 
that whose surface is a maximum, is equilateral. 

For, so long as any two adjacent sides are not equal, the 
surface may be augmented without increasing the perimeter. 

Remark. Nearly as in this theorem may it be proved 
that, of all triangles of equal heights, and of which the sum 
of the two sides is equal, that which is isosceles has the 
greatest base. And, of all triangles standing on the same 
base and having equal vertical angles, the isosceles one is 
the greatest. 

# 

THEOREM in. 

Of all right lines that can be drawn through a given pointy 
between two right lines given in position, that which is 
bisected by the given point forms with the other two lines 
the least triangle. 

Of all right lines on, ab, cd, that 
can be drawn through a given point 
p to cut the right lines ca, cd, given 
in position, that, ab, wnich is bi- 
sected by the given point r, forms 
with ca, co, the least triangle, abc 

For, let ee be drawn through a 
parallel to cd, meeting dg (produced if necessary) in e ; 
then the triangles p jo, pae, are manifestly equiangular ; and, 
since the corresponding sides pb, pa are equal, the triangles 
are equal also. Hence pbd will be less or greater than pag, 
according as cg is greater or less than ca. In the former 
case, let pacd, which is common, be added to both ; then 
will bac be less than dgc (ax. 4 Geona.). In the latter case, 
if pgcb bo added, dco will be greater than bac ; and conse- 
quently in this case also bac is less than dco. q. e. d. 

Cor. If ph and pn be drawn parallel to cb and ca re- 
spectively, the two triangles pam, pbn, will be equal, and 
these two taken together (since am = pn a mc) will be 
equal to the parallelogram pmcn : and consequently the 
parallelogram pmcn is equal to half abc, but less than half 
doc. From which it follows (consistently with both the al- 
gebraical and geometrical solution of prob. 8, Application of 



it denotes that the preceding quantity is less than the succeeding one : 
when, on the contrary, the separating character is > , it denotes that the 
preceding quantity is greater th*n the succeeding one. 




544 



KLXMEXTI OF UOPIUiaTBT. 



Algebra to Geometry), that a parallelogram is always lew 
than half a triangle in which it is inscribed, except when the 
base of the one is half the base of the other, or the height 
of the former half the height of the latter ; in which case the 
parallelogram is just half the triangle : this being the maxi- 
mum parallelogram inscribed in the triangle. 

Scho* urn. 

From the preceding corollary it might easily be shown, 
that the least triangle which can possibly be described about, 
and the greatest parallelogram which can be inscribed in, any 
curve concave to its axis, will be whenjL subtangent is equal 
to half the base of the triangle, or to » whole base of the 
% parallelogram : and tliat the two figures will be in the ratio 
of 2 to 1. But this is foreign to the present inquiry. 

THEOREM IV. 

Of all triangles in which two side* are given in magnitude, 
the greatest is that in which the two given sides are 
perpendicular to each other. 

For, assuming for base one of the given sides, the surface 
is proportional to the perpendicular let fall upon that side 
from the opposite extremity of the other given side : there- 
fore, the surface is the greatest, when that perpendicular is 
the greatest ; that is to say, when the other side is not in- 
clined to that perpendicular, but coincides with it : hence 
the surface is a maximum when the two given sides are per- 
pendicular to each other. 

Otherwise. Since the surface of a triangle, in which two 
sides are given, is proportional to the sine of the angle in- 
cluded between those two sides ; it follows, that the triangle 
is the greatest when that sine is the greatest : but the greatest 
sine is the sine total, or the sine of a quadrant ; therefore the 
two sides given make a quadrantal angle, or are perpendicular 
to each other* q. e. d. 

THEOREM V. 

Of all rectilinear figures in which all the sides except one are 
known, the greatest is that which may be inscribed in a 
semicircle whose diameter is that unknown side. 

For, if you suppose the contrary to be the case, then when- 
ever the figure made with the sides given, and the side un- 
known, is not inscribable in a semicircle of which this latter 



SURFACES. 



545 



i the diameter, viz. wtmever any one of the angles, formed 
by lines drawn from the extremities of the unknown side to 
one of the summits of the figure, is not u right angle ; we 
may make a figure greater than it, in which that angle shall 
be right, and which shall only differ from it in that respect : 
therefore, whenever all the angles, formed by right lines 
drawn from the several vertices of the figure to the extre- 
mities of the unknown line, are not right angles, or do not 
fall in the circumference of a semicircle, the figure is not in 
its maximum state, q. e. d. 

. THEOREM VI. 

I Of all figures made £th sides given in number and mag. 
' nitude, that which may be inscribed in a circle is the 
greatest. 

Let abcdefg be 
the polygon inscrib- 
ed, and abcdefg a 
polygon with equal 
sides, but not inscri- 
bable in a circle ; so 
that ab = <i6, bc = 
be, die. ; it is affirm, 
ed that the polygon 
abcdbfg is greater than the polygon abedefg. 

Draw the diameter bp ; join ai\ ph ; upon ab = ab ma'ce 
the triangle abp, equal in all respects to abp ; and join ep. 
Then, ofthe two figures edebp, pag fe> one at least is not (by 
hyp.) inscribable, in the semicircle of which ep is the diame- 
ter. Consequently, one at least of these two figures is smaller 
than the corresponding part of the figure apbcdrfg (th. 6). 
Therefore the figure aprcdrfo is greater than the figure 
apbc 'efg : and. if from these there be taken away the re- 
spective triangles apb, apb, which are equal by construction, 
there will remain (ax. 5 Geom.) the polygon abcdefg greater 
than the polygon abcdefg. q,. e. d. 




THEOREM VII. 

The magnitude of the greatest polygon which can be con- 
tained under any number of unequal sides, does not at all 
depend on the order in which those lines are connected 
with each other. 

For, since the polygon is a maximum under given side*, it 
is inscribable in a circle (th. 6). And this inscribed polygon 
is constituted of as many isosceles triangles as it has sides, 
those sides forming the bases of the respective triangles, the 

Vol. I. 70 



646 



ELEMENTS OF ISOmRIMJCTRY. 



other rides of all the triangle* being radii of the circle, and 
their common summit the centre of the circle. ConJu»quenujr 
the magnitude of tho polygon, tbmt is, of the asscmbliige of 
these triangles, does not at all depend on their d i spos it ion , 
or arrangement around tho common centre, a- k. d. 

