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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.
NEWYORK: .
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/ JrVwTer*, 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 eopyrights."
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 selfevident 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, onehalf, twothirds, 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 righthand side to other figures, it in
creases their value in the same tenfold 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 halfperiods, each halfperiod con.
sisting of three figures ; the name of the first period being
units ; of the secoud, millions ; of the third, millions of
millions, or bimillions, contracted to billions ; of the fourth,
millions of millions of millions, or trimillions, 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.
Halfper.
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 halfperiods ; then begin at the
lefthand 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 :
Fiftyseven.
Two hundred eightysix.
Nine thousand two hundred and ten.
Twentyseven thousand five hundred and ninetyfour.
Six hundred and forty thousand, four hundred and eightyone.
Three millions, two hundred sixty thousand, one hundred
and six.
NOTATION AND NUMXBATION.
Four hundred and eight millions, two hundred and fiftyfive
thousand, one hundred and ninetytwo.
Twentyseven thousand and eight millions, ninetysix thou
sand two hundred and four.
Two hundred thousand and five hundred and fifty millions,
one hundred and ten thousand, and sixteen.
Twentyone 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 righthand, 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 righthand 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) 45. 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 lighthorse, 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
10sl20
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 righthand
place, of the multiplicand, and draw a line below it. — Then,
beginning at the righthand, 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 righthand of the numbers ; mul
tiply the other figures only, and annex as many ciphers to
the righthand 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, beforementioned. — 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 onefourth 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 righthand 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 13fr
Silver.
Value,
s d
A Crown 5
Halfcrown 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^$Ujs of a pound,
which is equal to pennyweights, or equal to 5 dwt. 3^ gr. very
nearly, as stated in the preceding table. And multiples and parts of the
sovereign and shilling in their several proportions.
* The original of all weights used in England, was a grain or corn
of wheat, gathered out of the middle of the ear, and, being well dried,
SI of them were to make one pennyweight, 20 penny weigjhU out
Vol. I. 5
26
ARITHMETIC.
dr oz
16 = 1 lb
256 = 16 = 1 qr
7168 = 448 = 28= 1 ewi
28672 = 1752 = 112 = 4 = 1 Urn
573440 = 35840 = 2240 = 80 = 20 = 1
By this weight are weighed all things of a coarse or drossy
nature, as Corn, Bread, Butter, Cheese, Flesh, Grocery
Wares, and some Liquids ; also all metals, except Silver
and Gold.
oz dwt gr
Note, that lib Avoirdupois = 14 11 15J Troy
\oz   = 18
Idr .  =013}
LONG MEASURE.
2 Barleycorns make
12 Inches 
3 Feet
6 Feet 
5 Yards and a half
40 Poles
8 Furlongs
3 Miles
69 iV Miles nearly
1 Inch 
1 Foot 
1 Yard 
1 Fathom
1 Pole or Rod
1 Furlong
1 Mile
1 League
1 Degree 
In Ft
12= 1
36= 3 =
V98 = 16} =
In
Ft
Yd '
Fth
PI
Fur
Mile
Lea
Deg or
Yd
1 PI
5} = 1 Far
7920 = 660" = 220 = 40= 1 Mile
63360 = 5280 = 1760 = 320 = 8 = 1
CLOTH MEASURE.
2 Inches and a quarter make
4 Nails
3 Quarters
4 Quarters
6 Quarters 
4 Quarters l£ Inch
1 Nail   Nl
1 Quarter of a Yard Qr
1 Ell Flemish  EF
1 Yard .  Yd
1 Ell English  EE
1 Ell Scotch  E S
ounce, and 12 ounces one pound. But in later times it was thought
sufficient to divide the same pennyweight into 24 equal parts* still,
called grains, being the least weight now in common use ; and from
thence the rest are computed, as in the Tables above.
TABLES OP HEASURE8.
27
SQUARE MEA8URB.
144 Square Inches make 1 Sq Foot  Ft
9 Square Feet . 1 Sq Yard  Yd
30$ 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 drymeasure 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
43*
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
101704
H
H
To convert new ) f ,*« . .
measures to old. \ 1H)3IM
120032
•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 barleycorns 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
lOi
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 tallowchandler 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
QuarterMaster
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
■if!
■ 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
righthand, 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* 4c/ ?
Ans. 26/ 7s 7J<*.
7. Suppose that my rent for half a year is 20/ 12s, and
that I have laid out for the landtax 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 bookdebts 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 lefthand, 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 halfaguinea 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.
31. 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 Jandta.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 3fourths.
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 righthand 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 righthand figures of any number be divisible by 4, the
whole is divisible by 4. And if ihe three righthand 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 righthand 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 = 514 — 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 1273=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 1547=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 =63r7=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 { halfcrown 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 fj 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 gf = 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.
OrX*=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 = V4}.
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 bankstock, 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 righthand 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 lefthand, 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 lefthand, 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 290146, and 3146*5, and 2 J 09, tod
03417, and 1416.
29 0146
81465
2109
•62417
1416
EXAMPLES.
529929877 the Sum.
6S
ABXTHXSTIC*
2. What is the sum of 276, 39213, 720149, 417, and
50821 Ans. 77770113.
3. What is the sum of 7530, 16 201, 30142, 957*13,
6*72119 and 03014 ? Ans. 851309653.
4. What is the sum of 31209, 35711, 71956, 71498,
9739215, 179, and 0027 ? Ans. 175009718.
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 righthand, 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 79347 by 2315. Ans, 183688305.
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 righthand, as there are ciphers in the
multiplier ; and subjoining ciphers if need be.
EXAMPLES.
1. The product of 513 and 1000 is 51300.
2. The product of 2714 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 righthand 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 2714986 by 9241035, so as to retain only
four places of decimals in the product
Contracted Way.
27 14986
5301429
Common Way.
2714986
9241035
24434874
13
574930
542997
81
44958
108599
2714
986
2715
108599
44
81
542997
2
14
24434874
25089280
25089280
650510
2. Multiply 480*14936 by 272416, retaining only four
decimals in the product.
3. Multiply 24903048 by 5*73286, retaining only five
decimals in the product.
4. Multiply 325701428 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 12370536 by 54 25.
4. Divide 12 by 7854.
5. Divide 419568 by 100.
6. Divide 8297592 by 153.
1.
n 2700000 (1023114
6100
8220
3030
3910
12710
2154
Ans. 22802.
Ans. 15278.
Ans. 419568.
Ans. 54232.
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 455 by 2100.
2100) 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, 2173 r 100 = 2 173 Ans. 419 10 =
And 516 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 righthand 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 250892800 by 92*41035, so as to have only
four decimals in the quotient, in which case the quotient will
contain six figures.
Contracted.
924103,5) 2608928,09(971498
660721
13849
4608
912
80
6 •
Coitittum.
92.4103,5) 2608928,06 (271498
66072106
18848610
46075750
91116100
79467850
5639570
2. Divide 4109*2351 by 230409, so that the quotient may
contain only four decimals. Ans. 17*8345.
REDUCTION OP DECIMALS.
78
3. Divide 37 10438 by 571396, that the quotient may
contain only five decimals. Ans. '00649.
4. Divide 913 08 by 21372, 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 righthand, 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
lefthand 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 15500
12
4 60Q0
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 3242.
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 598848 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. 134229 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
12 005 os
1 0004166 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 215 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 12 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 lefthand, 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 righthand 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
12 975
20 J 178125
£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
#660666 &c.
