# Full text of "A course of mathematics : for the use of academies as well as private tuition : in two volumes"

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About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at http : / /books . qooqle . com/ k COURSE OF MATHEMATICS; FOR THB USE OF ACADEMIES AS WILL AS PRIVATE TUITION. IN TWO VOLUMES* CHARLj^g; IWTTON,. LUR JFJtS. fcATS PROFESSOR OF ajA&XMATICS IN THE WfAL VlMTART ACADEMY . THE FIFTH AMENlfeKlV FROM THE NINTH WITH MANY COR.RjplC£i0r;S 4 AND IMPROVEMENTS. RV OUNTHUS GREGORY, LL.D. -CorretponSing Associate of the Academy of Dijon. Honorary Member of the Literary and Philosophical Society of New- York, of the New- York Historical Society, of the Literary and Philosophical, and the Antiquarian Societies of Newcastle opoa Tyoa, of the Cambridge Philosophical Society, of tho Institution of Civil Engineers, ate. fce. Secretary to the Astronomical Society of London, and Professor of TTithamatioa in wfaa Royal Military Academy. WITH THE ADDITIONS or ROBERT ADRAIN, LL.D. F A. P S. F.A.A.S, lie. And Professor of Mathematics and Natural Philosophy. THE WHOLE CORRECTED AND IMPROVED. VOL. I. NEW-YORK: . W. JL DZAJf, PR UTTER.' r. AMD I. SW0RD8; T. A. RONALDS ; COLLINS AND CO. ; COLLINS iNb RAN* NAT J H. AND C. AND H. CARVILL; WHITE, OALLARERj AND WHITE f a A. ROORBACHf AND M'CLRATB AND BANGS. 1831. SiutiUr* JHrtrict o/ JrVw-Ter*, to wit I BE IT REMEMBERED, That oatb« 224 day of February, Ana© Domini ISSt, W. E. DEAN, of the Mid district, bath j i r w i ti4 U «kU edbce the title of a book, the title ef which U in the word* following, to wit: • A Count of Mathematics ; for the nse of Academies as well eg private toitioo. la Two Volomes. By Charles Hottoo, LL.D. F.R.S„ Uto Profeieor of Mathematics in the Royal Military Academy. The Fifth American from the Ninth London Edition, with many corrections and improvements. By Olinthus Gregory, LL.D. Correspond* Ing Associate of the Academy of Dijon, Honorary Member of the Literary and Philo- sophical Society of New- Fork, of the New- York Historical Society, of the Literary and Philosophical, and the Antiquarian Societies of Newcastle upon Tyno, of the Cambridge Philosophical Society, of the Institution of Clril Engineers, dec dec Secre- tary to the Astronomical Society of London, and Professor of Mathematics in the Ro- yal Military Academy. With the Additions of Robert Adrain, LL.D. F.A.P.8. F.A. AA, dec and Professor of Mathematics and Natural Philosophy. The whole correct* od and improved." Om right whereof ho claim as proprietor. In conformity with an Aat of Congress, e*> ttttt « A* Act to sjmb4 the atreral Acts respecting eopy-rights." FRED. J. BETTS, Cfcr* «/ we fiaHws Dirtrtrt e/AWFerfc . PREFACE. The present American edition is in part a re- print of the Ninth English edition by Dr. Oun- thus Gregory, with 'mQst.pf, the improvements introduced into former Al^ejricaa editions by Dr. Adrain, together with mich modifications of the English editions as appeared calculated to in- crease the general usefulness of the work. At the same time two or three Chapters, devoted to subjects of no great value at present to the American student, have been omitted, to leave room for matter of more interest and importance. CONTENTS OF VOL. I. Gin iRAt. Preliminary Principles I ] ARITHMETIC, ffotaiion and Numeration *j Roman Notation - . 7 Addition g Subtraction - . - ][ Multiplication - 13 Division - . - IS Reduction - Compound Addition • 32 Commissioned Officers' Regimen talPay . Compound Subtraction * Multiplication Division - 1 Rule, or Rule of Thr** ■ r _ jnd Proportion Vulgar Fractions Reduction of Vulgar Fractions Addition of Vulgar Fractions Subtra ct ion of VwTgar Fractions Multiplication t*f Vol fat Frtfst&ns'. ib. Di v i s ion of Vu *r*c«loas ? (ft Roto of Three ia Vulgar Fractions 65 imnl Fractions * .«HSe Involution by Logarithms Evolution by Logvrtthntt ALGEBRA,. Definitions and ] Addition Subtraction Multiplication Division - Fractious Involution Evolution Surds * Arithmaticiil Pa. 159 160 Decimal . . Su1*ra™m ofDectmftlf. *- : " A i . ^ Multiplication of Decimal*- - . ,.ib* Division of Decimals* m -l *' ^.55 Duodecimals - - 77 Involution - - - 70 Evolution - - 60 To extract the Square Root - ft I To extract »he Cube Root - 85 VTo extract any Root whatever • 8ti Table nf Powers and Roots - 9] Ratios, Proportions, end Progres- sion* - - . - Hi Ariihmetical Proportion - 112 Geometrical Proportion - - 116 Harmonical Proportion - 181 Fellowship, or Partnership - 132 Single Fellowship * - ib. Double Fellowship * - 125 Simple Interest - - 127 Compound Interest - - 13Q 13* - 119 136 - las 141 - 16* 166 - 170 172 - 175 179 - 189 192 - 196 id Pro- - 203 207 Geometrical Proportion and Pro- gression - - - 212 Infinite Series, and their Summa- tion * 214 Simple. Equations * • 231 Q r vulca(jc Equal iona - . 249 Xu&cTiftitl Higher Equations- 256 . Sixfiih inlerest - - - 266 ,Coni pound Interest * - 267 - 270 gression Piles of Shot and Shells frfinhloas - Alligation Alternate Single Position Double Position * Practical Questions LOGARITHM*. Definition and Properties of Loga- rithms • - - .146 To compute Logarithms 149 Description and Use of Logarithms 15 Multiplication by Logarithms - 157 Division by Logarithms 15* GEOtt£TRT* - 275 281 - ib. Defi- .318 Theorems <- 320 Of Plane* and Sollds^Deftriitions 336 Theorems - , . 340 Problems * . 355 Application of Algebra to Geome- iry - , . .371 Problems » - 372 Plane Trigonometry * - 378 Trigonometrical Formula - 393 Heights and Distances - - 396 Mensuration otPianes or Areas 405 Mensuration of Solids - * 420 Land Surveying - • 430 ArtifuerV Worls - - 459 Timber Measuring - . 468 Conic SecUoM - - -472 Of the Ellipse - - 476 Of the Hyperbole . -494 Of lhe Parabola - - 518- Problems, &c, in Conic Sections - 534 Equations of the Curve - 536 Element* of Isoperiim try * - 539 Surfaces - - - 541 Solids • - - .551 PracticaT Qiie^cmiinMantiiratlOB 562 Logarithms of Numbers * • 571 Table of Lorarithznic Sines, and 590 A COURSE OF MATHEMATICS, & c . GENERAL PRINCIPLES. 1. Quantity, or Magnitude, is any thing that will admit of increase or decrease ; or that is capable of any sort of calculation or mensuration ; such as numbers, lines, space, time, motion, weight, &c. - 2. Mathematics is the science which treats of all kinds of quantity whatever, that can be numbered or measured.— That part which treats of numbering is called Arithmetic ; and that which concerns measuring, or figured extension, is called Geometry. — Not only these two, but Algebra and Fluxions, which are conversant about multitude, magnitude, form, and motion, being the foundation of all the other parts, are called Pure or Abstract Mathematics ; because they investigate and demonstrate the properties of abstract numbers and magnitudes of all sorts. And when these two parts are applied to particular or practical subjects, they constitute the branches or parts called Mixed Mathematics* — Mathematics is also distinguished into Speculative and Practical: viz. Speculative, when it is concerned in dis- covering properties and relations ; and Practical, when applied to practice and real use concerning physical objects. The peculiar topics of investigation in the four prmci^ %1 departments of pure mathematics may be indicated Vw fo\u Vol. L 2 2 OEIfEKAL PRINCIPLES. words : viz. arithmetic by number, geometry by form, algtbrm by generality, fluxions by motion. 3. In mathematics are several general terms or principles ? such as, Definitions, Axioms, Propositions, Theorems, Pro- blems, Lemmas, Corollaries, Scholia, &c. 4. A Definition is the explication of any term or word in a science ; showing the 'sense and meaning in which the term is employed. — Every Definition ought to be clear, and ex* pressed in words that are common and perfectly well under* stood. 5. A Proposition is something proposed to be demon- strated, or something required to be done ; and is accordingly either a Theorem or a Problem. 6. A Theorem is a demonstrative Proposition ; in which some property is asserted, and the truth of it required to be proved. Thus, wpea R is said that, The sum of the three angles of a plane triangle is equal to two right angles, that is a Theorem, the truth of which is demonstrated by Geometry* — A set or collection of such Theorems constitutes a Theory, 7. A Problem is a proposition or a question requiring something to be done ; either to investigate some truth or property, or to perform some operation. As, to find out the quantity or sum of all the three angles of any triangle, or to draw one line perpendicular to another. — A Limited Pro* blem is that which has but one answer or solution. An Un- limited Problem is that which has innumerable answers. And a Determinate Problem is that which has a certain nun*, her of answer*. 8. Solution of a Problem, is the resolution or answer given to it. A Numerical or Numeral Solution, is the an. 89 ex given in numbers. A Geometrical Solution, is the an- swer even by tl*e principles of Geometry. And a Mechani- cal Solution, is one whicji is gained by trials. 9. A Lemma is a preparatory proposition, lajd down in ojrder U> shorten the demonstration of the main proposition which follows it. ip. A Corollary, or Conseetary, is a consequence draws imq^aWy from some proposition or other premises. 11. A Scholium is a remark or observation nade upon some foregping proposition or premises. 12. MMim,Qr Masim, is a self-evident prejwwuon ; requiring no formal deopnsteation to prove its truth ; but received and assented to as soon as mentioned. Such as, Tb* wlpole of any thing is. greater than a part of it ; or, The wjhoJ* i* equal to a)| its parts taken together ; or, Two quan. tides that are each of them equal to a third quantity, are equal to eaqh other, OSNK&AL AINCIPLE8. 3 13. A Postulate, or Petition, is something required to be done, which is so easy and evident that no person will hesi- tate to allow it. 14. An Hypothesis is * stipulation assumed to be true, in order to argue from, or to found upon it the reasoning and demonstration of some proposition. 19k Demonstration is the* collecting the several torments alri proofs, 4nd ttryiilg them together in proper order to shbifr the trtitt of the proposition under consideration, 10» A Ihtreet, TbsUloe, of Affirmative D&ntviishiitiun', is thai *hr6h cotidrides with the direct and certain pttibf df th* prtjpoeitioh in hand. IT. 4ji Indira*, or irt&trtttxr Beinortstrdtioh, is that whicti shows a proposition to be true, by proving that some ab- aofdlry would hecessanly fblloW if the proposition advanced were false* This is also sometitties called Redbctio ad Afr- HftNAdSi ; because it shows the absurdity arid falsehood of alt suppositions cbritrarjr to that contained in the proposi- tion. 18. Method is the art of disposing a train of arguments in a proper order, to investigate either the truth or falsity of a proposition, df to demons tra te it to others when it has been (bund oat.— 'this is either Analytical or Synthetical. 19. Analysis or the Analytic Method, is the art or mode of finding out the truth of a proposition, by first supposing the thing to be done, and then reasoning back, step by step, till we arrive at some known truth. This is also called the Method of Invention, or Resolution ; and is that which is com- monly used in Algebra. 20. Synthesis, or the Synthetic Method, is the searching out truth, by first laying down some simple and easy prin- ciples, and then pursuing the consequences flowing from them till we arrive at the conclusion. — This is also called the Method of Composition ; and is the reverse of the Ana* lytic method, as this proceeds from known principles to an unknown conclusion ; while the other goes in a retrograde order, from the thing sought, considered as if it were true, to some known principle or fact. Therefore, when any troth has been found out by the Analytic method, it may be demonstrated by a process in the contrary order, by Syn- thesis : and in the solution of geometrical propositions, it is Vety instructive to carry through both the analysis and the synthesis* 4 ARITHMETIC. Arithmetic is the art or science of numbering ; being that branch of Mathematics which treats of the nature and properties of numbers. — When it treats of whole numbers, it is called Vulgar, or Common Arithmetic ; but when of broken numbers, or parts of numbers, it is called Fractions. Unity, or an Unit, is that by which every thing is called one ; being the beginning of number ; as, one man, one ball, one gun. Number is either simply one, or a compound of several units ; as, one man, three men, ten men. An Integer, or Whole Number, is some certain precise quantity of units ; as, one, three, ten. — These are so called as distinguished from Fractions, which are broken num- bers, or parts of numbers ; as, one-half, two-thirds, or three- fourths. A Prime Number is one which has no other divisor than unity ; as 2, 3, 5, 7, 17, 19, &c. A Composite Number is one which is the product of two or more numbers ; as, 4* 6, 8, 9, 28, &c. NOTATION AND NUMERATION. These rules teach how to denote or express any pro* posed number, either by words or characters : or to read and write down any sum or number. The Numbers in Arithmetic are expressed by the follow- ing ten digits, or Arabic numeral figures, which were intro- duced into Europe by the Moors, about eight or nine hundred years since ; viz. 1 one, 2 two, 3 three, 4 four, 5 five, 6 six, 7 seven, 8 eight, 9 nine, cipher, or nothing. These characters or figures were formerly all called by the general name of Ciphers ; whence it came to pass that the art of Arithmetic was then often called Ciphering. The first nine are called Significant Figures, as distinguished from the cipher, which is of itself quite insignificant. Besides this value of those figures, they have also another, which depends on the place they stand in when joined to- gether ; as in the following table : • NOTATION AND KUXBBAITON. s -3 6c. '9 8 7 6 5 4 3 2 1 98 765432 9 8 7 6 5 4 3 9 8 7 6 5 4 9 8 7 6 5 9 8 7 6 9 8 7 9 8 9 Here, any figure in the first place, reckoning from right to left, denotes only its own simple value ; but that in the second place, denotes ten times its simple value ; and that in the third place, a hundred times its simple value ; and so on : the value of any figure, in each successive place, be- ing always ten times its former value. Thus, in the number 1796, the 6 in the first place denotes only six units, or simply six ; 9 in the second place signifies nine tens, or ninety ; 7 in the third place, seven hundred ; and the 1 in the fourth place, one thousand : so that the whole number is read thus, one thousand seven hundred and ninety- six. As to the cipher, 0, though it signify nothing of itself, yet being joined on the right-hand side to other figures, it in- creases their value in the same ten-fold proportion : thus, 5 signifies only five ; but 50 denotes 5 tens, or fifty ; and 500 is five hundred ; and so on. For the more easily reading of large numbers, they are divided into periods and half-periods, each half-period con. sisting of three figures ; the name of the first period being units ; of the secoud, millions ; of the third, millions of millions, or bi-millions, contracted to billions ; of the fourth, millions of millions of millions, or tri-millions, contracted to trillions, and so on. Also the first part of any period is so many units of it, and the latter part so many thousands. 6 AUOTonnno. The following Table contains a summary of the whole doctrine. Periods. Quadril. ; Trillions ; Billions ; Millions ; Units. Half-per. th. un. th. un. th. un. th. un. th. un. Figures. (123,456 ; 789,098 ; 765,432 ; 101,234 ; 567,890. Numeration is the reading of any number in words that is proposed or set down in figures ; which will be easily done by help of the following rule, deduced from the foregoing tables and observations — viz. Divide the figures in the proposed number, as in the sum- mary, above, into periods and half-periods ; then begin at the left-hand side, and read the figures with the names set to them in the two foregoing tables. EXAMPLXfi. Express in words the following numbers ; viz. 34 96 380 704 6134 9028 15080 72003 109026 483500 2500639 7523000 13405670 47050023 309025600 4723507689 274856390000 6578600307024 Notation is the setting down in figures any number proposed in words ; which is done by setting down the figures instead of the words or names belonging to them in the sum- mary above ; supplying the vacant places with ciphers where any words do not occur. EXAMPLES. Set down in figures the following numbers : Fifty-seven. Two hundred eighty-six. Nine thousand two hundred and ten. Twenty-seven thousand five hundred and ninety-four. Six hundred and forty thousand, four hundred and eighty-one. Three millions, two hundred sixty thousand, one hundred and six. NOTATION AND NUMXBATION. Four hundred and eight millions, two hundred and fifty-five thousand, one hundred and ninety-two. Twenty-seven thousand and eight millions, ninety-six thou- sand two hundred and four. Two hundred thousand and five hundred and fifty millions, one hundred and ten thousand, and sixteen. Twenty-one billions, eight hundred and ten millions, sixty, four thousand, one hundred and fifty. OF THE BOKAN NOTATION. The Romans, like several other nations, expressed their numbers by certain letters of the alphabet. The Romans used only seven numeral letters, being the seven following capitals : viz. i for one ; v for five ; x for ten ; l for fifty ; c for an hundred ; d for five hundred ; m for a thousand ; The other numbers they expressed by various repetitions and combinations of these, after the following manner : 1 = i 2 = ii As often as any character is re- 3 = iii peated, so many times is its value repeated. 4 = mi or iv A less character before a greater 5 = V diminishes its value. 6 = VI A less character after a greater 7 = VII increases its value. 8 = VIII 9 = IX 10 = X 60 = L 100 = C 500 = d or io For every o annexed, this be- comes 10 times as many. 1000 = m or cu For every c and o, placed one 2000 = MM at each end, it becomes 10 times as much. 6000 = v or loo A bar over any number in- 6000 = VI creases it 1000 fold. 10000 = x or ccioo 50000 = | l or 1003 00000 = f- 100000 = ■ < c or ccciooo 1000000 = nor CCCCI0O33 2000000 = ra 9 ARITHMETIC. EXPLANATION OF CERTAIN CHARACTERS* There are various characters or marks used in Arithmetic, and Algebra, to denote several of the operations and propo- sitions ; the chief of which are as follow : + signifies plus, or addition. — - . minus, or subtraction. X or - multiplication. H- - division, t :: : - proportion. «=* - T equality. «v/ - - square root. %/ - - cube root, &c. *r . . diff. between two numbers when it is not known which is the greater. Thus, 5 + 3, denotes that 3 is to be added to 5. 6 — 2, denotes that 2 is to be taken from 6. 7 X 3, or 7 • 3, denotes that 7 is to be multiplied by 3. 8 -r- 4, denotes that 8 is to be divided by 4. 2 : 3 : : 4 : 6, shows that 2 is to 3 as 4 is to 6. 6 + 4 = 10, shows that the sum of 6 and 4 is equal to 10. or 3^, denotes the square root of the number 3. $/5, or 5^, denotes the cube root of the number 5. 7 s , denotes that the number 7 is to be squared. S 3 , denotes that the number 8 is to be cubed, dec. OF ADDITION. Addition is the collecting or pitting of several numbers together, in order to find their sum, or the total amount of the whole. This is done as follows : Set or place the numbers under each other, so that each figure may stand exactly under the figures of the same value, ADDITION* 9 that is, units under units, tens under tens, hundreds under hundreds, dec. and draw a line under the lowest number, to separate the given numbers from their sum, when it is found. — Then add up the figures in the column or row of units, and find how many tens are contained in that sum. — Set down exactly below, what remains more than those tens, or if nothing remains, a cipher, and carry as many ones to the next row as there are tens. — Next add up the second row, together with the number carried, in the same manner as the first. And thus proceed till the whole is finished, setting down the total amount of the last row. TO PROVE ADDITION. First Method. — Begin at the top, and add together all the rows of numbers downwards, in the same manner as they were before added upwards ; then if the two sums agree, it may be presumed the work is right. — This method of proof is only doing the same work twice over, a little varied. Second Method. — Draw a line below the uppermost num. ber, and suppose it cut off. — Then add all the rest of the numbers together in the usual way, and set their sum under the number to be proved. — Lastly, add this last found num- ber and the uppermost line together ; then if their sum be the same as that found by the first addition, it may be pre- sumed the work is right. — This method of proof is founded on the plain axiom, that " The whole is equal to all its parts taken together." Third Method. — Add the figures in the uppermost line together, and find example i. how many nines are contained in their sum. — Reject those nines, and 3497 g 5 set down the remainder towards the 6512 .g 5 right hand directly even with the 8295 £ 6 figures in the line, as in the annexed — — o example. — Do the same with each of 18304 8 7 the proposed lines of numbers, set- — — g — ting all these excesses of nines in a co- W lumn on the right-hand, as here 5, 5, 0. Then, if the excess of 9's in this sum, found as before, be equal to the excess of 9's in the total sum 18304, the work is probably right. — Thus, the sum of the right-hand column, 5, 5, 6, is 16, the excess of which above 9 is 7. Also the sum of the figure* V& Vol. I. 3 10 ABZTHMSTXC. the sum total 18304, is 16, the excess of which above 9 m also 7> the same as die former*. OTHER EXAHPLES. 2. 3. 4. 12345 12345 12345 87890 67890 876 98765 9876 9087 432)0 543 56 12345 21 234 67890 9 1012 302445 90684 23610 290100 78339 11265 302445 90684 23610 Ex. 5. Add 3426 ; 9024 ; 5106 ; 8890 ; 1204, together. Ans. 27650. 6. Add 509267; 235809; 72920 ; 8392; 420 ; 21; and 9, together. Ans. 82683a. * This method of proof depends on a property of the number 9, which, except the number 3 ; belongs to no other digit whatever ; namely* that 44 any number divided by 9, will leave the same remain- der as the sum of its figures are digits divided by 9:" which may be demonstrated in this manner. Demonstration. Let there be any number proposed, as 4668. This, separated into its several parts, becomes, 4000 + 600+60 4- 8. Bnt 4000 = 4 X 1000 = 4 X (999 + 1) = (4 X 999) + 4. In like man- ner 600 = (6 X 99) +6 ; and 50 = (5 X 9) 4-5. Therefore the gi- ven number 4658 = (4 X 999) + 4 + (6 X 99) +6 +(5 X9) + 6 + 8 = (4 X 999) + (6 X 99) + (5 X 9) + 4 + 6 + 5 + 8; and 4658 + 9 = (4 X 999+6 X 99 + 5 X 9 + 4+ 6 + 5 +8) + 9. But (4 X 999) + (6 X 99; + (5 X 9) is evidently divisible by 9, without a remainder ; therefore if the given number 4658 be divided by 9, it will leave the same remainder as4+6+6+6 divided by 9. And the same, it is evident, will hold for any other number whatever. In like manner, the same property may be shown to belong to the number 3 ; but the preference is usually given to the number 9, on ac- count of its being more convenient in practice. Now, from the demonstration above given, the reason of the rale it- self is evident : for the eicess of 9's in two or more numbers being taken separately, and the excess of 9's taken also out of the sum of the former excesses, it is plain that this last excess must be equal to the ex- cess of 9's contained in the total sum of all these numbers ; all the parts taken together being equal to the whole.— This rule was first given by Dr. Waflis in his Arithmetic, published in the year 1657. •mmuonoN. 11 7. Add 2; 19; 817; 4298 ; 50916 ; 730205 ; 91806% together. Ans. 9966891. 8. How many days are in the twelve calendar months ? Ana. 965. 9. How many days are there from the. 15th day of April to die 24th day of November, both days included f Ans, 224 % 10. An army consisting of 52714 infantry*, or fbot, 51 HI horse, 6250 dragoons, 3927 light-horse, 928 artillery, of pinners, 1410 pioneers, 250 sappers, and 406 miners : what is the whole number of men ? Ans. 70995. OF SUBTRACTION. Subtraction teaches to find how much one number ex- ceeds another, called their difference, or the remainder, by taking the less from the greater. The method of doing which is as follows: Place the less number under the greater, in the same man- ner as in Addition, that is, units under units, tens under tens, and so on ; and draw a line below them. — Begin at the right hand, and take each figure in the lower line, or number, from the figure above it, setting down the remainder below it. — But if the figure in the lower line be greater than that above it, first borrow, or add, 10 to the upper one, and then take the lower figure from that sum, setting down the remain- der, and carrying 1, for what was borrowed, to the next lower figure, with which proceed as before ; and so on till the whole is finished. * The whole body of foot soldiers is denoted by the word htfanlrg; and all those that charge on horseback by the word Cavalry. — Some authors conjecture that the term infantry is derived from a certain In- fanta of Spain, who, finding that the army commanded by the king her lather had been defeated by the Moors, assembled n body of the people together on foot, with which she engaged and totally rooted the enemy. In honour of this event, and to distinguish the foot soldiers, who were not before held In much estimation, they received the name of Infantry, from her own title of Infanta. 4 12 AxrrBXxnc. TO PSOVE SUBTRACTION. Add the remainder to the less number, or that which is just above it ; and if the sum be equal to the greater or up- permost number, the work is right*. EXAMPLES. From 5386427 Take 2164315 From 5386427 Take 4258792 From 1234567 Take 702973 Rem. 3222112 Rem. 1127635 Rem. 531594 Proof. 5366427 Proof. 5386427 Proof. 1234567 4. From 5331806 take 5073918. Ans. 257888. 5. From 7020974 take 2766809. Ans. 4254 165. 6. From 8503402 take 574271. Ans. 7929131. 7. Sir Isaac Newton was born in the year 1642, and he died in 1727 : how old was he at the time of his decease ? Ans. 85 years. 8. Homer was born 2560 years ago, and Christ 1827 years ago : then how long before Christ was the birth of Homer ? Ans. 733 years. 9. Noah's flood happened about the year of the world 1656, and the birth of Christ about the year 4000 : then how long was the flood before Christ ? Ans. 2344 years. 10. The Arabian or Indian method of notation was first known in England about the year 1150: then how long is it since to this present year 1827 1 Ans. 677 years. 11. Gunpowder was invented in the year 1330 : how long was that before the invention of printing, which was in 1441 ? Ans. Ill years. 12. The mariner's compass was invented in Europe in the year 1302 : how long was that before the discovery of Ame- rica by Columbus, which happened in 1492 ? Ans. 190 years. * The reason of this method of proof is evident; for if the difference of two jpombers be added to the less, it most manifestly make up a sum equal to the greater. jcttltolicatioh. 18 OF MULTIPLICATION. Multiplication is a compendious method of Addition, teaching how to find the amount of any given number when repeated a certain number of times ; as, 4 times 6, which is 24. The number to be multiplied, or repeated, is called the Multiplicand. — The number you multiply by, or the number of repetitions, is the Multiplier. — And the number found, being the total amount, is called the Product. — Also, both the multiplier and multiplicand are, in general, named the Terms or Factors. Before proceeding to any operations in this rule, it is necessary to learn off very perfectly the following Table, of all the products of the first 12 numbers, commonly called the Multiplication Table, or sometimes Pythagoras's Table, from its inventor. MULTIPLICATION TABLE. JJ 2| 3 4 5 6 71 8 3 11 12 2 4 6 8 JO 12 14 LO 18 20 S3 24 3 6 9 15 18 21 34 27 30 33 361 4 8 12 16 20 24 28 32 36 40 44 48 5 10 15 1 20 25 30 35 40 45 50 55 60 6 12 18 24 30 36 42 48 54 60 66 72 7 14 21 28 35 42 49 56 63 70 77 84 8 16 24 32 40 48 56 64 72 80 88 96 9 18 27 36 45 54 63 72 81 130 99 108 10 20 30 40 50 60 70 80 00 100 110 120 11 22 33 44 55 66 77 88 99 110 121 132 12 24 36 48 00 72 84 96 10s|l20 132 144; 14 ARITHMETIC* To multiply any Given Number by a Single Figure, or by any Number not exceeding 12. * Set the multiplier under the unite 9 figure or right-hand place, of the multiplicand, and draw a line below it. — Then, beginning at the right-hand, multiply every figure in this by the multiplier. — Count how many tens there are in the product of every single figure, and set down the remainder directly under the figure that is multiplied ; and if nothing remains, set down a cipher. — Carry as many units or ones as mere are tens ' counted, to the product of the next figures ; and proceed in the same manner till the whole is finished. EXAMPLE* Multiply 9876543210 the Multiplicand. By 2 the Multiplier. 19753066420 To multiply by a Number consisting of Several Figures. f Set the multiplier below the multiplicand, placing them as in Addition, namely, units under units, tens under tens, &c. drawing a line below it. —Multiply the whole of the multi- plicand by each figure of the multiplier, as in the last article ; 6678 * The reason of this rale is the same as for 4 the process in Addition, in which 1 is car- — — ried for every 10, to the next place, gradu- 32 = 8X4 ally as the several products are produced 280 = 70 X 4 one after another, instead of setting them all 2400 = 600 X 4 down below each other, as in the annexed ex- 20000 = 6000 X 4 ample. ■ 22712 = 5678 X 4 t After having found the product of the multiplicand by the first figure of the multiplier, as in the former case, the multiplier is supposed to be divided into parts, and the product is found for the second figure in the same manner: but as this figure stands in the place of tens, the product must be ten times <jts simple value ; and therefore the first figure of this product must be set in the place of tens ; or, which is the same thing, directly under the figure multiplying by. And proceeding MULTIPLICATION. 15 setting down a line of products for each figure in the multi- plier, oo as that the first figure of each line may stand straight under the figure multiplying by. Add all the lines of pro* ducts together, in the order in which they stand, and their sum will be the answer or whole product required. TO PROVE MULTIPLICATION. There are three different ways of proving multiplication, which are as below : First Method. — Make the multiplicand and multiplier change places, and multiply the latter by the former in the same manner as before. Then if the product found in this way be the same as tho former, the number is right. Second Method.