THEOREM VIU. 

If n polygon inscribed in a circle have all its sides equal, all 
its angles are likewise equal, or it is a regular polygon. 

For, if lines be drawn from the several angles of the poly, 
gon, to the centre of the circumscribing circle, they will 
divide the polygon into as many iuuawles triangles as it has! 
sides ; nnd each of these isosceles triangles will be equal to™ 
either of the others in all respects, and of course they: will 
have the angles at their bases all equal : consequently, .the 
angles of the polygon, which are each made up of two angles 
at the bases of two contiguous isosceles triangles, will be equal 
to one another, a. k. d. 

THEOREM IX. 

Of all figures having the same number of sides and sequel 
perimeters, the greatest is regular. 

For, the greatest figure under the given conditions has 
all its sides equal (tli. 2. cor.). But since the sum of the 
sides and the number of them are given, each of them is 
given : therefore (th. 6\ the figure is inscribable in a circle: 
nnd consequently (th. 6) all its angles ore equal ; that is, it 
is regular, a. e. i>. 

Cor. Hence we see that regular polygons possess the pro- 
perty of a maximum of surface, when compared with any 
okher figures of the same name and with equal perimeters. 

THEOREM X. 

A regular polygon has a smaller perimeter than an irregular 
one equal to it in surface, and having the same number 
cf sides. 

This is the converse of the preceding theorem, and may 
be demonstrated thus : Let r and I be two figures equal in 
surface, and having the same number of sides, of which** is 
regular, i irregular : let also h' be a regular figure similar to 
r, and having a perimeter equal to that of i. Then (th. 9) 
r' > i ; but \ =• a; vYu&tetac* ^ > ^ wuijt are si- 



947 




stoHar ; o o aja oq ae n fly, perimeter, of b' > perimeter of a ; while 
per. a' as per. 1 (by hyp.). Hence, per. i > per. a. q. b. d. 

THBORKW XI* 

The surfaces of polygons, circumscribed about the same or 
equal circles, are respectively as their perimeters*. 

D 

Let the polygon abcd hi! circumscribed 
about theeircte vfoh ; and let this polygon 
be divided into triangles, by lines drawn 
pfrom its several angles to the centre o of 
the circle. Then, since each of the tan- 
gents ab, nc, dec. is perpendicular to its -A. 
eomspondtnf radius, or, op, &c, drawn to the point of con- 
tact (th. 46Geom.) ; and since the area of a triangle is equal 
lb the rectangle of the perpendicular and half the base (Mens, 
•f Surface*, pr. 3); it follows, that the area of each of the 
triangles abo, bco, Sic. is equal to the rectangle of the radius 
of the Circle and half the corresponding side ab, bc, &c. ; and 
consequently, the area of the polygon abcd, circumscribing 
the circle, will he equal to the rectangle of the radius of the 
circle and half the perimeter of the polygon. But, the sur- 
face of the circle is equal to the rectangle of the radius and 
half the circumference (th. 04 Geom.). Therefore, the sur- 
face of the circle, is to that of the polygon, as half the cir- 
cumference of the former, to half the perimeter of the latter ; 
or, as the circumference of the former, to the perimeter of 
the latter. Now, let p and p' be any two polygons circum- 
scribing a circle c : then, by the foregoing, we htive 

surf, c : surf, p : : circum. c : perim. p. 

surf, c : surf, p' :; circum. c : perim. p\ 
But, since the antecedents of the ratios in both these proper- 
tions, are equal, the conaequents are proportional : that is, 
aurf. p : surf, p' : : perim. p : perim. p'« 4. k. d. 

Cor. 1. Any one of the triangular portions abo, of a po- 
lygon circumscribing a circle, is to the corresponding circular 
sector, as the side ab of the polygon, to the arc of the circle 
included between ao and bo. 



•This theorem, together with tho analogous onei re«prcling hodips 
ctaaoiferihMg cylinders nnd spheres, were piveu by Eme.-son in his 
tfrometry, and their u*e4n tan theory of lso|M»rimetm was just suggest- 
ed : but the full *ppKc*Uua ef them te that theory is due to Siioee 
Uuillier. 



548 



xunanm of hokeikbtet. 



Car. 2. Every circular arc is greater than Ha chord, aad 
leas than the sum of the two tangents drawn from its extremi 
ties and produced till they meet. 

The first part of this corollary is evident, because a r bt 
line is the shortest distance between two given points. T <e 
second part follows at once from this proposition : for ea f 
ah being to the arch kih, as the quadrangle a boh to the c r 
cular sector hieo ; and the quadrangle being greater than the 
sector, because it contains it ; it follows that ea + ah if 
greater than the arch kih*. 

Cor. 8. Hence also, any single tangent sa, is greater than 
its corresponding arc ki. ^ 

THEOREM XII. 

If a circle and a polygon, circumscribahle about another 

circle, are isoperimeterc, the surface of the circle is a 
geometrical mean proportional between that polygon and 
a similar polygon (regular or irregular) circumscribed 
about that circle. 

Let c be a circle, p a polygon isoperimetrical to that circle, 
and circumscribahle about some other circle, and p' a polygon 
similar to p and circumscribahle about the circle c : it is 
affirmed that p : c : : c : p'. 

For, p : p' : : peri nr. p : perim a . p' : : circum*. c : perim*. t 

by tli. 89, geom. and the hypothesis. 
But (th. 1 1) p ' : c : : per. p' : cir. c ; : per 3 , p' : per. p' x cir.c 
Therefore p : c : : .... c-ir 9 . c : per. P x cir. c. 
: : cir. c : per. p' : : c : p'. Q. e. ». 

THEOREM XIII. 

If a circle and a polygon, circumscribahle about another 
circle, are equal in surface, the perimeter of that figure 
is a geometrical mean proportional between the circum- 
ference of the tiret circle and the perimeter of a similar 
polygon circumscribed about it. 

Let c = p, and let p' be circumscribed about c and similar 
to c : then it is affirmed that cir. c : per. p : : per. p : per. p'. 