^ = 8125 12
Ans. 6* Bd. d 799099 &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 righthand 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 35 ? Ans. 42575,
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 lefthand 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 jft, 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°.ajft .^aftjfts
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 lefthand,
and set its root on the righthand 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
righthand 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 (141421356 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 173056?
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. 1732050.
Ans. 2*236068.
Ans. 2*449489.
Ans. 2645751.
Ans. 3162277.
Ads. 3316624.
Ans. 3464101.
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. 0866025.
Ans. 0645497
Ans. 416S3&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
210358000 ( 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 9741 ?
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 6f 3a6« f b* ; where
a \n the root of the first part a3 ; the resolvend is 3a»6f 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
48228544 ( 364 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 57148219.
Ex. 3. Extract the cube root of 16281582.
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 210348.
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 210358
2 2
39366 420716
210358 19083
As 604018 : 617546 : : 27 : 276047.
27
4322822
1235092
604018) 16673742 (276047 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
21035318645155823, and the process by the* latter method
will be thus :
21035118645 &c.
42070637290 210358
,210358 21035318645 &c.
As 6310643729 : difT. 481355 : : 276017 :
thediff. 0002^ 0560.
conseq. the root req. is 27604910560.
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 41) a + (nl)r: (w+1) p. H(nl) 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 73, its 5th power is 20730 71503. Hence we have
r « 210358, n = 5, r = 73, and a = 2073071593 ; then
» + 1 . £a f n — 1 . £p : p ^ a : : r : r «^ r, that is,
8 X2073071593 + 2 x 210538 : 3050*4 : : 7 3 : 0213005
3 2 73
62192 14779
420716
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. 1259921.
2. What is tho 3d root of 3214 ? Ans. 1475758.
8. What is the 4th root of 2 ? Ans. 1 189207.
1 4. Whnt is the 4th root of 9741 ? Ans. 3 1415909.
5. Whnt is the 5th root of 2 ? Ans. 1148699.
6. What is the 6th mot of 210358 ? Ans. 5254037.
7. What is the 6th root of 2 ? Ans. 1122462.
8. What is tho 7th root of 210358 ? Ans. 4145392.
0. What is the 7th root of 2 ? Ans. 1104080.
10. What is the 8ih root of 210358 ? Ans. 3470328.
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. 1080059.
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
546 298116 162771330
'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
17320508
1442250
4
16
64
2 0000000
1587401
5
25
125
2 2360680
1 709976
6
36
216
24494397
1 817121
7
49
343
2 6457513
1912931
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
2S43867
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
3332222
38
1444
54S72
6 16 14140
3 361975
39
1521
59319
62449980
3 391211
40
1600
61000
63245553
3 419952
41
1681
69921
64031242
3 448217
42
1764
7408*
64807407
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
07823300
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
3708430
52
2704
11 CO US
72111026
3 732511
53
2300
14S877
72801099
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
3957883
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
4081655
69
4761
3285G9
83066239
4101566
70'
4900
343000
83666003
4 121285
71
son
357911
842G149S
4 140818
72
5184
373248
84852814
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
4497941
92
64JG4
77S6SS
9 5916630
4514357
93
S649
S04357
96436508
4530655
94
8836
8305S4
96953597
4 546836
95
9025
857375
97467943
4 562903
96
9216
884736
9*7979590
4 578857
97
9409
912673
98488578
4594701
98
9604
941192
9S994949
4610436
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
4717694
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
115758369
5 117230
135
18225
2460375
11'6189500
5 129928
136
18496
251545G
116619038
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
52S<ft&n
149
22201
3307949
122065556
150 J
22500 j
3375000 j
1224744S7
^» 1 rTYiltar
Jl ^ LI 114 Ucl .
Souafe
Cube,
Snuare Root
CuhA Rrmt
VUMC HWl.
151
22801
3442951
12 2882037
5325074
15^
23104
35 11808
123288280
5336303
153
23409
3581577
12 3693169
5 348481
154
23716
3652264
124006736
5360108 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
5592770
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
133790882
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
152970585
153297097
153622915
153948043
15 4272486
15 4596248
154919334
155241747
155563492
15 5884573
156204994
156524758
156843871
157162336
157480157
157797338
153113883
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
165831240
6 5029^6
J76
76176
21 024576
lb'6 132477
6 510^30
277
76729
it,/; IQ'JtTn
iD'OldiJl iKJ
b a 1 ob34
275
772a 4
4Iio4yD2
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
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70*194
1 ,fi 7Q9Q^1A
d Do 4 b7dt
283
80089
226J51S7
168226038
6 565415
234
80656
22906304
168522995
6573139
285
81225
23149125
16 8819430
6580844
286
81796
23393656
169115345
6588532
287
82369
23639903
169410743
6596202
288
82944
23987872
169705627
6603354
289
53521
24137560
170000000
6 611489
290
84100
24389000
170293864
6 6191 00
291
84631
24642171
170587221
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
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17 lA<".^iQ'i , > 1
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b /o;*.jl
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1 7'J.H !•> l0> '
1 l 4b i^ii'Z 
b / »3 1 1 b
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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
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1 / 00 ) 1 y^i 1
rfi^T^ 1 ftft
oiZ
y/o44
o\K57132S
J 7 bbo >J1 / .
b /S24zd
«ll *1
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oUob4297
17 oU LNUbU
ioybbl
ol4
QDSnA
yoosib
*J A ft X 1 1 1 < ■<
ousjoy 14 *
1 / / JUO l.ll
b #ub.?v±
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olo
yy^zo
OlJOO.S70
17 /4>Jo:M 
b olMUyj
olo
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31554496
1 TT^TlIM wCki 1
1 / / / OoT^*) '
/••CI 1 .'JC 4
boll Jh4
1 An 4 co
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0I000U 1 .3
1 7ViA < iO*>tt
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b ^4b J
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1U1 iz4
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b oj..)bJ4
olU
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101 / Ol
tSJ In 17o9
1 "1 \ 1 1 1
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b 80 J / 7 1
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o^u
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1 A'<AJ. 1
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1 / f 1 bt / — J
17 no 1
b ni tv)jz\
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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 &
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1 A i A"? ft
'1 1 A 1 OOO 1
1 QlUli'lfiMllM
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1 ~ « J« # I ... 1 1
r: w7 1 .4
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1 nfW7<:
A UU^ f U
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b *"io(33
f
1 w As;:; Hl'l
ftK^QJ 1 O
b o^y*i 1 ;
1 rt7%ftJ.
1 V/ 1 »JO±
tcii ii/70'?
l»p IM'// yjt*
b «M/bjoo
ozy
'\~xfl 1 1 OCA
b JUHOQ
1 option
*>017( win
ivirrVK]')' '
1 A IOQ
b i' 1 uj^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
184390SS9 i
6 97"532
341
116281
3965 1S21
18 4661S53
6986368
342
116964
40001 6 ss
Is : 93:120 !
6993191
343
11 761 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
186547581 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
187882942
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
7311448
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
7240504
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
7274786
386
14S996
57512456
196468827
7281079
397
149769
57960603
19 6723156
7287362
388
150544
5841 J 072
19 6977156
7293633
3S9
151321
58S63869
197230829
7299894
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
7331037
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
7404720
407
165649
67419143
20 1742410
7 410795
408
166464
U7''l 1131 2
20 1 990099
7 41 6859
409
167291
68417929
SO 9237481
7422 r na.