—* Cast all the 9's out of the sum of the figures in each of the two factors, as in Addition, and set down the remainders. Multiply these two remainders to- gether, and cast the 9's out of the product, as also out of the whole product or answer of the question, reserving the re- mainders of these last two, which remainders must be equal when the work is right. — Note, It is common to set the four remainders within the four angular spaces of a cross, as in the example below. In this manner separately with all the 1234567 the multiplicand, figures of the multiplier, it is evident 4567 that we shall multiply all the parts of the multiplicand by all the parts of 8641969— 7 times the mult, the multiplier, or the whole of the 7407402 = 60 times ditto, multiplicand by the whole of the mul- 6172b35 = 500 times ditto, tiplier: therefore these several pro- 4938268 =4000 times ditto. ducts being added together, will be equal to the whole required product ; 6638267489— 4667 times ditto, as in the example annexed. * This method of proof is derived from the peculiar property of the number 9, mentioned in the proof of Addition, and the reason for the one includes that of the other. Another more ample demonstration of this rule may, however, be as follows : — Let p and q denote the number of 9's in the factors to be multiplied, and a and 6 what remain ; then 9p -4- a and 9q + 6 will be the numbers themselves, and their product ia (9r X 9q) + (9p X b) + (9q X fl) + (« v 6 J ; but the first three of these products are each a precise number of 9's, because their factors are so, either one or both : these therefore being cast away, there remains only II X 6; and if the 9's also be cast out of this, the excess is the excess of 9's in the total product : but a and b are the eicesses in the factors themselves, and a X b is their product ; therefore the rule is true. This mode of proof, however, is not an ample check against the errors that night arise from a transposition of figures. 16 ARITHMETIC. Third Method. — Multiplication is also very naturally prov- ed by Division ; for the product divided by either of the fac- tors, will evidently give the other. But this cannot be prac- tised till the rule of division is learned. EXAMPLES* Mult 3542 or Mult. 6190 by 6196 Proof. by 3542 21252 \ / 12392 31878 \2/ 24784 3542 30980 21252 X *\ 18588 21946232 21946232 Proof! OTHER EXAMPLES. Multiply 123456789 Multiply 123456789 Multiply 123456789 Multiply 123456789 Multiply 123456789 Multiply 123456789 Multiply 1234567b9 Multiply 123456789 Multiply 123456789 Multiply 302914603 Multiply 273580961 Multiply 402097316 Multiply 82164973 Multiply 7564900 Multiply 8496427 Multiply 2760625 by 3. by 4. by 5. by 6. by 7. by 8. by 9. by 11. by 12. by 16. by 23. by 195. by 3027. by 579. by 874359. by 37072. Ans. 370370367. Ans. 493827156. Ans. 617288945. Ans. 740740734; Ans. 864197523. Ans. 987654312. Ans. 1111111101. Ans. 1358024679. Ans. 1481481468. Ans. 4846633648. Ans. 6292362103. Ans. 78408976620. Ans. 248713373271. Ans. 4380077100. Ans. 7428927415293. Ans. 102330768400. CONTRACTIONS IN MULTIPLICATION. I. When there are Ciphers in the Factors. If the ciphers be at the right-hand of the numbers ; mul- tiply the other figures only, and annex as many ciphers to the right-hand of the whole product, as are in both the fac- tors. — When the ciphers are in the middle parts of the mul- tiplier ; neglect them as before, oniy taking care to place MULTIPLICATION. 17 the first figure of every line of products exactly under the figure you are multiplying with. EXAMPLES. 1. Mult. 9001035 by - 70100 9001635 63011445 2. Mult. 390720400 by . 406000 23443224 15628816 631014613500 Products 158632482400000 3. Multiply 81503600 by 7030. Ans. 572970308000. 4. Multiply 9030100 by 2100. Ans. 18963210000. 5. Multiply 8057069 by 70050. Ans. 564397683450. II. When the Multiplier is the Product of iwo or more Num- bers in the Table ; then * Multiply by each of those parts separately, instead of the whole number at once. EXAMPLES. 1. Multiply 51307298 by 56, or 7 times 8. 51307298 7 359151086 8 2873208088 2. Multiply 31704592 by 3. Multiply 29753804 by 4. Multiply 7128368 by 5. Multiply 160430800 by 6. Multiply 61835720 by 36. Ans. 1141365312. 72. Ans. 2142273888. 96. Ans. 684323328. 108. Ans. 17326526400. 1320. Ans. 81623150400. * The reason of this rule is obvious enough ; for any number multi- plied by the component parts of another, must give the same product as if it were multiplied by that number at once. Thus, in the 1st ex- ample, 7 times the product of 8 by the given number, makes 66 time! the same number, as plainly as 7 times 8 make 5o\ Vol. I. 4 is ARITHMETIC. 7. There was an army composed of 104 4 battalions, eacb consisting of 500 men ; what was the number of men con- tained in the whole ? Ans. 52000. 8. A convoy of ammunition f bread, consisting of 250 waggons, and each waggon containing 320 loaves, having been intercepted and taken by the enemy, what is the num- ber of loaves lost ? Ans. 80000. OF DIVISION. Division is a kind of compendious method of Subtrac- tion, teaching to find how often one number is contained in another, or may be taken out of it : which is the same thing. The number to be divided is called the Dividend*— The number to divide by, is the Divisor. — And the number of times the dividend contains the divisor, is called the Quotient. — Sometimes 'there is a Remainder left, after the division is finished. The usual manner of placing the terms, is, the dividend in the middle, having the divisor on the left hand, and the quotient on the right, each separated by a curve line; as, to divide . 12 by 4, the quotient is 3, Dividend 12 Divisor 4) 12 (3 Quotient; 4subtr. showing that the number 4 is 3 times — contained in 12, or may be 3 times 8 subtracted out of it, as in the margin. 4 subtr. % Ride.— Having placed the divisor — before the dividend, as above directed, 4 find how often the divisor is contained 4 subtr. in as many figures of the dividend as — are just necessary, and place the num- ber on the right in the quotient. Mul- — * A battalion is a body of foot, consisting of 500, or 000, or 700 men, more or less. t The ammunition bread, is that which is provided for, and distribut- ed to, the soldiers ; the usual allowance being a loaf of 6 pounds to •vary soldier, once in 4 days. X In this way the dividend is resolved into parts, and by trial is found how often the divisor is contained in each of those parti, one after an- other, arranging the several figures of the quotient one after another, into one number. 19 tiply the divisor by this number, and set the product under the figures of the dividend, before-mentioned. — Subtract this product from that part of the dividend under which it stands, and bring down the next figure of the dividend, or more if necessary, to join oh the right of the remainder.— Divide this numbef, so increased, in the same manner as before ; and so on, till all the figures are brought down and used. Note* If it be necessary to bring down more figures than one to any remainder, in order to make it as large as the di- visor, or larger, a cipher must be set in the quotient for every figure so brought down more than one. TO PROVE DIVISION. * Multiply the quotient by the divisor ; to this product add the remainder, if there bo any ; then the sum will be equal to the dividend, when the work is right. When there is no remainder to a division, the quotient is the whole and perfect answer to the question. But when there is a remainder, it res so much towards another time, as it approaches to the divisor : so, the remainder be half the divisor, it will go the half of a time more ; if the 4th part of the divisor, it will go one-fourth of a time more ; and so on. Therefore, to complete the quotient, set the remainder at (he end of it, above a small line, and the divisor below it, thus forming a fractional part of the whole quotient. * This method of proof is plain enough: for since the quotient is the number of times the dividend contains the divisor, the quotient multi- plied by the divisor must evidently be equal to the dividend. There are several other methods sometimes used for proving Divi- sion, some of the most useful of which are as follow: Second Method Subtract the remainder from the dividend, and di- vide what is left by the quotient ; so shall tbe new quotient from this last division be equal to the former divisor, when the work is right. Third Method. — Add together the remainder and all the products of the several quotient figures by the divisor, according to the order in which they stand in the work ; and the sum will be equal to the divi- dend, when the work is right to abithottic. examples. 1 QuoU 3) 1234567 ( 411522 12 mult. 3 3 3 15 15 1234566 add 1 4 1234567 3 Proof. 2 Quot, 37) 12345678 ( 333666 111 37 124 111 135 111 2335662 1000996 rem. 36 12345G78 246 Proof. 222 6 6 247 222 7 6 258 222 Rem. 1 Rem. 36 Divide 73146085 by 4. Divide 53179b6027 by 7. Divide 570196382 by 12. Divide 74638105 Divide 137896254 Divide 35821649 Divide 72091365 Ana. 18286521J. Ans. 759712289$. Ans. 47516365/,. Ans. 2017246,^. Ans. 142I6l6f$. Ans. 46886}Jf . An*. 13*61 § y±\, by 37. by 97. by 764. by 5201. 10. Divide 4637064283 by 57606. Ans. 80496ftf£f 11. Suppose 471 men are formed into ranks of 3 deep, what is the number in each rank? Ans. 157. 12. A party, at the distance of 378 miles .from the head quarters, receive orders to join their corps in 18 days : what number of miles must they march each day to obey their orders? Ans. 21. 13. The annual revenue of a nobleman being 37960Z. ; how much per day is that equivalent to, there being 365 days in the year ? Ans. 104Z. CONTRACTIONS IN DIVISION. There are certain contractions in Division, by which the operation in particular cases may be performed in a shorter manner : as follows : DIVISION. 21 I. Division by any Small Number, not greater than 12, may- be expeditiously performed, by multiplying and subtracting mentally, omitting to set down the work except only the quo- tient immediately below the dividend. BXAMPLXS. 3) 56103061 4)52610675 5) 1370192 Quot. 18701320$ 6) 38672040 7) 81306627 8) 23718020 9) 43081062 11) 576 14230 12) 27080373 II. * When Ciphers are annexed to the Divisor'; cut off those ciphers from it, and cut off the same number of figures from the right-hand of the dividend ; then divide with the re- maining figures, as usual. And if there be any thing remain- ing after this division, place the figures cut off from the di- vidend to the right of it, and the whole will be the true re- mainder ; otherwise, the figures cut off only will be the re- mainder. EXAMPLES. 1. Divide 3704196 by 20. 2. Divide 31086901 by 7100, 2,0) 3*/04l9,6 . 71,00) 310869,01 (4378$Hft. 284 Quot. 185209$} 268 213 556 497 599 568 31 * This method serves to avoid a needless repetition of ciphers, wVncb would happen in the common way. And the truth of the pnivdpYt ou 82 ARITHMETIC 3. Divide 7380064 by 23000. 4. Divide 2304109 by 5800. Ana. 320}f£f Ant. 397ifiJ. III. When the Divisor is the exact Product of two or more of the small Numbers not greater than 12 : * Divide by each of those numbers separately, instead of the whole divisor at once. Note. There are commonly several remainders in work- ing by this rule, one to each division ; and to find the true or whole remainder, the same as if the division had been per- formed all at once, proceed as follows : Multiply the last re. mainder by the preceding divisor, or last but one, and to the product add the preceding remainder ; multiply this sum by the next preceding divisor, and to the product add the next preceding remainder ; and so on till you have gone backward through all the divisors and remainders to the first. As in the example following : EXAMPLES. 1. Divide 31046835 by 56 or 7 times 8. 7) 31046835 8) 4435262—1 first rem. 554407 — 6 second rem. Ans. 554407}} 6 the last rem. mult. 7 preced. divisor. 42 add 1 to the 1st rem. 43 whole rem. 2. Divide 7014596 by 72. 3. Divide 5130652 by 132. 4. Divide 83016572 by 240. Ans. 974244f. Ans. 38808 T W- Ans. 345902,VV- which it is founded, is evident ; for, cutting off the same number of ciphers, or figures, from each, is (he same as dividing each of them by 10, or 100, or 1000, &c. according to the number of ciphers cut off; and it is evident, that as often as the whole divisor is contained in the whole dividend, so often must any part of the former be contained in a like part of the latter. * This follows from the second contraction in Multiplication, being only the converse of it ; for the half of the third part of any thing, is evidently the same as the sixth part of the whole ; and so of any other numbers. — The reason of the method of finding the whole remainder from the Several particular ones, will best appear from the nature of Vulgar Fractions. Thus, in the first example above, the first remainder being 1, when the divisor is 7, makes ; this must be added to the se- cond remainder, 6, making 6+ to the divisor 8, or to be divided by 8. Bat 6f =^|±?= y; and this divided by 8 gives ^ g -^- REDUCTION. 28 • IV. Common Division may be performed more concisely, by omitting the several products, and setting down only the remainders; namely, multiply the divisor by the quotient figures as before, and, without setting down the product, subtract each figure of it from the dividend, as it is produced ; always remembering to carry as many to the next figure as were borrowed before. EXAMPLES. 1. Divide 3104679 by 833. 833) 3104679 (3727# 5 . 6050 2257 5919 2. Diyide 79165238 by 238. 3. Divide 29137062 by 5317. 4. Divide 62015735 by 7803. Ans. 832627,^. Ans. 5479f|{J. Ans. 79474f f f. OF REDUCTION. Reduction is the changing of numbers from one name or denomination to another, without altering their value. — This is chiefly concerned in reducing money, weights, and measures. When the numbers are to be reduced from a higher name to a lower, it is called Reduction descending ; but when, contrarywise, from a lower name to a higher, it is Reduction ascending. Before we proceed to the rules and questions of Reduction, it will be proper to set down the usual tables of money, weights, and measures, which are as follow : OF MONEY, WEIGHTS, AND MEASURES. TABLES OF MONET. 2 Farthings = 1 Halfpenny | 4 Farthings = 1 Penny d 12 Pence = 1 Shilling s 20 Shillings = 1 Pound £ qrs d 4=1 s 48 = 12 = 1 £ 960 = 240 = 20 = \ 24 ASXTHMBTIC. PBNCB TABLE. SHILLINGS TABLE* d 8 d t d 20 is 1 8 1 is 12 30 2 6 2 24 40 3 4 3 36 50 4 2 4 48 60 5 5 60 70 5 10 6 72 80 6 8 7 84 90 7 6 8 96 100 8 4 9 108 110 9 2 10 120 120 10 11 132 /Vote.— £ denotes pounds, « shillings, and d denotes pence. | denotes 1 farthing, or one quarter of any thing. £ denotes a halfpenny, or the half of any thing. } denotes 3 farthings or three quarters of any thing. The fall weight and value of the English gold and silver coin, old and new, are as here below. both Gold. Guinea Half do. Third do. Double Sov. Sovereign Half do. Value £ $ d 1 1 10 7 2 1 10 Weight, dwtgr '6 h 2 16} 1 l*t 10 6+t 5 6A 2 13-fr Silver. Value, s d A Crown 5 Half-crown 2 Shilling 1 Sixpence Old Wt dwt gr 19 8* 9 I6t 3 21 1 22A New Wt. dwt gr. 18 4^ 9 2^ 3 15* 119# The usual value of gold is nearly 41 an ounce, or 2d a grain ; and that of silver is nearly 5* an ounce. Also the value of any quantity of gold, was to the value of the same weight of standard silver, as l^rir to 1, in the old coin ; bdt in the new coin they are as 14^ to 1. Pure gold, free from mixture with other metals, usually called fine Sold, is of so pure a nature, that it will endure the fire without wasting, lough it be kept continually melted. But silver, not having the purity of gold, will not endure the fire like it : yet fine silver will waste but very little by being in the fire any moderate time ; whereas copper, tin, lead, kc. will n#t only waste, but may be calcined, or burnt to a powder. Both gold and silver, in their purity, are so soft and flexible (like new lead, &c.) that they are not so useful, either in coin or otherwise (ex- cept to beat into leaf gold or silver), as when they are alloyed, or mix- ed and hardened with copper or brass. And though most nations differ, more or less, in the quantity of such alloy, as well as in the same place at different times, yet in England the standard for gold and silver coin has been for a long time as follows— viz. That 22 parts of fine gold, and 2 parts of copper, being melted together, shall be esteemed the true standard for sold coin : And that 11 ounces and 2 pennyweights of fine silver, and 18 pennyweights of copper, being melted together, be esteemed the true standard for silver coin, called Sterling silver. In the old coin the pound of sterling gold was coined into 42| gui- neas, of 21 shillkip each, of which the pound of sterling silver was di- vided into 02. Tie new coin is also of the same quality or degree of TABUS OF WEIGHTS. 2$ TKOY WEIGHT*. Grains - - marked gr 24 Grains make 1 Pennyweight dwt 20 Pennyweightsi Ounce oz 12 Ounces 1 Pound lb fr dwt 4= I oz 480= 20= 1 lb 5760=240=12=1 By this weight are weighed Gold, Silver, and Jewels. APOTHECARIES' WEIGHT. Grains - - marked gr , 20 Grains make 1 Scruple sc or 3 3 Scruples 1 Dram dr or 3 8 Drams 1 Ounce oz or J 12 Ounces 1 Pound lb or ft 8C 20 = 1 dr 60 = 3 = 1 02? 480 = 24 = 8 = 1 lb 5760 = 288 = 96 = 12 = 1 This is the same as Troy weight, only having some dif- ferent divisions. Apothecaries make use of this weight in compounding their medicines ; but they buy and sell their Drugs by Avoirdupois weight. AVOIHDUPOIS WEIGHT. Drams 16 Drams 16 Ounces 28 Pounds 4 Quarters make 1 Ounce - 1 Pound - 1 Quarter - 1 Hundred weight marked dr oz lb cwt 20 Hundred Weight 1 Ton ton fineness with that of the old sterling gold and silver above described, but divided into pieces of other names or values ; viz. the pound of the sil- ver into 66 shillings, of course each shilling is the 66th part of a pound ; and 20 pounds of the gold into 934£ pieces called sovereigns, or the pound weight into 4Gf« sovereigns, each equal to 20 of the new shil- lings. So that the weight of the sovereign is 46^$Ujs of a pound, which is equal to pennyweights, or equal to 5 dwt. 3^ gr. very nearly, as stated in the preceding table. And multiples and parts of the sovereign and shilling in their several proportions. * The original of all weights used in England, was a grain or corn of wheat, gathered out of the middle of the ear, and, being well dried, SI of them were to make one pennyweight, 20 penny weigjhU out Vol. I. 5 26 ARITHMETIC. dr oz 16 = 1 lb 256 = 16 = 1 qr 7168 = 448 = 28= 1 ewi 28672 = 1752 = 112 = 4 = 1 Urn 573440 = 35840 = 2240 = 80 = 20 = 1 By this weight are weighed all things of a coarse or drossy nature, as Corn, Bread, Butter, Cheese, Flesh, Grocery Wares, and some Liquids ; also all metals, except Silver and Gold. oz dwt gr Note, that lib Avoirdupois = 14 11 15J Troy \oz - - = 18 Idr . - =013} LONG MEASURE. 2 Barley-corns make 12 Inches - 3 Feet 6 Feet - 5 Yards and a half 40 Poles 8 Furlongs 3 Miles 69 iV Miles nearly 1 Inch - 1 Foot - 1 Yard - 1 Fathom 1 Pole or Rod 1 Furlong 1 Mile 1 League 1 Degree - In Ft 12= 1 36= 3 = V98 = 16} = In Ft Yd ' Fth PI Fur Mile Lea Deg or Yd 1 PI 5} = 1 Far 7920 = 660" = 220 = 40= 1 Mile 63360 = 5280 = 1760 = 320 = 8 = 1 CLOTH MEASURE. 2 Inches and a quarter make 4 Nails 3 Quarters 4 Quarters 6 Quarters - 4 Quarters l£ Inch 1 Nail - - Nl 1 Quarter of a Yard Qr 1 Ell Flemish - EF 1 Yard . - Yd 1 Ell English - EE 1 Ell Scotch - E S ounce, and 12 ounces one pound. But in later times it was thought sufficient to divide the same pennyweight into 24 equal parts* still, called grains, being the least weight now in common use ; and from thence the rest are computed, as in the Tables above. TABLES OP HEASURE8. 27 SQUARE MEA8URB. 144 Square Inches make 1 Sq Foot - Ft 9 Square Feet . 1 Sq Yard - Yd 30$ Square Yards - 1 Sq Pole - Pole 40 Square Poles - 1 Rood - Rd 4 Roods - - 1 Acre - Act Sq Inc SqFt 144 = 1 Sq Yd 1296= 9 = 1 SqPl 39204 = 272± = 30}= 1 Rd 1568160 = 10890 = 1210 = 40 = 1 Acr 6272640 = 43560 = 4840 = 160 = 4 = 1 By this measure, Land, and Husbandmen and Gardeners' work are measured ; also Artificers' work, such as Board, Glass, Pavements, Plastering, Wainscoting, Tiling, Floor- ing, and every dimension of length and breadth only. When three dimensions are concerned, namely, length, breadth, and depth or thickness, it is called cubic or solid measure, which is used to measure Timber, Stone, dec. The cubic or solid Foot, which is 12 inches in length and breadth and thickness, contains 1728 cubic or solid inches, and 27 solid feet make one solid yard. DRY, OR CORN MEASURE* 2 Pints make 1 Quart - - Qt 2 Quarts - 1 Pottle - - Pot 2 Pottles - 1 Gallon - - Gal 2 Gallons - 1 Peck - - Pec 4 Pecks - 1 Bushel - - Bu 8 Bushels 1 Quarter - - Qr 5 Quarters 1 Wey, Load, or Ton Wey 2 Weys - 1 Last - - Last Put Gal 8=1 Pec 16= 2= 1 Bu 64= 8= 4= 1 Qr 512= 64 = 32 = 8= 1 Wey 2560 = 320 = 160 = 40= 5=1 Last 5120 = 640=320 = 80 = 10 = 2=1 28 ▲uiTMjcmc* By this are measured all dry wares, as, Corn, Seeds, Roots, Fruits, Salt, Coals, Sand, Oysters, <Scc. The standard Gallon dry-measure contained 268$ cubic or solid inches, and the corn or Winchester bushel 2150} cubic inches ; for the dimensions of the Winchester bushel, by the old Statute, - were 8 inches deep, and 18£ inches wide or in diameter. But the Coal bushel was to be 19£ inches in dia- meter ; and 36 bushels, heaped up, made a London chaldron of coals, the weight of which was 3136 lb Avoirdupois, or 1 ton 8 cwt nearly. See, however, page 29. ALE AND BEER MEASURE. 2 Pints make 1 Quart Qt 4 Quarts 1 Gallon . Gal 36 Gallons - 1 Barrel Bar 1 Barrel and a half 1 Hogshead - Hhd 2 Barrels - 1 Puncheon ~ Pun 2 Hogsheads 1 Butt Butt 2 Butts 1 Tun Tun Pfc Qt 2= 1 Gal 8 = 4 = 1 Bar 288 = 144 = 36 = 1 Hhd 432 = 216 = 54 = 1}= 1 Butt 864= 432 = 108 = 3=2=1 Nate. The Ale Gallon contained 282 cubic or solid inches, by which also milk was measured. WISE MEASURE. 2 Pints make 1 Quart Qt 4 Quarts 1 Gallon Gal 42 Gallons, . 1 Tierce Tier 63 Gallons or 1£ Tierces 1 Hogshead • Hhd 2 Tierces - 1 Puncheon - Pun 2 Hogsheads 1 Pipe or Butt Pi 2 Pipes or 4 Hhds 1 Tun Tun Pts Qt 2 = 1 Gal 8= 4= 1 Tier 336 = 168 = 42 = 1 Hhd 504 = 252 = 63 = H = 1 Pun 672 = 336 = 84 = 2 = 1*= 1 Pi 1008 = 504 = 126 = 3 -4-3*- 1 Tun 2016 = 1008 = 252 = 6 2 = 1 TABLES OF MEASURES AITD TIME. 29 Note, by this are measured all Wines, Spirits, Strong- waters, Cyder, Mead, Perry, Vinegar, Oil, Honey, &c. The old Wine Gallon contained 231 cubic or solid inches. And it is remarkable that these Wine and Ale Gallons have the same proportion to each other, as the Troy and Avoir- dupois Pounds have ; that is, as one Pound Troy is to one Pound Avoirdupois, so is one Wine Gallon to one Ale Gallon. OF 60 Seconds or 60" make - 1 Minute - ilfor' 60 Minutes - 1 Hour - Hr 24 Hours • 1 Day Day 7 Days . lWeek - »* 4 Weeks . 1 Month - Mo 13 Months 1 Day 6 Hours, ) or 365 Days 6 Hours $ 1 Julian Year Yr Sec 60 = 3600 = 86400 = Min 1 60 *= 1440 = Hr 1 Day 24 = 1 Wk 604800 = 10080 = 168 = 7 = 1 Mo 2419200 = 40320 = 672 = 28 = 4 = 1 31557600 == 525960 = 8766 = 365J = 1 Year Wk Da Hr Mb Or 52 1 6 = 13 Da Hr M Sec But 365 5 48 - 45£ = Solar Year Da Hr 1 6=1 Julian Year IMPERIAL MEASURES. By the late Act of Parliament for Uniformity of Weights and Measures, which commenced its operation on the 1st of January, 1826, the chief part of the weightsjand measures are allowed to remain as they were ; the Act simply pre- scribing scientific modes of determining them, in case they should be lost. The pound troy contains 5760 grains. The pound avoirdupois contains 7000 grains. The imperial gallon contains 277*274 cubic inches. The corn bushel, eight times the above. 30 ARITHMETIC. Henco, with respect to Ale, Wine, and Corn, it will be expedient to possess a TABLE OF FACTORS, For converting old measures into new, and the contrary. By decimals. By vulgar frac- tions nearly. Corn Measure. Wine Measure. Ale Measure. Corn Mea- sure. Wine Mea- sure. Ale Mm. sure. To convert old j measures to neW. 1 •96943 •83311 1-01704 H H To convert new ) f ,*« . . measures to old. \ 1H)3IM 1-20032 •98324 ! H * N. B. For reducing the prices, these numbers must all be reversed. RULES FOR REDUCTION. I. When the Numbers are to be reduced from a Higher Denomination to a Lower : Multiply the number in the highest denomination by as many of the next lower as make an integer, or 1, in that higher ; to this product add the number, if any, which was in this lower denomination before, and set down the amount. Reduce this amount in like manner, by multiplying it by as many of the next lower as make an integer of this, taking in the odd parts of this lower, as before. And so proceed through all the denominations to the lowest ; so shall the number last found be the value of all the numbers which were in the higher denominations, taken together*. * The reason of this rule is very evident; for pounds are brought into shillings by multiplying them by 20 ; shillings into pence, by mul- tiplying them by 12 ; and pence into farthings, by multiplying* by 4; and the reverse of this rule by division.— And the same, it is evident, will be true in the reduction of numbers consisting of any denomina- tions whatever. HOLES FOR SEDUCTION. 31 EXAMPLE. 1. Ill 12342 15* 7d, how many farthings ? I 8 d 1234 15 7 24095 Shillings 12 206347 Pence 4 Answer 1 1 85388 Farthings. II. When the Numbers are to be reduced from a Lower De- nomination to a Higher : Divide the given number by as many of that denomina- tion as make 1 of the next higher, and set down what re- mains, as well as the quotient. Divide the quotient by as many of this denomination as make 1 of the next higher ; setting down the now quotient, and remainder, as before. Proceed in the same manner through all the denomina- tions to the highest ; and the quotient last found, together with the several remainders, if any, will be of the same value as the first number proposed. EXAMPLES. 2. Reduce 1185388 farthings into pounds, shillings, and pence. 4) 1185388 12) 2963474 2,0) 2469,5*— Id Answer 1234Z 15* Id 3. Reduce 24Z to farthings. Ans. 23040. 4. Reduce 337587 farthings to pounds, &c. Ans. 3511 \3s 83 .ARITHMETIC* 5. How many farthings are in 36 guineas ? Ans. 36288. 6. In 36288 farthings how many guineas ? Ans. 36. 7. In 69 lb 13 dwts 5 gr. how many grains ? Ans. 340157. 8. In 8012131 grains how many pounds, d&c. Ans. 1390 lb 11 oz 18 dwfc 19 gr. 9. In 35 ton 17 cwt 1 qr 23 lb 7 oz 13 dr how many drams ? Ans. 20571005. 10. How many barley-corns will reach round the earth, supposing it, according to the best calculations, to be 25000 miles? Ans. 4752000000. 11. How many seconds are in a solar year, or 365 days 5 hrs 48 min. 45} sec ? Ans. 31556925}. 12. In a lunar month, or 29 ds 12 hrs 44 min 3 sec, how many seconds? Ans. 2551443. COMPOUND ADDITION. Compound Addition shows how to add or collect several numbers of different denominations into one sum. Rule. — Place the numbers so, that those of the same de- nomination may stand directly under each other, and draw a line below them. Add up the figures in the lowest deno- mination, and find, by Reduction, how many units, or ones, of the next higher denomination are contained in their sum. —Set down the remainder below its proper column, and carry those units or ones to the next denomination, which add up in the same manner as before. — Proceed thus through all the denominations, to the highest, whose sum, together with the several remainders, will give the answer sought. The method of proof is the same as in Simple Addition. COMPOUND ADDITION. SI EXAMPLES OF MONEY*. 1. 2. 3. 4. I s d / 1 <* / 8 d Z d 7 13 3 14 7 5 15 17 10 53 14 8 3 5 lO-i 8 19 2f 3 14 6 5 10 Sf 6 18 7 7 8 1} 23 6 *1 93 11 6 2 5j 21 2 9 14 9 *i 7 5 4 3 7 16 8J- 15 6 4 13 2 5 17 15 *i 4 3 6 12 18 7 99 15 9 ? 32 2 6j 39 15 9J 5. 6. I s d lad 14 7J 37 15 8 8 15 3 14 12 9; 62 4 7 17 14 9 4 17 8 23 10 9J 23 4f 8 6 6 6 7 14 5J 91 10{ 54 2 l\ 7. 8. 1 8 d / 8 d 61 3 2£ 472 15 3 7 16 8 9 2 29 13 lOf 27 12 si 12 16 2 370 16 2 1 7 5J 13 7 4 24 13 6 10 5f 5 10? 30 11* Exam. 9. A nobleman, going out of town, is informed by his steward, that his butcher's bill comes to 197/ 13* 7\d ; his baker's to 592 5* 2Jd ; his brewer's to 85/ ; his wine* merchant's to 103/ 13* ; to his corn chandler is due 75/ 3d ; to his tallow-chandler and cheesemonger, 27/ 15* 11{</; and to his tailor 55/ 3* 5f d ; also for rent, servants' wages, and other charges, 127/ 3* : Now, supposing he would take 100/ with him, to defray his charges on the road, for what sum must he send to -bis bankor ? Ans. 8301 14s $Vd Vol. I. 6 34 ARITHMETIC. 10. The strength of a regiment of foot, of 10 companies, and the amount of their subsistence*, for a month of SO days, according to the annexed Table, are required ? Numb. Rank. Subsistence for a Month. • 1 Colonel £27 1 Lieutenant Colonel 19 10 1 Major 17 5 7 Captains 78 15 11 Lieutenants 57 15 9 Ensigns 40 10 1 Chaplain 7 10 1 Adjutant 4 10 1 Quarter-Master 5 5 1 Surgeon 4 10 1 Surgeon's Mate 4 10 30 Serjeants 45 30 Corporals 30 20 Drummers 20 2 Fifers 2 390 Private Men 292 10 507 Total. £656 10 * Subsistence Money, is the money paid to the soldiers weekly ; which is short of their fall pay, because their clothes, accoutrements, Ac. are to be accounted for. It is likewise the money advanced to officers till their accounts are made up, which is commonly once a year, when they are paid their arrears. The following table shows the foil pay and subsistence of each rank on the English establishment. COMPOUND ADDITION. 35 1*4 a in 7 4 « — « _ o ~ no< d _ t t- ' | o as " o '<= o * - » if 3 i £3 CS=J CI Q w 2 . , O Q . , , 1W - Q 3 O O . .QQ - . . 5*^> - - - -» cor-' + vo | 2* * n cc*fl ,«t^ . , — l- *"- QQ ' 5 '3000 , ( QO « o o a v o ^ a, 6 -a »~ — ^ £ S£ ** *o *^ » <c . . — * r* O w i fi ^ g o " : " ii ■ ^ , Ira Jtuoq fg "TV r»™ <wj w» n .irfj Pin .>i!.|..„[ *a Pi Jiff 4#J .uinj.ttdE -jy ptrm liv j wajQ U mi Lii.ifi r ail ,rjtu)*Min 3IJ] Mtl JUJCM] STT -(V pin Vq,[ twijij 2 * *S ■ ■ & « c ^ r«S * « « 7J . <£> . T . O i c .3. K 2 - "2 • : ' a ■ r 9 - O O . O O 3* 1 f s 1 1 ■i|f! ■ c 2U - c - £ = in ill ill if 5 HI ■ Mi * 111 5 I : 1 t5S c - — *^ _ — | • I. 1 ? ti: * r- 51 36 ARITHMETIC. EXAMPLES OF WEIGHTS. MEASURES, <f . TROT WEIGHT. APOTHEC ARIES' WEIGHT. 1. 2. 3. 4. lb oz dwt oz dwt gr lb oz dr sc oz dr sc 17 3 15 37 9 3 3 5 7 2 3 5 1 17 7 9 4 9 5 3 13 7 3 7 3 2 5 10 7 8 12 12 19 10 6 2 16 7 12 9 5 17 7 8 9 1 2 7 3 2 9 176 2 17 5 9 36 3 5 4 1 2 18 23 11 12 3 19 5 8 6 1 36 4 1 14 AVOIRDUPOIS WEIGHT. LONG MEASURE. 5. 6. 7. 8. lb oz dr cwt qr lb mis fur pis yds feet inc 17 10 13 15 2 15 29 3 14 127 1 5 5 14 8 6 3 2 4 19 6 29 12 2 9 12 9 18 9 1 14 7 24 10 10 27 1 6 9 1 17 9 1 37 54 1 11 040 10 2 6 703 527 6 14 10 3 3 4 5 9 23 5 CTOTH MEASURE. LAND MEASURE. 9. 10 11. 12. yds qr nls el en qrs nls ac ro |> ac ro p 26 3 1 270 1 225 3 37 19 16 13 1 2 57 4 3 16 1 25 270 3 29 9 1 2 18 1 2 7 2 18 6 3 13 217 3 3 2 4 2 9 23 34 9 1 10 1 42 1 19 7 2 16 55 3 1 4 4 1 7 6 75 23 WINE .MEASURE. ALF. AND BEER MEASURE. 13. 14. 15. 16. t hds gal hds gal pts hds gal pts hds gal pts 13 3 15 15 61 5 17 37 3 29 43 5 8 1 37 17 14 13 9 10 15 12 19 7 14 1 20 29 23 7 3 6 2 14 16 6 25 12 3 15 1 5 14 6 8 1 3 1 9 16 8 12 .9 6 57 13 4 72 3 21 4 96 (5 8 42 4 5 6 COMPOUND SUBTRACTION. 