* This second enrol In r v is introduced, not l>ecfui«* (if its immrdiftle 
connexion with the subject nndr discission, but because, nottrith- 
standing its timpVuAly, *v\\Wr% V\wv« employed whole pegetia 
attempting Us demoi\&Vt*V\w\, 



SURFACES* 



For cir. c : per. p' : : c : p' : : p : p' : : per*, p : per 1 . p'. 
A Up, per. p' : per. p - : : per*, p' : per. p X per. p'. 

Tbi reforo, cir. c : per. p : : per", p : per. p x per p'. 

: : per. p : : per. p. q,. b. d. ' 

THEOREM XIV. 

The circle is greater than any rectilinear figure of the same 
perimeter : and it has a perimeter smaller tha » ai»y recti- 
linear figure of the same surface. 

For, in the proportion, p : c :: c : p (th. 12), since c < p', 

therefore p < c. 
And, in the propor. cir. c . per. p : : per. p : per. p' (th* 13), 
or, cir. c : per. p' : : cir 3 . c : per*, p, 
and cir. c < per. p' : 
therefore, cir*. c < per 1 , p, or cir. c < per. p. q. e. d. 

Cor. 1. It follows at once, from this and the two pre- 
ceding theorems, that rectilinear figures which are isoperi* 
metera, and each circumscribable about a circle, are re- 
spectively in the inverse ratio of the perimeters, or of the 
surfaces, of figures similar to them, and both circumscribed 
about one and the same circle. And that the perimeters of 
equal rectilineal figures, each circumscribable about a circle, 
are respectively in the subduplicate ratio of the perimeters, 
or of the surfaces, of figures similar to them, ana both cir- 
cumscribed about one and the same circle. 

Cor. 2. Therefore, the comparison of the perimeters of 
equal regular figures, having different numbers of sides, and 
that of the surfaces of regular isoperimetrical figures, is re- 
duced to the comparison of the perimeters, or of the surfaces 
of regular figures respectively similar to them, and circum- 
scribable about one and the same circle. 

Lemma 1. 

If an acu*o anglo of a right-angled triangle be divided 
into any number of equal parts, the side of the triangle 
opposite to that acute angle is divided into unequal parts, 
which are greater as they are more remote from the right 
angles. 

Let the acute angle c, of the right- Cfev 
angled triangle acp, be divided into equal V\^V 
parts, by the lines rc, cd, ce, drawn from \\N\ 
that angle to the opposite side; then shall \ \ 
the parts ab, bd, &c. intercepted by the A B J> E 7 



650 



ELEMENTS OF nOPEBJXETRY. 



lines drawn from c, be successively longer us they are more 
lemofe from the right angle a. 

For, the tingles acd, bce. &c. being bisected by c», CD, 
d&c. iherefure by thcor. 83 Geom. ac : cd : : ab : bd, and 
rc : ce : : bu : de, and dc : cf : : de : ef. And by th. 21 
Geom. cd > ca, ce > cb, cf > cc, and so on : whence it 
follows, that db > ab, de > db, and soon. u. e. d. 

Cor. Hence it is ohviius that, if the part the most remote 
from the right angle a, be repeated a number of times equal 
to that into which the acute angle is divided, there will re. 
suit a quantity greater than the side opposi e to the divided 
angle. 

THEOREM XV. 

If two reg liar figures, circumscribed about the same circle, 
differ in their number of sides* by unity, that which has 
the greatest number of sides shall have the smallest peri- 
meter. 

Let ca be the radius of a circle, and ab, ad, the haff sides 
of two regular polygons circumscribed about thai circle., of 
which the. number of sides differ by unity, being C 
respectively n + 1 und n. The angles acb, acd, 

therefore arc respectively the ~^ and the ^ th 

part of two right angles : consequently these <A- I^D 
angles ar* as n and n + 1 : and hence, the angle may be 
conceived divided into n + 1 equal purls, of which bcd is 
one. Consequently, (cor. to the lemma) (n -f 1) no > ad. 
Taking, then, unequal quantities from equal quantities, we 
shall have 

(n + 1} ad - (n + 1) bd < (n + 1) ad - ad, 
or(n t!)ab< n . ad. 
That is, the scmiperimeter of the polygon whose half side is 
ab, is smaller than the scmiperimeter of the polygon whose 
half side is ad : whence the proposition is manifest. 

Cor. Hence, augmenting successively by unity the num. 
ber of sides, it follows generally, that the perimeters of 
polygons circumscribed about any proposed circle, become 
smaller as the number of their sides become greater. 




THEOREM XVI. 

The surfaces ef regular isoperimetrical figures are grAriter 
as the number of their sides is greater : and the peri- 
meters of equal ve^ulur figures are smaller as the number 
of their sides is gtuutet. 



SOLIDS. r,$| 

For, 1st. Regular isoperimetrical figures are (cor. 1. th. 
14) in the inverse ratio of figures similar to tliem circum- 
scribed about the same circle. And (th. 15) these latter ara 
smnller when their .number of sides is greater : therefore, on 
the contrary, the foifaer become greater as they have more 
sides. 

2dly. The perimeters of equal regular figures are (cor. 1 
th. 14; in the subduplicute ratio of the perimeters of similar 
figures otptamscrib$d .about the same circle : and (th. 15) 
these latter are smaller as they have more side's: therefore 
the perimtffsrs of the former also arc smaller when the num- 
ber of then* sides is greater, a. e. d. 



SECTION II. 
4 SOLIDS. 

THEOREM XV1T. 

Of all prisms of the same altitude, whose base is given in 
magnitude and species, or figure, or shape, the right 
prism has the smallest surface. 

For, the area of each face of the prism is proportional to 
its height ; therefore the area of each face is the smallest 
when its height is the smallest, that is to say, when it is equal 
to the altitude of the prism itself: and iu that case the prism 
is evidently a right prism, q. k. d. 

TIIEOREM xviii. 

Of all prisms whose base is given in magnitude and species, 
and whose lateral surface is the same, tho right prism has 
the greatest altitude, or the greatest capacity. 

This is the converse of the preceding theorem, and may 
readily be proved after the manner of theorem 2. 

TIIEOREM XW. 

Of all right prisms of the same altitude, whose liases are 4 
given in magnitude and of a given number of sides, that 
whose base is a regular figure has the smallest surface. 