 1 Jd H.' J 7t
410
168100
74 28050
411
163921
60426 531
7434994
412
169744
69934 52S
20 2Q77A3I
7 44 1019
413
170569
70444997
2032240 14
7447034
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
2059 1 2603
7 512571
425
180625
76765625
206)55281
7 51S473
426
181476
77308776
20 6397674
7 524365
427
182329
77854483
20 6639783
7530248
423
183184
78402752
20 6Stf 1609
7 536 J 21
429
18404 I
78953589
20 7123152
7 541986
430
184900
79507000
20 7364414
7547^42
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
211896201
450 I
202500 f
91125Q0Q 
21 2132034 \ TWWM
trrx
\
100
ARITHMETIC.
iNJ 11 m nor
Cube.
451
203401
91733851
21 2367606
7668766
452
204304
9234540b
212602916
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
7702625
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
7736188
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
•7758402
468
219024
102503232
216333077
7763936
469
219961
103161709
21 6564078
7769462
470
220900
103823000
216794834
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
7802454
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
219089023
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
222036033
7899792
494
244036
120553784
22 2261108
7905129
495
245025
121287375
222485955
7910460
496
246016
122023936
222710575
7915783
497
247009
122763473
222934968
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
7952848
504
254016
128024064
224499443
7958114
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
8148276
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
234520788
819**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
236008474
8227825
558
311364
173741112
23 6220236
8 232746
559
312481
174676879
236431303
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
238327506
8281635
569
323761
184220009
23 5537209
3236493
570
324900
135193000
238746728
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
8368209
587
344569
202262003
24 2230829
8372967
588
345744
203297472
24 2487113
8 377719
589
346921
204336469
24 2693222
8382465
590
348100
205379000
242899156
8387206
591
349281
206425071
24 3104916
3391942
592
350464
207474688
24 3310501
8396673
593
351649
208527857
243515913
8401393
594
352836
209584584
243721152
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
245967473
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
247386338
8490 185
613
375769
230346397
247538363
8 4 94 806
614
376996
23 1475544
247790234
8 499423
615
378225
232608375
247991935
ft 504035
616
379456
233744396
2431 93473
8508642
617
330639
234885113
248394847
8513243
618
381924
236029032
24 3596053
8 517840
619
3S3161
237176659
24 3797106
3522432
620
384400
236328000
248997992
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
8549379
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
8590724
635
4032*5
256047875
25 1992063
8595238
636
404496
257259456
252190404
8599747
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
253377189
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
8644585
647
418609
570340023
25 4361947
3649044
648
419904
272097792
254558441
649
421201 :
273359449 j
254754784
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
8675697*
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
259229628
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'ilA
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 rtflif
2b 1 1 J 2*17
8 r>02272
683
46G489
31861 1987
2b 1342GS7
S MJ6572
JO Lj.j...ipji
o OI U^DEs
685
469225
321419125
261725047
8S15160
BSC
470596
322828856
261916017
8819447
687
471969
324242703
262106848
8 823731
683
473344
325660672
262297541
8S2S009
689
474721
3270827G9
262488095
8832285
690
476100
32850900G !
26 267851 1
8836556
691
4774S1
329939371
262868789
S840823
692
478864
331373S3S
26 3058929
8845085
693
480249
332812557
263248932
8849344
694
481636
3342553S4
263438797
8S53598
695
483025
335702375
263628527
8857849
696
484416
33715353G
263818119
8862095
697
485809
338608873
264007576
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
8908538 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
2690724H1 1
8979376
725
525625
381078125
26 9258240
8983509
726
527076
382657 176
269443872
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 ,
9032802
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
\ 9QSU&&
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
9133803
763
582169
444194947
27 6224546
9*137797
764
58369b
445943744
276405499
9*141788
765
585225
447697125
27 b 586334
9' 145774
766
586 7a6
449455096
276767050
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
277308492
9161686
770
r fill nriA
592900
456533000
27 7438739
9' 165656
771
594441
458314011
277668868
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
jfc 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
283543933
28 3725210
233901391
28 4077454
234253408
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
9375096
937S887
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
94WMA
94*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
9483813
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
295127091
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
9582840
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
9604570
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
835396
837225
839056
840889
842724
844561
846400
848241
850084
851929
853776
855625
857476
S59329
861 184
863041
864900
866761
868624
870489
872356
874225
876096
877969
979844
881721
883600
8*548]
887364
889249
891136
893025
S9491C
896809
89S704
900601 j
902500 /
731432701
733870808
736314327
73S763264
741217625
743677416
746142643
748613312
751039429
753571000
756058031
758550528
761048497
763551944
766060375
768575296
771095213
773620632
776151559
778688000
791229961
783777448
786330467
788389024
791453125
794022776
796597983
799178752
301765039
804357000
806954491
809557568
812166237
314730504
817400375
820025856
822656953
825293672
827936019
830584000
833237621
83589G88S
838561807
841232384
843908625
846590536
849278123
f S5J971BB2
954670349
857375000
Square Hoot* Cube Root
300166620
30 0333148
300499584
300665928
30 0832179
30 0993339
30 1164407
30 1330393
30 1496269
30 1662063
30 1827765
30 1993377
30 2158899
302324329
302439669
30 2654919
302820079
302935148
303150123
303315018
30 3479818
303644529
30 3309151
303973633
304138127
30 4302431
30 4466747
304630924
30 4795013
304959014
305122926
305296750
30 5450487
30 5614136
305777697
30 5941171
30 6104557
306267857
30 6431069
306594194
306757233
30G920185
30 7083051
307245830
307408523
30 7571130
307733651
3078960S6
30805343&
30*8220700
9658469
9662040
9665609
9 669176
9672740
9676302
9 679860
9 683416
9686970
9690521
9694069
9697615
9 701158
9704699
9708237
9 711772
9 715305
9718335
9722363
9725888
9 729411
9 732931
9736448
9739963
9743476
9 746986
9750493
9753998
9757500
9 761000
9764497
9767992
9 771484
9 774974
9778462
9782946
9785429
9788909
9 792386
9795361
9799334
9802904
9 806271
9809736
9313199
9S1665*
110
ARITHMETIC
951
952
955
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
WOO /
Square. Cube. Square Root. Cube Root
904401
906304
910116
912025
913936
915849
! 917764
9196S1
921600
923521
925444
927369
929296
931225
933156
935089
937024
938961
940900
942841
944784
946729
948676
950625
954529
956484
958441
960400
962361
964324
966289
968256
970225
972196
974169
976144
978121
980100
9S2081
934064
986049
986036
990025
992016
994009
( 996004
998001
1000000
860085351
862801408
865523177
868250664
8709S3375
873722816
376467493
879217912
891974079
884736000
887503681
890277128
893056347
895841344
898632125
90142S696
904231063
907039232
909853209
912673000
915498611
918330048
921167317
924010424
926859375
929714176
932574833
935441352
938313739
941192000
944076141
946966168
949862087
952763904
955671625
958585256
961504903
964430272
967961669
970299000
973242271
976191488
979146657
982107784
985074875
983047936
991026973
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
312089731
31 2249900
31
31
313729915
31 2889757
31 3049517
31 3209195
31 3368792
31 3528303
31 3687743
31 3347097
31 4006369
314165561
314324673
31 4483704
31 4642654
314801525
31 4960315
31 5119025
31 5277655
31 5436206
31 5594677
31 6753068
9833924
9837369
9840813
9344254
9847692
9951128
9854562
9857993
9 361422
9 864348
9868272
9 871694
9875113
9878530
9 881945
9885357
9888767
9892175
9895580
9898983
9902333
9905782
9909178
9912571
9915962
9919351
9922738
9 926122
9929504
9932884
9936261
9939636
9943009
9946380
9949748
9 953114
9*956477
9959839
9 963198
9966555
9969909
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 42 = 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 ' 9T ¥
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:: 14t2:7; 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 3015 „
= — and = . Hence
6 ± 3 3 6+3 63
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,
10242
(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
*Cuare 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 9612=48, <fc 48^2=24, & 24^2=12, & 1252=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 : : 86 : 128,
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 : : 86 : 1812,
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
440
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 :: 54775
Or 20 : 1 :: 547*75 : 273875 interest for 1 year.