37 COMPOUND SUBTRACTION. Compound Subtraction shows how to find the difference between any two numbers of different denominations. To perform which, observe the following Rule. * Place the less number below the greater, so that the parts of the same denomination may stand directly under each other ; and draw a line below them. — Begin at the right-hand, and subtract each number or part in the lower line, from the one just above it, and set the remainder straight below it. — But if any number in the lower line.be greater than that above it, add as many to the upper number as make 1 of the next higher denomination ; then take the lower number from the upper one thus increased, and set down the remainder. Carry the unit borrowed to the next number in the lower line ; after which subtract this number from the one above it, as before ; and so proceed till the whole is finished. Then the several remainders, taken to- gether, will be the whole difference sought. The method of proof is the same as in Simple Subtraction. EXAMPLES OP MONKF. 1. 2. 3. 4. 1 s d 1 * d 1 s d 1 s d From 79 17 8} 103 3 2£ 81 10 11 254 12 Take 35 12 4' 71 12 5j 29 13 3| 37 9 4f. Rem. 44 5 4£ 31 10 83 Proof 79 17 8J- 103 3 2£ 5. What is the difference between 73/ 5irf and 19/ 13s 10J ? Ans. 53Zt>*7irf. * The reason of this Rule will easily appear from what has been said in Simple Subtraction ; for the borrowing depends on the same princi- ple, and is only different as the numbers to be subtracted are of differ- ent denominations. 38 ARITHMETIC Ex. 6. a lends to b 100/, how much is b in debt after a has taken goods of him to the amount of 73/ 12* 4|c/ ? Ans. 26/ 7s 7J<*. 7. Suppose that my rent for half a year is 20/ 12s, and that I have laid out for the land-tax 14s 6</, and for several repairs 1/3* 3*d, what have I to pay of ray half- year's rent? Ans. \81Us2$d. 8. A trader, failing, owes to a 35/ 7s 6d, to b 91/ 13s \d, to c 53/ 7» cZ, to d 87/ 5*, and to e 111/ 2s 5$d. When this happened, he had by him in cash 23/ 7s bd, in wares 53/11* 10{d f in household furniture 63/ 17* 7£</, and in recoverable book-debts 25/ 7s 5d. What will his creditors lose by him, supposing these things delivered to them ? Ans. 212/ 5* 3}<*. EXAMPLES OF WEIGHTS, MEASURES, 6fC. TROT WEIGHT. APOTHECARIES 1 WEIGHT. 1. 2. 3. lb oz dwt gr lb oz dwt gr lb oz dr scr gr From 9 2 12 10 7 10 4 17 73 4 7 14 Take 5 4 6 17 3 7 16 12 29 5 3 4 19 Rem. Proof AVOIRDUPOIS WEIGHT. LONG MEASURE. 4. 5. 6. 7. c qrs lb lb oz dr m fu pi vd ft in From 5 17 71 5 9 14 3 17 96 4 Take 2 3 10 17 9 18 7 6 11 72 2 9 Rem. Proof CLOTH MEASURE. LAND MEASURE. 8. ». 10. 11. yd qr nl vd qr nl ac ro p ac ro p From 17 2 1 9 2 17 1 14 57 1 16 Take 9 2 7 2 1 16 2 8 22 3 29 Rem. Proof COMPOUND MULTIPLICATION. 39 WI5* MEASURE. ALE AND BEER MEASURE. 12. 13. 14. 15. t hd gal hd gal pt hd gal pt hd gal pt From 17 2 23 5 4 14 29 3 71 16 5 Take 9 1 36 2 12 6 9 35 7 19 7 I Rem. Proof DRY MEASURE. TIME.. 16. 17. 18. 19. la qr bu bu gal pt mo we da ds hrs min From 9 4 7 13 7 1 71 2 5 114 17 26 Take 6 3 5 9 2 7 17 1 6 72 10 37 Rem. Proof 20. The line of defence in a certain polygon being 236 yards, and that part of it which i9 terminated by the curtain and shoulder being 146 yards 1 foot 4 inches ; what then was the length of the face of the bastion ? Ans. 89 yds 1 ft 8 in. COMPOUND MULTIPLICATION. Compound Multiplication shows how to find the amount of any given number of different denominations repeated a certain proposed number of times ; which is performed by the following rule. Set the multiplier under the lowest denomination of the multiplicand, and draw a line below it. — Multiply the num- ber in the lowest denomination by the multiplier, and find how many units of the next higher denomination arc con- tained in the product, setting down what remains. — In like manner, multiply the number in the next denomination, and to the product carry or add the units, before found, and dud how many units of the next higher denomination arc m VVvvs 40 ARITHMETIC amount, which carry in like manner to the next product, setting down the overplus. — Proceed thus to the highest de- nomination proposed : so shall the last product, with the se- veral remainders, taken as one compound number, be the whole amount required. — The method of Proof, and the rea- son of the Rule, are the same as in Simple Multiplication. EXAMPLES OF MONF.Y. 1. To find the amount of 8 lb of Tea, at 5*. 8hd. per lb. s d 5 Si £2 5 8 Answer. / -v d 2. 4 lb of Tea, at 7s 8d per lb. Ans. 1 10 8 3. 6 lb of Butter, at $\d per lb. Ans. 4 9 4. 7 lb of Tobacco, at 1* 8$d per lb. Ans. 11 1H 5. 8 stone of Beef, at 2s l\d per st. Ans. 110 6. 10 cwt cheese, at 2Z 17* lOcZ per cwt. Ans. 28 18 4 7. 12 cwt of Sugar, at 3Z Is 4d per cwt. Ans. 40 8 CONTRACTIONS. I. If the multiplier exceed 12, multiply successively by its component parts, instead of the whole number at once. EXAMl'l.KS. 1. 15 cwt of Cheese, at 17* tod per cwt. / * d 17 6 3 2 12 5 13 t 6 Answer. I s d 2. 20 cwt of Hops, at 41 7* 2d per cwt. Ans. 87 3 4 3. 24 tons of Hay, at 3J 7* 672 per ton. Ans. 81 4. 46 eUs of Cloth, at 1* M per ell. Ans. 3 7 6 COMPOUND HULTOUCATION. 41 I sd Ex. 5. 63 gallons of Oil, at 2s 3d per gall. Ana. 7 19 & 70 barrels of Ale, at 11 4* per barrel. Ans. 84 7. 84 quarters of Oats, at 1/ 12* Sd per qr. Ans. 137 4 8. 96quartersofBarley,atH3*4<2perqr. Ans.112 9. 120 days' Wages, at 5s 9d per day. Ans. 34* 10 10. 144 reams of Paper, at 13s 4d per ream. Ans. 96 11. If the multiplier cannot be exactly produced by the multiplication of simple numbers, take the nearest number to it, either greater or less, which can be so produced, and multiply by its parts, as before. — Then multiply the given multiplicand by the difference between this assumed number and the multiplier, and add the product to that before found, when the assumed number is less than the multiplier, but subtract the same when it is greater. EXAMPLES. 1. 26 yards of Cloth, at 3* Of cZ per yard. I s d 3 0; 5 15 3J 5 3 16 Of 3 0| add £3 19 7* Answer. EXAMPLES OF WEIGHTS AND MEASURES. 2. 29 quarters of Corn, at 21 5s 3{d per qr. Ans. 65 12 lOj 3. 53 loads of Hay, at 3Z 15* 2d per Id. Ans. 199 3 10 4. 79 bushels of Wheat, at 11* 5fd per bush. Ans. 45 6 10* 5. 97 casks of Beer, at 12* 2d per cask. Ans. 59 2 C. 114 stone of Meat, at 15* 3jrf per st. Ans. 87 5 7} 1. 2. 3. lb oz dwt gr lb oz dr sc gr cwt qr lb oz 28 7 14 10 2 6 3 2 10 29 2 16 14 5 8 12 Vol. L 7 48 rote fit . pis yds 22 5 20 .6 4 5. yds qra na 126 3 1 7 6. EC 10 pO 28 8 27 9 7, . 8. 9. luns hhd gal pts we qr bu pe mo we da ho mm 20 2 26 2 24 2 5 3 172 3 6 16 49 3 6 10 COMPOUND DIVISION. Compound Division teaches how to divide a number of several denominations by any given number, or into any number of equal parts ; as follows : Place the divisor on the left of the dividend, as in Simple Divis ion. — Begin at the left-hand, and divide the number of jflfehighest denomination by the divisor, setting down the ^Hrent in its proper place. — If there be any remainder after this division, reduce it to the next lower denomination, which add to the number, if any, belonging to that denomination, and* divide the sum by the divisor. — Set down again this quo- tient, reduce its remainder to the next lower denomination again, and so on through all the denominations to the last. EXAMPLE* OF MONEY. Divide 237Z 8* 6d by 2. I 8 d 2) 237 8 6 £118 14 3 the Quotient. conform Division. I 9 d I 9 d 2. Divide 482 12 1} by 3. Ans. 144 4 Oj 3. Divide 507 3 5 by 4. Ana. 126 15 10' 4. Divide 032 7 0} by 5. Ans. 126 9 6 5. Divide 090 14 3fby6. Ana. 115 2 4* 6. Divide 705 10 2 by 7. Ans. 100 15 8} 7. Divide 760 5 6 by a Ans. 95 s\ 8. Divide 761 5 7} by 9. Ans. 84 11 8} 9. Divide 829 17 10 by 10. Ans. 82 19 9£ 10. Divide 937 8 8} by 11. Ans. 85 4 5 11. Divide 1145 11 4} by 12. Ans. 95 9 3J COATKACTlOlfS. i I. If the divisor eieeed 12, find what simple numbers, multiplied together, will produce it, and divide by them separately, as in Simple Division, as below. EXAXFLES. 1. What is Cheese per cwt, if 16 cwt cost 252 14i Sdl I 9 d 4) 25 14 8 4) 6 8 8 £ 1 12 2 the Answer. I $ d 2. If 20 cwt of Tobacco come to ) k 7 in ± 1501 6s 8rf, what is that per cwt ? \ 3. Divide 982 8s by 36. Ans. 2 14 8 4. Divide 712 13s \0d by 56. Ans. 1 5 1{ 5. Divide 442 4s by 96. Ans. 9 2} 6. At 312 10s per cwt, how much per lb ? Ans. 5 7} II. If the divisor cannot be produced by the multiplica- tion of small numbers, divide by the whole divisor at once, after the mariner oT Long division, as follows. 44 ARITHMETIC* EXAMPLES. 1. Divide 59Z 6* 3fd by 19. I s d ltd 19) 59 6 3} '(3 2 5j Ana. 57 09 (5 95 ' ~4 4 19 (i I s d I s d 2. Divide 89 14 5} by 57. Ans. 13 llj 3. Divide 125 4 9 by 43. Ans. 2 18 3 4. Divide 542 7 10 by 97. Ans. 5 11 10 5. Divide 123 11 2 J by 127. Ans. 19 5* EXAMPLES OF WEIGHTS AND XEASUBS8. 1. Divide 17 lb 9 oz dwts 2 gr by 7. Ans. 2 lb 6 oz 8 dwts 14 gr. 2. Divide 17 lb 5 oz 2 dr 1 scr 4 gr by 12. Ans. 1 lb 5 oz 3 dr 1 scr 12 gr. 3. Divide 178 cwt 3 qrs 14 lb by 53. Ans. 3 cwt 1 qr 14 lb. 4. Divide 144 mi 4 fur 20po 1 yd 2ft in by 39. Ans. 3 mi 5 fur 26 po yds 2 ft 8 in. 5. Divide 534 yds 2 qrs 2 na by 47. Ans. 11 yds 1 qr 2 na. 6. Divide 77 ac 1 ro 33 po by 51. Ans. 1 ac 2 ro 3 po. 7. Divide 2 tu hhds 47 gal 7 pi by 65. Ans. 27 gal 7 pi. 8. Divide 387 la 9 qr by 72. Ans. 5 la 3 qrs 7 bu. 9. Divide 206 mo 4 da by 26. Ans. 7 mo 3 we 5 ds. BULK OF THREE. 45 THE GOLDEN RULE, OR RULE OF THREE. The Rule of Thus- teaches how to* find a fourth propor- tional to three numbers given : for which reason it is some, times called the Rule of Proportion. It is called the Rule of Three, because three terms or numbers are given, to find a fourth. And because of its great and extensive usefulness, it is often called the Golden Rule. This Rule is usually by practical men considered as of two kinds, namely, Direct and Inverse. The distinction, however, as well as the man- ner of stating, though retained here for practical purposes, does not well accord with the principles of proportion ; as will be shown farther on. The Rule of Three Direct is that in which more requires more, or less requires less. As in this ; if three men dig 21 yards of trench in a certain time, how much will six men dig in the same time ? Here more requires more, that is, 6 men, which are more than three men, will also perform more work, in the same time. Or when it is thus : if 6 men dig 42 yards, how much will 3 men dig in the same time ? Here then, less requires less, or 3 men will perform proportionately less work than 6 men, in the same time. In both these cases then, the Rule, or the Proportion, is Direct ; and the stating must be thus, as 3 : 21 : : 6 : 42, or as 3 : 6 : : 21 : 42. And, as 6 : 42 : : 3 : 21, or as 6 : 3 : : 42 : 21. But the Rule of Three Inverse, is when more requires less, or less requires more. As in this : if 3 men dig a certain quantity of trench in 14 hours, in how many hours will 6 men dig the like quantity ? Here it is evident that 6 men, being more than 3, will perform an equal quantity of work in less time, or fewer hours. Or thus : if 6 men perform a certain quantity of work in 7 hours, in how many hours will 3 men perform the same ? Here less requires more, for 3 men will take more hours than 6 to perform the same work. In both these cases then the Rule, or the Proportion, is In- verse ; and the stating must be thus, as 6 : 14 : : 3 : 7, or as 6 : 3 : : 14 : 7. And, as 3 : 7 : : 6 : 14, or as 3 : 6 : : 7 : 14. And in all these statings, the fourth term is found, by mul- tiplying the 2d and 3d terms together, and dividing the pro* duct by the 1st term. Of the three given numbers : two of them contain the sup- position, and the third a demand. And for stating and work- ing questions of these kinds, observe the following general Rule: 46 ARITHMETIC. State the question by setting down in a straight line the three given numbers, in the following manner, viz. so that the 2nd term be that number of supposition which is of the same kind that the answer or 4th term is to be ; majtiog the other number of supposition the 1st term, and the demanding number the 3d term, when the question is in direct propor- tion ; but contrariwise, the other number of supposition the 3d term, and the demanding number the 1st term, when the question has inverse proportion. Thon, in both cases, multiply the 2d and 3d terms together, and divide the product by the 1st, which will give the answer, or 4th term sought, viz. of the same denomination as the second term. Note, If the first and third terms consist of different deno- minations, reduce them both to the same : and if the second term be a compound number, it is mostly convenient to re* duce it to the lowest denomination mentioned. — If, after division, there be any remainder, reduce it to the next lower denomination, and divide by the same divisor as before, and the quotient will be of this last denomination. Proceed in the same manner with all the remainders, till they be. reduc- ed to the lowest denomination which the second admits of, and the several quotients taken together will be the answer required. Note also, The reason for the foregoing Rules will appear, when we come to treat of the nature of Proportions. — Some- times two or more statings are necessary, which may always be known from the nature of the question. EXAMPLES. 1. If 8 yards of Cloth cost 11 4*, what will 96 yards cost? yds 1 s yds 1 s As 8 : 1 4 : : 96 : 14 8 the Answer. 20 24 96 144 216 8)2304 2,0) 28,8* £14 8 Answer. BULB OF THREE. 47 Ex. 2. An engineer having raised 100 yards of a certain work in 34 days with 5 men ; how many men must he em* ploy to finish a like quantity of work in 15 days ? ds men ds men As 15 : 5 : : 24 : 8 Ans. 5 15) 120 (8 Answer. 120 3. What will 72 yards of cloth cost, at the rate of 9 yards for 52 12* ? Ans. 44/ 16*. 4. A person's annual income being 146/ ; how much is that per day ? Ans. 8*. 5. If 3 paces or common steps of a certain person be equal to 2 yards, how many yards will 160 of his paces make ? Ans. 106 yds 2 ft. 6. What length must be cut off a board, that is 9 inches broad, to make a square foot, or as much as 12 inches in length and 12 in breadth contains ? Ans. 16 inches. 7. If 750 men require 22500 rations of bread for a month ; how many rations will a garrison of 1000 men re. quire ? Ans. 36000. 8. If 7 cwt 1 qr. of sugar cost 26/ 10* 4d ; what will be the price of 43 cwt 2 qrs ? Ans. 159/ 2*. 9. The clothing of a regiment of foot of 750 men amount- ing to 28312 5* ; what will the clothing of a body of 3500 men amount to ? Ans. 13212/ 10*. 10. How many yards of matting, that is 3 ft broad, will cover a floor that is 27 feet long and 20 feet broad ? Ans. 60 yards. 11. What is the value of six bushels of coals, at the rate of 1/ 14*. 6d the chaldron ? Ans. 5* \)d. 12. If 6352 stones of 3 feet long complete a certain quan- tity of walling ; how many stones of 2 feet long will raise a like quantity ? Ans. 9528. 13. What must be given for a piece of silver weighing 73 lb 5 oz 15 dwts, at the rate of 5* 9d per ounce ? Ans. 253/ 10* 0f<Z. 14. A garrison of 536 men having provision for 12 months ; how long will those provisions last, if the garrison be increased to 1124 men ? Ans. 174 days and T f | T . 15. What will be the tax upon 763/ 15* at the rate of 3t Qd per pound sterling ? Ans. 1331 \3s \\d. 46 ARITHMETIC. 16. A certain work being raised in 12 days, by working 4 hours each day ; how long would it nave been in raising by working 6 hours per day ? Ans. 8 days. 17. What quantity of corn can I buy for 90 guineas, at the rate of 6* the bushel ? Ans. 39 qra 3 bu. 18. A person, failing in trade, owes in all 9772 ; at which time he has, in money, goods, and recoverable debts, 420Z 6* 3JcZ ; now supposing these things delivered to his creditors, - how much will they get per pound? Ans. 8* 7\d. 19. A plain of a certain extent having supplied a body of 3000 horse with forage for 18 days ; then how many days would the same plain have supplied a body of 2000 horse ? Ans. 27 days. 20. Suppose a gentleman's income is 600 guineas a year, and that he spends 25* 6d per day, one day with another ; how much will he have saved at the year's end ? Ans. 164/ 12* 6c*. 21. What cost 30 pieces of lead, each weighing 1 cwt 121b. at the rate of 16* 4d the cwt ? Ans. 27Z 2s 6d, 22. The governor of a besieged place having provision for 54 days, at the rate of l£lb of bread ; but being desirous to prolong the siege to 80 days, in expectation of succour, in that case what must the ration of bread be ? Ans. 1 ^ ¥ lb. 23. At half-a-guinea per week, how long can I be boarded for 20 pounds ? Ans. 38 T <& wks. 24. How much will 75 chaldrons 7 bushels of coals come to, at the rate of 11 13* 6d per chaldron ? Ans. 125Z 19* 0j<*. 25. If the penny loaf weigh 8 ounces when the bushel of wheat costs 7* 3d, what ought the penny loaf to weigh when the wheat is at 8* 4d ? Ans. 6 oz 15 ffo dr. 26. How much a year will 173 acres 2 roods 14 poles of land give, at the rate of 11 Is 8d per acre ? Ans. 240Z 2* 7&d. 27 To how much amounts 73 pieces of lead, each weigh- ing 1 cwt 3 qrs 7 lb, at 10Z 4* per fother of 19| cwt ? Ans. 69Z4*2d l^fq. 28. How many yards of stuff, of 3 qrs wide, will line a cloak that is 1 J yards in length and 3J yards wide ? Ans. 8 yds Oqrs 2$ nl. 29. If 5 yards of cloth cost 14* 2d, what must be given for 9 pieces, containing each 21 yards 1 quarter ? Ans. 27Z 1* 30. If a gentleman's estate be worth 2107Z 12* a year ; what may he spend per day, to save 500Z in the year ? Ans. 4Z 8s ljftd. RULE OF TIIHKE. 49 31. Wanting just an acre of land cut off from a piece which is 131 poles in breadth, what length must the picco be ? Ans. 11 po 4 yds 2 ft Otf in. 32. At 7s 9' d per yard, what is the value of a piece of cloth containing 53 ells English 1 qr ? Ans. 25/ ISs 1 J<J. 33. If the carriage of 5 cwt 11 lb for 90 miles he 1/ 12s (yd; how fur may I have 3 cwt 1 qr carried for the same money? Ans. 151 m 3 fur 3,-^ pol. 3-1. Bought a silver tankard, weighing 1 lb 7 oz 14 dwts ; what did it cost me at Hs 4d the ounce ? Ans. (U 4s 9jd. 35. What is tho half 3 T ear's rent jf 547 acres of land, at 15s Oct the acre ? Ans. 211/ 19* 3d. 36. A wall that is to be built to the height of 30 feet, was raised feet high by 10 men in days ; then how many men must be employed to finish the wall in 4 days, at the same rate of working / Ans. 72 men. 37. What will be the charge of keeping 20 horses for a year, at the rate of 14 id pur dav for each horse? Ans. 441/ 0* lOd. 38. If 18 ells of stuff that is J yard wide, cost 39s M ; what will 50 oils, of the same goodness, cost, being yard wide? Ans. 11 i\s 3J{<#. 39. How many yards of paper that is 30 inches wide, will hang a room that is 20 yards in circuit and 9 feet high ? Ans. 72 yards. 40. If a gentleman's estate be worth 384/ 1 0* a year, and the Jand-ta.x be assessed at 2s \)} A d per pound, what is his net annual income ? Ans. 331/ Is 9£<Z. 41. The circumference of the earth is about 25000 miles ; at what rate per hour is a person at the middle of its surface carried round, one whole rotation being made in 23 hours 56 minutes ? Ans. 1044 T s Afe miles. 42. If a person drink 20 bottles of wine per month, when it costs 8s. a gall ; how many bottles per month may ho drink, without increasing the expense, when wine costs 10s he gallon ? Ans. 16 bottles. 43. What cost 43 nrs 5 bushels of corn, at 1/ 8s 6*2 the quarter ? * Ans. 62/ 3s 3J<Z. 44. How T many yards of canvas that is ell wide will line 50 yards of say that is 3 quartern wide ? Ans. 30 yds. 45. If an ounce of gold cost 4 guineas, what is the value of a grain ? Ans. 2 T Vd. 40. If 3 cwt of tea cost 10/ 12s ; at how much a pound must it L9 rolailH, to gain 10/ by the whole \ Ans. 3^$. Vol. I. 50 COMPOUND PROPORTION. Compound Pbofostion is a rule by means of which the student may resolve such questions as require two or more stsiings in simple proportion. The general rule for questions of this kind may be ex. hibited in the following precepts : viz. 1. Set down the terms that express the conditions of the question in one line. 2. Under each conditional term, set its corresponding one, in another line, putting the letter a in the (otherwise) blank place of the term required. 3. Multiply the producing terms of one line, and the pro* duced terms of the other line, continually, and take the re. suit for a dividend. 4. Multiply the remaining terms continually, and let the product be a divisor. 5. The quotient of this division will be q, the term re. quired.** Note. By producing terms are here meant whatever ne- cessarily and jointly produce any effect ; as the cause and the time ; length, breadth, and depth ; buyer and his mo- ney ; things carried, and their distance, dec. all necessarily inseparable in producing their several effects. In a question where a term is only understood, and not ex- pressed, that term may always be expressed by unity. A quotient is represented by the dividend put above a line, and the divisor put below it. EXAMPLES. 1. How many men can complete a trench of 135 yards long in 8 days, when 16 men can dig 54 yards of the same trench in 6 days ? M D Yds 16 • ... 6 ... • 54 a .... 8 .... 135 * This rale, which is as applicable to Simple as to Compmmd Propor- tion, was given, in 1706, by W. Jones, Esq. F.R.S., the father of the late Sir W. Jones. COMPOUND PROPORTION. 51 Here 16 men and 6 days, are the producing terms of the first line, and 185 yards, the produced term of the other. Therefore, by the rule, 16X6X135 2X135 m UCa -8X54- 9 * the number of men required. ANOTHER question. If a garrison of 3600 men have bread for 35 days, at 94 os each a day : How much a day must be allowed to 4800 men, each for 45 days, that the same quantity of bread may serve? men ok days bread 3600 . . 24 . . 35 . . 1 4800 . . a . . 45 . . 1 3600X24X35 AN EXAJfPLE IN SIMPLE PROPORTION. If 14 yards of cloth cost 21Z, how many yards may be bought for 73/ 10f? man £ yds. 1 .... 21 .... 14 1 .... 73* .... q a = = { of 73* = 49 yards, Answer. «1 2. If 1002 in one year gain 51 interest, what will be the interest of 7501 for seven years ? Ans. 2622 10*. 3. If a family of 8 peraons expend 200/ in 9 months ; how much will serve a family of 18 people 12 months ? Ans. 600/. 4. If 27# be the wages of 4 men for 7 days ; what will be the wages of 14 men for 10 days ? Ans. 6/ 15*. If a footman travel 130 miles in 3 days, when the days are 12 hours long ; in how many days, of 10 hours each, may he travel 360 miles 7 Ans. 9j j days. G. If 120 bushels of corn can serve 14 horses 56 days ; how many days will 94 bushels serve 6 horses ? Ans. VXL\\ day*. 52 ARITHMETIC. 7. If 8000 lbs of beef serve 340 men 15 days ; how many lbs will serve 120 men for 25 days ? Ans. 1764 lb 1 1 ^ oz. 8. If a barrel of beer be sufficient to last a family of 8 persons 12 days ; how many barrels will be drank by 16 persons in the space of a year 7 Ans. 60 J barrels. 9. If 180 men, in six days, of 10 hours each, can dig a trench 200 yards long, 3 wide, and 2 deep ; in how many days of 8 hours long, will 100 men dig a trench of 360 yards long, 4 wide, and 3 deep ? Ans. 48} days. OP VULGAR FRACTIONS. A Fraction, or broken number, is an expression of a part, or some parts, of something considered as a whole. It is denoted by two numbers, placed one below tho other, with a line between them : Thus, JL I j umer * tor \ which is named 3-fourths. 4 denominator > The denominator, or number placed below the line, shows how many equal parts the whole quantity is divided into ; and it represents the Divisor in Division. — And the Nu- merator, or number set above the line, shows how many of these parts are expressed by the Fraction : being the re- mainder after division. — Also, both these numbers are in general named the Terras of the Fraction. Fractions are either Proper, Improper, Simple, Compound, Mixed, or Complex. A Proper Fraction, is when the numerator is less than the denominator ; as, £, or |, or £, &c. An Improper Fraction, is when the numerator is equal to, or exceeds, the denominator ; as, f , or f , or }, &c. In these cases the fraction is called Improper, because it is equal to, or exceeds unity. A Simple Fraction, is a single expression, denoting any number of parts of the integer ; as, f , or |. A Compound Fraction, is the fraction of a fraction, or two or more fractions connected with the word of between them ; as, j of §, or $ of £ of 3, dec. A Mixed Number, is composed of a whole number and a fraction together ; as, 3|, or 12f , dec. A Complex Fraction, is one that has a fraction or a mixed number for its numerator, or its denominator, or both ; i 2 I 3 * jt **' "t"' or 7' 01 4 ' or 4 9 &c ' REDUCTION OF VULGAR FRACTIONS. 53 A whole or integer number may be expressed like a frac- tion, by writing 1 below it, as a denominator ; so 3 is f , or 4 is f , Ac. A fraction denotes division ; and its value is equal to the quotient obtained by dividing the numerator by the deno- minator : so y is equal to 3, and V ' 8 equal to 4}. Hence then, if the numerator be less than the denominator, the value of the fraction is less than 1. But if the numerator be the same as the denominator, the fraction is just equal to 1. And. if the numerator be greater than the denominator, the fraction is greater than 1. REDUCTION OF VULGAR FRACTIONS. Reduction of Vulgar Fractions, is the bringing them out of one form or denomination into another ; commonly to pre- pare them for the operations of Addition, Subtraction, dec. ; of which there are several cases. PROBLEM. To find the Greatest Common Measure of Two or more Numbers. The Common Measure of two or more numbers, is that number which will divide them all without remainder ; so, 3 is a common measure of 18 and 24; the quotient of the former being 6, and of the latter 8. And the greatest num- ber that will do this, is the greatest common measure : so 6 is the greatest common measure of 18 and 24 ; the quotient of the former being 3, and of the latter 4, which will not both divide further. RULE. If there be two numbers only, divide the greater by the less ; then divide the divisor by the remainder ; and so on, dividing always the last divisor by the last remainder, till nothing remains ; so shall the last divisor of all be the great, est common measure sought. When there are more than two numbers, find the greatest common measure of two of them, as before ; then do tho same for that common measure and another of lYifc uronfofct*% 54 ARITHMETIC and so on, through all the numbers ; so will the greatest com- mon measure last found be the answer. If it happen that the common measure thus found is 1 ; then the numbers are said to be incommensurable, or not to have any common measure, or they are said to be prime to each other. examples. 1. To find the greatest common measure of 1908, 086, and 030. 036 ) 1908 ( 2 So that 36 is the greatest common 1872 measure of 1908 and 930. 36 ) 936 ( 26. Hence 36 ) 630 ( 17 7© 36 216 270 216 252 18) 36 (2 36 m Hence 18 is the answer required. 2. What is the greatest common measure of 246 and 372 ? An?. 6. 3. What is the greatest common measure of 324, 612, and 1032? Ans. 12. CASE I. To Abbreviate or Reduce Fractions to their Lowest Terms. * Divide the terms of the given fraction by any number that will divide them without a remainder ; then divide these * That dividing both the terms of the fraction by the same number, whatever it be, will give another fraction equal to the former, is evi- dent. And when these divisions are performed as often as can be done, or when the common divisor is the greatest possible, the terms of the resulting fraction must be the least possible Note. 1. Any number ending with an even number, or a cipher, is divisible, or can be divided, by 2. 2. Any number, ending with 6, or 0, is divisible by 6. 3. If the right-hand place of any number be 0, the whole is divisible by 10 ; if there be two ciphers, it is divisible by 100 ; if three ciphers, by 1000 : and so on ; which if only cutting off those ciphers. REDUCTION OF WLQAJL FRACTIONS* 35 quotients again in the same manner ; and so on, till it appears that there is no number greater than 1 which will divide them ; then the fraction will be in its lowest terms. Or, divide both the terms of the fraction by their greatest common measure at once, and the quotients will be the terms of the fraction required, of the same value as at first. EXAMPLES. 1. Reduce }jf to its least terms. ttt = tt = » = tt = f = h the answer. Or thus : 216) 288 (1 Therefore 72 is the greatest common 216 measure ; and 72) |jj = J the Answer, the same as before. 72) 216 (3 216 2. Reduce t0 iis ^west terms. Ans. J. 3. Reduce J$f to its lowest terms. Ans. §. 4. Reduce £f$ to its lowest terms. Ans. |. 4. If the two right-hand figures of any number be divisible by 4, the whole is divisible by 4. And if ihe three right-hand figures be divisible by 8, the whole is divisible by 8. And so on. 5. If the sum of the digits in any number be divisible by 3, or by 9, the whole is divisible by 3, or by 9. 6. If the right-hand digit be even, and the sum of all the digits be di- visible by 6, then the whole is divisible by 6. 7. A number is divisible by 11, when the sum of the 1st, 3d, 5th, &c. or all the odd places, is equal to the sum of the 2d, 4th, 6th, Ac. or of all the even places of digits. 8. If a number cannot be divided by some quantity less than the square root of the same, that number is a prime, or cannot be divided by any number whatever. 9. All prime numbers, except 2 and 5, have either 1, 3, 7, or 9, in the place of units; and all other numbers are composite, or can be divided. 10. When numbers, with the sign of addition or subtraction between them, are to be divided by any number, then each of those numbers must be divided by it. Thus l^iAlli = 5-1-4 — 2 = 7. 11. But if the numbers have the sign of multiplication between them, only one of them must be divided. Thus, 10X8X3 _ 10 X 4 X 3 _ 10 X 4 X 1 = 10 X 2 X 1 20 6X2" 6X1 2X1 1X1 1 56 AKITHIUETIC. CASE II. To Reduce a Mixed Number to its Equivalent Improper Frac- tion* * Multiply the integer or whole number by the deno- minator of the fraction, and to the product add the numera- tor ; then set that sum above the denominator for the fraction required. EXAMPLES. 1. Reduce 23} to a fraction. 23 5 115 Or, thus, 2 (23X5)+2 117 t . _ ~ = the Answer. 117 5 2. Reduce 12 J to a fraction. Ans. 1 j 5 . 3. Reduce 14 T \ to a fraction. Ans. y 7 7 . 4. Reduce 183 f s r to a fraction. Ans. 3 £-f ». CASE III. To Reduce an Improper Fraction to its Equivalent Whole or Mixed Number. f Divide the numerator by the denominator, and the quo- tient will be the whole or mixed number sought. EXAMPLES. 1. Reduce ^ to its equivalent number. Here y or 12-7-3=4, the Answer. * This is no more than first multiplying a Quantity by some number, and then dividing the result back again by the same : which it is evi- dent does not alter the value ; for any fraction represents a division of the numerator by the denominator. t This rule is evidently the reverse of the former ; aod the reason of it is manifest from the nature of Common Division. .msBUonozr or vulgab fkactiojci. 