For, the surface of a right prism of given altitude, and 
base giveu in magnitude, is evidently proportional to the 
perimeter of its base. But (th. 10) the base being given in 
magnitude, and having a given nnmber of sides, its netu 



f 

552 elements or nornuf cm. 

meter ■ smallest when it is regular : whence, the truth of 
the pr position is manifest. 

THEOREM XX. 

Of two right prisms of the same altitude, and with ir.*egu!ar 
bases equal in surface, that whose base has the ^-eatest 
number of sides has the smallest surface ; aj^, in par- 
ticular, the right cylinder has a smaller surface than any 
prixm of the same altitude and the same capacity. 

The demonstration is analogous to that of the preceding 
theorem, being at once deduciblc from theorems 16 and 14. 

THEOREM XXI. 

Of all right prisms whose altitudes and whose whole sur- 
faces are equal, and whose bases have a given number 
of sides ; that whose base is a regular figure is the 
greatest. 

Let i', p\ be two right prisms of the same name, equa\ in 
altitude, and equal whole surface, the first of these having a 
regular, the second an irregular bnse ; then is the base of 
the prism p', less than the base of the prism p. 

For, let p" he a prism of equal altitude, and whose base 
is equal to that of the prism p' and similar to that of the 
prism p. Then, the lateral surface of the prism p" is smaller 
than the lateral surface of the prism p' (th. It)) : hence, the 
total surface of p" is smaller than the total surface of p', and 
therefore (by hyp.) smaller than the whole surface of p. But 
the prisms p" and v have equal altitudes, and similar bases ; 
therefore the dimensions of the base of v" are smaller than 
the dimensions of the base of p. Consequently the base of 
p% or that of p', is less than the base of v ; or the base of r 
greater than that of p'. q. k. d. 

THEOREM XXII. 

Of two right prisms, having equal altitudes, equal total 
surfaces, and regular bases, that whose base has the 
greatest number of sides, has the greatest capacity. And, 
in particular, a right cylinder is greater than any right 
prism of equal altitude and equal total surface. 

The demonstration of this is similar to that of the pre- 
ceding theorem, and flows from th. 20. 



tOLIB*. 



THEOREM XXm. 

The greatest parallelopiped which can be contained under 
the three parts of a given line, any way taken, will he 
that constituted of equal length, breadth, and depth. 

For, let ab be the given line, and, 
if possible, let two parts ae, kd, be 1 | \ 

unequal. Bisect ad and c, then will A C B D B 
the rectangle under ab (= ac+ce) 

and ed (= ac — ce), be less than ac 3 , or than ac . cd, by 
the square of ce (th. 33 Geom.). Consequently, the solid 
ab . ed . db, will be less than the solid ac . cd . db 4 which 
is repugnant to the hypothesis. 

Cor. Hence, of all the rectangular parallelopipeds, having 
the sum of their three dimensions the same, the cube is the 
greatest. 

THSOBEM XXIV. 

The greatest parallelopiped thai can possibly be contained 
under the square of one part of a given line, and the 
other part, any way taken, will be when the former pari 
is the double of the latter. 

Let ab be a given line, and t t t , 

ac = 2cb, then is ac* . cb the " I J LI 
greatest possible. D D CTC B 

For, .let ac and cb be any other parts into which the 
given line ab may be divided ; and let ac, ac be bisected 
in dd', respectively. Then shall ac 9 . cb = 4ai> • dc • cb 
(cor. to theor. 31 Geom. ) > 4ad' . dc . cb, or greater than 
its equal c'a 9 . c b, by the preceding theorem. 

THEOREM XXV. 

Of all right parallelopipeds given in magnitude, that which 
has the smallest surface has all its faces squares, or is a 
cube. And reciprocally, of all parallelopipeds of equal 
surface, the greatest is a cube. 

For, by theorems 19 and 21, the right parallelopiped 
having the smallest surface with the same capacity, or the 
greatest capacity with the same surface, has a square for its 
base. But, any face whatever may be taken for base : there- 
fore, in the parallelopiped whose surface is the smallest with 
the same capacity, or whose capacity is the greatest with the 
same surface, any two opposite faces whatever axe squares ; 
consequently, this parallelopiped is a cube. 

Vol. I. 71 



5M 



BLtMBffTs of uorxxmraT. 



THEOREM XXVI. 

The capacities; of prisms circumscribing the same right 
cylinder, are respectively as their surfaces, whether total 
or lateral. 

For, the capacities are respectively as the bases of the 
prisms; that is to say (th. 11), as the perimeters of their 
bases ; and these arc manifestly as the lateral surfaces : 
whence the proposition is evident. 

Cor. The surface of a right prism circumscribing a 
cylinder, is to the surface of that cylinder, as the capacity of 
the former, to the capacity of the latter. 

Def. The Archimedean cylinder is that which circum- 
scribes a sphere, or whose altitude is equal to the diameter 
of its base. 

THEOREM XXVII. 

The Archimedean cylinder has a smaller surface than any 
other right cylinder of equal capacity ; and it is greater 
than any other right cylinder of equal surface. 

Let v and c denote two right cylinders, of which the first 
is Archimedean, the other not : then, 

1st, If . . . r = c', surf. c<surf. c' : 
«Mly, if surf, c = surf, c', c>c'. 

For, having circumscribed about the cylinders c, c', the 
right prisms i\ p', with square bases, the former will be a 
cube, the second not : and the following series of equal ra- 
tios will obtain, viz. c : v : : surf, c : surf, p : : base c : base 
v : : base c' : base v : : r' : p' : : surf, c' : surf. r'. 

Then, 1st : when c = c'. Since c : r : : c' : r', it follows 
that p = r' ; and therefore (th. 25) surf. r<surf. r'. But, 
surf, c : surf, p : : surf, c' : surf, r' ; consequently surf. c< 
surf. c'. a. r.. Id. 

2dlv : when surf, c =- surf. c/. Then, since surf, c : surf, 
p : : surf, c' : surf, p', it follows that surf, p = surf, p' ; and 
therefore (th. 25) p > p'. But c : r : : c' : r' ; consequently 
c>c'. a. i:. 2i>. 

THEOREM XXVIII. 

Of all right prisms whose bases are circumscribable about 
circles, and given in soecies, that whose altitude is double 
the radius of \n© cwta \nBttrta«& v& 



SOLIDS. 



555 



smallest surface with the some capacity, and the greatest 
capacity with the same surface. 