3
I 821625 ditto for 3 years.
20
s 32500
12
<2 300 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: 945: : 25: 2
41 or 73:945:: 5 : 6472
5
340
105 73) 4725 (6472
345
945 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 439597
20
* 191940
12
d 23280
4 Ans. 432 19* 2\d.
q 13120
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. 105 *
88 1st yr*s interest. , tnCTe
J 1 1026 2d power of it.
20)756 2d yr*s princip. 11025
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 halfyears,
at 5 per cent, per annum, compound interest, the interest
payable halfyearly. 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 (fff6fc)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
TrialandError, 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 onethird of mine ; but 5 years
ago, your age was only onefourth 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 cannonball, 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 cannonball, 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 paradestep 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 quicktime 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 onefifth 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 91890&
▼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 powerof 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 X3102277?)=56234132 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 74989422 is 875.
Fourthly, the log.of 10 is 1, and the log. of 74989422 is 875;
theref. (1 + 875) r 2 =9375 is the arithmetical mean,
and v/(10 X 74989422) = 86596431 the geom. mean ;
hence the log. of 86596431 is 9375.
Fifthly, the log. of 10 is 1, and the log. of 86596431 is 9375 ;
theref. (l+9375)i2=96875 is the arithmetical mean,
and v/( 10 X 86596431) = 93057204 the geom. mean ;
hence the log. of 93057204 is 96875.
Sixthly, the log. oT 86596431 is 9375, and the log. of
93057204 is 96875.
theref. ( 9375+ 96875) r2= 953 125 is the arithmean,
and ^(86596431 X 93057204) = 8*9768713 the geo
metric mean ;
hence the log. of 89768713 is 953125.
And proceeding in this manner, after 25 extractions, it
will be found that the logarithm of 89999998 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
•808588901
•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 1079181246
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 3423410, 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
2 051
2 6 51
2651
•2 6 5 1
•0 2 6 5 1
•0 2 6 5 1
34 2 3 4 1
24 2 3 4 1
14 2 3 4 1
4 2 3 4 1
14 2 3 4 1
24 2 3 4 1
34 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
aasderto 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 lefthand 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 0G9S970. 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
9409 is 1532627, 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 340926.
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, 1532660 for the log. of 340926.
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  1568202
Log. of 94  1973128
Dif. log. of» J —1595074
Where the index 1 is negative
2. To find the log. of 17$f •
First, 17j = Then,
Log. of 405 . 2807455
Log. of 2ft  1361728
Difl ]p[. of 17iJ 1245727
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 3411.
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 1532708, 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 2314 by
5062
Numbers. Logs.
23 14 . 1364363
5 062 . 0704322
2. To multiply 2581926
by 8«457291
Numbers. Logs.
2 581920 . 0411944
3 457291 . 0538786
**rodoct 1171347 2 068685
Prod. 892648 . 0950680
158
DIVISION BY &OGABXTHK8.
3. To mult. 3902 and 597 16
and 0314726 all together.
Numbers.
3902
597 16
•0314728
059F287 !
2776091
2497935
Prod. 733333 . 1865313
Here the — 2 cancels the 2,
and the 1 to carry from the
decimals is set down.
4. To mult. 3586, and 21046,
and 08372, and 00294 all
together.
Numbers. Logs.
3 586  0554610
21046  323170
08372 , 1922829
00294 2468347
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 lefthand 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  3659631
2. To divide 37 1 49 by 52376.
Numbers. Logs.
Dividend 37149  1569947
Divisor 52376 . 2 719132
Quot. 529078 0723520
Quot. 0709275285081*
UfTOLUTIOM BY LOGARITHMS.
150
8. Divide 06314 by 007241
Numbers. Logs.
Dividend 063142800305
Divisor 0072413859799
4. Todivide7438by 129476.
Numbers. Logs*
Divid. 7438 1871456
Divisor 129476 1112169
Quo*. 871979 0940506 Quot. 057447  2750267
. '•' 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
25791.
Numb. Log.
Root 25791   0411468
The index   2
Power 665174 0822936
2. To find the cube of
307146.
Numb. Log.
Root 3 07146   0487345
The index   3
Power 289758 1462035
■VOLUTION BY LOGARITHMS.
3. To raise •09163 to the
4th power.
Numb. Log.
Root •09163 —2962038
The index   4
Pow. 000070494— 5848152
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   0001950
Thejndex  . 365
9750
■ 11700
5850
0711750
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 lefthand 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) 2562293
Root 1910496 1281146£
2. To find the 3d root of
12345.
Numb. Log.
Power 12345 3)4 091491
Root 231116 1363830}
3. To find the 10th root of
2.
Numb. Log.
Power 2  10) 0301030
Root 1071773 0030103
4. To find the 365th root of
1045.
Numb. Log.
Power 1 045 365) 019116
Root 1000121 0000052
EVOLUTION BY tOGA
161
5 To find </ <m.
Numb. Log*
Power 093 2) —2968483
Root 304959 —148424!*
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)— 4681241
Root 0782973 —2893747
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 31416 X 82 X ff
8. To find 029IG X 7513 X fl T
9. Afl 7241 : 358 : : 2046 : ?
10, : v/ff : : 6927 : ?
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 Coefficients.
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 Coefficient 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 + 3aftc 3 =36 + 9016 = 110.
2. And 2a*3a 2 b + c 3 = 432540 + (54 =  44.
3. And a 3 x(a+6)2ate = 36 X 11240=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 Sav/^+^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& + cd = 183.
11. And0a6106 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' = 44936249.
25. And 3ac» + i/a 3 — V = 292497942.
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 coefficients 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 coefficient 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^coefficients 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 coefficient,
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*
HlOv/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
2v/xy + 14x
3x + 2y
—9 + 3v/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+263c10 to 36 4a + 5c + 10 and 56 c.
Add a +6 and a — 6 together.
Add 3a + 610 to>c<*a and — 4c + 2a 36 7.
Add 3a a +6 3 cto 2rt63« 2 +6c— 6.
Add a 3 +6 3 c6 2 to a6 3 ~a6c+6 3 .
Add 9a86 f 10x6a*7c + 50 to $r3a5c +46 +6d
10.
SUBTRACTION.
Set down in one line the first quantities from which the
subtraction is to be made ; and underneath them place all the
other quantities composing the subtrahend ; ranging the like
quantities under each sifter, as in Addition.
Then change all the signs (+ and — ) of the lower line,
or conceive them to be changed ; after which, collect all the
terms together as in the cases of Addition*.
* This rule is founded on (he consideration, that addition and sub
traction are opposite to each other in their nature and operation, as are
the signs {and — , by which they are expressed and represented. So
that, since to unite a negative quantity with a positive one of the same
kind, has the effect of diminishing it, or subducting an equal positive
•UBTRAOTIO.N.