57 3. Reduce y to its equivalent number. Here y or 15-4-7=24, the Answer. & Reduce 7 T y to its equivalent number. Thus, 17 ) 749 ( 4 T V4 68 69 So that \y =44 jV, the Answer. 68 1 4. Reduce y to its equivalent number. Ans. 8. 5. Reduce to its equivalent number. Ans. 54}}. 6. Reduce *ff • to its equivalent number. Ans. 171ff. CASE IV. To Reduce a Whole Number to an Equivalent Fraction, hav- ing a Given Denominator. * Multiply the whole number by the given denominator ; then set the product over the said denominator, and it will form the fraction required. EXAMPLES. 1. Reduce 9 to a fraction whose denominator shall be 7. Here 9X7=63: then V is the Answer; For y =63-r-7=9, the Proof. 2. Reduce 12 to a fraction whose denominator shall be 13. Ans. yj. 3. Reduce 27 to a fraction whose denominator shall be 11. Ans. yp. CASE V. 7b Reduce a Compound Fraction to an Equivalent Simple one. f Multiply all the numerators together for a numerator, . and all the denominators together for a denominator, and they will form the simple fraction sought. • Multiplication and Division being here equally osetf, the result mst be the same as the quantity first proposed. f The troth of this rule may be shown as follows : Let the compound faction be | of Now } of ^ is which is ^ \ tonaftqtMutty Vol, I 9 58 AR I T H METIC, When part of the compound fraction is a whole or mixed number, it must first be reduced to a fraction by one of the former cases. And, when it can be done, any two terms of the fraction may be divided by the same number, and the quotients used instead of them. Or, when there are terms that are com* mon, they may be omitted, or cancelled. EXAMPLES. 1. Reduce J of f of f to a simple fraction. _ 1X2X3 6 1 , . Here 2^3X4 = 24 = 4* *• An8Wer ' ♦ Or, ^^| = 1, by cancelling the 2*s and 3's. 2. Reduce | of J of H to a simple fraction. 2X3X10 60 12 4 . . Here 3 X5X11 = 165 = 33 = TP the Answer ' 2 _ 2X£X20 4 . . _ . ^XjgXll = IP same as tofore, by cancelling the 3's, and dividing by 5's. 3. Reduce ^ of f to a simple fraction. Ana. 4. Reduce $ of } of $ to a simple fraction. Ans. }• 5. Reduce f pf f of 3} to a simple fraction. Ans. }. 6. Reduce f of \ of } of 4 to a simple fraction. Ans. f • 7. Reduce 2 and f of £ to a fraction. Ans. 2. CASE vi. Tb Reduce Fraction* of Different Denominations to Equivalent Fractions having a Common Denominator. * Multiply each numerator by all the denominators ex* cept its own for the new numerators : and multiply all the denominators together for a common denominator. f of f will be ^X2 or ^ ; that is, the numerators are multiplied to- gether, and also the denominators, as in the Role. When the compound fraction consists of more than two single ones ; having first reduced two of them as above, then the resulting fraction and a third will be the same as a compound fraction of two parts ; and so on to the last of all. * This is evidently no more than multiplying each numerator and its- denominator by the same quantity r and consequently the value of the* fraction ti not altered. mttcnoK or vttloae fractions. 59 Male, It it evident, that in this and several other operations, 'when any of the proposed quantities are integers, or mixed tratnbersy or oompoond fractions, they must first be reduced, by their proper Rules, to the form of simple fractions. EXAMPLES. 1. Reduce *, }, and }, to a common denominator, 1 X 3 X 4 = 12 the new numerator for £. 2X2X4 = 16 ditto |. 3 X 2 X 3 =* 18 ditto f . 2 X 3 X 4 = 24 the common denominator. Therefore the equivalent fractions are £}, £f , and J }. Or the whole operation of multiplying may often be per- formed mentally, only setting down the results and given ^ 0M ««*»^tt|» = «,Jf,if = ^A.A»by abbre. vianon. 2. Reduce f and | to fractions of a common denominator. Ans - Ih *f • 3. Reduce |, }, and f to a common denominator, Ans. f$, |f, 4. Reduce |, 2}, and 4 to a common denominator. Ans.H,}f,W- iVsfe 1, When the denominators of two given fractions have a common measure, let them be divided by it ; then multiply the terms of each given fraction by the quotient arising from the other's denominator. Ex. ft and ft = {ft {ft, by multiplying the former 5 7 by 7 and the latter by 5. 2. When the less denominator of two fractions exactly divides the greater, multiply the terms of that which has the less denominator by the quotient. Ex. 4 and ft = ft and ft, by mult, the former by 2.* 2 3. When more than two fractions are proposed, h is some- times convenient, first to reduce two of them to a common denominator ; then these and a third ; and so on till they he -all reduced to their least common denominator. Ex. | and J and J = J and f and \ =» \\ and \\ and \\. CASK VII. 7b reduce Complex Fractions to single ones. Reduce the two parts both to simple fractions ; then mul. tiply the numerator of each by the denominator of the other ; which is in fact only increasing each part by equal multi- fft Asrrusnc. pUCatjons, which makes no difference in the value of the ▼hole. 6' And = A lso?i = V : 4 12 4J- f 17 v 2 _ 34 ~5 * 9 ~ 45' CASE Tin. 3b find the valve of a Fraction in Parte of the Integer. Multiply the integer by the numerator, and divide the product by the denominator, by Compound Multiplication and Division, if the integer be a compound quantity. Or, if it be a single integer, multiply the numerator by the parts in the next inferior denomination, and divide the pro- duct by the denominator. Then, if any thing remains, mul- tiply it by the parts in the next inferior denomination, and divide by the denominator, as before ; and so on as far as ne- cessary ; so shall the quotients, placed in order, be the value of the fraction required.* EXAMPLES. 1. What is the \ of 2J6*? By the former part of the Rule 2/6* 4 5) 9 4 11 \QsQd2iq. 2. What is the value of |of III By the 2d part of the Rule, 2 20 3) 40 (13* 4d Ans. 1 12 3) 12 (4d 3. Find the value of } of a pound sterling. Ans. lit W. 4. What is the value of } of a guinea ? Ads. As 8d. & What is the value of J of a half crown ? 6. What is the value of} of 4* \0dl 7. What is the value off lb troy ? 8. What is the value of ft of a cwt ? 9. What is the value of J of an acre ? 10. What is the value of ft of a day? Ans 1* lOjrf. Ans. Ullfd. Ans. 9 oz 12 dwts. Ans. 1 qr 7 lb. Ans. 3 ro 20 po. Ans. 7 hrs 12 min. * The numerator of a fraction being considered as a remainder, in . Division, and the denominator as the divisor, this rule is of ibe same aature as Compound Division, or the valuation of remainders in the Hole of Three, before explained. mwcTioN op tom^r nuonoics. tt cabs is. ft 7> JMpMtf a JVaefofi ,/r©ii one JDe^onifiafiofi to another. * Consider how many of the leas denomination make one of the greater ; then multiply the numerator by that number, if the reduction be to a leas name, but multiply the denominator, if to a greater. 1. Reduce } of a pound to the fraction of a penny. f X Y * Y = T = l Vt Ae Answer. 2. Reduce 4 of a penny to the fraction of a pound. # x tV * 1V 88 ^ Answer. $. Reduce ftl to the fraction of a penny. Ana. *fd. 4. Reduce }g to the fraction of a pound. Ana. t*Vt* 5. Reduce f cwt to the fraction of a lb. Ana. y . 6. Reduce ] dwt to the fraction of a lb troy. Ana. T £ r . 7. Reduce f crown to the fraction of a guinea. Ana. fy, 8. Reduce { half-crown to the fract. of a shilling. Ana. . 9. Reduce 2* 6d to the fraction of a £. Ana. \. 10. Reduce 17* Id 3{? to the fraction of a £. Ans. fiff. ADDITION OF VULGAR FRACTIONS. If the fractions have a common denominator ; add all the numerators together, then place the sum over the common denominator, and that will be the sum of the fractions re* quired. f If the proposed fractions have not a common denomina- tor, they must be reduced to one. Also compound fractions * This is the same as the Rale of Reduction in whole numbers from one denomination to another. \ Before fractions are reduced to a common denominator, they are quite dissimilar, as much as shillings and pence are, and therefore can- not be incorporated with one another, any more than these can. But when they are reduced to a common denominator, and made parts of the same thing, their sum, or difference, may then be as properly ex- pressed by the sum or difference of the numerators, as the sum or dif- ference of any two quantities whatever, by the sum or ditfemw* <A 62 AKrr&wftncr must be reduced to simple ones, and fractions of different denominations to -those |>f the same denomination. Then add the numerators, as before. As to mixed, numbers, they may either be reduced to improper fractions, and so added with the others ; or else the fractional parts only added, and the integers united afterwards. ' examples. 1. To add | and £ together. Here }+| = J = If, the Answer. 2. To add } and | together. * + * = it + tt = it = Hi* Answer. 3. To add J and 7\ and | of f together. i+7i+i off = i+.V+i « i+V+t = V 4. To add ^ and 4 together. Ans. If. 5. To add $ and f together. Ans. 6. Add \ and ft together. Ans. ft. their individuals. Whence the reason of the Role is manifest, both for Addition and Sob traction. When several fractions are to be collected, it is commonly best first to add two of them together that most easily reduce to a common de- nominator; then add their sum and a third, and so on. Note 2. Taking any two fractions whatever, ft and f-j- t for example, after reducing them to a common denominator, we judge whether they are equal or unequal, by observing whether the products 35 X 11, and 7 X 65, which constitute the new numerators, are equal or unequal. If, therefore, we have two equal products 35x11=7X56, we may compose from them two equal fractions, as -g-f- = ft, or = ^ . If, then, we take two equal fractions, such as ft and we shall have 36 X 11 = 7 X 56 ; taking from each of these 7 X 11, there will OK J remain (35 — 7) X 11 = (65 — 11) X 7, whence we have ^ = 55 — 11 In like manner, if the terms of ft were respectively added to those of ff, we should have = tf = ft. Or, generally, if = ^, it may in a similar way be shown, that _ tf _ c Hence, when two fractions are of equal valuta the fraction formed by ta- king the sum (or Ike diffirence) of their numerators respctireJy, and of their denominators respecticelij, it a fraction equal in value to each of the original fractions. This proposition will be found useful in the doctrine of pro- portions. MULTlFJJCATIOlf OF VULGAR FRACTIOUS. 68 • 7. What is the sum of f and f and 4 ! Ans. l}ff . & What is the sum of f and | and ? Ans. 3f|. 9. What is the sum off and \ of |, and 9ft ? Ans. 10^. 10. What is the sum of } of a pound and $ of a shilling ? Ans. l |»t or 13* MM 2}?. 11: What is the sum of } of a shilling and ft of a penny 1 Ans. y/dor7d Hftf- 12. What is the* sum of } of a pound, and f of a shilling, and ft of a penny ? Ans. fiff* or 3 * ld Hffl- SUBTRACTION OF VULGAR FRACTIONS. Prepare the fractions the same as for Addition, when necessary ; then subtract the one numerator from the other, and set the remainder over the common denominator, for the difference of the fractions sought. EXAMPLES. 1. To find the difference between £ and £. Here $ — £ = $ = |, the Answer. 2. To find the difference between J and f . ? — i = H — H = *V Answer. 3. What is the difference between ft and ft 1 Ans. 4. What is the difference between ft and ft ? Ans. ft. 5. What is the difference between ft and ft ? Ans. T \fr. 6. What is the diff. between 5} and 4 of 4$ ? Ans. 4^- 7. What is the difference between £ of a pound, and | of I of a shilling ? Ans. 'ft 8 or 10* Id \\q. 8. What is the difference between $ of 5£ of a pound, and } of a shilling. Ans. or 11 Ss llftd. MULTIPLICATION OF VULGAR FRACTIONS. ^ * Reduce mixed numbers, if there be any, to equivalent ♦ Multiplication of any thing by a fraction, implies the taking some put or parts of the thing; it may therefore be truly eipresaadb^ * ▲RlTHttBTft/* fractions ; then multiply all the numerators together for a numerator, and all the denominators together for a denomi- nator, which will give the product required. BXAMPLS8. 1. Required the product of } and f . Here J Xf =^=^, the Answer. Or|X*=iX*=*. 2. Required the continued product of |, 3£, 5, and f of j. Heie y x T T T 5 r "4xf""I"'^ 3. Requited the product of f and f . Ans. 4. Required the product of <fo and Ana. ^ . 5. Required the product of f, }|. Ans. 6. Required the product of }, }, and 3. Ans. 1. 7. Required the product of }, f , and 4^. Ans. 3^. 8. Required the product of f , and } of f . Ans. 9. Required the product of 6, and } of 5. Ans. 20. 10. Required the product off of j, and f of 3f. Ans. |{. 11. Required the product of 8} and 4}}. Ans. 14|}f • 12. Required the product of 5, f, f of}, and 4}. Ans. 2^-. DIVISION OP VULGAR FRACTIONS. * Prepabb the fractions as before in Multiplication : then divide the numerator by the numerator, and the denominator by the denominator, if they will exactly divide : but if not, compound fraction ; which is resolved by multiplying together the numerators and the denominators. Ad/a. A Fraction b best multiplied by an integer, by dividing the denominator by it ; bat if it will not exactly divide, then multiply the numerator by ft. • Division being the reverse of Multiplication, the reason of the rale is evident. Acta, A fraction is best divided by an integer, by dividing the nume- rator by It ; bat if it will not exactly divide, then multiply the denorni- nstor by it. ftULE Or THXE1 IN VULGAR FRACTIONS. 65 invert the terms of the divisor, and multiply the dividend by K, as in Multiplication. • EXAMPLES. 1. Divide V by |. Here y 4. f = a = If, by the first method. 2. Divide f by ft. Heref^A=*Xy=|Xf = V-4}. 3. It is required to divide by |. Ans. f . 4. It is required to divide ^ by }. Ans. fa 5. It is required to divide y by J. Ans. 1 J. 6. It is required to divide { by y . Ans. y^. 7. It is required to divide by |. Ans. 4. 8. It is required to divide ij- by }. Ans. . 9. It is required to divide ft by 3. Ans. fa 10. It is required to divide } by 2. Ans. fa 11. It is required to divide 7± by 9|. Ans. }|. 12. It is required to divide f of } by | of 7|. Ans. jJt. RULE OF THREE IN VULGAR FRACTIONS. Make the necessary preparations as before directed ; then multiply continually together, the second and third terms, and the first with its parts inverted as in Division, for the answer*. EXAMPLES. 1. If } of a yard of velvet cost f of a pound sterling ; what will -f € of a yard cost ? 2. What will 3f oz. of silver cost, at 6s 4d an ounce ? Ans. I/ I* 4J<*. * This Is only multiplying the 2d And Sd terms together, and divld- bftbe product by the first, as in the Rale of Three in whole nunbtm Vol. I. 10 AJtlTHXBTIC. a If A of a ihip be worth 27813* 64; what are A of her worth? • An* 8871 12$ Id. 4. What is the purchase of 12901 bank-stock, at 108f per cent.? Ana. 133«1# W. 5. What is the interest of 2731 15* for a year, at 3£ per cent.? Ans. 8Z17* 11J<*. 6. If | of a ship be worth 731 Is 3d ; what part of her is worth 250Z 10* ? Ans. J. 7. What length must be cut off a board that is 7| inches broad, to contain a square foot, or as touch as another piece of 12 inches long and 12 broad ? Ans. 18jf inches* 8. What quantity of shalloon that is J of a yard wide, will line flivards of cloth, that is 2} yards wide? Ans. 31 J yds. fc' if the penny loaf weigh 6^ 0z. when the price of wheat is 5* the bushel ; what ought it to weigh when the wheat is 8a 6d the bushel ? Ans. 4^ oz. 10. How much in length, of a piece of land that is 11{4 ptoles broad, will make an acre of land, or as much as 40 pdlei in length and 4 in breadth ? Ans. 13 T y T poles. Hi If a courier perform a certain journey in 35$ days, travelling 13f hours a day ; how long would he be in per- forming the same, travelling only 11 -ft hours a day ? Ans. 40f)f days. 12. A regiment of soldiers, consisting of 976 men, are to be new clothed ; each coat to contain 2J yards of cloth that is If yard wide, and lined with shalloon J yard wide : how many yards of shalloon will line them ? Ans. 4531 yds 1 qr 2f nails. DECIMAL FRACTIONS. A Decimal Fraction is that which has for its deno- minator an unit (1), with as many ciphers annexed as the numerator has places ; and it is usually expressed by setting down thgfiumetator only, with a point before it, on the left, hand. Thus, & is -4, and ffr is -24, and rifo is *074, and ttVvVt i* '00124 ; where ciphers are prefixed to make up as many placet ma are ciphers in the denominator, when there is a deficiency in the figures. A mixpd numbejr is made up of a whole number with some .decimal fraction, the one being separated from the other by a fcoint. Thus, 3*25 is the same as or f| f . Ciphers on the right-hand of decimals make no alteration in their value ; for -4, or '40, or *400 are decimals having all IbeMne value, each being = ^ or f . Bur when they are ADDlfHMI W MCDIAIS. placed m the left-hand, they decrease the value in m ten4bld proportion : That, -4 is ^ or 4 tenths ; but -04 is only t4y> or4himfw>dths > and -004 is «nly y/^, or 4 thousandths. Ia sWrknalfr as well as in whole numbers, the vahief .of tfcefiaee* increase towards Joe left-hand, and 4e&*ase to. wards die right, both in the same tenfold proportion $ s# in the following Scale or Table of Notation. 3333338*333939 ADDITION OF DECIMALS. Sbt the numbers under each other according to the value of their places, as in whole numbers ; in which state the decimal separating points will stand all exactly under each other. Then, beginning at the right hand, add up all the columns of numbers as in integers ; and point off as many places for decimals, as are in the greatest number of decimal places in any of the lines that are added ; or pla<?e the point directly below all the other points. 1. To add together 29-0146, and 3146*5, and 2 J 09, tod 03417, and 14-16. 29 0146 8146-5 2109- •62417 1416 EXAMPLES. 5299-29877 the Sum. 6S ABXTHXSTIC* 2. What is the sum of 276, 39-213, 72014-9, 417, -and 50821 Ans. 77770-113. 3. What is the sum of 7530, 16 201, 3-0142, 957*13, 6*72119 and -03014 ? Ans. 8513-09653. 4. What is the sum of 312-09, 3-5711, 7195-6, 71-498, 9739-215, 179, and -0027 ? Ans. 17500-9718. SUBTRACTION OF DECIMALS. Place, the numbers under each other according to the value of their places, as in the last Rule. Then, beginning at the right-hand, subtract as in whole numbers, and point off the decimals as in Addition. EXAMPLES. 1. To find the difference between 91*73 and 2*138. 91*73 2*138 Ans. 89.592 the Difference. 2. Find the diff. between 1 -9185 and 2*73. Ans. 0*81 15. 3. To subtract 4*90142 from 214*81. Ans. 209*90858. 4. Find the diff. between 2714 and -916. Ans. 2713*084. MULTIPLICATION OF DECIMALS. * Place the factors, and multiply them together the same as if they were whole numbers. — Then point off in the pro- duct just as many places of decimals as there are decimals in both the factors. But if there be not so many figures in the product, then supply the defect by prefixing ciphers. * The rule will be evident from this example Let it be required to multiply -12 by '961 ; these numbers are equivalent to -jfo and ftftfc ; the product of which is v \ g g g = 04432, by the nature of Notation, which consists of as many places as there are ciphers, that is v of as many places as there are in both numbers. And in like manner for any other numbers. HULTIPLICATfOll OF DECIMAL!. 69 EXAMPLES* 1. Multiply -321006 by -2465 1605480 1926576 1284384 642192 Ana.' •0791501640 the Product. 2. Multiply 79-347 by 23-15. Ans, 1836-88305. 3. Multiply -63478 by -8204. Ana. -520773512. 4. Multiply -385746 by -00464. Ana. -00178986144. CONTRACTION I. To multiply Decimals by 1 with any Number of Ciphers, as by 10, or 100, or 1000, $c. This is done by only removing the decimal point so many places farther to the right-hand, as there are ciphers in the multiplier ; and subjoining ciphers if need be. EXAMPLES. 1. The product of 51-3 and 1000 is 51300. 2. The product of 2-714 and 100 is 8. The product of -916 and 1000 is 4. The product of 2i»31 and 10000 is CONTRACTION II. To contract the Operation so as to retain only as many Deci- mals in the Product as may be thought necessary, when the Product would naturally contain several more Places. Set the unit's place of the multiplier under the figure of the multiplicand whose place is the same as is to be retained for the last in the product ; and dispose of the rest of the figures in the inverted or contrary order to what they are usually placed in. — Then, in multiplying, reject all the figures that are more to the right-hand than each multiplying figure, and set down the products, so that their njVv\-\i«iA 70 figures may fall in a column straight below each other ; but observe to increase the first figure of every line with what would arise from the figures omitted, in this manner namely 1 from 5 to 14, 2 from 15 to 24, 3 from 25 to 34, dec. ; and the sum of all the lines will be the product as required, com- monly to the nearest unit in the last figure. EXAMPLES. 1. To multiply 27-14986 by 92-41035, so as to retain only four places of decimals in the product Contracted Way. 27 14986 53014-29 Common Way. 27-14986 92-41035 24434874 13 574930 542997 81 44958 108599 2714 986 2715 108599 44 81 542997 2 14 24434874 2508-9280 2508-9280 650510 2. Multiply 480*14936 by 2-72416, retaining only four decimals in the product. 3. Multiply 2490-3048 by -5*73286, retaining only five decimals in the product. 4. Multiply 325-701428 by -7218393, retaining only three decimals in the product. DIVISION OF DECIMALS. Divide as in whole numbers ; and point off in the quo- tient as many places for decimals, as the decimal places in the dividend exceed those in the divisor*. * The mason of this Rale is evident; for, since the divisor multiplied by the quotient gives the dividend, therefore the Dumber of deoimal places iu the dividend, is equal to those in the divisor and quotient, taken together, by the nature of Multiplication ; and consequently the quotient itself mast contain as many as the dividend eieeeds the divisor. DIVISION OF 9MQULLL8. 71 Another way to know the place for the decimal point is this : The first figure of the quotient must be made to occupy the same place, of integers or decimals, as that figure of the dividend which stands over the unit's figure of the first pro- When the places of the quotient are not so many as) the Rule requires, the defect is to be supplied by prefixing ciphers. When there happens to be a remainder after the division ; or when the decimal places in the divisor are more than those in the dividend ; then ciphers may be annexed to the divi- dend, and the quotient carried on as far as required. EXAMPLES. 00272589 -2639 1. 178) "48580998 ( 1392 460 1049 1599 1758 156 3. Divide 123-70536 by 54 25. 4. Divide 12 by -7854. 5. Divide 4195-68 by 100. 6. Divide -8297592 by -153. 1. n 27-00000 (102-3114 6100 8220 3030 3910 12710 2154 Ans. 2-2802. Ans. 15-278. Ans. 41-9568. Ans. 5-4232. CONTRACTION I. When the divisor is an integer, with any number of ciphers annexed : cut off those ciphers, and remove the decimal point in the dividend as many places farther to the left as there are ciphers cut off, prefixing ciphers, if need be ; then proceed as before. EXAMPLES. 1. Divide 45-5 by 2100. 21-00) -455 ( 0216, <&c. 35 140 14 3. Divide 41020 by 32000. & Divide 953 by 21600. 4. Divide 61 by 79000. 73 ARITHMETIC. CONTRACTION II. Hence, if the divisor be 1 with ciphers, as 10, 100, or 1000, &c ; then the quotient will be found by merely mov- ing the decimal point in the dividend so many places farther to the left, as the divisor hath ciphers ; prefixing ciphers if need be. EXAMPLES. So, 217-3 -r 100 = 2 173 Ans. 419 10 = And 5-16 -f- 100= Ans. -21 -f- 1000 = CONTRACTION HI. When there are many figures in the divisor ; or when only a certain number of decimals are necessary to be retained in the quotient ; then take only as many figures of the divi- sor as will be equal to the number of figures, both integers and decimals, to be in the quotient, and find how many times they may be contained in the first figures of the dividend, as usual. Let each remainder be a new dividend ; and for every such dividend, leave out one figure more on the right-hand side of the divisor ; remembering to carry for the increase of the figures cut off, as in the 2d contraction in Multiplication. Note. When there are not so many figures in the divisor as are required to be in the quotient, begin the operation with all the figures, and continue it as usual till the number of figures in the divisor be equal to those remaining to be found in the quotient ; after which begin the contraction. examples. 1. Divide 2508-92800 by 92*41035, so as to have only four decimals in the quotient, in which case the quotient will contain six figures. Contracted. 92-4103,5) 2608-928,09(971498 660721 13849 4608 912 80 6 • Coitittum. 92.4103,5) 2608-928,06 (27-1498 66072106 18848610 46075750 91116100 79467850 5639570 2. Divide 4109*2351 by 230-409, so that the quotient may contain only four decimals. Ans. 17*8345. REDUCTION OP DECIMALS. 78 3. Divide 37- 10438 by 5713-96, that the quotient may contain only five decimals. Ans. '00649. 4. Divide 913 08 by 2137-2, that the quotient may contain only three decimals. REDUCTION OF DECIMALS. CASE I, To reduce a Vulgar Fraction to its equivalent Decimal. Divide the numerator by the denominator, as in Division ef Decimals, annexing ciphers to the numerator as far as necessary ; so shall the quotient be the decimal required*. * The following method of throwing a vulgar fraction, whose de- nominator is a prime number, into a decimal consisting of a great num- ber of figures, is given by Mr. Cohan in page 162 of Sir Isaac Newton's Fluxions. [ EXAMPLE. Let ^ be the fraction which is to be converted into an equivalent decimal. Then, by dividing in the common way till the remainder becomes a tingle figure, we shall have -fy — -03448^ for the complete quotient^ and this equation being multiplied by the numerator 8, wiJJ give -f^ == 275g4ji4. f ( ,r rather u % ~ 27586\/ ff : and if this be substituted instead of the friction in the first equation, it will make ^ = -0344827586^. Again, let this eqaation be multiplied by 6, and it will give A = '206*8965517^ ; and then by substituting as before -if = -034482758610689605175^ ; and so on, as far as may he thought proper; each fresh multiplication doubling the number of figures in the decimal value of the fraction. In the present instance the decimal circulates in a complete period of 28 figures, i. e. one less than the denominator of the fraction. This, again, may be divided into equal periods, each ot 14 figures, as below : •03448275862068 •96551724137931 in which it will be found that each figure with the figure vertically be- low it makes 9; + 9 = 9; 3 -f 6 ^ 9 ; and so on. This circulate also comprehends all the separate values of &c. in correspond- ing circulates of 28 figures, only each beginning in a distinct place, easi- ly ascertainable. Thus, ^ — 06896, &c. beginning at the 12th place of the primitive circulate. ^ = 103448, &c. beginning at the 28th plnce. So that, in fact, this circle includes 28 complete circles. 8ee, on this curious subject, Mr. Goodwyn's Tables of Decimal Cir- cles, and the Ladies' Diary for 1824. Vol. I. 11 AKTEOtBTICV BX AMPLE** I. Reduce yV to a decimal. 24 = 4 X 6. Then 4) 7- 6) 1*750000 •291606 dec. 2. Reduce J, and J, and J, to decimals. 3» Reduce f to a decimal. 4. Reduce ^ to a decimal. 5* Reduce T J ¥ to a decimal. 6\ Reduce to a decimah. Ans* *25, and # 5, and *75v Ana. -625. Ana. -12. Ana. -08135. Ana. -14*154 IK- CASE n. To Jmd the Yakut of a Decimal in terms of the Inferior Denominations. Multiply the decimal by the number of parts in the next lower denomination ; and cut off as many places for a remainder to the right-hand, as there are places in the given decimal. Multiply that remainder by the parts in the next lower denomination again, cutting off for another remainder as before. Proceed in the same manner through all the parts of the integer ; then the several denominations separated on the left-hand will make up the answer. Note, This operation is the same as Reduction Descending in whole numbers. 1. Required to find the value of -775 pounds sterling. EXAMPLE?. •775 20 s 15-500 12 4 6-0Q0 Ans. 15* 6rf» ■SDUCTlOlf OF BECfMALS. 76 % What is the value of -625 shil ? fc Ans. 74*. 3. What is the value of -86352 ? Ans. 17s 3-242. 4. What is the value of -0125 lb troy? Ans. 3 dwts. & What is the value of -4694 lb troy ? Ans. 5 or. 12 dwts 15*744 gr. & What is the value of -625 cwt ? Ans. 2 qr 14 lb. 7. What is the value of -009943 miles? Ans. 17 yd 1 ft 5-98848 inc. & What is the value of -6875 yd ? Ans. 2 qr 3 nls. 9. What is the value of -3375 acr ? Ans. 1 rd 14 poles* 19. What is the value of -2083 hhd of wine? Ans. 13-4229 gaL CASS III. lb reduce Integer* or Decimals to Equivalent Decimal* of Higher Denominations. Divide by the number of parts in the next higher deno- ounation ; continuing; the operation to as many higher de- aominations as may be necessary, the same as in Reduction Ascending of whole numbers. EXAMPLES. 1. Reduce 1 dwt to the decimal of a pound troy* 20 \ 1 dwt 12 0-05 os 1 0-004166 dec, lb. Ans. 2. Reduce 9d to the decimal of a pound. Ans/ 0*3757. 3. Reduce 7 drams to the decimal of a pound avoird. Ans. -027343751b. 4. Reduce -264 to the decimal of a I. Ans. 0910833 6zc. /. 5. Reduce 2-15 lb to the decimal of a cwt. Ans. -019196 + cwt. 6. Reduce 24 yards to the decimal of a mile. Ans. 013636 &c. mile. 7. Reduce -056 pole to the decimal of an acre. Ans. O0035 ac. 8. Reduce 1-2 pint of wine to the decimal of a hhd. Ans. -00238 + hhd. 3. Reduce 14 minutes to the decimal of a day. Ans. -009722 <fcc. da* JO. Reduce -21 pint to the decimal of a peck. Ans. -031325 pec. 11. Reduce 28" 12" to the decimal of a minute. 79 AKIfflMlG* Hon, When (here to Vke ieemdl of the highest ; Set the given numbers directly under each other, for di- vidends* proceeding orderly from the lowest denomination to the highest. Opposite to each dividend, on the left-hand, set such a number for a divisor as will bring it to the next higher name ; drawing a perpendicular line between all the divisors and dividends ' Begin at the uppermost, and perform all the divisions : only observing to set the quotient of each division, as deci- mal parts, on the right-hand of the dividend next below it ; so shall the last quotient be the decimal required. EXAMPLES. 1. Seduce 17* 9f J to the decimal of a pound. 4 1 3- 12 9-75 20 J 17-8125 £0 890625 Aits. 2. Beduce 191 17s 3}d to a I. Ans. 19*86354166 &c. I. 3. Reduce 15* 6d to the decimal of a I. Ans., -775/. 4. Reduce l\d to the decimal of a shilling. Ans. *625*. 