This may be demonstrated exactly as the preceding theo- 
rem, by supposing cylinders inscribed in the prisms. 

Scholium, 

If the base cannot be circumscribed about a circle, the 
right prism which has the minimum surface, or the maximum 
capacity, is that whose lateral surface is quadruple of the 
surface of one end, or that whose lateral surface is two thirds 
of the total surface. This is manifestly the case with the 
Archimedean cylinder ; and the extension of the property 
depends solely on the mutual connexion subsisting between 
the properties of the cylinder, and those of circumscribing 
prisms* 

THEOSEX XXIX. 

The surfaces of right cones circumscribed about a sphere, 
are as their solidities. 

For, it may bo demonstrated, in a manner analogous to the 
demonstrations of theorems 11 and*26, that these cones are 
equal to right cones whose altitude is equal to the radius of 
the inscribed sphere, and whose bases are equal to the total 
surfaces of the cones : therefore the surfaces ai\d solidities 
are proportional. 

THEOREM XXX. 

The surface or the solidity of a right cone circumscribed 
about a sphere is directly as the square of the cone's 
altitude, and inversely as the excess of that altitude over 
the diameter of the sphere. 

Let vat be a right-angled triangle which, 
by its rotation upon va as an axis, generates a 
right cone ; and bda the semicircle which by 
a like rotation upon va forms the inscribed 
sphere : then, the surface or the solidity of 

the cone vanes as — . 

VB 

For, draw the radius cd to the point of contact of the 
semicircle and vt. Then, because the triangles vat, vdc, 
are similar, it is at : vr : : cd : vc. 
And, by compos, at : at + vt : : cd : cd + cv = va ; 




556 ELEMENTS OF nOPEHlMSTIT. 

Therefore at* : (at + vt) at : : cd : va, by multiply* 

ing the terms of the first ratio by at. 

But, because vb, vd, va, are continued proportionals, 

it is vb : va : : vn 3 : va' : : cd 3 : at 3 by aim. triangles. 

But cd : va s : at 3 : (at + vt) at by the last : and thest 

mult, give cd . vb : va 3 : ; cd 3 : (at + vt) at, 

or v» : cd : : va 3 : (at + vt) at = cd . — . 

But the surface of the cone, which is denoted by «r . at* + 
* . at . vt*, is manifestly proportional to the first member 
of this equation, is also proportional to the second member, 

or, since cd is constant, it is proportional to i^, or to a third 

proportional to bv and av. And, since the capacities of these 
circumscribing cones are as their surfaces (th. 29), the troth 
of the whole proposition is evident. 

Lemma 2. 

The difference of two right lines being given, the third 
proportional to the less and the greater of them is a minimum 
when the greater of those lines is double the other. 

Let av and bv he two right 

lines, whose difference ab is x , 

given, and let ap be a third 7 i _L p 
proportional to bv and av ; ^ * 

then is ap a minimum when av = 2bv. 

For, since ap : av : : av : bv ; 

By division ap : ap — av : : av : av — bv ; 

That is, ap : vp : : av : ab. 

Hence, vp . av=ap . ab. 
But vp . av is cither = or < Jap 2 (cor. to th. 31 Geooi. 
and th. 23 of this chapter). 

Therefore ap . ab < {ap 3 : whence 4ab <~ ap, or ap > 4ab. 
Consequently the minimum value of a p is the quadruple of 
ab ; and in that case rv .= va ~ 2ab. q. e. of. 



• w being rr- 3 141593. See Vol. i. p. 422. 

t Though the evidence of n single demonstration, conducted on sound 
mathematical principles, is really irresistible, and then-fore needs no 
corroboration ; yet it is frequently conducive as well I omen'al improve- 
ment, as to mental delight, to obtain like results from different processes. 
In this view it will be advantageous to the student, to confirm the truth 
of several i»{ \\\e pro\io*\\\wia\ii \\\\% Ocv*\»Vt*\\^ w\*tvw.^ ^ fluxions I 
Mialyth. Let the Vn&vb «nwvrcw*X»& \^TOft^feaH«\KVN&%\A v&wotVst'vt. 



■OUDfe 



THEOREM XXXI. 

Of all right cones circumscribed about the same sphere, the 
smallest is that whose altitude is double the diameter of 
tho sphere. 

For, by th. 30, the solidity varies as (see the fig. to 

that theorem) : and, by lemma 2, since va — vb is given, the 
va" 

third proportional — is a minimum when va = Sab. q. k. d* 

Cor. 1. Hence, the distanco from the centre of the sphere 
to the vertex of the least circumscribing cone, is triple the 
radius of the sphere* 

Cor. 2. Hence also, the side of such cone is triple the 
radius of its base. 

THEOREM XXXII. 

The whole surface of a right cone being given, the inscribed 
sphere is the greatest when the slant side of the cone is 
triple the radius of its base. 

* For, let c and c' be two right cones of equal whole sun 
face, the radii of their respective inscribod spheres being 
denoted by r and r' ; let the side of the cone c be triple the 
radius of its base, the same ratio not obtaining in c ; and 
let c" be a cone similar to c, and circumscribed about the 
same sphere with c. Then, (by th. 31) surf, c" < surf, c' ; 
therefore surf, c" < surf. c. But c" and c are similar, therefore 
all the dimensions of c" are less than the corresponding 
dimensions of c : and consequently the radius r' of the sphere 
inscribed in c" or in c', is less than the radius s of the sphere 
inscribed in c, or r > a', q. e. d. 

Cor. The capacity of a right cone being given, the in- 
scribed sphere is the greatest when the side of the cone is 
triple the radius of its base. 



example ; and let ab be denoted by a, av by x, bv by x — a. Then we 
shall have x — a : x : t x : the third proportional ; which is to be a 
minimum. Hence, the fluxion of this frnctfon will he equal to sero 
(Flux. art. 57). That is, (Flui.arts. 19 and 35), X J^~L =0. Cost 
sequently **— 8oz=0, and x-2«, or av=2ab, at above. 



558 



ELEMENTS OP ISOFESIXETRY. 



For the capacities of such cones vary as their surfaces 
(th. 29). 



THEOREM XXXIII. 

Of all right cones of equal whole surface, the greatest is that 
whose side is triple the radius of its base : and recipro- 
cally, of all right cones of equal capacity, that whose side 
is triple the radius of its base has the least surface. 