1T1
XXAXPLK3.
From la % — 36 9x* ~ 4y + 8 8xy — 3 + 6z — y
Take 2a a — 86 bV+5y4 4xy — 7 — 6x — 4y
*l*y + 4 + 12x+3y
— 20 — 6x— 5xy
3xy — 9x X 8 — 2ay
Rem. 7xy— 12 2$f£%fc — 8 r28+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
^*omfto— , 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 coefficienti .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 fa 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 4c ; 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'bc3bc l + 56c
12x2ac
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 — 7xyl(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
— abb 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 lefthand 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 lefthand 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 coefficient 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 ctc =  = 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 )H26,or^ a =^ = *a + i6.
2. (10*6 + 15a*) s 5a, or 10fl6 + 15fl * =26 + 3*.
3. (30az>48z) f *, or 30g *~ 48 * = 30a48.
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 isf 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 — rr~ 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 — ^rr „»
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 — — rjr,.
a+2a646 1
FRACTIONS. 188
a*+2ab+P) a 3 — <io 3 (a
— tab— 2ab 2 ) a 2 4 2*6 + 6 3
or af 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.rr —r—7r 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
Sbc . ,
2. To reduce rrr. . ... 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 arxor
a 1 — o 4 . ,
5. Reduce = — 5^70—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 acd46W+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
. . , 2abx3cx* . .
= 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.
BetogJt5 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 41 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* _
3a26 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 aM6> 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 Coefficients. 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 faa or — = as— a — a .
as
* This rule, expressed in general terms, is as follows:
<.+x)*=a^». aix+n . "^a^ajn. 'I" 1 . "^v ^
<«^)»^«.a»ix4n . *^ l a*~ z"n. *~ l . n ^ 2 a y Ac.
/toil. Xb* ma of the coefficienU, 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 coefficient is al
ways 1, and the second is the same as the index of the in
tended power ; to find the 3d coefficient, 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
coefficient 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 coefficient of the
term next following ; which rule will find all the coefficients,
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 coefficients 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 coefficients, by the 1st rule,
will be
a 5 , a 4 *, aV, aV, or 4 , **,
and the coefficients, 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 coefficients 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 coefficients will be
a 8 — 6a 5 * + 15aV20aV + 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 coefficient, 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 577. 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 Cornpound 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 jctos
moltiplpd 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 recriired.
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* LIIS.
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^/aft 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 commonsurd, 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 $/256 and ]/32. Ans 2^4.
5. Required the difference of y/% and Ans. Jv'S.
6. Find the difference of t/ and yf. Ans. ^^/^
7. Required the difference of y f and (/V* Ans.fj^/75.
8. Find the difference of v/24a*o a and ^/M'* 4 .
Ans. ^/(36 3 — 206)^/6.
PROBLEM VI.
To multiply Surd Quantities together.
Reduce the surds to the same index, if necessary ; next
multiply the rational quantities together, and the surds to
gether ; then annex the one product to the other for the
whole product required ; which may be reduced to more
simple terms if necessary.
EXAMPLES.
1. Required to find the product of 4y/\2 and 3^2.
Here, 4*3X y/ 12x^2 = 12 v /(12v2) = 12 v /24=12 v /(4x6)
= 12X2X^/6=24^/6, the product required.
2. Required to multiply \%/} by .
Here lXtf/f xyt^iV* V*=rV Xl/«=A X * X ^ 18
=J r ^/18, the product required.
3. Required the product of SyZ and 2^/8. Ans. 24.
4. Required the product of 1^/4 and jyl2. Ans.
5. To find the product of f and rWi* Ans. jtV15.
6. Required the product of 2^/14 and 3y4. Ans. 12^/7.
7. Required the product of 2a* and a 3 . Ans. 2n f .
8. Required the product of (a+A)^ and (a+&)^.
Ml
9. Required the product of 2x+^b and 2x  y/b.
10. Required the product of (af 2y/6)*, and
11. Required the product of 2x* and 3x'»
12. Required the product of 4x* and2y".
PSOBLEH VII.
7b divide one Surd Quantity by another.
Reduce the surds to the same index, if necessary ; then
take the quotient of the rational quantities, and annex it to
the quotient of the surds, and it will give the whole quotient
required ; which may be reduced to more simple terms if
requisite.
EXAMPLES.
1. Required to divide 6 1/ 90 by 3^/8.
Here6r 3.^(96 + 8) = 2^12 =2 /(4X3) =2 X2v/3
= 4v/3, the quotient required.
2. Required to divide 12^/280 by 3$/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»04l= 1 or 1M be sum of 1 term,
143= 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
equidistant 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 Vihn 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 reenterine 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, arfa: 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 21 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 — pr 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(lf + ^ + &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  — rrr, 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 + ,Vr5F &c.
3. Expand ^/(l — 1) into an infinite scries.
Ans. iyV— 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.
mn 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 14 i« V 36* t . ...
Hence a + ^  g — + jg  &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 ara 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 rri 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 n2 n3 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' 87283366 23516 1QA
3 5 87306882 23390 ~ifl 1
3 6 87330272 23263 ~\ZL — *
3 7 87353535 23140 ~ KM
3 8 87876675
Here x "= (3*6' 15' 3° 4' = 2' 15' = ) f = the distance
of the term y, to be interpolated; a = 87283366, d' =
23516, d" =  126, d'" = 1, and y = a + <rd' + *.
2=1' * + x ?=1. . = (a + K + Hd' +
M*" = ) 87283366 + 0052911 — 00001771875 +
•0000000117 = 873360999296, 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. 82537533.
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 = ) 33166248
c = (v/12=) 84641016
d = L/12 = ) 36055512
/ = (^15=) 88729833
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 = — 31622776 + (5 X 33166248)
— (10 x 34641016) + (10 X 36055512) + 38729833 =*
565116193— 378032936=187088257, and e= 18 7083257
5
ss 374166514 = the root, nearly.
2. Given the square roots of 37, 38, 39, 41, and 42, to
find the square root of 40. Ans. 632455532.
3. Given the cube roots of 45, 46, 47, 48, and 49, to find
the cube root of 50. Ans. 3684033.
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
coefficients 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 Wac
whence) c = ( = )  — 5  — ; J> = ( — 
&bAC+bi?+3cA a B+dA* . 5abc5b 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*+
nl n2 n3 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 ? f2n 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 r3 + 2li + ro + &c  e= *
INFINITE SBBXE8.
EXAMPLE 4.
Find the s»m of the series ^ + jig + — + &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 coefficient ; 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 — $6.
Again, if fx — f = V — f 
Then, taking away the £, it becomes fx = f s ° ;
And taking away the 3's, it is 2x = 10 ;
Lastly, dividing by 2, gives x = 5.
SIMPLE EQUATIONS.
286
MISCELLANEOUS EXAMPLES.
1. Given 7* — 18 = 4x + 6 ; to find the value of x.
Firsts transposing 18 and 4x gives 3x = 24 ;
Then dividing by 3, gives x = 8.
2. Given 20 — 4* — 12 = 02 — 10* ; to find x.
First, transposing 20 and 12 and lOx, gives 6x = 84 ;
Then dividing by 6, gives x = 14.
3. Let 4ax — 5b = Sdx + 2c be given : to find x.
First, by trans. 56 and 9dx 9 it is 4ax — 3cfx = 55 + 2c :
Then dividing by 4a — 3d, gives x = ~~~^»
4. Let 5x" — 12x = 9x + 2X 3 be given ; to find x.
First, by dividing by x, it is 5x — 12 = 9 f 2x ;
Then transposing 12 and 2x, gives 3x=21 ;
Lastly, dividing by 3, gives x =■• 7.