5. Reduce 5 oz 12 dwts 16 gr to lb. Ans. -46944 dec. lb. RULE OP THREE IN DECIMALS. • Pbbpaxe the terms, by reducing the vulgar fractions to decimals, and any compound number either to decimals of the higher denominations, or. to integers of the lower, also the first and third terms to the same name : Then multiply and divide as in whole numbers. Note. Any of the convenient Examples in the Rule of Three or Rule of Five in Integers, or Vulgar Fractions, may be taken as proper examples to the same rules in Decimals. —The following example, which is the first in Vulgar Frac- tions, is wrought out here, to show the method. DUODKCTJLAXt* 77 If I of a yard of relvet cost fJ, what will ^ yd cost? yd I yd I 9 d } = -375 -375 : -4 : : -3125 : -333 dec. or 6 8 •4 J = -4 -375) -12500 (-333333 fcc. 1250 20 125 #6-60666 &c. ^ = -8125 12 Ans. 6* Bd. d 7-99099 &c. = 8d. DUODECIMALS. Duodecimals, or Cross Multiplication, is a rule used by workmen and artificers, in computing the contents of their works. ♦ Dimensions are usually taken in feet, inches, and quarters ; any parts smaller than these being neglected as of no con- sequence. And the same in multiplying them together, or computing the contents. The method is as follows. Set down the two dimensions to be multiplied together, one under the other, so that feet may stand under feet, inches under inches, &c. Multiply each term in the multiplicand, beginning at the lowest, by the feet in the multiplier, and set the result of each straight under its corresponding term, observing to car- ry 1 for every 12, from the inches to the feet. In like manner, multiply all the multiplicand by the inches and parts of the multiplier, and set the result of each term one place removed to the right-hand of those in the mul- tiplicand ; omitting, however, what is below parts of inches, only carrying to these the proper numbers of units from the lowest denomination. Or, instead of multiplying by the inches, take such parts of the multiplicand as these are of a foot. Then add the two lines together, after the manner of Compound Addition, carrying 1 to the feet for every 12 inches, when these come to so many. 78 ▲JEtlTHMBTIC. EXAMPLES. 1. Multiply 4 f 7 inc. 2. Multiply 14 f 9 inc. by 6 4 by 4 6 27 6 59 1 <** 7 Ans. 29 °* Ans. 66 4* 3. Multiply 5 feet 7 inches by 9 f 6 inc. Ans. 43 f 6 J. inc. 4. Multiply 12 f 5 inc by 6 f 8 inc. Ans. 82 9} 5. Multiply 35 f 4£ inc by 12 f 3 inc. Ans. 433 4| 6. Multiply 64 f 6 inc by 8 f 9i inc. Ans. 565 8f Nate, The denomination which occupies the place of inches in these products, means not square inches, but recf- angles of an inch broad and a foot long. Thus, the answer to the first example is 29 sq. feet, 4 sq. inches ; to the second 66 sq. feet, 54 sq. inches* INVOLUTION. Involution is the raising of Powers from any given num. ber, as a root. A Power is a quantity produced by multiplying any given number, called the Root, a certain number of times conti- nually by itself. Thus, 2 = 2 is the root, or 1st power of 2. 2X2 = 4 is the 2d power, or square of 2. 2X2X2= 8 is the 3d power, or cube of 2. ' 2X2X2X2 = 16 is the 4th power of 2, &c. And in this manner may be calculated the following Table of the first nine powers of the first 9 numbers* nrroLtmozr. 79 TABLE OT THE FIRST NINE POWERS OF NUMBERS. 1 5 3d 4th 5 th 6th 7th 8th 9ih " 1 2 3 4 5 8 7 8 9 I 1 1 1 1 1 1 1 1 8 16 32 64 128 256 512 27 81 243 729 2187 6561 19683 16 64 256 1024 409G 16384 65536 262144 25 125 625 31^5 15620 78125 390625 1953125 m 216 1296 7776 46656 279936 1679616 10077696 id 343 2401 16807 117649 823543 5764801 40353607 H si 512 721) 4006(32768 6561 59049 262144 2097152 16777216 134217728 [6814*] 4782969 43046721 387420489 The Index or Exponent of a Pnwer, is the number de- noting the height or degree of that power ; and it is 1 more than the number of multiplications used in producing the same. So 1 is the index or exponent of the 1st power or root, 2 of the 2d power or square, 8 of the third power or cube, 4 of the 4th power, and so on. Powers, that are to be raised, are usually denoted by placing the index above the root or first power. So 2 s = 4 is the 2d power of 2. 2 3 = 8 is the 3d power of 2. 2 4 = 16 is the 4th power of 2. 540* is the 4th power of 540, ozc. When two or more powers are multiplied together, their .product is that power whose index is the sum of the expo- nent of the factors or powers multiplied. Or the multiplica- tion of the powers, answers to the addition of the indices. Thus, in the following powers of 2, 1st 2d 3d 4th 5th 6th 7th 8th 9th 10th 2 4 8 16 32 64 128 256 512 1024 orS 1 2* 2 s 2* 2 8 2 s 2 7 2 s 2 B * 10 80 ▲XtTHXETIC. Here, 4 X 4 = 16, and 2 + 2 = 4 its index ; and 8 x 16= 128, and 3 + 4= 7 its index; also 16 X 64 — 1024, and 4 + 6 = 10 its index. OTHER EXAMPLES. 1. What is the 2d power of 45 1 Ans. 2025. 2. What is the square of 4*16 ? Ans. 17*8056. 3. What is the 3d power of 3-5 ? Ans. 42-575, 4. What is the 5th power of -029 ? Ans. -00000002051 1149. 5. What is the square of f ? Ans. f . 6. What is the 3d power of f ? Ans. 7. What is the 4th power of } ? Ans. ,VV- EVOLUTION. Evolution, or the reverse of Involution, is the extracting or finding the roots of any given powers. The root of any number, or power, is such a number, as being multiplied into itself a certain number of times, will produce that power. Thus, 2 is the square root, or 2d root of 4, because 2 a = 2 x 2 == 4 ; and 3 is the cube root or 3d root of 27, because 3 3 == 3 X 3 X 3 = 27. Any power of a given number or root may be found ex- actly, namely, by multiplying the number continually into itself. But there are many numbers of which a proposed root can never be exactly found. Yet, by means of deci- mals, we may approximate or approach towards the root, to any degree of exactness. Those roots which only approximate, are called Surd Roots ; but those which can be found quite exact, are called Rational Roots. Thus, the square root of 3 is a surd root ; but the square root of 4 is a rational root, being equal to 2 : also the cube root of 8 is rational, being equal to 2 ; but the cube root of 9 is surd or irrational. Roots' are sometimes denoted by writing the character •/ before the power, with the index of the root against it. Thus, the 3d root of 20 is expressed by f/JJO ; and the square SQUARE ROOT. SI root or 2d root of it is ^/20, the index 2 being always omit- ted, when only the square root is designed. When the power is expressed *»y several numbers, with the sign + or — beiwxeu them, a line is drawn from the top of the sign over all the parts of it : thus the third root of -15 — 12 is \/ 45 — 12, or thus, y(45— - 12), inclosing the numbers in parentheses. But all roots arc now often designed like powers, with i fractional indices ; thus, the square root of 8 is 8 3 , the cube root of 25 is 25», and the 1th root of 4i> — 18 is (45 - 18)*. TO EXTRACT THE SQUARE ROOT. * Divide the given number into period" of two figures each, by setting a point over the place of units, another over the place of hundreds, a.id so on, over every second figure, both to the left-hand in integers, and to the right in deci- mals. * The reason for separating the figures of the dividend into periods or portions of two places each, U, that the square of any single figure never consists of more than two places; t tic square of a number of two figures, <if nmi more than four places, and so on. So that there will he as many figures in the root us the given number contains periods so di- vided or parted off. And the reason of the several steps hi the operation appears from the ilgchraic form of the -quare of any number of terms, whether two or three or more. Thus («- h)2 . a-* -•- 2ntt-\- fta _- aa \-( % la f ft) ft, the square of two terms ; wuerc it appears that a i* the first term of the root, and ft the second term ; al«*o a the first divisor, and 'lie new divisor i« 2a j-ft, or double the first term i». creased by the second. And hence the manner of ex- traction is thus : h: divisor a) »*2 -f Sift -|- ft (a j- ft the root. ?.a di-;$or°.a-j-ft .^aft-j-fts ft I ft :-fta Again, for a root of three P'.Im, ft. r f thus. (a \-b-\-c)i -■•|3"-2rtft bi -' r %ar-\-%bc j /8 a- -i ( m 2a ; h)h ■: (2/*' r *2ft j r)c, the square of three terms, where a is too fuM term of the root, ft the «»rond, and c the third term ; also a the fu?t divisor, 2.i- r ft the second, and &r -| 'lb --c the third, each consi.-ting of the double of the root increased by thu next term of 'lie -»»m. . And the mode of extr.ictifui agrees with iiilu. r?«.e far'.iier, Case 2. of Kvolution ii. ire. Algebra. hi' -■; "»h , . I* or an approximation observe that Vul± v - a. -— ---nearly in All cases where <i it small in respect of a. Vol. J. l'J 82 Find the greatest square in the first period en the left-hand, and set its root on the right-hand of the given number, after the manner of a quotient figure in Division. Subtract the square thus found from the said period, and to the remainder annex the two figures of the next following period, for a dividend. Double the root above mentioned for a divisor ; and find how often it is contained in the said dividend, exclusive of its right-hand figure ; and set that quotient figure both in the quotient and divisor. Multiply the whole augmented divisor by this last quotient figure, and subtract the product from the said dividend, bring- ing down to it the next period of the given number, for a new dividend. Repeat the same process over again, viz. find another new divisor, by doubling; all the figures now found in the root ; from which, and the last dividend, find the next figure of the root as before ; and so on through all the periods, to the last. iVbfe, The best way of doubling the root, to form the new divisors, is by adding the last figure always to the last divi- sor, as appears in the following examples. — Also, after the figures belonging to the given number are all exhausted, the operation may be continued into decimals at pleasure, by add- ing any number of periods of ciphers, two in each period. EXAMPLES. 1. To find the square root of 20506624. 20506624 (5432 the root. 25 104 450 4 416 1083 3466 3 3249 10862 21724 2 21724 SQUABR ROOT* W Nero, When the ml it to he extracted to many places of figures, the work may he considerably shortened, thus y . Having proceeded in the extractioa after the common method, till there be found half the required number of figures in the root, or one figure more ; then, for the retf, divide the last remainder by its corresponding divisor, after the manner of the third contraction in Division of Deci- mals; thus, 2. To find the root of 2 to nine places of figures. 2 (1-41421356 the root. 1 24 4 100 281 | 1 I 400 281 2824 4 11900 11206 60400 56564 3. What 4. What 5. What 6. What 7. What 8. What 0. What 10. What 11. What 12. What 28284) is the square is the square is the square is the square is the square is the square is the square is the square is the square is the square 31936 (1356 1008 160 19 2 root of 2025? root of 17-3056? root of -000729? root of 3? root of 5? root of 6 ? root of 7 ? root of 10? root of 11? root of 12? Ads. 45« Ads. 416. Ans. -027. Ans. 1-732050. Ans. 2*236068. Ans. 2*449489. Ans. 2-645751. Ans. 3-162277. Ads. 3-316624. Ans. 3-464101. RULES FOR THE SQUARE ROOTS OF VULGAR FRACTIONS AND MIXED NUMBERS. First prepare all vulgar fractions, by reducing \Y&m to their least terms, both for this and all other roots. Then 84 ARITHMETIC* 1. Take the root of the numerator and of the denominator for the respective terms of the root required ; which is the best way if the denominator be a complete power : but if it be not, then 2. Multiply the numerator and denominator together ; take the root of the product : this root being made the nu. merator to the denominator of the given fraction, or made the denominator to the numerator of it, will form the frac- tional root required. That is, ^ « ^ = / flft ---L- b ~~ x/b "~ b ~~ i/ab This rule will serve, whether the root be finite or infinite. 3. Or reduce the vulgar fraction to a decimal, and extract its root. 4. Mixed numbers may be either reduced to improper fractions, and extracted by the first or second rule, or the vulgar fraction may be reduced to a decimal, then joined to the integer, and the root of the whole extracted. EXAMPLES. 1. What is the root of |§? 2. What is the root of T y T ? 3. What is the root of 7 \ ? 4. What is the root of ? 5. What is the root of 17} ? Ans. | Ans. if. Ans. 0-866025. Ans. 0-645497 Ans. 4-16S3&3. By means of the square root also may readily be found the 4th root, or the 8th root, or the 16th root, <fcc. that is, the root of any power whose index is some power of the number 2 ; namely, by extruding so oleen the square root as is de- noted by that power of 2 ; that is ; two extractions for the 4th root, three for the 8th root, and so on. So, to find the 4th root of the number 21035*8, extract the square root two times as follows : CUBE BOOT* 85 21035-8000 ( 145 037237 ( 12 0431407 the 4th root. 1 1 24 4 110 06 45 44 285 5 1435 1425 2404 4 10372 0610 29003 108000 24083 87009 3 20991 (7237 6*7 107 75637 72249 3388 ( 1407 980 17 Ex. 2. What is the 4th root of 97-41 ? TO EXTRACT THB CUBE ROOT. I. By tint Common Rule*. 1. Having divided the given number into periods of three figures each (by setting a point over the place of units, and also over every third figure, from thence, to the left hand in whole numbers, and to the right in decimals), find the nearest less cube to the first period ; set its root in the quotient, and subtract the said cube from the first period ; to the remainder bring down the second period, and cull this the resolvend. 2. To three times the square of the root, just found, add three times the root itself, setting this one place more to the right than the former, and call this sum the divisor. Then divide the resolvend, wanting the last figure, by the divisor, for the next figure of the root, which annex to the former ; * The reason for pointing the given number into periods of three fi- gures each, is because the cube of one figure never amounts to more than three places. And, for a similar reason, a given number is point- ed into periods of four figures for the 4th root, of five figures for the 5th root, and so on. The reason for the other parts of the rule depends on the algebraic formation of a cube : for, if the root consist of the two parts a r I then its cube is as follows: («-{-fc)3 - aa -j- 3a 6-f- 3a6« -f b* ; where a \n the root of the first part a3 ; the resolvend is 3a»6-f 3a^2 -f 63 ; which is also the same as the three parts of the subtrahend ; also the divisor is 3a -7- 3a, by which dividing the first two terms of the resolv- end 3aa 6 + ab* , gives b for the second part of the rool \ and w> on. 86 ARITHMETIC. calling this last figure e, and the part of the root before found let be called a. 3. Add all together these three products, namely, thrice a square multiplied by c, thrice a multiplied by e square, and e cube, setting each of them one place more to the right than the former, and call the sum the subtrahend ; which must not exceed the resolvend ; but if it does, then make the last figure e less, and repeat the operation for finding the subtra- hend, till it be less than the resolvend. 4. From the resolvend take the subtrahend, and to the remainder join the next period of the given number for a new resolvend ; to which form a new divisor from the whole root now found ; and from thence another figure of the root, as directed in article 2, and so on. EXAMPLE* To extract the cube root of 48228*544. 3 X 3» = 27 3 X 3 = 09 Divisor 279 48228-544 ( 36-4 root. 27 21228 resolvend. 3 X 3* X 6 =162 3X3 X6>= 324 ) add 6» = 216 f 3 x 36 J =; 3 X 36 = 108 38988 19656 subtrahend. 1572544 resolvend. 3 X 36» X 4 = 15552 ) 3 X 36 X 4* = 1728 > add 4 3 = 64 J 1572544 subtrahend. 0000000 remainder. Ex. 2. Extract the cube root of 571482-19. Ex. 3. Extract the cube root of 1628-1582. Ex. 4. Extract the cube root of 1332. CUBE SOOT. 87 II. Ib extract the Cube Root by a short Way*. 1. By trials, or by the table of roots at p. 93, dec. take the nearest rational cube to the given number, whether it be greater or less ; and call it the assumed cube. 2. Then say, by the Rule of Three, As the sum of the given number, and double the assumed cube, is to the sum of the assumed cube and double the given number, so is the root of the assumed cube, to the root required, nearly. Or, As the first sum is to the difference of the given and assumed cube, so is the assumed root to the difference of the roots nearly. 3. Again, by using, in like manner, the cuhe root of the last found as a new assumed cube, another root will be ob- tained still nearer. And so on as far as we please ; using always the cube of the last found root, for the assumed cube. EXAMPLE. To find the cube root of 21034-8. Here we soon find that the root lies between 20 and 30, and then between 27 and 28. Taking therefore 27, its cube is 19683, which is the assumed cube. Then 19683 21035-8 2 2 39366 42071-6 21035-8 19083 As 60401-8 : 61754-6 : : 27 : 27-6047. 27 4322822 1235092 60401-8) 1667374-2 (27-6047 the root nearly, 459338 36525 284 42 * The method usually given for extracting the cube root, Vi so ev ceedingly tedious, and difficult to be remembered, that \w\ov\* oXtat 88 ARITHMETIC. Again, for a second operation, the cube of this root is 21035-318645155823, and the process by the* latter method will be thus : 21035-118645 &c. 42070-637290 21035-8 ,21035-8 21035-318645 &c. As 63106-43729 : difT. -481355 : : 27-6017 : thediff. -0002^ 0560. conseq. the root req. is 27-604910560. Ex. 2. To extract the cube root of -67. Ex. 3. To extract the cube root of -01. TO EXTRACT ANY ROOT WJIATKVER • Let p be the given power or number, ?* the index of the power, a the assumed power, r its root, r the required root of p. Then say, As the sum of n + 1 times a and n — 1 times p, is to the sum of n + 1 times p and n — 1 times a ; so is the assumed root r, to the required root r, Or, as half the said sum of n + 1 times a and n — 1 times r, is to the difference between the given and assumed powers, approximating rules have been invented, viz. by Newton, Rnphson, Halley, De Lagny, Simpson, Emerson, and several other mathemati- cians ; but no one that I have yet seen is so simple in it form, or seems so well adapted for general use, as that above given. This rule is the same in effect as Dr. Halley's rational formula, but more commodious- ly expressed ; and the first investigation of it was given in my Tracts, p. 49. The algebraic form of it is this: As p j - 2a : a -|- 2r : : r : jr. Or, A* p -(- 2a : p *r a : : r : k *r r ; where p is the given number, a is the assumed nearest cube, r the cube root of a, and n the root of p sought. * This is a very general approximating rule, of which that for the cube root is a particular ca*e, and is the best adapted for practice, and for memory, of any that I have yet seen. It was first discovered in this form by myself, and the investigation and use of it were given at large in my Tracts, p. 45, &c GENERAL ROOTS. 80 so is tlss^sjhjuMd root r, to the difference between the true end iMMn roots ; which difference, added or subtracted, ee the case requires, gives the true root nearly. That ie, (n 4-1) a + (n-l)r: (w+1) p. H-(n-l) a: :r:a- Or, (n + 1) Aa + (n — 1) : p^a : : r : it ^ r. And the operation may be repeated as often as we please, by using always the lust found root for the assumed root, and its nth power for the assumed power a. example. To extract the 5th root of 21035& Here it appears that the 5th root is between 7 3 and 7*4* Taking 7-3, its 5th power is 20730 71503. Hence we have r « 21035-8, n = 5, r = 7-3, and a = 20730-71593 ; then » + 1 . £a -f- n — 1 . £p : p ^ a : : r : r «^ r, that is, 8 X20730-71593 + 2 x 21053-8 : 305-0*4 : : 7 3 : -0213005 3 2 7-3 62192 14779 42071-6 104263 74779 42071 -(J 915252 2135568 2227 1 13 1( -0213605=11^ 7 3=r, add 7 321360- r, true to the last figure. OTHER EXAMPLES. 1. What is tho 3d root of 2 ? Ans. 1-259921. 2. What is tho 3d root of 3214 ? Ans. 14-75758. 8. What is the 4th root of 2 ? Ans. 1 189207. 1 4. Whnt is the 4th root of 97-41 ? Ans. 3 1415909. 5. Whnt is the 5th root of 2 ? Ans. 1-148699. 6. What is the 6th mot of 21035-8 ? Ans. 5-254037. 7. What is the 6th root of 2 ? Ans. 1-122462. 8. What is tho 7th root of 21035-8 ? Ans. 4-145392. 0. What is the 7th root of 2 ? Ans. 1-104080. 10. What is the 8ih root of 21035-8 ? Ans. 3-470328. Vol. I. 13 00 AStTHWCTIC. 1 1. What is the 8th root of 2 ? • • An. l^HK 12. What is the 9th root of 21035*8 1 Ana. S^tHS9. 13. What is the 9th root of 2 ? Ans. 1-080059. The following is a Table of squares and cubes, and also the square roots and cube roots, of all numbers from 1 to 1000, which will be found very useful on many occasions, in nu- meral calculations, when roots or powers are concerned. The use of this table may be greatly extended, either by the addition of ciphers, or by changing the places of the separating points. The following examples will suffice to suggest the method. Root. Square. Cube. 36- 12<J6* 46656- 360- 129600- 46656000- 3600- 12960000- 46656600000- 546- 298116- 162771336- 54-6 2981-16 162771-330 '546 -298116 -162771336 For a simple and ingenious method of constructing tables of square and cube roots, and the reciprocals of numbers, see Dr. Hutton's Tracts on Matheirtotical and Philosophical Subjects, vol, i. Tract 24, pa. 459. A TABU OF 4QUAAU, CUBES, AND ROOTS. 91 square. Square ttooi. ouce ttoot* 1 " 1 1 1 0000000 1 000000 i 4 8 14142136 1 259921 9 27 1-7320508 1-442250 4 16 64 2 0000000 1-587401 5 25 125 2 2360680 1 709976 6 36 216 24494397 1 817121 7 49 343 2 6457513 1-912931 8 64 512 28284271 2 000000 9 81 729 30000000 2 080084 10 100 1000 3 1622777 2 154435 11 121 1331 33166248 2 223990 12 144 1728 3 4641016 2289428 13 169 2197 3 6055513 2 351335 14 196 2744 37416574 2410142 15 225 3375 38729833 2 466212 16 256 4096 40000000 2519842 17 289 4913 4 1231056 2 571282 IS 324 5832 4 2426407 2 620741 10 861 6359 4 35SS9S9 2 668402 30 400 SO 00 4 4721360 2714418 21 44 1 9261 4 5825757 2 758924 22 434 10643 4 6904158 2 802039 23 529 12107 4 795S315 2-S43867 24 576 13S24 4 8989795 2 884499 25 625 15625 50000000 2 924018 SO 676 17576 5 0990195 29G2496 27 729 19683 5 1961524 3 000000 28 7S4 21952 5 2915026 3036589 29 841 24389 5 385164S 307231 7 30 900 27000 54772256 3 107232 31 9G1 29791 5 5677644 3 141381 39 1024 32768 5656S542 3 174802 33 1089 35937 5 7445626 3207534 » j'l 1 1 'iJi 1 1<JD 35 1225 42875 59160798 3271066 36 12D6 46656 60000000 3 301927 37 1369 50653 60827G25 3-332222 38 1444 54S72 6 16 14140 3 361975 39 1521 59319 62449980 3 391211 40 1600 61000 6-3245553 3 419952 41 1681 69921 64031242 3 448217 42 1764 7408* 6-4807407 3476027 43 1849 79507 6 5574385 3 503398 44 i<m 85184 6 0332496 3 530348 45 2025 91125 670S2039 3 556893 46 2116 97336 0-7823300 3 583043 47 2209 103823 68556546 3603826 48 2304 110592 , 6 9282032 M02 117649 70000000 ^ si 2500 / 125000 70710678 \ 3 AJUIBMETtC. Number. Square. Cube. Square EooL , Cube Root. 51 260] 132651 7 1414284 3-708430 52 2704 1-1 CO US 72111026 3 732511 53 2300 14S877 7-2801099 3756286 54 2916 157464 73484692 3779763 55 3025 16.G375 7 4161985 3802953 56 3136 i 175616 74833148 3 825862 57 3249 185193 75498344 3 848501 58 3364 195)12 7 6157731 3870877 50 34*1 205379 7 681)457 3 892996 60 3600 216C0D 77459667 3 914863 61 3721 226981 78102497 3936497 62 3844 238328 76740079 3-957883 63 3969 250047 7*9372539 3979057 64 4096 262144 800C00OO 4 000000 w 4225' 274625 80622577 4 020726 60 4356 287496 S12403S4 4041240 67 4489 300763 8 1 85352S 4061548 6S 4624 314432 8 2462113 4-081655 69 4761 3285G9 83066239 4101566 70' 4900 343000 83666003 4 121285 71 son 357911 842G149S 4 140818 72 5184 373248 8-4852814 4 160153 73 5329 399017 8*5440037 4 179339 74 5476 405224 86023253 4 198336 75 5625 421875 86602540 4217163 76 5776 438976 87177979 4 235824 77 5929 456533 S 7749644 4 25432J 78 6084 474552 SS317609 4 272659 79 6241 493039 8 8881944 4290841 80 6400 512000 S 9442719 4 308870 61 6561 531441 90000000 4 326749 82 6724 551368 90553S51 4 344431 83 0889 571787 & 11 04336 4 362071 84 7C56 51)2704 Ai 1 PSl £ 1 4 4*379519 65 7225 614125 92195445 4396830 86 7396 636056 2736185 4414005 87 7569 658503 9 3273791 4431047 88 7744 6S1472 936CS315 4447960 89 7921 704969 9 4339811 4464745 90 8100 720000 9486S:^30 4481405 91 £281 753571 95393920 4-497941 92 64JG4 77S6SS 9 5916630 4-514357 93 S649 S04357 9-6436508 4530655 94 8836 8305S4 96953597 4 546836 95 9025 857375 97467943 4 562903 96 9216 884736 9*7979590 4 578857 97 9409 912673 98488578 4-594701 98 9604 941192 9S994949 4-610436 99 / 9S01 970299 L 9949S744 . 4626065 100 1 10000 10000GO \ 10 moggou •WAftlS, CUUS, AlfD BOOTf . 93 Number. Square. Cube. Square Root. Cube Root. 101 10201 1030301 10' 04 98756 4657010 102 10404 106120S 100995049 4 672329 103 lCGGO 1092727 10 1488916 4 687548 104" 10813 1124>G4 1 /> i AC) Art 10' 1930390 4 702t)b9 105 1 1 02 1157625 1 J"l J Aft t J lO 10 2469508 4-717694 100 I12CG 11010IG 10 23o6301 4*732624 107 11449 1225043 JO 3 I40J50* 4 747459 loa 11064 1250712 10 3923048 4762203 109 11 SSI 1295029 104403065 4 77685b 110 12100 1331000 10 4880885 4 791420 111 12321 1367631 10 5356536 4 HO 5896 112 12544 1404928 105830052 j .a tin ao * 4 82 0284 113 12700 1442S97 10 6301459 A Q'li CQQ 4 Wo45S9 114 12996 1481544 10 67707S3 4o4SS0S 115 13225 1520875 10 7239053 4 00^944 116 13456 1560896 1 A^mv aaa nit 1 07 703 2 06 J V ""t" ■ 1 1 1 f 1 117 13698 1601613 10 8166538 4 3905173 11S 13924 1643032 10 8627805 4 yu4obo 119 14161 1685159 10'90S7121 j,ni bjjqe 4 y 1 ooSo 120 14400 1728000 109544512 121 14641 1771561 i rooooooo 4 946088 122 149S4 181584S 186CS67 11 0453610 4 959676 123 ; 15120 I 1 .Anne 'ic r I I 0905365 j A^At AA 4 973 19U 4 98663 1 124 15376 1 A A** Cm 1906624 11 1 r\r rnriH 11 1355287 15625 1953125 T 1 . i cm a a a a 1 1 1803399 o uuuuuu 126 15&7G 2000376 1 I 2249722 o U ] a/Ho 127 16129 2048383 1 1 2694277 t o OZOfr^b 129 16331 2007152 11 '3137085 U3Ubb4 129 16641 2146689 I IOC TO i c T I I 3o7SI67 o 05^774 139 16900 2107000 11 4017543 065797 131 17161 2248091 11 j j *~ ^ A A 1 11*4455231 5 078753 132 17424 2299908 1 1 4891253 c a a l i* m a 5 091643 133 17639 2352637 1 lo325626 5 104469 134 17956 2406104 11-5758369 5 117230 135 18225 2460375 11'6189500 5 129928 136 18496 251545G 11-6619038 5 142563 137 18760 2571353 11 7046999 5 155137 138 19044 2628072 11 7473444 5 167649 130 19321 2685610 11*7898261 5 1S0101 140 19600 2744000 11 8321596 5 192494 111 198S1 2803221 U 8743421 5 204828 142 20164 2863283 , 11 9163753 5217103 143 20440 2924207 i 11 9582607 5 220321 144 20736 23859S4 12 0000000 5241483 145 21025 3048625 V2 0415946 5 253583 146 21316 3112136 120830460 5 265637 147 21609 3176523 12 1243557 5277632 148 21904 3241792 12 1655251 5-2S<ft&n 149 22201 3307949 122065556 150 J 22500 j 3375000 j 12-24744S7 ^» 1 rTYiltar Jl ^ LI 114 Ucl . Souafe Cube, Snuare Root CuhA Rrmt VUMC HWl. 151 22801 3442951 12 2882037 5325074 15^ 23104 35 11808 12-3288280 5336303 153 23409 3581577 12 3693169 5 348481 154 23716 3652264 124006736 5-360108 l 155 24025 3723875 12449S996 5 371685 156 24336 3796416 12 4899960 5383213 157 24649 3809S93 12 5299641 5 394691 158 24964 3944312 125693051 5406120 159 25281 4019679 12 6095202 5 417501 160 25600 4096000 126491106 5428835 Ml 26921 4173281 126885776 5440122 16a 26244 425152S 12 7279221 5 451362 163 26569 4330747 127671453 5462556 164 26896 4410944 12 8062485 5473704 165 27225 4492125 128452326 5494806 166 27566 4574296 12 8840987 5495865 167 27SS9 46*7463 12 922S480 5506879 168 4741632 129614814 5517S4S 169 2S561 4826309 13 0000000 552S775 170 28900 4913300 13 0384048 5 539658 171 29241 5000211 130766968 5550499 172 29584 5088448 13 1148770 5 561298 173 29929 5177717 13 1529464 5572055 174 30276 5268024 13 1909060 5-592770 175 30625 6359375 132287566, 5 593445 176 30976 5451776 13 2664992 5604079 177 31329 5545233 133041347 5 614673 178 31684 5639752 13 3416641 5625226 179 32041 6735339 13-3790882 5 635741 180 32*00 5932000 13 4164079 5 646216 181 ,32761 5929741 134536240 5656653 182 33124 6028568 134907376 5667051 183 33439 6128487 13 5277493 5 677411 194 0^29,0 04 Jo Ot>4bbUU 5 687734 185 34225 6331625 13 6014705 5699019 186 34596 6434856 136381817 5 708267 187 34969 6539203 136747943 5718479 188 35344 6644672 13 7113092 5 729654 189 35721 6751269 13 7477271 5738794 190 3G100 6959000 137840488 5749S07 191 36481 6967871 138202750 5758965 192 36864 70778S8 13 8564065 5768998 193 37249 7189057 13 8924440 5 778996 194 37636 7301384 139283883 5 789960 195 38025 7414875 139642400 5 798890 196 38416 7529536 1400000(10 5S08786 197 3S809 7645373 140356688 5 818648 198 | 39204 7762302 140712473 5828476 199 * 39601 7880599 \ vmm 2W / 40000 j 8000000 ■40AAIS, CUBES, AND ROOT*. /Number. 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 349 Square. / 40401 40804 41209 41616 42025 42436 42849 43264 43661 44100 44521 44944 45369 45796 46225 46656 47089 47524 47961 48400 48841 49264 49729 50176 50625 51076 51529 51984 52441 52900 53361 53S24 54289 54756 55225 55696 56169 56644 57121 57600 58061 58564 59049 59536 60025 60516 61009 61504 62001 62500 Cube. 8120601 8242408 8365427 8489664 8615125 8741816 8869743 8998912 9123329 9261000 9393931 9528128 9663597 9800344 9938375 10077696 10218313 10360232 10503459 10648000 10793861 10941048 11089567 11239424 11390625 11543176 11697063 11852352 12008989 12167000 12326391 12487168 12649337 12S12904 12977875 13144256 13312053 13481272 13651919 13824000 13997521 1417248S 14348907 14526789 14706125 14886936 15069223 15252992 15438249 15625000 1 Square Root. Cube Rogt 141774469 142126704 142478068 142828569 14 3178211 143527001 143874946 144222051 14 4568323 144913767 145258390 14 5602198 14 5945195 146287388 146628783 146969385 14 7309199 147648231 147986486 14 8323970 148660687 148996644 14 9331845 149666295 15 0000000 15 0332964 150665192 15 0996689 15 1327460 15- 1657509 15 1986842 152315462 15 2643375 15-2970585 153297097 15-3622915 15-3948043 15 4272486 15 4596248 15-4919334 15-5241747 155563492 15 5884573 156204994 156524758 156843871 157162336 157480157 157797338 15-3113883 5857766 5 867464 5877130 5 886765 5896368 5 905941 5915482 5924992 5 934473 5 943922 5953342 5962731 5972091 5 981426 5990727 6000000 6009244 6018463 6027650 6 036811 6045943 6055048 6064126 6073178 6082201 6 091199 6 100170 6 109115 6 118033 .6 126925 6 135792 6 144634 6 153449 6 162239 6 171005 6 179747 6- 188463 6 197154 6205822 6 214465 6223084 6231679 6240251 6248800 6257325 6265826 M Number. Square, Cube. Square Root. 1 Cube Root. 251 ■ 1 63001 15813251 15 r S429 795 1 b o07994 252 63504 16003G08 15 8745079 6'31 6359 253 64009 1 C -1 " 1 t\ J a 1 -*t*f 16134277 1 5' 059737 024704 254 64516 1 1 f r>fi A a j 16387064 15 9373775 b 3 302 6 net 65025 1 CCDl 'ITS lbo3 137o 159b37194 b 44 1*3 2 b 256 65536 lb77721b lb 0000000 b <H9b04 257 06049 loy7459o i c . ao 1 a 1 a el 1 b Oo 1 2) 9o b o57ool 2 DO 66564 1 C AC ill 1 2 4 lb Ug JJ734 o t5bouyo 259 67081 17d*3y79 lb 0934769 b i>74o 1 1 260 67600 17576000 lb 1245155 332504 261 68121 17779591 16 1554944 6 390676 262 68644 17934729 161964141 6 39S829 263 69169 1S191447 16 2172747 6 406959 264 69696 18399744 162480768 6 415003 265 70225 1360962:> 16*2788206 6 423153 266 70756 18821096 ^ n a a p a n a 163095064 a , t a t n ftn b 43122S 267 nr 1 nan 71289 19034163 16 3401346 G 4392 / * 268 71824 1 A A *. O a !i »i 19249332 16 p37070o5 6 447305 269 72361 19465109 te 1 Ai i rtr 164012195 6 455315 OTA lybSJOOO 164316767 ( 6 463304 271 73441 iyyo25i i 16 4620776 6 4712i4 272 73 934 20123643 16 4924225 6 479224 273 74529 20046417 16 52271 16 6 4S7154 274 75076 2O570S24 lb 5529454 6 495065 275 75625 207 9 6,3 7o 16-5831240 6 5029^6 J76 76176 21 024576 lb'6 132477 6 510^30 277 76729 it,/; IQ'JtTn iD'O-ldiJl iKJ b a 1 ob34 275 772a 4 4I-io4yD2 1 D O r ooo20 b *>2b519 279 77341 Ol Tl 7KQQ ID / JJ b Qd4iJo5 230 • 7Q 4 A A 7&4i>u 41 4 J4UUU 1 ifi . "i " J fl A A tl 1 (W*5d2O0O 542133 43 1 99 1 QCtHi 1 ID f OoUO-^O i? - c j* A,A 1 n - ~ 70*194 1 ,fi 7Q9Q^1A d Do 4 b7dt 283 80089 226J51S7 168226038 6 565415 234 80656 22906304 16-8522995 6573139 285 81225 23149125 16 8819430 6-580844 286 81796 23393656 16-9115345 6588532 287 82369 23639903 16-9410743 6-596202 288 82944 23987872 169705627 6603354 289 53521 24137560 17-0000000 6 611489 290 84100 24389000 17-0293864 6 6191 00 291 84631 24642171 17-0587221 6 626705 292 85264 24897088 1708S0075 6634287 293 85849 25153757 171172428 6641852 294 86436 25412184 1T1464282 6649399 2&5 87025 25672375 171755640 296 87616 25934336 172046505 297 SS209 26198073 172336S79 6671940 S&S / 3SS04 26463592 at* 89401 26730S99 300 / 90000 27000000 SQ.UARE8, CUBES, AND ROOTS. 