For, by th. 20, the capacity of a right cone is in the com- 
pound ratio of its whole surface and the radius of its in- 
scribed sphere. Therefore, the whole surface being given, 
the capacity is proportional to the radius of the inscribed 
sphere : and consequently is a maximum when the radius of 
the inscribed sphere is such ; that is, (th. 32) when the side 
of the cone is triple the radius of the base*. 

Again, reciprocally, the capacity being given, the surface 
is in the inverse ratio of the sphere inscribed : therefore, it 



• Here again a similar result mny easily be deduced from the method 
of fluxion*. Let the radius of the Iirsc he denoted hy x, the slant side 
of the cone hy its whole surface hy a\ and 3*14159$ hy r. Then the 
circumference of the cone's base will be 2*7, iU area rx-\ and the con- 
vex surface ?n:. The whole surface is, therefore, — ir* 1 -f- *z= : and 

this being — a\ we have s — — T - But the altitude of the cone b 

equal to the square root of the difference of ti c squares of the side and 

of the radius of the base ; that is, it is .-_ v(~ — ). And this mnl- 

tiplied into \ of the area of the base, viz. by \vi\ gives \tx2 V{— — — \ 

for the capacity of the cone. Now, this being a maximum, its square 

must be so likewise (Flux. art. 58), that h, ^JLZ ^**'* 1 , or rejecting the 

denominator, as constant, ah:- — 2t« 7 x' must be a maximum. This, in 
fluiions, is 2a*xi — ti*a"i<x 0; whence wc have a n - — 4rJ 3 0, and 

consequently x ._ \ — ; and a? - AvzK Substituting this value of «* 

for it, in the value of z above given, there results z — — — x = 4 * J * 

■—z - Ax — x - Therefore, the side of the cone is triple the ra- 
dius of its base. Or, the squan* of the altitude is to the square of the 
radius of the base, hs ft \o A . \u vV& v^&ax^ &&so&t&c of the bast, 
as 2 to 1. 



SOLUM. 



MO 



is the smallest when that radius is the greatest; that is (th. 
32) when the side of the cone is triple the radius of its base. 

Q. E. D. 

THEOREM XXXIV. 

The surfaces, whether total or lateral, of pyramids circum- 
scribed about the same right cone, are respectively as their 
solidities. And, in particular, the surface of a pyramid 
circumscribed about a cone, is to the surface of that cone, 
as the solidity of the pyramid is to the solidity of the cone ; 
and these ratios are equal to those of the surfaces or the 
perimeters of the bases. 

For, the capacities of the several solids are respectively as 
their bases ; and their surfaces are as the perimeters of those 
bases : so that the proposition may manifestly be demonstrat- 
ed by a chain of reasoning exactly like that adopted in theo- 
rem 11. 

THEOREM XXXV. 

The base of a right pyramid being given in species, the capa- 
city of that pyramid is a maximum with the same surface, 
and, on the contrary, the surface is a minimum with the 
same capacity, when the height of one face is triple the 
radius of the circle inscribed in the base. 

Let f and p' be two right pyramids with similar bases, the 
height of one lateral face of p being triple the radius of the 
circle inscribed in the base, but this proportion not obtaining 
with regard to p' : then 

1st. If surf, p = surf, p', p > p'. 

2dly. If . . p = . . p', surf, p < surf. p'. 

For, let c and c' be right comes inscribed within the pyra- 
mids p and p' : then, in the cone c, the slant side is triple the 
radius of its base, while this is not the case with respect to 
the cone c'. Therefore, if c = c', surf, c < surf, c' ; and, if 
surf, c = surf, c', c > c (th. 33). 

But, 1st. surf, p :• surf, c : : surf, p' : surf, c ; 
whence, if surf, p = surf, p', surf, c = surf, c ; 
therefore c > c\ But p : c : : p' : c'. Therefore p > p'. 

2dly. p:c::p':c'. Theref. if p=p, c=c' : consequently 
surf, c < surf, c . But surf, p : surf, c • : surf, p' : surf. c\ 
Whence, surf, p < surf. p. 

Cor. The regular tetraedron possesses the property of the 
minimum surface with the same capacity, and of tha toAai* 



MO 



ELEMENTS OF ISOFEIUXETEY. 



mum capacity with the same surface, relatively to all right 
pyramids with equilateral triangular bases, and, a fortiori, 
relatively to every other triangular pyramid. 

THEOREM XXXVI. 

A sphere is to any circumscribing solid, bounded by plane 
surfaces, as the surface of the sphere to that of the cir- 
cumscribing solid. 

For, since all the planes touch the sphere, the radius drawn 
to each point of contact will be perpendicular to each re- 
spective plane. So that, if planes be drawn through the cen- 
tre of the sphere and through all the edges of the body, the 
body will be divided into pyramids whose bases are the re- 
spective planes, and their common altitude the radius of the 
sphere. Hence, the sum of all these pyramids, or the whole 
circumscribing solid, is equal to a pyramid or a cone whose 
base is equal to the whole surface of that solid, and altitude 
equal to the radius of the sphere. But the capacity of the 
sphere is equal to that of a cone whose base is equal to the 
surface of the sphere, and altitude equal to its radius. Con- 
sequently, the capacity of the sphere, is to that of the circum- 
scribing solid, as the surface of the former to the surface of 
the hitter : both having, in this mode of considering them, a 
common altitude, q. k. p. 

Cor. 1. All circumscribing cylinders, cones, &c. are to 
the sphere they circumscribe, as their respective surfaces. 

For the same proportion will subsist between their inde- 
finitely small corresponding segments, and therefore between 
their wholes. 

Cor. 2. All bodies circumscribing the same sphere, are 
respectively as their surfaces. 

THEOREM XXXVII. 

The sphere is greater than any polyedron of equal surface. 

For, first it may be demonstrated, by a process similar to 
that adopted in theorem 9, that a regular polyedron has a 
greater capacity than any other polyedron of equal surface. 
Let p, therefore, be a regular polyedron of equal surface to 
a sphere s. Then r must either circumscribe s, or fall partly 
within it and partly without it, or fall entirely within it. The 
first of these sud^siVvoti* \a cxrotaurg \^vV, V\^<*thesis of the 
proposition, because m c*s^<v\fc 



SOLIDS. 