5. Given 9ax 3 — 15abx* = Gar* + 12ax 3 ; to find x.
First, dividing by Sax 3 , gives 3x — 56 = 2x + 4 ;
Then transposing 56 and 2x, gives x=56 fi.
— XXX
6* Let  —  + ^ =: 2 be given, to find x.
First, multiplying by 3, gives x — f rf 3 ! x  6 ;
Then, multiplying by 4, gives x + V*x =24.
Also multiplying by 5, gives 17x = 120 :
Lastly, dividing by 17, gives x = 7y T .
~ ^ x — 5 , x , _ x — 10 „ ,
7. Given — — f  = 12 — ; to find x.
First, mult, by 3, gives x — 5 + £r=3G — x + 10 ;
Then transposing 5 and x, gives *2» , 4 , ;r=51 ;
And multiplying by 2, gives 7x = 102 ;
Lastly, dividing by 7, gives x=l 1$.
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 = 305x+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 /(4fx)=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
172x
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 coefficient 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 twopence, 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>fri
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 coefficient 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 coefficient 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(6f 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 — §x4J=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 4o ;
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?— 627 = 33 ; to find 2. Ans. 2 = 10 or4.
* 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 48** + 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 coefficient, 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 coefficient of the
unknown square.
Add the square of the coefficient 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 80x416* = 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 coefficient, 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 —20611000.
8. Given x 8 + 4x* = 12 ; to find x.
Ans. x = = 1259921, or y — 6 =  1817121.
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 = 41;
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 = 26180 +, and
y= 16180 +.
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, xf 2y, xfc3y, 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 rn 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 abovementioned powers
or equations. But they may be all readily solved by the
following easy rule of Double Position, sometimes called
Trial andE 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 43,
and repeat the work aa fol
lows :
155
100
1st Sup.
42
1704
74 088
05928
100
x
x 2
X 3
sums
2d Sup.
43
. 1849
. 79507
102297
100
072 errors +2297
the sum of which is &l
! As 0309 :l : : 2297: 0036
I This taken from  4300
loaves x nearly
= 4264
CUBIC, Ac. EQUATIONS.
250
Again, suppose 4*264, and 4*265, and work as follows:
4*264
X
4*265
18181696
X*
18*100225
77526752
x 1
77*581310
99972448
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 = 42644209
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  1331
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 103 nearly.
Again, suppose the two num.
bers 1*03 and 102, &c. as
follows :
1*03 . i  102
64*89  63* 64*26
—15*9135 IS* 3 — 15*6060
1092727 x 3 1 061208
50 069227 sums 49715208
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 — 102804 ) s 3  15r» + 63* — 50 ( x*  1397196* +
4863627 = 0.
Then the two roots of this quadratic equation, or   
X s — 1397196* = — 4863627, by completing the square,
are 657653 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 coefficient, 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 657653
and 739543
sum 1500000
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.
47906 . /[144* 1 — («» + 20)n  46476
65384 . v/flOO* 1 — (x a + 24)«]
113 290
114 000
—0710
+1759
 the sums of these
 the true number
• the two errors
115750
114000
+1750
As 2469 : 1
0710 : 02 nearly.
70
Therefore x = 72 nearly.
Suppose again x = 72, and then, because it turns out too
great, suppose x also = 71, dec. as follows :
Supp. x = 72,
47990 . v /[144x a — (s 8 + 20) 2 ]
66402 . ^[196**— (x* + 24) 2 ]
414*392  the sums of these
114000  the true number
+0392
0123
the two errors
Supp. * = 7l,
 47973
. 65904
113 877
114 000
—0128
As 515 : 1
*123 : 024 the correction,
7100 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 = 35. .
Log. of 35 = 0544068
then 35 Xlog.3 5=1 904238
Second Supp. x = 3*6.
Log. of 3 6 = 0*556303
then 36 X log. 3 6=2002689
the true number 2000000 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 360000
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 — 10715124 = 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 8640026 the value of x,
which is true to the last figure, the value given by Dr. Hut*
ton being 86400268.
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 = 41 179857.
Ex. 6. To find the value of x in the equation x 3 — 2x=50.
Ans. 38648854.
Ex. 7. To find the value of x in the equation x 3 + 2x 2 —
23x = 70. Ans. x = 513457.
Ex. 8. To find the value of x in the equation x 3 — 17x 2
+ 54x = 350. Ans. x = 1495407.
Ex. 9. To find the value of x in the equation x 4 — 3x* —
75x = 10000. Ans. x = 102609.
Ex. 10. To find the value of x in the equation 2x 4 — 16x 3
+ 40X 3 — 30x = — 1. Ans. x = 1284724.
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 = 18360877.
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
coefficient 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 coefficient 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 coefficient 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 coefficient 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+^(4A)]«l/(2+^ W)^
and Vt^^+c 3 )] ^/[2v/(4A)]l/(2^ W)»
3/(2 y v /3)=042265 ^
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 21 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 = ytheratio; 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. 105 * 021189
142067 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
280701
234498
201488
176730
157473 h
142067 1
118957?
102448
9 0065
8 0432
72725
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
10350
10400
10450
10500
10600
2
10609
10712
10816
10920
11025
11236
3
10927
11087
11249
11412
11576
11910
4
11255
11475
11699
11925
12155
12625
*
11593
11877
12167
12462
12763
13382
6
11948
12293
12653
13023
13401
14185
7
12299
12723
13159
13609
14071
15036
8
12668
13168
13686
1«4221
14775
15939
9
13048
13629
14233
14861
15513
16895
10
13439
14106
14802
15530
16289
17909
11
13842
14600
15895
16229
17103
18983
12
14258
15111
16010
16959
1.7959
20122
13
14685
15640
16651
17722
18856
21329
14
15126
16187
17317
18519
19799
22609
15
15580
16753
18009
19353
20789
23966
16
16047
17340
18730
2 0224
21829
25404
17
16528
17947
19479
21134
22920
26928
18
17024
18575
2 0258
22085
24066
28543
19
17535
19225
21068
23079
25270
30256
20
18061
19828
21911
24117
26533
32071
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   10872647
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
42149
42465
42782
43101
43746
5
5 3091
53625
5 4163
54707
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
105828
108021
110266
11 4913
10
114639
117314
120061
122882
12 5779
131808
11
128078
131420
13 4864
138412
142068
149716
12
14 1920
146020
150258
154640
159171
168699
13
156178
16 1130
16 6268
171599
177130
18*8821
14
170863
17 6770
182919
189321
19 5986
21 0151
15
185989
19 2957
203236
20 7841
21 5786
232760
16
20 1569
209710
21 8245
22 7193
236575
256725
17
21 7616
227050
236975
247417i 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
18997
1*8861
3
28286
2*8016
27751
4
37171
36731
36299
5
45797
4*5151
44518
6
54172
5*3286
52421
7
62303
61145
60020
8
70197
6*8740
77327
9
77861
76077
74353
10
8*5302
8*3166
8 1109
11
95256
90016
87605
12
99540
96633
93851
13
106350
10 3027
99857
14
11 2961
10*9205
105631
15
11 9379
11 5174
11 1181!
16
12 5611! 120941
1165231
17
131661
126513
12 1657
18
137535
13 1S97
12659?!