97 Number. Square. Cube. j Cubo Root. oUX one a 1 yUbUl 0*70^1 IftAl 1 ■ *> 1 a •* "-\ i ^ 1 1 u 1 / oy aiKS 1* J ZVJ4 Z 7 04obUS 1 / »5 / ■*» I -1 / — l*«-7AQ 1 7*i b tyiji to QUO' Ql A AO •>7tt 1 4il 07 1 17 lA<".^iQ'i , > 1 n-7 1 a !%7n 0114 9Z4XO *r>Uy4-l )4 ! b /-o;*.jl OUO 1 7'J.H !•> l0-> ' 1 l 4b i^-ii'Z - b / »3 1 1 b OUU 0OO0O z*>b^-5i.> I b 1 7- IOOk'%%7 b / Oobbo QAT 94Z4;f OwO'i 4 i t M 1 1 •)£ 1 •! 1 O'J b HOyy/ OAO •JUO y4oo4 2'JJISI 12 i*.r: ((HIOh:Q 1 / t\t — • ^ b / o»5o 1 QAA ouy OI% 4 Q 1 y»481 zl'olMojy b /b0bi4 olO 96100 2979 1 00(1 1 / bUOSJ oil b /b/syy Ql 1 ol 1 o(JU802.>l 1 / 00 ) 1 y^i 1 rfi^T^ 1 ftft oiZ y/o44 o\K57132S J 7 bbo >J1 / . b /S24zd «ll *1 Old oUob4297 17 oU LNUbU ioybbl ol4 QDSnA yoosib *J A ft X 1 1 1 < ■< ousjoy 14 * 1 / / JUO l.ll b #ub.-?v± QIC olo yy^zo OlJOO.S70 17 /4->Jo:M | b olMUyj olo yyoob 31554496 1 T-T^TlIM wCki 1 1 / / / Oo-T^*) ' /••CI 1 .'JC 4 boll Jh4 1 An 4 co Qlti': 1 •> 0I000U 1 .3 1 7-ViA < iO*>tt /< vim b -^4b J 010 1U1 iz4 o.)i KT 1 -JO 1 / OOJO.) 1.) b oj..)bJ4 olU 1 A I T/l 1 101 / Ol tSJ In 17o9 1 "1 \ 1 1 1 J / nUH.) /II b 80 J / 7 1 oon o^u / b800l» 0^1 1 A'<AJ. 1 00U / b 1 u L 1 / f 1 bt / — J 17 no 1 b n-i tv)jz\ OZJ i 'MU-.H od-*r*bi - s 1*7 fl « 1 '> \<< » b r».J'l 1 J4 100 O40 1 A 1 "*Oft odbii^vtb # 1 / .» / iil.'U? b nO \£l & oZ4 1 A i A"? ft '1 1 A 1 OOO 1 1 QlUli'lfiMllM *io % 0^0 l v'OO JtO > •.••iOj.il-,-) 1 ~ « J«- # I ... 1 1 r: w7 1 .4 '•ton 1 nfW7<: A UU^ f U O iU-JO.I / b b *"io(33 f 1 w As;:; Hl'l ft-K^QJ 1 O b o^y*i 1 ;| 1 rt7%ftJ. 1 V/ 1 »JO-± tc-ii ii/70'? l»p IM'// yjt* b «M/b-joo ozy '\~xfl 1 1 OCA b JU-HOQ 1 option -*>017( win ivirrVK]')' ' 1 A IOQ b i' 1 u-j^o 109561 Ml'i'^fi 0"»<i 1 ] s* 1934054 #VA 1 7*-lQA oo<s 1 1022 l yvf'/J I • J li ~ fi QO fJ^X 333 1 lOSsO 18'24S2S76 • i;n*i 1 oni «'•)[ 1 334 111556 37259704 18 2756669 ' 6 938232 335 11222.3 37595375 18 303U052 , 6 945149 336 112896 37933056 1S 330302S ; 6 952053 337 113.i69 3S272753 18 357559S ; 6 958943 338 11 1214 38614472 jS 3s 17763 ! 6 965819 339 114921 3.S95S219 i s 4 H 9526 ! 6 972683 340 115600 39304000 18-4390SS9 i 6 97"532 341 116281 3965 1S21 18 4661S53 6986368 342 116964 40001 6 ss Is : 93:120 ! 6993191 343 11 76-1 9 40353**0? 18 52;:::592 j 7O00000 344 118.336 4070 : .^4 18 5172370 i 7006796 345 119025 41063i>25 18 5741756 | 7013579 346 119716 41421736 18 6010752 j 7020349 347 120409 4 178 1923 18 6279360 j 7 02710ft 348 121104 18-6547581 1 849 121*01 4250S549 18 6S15417 \ 7-^V)o^\ 850 I 122600 : 42875000 18 70S2S69 Vol. I Number. Cube. Square Root. Cube Root. ! 351 123201 43243551 18*7349940 7054004 352 123904 43614208 187616630. 7060696 353 124609 43936977 18-7882942 7067376 354 , 125316 44361964 188148S77 7074044 355 126025 44739875 18 8414437 7080609 356 126736 45118016 1B8679623 7 087341 357 127449 45499293 1SS944436 7093971 35S 128164 45882712 IS 9208879 7100588 359 128881 46268279 189472953 7 107194 360 129600 46656000 18 9736660 7 113786 361 i 130321 47045881 19 0000000 7 120367 362 131044 4743792S 19 0262976 7 126936 363 131769 47832147 19 0525589 7133492 364 132496 48228544 19 0787840 7*140037 365 133225 4S627125 19' 1049732 7146569 366 133956 49027896 19 1311265 7 153090 367 134689 49430863 19 1572441 7 159599 363 135424 49836032 19 1333261 7166096 369 136161 50243409 19 2093727 7172580 370 136900 50653000 19*2353841 7179054 371 137641 51064811 19 2613603 7185516 372 138384 51478S48 19 2873015 7 191966 373 139129 51895117 193132079 7 198405 374 139S7G 52313624 19 3390796 7 204832 375 140625 52734375 193649167 7-311448 376 141376 53157376 19 3907194 7 217652 377 * 142129 535S2633 19*4164878 7 224045 ; 378 142884 54010152 19 4422221 7230427 379 143641 54439939 194679223 7236797 380 144400 54872000 194935887 7 243156 381 145161 55306341 19 5192213 7-240504 3S2 145924 55742968 19 5448203 7255841 383 146689 147456 561S1S87 56623104 195703858 7 262167 384 19 5959179 T2694&2 3S5 14S225 19 0214169 * & **** i'lVf 7-274786 386 14S996 57512456 196468827 7281079 397 149769 57960603 19 6723156 7287362 388 150544 5841 J 072 19 6977156 7293633 3S9 151321 58S63869 197230829 7-299894 390 152100 59319000 197484177 7 306143 391 152S81 59776471 19T737199 7 312383 392 153664 60236289 19 7989899 7318611 393 154449 60698457 198242276 7324829 394 155236 61162984 19 8494332 7-331037 395 156025 61629875 198746069 7 337234 396 156S16 62099136 198997487 7*343420 3t)7 157609 62570773 199248588 7349597 39? 158404 63044792 19 9499373 7355762 399 1 159201 6352 1199 19 9749844 7361918 400 j leoooo 64000000 200000000 SQUARES, CUBES, AND ROOTS. 99 uTTkuer, Square. v>ube< Square Root. Cube Root, 401 160801 64481201 20 0249844 7 37419S 401 161604 64964808 200499377 403 404 162409 65450827 20 0748599 7386437 163216 05939264 20 0997512 7 392542 405 164025 66430125 20 1246118 7 39H636 406 16493G 669234 16 20 1494417 7-404720 407 165649 67419143 20 1742410 7 410795 408 166464 U7''l 1131 2 20 1 990099 7 41 6859 409 167291 68417929 SO 9237481 7-422 r na. | 1 Jd H.' J 7t 410 168100 7-4 28050 411 163921 60426 531 7-434994 412 169744 69934 52S 20 2Q77A3I 7 44 1019 413 170569 70444997 2032240 14 7-447034 414 171396 70957944 20 3469899 7453040 415 172225 71473H75 20 3715488 7 459036 416 173056 7' 4 65022 417 173889 7251 1713 20 4205779 7 470999 418 174724 73034632 20 4450483 7 476966 419 175561 73560059 204694895 7 4S2924 420 176400 740SS000 20 4939015 7 4B8H72 421 177241 74618461 205182845 7 49481 1 422 17S0S4 75151448 20 5426380 7 500741 423 17S929 75G969C7 7 506 tiG 1 424 179776 76225024 20-59 1 2603 7 512571 425 180625 76765625 20-6)55281 7 51S473 426 181476 77308776 20 6397674 7 524365 427 182329 77854483 20 6639783 7-530248 423 183184 78402752 20 6Stf 1609 7 536 J 21 429 18404 I 78953589 20 7123152 7 541986 430 184900 79507000 20 7364414 7-547^42 431 185761 80062991 20 7605395 7 553688 432 186624 80621568 207846097 7559526 433 187489 81182737 208086520 7 565355 434 188356 61746504 208326667 7'57 1174 435 189225 82312875 20S566530 7 5769*5 436 190096 82381856 208806130 7582786 437 190969 83453453 20 9045450 75wy579 438 191844 84027672 20 9284495 7594363 439 192721 846045 1 9 209523268 760013S 440 193600 85184000 209761770 7605905 441 194481 85766121 21 0000000 7611662 442 195364 863503S8 21 0237960 7 617412 443 196249 86938307 21 0475652 7623152 444 197136 87528384 210713075 7 628884 445 198025 88121125 21 0950231 7634607 446 198916 88716536 211187121 7640321 447 199809 89314623 21 1423745 7646027 448 200704 21 1660105 \ tmviw 449 201601 90513849 21-1896201 450 I 202500 f 91125Q0Q | 21 2132034 \ TWWM t-rrx- \ 100 ARITHMETIC. iNJ 11 m nor Cube. 451 203401 91733851 21 2367606 7668766 452 204304 9234540b 21-2602916 7674430 453 205209 92959677 21 2837967 7 680086 454 206106 93576664 21 3072758 7685733 455 207025 94196375 21 3307290 7691372 456 207936 94818S16 21 3541565 7697002 457 208849 95443993 21 3775583 7-702625 458 209764 96071912 21 4009346 7 708239 459 210681 96702579 21 4242853 7713845 460 211600 97336000 214476106 7 719442 461 212521 97972181 21 4709106 7725032 462 213444 9S611128 21 4941853 7 730614 463 214369 99252847 21 517434S 7-736188 464 215296 99897344 21 5406592 7741753 465 216225 100544625 21 5638587 7 747311 466 217156 101194696 21 5870331 7 752861 467 218089 101847563 21 6101828 •7-758402 468 219024 102503232 21-6333077 7763936 469 219961 103161709 21 6564078 7769462 470 220900 103823000 21-6794834 7774980 471 221841 104487111 21 7025344 7780490 472 222784 105154048 21 7255610 7785993 473 223729 105823817 217485632 7791487 474 224676 106496424 21 7715411 7796974 475 225625 107171875 21 7944947 7-802454 476 226576 107850176 21 8174242 7807925 477 227529 108531333 21 8403297 7813389 478 22*184 109215352 21 8632111 7818846 479 229411 109902239 21 88606S6 7S24294 480 230400 110592000 21-9089023 7829735 481 231361 111284641 21 9317122 7835169 482 232324 111980168 21 9544984 7840595 483 233289 11267S587 21 9772610 7 846013 484 234254 113379904 22 0000000 7851424 485 235225 114084125 220227155 7856828 4S6 236196 114791256 220454077 7862224 487 237169 115501303 220680765 7 867613 488 238144 116214272 220907220 7872994 489 239121 116930169 22 1133444 7878368 490 240100 117649000 22 1359436 7883735 491 241081 118370771 22- 1585 198 7889095 492 212064 119095488 22 1810730 7894447 493 243049 119823157 22-2036033 7899792 494 244036 120553784 22 2261108 7905129 495 245025 121287375 222485955 7-910460 496 246016 122023936 222710575 7-915783 497 247009 122763473 22-2934968 7 921100 498 24S004 123505992 223159 Vifc 499 / 249001 124251499 oOO l 250000 1 125000000 \ SQUARES, CUBES, AND ROOTS. 101 Number. Square. — Cube. Square Root. Cube Root. u 501 251001 125751501 223330293 7942293 502 252004 126506008 224053565 7947574 503 253009 127263527 224276615 7-952848 504 254016 128024064 224499443 7-958114 505 255025 12S7t*7625 224722051 7963374 506 256036 129554216 224944438 7968627 507 257049 130323843 225166605 7973873 50S 25S064 131096512 225333553 7 979112 509 259081 131872229 22 5610283 7984344 510 260100 132651000 22 5831796 7989570 511 261121 133432831 226053091 7994788 512 262144 134217728 226274170 8000000 513 263169 135005697 226495033 8005205 514 264196 135796744 22 6715681 8010403 515 265225 136590875 22 6936114 8015595 516 266256 137388096 22 7156334 8020779 517 267289 138188413 22 7376340 8025957 518 26S324 138991832 227596134 8 031129 519 269361 139798359 227815715 8036293 520 270400 140608000 228035085 8041451 521 271441 141420761 228254244 8046603 522 272484 142236643 228473193 8051748 523 273529 143055667 228691933 8056886 524 274576 143877824 22 8910463 8062018 525 275625 144703125 22 9128785 8067143 526 276676 145531576 229346899 8072262 527 277729 146363183 22 9564806 8077374 528 278784 147197952 22 9782500 8082480 529 279841 148035889 230000006 8087579 530 280900 148877000 23 0217289 8092672 531 281961 149721291 23 0434372 8097759 532 283024 15056S768 230651 252 8 102839 533 284089 151419437 230867928 8 107913 1 •J^Z / OOUi Q.i 1 OOQA 535 286225 153130375 23- 1300670 8 118041 536 287296 153990656 231516738 8123096 537 286369 154854153 23 1732605 8128145 538 289444 155720872 23 1948270 8 133187 539 290521 156590S19 23 2163735 8138223 540 291600 157464000 232379001 8 143253 541 292681 158340121 232594067 8-148276 542 293764 159220088 232808935 8 153294 543 294849 160103007 233023604 8- 158305 544 295936 160989184 233238076 8 163310 545 297025 161878625 23 3452351 8 168309 546 298116 162771336 233666429 8- 173302 547 299209. 163667323 23 3880311 8- 178289 54S 300304 164566592 234093998 549 301401 165469149 23 4307490 S-\88*44 550 1 302500 I 166375000 23-4520788 8-19**1* ARITHMETIC. Number. Square. Cube. Square Root* Cube Root 551 303601 167284151 23 4733892 8 198175 553 304704 168196608 234946802 8 203132 553 305309 169112377 23 5159520 8208082 554 306916 170031464 23 5372046 8 213027 555 308025 170953875 235584380 8 217966 556 309136 171879616 235796522 8222898 557 310249 172308693 23-6008474 8227825 558 311364 173741112 23 6220236 8 232746 559 312481 174676879 23-6431303 8 237661 560 313600 175616000 23 6643191 8242571 561 314721 176553481 23 6854386 8247474 562 315844 177504328 237065392 8252371 563 316969 178453547 23 7276210 8257263 564 318096 179406144 237486842 8262149 565 319225 180362125 237697236 8 267029 566 320356 181321496 237907545 8 271904 567 321439 182284263 23 8117618 8 276773 568 322624 133250432 23-8327506 8-281635 569 323761 184220009 23 5537209 3236493 570 324900 135193000 23-8746728 8 291344 571 326041 186169411 2389560(53 8 296190 572 327184 187149248 23 9165215 8 301030 573 323329 183132517 23 9374184 8305865 574 329476 189119224 239532971 8 310694 575 330625 190109375 23 9791576 8315517 576 331776 191102976 240000000 3320335 577 332929 192100033 240208243 3 325147 578 3340S4 193100552 24 04 16306 8 329954 579 335241 194104539 24 0624183 8 334755 580 336400 195112000 24 0831892 8339551 581 337561 196122941 24 1039416 8 344341 582 333724 197137368 24 1246762 8 349126 583 339889 198155287 24 1453929 8353905 584 341056 1% lbbU919 W d&3o78 585 342225 200201625 24 1867732 8363446 586 343396 201230056 24 2074369 8-368209 587 344569 202262003 24 2230829 8372967 588 345744 203297472 24 2487113 8 377719 589 346921 204336469 24 2693222 8-382465 590 348100 205379000 242899156 8387206 591 349281 206425071 24 3104916 3-391942 592 350464 207474688 24 3310501 8396673 593 351649 208527857 243515913 8-401393 594 352836 209584584 24-3721152 8406118 595 354025 210644875 24 3926218 8 410833 596 355216 2 11 703736 244131112 8 415542 597 356409 212776173 24 4335334 8420246 598 357604 | 213847192 244540335 . 8 424945 599 f 358801 214921799 24 47447^ 600 I 360000 \ 216000000 24*M4fiOT4 SQUARES, CUBES, AND ROOTS. 103 Numbtf. Square. Square KooL Cube Hoot, 601 361201 217081801 24 5153013 | 3 439010 602 362404 218167208 24 5356333 8 443688 603 363609 919256227 24 5560583 8 448 360 604 364816 220348864 24 5764115 8 453028 ! 605 366025 221445L25 24-5967473 3 457691 60S 367236 222545016 24 6170673 8 462343 607 36S449 223648543 24 6373700 8 467000 60S 369664 224755712 24 6576560 8 471647 609 37088 1 225366529 24 6779254 8 476289 610 372 1 00 226 9S 1000 24698 1781 3480426 611 373321 228099131 24 -7 134 142 8 43 55 58 612 374544 229220928 24-7386338 8490 185 613 375769 230346397 24-7538363 8 4 94 806 614 376996 23 1475544 24-7790234 8 499423 615 378225 232608375 247991935 ft 504035 616 379456 233744396 24-31 93473 8508642 617 330639 234885113 24-8394847 8-513243 618 381924 236029032 24 3596053 8 517840 619 3S3161 237176659 24 3797106 3-522432 620 384400 236328000 24-8997992 8 527019 6 + >l 38564 1 239483061 24 9198716 8 531601 622 336884 240641848 24 9399278 8 536178 623 388129 241804367 24 9599679 8510750 624 3S9376 24297QG24 24 9799920 8 545317 625 390625 244140625 25 0000000 8-549379 626 391876 245314376 25 0199920 8554437 627 393129 246491833 25 0399681 8 558990 623 394384 247673152 25 0599282 8563533 629 395641 243858189 25079S724 8 563081 630 396900 250047000 25 0993008 85726 19 631 398161 251239591 25 1197134 3 577152 632 399424 252435963 25 1396102 8531631 633 400689 253636137 25 1594913 8 536205 634 401956 254840104 25 1793566 8-590724 635 4032*5 256047875 25 1992063 8595238 636 404496 257259456 252190404 8-599747 637 405769 25S474S53 252388539 3604252 638 407044 259694072 25 2536GJ9 8608753 639 408321 260917119 252784493 8613248 640 409600 262144000 25 2982213 8 617739 641 410881 263374721 25 3179778 8622225 642 412164 2646092S8 25-3377189 8626706 643 413449 265347707 253574447 8631183 644 1 414736 2670399*4 253771551 8635655 645 416025 263336125 25 3968502 8 640123 646 417316 269536136 25 4165301 8-644585 647 418609 570340023 25 4361947 3649044 648 419904 272097792 25-4558441 649 421201 : 273359449 j 25-4754784 I 8657946 050 j 422500 j 274625000 j 25 4950976 \ 8 Wt 104 ARITHMETIC. Number. Square, Cube. Square Root. Cube Root. 651 423S01 275894451 25 5147016 8666831 652 425104 277167808 25 5342907 8671266 653 426409 278445077 25 5 538647 8-675697* 654 427716 279726264 (CtC C.TO #n*l»w 25 5734 237 S 680 124 655 429025 £\<Cjr 1 111 1 O T £ 28101 14 7o 25 5929678 8 684540 656 430336 282300416 25 6124969 8 68y96i> 657 431649 25 63201 12 8 69337b 653 432964 2848903 12 256515107 8 6SJ7784 659 434291 2obl91 1 TSJ Ac,;' nfj'i^v ftco 8 7l>£ loo 660 4d5h00 not i ncAAn 2a7496000 256904652 b r Ubao 7 661 436921 2 8 88 04781 257099203 8 7109S3 662 663 43S244 290117528 25 7203607 8 715373 439569 fin l ^ n j a j pv 291434247 25 7487864 8 7197^9 664 a. incinc *4U$Ub 292754944 29407962a 25 7681975 8724141 665 A A a aa r 442225 25 7875939 25S069758 8 728518 G66 443556 295408296 8 732892 667 444889 296740963 25 8263431 8737260 GtiH 446224 298077632 258456960 8 741624 669 447561 299418309 25 S6 50343 O wf 4 t A£5 ^ 8 74 59 So 670 A A D AA/\ 44S900 300763000 25 8o43582 O'7o0o40 671 450241 3021 1 171 1 25 9036677 A f E J /■fill 8 754691 672 45158! 3 034 G 444 8 25-9229628 s- ?5903S 673 452929 304821217 25 9422435 8 76oo81 674 454276 306182024 25961 51 00 8 767719 675 455625 307546875 25 9307621 S 772053 676 456976 308915776 2G 0000000 S 7/6383 677 45S329 3 1 0288733 2» 0192237 8 780708 67S 459684 31 16G5752 2t> Uo^4*id 1 Q .FT U & f\ A A 679 461041 h 1 lid i ni'i-lA 313046839 2605762S4 □ .frtfin j j ■ 680 462400 o 1 44 o2trO(i ZD u /onUiJb S /HoboU 6S1 4bi37b 1 •> 1 COO 1 O IT 6S2 465124 *5 1 72 14568 nf i i Ei rtfli-f 2b 1 1 J 2*17 8 r>02272 683 46G489 31861 1987 2b 1342GS7 S MJ6572 JO L-j.j...ipji o OI U^DEs 685 469225 321419125 261725047 8S15160 BSC 470596 322828856 26-1916017 8819447 687 471969 324242703 262106848 8 823731 683 473344 325660672 262297541 8S2S009 689 474721 3270827G9 262488095 8-832285 690 476100 32850900G ! 26 267851 1 8836556 691 4774S1 329939371 262868789 S-840823 692 478864 331373S3S 26 3058929 8-845085 693 480249 332812557 263248932 8849344 694 481636 3342553S4 26-3438797 8S53598 695 483025 335702375 26-3628527 8-857849 696 484416 33715353G 26-3818119 8862095 697 485809 338608873 26-4007576 SS66337 698 487204 340068392 26*4196896 8870576 699 488601 341632099 TOO j 490000 343000000 \ MA^bm SQUARES, CUBES, AND ROOTS. 105 Cube. | Square Foot j CuLe Root. 701 491401 344472101 I 264764046 ! 8883266 702 492804 345948088 26 4052826 , 8887488 703 494209 347428927 i 265141472 8891706 704 495616 348913664 35041)2625 265329988 8895920 705 497025 265518361 | 8900130 706 498436 351895816 26 5706605 j 8904336 707 499849 353303243 26 5894716 j 8-908538 i i 708 501264 354894912 26 6082694 8 912737 709 502681 356400*29 26 6270539 8916931 710 504 1 00 357911000 26 6458252 > 8921121 711 505521 359425431 26 6645833 8925308 712 506944 360944 12^ 266833281 8929490 713 50S369 362467097 267020598 8933668 714 509796 363994344 267207784 8*937843 715 511225 365525875 26 7394839 8 942014 716 512656 367061696 267581763 8946181 717 514089 368601813 26 7768557 8950344 718 515524 370146232 267955220 8954503 i 719 516961 371694959 268141754 8958658 j 720 518400 373248000 26 8328157 8962809 | 721 519841 374805361 26 8514432 8966957 ! 722 37636704 H 26 8700577 8 971101 ! 723 522729 524176 377933067 26 8886593 8975240 724 379503424 26-90724H1 1 8979376 725 525625 381078125 26 9258240 8-983509 726 527076 382657 176 26-9443872 8 9*7637 727 52S529 384240583 . 269629375 S 991762 728 529984 3S5828352 269814751 8 995883 729 531441 3S7420489 270000000 9 000000 730 532900 389017000 270185122 9 004113 731 534361 390617SUI 27 0370117 9008223 732 535S24 392223168 270554985 9*012328 733 537289 £.1 U tOiJ 1 41 o.m £1 A M 1 \i i) I b4ol 734 538756 395446904 27 0924 344 9 020529 735 540225 397065375 27 1108834 9024624 736 541696 398688256 27 1293199 9 028715 737 543169 400315553 27 1477439 , 9-032802 738 544644 401947272 27 1661554 9036886 739 546121 403583419 27 1S45514 9040965 740 547600 405224000 27 2029410 9 045041 741 549081 406869021 27 2213152 9 049114 742 550564 40851848H 272396769 9053183 743 552049 410172407 27 2580263 9057248 744 553536 4118307*4 27 2763634 j 9061310 745 555025 113403625 27 2946SS1 9065367 746 556516 i 415160936 27 3130006 : 9 069422 747 558009 416832723 27 3313007 « 9073473 748 559504 418508992 ! 273495K87 / 749 1 561001 . ' 4201S9749 27 367S644 \ 9-QSU-&& L_750 I 562500 I 421875000 j 27 3861270 Vol. I. 15 106 ARITHMETIC. - . — * t ■■ ■ * Number. Square. Cube. Square Root. Cube Root. 751 564001 42o0b47ol 274043792 - — - — ■ 9 089639 752 565504 42525900H 27 42261S4 9 093672 753 567009 426957777 2744084 55 9 + 09770 1 754 568516 428661064 27 4590604 9 101726 755 570025 430368875 27 "4772633 9 1 105748 756 571536 432081216 27" 4954542 9109766 757 573049 433798093 MH jr ill ft A A A #V 27 5136330 9113781 758 574564 I ■ I • - 1 , , - 1 l L^nif ft 1 PV ft n fi 27 5317998 9" 117793 759 576081 437245479 Afv E A n H E J 27 5499 546 9121801 760 577600 43a!J7600U Af¥~. E Jf* OftftPV f 27*5680975 9125S05 761 579121 4 * ATT 1 jn o i 440711081 2 7- 5862284 9 129606 762 580644 44245Q72S 27 H 6043475 9-133803 763 582169 444194947 27 6224546 9*137797 764 58369b 445943744 27-6405499 9*141788 765 585225 447697125 27 b 586334 9' 145774 766 586 7a6 449455096 27-6767050 9' 149757 767 588289 Jt(£TA*tWji.jtA 451217663 276947648 9 153737 76S f£!}ft£?A J 589824 452984832 27 7128129 9157714 769 591361 454756609 27-7308492 9161686 770 r fill nriA 592900 456533000 27- 7438739 9' 165656 771 594441 458314011 27-7668868 9 1 169622 772 595984 460099648 27- 7848880 n, -ft Pwn jp ja. n 9- 173585 773 597529 461889917 27 8028775 9 177544 774 599076 463bS4H24 27S20S555 9181500 775 600625 465484375 27*8388218 9 185453 776 602176 467288576 A*V A f nfVM n ^1 27 8567766 9- 189402 777 603729 A £3 A J"ki i\M j nil 469097433 27 8747197 9 193347 778 605284 4709 1 0952 27 8 9265 14 9' 197239 779 bU6341 47272^139 £)■* Ai AFrVi r- 279105715 9201229 780 608400 47 4 DO 2000 Ilk ft AO J 4 2 7 + 9284801 9 205 164 781 bU99bl 4ro379541 279463772 9 209096 782 fill 524 47821 1768 279642629 9 213025 783 j-fc i A AAA 613089 480048687 279921 372 1 9216950 784 614656 481890304 280000000 9 '220873 785 616225 483736625 280178515 9 224791 786 617796 485587656 28 0356915 9228707 787 619369 487443403 280535203 9 232619 789 620944 489303872 28 0713377 9 237528 789 622521 491169069 28 0891438 9240433 790 624100 493039000 28 1069386 9 244335 791 625681 494913671 28 1247222 9243234 792 627264 496793088 28 1424946 9252130 793 628849 498677257 28 1602557 9256022 794 630436 500566184 28 1780056 9 259911 795 632025 502459875 28 1957444 9263797 796 633616 504358336 23 2134720 9*267680 797 635209 506261573 28 231 1884 28 2488938 9271559 9 275435 798 ' 636S04 5031 69592 799 63S401 510082399 40b0881 9279308 800 j 640000 | 512000000 BQVAftES, CUBES, AND ROOTS. 107 Number, 801 803 803 804 805 806 807 810 811 812 813 814 815 816 817 818 810 820 831 822 823 824 825 826 837 823 820 830 831 832 835 837 840 841 842 843 844 845 846 847 848 849 Square. 6416Q1 643304 644300 646416 643025 649636 651249 654481 656100 657721 659344 660960 662596 664225 665956 6674S9 669124 670761 672400 674041 675684 677329 678976 680625 682276 683929 6S5584 637241 688900 690561 697225 848 I 850 I S 700569 702244 703921 705600 707281 708964 710649 712336 714025 715716 717409 719104 720601 j 722500 / Cube. 513922401 515849608 517781627 519718464 521660125 623606616 525557943 527514112 529475120 531441000 533411731 635387328 537366797 530353144 541343375 543338496 545336513 547343432 540353250 55136S000 553387661 555412248 557441767 559476224 561515625 563559976 565600283 567663552 569722780 571787000 573S5G191 575030368 578009537 580093704 5821S2875 584277056 586376253 5884S0472 590589719 592704000 594823321 596947683 599077107 601211584 603351125 605495736 607645423 ! 609800192 611960049 614125000 Square Root. Cube Root. 28 3019434 23 3106045 283372546 28-3543933 28 3725210 233901391 28 4077454 23-4253408 284420253 284604989 23 4780617 28 4956137 285131549 285306852 23 5482048 28 5657137 23 5832119 28- 6006993 28 6181760 236356421 28*6530976 28 6705424 23 6370766 287054002 28 7228132 287402157 28 7576077 28 7749891 28 7923601 238097206 288270706 28 8444102 288617394 288790582 28 8063666 28 9136646 289309523 28 9482297 289654967 28 9827535 290000000 200172363 29 0344623 29 0516781 290688837 29 0860791 29 1032644 29 1204396 29*1376046 29- 1547595 0287044 0290907 9294767 9298624 9302477 9 306328 9310175 9314019 317860 321697 9325532 9329363 9 333192 337017 9*340838 0344657 0348473 9352286 0356095 9359902 9 363705 9367505 9 371302 9-375096 9-37S887 9382675 9386460 9300242 9394020 0397796 9401569 9405339 9 409105 9 412369 9 416630 9 420337 9424142 9427894 9431642 9 435388 9439131 9 442870 944C607 9450341 9454072 945TOQQ 9-4-WMA 9-4*ra&9A 108 ARITHMETIC. rHUIIlUci. Cube. SnunTP Root Cube Root 851 724201 616295051 29 1719043 9476395 852 725904 618470208 29 1690390 9480106 853 727609 620650477 292061637 9-483813 854 729316 622835864 292232784 9487518 855 731025 625026375 292403830 9491220 856 732736 627222016 292574777 9494919 857 734449 629422793 292745623 9 498615 858 736164 631628712 292916370 9502308 859 737881 633839779 293087018 9505998 860 739600 636056000 293257566 9509685 861 741321 638277381 293428015 9513370 862 743044 640503928 293598365 9517051 863 744769 642735647 293768616 9520730 864 746496 644972544 293938769 9524406 865 748225 647214625 294108823 9528079 866 749956 649461896 294278779 9531749 867 751689 651714363 294446637 9 535417 868 753424 653972032 29 4618397 9539082 869 755161 656234909 294788059 9542744 870 756900 658503000 294957624 9546403 871 758641 660776311 29-5127091 9550059 872 760384 663054848 295296461 9553712 873 762129 665338617 295465734 9557363 874 763876 667627624 295634910 9561011 875 765625 669921875 295803989 9564656 876 767376 672221376 295972972 9568298 877 769129 674526133 296141858 9571938 878 770884 676836152 29 6310648 9575574 879 772641 679151439 296479325 9579208 880 774400 681472000 296647939 9-582840 881 776161 683797841 296616442 9586468 S82 777924 686128968 296984848 9590094 883 779689 688465387 297153159 9593716 884 781456 690807104 297321375 9597337 885 783225 693154125 297489496 9600955 886 784996 695506456 29 7657521 9-604570 8S7 786769 697864103 297825452 9608182 888 788544 700227072 297993289 9611791 889 790321 702595369 29 8161030 9615398 890 792100 704969000 298328678 9619002 891 793881 707347971 29 8496231 9622603 892 795664 7097322S8 298663690 9626201 893 797449 712121957 298831056 9629797 894 799236 714516984 298998328 9633390 895 801025 716917375 299165506 9636981 896 802816 719323136 299332591 9640569 897 804609 721734273 299499583 9 644154 898 I 806404 724150792 I 9 647737 899 808201 726572699 900 / 810000 729000000 \ 30 OOOOWto \ 9 •QUARKS, CUBES, AND ROOTS. 109 Number. Square, Cube 811801 813604 815409 817316 819025 H2083G 826281 828100 829921 831744 833569 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 30-2439669 30 2654919 30-2820079 302935148 303150123 303315018 30 3479818 303644529 30 3309151 303973633 304138127 30 4302431 30 4466747 304630924 30 4795013 304959014 305122926 30-5296750 30 5450487 30 5614136 305777697 30 5941171 30 6104557 306267857 30 6431069 306594194 306757233 30G920185 30 7083051 307245830 307408523 30 7571130 307733651 3078960S6 30-805343& 30*8220700 9658469 9662040 9665609 9 669176 9672740 9-676302 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 9-795361 9799334 9802904 9 806271 9-809736 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 31-4801525 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 9-885357 9888767 9892175 9895580 9898983 9902333 9-905782 9909178 9-912571 9915962 9919351 9922738 9 926122 9929504 9932884 9936261 9939636 9943009 9946380 9949748 9 953114 9*956477 9959839 9 963198 9-966555 9-969909 9 973262 9 976612 9979960 9983305 9986649 9989990 Ill OF RATIOS, PROPORTIONS, AND PRO- GRESSIONS. Numbers are compared to each other in two different ways : the one comparison considers the difference of the two numbers, and is named Arithmetical Relation ; and the dif- ference sometimes the Arithmetical Ratio : the other con- siders their quotient, which is called Geometrical Relation ; and the quotient is the Geometrical Ratio. So, of these two numbers 6 and 3, the difference, or arithmetical ratio is 6—8 or 8, but the geometrical ratio is f or 2. There must be two numbers to form a comparison : the number which is compared, being placed first, is called the Antecedent : and that to which it is compared, the Con- sequent. So, in the two number* above, 6 is the antecedent, and 3 the consequent. If two or more couplets of numbers have equal ratios, or equal differences, the equality is named Proportion, and the terms of the ratios Proportionals. So, the two couplets, 4, 2 and 8, 6, are arithmetical proportionals, because 4-2 = 8 —6 = 2; and the two couplets 4, 2 and 0, 3, are geome- trical proportions, because | = f = 2, the some ratio. To denote numbers as being geometrically proportional, a colon is set between the terms of each couplet, to denote their ratio ; and a double colon, or else a mark of equality, between the couplets or ratios. So, the four proportionals, 4, 2, 6, 3 are set thus, 4 : 2 : : 6 : 3, which means, that 4 is to 2 as 6 is to 3 ; or thus, 4:2 = 6:3, or thus, | = |, both which mean, that the ratio of 4 to 2, is equal to the ratio of 6 to 3. Proportion is distinguished into Continued and Discon- tinued. When the difference or ratio of the consequent of one couplet, and the antecedent of the next couplet, is not the same as the common difference or ratio of the couplets, the proportion is discontinued. So, 4, 2, 8, 6, are in discon- tinued arithmetical proportion, because 4—2=8 — 6=2, whereas 8 — 2=6: and 4, 2, 6, 3 are in discontinued geo- metrical proportion, because f = §■ = 2, but -J = 3, which is not the same. But when the difference or ratio of every two succeeding terms is the same quantity, the proportion is said to be Con- tinued, and the numbers themselves make a scrisa ot Cwv- 112 ARITHMETIC. Untied Proportionals, or a progression. So 2, 4, 6, 8 form an arithmetical progression, because 4 — 2 = — 4 = 8 — 6 = 2, all the same common difference ; and 2, 4, 8, 16, a geometrical progression, because f = } = Y = 2, all the same ratio. When the following terms of a progression increase, or exceed each other, it is called an Ascending Progression, or Series ; but when the terms decrease, it is a descending one. So, 0, 1, 2, 3, 4, &c. is an ascending arithmetical progression, but 9, 7, 5, 3, 1, dec. is a descending arithmetical progression. Also 1 ,2, 4, 8, 16, dec. is an ascending geometrical progression, and 16, 8, 4, 2, 1 , dec. is a descending geometrical progression* ARITHMETICAL PROPORTION AND PROGRESSION. In Arithmetical Progression, the numbers or terms have all the same common difference. Also, the first and last terms of a Progression, are called the Extremes ; and the other terms, lying between them, the Means. The most useful part of arithmetical proportion, is contained in the following theorems : Theorem 1. When four quantities are in arithmetical proportion, the sum of the two extremes is equal to the sum of the two means. Thus, of the four 2, 4, 6, 8, here 2 + 8 = 4 + 6=10. Theorem 2. In any continued arithmetical progression, the sum of the two extremes is equal to the sum of any two means that are equally distant from them, or equal to double the middle term when there is an uneven number of terms. Thus, in the terms 1, 3, 5, it is 1 + 5 = 3 + 3 = 6. And in the series 2, 4, 6, 8, 10, 12, 14, it is 2 + 14 = 4 + 12 =6 + 10 = 8 + 8 = 10. Theorem 3. The difference between the extreme terms of an arithmetical progression, is equal to the common dif- ference of the series multiplied by one less than the number of the terms. So, of the ten terms, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, the common difference is 2, and one less than the number of terms 9 ; then the difference of the extremes is 20 - 2 = 18, and 2 X 9 = 18 also. ARITHMETICAL PHOFORTION. 