Ml 



be equal to that of s. Either the 2d or 3d supposition there, 
fore must obtain ; and then each plane of the surface of p 
roust fall either partly or wholly within the sphere s : which- 
ever of these be the case, the perpendiculars demitted from 
the centre of s upon the planes, will be each less than the 
radius of that sphere : and consequently the polyedron r 
must be less than the sphere s, because it has an equal base, 
but a less altitude, q. e. d. 

Cor. If a prism, a cylinder, a pyramid, or a cone, be equal 
to a sphere either in capacity, or in surface ; in the first case, 
the surface of the sphere is less* than the surface of any of 
those solids ; in the second, the capacity of the sphere is 
greater than that of either of those solids. 

The theorems in this chapter will suggest a variety of 
practical examples to exercise the student in computation. 
A few such are given below. 

EXERCISES. 

Ex. 1. Find the areas of an equilateral triangle, a square, 
a hexagon, a dodecadoo, and a circle, the perimeter of each 
being 30. 

Ex. 2. Find the difference between the area of a triangle 
whose sides are 3, 4, and 5, and of an equilateral triangle of 
equal perimeter. 

Er. 3. What is the area of the greatest triangle which 
can be constituted with two given sides 8 and 11 ; and what 
will be the length of its third side ? 

Ex. 4. The circumference of a circle is 12, and the pe- 
rimeter of an irregular polygon which circumscribes it is 15 : 
what are their respective areas ? 

Ex. 5. Required the surface and the atiMfty of the 
greatest parallelopiped, whose length, breadth; and depth, 
together make 18? 

Ex. 6. The surface of a square prism is 546 : what is its 
solidity when a maximum ? 

Ex.1. The content of a cylinder is 169-645068: what 
is its surface when a minimum ? 

Ex. 8. The whole surface of a right cone is 201 -061952 : 
what is its solidity when a maximum ? 

Ex. 9. The surface of a triangular pyramid is 43*30127: 
what is its capacity when a maximum ? 

Ex. 10. The radius of a sphere is 10. Required the so- 

Vol. I. 72 



663 

• 



questions in 



lidities of this sphere, of its circumscribed equilateral cone, 
and of its circumscribed cylinder. 

Ex. 11. "the surface of a sphere is 28*274887, and of an 
irregular polycdron circumscribed about it 85 : what are their 
respective solidities ? 

Ex. 12. The solidity of a sphere, equilateral cone, sad 
Archimedean cylinder, are each 500 : what are the surfaces 
and respective dimensions of each ? 

Ex. 13. If the surface of a sphere be represented by the 
number 4, the circumscribed cylinder's eonvex surface and 
whole surface will be 4 and 0, and the circumscribed equila- 
teral cone's convex and whole surface, 6 and respectively. 
Show how these numbers are deduced. 

Ex. 14. The solidity of a sphere, circumscribed cylinder, 
and circumscribed equilateral cone, are as the numbers 4, 6, 
and 0. Required the proof. 



PRACTICAL EXERCISES IN MENSURATION. 

Quest. 1. Wiiat difference is there between a floor 28 
feet long by 20 broad, and two others, each of half the 
dimensions ; and what do all three come to at 45*. per 
square, or 100 square feet ? 

Ans. dif. 280 sq. feet. Amount 18 guineas. 

Quest. 2. An elm plank is 14 feet 3 inches long, and I 
would have just a square yard slit off it ; at what distance 
from the edge must the line be struck ? Ana. 7f| inches. 

Quest. 3. A ceiling contains 114 yards 6 feet of plaster, 
ing, and the room is 28 feet broad ; what is the length of itf 

Ana. 36$ feet. 

Quest. 4. A common joist is 7 inches deep, and 2J 
thick ; but I want a scantling just as big again, that shall 
be 3 inches thick ; what will the other dimension be ? 

Ans. 11} inches. 

Quest. 5. A wooden trough cost me 3*. 2d. painting 
within, at 6d. per yard ; the length of it was 102 inches, 
and the depth 21 inches ; what was the width ? 

Ans. 27} inches. 

Quest. 6. If my court-yard be 47 feet 9 inches square, 
and I have laid a' foot-path with Purbeck-stone, of 4 feet 
*ide, along one side of it ; what will paving the rest with 
flints come to, at 6d. per square yard ? Ans. bl. 16#. 0\<L 

Quest. 7. lL ladder, 36 feet long, may be so planted. 



XSKSUSATIOX. 



568 



that it shall reach a window 30*7 feet from the ground on 
one side of the street ; and, by only turning it over, without 
moving the foot out of its place, it will do the same by a 
window 18-9 feet high on the other side : what is the breadth 
of the street ? Ans. 50 084 feet. 

Qubst. 8. The paving of a triangular court, at 18d. per 
foot, came to 1007. ; the longest of the three sides was 
88 feet ; required the sum of the other two equal sides ? 

Ans. 10-085 /eet. 

Quest. 9. The perambulator, or surveying wheel, is so 
contrived, as to turn just twice in the length of a pole, or 
16^ feet ; required the diameter ? Ans. 2*626 feet. 

Quest. 10. In turning a one- horse chuise within a ring of 
a certain diameter, it was observed, that the outer wheel 
made two turns, while the inner made but one : the wheels 
were both 4 feet high ; and, supposing them fixed at the 
statutable distance of 5 feet asunder on the axle-tree, what 
was the circumference of the track described by the outer 
wheel ? Ans. 62-832 feet. 

Quest. 11. What is the side of that equilateral triangle, 
whose area cost as much paving at 8d. a foot, as the palli. 
sading the three sides did at a guinea a yard ? 

Ans. 72-746 feet. 

Quest. 12. A roof, which is 24 feet 8 inches by 14 feet 
6 inches, is to be covered with lead at 81b. per square foot : 
what will it come to at 18*. per cwt. ? Ans. 22/. 19*. 10><i. 

Quest. 13. Having a rectangular marble slab, 58 inches 
by 27, I would have a square foot cut off parallel to the 
snorter edge ; I would then have the like quantity divided 
from the remainder parallel to the longer side ; and this 
alternately repeated, till there shall not be the quantity of 
a foot left : what will be the dimensions of the remaining 
piece? Ans. 20*7 inches by 6-086. 