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
72688i
7*9127
8*5289'
91 186!
9*6S29
10 2228:
107396
11 2340.
11 7072:
12 16001
1259331
13 0079
13 4047'
09524
18594
27233
35460
4 :)23b
50757
57864
64632
7* 1078
7 7217
83G54
88633
93S3S
989S6
103797
6 per c.
95. 4
2 (>7oC
34651
42124
49173
55824
62095
68017
73601
78869
83838
88527
92950
971*3
10 8378: 10 1069
112741
11 6S96
120853
104773
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 282797 X 50, gives 14139851 =
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
142124; therefore 142124 X 50 = 710022 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 59183
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 Rightangled Triangle is that which
has one right angle. ^
28. Other triangles are Obliqueangled, and
are either obtuse or acute.
20. An Obtuseangled Triangle has one ob
tuse angle.
30. An Acuteangled 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 obliqueangled 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 rightangled 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 rightlined 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 rightlined 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 rightlined 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 rightlined 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 iltnlical, 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)(ft6):=«*— 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 rightangled 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 rightangled 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 obtuseangled 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 rightangled 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 siles 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 termDivided, 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 ijfl, and b } 6, or, in other
words, if the equation a+o = b6 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 rightangled 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 rightangled 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
rightlined 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 rightangled 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 twothirds 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 rightangled 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 twothirds ; or the whole sphere efgh equal to
twothirds 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^JLJK*
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 rightangled 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
rightangled 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 XJsA )
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 rightangled 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 onethird of the'
FB0BLEMS.
987
whole, or of two right angles (th. 17), or
onesixth 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 rightangled triangle aog, being equal
to the two sides af, ao, of the rightangled 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 rightangled 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
rightangled 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 rightangled 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, rightangled 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
rightnngled 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
rightangled 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
af. & + 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 rightangled 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 rightangled 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 rightangled 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 rightangled 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 rightangled 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 rightangled 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 rightangled 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. wa+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 rightangled 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 rightangled 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 rightangled
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 rightangled triangle chf or cbk. As also the
radius, cotangent, and cosecant, another rightangled triangle
cdl. And all these rightangled 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
314169^653589703 &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 + 34720 + &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 rightangled 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. 23654880
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. 97827956
 ' 23654*80
9*6580371
9*9908291
'84407*93
25735218
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 11265 feet
lor 1647 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  99888903
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. 095518S0
To its ou. side ar 345    253781M
Sown. ofZ a37 2ff •   07827U58
To its op. side bc 232  •  23054880
EXAMPLE n.
In the plane triangle abc,
( ab 365 poles ( bc 80086
Given I ac 15433 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 11265
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 rightangled 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 rightangled 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 rigtaangled 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 17407
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 : 6818 = ap — bp,
its half is  34 09
the half base is 17250
the sum of these is 20659 = ap.
and their diff. is 138*41 = bp.
Then, in the triangle apc, rightangled at p,
As the side ac   232  log. 23054880
To sin. op. iiP . . 90 s .  10*0000000
So is the side ap .  20859 . 23151093
To sin. op. L acp .  62° 56' . 994t6213
which taken from  90 00
leaves the Lh. 27 04
990
plaice naooHoxmr.
Again, in the triangle bpc, rightangled at p 9
As the aide bc . . 17407  log. 22407230
Toain. op. £p  00" . . 100000000
8oiaaide bp   18841 . . 21411675
To ein op. L bcp   52° 40' . . 90004486
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'
ssissa Has 
To find the angles
exampw in.
r; w «n (ab120 (^a 57*27' 0'
\ AC 11265 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 rightangled 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 Rightangled ; 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 *
rightangled 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 rightangled 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 22005150
So tang. 53° 7 48" 101240871
TolegBc 216  SS&M&fllY
S03
PLANE raieONOMBTRY.
So secant Lk  53° T 48 7  10 2218477
To hyp. ac 270 . 24318627
EXAMPLE H.
In the rightangled triangle abc,
n . 5 the leg ab 180 . (ac 3020146
UlTcn ) the Z a 62 40' An8, I bc 3482464
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 rightangled
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 rightangled 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(a5)
= tan. £(a+6) (cos. a+cos. b)
22. sin. a sin. b = tan. ^(a6) 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 rightangled 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 suchlike 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 plumbliner
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  100000000
To ac 200   23010300
So tang. L c 47° 80' 100379475
To ab 21826 required 23380775
Or, by the nat. tangents, we have ac X tan. bca =
200 X 1091308 = 2182616 = 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° ... 06855712
To op. side ab 880  29444827
So sin. Lk 35°  97585913
To op. side bc 1041125 30175028
AND DISTANCES.
399
As sin. ^acb 29°  .  96855712
To op. side ab 880 ... 29444827
So sin. Z.B 1 16° or 64°  99536602
To op. side ac 1631442   32125717
And in the triangle bcd, (
As sin. L d 90°    100000000
To op. side bc 1041125  . 30175028
So sin. L b 64°    09536602
To op. side cd 935757   29711630
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
94720856
17781513.
96050320
i9\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
41488 . 16179225
60* 3C  104272623
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 57623   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 rightangled triangle abi ,
As radius  100000000
To ab . 190 ... 4i 2 0791812
Sotaug. /I bac 33 l »   . 98125174
Tob<; . 77«»   18916980
ASD DISTANCES.
401
Then in the rightangled triangl* abd,
As radius 10 'OOOOOOO
To ab •  120 . 2 0701812
So tang. L bad  64° 3a   10*3215089
To bd • 251586  . 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  = 17305632
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 6964,
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. 56649 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. 22155 feet
Exam. xi. From the top of a toner, by the seaside, 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. 20422 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. 28628 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 30753
An8# } distance 19815
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. 9333148 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 25079
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 5726
Ans " I distance 15056
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 plumbline
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. 185943 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 *' {229605
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 bybe 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. 52048 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 eastby .south ; and after sailing northeast 12
miles, it was found to bear south.eastbyeast. It is required
to determine the place of that headland, and the ship's dis
tance from it at the last observation ? Ans. 260728 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. 9595866 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 53681
n ' \ bo 50047
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 65580
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 7103
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 620 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 eqial 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. 0684887
600 2 778151
Anawer 2904276 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. 2086047.
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. J6f — — = — = the segment ap ;
hence V^ao 1 — ap*) = the perp. cr, that is,
CP.
66 + «« — «»
V(as(— ^ ))= 
Wb>~a< +2b€i — »if g«y — f i _
V 466 "
Bot Jab X cp is the area, that is,
16 X cp = V J — jg 1
_ vi mabbcc + 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 2904737, 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 172047737.
Here 25 X 5 = 125 is the perimeter.
And 172047737 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.
04330127
05778503
10000000
07071068
17204774
08506508
25980762
10000000
: 6339124
1 1523824
4828*271
13065628
61818242
14619022
76942088
16180340
93656399
17747324
11 1961524
19318517
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. 17204774 X 625 = 1075298375, as before.
Ex. 2. To find the area of the trigon or equilateral tri
angle, whose side is 20. Ans. 17320508.
Ex. &• To find the area of the hexagon whose side is 20.
Ans. 103923048.
Ex. 4. To find the area of an octagon whose side is 20.
Ans. 193137084.
Ex. 5. To find the area of a decagon whose side is 20.
Ans. 307768352.