113 Consequently the greatest term is equal to the least term added to the product of the commoo difference multiplied by 1 less than the number of terms. Theorem 4. The sum of all the terms, of any arith- metical progression, is equal to the sum of the two extremes multiplied by the number of terms, and divided by 2 ; or the sum of the two extremes multiplied by the number of the terms, gives double the sum of all the terms in the series. This is made evident by setting the terms of the series in an inverted order, under the same series in a direct order, and adding the corresponding terms together in that order. Thus, in the series 1, 3, 5, 7, 9, 11, 13, 15; ditto inverted 15, 13, 11, 9, 7, 5, 3, 1 ; the sums are 16 +10 + 16 + 16 + 16 + 16 + 16 -f- 16, which must be double the sum of the single series, and is equal to the sum of the extremes repeated as often as are the number of the terms. From these theorems may readily be found any one of these five parts ; the two extremes, the number of terms, the common difference, and the sum of all the terms, when any three of them are given ; as in the following problems : PROBLEM I. Given the Extremes, and the Number of Terms, to find the Sum of all the Terms, Add the extremes together, multiply the sum by the number of terms, and divide by 2. EXAMPLES. 1. The extremes being 3 and 19, and the number of terms 9 ; required the sum of the terms ? 19 3 22 2) 198 Ans. 99 2. It is required to find the number of all the strokes a common clock strikes in one whole revolution of the index, or in 12 hours. Aua.nfc. Vox. 1. 1G = ^X9=11 X 9 = 99, the same answer. 114 ▲xinufsnc. Ex. 3. How many strokes do the clocks of Venice strike in the compass of the day, which go continually on from 1 to 24 o'clock ? Ana. 300. 4. What debt can be discharged in a year, by weekly payments in arithmetical progression, the first payment being U y and the last or 52d payment 5i 3t ? Ans. 1851 4*. PROBLRX n. Given the Extremes, and the Number of Terms ; t^findthe Common Difference. Subtract the less extreme from the greater, and divide the remainder by 1 less than the number of terms, for the common difference. EXAMPLES. 1. The extremes being 3 and 19, and the number of terms 9 ; required the common difference ? 19 Ur ' 9-T ¥ 2. If the extremes be 10 and 70, and the number of terms 21 ; what is the common difference, and the sum of the series ? Ans. the com. diff. is 3, and the sum is 840/ 3. A certain debt can be discharged in one year, by weekly payments in arithmetical progression, the first payment being 1*, and the last 51 Ss ; what is the common difference of the terms? Ans. 2. PROBLEM III. v Given one of the Extremes, the Common Difference, and the Number of Terms ; to find the other Extreme, and the Sum of the Series. Multiply the common difference by 1 less than the number of terms, and the product will be the difference of the extremes : Therefore add the product to the less ex- treme to give the greater ; or substract it from the greater, to give the less extreme. ARITHMETICAL PROGRESSION. 115 EXAMPLES. 1. Given the least term 3, the common difference 2, of an arithmetical aeries of 9 terms ; to find the greatest term, and the sum of the series. Here 2 X (9 — 1) + 3 = 19, the greatest terra. Theref. (10 +3)| = »}• =99, the sum of the series. 2. If the greatest term be 70, the common difference 3, and the number of terms 21, what is the least term, and the sum of the series ? Ans. The least term is 10, and the sum is 840. 8. A debt can be discharged in a year, by paying 1 shilling the first week, 8 shillings the second, and so on, always 2 shillings more every week ; what is the debt, and what will the last payment be ? Ans. The last payment will be 5* 3*, and the debt is 1357 4s. PROBLEM IV. To find an Arithmetical Mean Proportional between two given terms. Add the two given extremes or terms together, and take half their sum tor the arithmetical mean required. EXAMPLE. To find an arithmetical mean between the two numbers 4 and 14. Here 14 4 2) 18 Ans. 9 the mean required. problem v. To find two Arithmetical Means between two given Extremes. Subtract the less extreme from the greater, and divide the dMfeienee by 8, so will the quotient be the coranKfli&i- 116 ARITHMETIC. ference ; which being continually added to the less extreme, or taken from the greater, will give the means. EXAMPLE. To find two arithmetical means between 2 and 3. Here 8 2 3) 6 Then 2 + 2 = 4 the one mean. — — and 4 + 2 = 6 the other mean, com. dif. 2 PROBLEM VI. To find any Number of Arithmetical Means between two given Terms or Extremes. Subtract the less extreme from the greater, and divide the difference by 1 more than the number of means required to be found, which will give the common difference ; then this being added continually to the least term, or subtracted from the greatest, will give the mean terms required. EXAMPLE. To find five arithmetical means between 2 and 14. Here 14 2 6) 12 Then by adding this com. dif. continually, the means are found 4, 6, 8, 10, 12. com. dif. 2 See more of Arithmetical progression in the Algebra. GEOMETRICAL PROPORTION AND PRO- GRESSION. If there be taken two ratios, as those of 6 to 3, and 14 to 7, which, by what has been already said (p. 113), may GEOMETRICAL tfftOGBBMIO*. 117 be expressed fractionally, } and \f ; to judge whether they are equal or unequal, we must reduce them to a common denominator, and we shall have 6X7, and 14 X 3 for the two numerators. If these are equal, the fractions or ratios are equal. Therefore, Theorem i. If four quantities be in geometrical propor- tion, the product of the two extremes will be equal to the product of the two means. And hence, if the product of the two means be divided by one of the extremes, the quotient will give the other ex- treme. So, of the above numbers, if the product of the means 42 be divided by 6, the quotient 7 is the other extreme ; and if 42 be divided by 7, the quotient 6 is the first ex- treme. This is the foundation of the practice in the Rule of Three. We see, also, that if we have four numbers, 6, 3, 14, 7, such, that the products of the means and of the extremes are equal, we may hence infer the equality of the ratios $ =y , or the existence of the proportion 6 : 3 : : 14 : 7. Hence Theorem n. We may always form a proportion of the factors of two equal products. If the two means are equal, as in the terms 3, 6, 6, 12, their product becomes a square. Hence Theorem hi. The mean proportional between two num- bers is the square root of their product. We may, without destroying the accuracy of a proportion, S've to its various terms all the changes which do not affect e equality of the products of the means and extremes. Thus, with respect to the proportion 6 : 3 : : 14 : 7, which gives 6X7 = 3X1 4, we may displace the extremes, or the means, an operation which is denoted by the word AMcrnando. This will give 6 : 14 : : 3:7 or 7 : 3 : : 14 : 6 or 7 : 14 : : 3:6 Or, 2dly, we may put the extremes in the places of the means, oalled lnvertendo. Thus 3 : 6 : : 7 : 14. Or, 3dly, we may multiply or divide the two antecedents, or the two consequents, by the same number, when propor- tionality will subsist. 118 ARITHMETIC. As 6 X 4 : 3 : : 14 X 4 : 7 ; viz. 24 : 3 : : 66 : 7 and6^-2:3:: 14-t-2:7; viz. 3:3:: 7:7. Also, applying the proposition in note 2, Addition of Vulgar Fractions, to the terms of a proportion, such as 30 : 6 : : 15 : 3, or y = y, we shall have 30±15 15 . 30+15 30-15 „ = — and = . Hence 6 ± 3 3 6+3 6-3 Theorem iv. The sum or the difference of the antece- dents, is to that of the consequents, as any one of the ante- cedents is to its consequent. Theorem v. The sum of the antecedents is to their difference, as the sum of the consequents is* to their dif- ference. In like manner, if there he a series of equal ratios, 1 = V = V 4 = i J 5 we have 6+10+14+30 = 14 = 30 3+5+7+15 7 15 *nereiure, Theorem vi. In any series of equal ratios, the sum of the antecedents is to that of the consequents, as any one an- tecedent is to its consequent. Theorem vii. If two proportions are multiplied, term by term, the products will constitute a proportional. Thus, if 30 : 15 :: 6 : 3 and 2 : 3 : : 4 : 6. Then 30 X 2 : 15 X 3 :: 6 X 4 : 3 X 6 or 60 : 45 :: 24 : 18 ; or f J- = Theorem viii. If four quantities are in proportion, their squares, cubes, &c. will be in proportion. For this will evidently be nothing else than assuming the proportionality of the products, term by term, of two, three, or more identical proportions. The same properties hold with regard to surd or irrational expressions, Thus, v'TSO : ^80 :: ^567 : ^63 and -y/12 :-y/3 :: : -y/1. y/720 = y'O . 80 = 3 y/567 _ y/9X63 _ 3 y/80 "~ V ** -v/63 y/63 1 , 12 v/4 2 and ^3 c= ^-3 = — = r GEOMETRICAL PROGRESSION. 119 Theorem ix. The quotient of the extreme terms of a • geometrical progression is equal to the common ratio of the series raised to the power denoted by 1 less than the number of the terms. So, of the ten terms 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, the common ratio is 2, one less than the number of terms 9 ; then the quotient of the extremes is 1 °^ 4 = 512, and 2° = 512 also. Consequently the greatest term is equal to the least term multiplied by the said power of the ratio whose index is 1 less than the number of terms. Theorem x. The sum of all the terms, of any geome- trical progression, is found by adding the greatest term to the difference of the extremes divided by 1 less than the ratio. So, the sum of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 1024-2 (whose ratio is 2) is 1024 + -^-y- = 1024 + 1022 =2046. This subject will be resumed in the Algebraic part of this "work. A few examples may here be added. EXAMPLES. 1. The least of ten terms, in geometrical progression, being 1, and the ratio 2 ; what is the greatest term, and the sum of all the terms ? Ans. The greatest term is 512, and the sum 1023. 2. What debt may be discharged in a year, or 12 months, iy paying 1Z the first month, 21 the second, 4Z the third, and 40 on, each succeeding payment being double the last ; and what will the last payment be ? Ans. The debt 4095Z, and the last payment 2048/. PROBLEM I. To find one Geometrical Mean Proportional between any two Numbers. Multiply the two numbers together, and extract the *C|uare root of the product, which will give the mean propor- tional sought. 120 ARITHMETIC* EXAMPLE. To fiad a geometrical mean betweea the two numbers 3 and 12. 12 3 36 (6 the mean. 36 PROBLEM II. To find two Geometrical Mean Proportionals between any two Numbers. Divide the greater number by the less, and extract the cube root of the quotient, which will give the common ratio of the terms. Then multiply the least given term by the ratio for the first mean, and this mean again by the ratio for the second mean : or, divide the greater of the two given terms by the ratio for the greater mean, and divide this again by the ratio for the less mean. EXAMPLE. To find two geometrical means between 3 and 24. Here 3 ) 24 ( 8 ; its cube root 2 is the ratio. Then 3 X 2 = 6, and 6 X 2 = 12, the two means. Or 24 + 2 = 12, and 12 -r- 2 = 6, the same. That is, the two means between 3 and 24, are 6 and 12. problem in. To find any number of Geometrical Means between two Numbers. Divide the greater number by the less, and extract such root of the quotient whose index is 1 more than the number of means required ; that is, the 2d root for one mean, the 3d root for two means, the 4th root for three means, and so on ; and that root will be the common ratio of all the terms. OF HARJf OXICAL PROPORTION* 121 Then, with the ratio, multiply continually from the first term, or divide continually from the last or greatest term. EXAMPLE. To find four geometrical means between 3 and 96. Here 3) 96 (32 ; the 5th root of which is 2, the ratio. Then 3X2=6, dc 6X2=12, & t2 X 2=24, & 24 X2=48. Or 96-1-2=48, <fc 48-^2=24, & 24^-2=12, & 12-5-2=6. That is, 6, 12, 24, 48, are the four means between 3 and 96. f " OF HARMONIC AL PROPORTION. There is also a third kind of proportion, called Harmo- nical or musical, which being but of little or no common use, a very short accouut of it may here suffice. Musical Proportion is when, of three numbers, the first has the same proportion to the third, as the difference be- tween the first and second has to the difference between the second and third. As in these three, 6, 8, 12 ; where 6 : 12 : : 8-6 : 12-8, that is 6 : 12 : : 2 : 1. When four numbers are in musical proportion ; then the first has the same ratio to the fourth, as the difference be- tween the first and second has to the difference between the third and fourth. As in these, 6, 8, 12, 18 ; where 6 : 18 : : 8-6 : 18-12, that is 6 : 18 : : 2 : 6. When numbers are in musical progression, their recipro- cals are in arithmetical progression ; and the converse, that is, when numbers are in arithmetical progression, their reci- procals are in musical progression. So in these musicals 6, 8, 12, their reciprocals, £, |, yiy, are in arithmetical progression ; for ^ + = = i I and i + i = | = I ; that is, the sum of the extremes is equal to double the mean, which is the property of arithme- ticals. Vol. I. 17 122 ARITHMETIC. The method of finding out numbers in musical proportion ii best expressed by letters in Algebra. FELLOWSHIP, OR PARTNERSHIP. Fellowship is a rule, by which any sum or quantity may be divided into any number of parts, which shall be in any given proportion to one another. By this rule are adjusted the gains or loss or charges of partners in company ; or tho effects of bankrupts, or lega- cies in case of a deficiency of assets or effects ; or the shares of prizes ; or the numbers of men to form certain detach* ments ; or the division of waste lands among a number of proprietors. Fellowship is either Single or Double. It is single, when the shares or portions are to be proportional each to one single given number only ; as when the stocks of partners are all employed for the same time; and Double, when each portion is to be proportional to two or more numbers ; as when the stocks of partners are employed for different times. SINGLE FELLOWSHIP. GENERAL RULE. Add together the numbers that denote tho proportion of the shares. Then say, As the sum of the said proportional numbers, Is to the whole sum to be parted or divided, So is each several proportional number, To the corresponding share or part. Or, as the whole stock, is to the whole gain or loss, So is each man's particular stock. To his particular share of the gain or loss. To prove the Work. Add all the shares or parts to- gether, and the sum will be equal to the whole number to be shared, when the work is right. •XH6UB FBLLOWSHIP. 138 EXAMPLES. 1. To divide the number 240 into three such parts, as shall be in proportion to each other as the three numbers 1, Sand 3. Here 1 + 2 + 3 = 6, the sum of the numbers* Then, as 6 : 240 : : 1 : 40 the let part, and as 6 : 240 : : 2 : 80 the 2d part, also as 6 : 240 : : 8 : 120 the 8d part. Sum of all 240, the proof. 2. Three persons, a, b, c, freighted a ship with 840 tuns of wine, of which a loaded 100 tuns, n 97, and c the rest : in a storm the seamen were obliged to throw overboard 85 tuns ; how much must each person sustain of the loss ? Here 110 + 97 « 207 tuns, loaded by a and n ; theref. 340 — 207 = 133 tuns, loaded by c. Hence, as 840 : 85 : : 110 or as 4 : 1 : : 1 10 : 27| tuns = a's loss ; and as 4 : 1 : : 97 : 24| tuns = b's loss ; also as 4 : 1 : : 133 : 33} tuns = c's loss ; Sum 85 tuns, the proof. 8. Two merchants, c and d, made a stock of 120/ ; of irbich c contributed 757, and d the rest : by trading they gained 30Z ; what must each have of it ? Ans. c 18/ 15*, and d 11/ 5*. 4. Three merchants, b, f, g, make a stock of 700/, of ^vhich b contributed 128/, r 358/, and o the rest : by trading fihey gain 125/ 10* ; what must each have of it ? Ans. b must have 22/ Is Od 2fjq. p ... 64 3 8 Off. g ... 30 5 3 1,V- 5. A General imposing a contribution* of 700/ on four * Contribution it a Ui paid by provinces, towns, village! , Ac. to ex- xon them from being; plundered. It is paid in provisions or in money, *fcd sometimes in both. 134 ARITHMETIC. villages, to be paid in proportion to the number of inhabitants contained in each ; the first containing 250, the 2d 350, the 3d 400, and the 4th 500 persons ; what part must each vil- lage pay ? Ans. the 1st to pay 116/ 13s Ad the 2d - - 163 6 8 the 3d - - 186 13 4 the 4th - - 233 ' 6 8 6. A piece of ground, consisting of 37 ac 2 ro 14 ps, is to be divided among three persons, l, m, and n, in propor- tion to their estates : now if l's estate be worth 5002 a year, m's 320/, and n's 75/ ; what quantity of land must each one have ? Ans. l must have 20 ac 3 ro 39^} {ps. m - - - 13 1 30 T \V *. n - - - 3 23tf| 7. A person is indebted to o 57/ 15*, to r 108/ 3s 84, to a 22/ 10c/, and to r 73/ ; but at his decease, his effects are found to be worth no more than 170/ 14s ; how must it be divided among his creditors ? Ans. o must have 37/ 15* 5c/ 2^/^c/. p ... 70 15 2 2^V- q ... 14 8 4 2flftV- B - - - 47 14 11 2tWA. 8. A ship, worth 900/, being entirely lost, of which -J- be- longed to s, I to t, and the rest to v ; what loss will each sustain, supposing 540/ of her were insured ? Ans. s will lose 45/, t 907, and v 225/. 9. Four persons, w, x, y, and z, spent among them 25*, and agree that w shall pay £ of it, x |, y }, and z j ; that is, their shares are to be in proportion as £, |, J, and £ : what are their shares t Ans. w must pay 9s 8d 3%]q. x - - . 6 5 34$. y - - - 4 10 14$. z - . . 3 10 3,V- 10. A detachment, consisting of 5 companies, being sent into a garrison, in which the duty required 76 men a day ; what number of men must be furnished by each company, in proportion to their strength ; the 1st consisting of 54 men, DOVBU n&LOWlBXP. 136 » the 2d of 51 men, the 3d of 48 men, the 4th of 89, and the 5th of 36 men? Ana. The 1st must furnish 18, the 2d 17, the 3d 16, the 4th 13, and the 5th 12 men 41 . DOUBLE FELLOWSHIP. Doubub Fkllowbhip, aa has been said, is concerned in cases in which the stocks of partners are employed or con- tinued for different times. RuLsf. — Multiply each person's stock by the time of its continuance ; then divide the quantity, aa in Single Fellowship, into shares, in proportion to these products], oy saying, As the total sum of all the said products, Is to the whole gain or loss, or quantity to be parted, So is each particular product To the correspondent share of the gain or loss. EXAMPLES. 1. a had in company 50f for 4 months, and b had 60J for 6 months ; at the end of which time they find 242 gained : how roust it be divided between them ? Here 50 60 4 5 200 + 300 = 500 Then as 500 : 24 : : 200 : 9} = 91 12* = a's share, and as 500 : 24 : : 300 : 14f = 14 8 = b's share. * Questions of this nature frequently occurring in military service. General Haviland, an officer of great merit, contrived an ingenious in- strument, for more expeditiously resolving them ; which is distinguish- ed bv the name of I he inventor, being called a Haviland. t Th»* proof of this rule is as follows : When the times are equal, the slmres of the gain or loss are evidently as the stocks, as in Single Fellowship; and when the stocks are equal, the shares are as the times; therefore, when neither are equal, the shares must be esthete products. 196 ABTTH3CKTIC. 2. c and d hold a piece of ground in common, for which they are to pay 542. c put in 23 horses for 27 days, and d 21 horses for 30 days ; how much ought each man to pay of the rent ? . Ans. c must pay 23/ 5* 9d. d must pay 30 14 3. 3. Three persons, e, f, o, hold a pasture in common, for which they are to pay 30/ per annum ; into which e put 7 oxen for 3 months, f put 9 oxen for 5 months, and o put in 4 oxen for 12 months ; "how much must each person pay of the rent 1 Ans. e must pay 57 10* Gd lftq. f .. 11 16 10 0^V o - - 12 12 7 2ft. 4. A ship's company take a prize of 1000/, which they agree to divide among them according to thejr pay and the time they have been on board : now the officers and midship- men have been on board 6 months, and the sailors 3 months ; the office re have 40* a month, the midshipmen 30*, and the sailors 22* a month ; moreover,' there are 4 officers, 12 mid- shipmen, and 110 sailors ; what will each man's share be ? Ans. each officer must have 231 2s 5d Y y^q. each midshipman - 17 6 9 3^. each seaman - - 6 7 2 0fo\. 5. r, with a capital of 1000/, began trade the first of January, and, meeting with success in business, took in i as a partner, with a capital of 1500/, on the first of March fol- lowing. Three months after that they admit k as a third partner, who brought into stock 2K007. After trading toge- ther till the end of the year, they find there has been gained 1776/ 10s ; how must this be divided among the partners ? Ans. h must have 475/ 9s A\d j^q* i 571 16 8J k - . - 747 3 11J iff. 6. x, y, and z made a joint stock for 12 months ; x at first put in 20/, and 4 months after 20/ more ; y put in at first 30/, at the end of 3 months he put in 20/ more, and 2 months after he put in 407 more ; z put in at first 60/, and 5 months after he put in 10/ more, 1 month after which he took out 30/ ; during the 12 months they gained 50/ ; how much of it must each have ? Ans. x must have 10/ 18s 6d 3} fa. y ... 22 8 1 0ft. z ... 16 13 4 0. SIMPLE onraBST. SIMPLE INTEREST. Interest is the premium or sum allowed for the loan, or forbearance of money. The money lent, or forborn, is called the Principal ; and the sum of the principal and its x interest added together, is called the Amount. Interest is allowed at so much per cent, per annum ; which premium per cent, per annum, or interest of 100/ for a year, is 1 called the rale of interest : — So, When interest is at 3 per cent, the rate is 3 ; - 4 per cent. - - - 4 ; . 5 per cent. - - - 5 ; - 6 per cent. ... 6. But, by law, interest ought not to be taken higher than at the rate of 5 per cent. Interest is of two sorts ; Simple and Compound. Simple Interest is that which is allowed for the principal lent or forborn only, for the whole time of forbearance* As the interest of any sum, for any time, is directly pro- portional to the principal sum, and also to the time of con- tinuance ; hence arises the following general rule of calcu- lation. As 100Z is to the rate of interest, so is any given principal to its interest for one year. And again, As 1 year is to any given time, so is the interest for a year, just found, to the interest of the given sum for that time. Otherwise. Take the interest of 1 pound for a year, which multiply by the given principal, and this product again by the time of loan or forbearance, in years and parts, for the interest of the proposed sum for that time. Note. When there are certain parts of years in the time, as quarters, or months, or days : they may be worked for, either by taking the aliquot or like parts of the interest of a year, or by the Rule of Three, in the usual way. Also, to divide by 100, is done by only pointing off two figures Cot decimals. 138 ASfHDtBTXG. EXAKPLK8. 1. To find the interest of 2801 10«, for 1 year, at the rate of 4 per cent, per annum.„ Here, As 100 : 4 :: 230/ 10* : 97 4* 4Jd. 4 100) 0,22 20 4-40 12 4 80 t Ans. 01 4s 4f<*. 3*20 2. To find the interest of 547/ 15*, for 3 years, at 5 per cent, per annum. As 100 : 5 :: 547-75 Or 20 : 1 :: 547*75 : 27-3875 interest for 1 year. 3 I 82-1625 ditto for 3 years. 20 s 3-2500 12 <2 3-00 Ans. 82/ 3* 3d. 3. To find the interest of 200 guineas, for 4 years 7 months and 25 days, at 4$ per cent, per annum. SIMPLE INTEREST. 129 ds 2 ds < 8102 As 356: 9-45: : 25: 2 41 or 73:9-45:: 5 : -6472 5 340 105 73) 47-25 (-6472 345 9-45 interest for 1 yr. 530 4 19 37*80 ditto 4 years. 6 mo = \ 4*725 ditto 6 months. 1 mo = £ -7875 ditto 1 month. -6472 ditto 25 days. I 43-9597 20 * 191940 12 d 2-3280 4 Ans. 432 19* 2\d. q 1-3120 4» To find the interest of 450/, for a year, at 5 per cent, per annum. Ans. 222 10*. 5. To find the interest of 7152 12* 6c2, for a year, at 41 per cent, per annum. Ans. 322 4* 0}a. 6. To find the interest of 7202, for 3 years, at 5 per cent per annum. Ans. 1082. 7. To find the interest of 3552 15*, for 4 years, at 4 per cent per annum. Ans. 562 18* 4Jd. - 8. To find the interest of 322 5* Bd 9 for 7 years, at 4£ per cent per annum. Ans. 92 12* Id. 9. To find the interest of 1702, for 1} year, at. 5 per cent per annum. Ans. 122 15** 10. To find the insurance on 2052 15*, for J of a year, at 4 per cent, per annum. Ans. 22 1* lf<f. 11. To find the interest of 3192 6d, for 5? years, at 3| per cent, per annum. Ans. 682 14* 9|<2. 12. To find the insurance on 1072, for 117 days, at 4f per cent per annum. Ana. U \fe Id* Vol. I. 18 180 ARITHMETIC 13. To find the interest of 172 5*, for 117 days, at 4f per cent, per annum. Ans. 5* 3d. It To find the insurance on 7122 6s, for 8 months, at 71 per cent, per annum. Ans. 352 1£# 3J<£ Note. The Rules for Simple Interest, serve also to calcu- late Insurances, or the Purchase of Stocks, or any thing else that is rated at so much per cent. See also more on the subject of Interest, with the algebrai- cal expression and investigation of the rules at the«adof the Algebra. COMPOUND INTEREST. Compound Interest, called also Interest upon Interest, is that which arises from the principal and interest, taken together, as it becomes due, at the end of each 'stated time of payment. Though it be not lawful to lend money at Compound Interest, yet in purchasing annuities, pensions, or leases in reversion, it is usual to allow Compound Interest to the purchaser for his ready money. Rules. — 1. Find the amount of the given principal, for the time of the first payment, by Simple Interest. Their consider this amount as a new principal for the second pay- ment, whose amount calculate as before. And so on through all the payments to the last, always accounting the last amount as a new principal for the next payment. The reason of which is evident from the definition of Compound Interest* Or else, 2. Find the amount of 1 pound for the time of the first payment, and raise or involve it to the power whose index is denoted by tho number of payments. Then that power multiplied by the given principal, will produce the whole amount. From which the said principal being subtracted, leaves the Compound Interest of the same. As is evident from the first Rule. EXAMPLES. 1. To find the amount of 720/, for 4 years, at 5 per cent* per annum. Here 5 is the 20th part of 100, and the interest of H for a year is -fa or *95, and its amount 1*05. Therefore, ALLIGATION. 1. By ike lit Rtde. 2. By the 2d Rule. I s d 1*05 amount of 1Z. SO) 720 1st yr's princip. 1-05 * 88 1st yr*s interest. , tnCTe J 1 1026 2d power of it. 20)756 2d yr*s princip. 1-1025 97 16 2d yr's interest.-———-:^ , J 1 -21550625 4th power of it. SO) 703 16 3d yr's princip. 720 30 13 9 h 3d yr's interest y 20) 838 9 9J 4th yr's princip. 20 41 13 5} 4thfr' 8 int«e,t. ~~ £875 3 3} the whole amou 12 or ans. required. 2. To find the amount of 502 in 5 years, at 5 per cent, per annum, compound interest. Ans. 032 16* 8jtf. 3. To find the amount of 50Z in 5 years, or 10 half-years, at 5 per cent, per annum, compound interest, the interest payable half-yearly. Ans. 64Z 0* Id. 4 To find the amount of 50Z in 5 years, or 20 quarters, at 5 per cent, per annum, compound interest, the interest payable quarterly. Ans. 04/ 2s (){d. 5. To find the compound interest of 370Z forborn for 6 years, at 4 percent, per annum. Ans. 98Z 3* 4£<Z. 6. To find the compound interest of 41 OZ forborn for 2* years, at 4£ per cent, per annum, the interest payable half, yearly. Ans. 48Z 4* 11 {d* 7. To find the amount, at compound interest, of 217Z, for* born at 2{ years, at 5 per cent, per annum, the interest pay- able quarterly. Ans. 242/ 13* 4£rf. ALLIGATION. Alligation teaches how to compound or mix together several simples of different qualities, so that the composition •nay be of some intermediate quality, or rate. It is com- monly distinguished into two cases. Alligation Medial, and Alligation Alternate. 1*2 AS1THXXTIC. ^ ALLIGATION MEDIAL. Alligation Medial is the method of finding the rate or quality of the composition, from having* the quantities and rates or qualities of the several simples given. And it is thus performed : * Multiply the quantity of each ingredient by its rate or Siality ; then add all the products together, and add also all e Quantities together in another sum ; then divide the former sum by the latter, that is, the sum of the products by the sum of the quantities, and the quotient will be the rate or quality of the composition required. EXAMPLES. 1. If three sorts of gunpowder be mixed together, viz. 601b at I2d a pound, 441b at 9d, and 261b at 8d a pound ; how much a pound is the composition worth ? Here 50, 44, 26 are the quantities, and 12, 9, 8 the rates or qualities ; then 50 X 12 = 600 44 X 9 = 396 26 X 8 = 208 120) 1204 (lOrfr = 10 f V- Ans. The rate or price is lO^d the pound. • Demonstration. The Rule is thus proved by Algebra. Let a, 6, e be the quantities of the ingredients, and m, n, p their rates, or qualities, or prices ; then am, in, cp are their several values, and am + bn -\- cp the sum of their values, also « + b + e is the sum of the quantities, and if r denote the rate of the whole composition, then (ff-f-6-fc)Xr will be the value of the whole, conseq. {a -f b 4- c) X r = am + bn + cp, and r = (am -f bn + cp) (a + b -f- c), u hich is the Rule. Note. If an ounce or any other quantity of pure gold be reduced into 94 equal parts, these parts are called Caracts ; but gold is often mixed with some base metal, which is called the Alloy, and the mixture is said to be of so many caracts fine, according to the proportion of para Kid contained in it : thus, if 522 caracts of pure gold, and 2 of a) Joy mixed together, it is said to be 32 caracts fine. If any one of the simples be of little or no value with respect to the rail, its rate is supposed to be nothing ; as water mixed with wine, and alloy with gold and silver. AU.IGATION ALTERNATE. 138 2. A composition being made of 51b of tea at 7s per lb, 91b at 8* 64 per lb, and 14£lb at 5* lOd per lb ; what is a lb of it worth? Ana. 6* I0\d. 3. Mixed 4 gallons of wine at 4s lOd per gall, with 7 gal- Ions at 5# 3d per gall, and 9} gallons at 6* Sd per gall ; what is a gallon of this composition worth ? Ans. 5* 4{d» 4. Having melted together 7 oz of gold of 22 caracts fine, 12 1 oz of 21 caracts fine, and 17 oz of 19 caracts fine : I would know the fineness of the composition ? Ans. 2(ty$ caracts fine. ALLIGATION ALTERNATE. Alligation Alternate is the method of finding what quantity of any number of simples, whose rates are given, will compose a mixture of a given rate. So that it is the re- verse of Alligation Medial, and may be proved by it. RULE i*. 