N. B. This question may be solved neatly by an alge- 
braical process, as may be seen in the Ladies' Diary for 18V 3. 

Quest. 14. Given two sides of an obtuse -angled triangle, 
which are 20 and 40 poles ; required the third side, that the 
triangle may contain just an acre of land ? 

Ans. 58-876 or 23 099. 

Quest. 15. How many bricks will it take to build a wall, 
10 feet high, and 500 feet long, of a brick and half thick ; 
reckoning the brick 10 inches long, and 4 courses to the 
foot in height ? Ans. 72000. 

Quest. 16. How many bricks will build a square pyramid 
of 100 feet on each side at the base, and also 100 feet per- 



:ttodfinsnsisnsnf.a 

\ inches bog, 6 inch** broad, and 8 inches thick? 



Qmr. 17. IT, from a right-angled triangle, 
-is 1% and perpendicular 16 feet, a line be drawn parallel le 
Jar, 



ttii^off a triai^W whose araak M 
required the aides of this triangle? - 

Ana. 6, 8, and 10. 

' Qraev. 18. If a round pillar, 7 inches acroee* bare 4 feet 
«f Hon* in it; of what diameter ia the column, of equal 



tenth, that conlaint 10 tinea as nmeh f .jb, • 

Quest. 19. A circular fishpond ie to be m ad e inn gar* 
den, that shall take up just half an acre ; what innat be the 
length of the cord that striken the circle T s Ans* SYf^prda. 

QrasT. 20. When a roof is of a tree pitch,' the rafters 
are f of the breadth of the building : now supposing the 
eaves-board* to project 10 inches on a side, what will the 
new ripping a house cost, that measures 32 feet 9 inches 
long, by 22 feet 9 inches broad on the flat, at 15s. per 
square f Ans. 81. 15s. *\<L 

Quest. 21. A cable, which is 3 feet long, and inches 
in compass, weighs 221b. ; what will a fathom of that cable 
weigh, which measures a foot round ? Ans. 78jlb. 

Quest. 22. A plumber has put 281b. per square foot 
into a cistern, 74 inches and twice the thickness of the lead 
long, 26 inches broad, and 40 deep ; he has also put three 
stays across it within, 16 inches deep, of the same strength, 
and reckons 22s. per cwt. for work and materials. A mason 
has in return paved him a workshop, 22 feet 10 inches broad, 
with Purbeck stone, at Id. per foot ; and upon the balance 
finds there is 3*. 6d. due to the plumber ; what was the 
length of the workshop, supposing sheet lead T v v of an inch 
thick to weigh 5-809lbs. per foot ? Ans. 32-2825 feet 

Qckst. 23. The distance of the centres of two circles, 
whose diameters are each 50, being given, equal to 30; what 
is the area of the space inclosed by their circumferences ! 

Ans. 559*119. 

Quest. 24. If 20 feet of iron railing weigh half a ton, 
when the bars are an inch and quarter square ; what will 50 
feet come to at 3} d. per lb., the bars being but J of an inch 
square 1 Ans. 20Z. Of. 2d. 

Quest. 25. It is required to find the thickness of the 
lead in a pipe, of ttu vm& ami piaster bore, which weighs 



I 



MElfSUEJ^TIOlt. 



565 



141b. per yard in length ; the cubic foot of lead weighing 
11925 ounces ? Ans. -20737 inches. 



Quest. 26. Supposing the expense of paving a semicir- 
cular plot, at 2#. 4d. per foot, come to 102. ; what is the 
diameter of it ? Ans. 14*7737 feet. 

Quest. 27. What is the length of a chord which cuts off 
£ of the area from a circle whose diameter is 280 ? 



Quest. 28. My plumber has set me up a cistern, hit 
shop-book being burnt, he has no means of bringing in the 
charge, and I do not choose to take it down to have it 
weighed ; but by measure he finds it contains Mf\ square 
feet, and that it is precisely | of an inch in thickness. Lead 
was then wrought at 2/. per fother of 19} cwt. It is 
required from these items to make out the bill, allowing 
6J oz. for the weight of a cubic inch of lead ? 

Ans. 4/. \U.2d. 

Quest. 20. What will iho diameter of a globe be, when 
the solidity and superficial content are expressed by the 



Quest. 30. A sack, that would hold 3 bushels of corn, 
is 22^ inches broad when empty ; what will another sack 
contain, which, being of the same length, has twice its 
breadth or circumference ? Ans. 12 bushels. 

Quest. 31. A carpenter is to put an oaken curb to a 
round well, at 84. per foot square : the breadth of the curb 
is to be 8 inches, and the diameter within 3} feet : what 
will be the expense ? Ans. 61. 6|<f. 

Quest. 32. A gentleman has a garden 100 feet long, and 
80 feet broad ; and a gravel walk is to be made of an equal 
width half round it : what must the breadth of the walk be, 
to take up just half the ground ? Ans. 25*068 feet. 

Quest. 33. The top of a may-pole, being broken off by 
a blast of wind, struck the ground at 15 feet distance from 
the foot of the pole ; what was the height of the whole may- 
pole, supposing the length of the broken piece to be 39 
feet ? Ans. 75 feet. 

Quest. 34. Seven men bought a grinding-stone of 60 
inches diameter, each paying | part of the expenxe ; what 
part of the diameter must each grind down for his share ? 
Ans. the 1st 4-4508, 2d, 4-8400, 3d 5-3535, 4th 6 0705, 
5th 7-2079, 6th 9-3935, 7th 22-6778 inches. 

Quest. 35. A maltster has a kiln, that is 16 feet 6 inches 
square : but he wants to pull it down, and build a new one, 



Ans. 278-6716. 




Ans. 6. 



/aVytWeatawasasnehati 

what onnt te the length of its side T 
Ovist. 90. How assay 8 inch < 
12 inch cohe T 

Qukst* 91* How long must tho tether of a home he, Hist 
will allow him to grase, quite aroaaat, just an mere sf 
ground? An*. 30* jar*. 

QvasT. 38. What will the painting of a conical sane 




to at 84 per yard ; supposing the height to ha 118 
feet, and the circumference of the base 64 feet 7 

Ana. 112. On SjdL 

Quest. 89. The diameter of a standard corn