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 31410, 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(2f 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 rightangled triangle a CD,
it « ill be dc ^ V(ad  ac ) = V( ,i — 1) = V3 = 17320508U76, 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 :
^/ 393 1 85 165*5 :
^3 9828897227 :
^39957178165 :
v/39989291743 :
v/39997322757 :
v/39999330678 :
V39999832669 
19318516525
19828097227
19957178465
: 19939291743
1*9997322757
; 19999330G78
19999832669
° 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 248774 miles,
what is its diameter ?
By the 2d rule, as 31416 : 1 : : 248774 : 79187 nearly
the diameter.
By the 3d rule, as 355 : 113 : : 248774 : 79187 nearly.
i>; u.e4lhrule,asl : 318309 : : 24877 4 : 79187 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
00000167331 will tie the square of the chord of the said 1 546th part of
the periphery, and consequently the root \/O 00O0 167331 = 0OO40906 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 = 00040906 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)81 ; 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 6283178&
but less than 62831940,
and must therefore be nearly equal to £ their sum, <>r 62831854, which
% in fact is true to the last figure, which fhould be a 3, instead of the 4.
Hence the circumference heing 62831854 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. 471231.
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.
31416 7854 31416
10 10 3 = 100 31416
4)314 16 ~* 986965
7854 70 34 07958
7854
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" :
62831854: : 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
125664, 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. 114595.
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, 31416 X 3 = 04248, the circumference.
And 360 : 18 : : 04848 : 47124, the length of the arc.
Then 47124 X 3 ^4= 141372 + 4 « 35343, the area.
2; By the 2d Rule.
First, 7854 X 3" = 70686, 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. 8043086.
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 = 31416, the area of VX/
the whole circle. . / '
Therefore 360° : 7374 : : 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 = 163504, 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 =* 12870059
sin. 78° 44'}, to radius 1 = 9600010
2) 3270009
•1685084
whence, 1635034 X 10* = 1635034, 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. 636375.
Ex. 3. Required the area of the segment whose chord is
16, the diameter being* 20 ? Ans. 44728, or 269432.
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 82, 74, 92, 102, 86 ; and the whole
length 39 ; required the area.
8*2 352 sum.
86 39
2 ) 168 sum of the extremes. 8168
84 mean of the extremes. 1066
74 4 ) 13728
92 3432 An*.
102
362 sum.
Ex. 2. The length of an irregular figure being 84, and the
breadths at six equidistant places 17*4,206, 142, 165, 20*1,
244 ; what is the area? Ana. 155064.
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. 27489.
Ex. 2. To find the area of the oval whose two axes are 24
and 18. Ans. 3392928.
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. 16203.
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. 978425.
PROBLEM XVI.
To find the Area of a Parabola, or iU Segment.
MultiAy the base by the perpendicular height; then
take twothirds 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. 125664.
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. 607511.
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. 13153.
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. 38971143.
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. 275276.
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. 138564 feet.
Ex. 6. Required the content of a cone, its height being
lty feet, and the circumference of its base 9 feet,
Ans. 2256093.
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. (cfM 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 onethird 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 onethird 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. 931925 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. 527788P.
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. 464216.
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. 790613.
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 81416,
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 rylindtr.
And as this is every where the case, therefore the sums of any corres
ponding number of tlnse 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. 18095616.
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. 11875248 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. 7854 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
504 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 31416.
Then, \t is =. the base of the cylinder, or one gre«t circle of the
sphere ; and d is the height of the cylinder : thenfore \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. 41888.
• 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
figues 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
78*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 threelegged 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 twofoot 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 stationpoint 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 fieldbook,
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 clockwork.
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 stationstaff, 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 offsetstaff, 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 flowerdeluce. 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 Foursided 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 fieldbook, 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 flowerdeluce ; 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 flowerdeluce ; 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 flowerdeluce, 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 fieldbook,
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 fieldbook, 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 suchlike.
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
eyedraught 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 seacoasts, rivermouths,
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 fieldbook ; 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 flowerdeluce 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 footpaths, 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
fieldbook, 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 fieldbook 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 chainline ; 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 fieldbook following, p. 454.
The letter in the lefthand corner at the beginning of every
line, is the mark or place measured from; and that at the
righthand 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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458
plan from the measures in the fieldbook, 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 fieldbook, 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 fieldbook 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 fieldbook. 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 FIELDBOOK.
In surveying with the plain table, a fieldbook 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
footpath 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 fieldbook, 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 recomputed, 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 rightlined 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 horsehair,
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 whalebone, or suchlike 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 rightlined 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 blacklead 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 suchlike, 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 halfquartoM 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 1805, 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. 169753 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. 3208J 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. 521375 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 chimneypiece, 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 woodwork 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 Staircases, 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 handrail, and girt over its end till it meet the
top of the newelpost, for the one dimension ; and twice the
length of the baluster on the landing, with the girt of the
handrail, 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.
Windowshutters, 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 windowshutters 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 fireplace 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 fireplace 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 windowshutters 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 fireplace 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 275 feet broad ? Ans. llf
PAVEIt'* WORK.
407
Exam. 2. What will the glazing a triangular skylight
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 footpath, 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 courtyard, whose length is 63 feet, and breadth 45
feet ; in which there is laid a footpath 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 8lb. 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 Fourtided 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 104 feet ? Ana. 26} feet.
Exam. 3. Required the content of a piece of timber,
whose length is 2038 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 104, 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. 89922 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 615, in the fourth 4*74, and in the fifth 3*16 ?
Ans. 42519525.
[ 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 rightangled, 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
semiconjugate.
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 Semiconjugate 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 semiaxes, 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 semitransverse.
Carol. 2. The conjugate semiaxis 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 semiconjugate ;
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 rightangled 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. Betnil <« 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 rightangled 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 semiconjugate ;
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 rightangled 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 semitransverse
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 semiaxis 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? *. *. \ *xvnu
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 Semiaxes.
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 semiaxes, and the fncaldistance from
the centre, are the sides of a rightangled triangle c\a ; and
the distance \a is = cf the focal distance.
Corol. 2. The conjugate semiaxis 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 semitransverse.
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 rightangled 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
semitransverse axis a<\ perp. to which draw al or a/, which
will be equal to the semiconjugate, 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
460517018 : 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 rightangled 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^cdMdh^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 semitrans
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, andcn = 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.
JjffiMS. 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 seraiaxis 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 semidiamHter 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 semidiamater 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 setnidiameter 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 semidiameters 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 rightang. 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 TwoThirds 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 semiparabola acb, p . ap = pm 1 ,
also, in the semiparabola 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 semiordinates 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;
* sbh its conjugate ;
* = ak, any absciss, from the extremity of the diam.
y = ok. the correspondent ordinate : the two being jointly
denominated coordinates.
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 coordinates 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 coordinates 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 rightangled 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 lsoM»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 : : .... cir 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 rightangled 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 rightangled 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 thenfore 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 x2«, 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 169645068: 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
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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 courtyard be 47 feet 9 inches square,
and I have laid a' footpath with Purbeckstone, 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.
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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 189 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. 10085 /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 axletree, what
was the circumference of the track described by the outer
wheel ? Ans. 62832 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. 72746 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 6086.
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. 58876 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 rightangled 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
eavesboard* 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 5809lbs. per foot ? Ans. 322825 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
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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
shopbook 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 maypole, 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 grindingstone 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 44508, 2d, 48400, 3d 53535, 4th 6 0705,
5th 72079, 6th 93935, 7th 226778 inches.
Quest. 35. A maltster has a kiln, that is 16 feet 6 inches
square : but h