1. Set the rates of the simples in a column under each other. — 2. Connect, or link with a continued lino, the rate * Demonst. By connecting the less rate with the greater, and placing the difference between them and the rate alternately, the quantities re- sulting are such, that there is precisely as much gained by one quantity as is Inst by the other, and therefore the gain and loss upon the whole is equal, and is exactly the proponed rate : and the same will be true of any other two simples managed according to the Rule. In like manner, whatever the number of simples may be, and with how many soever every one is linked, since it is always a le^s with a greater than the mean price, there will be an equal balance of loss and gain between every two, and consequeutly an equal balance on the whole. s. d. It is obvious, from the Rule, that questions of this sort admit of a great variety of answers ; for, having found one answer, we may find as many more as we please, by only multiplying or dividing each of the quantities found, by 2, or 3, or 4, fee. : the reason of which is evi- dent: for, if two quantities, of two simples, make a balance of loss and gain, with respect to the mean price, so must also the double or treble, the 1 or | part, or any other ratio of these quantities, and so on ad w- Jutiltm. These kinds of questions are called by algebraists indeterminate or unlimited problems ; and by an analytical process, theorems may be raited that will give all the possible answers. 184 ASmULETlG. of each simple, which is less than that of the compound, with one, or any number, of those that are greater than the com- pound*; and each greater rate with one or any number of the less.— -3. Write the difference between the mixture rate, and that of each of the simples, opposite the rate with which they are linked. — 4. Then if only one difference stand against any rate, it will be the quantity belonging to that rate ; but if there be several, their sum will be the quantity. The examples may be proved by the rule for Alligation Medial. EXAMPLES. 1. A merchant would mix wines at 16*, at 18*, and at 22* per gallon, so as that the mixture may be worth 20* the gal- lon ; what quantity of each must be taken ? 1 2 at 16* Here 20 ?J8>v J2 at 18* (22j/4 + 2 = 6at 22* 2. How much sugar at 4d, at 6d, and at lid per lb, must be mixed together, so that the composition formed by them may be worth 7d per lb ? Ans. 1 lb, or 1 stone, or 1 cwt, or any other equal quan- tity of each sort. 3. How much corn at 2* 6d, 3* Sd y 4*, and 4* 8d per bushel must be mixed together, that the compound may be worth 3* lOd per bushel ? Ans. 2 at 2* 6d, 3 at 3* 8d, 3 at 4*, and 3 at 4* Sd. RULE II. When the whole composition is limited to a certain quan- tity : Find an answer as before by linking ; then say, as the sum of the quantities, or differences thus determined, is to the givm quantity ; so is each ingredient, found by link- ing, to the required quantity of each. EXAMPLE. 1. How much gold of 15, 17, 18, and 22 caracts fine, mutt be mixed together, to form a composition of 40 oz of 20 cap racta fino ? ALLIGATION AI/TlHUVATE. 135 - 2 Here 20? l^M )- . . 2 5 + 3+ 2=10 16 Then as 10 : 40 : : 2 : 5 and 16 : 40 : : 10 : 25 Ans. 5 oz of 15, of 17, and of 18 caracts fine, and 25 oz of 22 caracts fine*. rule inf. t When one of the ingredients is limited to a certain quan- tity ; Take the difference between each price, and the mean rate as before ; then say, As the difference of that simple, whose quantity is given, is to the rest of the differences severally ; so is the quantity given, to the several quantities required. * A great number of questions might be here given relating to the specific gravities of metals, &c. but one of the most curious may suf- fice. Hiero, king of Syracuse, gave orders for a crown to be made entire- ly of pure gold ; but suspecting the workmen had debased it by mixing it with silver or copper, be recommended the discovery of the fraud to the famous Archimedes, and desiied to know the exact quantity of alloy in the crown. Archimedes, in order to detect the imposition, procured two other masses, the one of pure gold, the other of. silvei or copper, and each ot the same weight with the former; and by putting each separately into a vessel full of water, the quantity of water expelled by them deter- mined their specific gravities : from which, and their givr»n weights, the exact quantities of gold and alloy in the crown may be determined. Suppose the weight of each crown to be 101b, and that the water ex- pelled by the copper or silver was 921b, by the gold 5*2 lb, and by the compound crown C4lb ; what will be the quantities of gold and alloy Id the crown ? The rates of the simples are 92 and 52, und of the compound 64 ; therefore CxA I 62 > 12 of copper °* I 62... ^28 of gold And the sum of these is 12 + 28 — 40, which should have been 10; therefore by the Rule, 40 : 10 :: 12 : 31b of copper „„,„,„, 40: 10: :28 : 71b of «olJ Jthe answer t Id- the very same manner questions may be wrought when several •iof the ingredients are limited to certain quantities, by finding first fur one limit, and then for another. The two last Rules can need do de- monstration, as they evidently result from the first, the reason ut has been already explained. 186 ABTTHXBTIC EXAMPLES. 1. How much wine at 6*, at 5* 64, and 6* the gallon, must be mixed with 3 gallons at 4* per gallon, so that the mixture - may be worth 5* 4d per gallon ? 16+4=20 Then 10 : 10 : : 3 : 3 10 : 20 : : 3 : 6 10:20::3:6 Ads. 3 gallons at 5*, 6 at 5s 6d 9 and 6 at 6*. 2. A grocer would mix teas at 12*, 10*, and 6* per lb, with 201b at 4s per lb : how much of each sort must he take to make the composition worth 8* per lb ? Ans. 201b at 4*, 101b at 6*, 101b at 10*, and 201b at 12*. Position is a rule for performing certain questions, which cannot be resolved by the common direct rules. It is some- times called False Position, or False Supposition, because it makes a supposition of false numbers, to work with the same as if they were the true ones, and by their means dis- covers the true numbers sought. It is sometimes also called i Trial-and-Error, because it proceeds by trials of f ilse num- j bers, and thence finds out the true ones by a comparison ^ of the errors. — Position is either Single or Double. Single Position is that by which a question is resolved by means of one supposition only. Questions which have their result proportional to their supposition, belong to Single Position : such as those which require the multiplier POSITION. SINGLE POSITION. SINGLE POSITION. 137 tion or division of the number sought by any proposed num- ber ; or when it is to be increased or diminished by itself, or any parts of itself, a certain proposed number, of times. The rule is as follows : Take or assume any number for that which is required, and perform the same operations with it, as are described or performed in the question. Then say, As the result of the said operation, is to the position, or number assumed ; so is the result in the question, to a fourth term, which will be the number sought*. EXAMPLES. 1. A person after spending J- and { of his money, has yet remaining 60/ ; what had he at first ? Suppose he had at first 1201. Proof. Now { of 120 is 40 £ of 144 is 48 J of it is 30 i of 144 is 36 their sum is 70 their sum 84 which taken from 120 taken from 1 14 leaves 50 leaves (50 as Then, 50 : 120 : : 60 : 144 the Answer. per question. 2. What number is that, which, being increased by », J, and i of itself, the sum shall be 75? Ans. 36. 3. A general, after sending out a foraging i and £ of his men, had yet remaining 1000; what number had he in command ? Ans. 6000. 4. A gentleman distributed 52 pence among a number of poor people, consisting of men, women, and children ; to each man he gave 6d, to each woman 4d, and to each child 2d : moreover there were twice as many women as men, and * The reason of this Kale is evident, because it is supposed that the results are proportional to the suppositions. Thus, na : a : : nz : s, — h — «fcc. n — m a or - : a : n a a or - - n — m and so on. Vol. I. 19 138 AJCXTHMCTTC. thrice as many children as women. How many were them of each ? Ana. 2 men, 4 women, and 12 children. ft. One being aaked his age, said, if f of the years I have lived, be multiplied by 7, and § of them be added to the product, the sum will be 210. What was his age ? Ana. 45 yeanr. DOUBLE POSITION. Double Position is the method of resolving certain ques- tions by means of two suppositions of false numbers. To the Double Rule of Position belong such questions as have their results not proportional to their positions : such are those in which the numbers sought, or their parts* or their multiples, are increased or diminished by some given absolute number, which is no known part of the number sought, bulk*. Take or assume any two convenient numbers, and pro- ceed with each of them separately, according to the con- * Dcmonstr. The Rule is founded on this supposition, namely, tbet the first error is to the second, as the difference between the true and first supposed number, is to the difference between the true end second supposed number: when that is not the case, the exact answer to the question cannot be found by this Rule.— That the Rule is true, accord- ing to the assumption, may be thus proved. Let a and 6 be the two suppositions, and A and a their results, pro- duced by similar operation ; also r and s their errors, or the differences between the results a and a from the true result a ; and let x denote the number sought, answeriog to the true result a of the question. Then is k — a = r, and k — b ~ #, or b — a = r— s. And r according to the supposition on which the Rule is founded, r : s :: x — a : x— 6 ; hence, by multiplying extremes and means, rx — rb = sx— as; then, by transposition, rx — #x = rb — $a ; and, by division, x = r ^~~* g = the number sought, which is the rule when the results are both too little. If the results be both too great, so that ▲ and a are both mater than h ; then h — a = — r, and a — a = — *, or r and s are both negative ; hence— r : — t : : x — a : x— b f but — r : — $ t r -f r : -4- *, there- fore r : s : : x—a : x—b \ and the rest wHI be exactly as in the for- mer case. But if one result a only be too little, and the other a too great, or one error r positive, and the other » negative, then the theorem be- comes x = ^4— t which is the rule in this case, or whea the errors are unlike. DOVBU POSITION. 180 Virions or the question, as in Single Position ; and find how enuch each remit is different from the result mentioned in the question, calling these differences the errors, noting also whether the results are too great or too little. Then multiply each of the said errors by the contrary supposition, namely, the first position by the second error, and the second position by the first error. Then, If the . errors are alike, divide the difference of the pro- ducts by the difference of the errors, and the quotient will lie the answer. But if the errors are unlike, divide the sum of the pro- ducts by the sum of the errors, for the answer. Note, The errors are said to be alike, when they are either both too great or both too little ; and unlike, when one is too great and the other too little. SXAMPLB. 1* What number is that, which being multiplied by 6, the product increased by 18, and the sum divided by 9, the quo- tient should be 20 ? Suppose the two numbers 18 and 30. Then, First Position. 8econd Position. 18 Suppose 30 6 mult. 6 ProoC 27 6 108 18 add 180 18 162 18 *) 126 div. 9) 198 -0) 180 14 20 results true res. 22 20 20 :2d pos. 80 errors unlike mult. -2 18 1st pos. Er. >2 jorsyC \ 180 \ 36 36 sum 8) 216 27 sum of products Answer sought. 140 ARITHMETIC. RULE II. Find, by trial, two numbers, as near the true number as convenient, and work with them as in the question ; mark- ing the errors which arise from each of them. Multiply the difference of the two numbers assumed, or found by trial, by one of the errors, and divide the product by the difference of the errors, when they are alike, but by their sum when they are unlike. Or thus, by proportion : As the difference of the errors, or of the results, (which is the same thing), is to the difference of the assumed numbers, so is either of the errors, to the correction of the assumed number belonging to that error. Add the quotient, or correction, last found, to the number belonging to the said error, when that number is too little, but subtract it when too great, and the result will give the true quantity sought *. EXAMPLES. 1. So, the foregoing example, worked by this 2d rule, will be as follows : 30 positions 18 ; their diff. 12 -2 errors +6 ; least error 2 sum of errors 8 ) 24 ( 3 subtr. from the position 30 leaves the answer 27 Or, as 22 - 14 : 30 - 18, or as 8 : 12 : : 2 : 3 the cor- rection, as above. 2. A son asking his father how old he was, received this answer : Your age is now one-third of mine ; but 5 years ago, your age was only one-fourth of mine. What then are their two ages ? Ans. 15 and 45. 3. A workman was hired for 20 days, at 3s per day, for every day he worked ; but with this condition, that for every day he did not work, he should forfeit 1*. Now it so hap. • For since, by the supposition, r : t :: x : — a:x — b, therefore by division, r — f : s :: b — a: z — 6, or as b — a : 6 — a : : s :x — 6, for b — a if = r — $ : which is the 2d Rule. PRACTICAL QUESTIONS. 141 pened, that upon the whole he had 27 4* to receive. How many of the days did he work ? Ans. 16* 4. a and b began to play together with equal sums of money: a first won 20 guineas, but afterwards lost back} of what he then had ; after which b had four times as much as a. What sum did each begin with ? Ans. 100 guineas. 5. Two persons, a and b, have both the same income, a saves } of his ; but b, by spending 60/ per annum more than a, at the end of 4 years finds himself 100/ in debt. "What does each receive and spend per annum ? Ans. They receive 125/ per annum ; also a spends 100/, and b spends 150/ per annum. PRACTICAL QUESTIONS IN ARITHMETIC. Quest. 1. The swiftest velocity of a cannon-ball, is about 2000 feet in a second of time. Then in what time, at that rate, would such a ball move from the earth to the sun, admitting the distance to be 100 millions of miles, and the year to contain 365 days 6 hours ? Ans. 8 tVtVV years. Quest. 2. What is the ratio of the velocity of light to that of a cannon-ball, which issues from the gun with a ve- locity of 1500 feet per second ; light passing from the sun to the earth in 7£ minutes ? Ans. the ratio of 782222$ to 1. Quest. 3. The slow or parade-step being 70 paces per minute, at 28 inches each pace, it is required to determine at what rate per hour that movement is ? Ans. miles. Qukht. 4. The quick-time or step, in marching, being 2 paces per second, or 120 per minute, at 28 inches each ; at what rate per hour does a troop march on a rout, and how long will they be in arriving at a garrison 20 miles distant, allowing a halt of one hour by the way to refresh ? . ) the rate is 3^ T miles an hour. ' \ and the time 7$ hr, or 7h 17| min. Quest. 5. A wall was to be built 700 yards long in 29 days. Now, after 12 men had been employed on it for 11 dtiys, it was found that they had completed only 229 yards of the wall. It is required to determine how many men must be* added to the former, that the whole number of them may just finish the wall in the time proposed, at the same rate of forking. Ans. 4 men to be 142 ARITHMETIC* Quest. 6. Determine how far 500 millions of guineas will reach, when laid down in a strait line touching one an- other ; supposing each guinea to be an inch in diameter, as it is very nearly. Ans. 7891 miles, 728 yds, 2 ft. 8 in. Quest. 7. Two persons, a and b, being on opposite sides of a wood, which is 536 yards about, they begin to go round it, both the same way, at the same instant of time ; a goes at the rate of 1 1 yards per minute, and b 34 yards in 3 mi. nutes ; the question is, how many times will the wood be gone round before the quicker overtake the slower 1 Ans. 17 times. Quest. 8. a can do a piece of work alone in 12 days, and b alone in 14 ; in what time will they both together per- form a like quantity of work ? Ans. 6 T T days. Quest. 9. A person who was possessed of a | share of a copper mine, sold f of his interest in it for 1800/ ; what was the reputed value of the whole at the same rate ? Ans. 4000/. Quest. 10. A person after spending 20/ more than J of his yearly income, had then remaining 30/ more than the half of it ; what was his income ? Ans. 200/. Quest. 11. The hour and minute hand of a clock are exactly together at 12 o'clock ; when are they next together T Ans. at l T y hr. or 1 hr. 5> r min. Quest. 12. If a gentleman whose annual income is 15002, spend 20 guineas a week ; whether will he save or run in debt, and how much in the year ? Ans. save 408/, Quest 13. A person bought 180 oranges at 2 a penny, and 180 more at 3 a penny ; after which, selling them out again at 5 for 2 pence, whether did he gain or lose by the bargain ? Ans. he lost 6 pence. Quest. 14. If a quantity of provisions serves 1500 men 12 weeks, at the rate of 20 ounces a day for each man ; how many men will the same provisions maintain for 20 weeks, at the rate of 8 ounces a day for each man ? Ans. 2250 men. Quest. 15. In the latitude of London, the distance round the earth, measured on the parallel of latitude, is about 15550 miles ; now as the earth turns round in 23 hours 56 minutes, at what rate per hour is the city of London carried by this motion from west to east ? Ans. 649$}} miles an hour. Quest. 16. A father left his son a fortune, ± of which he ran through in 8 months : $ of the remainder lasted him 12 months longer ; after which he had 820/ left. What sum did the father bequeath his son ? Ans. 1913/ 6s 8d. Quest. 17. If 1000 men, besieged in a town, with pro* PRACTICAL QUESTIONS. 143 visions for 5 weeks, allowing each man 16 ounces a day, be reinforced with 500 men more ; and supposing that they can- not be relieved till the end of 8 weeks, how many ounces a day must each man have, that tho provision may last that time ? Ans. 6} ounces. Qukst. 18. A younger brother received 8400/, which was just J of his elder brother's fortune : What was the lather worth at his death ? Ans. 19200/. Qukst. 19. A person, looking on his watch, was asked what was the time of the day, who answered, It is between 5 and 6 ; but a more particular answer being required, he said that the hour and minute hands were then exactly to- gether : What was the time ? Ans. 27^ min. past 5. Quest. 20. If 20 men can perforin a piece of work in 12 days, how many men will accomplish another thrice as large in one-fifth of the time ? Ans. 300. Quest. 21. A father devised T \ of his estate to one of his sons, and T \ of the residue to another, and the surplus to his relict for life. The children's legacies were found to be 514/ 6* 8d different : What money did he leave the widow the use of? Ans. 1270/ 1* 9^d. Quest. 22. A person, making his will, gave to one child of his estate, and the rest to another. When these legacies came to be paid, the one turned out 1200/ more thun the other : What did the testator die worth ? Ans. 4000/. Quest. 23. Two persons, a and b, travel between Lon- don and Lincoln, distant 100 miles, a from London, and b from Lincoln, at the same instant. After 7 hours they meet on the road, when it appeared that a had rode l£ miles an hour more than b. At what rate per hour then did each of the travellers ride ? Ans. a 7§J and b miles. Quest. 24. Two persons, a and n, travel between Lon- don and Exeter, a leaves Exeter at 8 o'clock in the morn- ing, and walks at the rate of 3 miles an hour, without inter- naission ; and b sets out from London at 4 o'clock the same evening, and walks for Exeter at the rate of 4 miles an hour constantly. Now, supposing the distance between the two cities to be 130 miles, whereabouts on the road will they xtieet ? Ans. 09} miles from Exeter. Quest. 25. One hundred eggs being placed on the ground, in a straight line, at the distance of a yard from each other : How far will a person travel who shall bring them one by one to a basket, which is placed at one yard from the first egg? Ans. 10100 yards, or 5 miles and 1300 yds. Quest. 26. The clocks of Italy go on to Yvoura \ 144 ARITHMETIC. Then how many strokes do they strike in one complete re- volution of the index ? Ans. 300. Quest. 27. One Sessa, an Indian, having invented the game of chess, showed it to his prince, who was so delighted with it, that he promised him any reward he should ask ; on which Sessa requested that he might be allowed one grain of wheat for the first square on the chess board, 2 for the second, 4 for the third, and so on, doubling continually, to 64, the whole number of squares. Now, supposing a pint to con- tain 7680 of these grains, and one quarter or 8 bushels to be worth 27^ 6tZ, it is required to compute- the value of all the corn ? Ans. 6450468916285/ 17* 3d -Sitffff. Quest. 28. A person increased his estate annually by 100/ more than the j- part of it ; and at the end of 4 years found that his estate amounted to 10342/ Ss 9d. What had he at first ? Ans. 4000/. Qukst. 29. Paid 1012/ 10* for a principal of 750/, taken in 7 years before : at what rate per cent, per annum did 1 pay interest ? Ans. 5 per cent. Quest. 30. Divide 1000/ among a, b, c ; so as to give a 120 more, and b 95 less than c. Ans. a 445, b 230, c 325. Quest. 31. A person being asked the hour of the day, said, the time past noon is equal to £ths of the time till mid- night. What was the time? Ans. 20 min. past 5. Quf.8T. 32. Suppose that I have T ^ of a ship worth 1200/; what part of her have I left after selling J of £ of my share, and what is it worth ? Ans. 5 y„, worth 185/. Quest. 33. Part 1200 acres of land among a, b, c ; bo that b may have 100 more than a, and c (>4 more than b. Ans. a 312, b 412, c 476. Quest. 34. What number is that, from which if there be taken f of }, and to the remainder be added T 8 5 - of \> the sum will be 10? Ans. O^J. Quest. 35. There is a number which, if multiplied by f of | of 1J, will produce 1 : what is the square of that number ? Ans. 1 TT . Quest. 36. What length must be cut off a board, 8} inches broad, to contain a square foot, or as much as 12 inches in length and 12 in breadth ? Ans. 16 f $ inches. Quest. 37. What sum of money will amount to 138/ 2s 6d, in 15 months, at 5 per cent, per annum simple interest ? Ans. 130/. Quest. 38. A father divided his fortune among his three nucncAft Sjussrnoifs. 145 •om, a, a, c, giving i 4 u often as b 3, and o 5 at often as b 6 ; what was the whole legacy, supposing a's share was 40002? Ans. 95002. Quest. 39. A young hare starts 40 yards before a grey, bound, and is not perceived by him till she has been up 40 seconds ; she scuds away at the rate of 10 miles an hour, and the dog, on view, makes after her at the rate of 18 : how long will the course hold, and what ground will be run over, counting from the outaetting of the dog ? Ans. 60^ sec. and 530 yards run. Quest. 40. Two young gentlemen, without private for- tune, obtain commissions at the same time, and at the age of 18. One thoughtlessly spends 102 a year more than his pay ; but, shocked at the idea of not paying his debts, gives his creditor a bond for the money, at the end of every year, and also insures his life for the amount ; each bond costs him 30 ■billings, besides the lawful interest of 5 per cent, and to in- sure his life costs him 6 per cent. The other, having a proper pride, is determined never to ran in debt ; and, that he may assist a friend in need, per. •everes in saving 102 every year, for which he obtains an interest of 5 per cent, which interest is every year added to his savings, and laid out, so as to answer the effect of com- pound interest. Suppose these two officers to meet at the age of 50, when each receives from Government 4002 per annum ; tha the one, seeing his past errors, is resolved in future to spend no more than he actually has, after paying the interest for what he owes, and the insurance on his life. The other, having now something beforehand, means in future to spend his full income, without increasing his stock* It is desirable to know how much each has to spend per annum, and what money the latter has by him to assist the distressed, or leave to those who deserve it ? Ana. The reformed officer has to spend 662 19# 1}*5389& per annum. The prudent officer has to spend 4372 12s 1 lj«4379cf. per annum, and The latter has saved, to dispose of, 7522 19s 9-1890& ▼ax. /. 20 146 LOGARITHMS. OF LOGARITHMS *. Logarithms are made to ftrinWfe troublesome calcu- lations in numbers. This they doy because they perform multiplication by only addition, and division by subtraction, and raising of powers by multiplying the logarithm by the index of the power, and extracting of roots by dividing the logarithm of the number by the index of the root For, logarithms are numbers so contrived, and adapted to other numbers, that the sums and differences of the former shall correspond to, and show the products and quotients of the latter, dec. Or, more generally, logarithms are the numerical expo- nents of ratios ; or they are a series of numbers in arith- * The invention of Logarithms is due to Lord Napier, Baron of Merchiston, in Scotland, and is properly considered as one of the most useful inventions of modern times. A table of these numbers was first published by the inventor at Edinburgh, in the year 1614, in a treaties entitled Canon Mirificum Logarithmorum ; which was eagerly received by all the learned throughout Europe. Mr. Henry Bnggs, then pro- fessor of geometry at Gresham College, soon after the discovery, went to visit the noble inventor ; after which, they jointly undertook the arduous task of computing new tables ou this subject, and reducing them to a more convenient form than that which was at first thought of. But Lord Napier dying soon after, the whole burden fell upon Mr. Briggs, who, with prodigious labour and great skill, made an entire Canon, according to the new form, for all numbers from 1 to 20000, and from 90000 to 101000, to 14 places of figures, and published it at London in the year 1624, in a treatise entitled Arithmetica Logarithmic*, with directions for supplying the intermediate parts. This Canon was again published in Holland by Adrian Vlacq, in the year 1628, together with the Logarithms of all the numbers which Mr. Briggs had omitted ; but he contracted them down to 10 places of de- cimals. Mr. Briggs also computed the Logarithms of the sines, tan- Sents, and secants, to every degree, and centestn, or 100th part of a egree, of the whole quadrant ; and annexed them to the natural sines* tangents, and secants, which lie had before computed, to fifteen placet of figures. These tables, with their construction and use, were first Eublished in the year 1633, after Air. Briggs 's death, by Mr. Henry Getti* rand, under the title of Trigouometria Britomiica. LOGARITHMS. 147 metical progression, answering to another series of numbers in geometrical progression. rp. $0,1,2,3, 4, 5, 0, Indices, or logarithms, inus I 1, 2, 4, 8, 10, 32, 04, Geometric progression, ft i 0, 1, 2, 3, 4, 5, 6, Indices, or logarithms, w £ 1, 3, 9, 27, 81, 243, 729, Geometric progression. n« $ 0, 1, 2, 3, 1, 5, Indices, or logs. w $ 1, 10, 100, 1000, 10000, 100000, Geom. progres. Where it is evident, that the same indices serve equally for any geometric aerie* ; and consequently there may be an endless variety of system of logarithms, to the same com- mon numbers, by only changing the second term, 2, 3, or 10, &c. of the geometrical series of whole numbers ; and by interpolation the whole system of numbers may be made to enter the geometric series, and receive their proportional loga- rithms, whether integers or decimals. It is also apparent, from the nature of these series, that if any two indices be added together, their sum will be the in- dex of that number which is equal to the product of the two terms, in the geometric progression, to which those in- dices belong. Thus the indices 2 and 3, being added toge- ther, make 5 ; and the numbers 4 and 8, or the terms cor- responding to those indices, being multiplied together, make 82, wnich is the number answering to the index 5. In like manner, if any one index be subtracted from an- other, the difference will be the index of that number which Benjamin Ursinus also gave "a Table of Napier's Logs, and of sines, to every 10 seconds. And Chr. Wolf, in his Mathematical Lexicon, says that one Van Loser had computed them to every single second, but Ms untimely death prevented their publication. Many other authors have treated on this subject ; but as their numbers are frequently in ac- curate and incommodiously disposed, they are now generally neglect- ed. The Tables in most repute at present, nrc those of Gardiner in 4to, first published in the* year 1742 ; and my own Tables in 8vo, first printed in the year 1785, where the Logarithms of all numbers may be easily found from 1 to 10800000 ; and those of the sines, tangents, and secants, to any degree of accuracy required. Mr. Michael Taylor's Tables in large 4to, containing the common logarithms, and the logarithmic sines and tangents to every second of the quadrant, are very valuable. And, in France, the new book of logarithms by Callet; the 2d edition of which, in 1795, has the tables still further extended, and are printed with what are called stereo- types, the types in each page beng soldered together into a solid mass or block. Dodson's Antiiogarithmic Canon is likewise a very elaborate work, and used for finding the numbers answering to any given Vogpx\\\k\a, neb to 11 places. 148 LOGARITHM. is equal to the quotient of the two feme to which thoee io« dices belong. Thai, the index 6, minus the index 4, is = 2 ; and the terms corresponding to those indices are 64 and 16, whose quotient is = 4, which is the number answering to the index 2. For the same reason, if the logarithm of any number be multiplied by the index of its power, the product will be equal to the logarithm of that power. Thus, the index or loga- rithm of 4, in the above series, is 2;. ind if this number be multiplied by 3, the product will be «6 j. which is the loga- rithm of 64, or the third power-of 4. And, if the logarithm of any number be divided by the index of its root, the quotient will be equal to the logarithm of that root. Thus, the index or logarithm of 64 is 6 ; and if this number be divided by 2, the quotient will be = 3 ; which is the logarithm of 8, or the square root of The logarithms most convenient for practice, are such as are adapted to a geometric series increasing in a tenfold pro- portion, as in the last of the above forms ; and are those which are to be found, at present, in most of the common tables on this subject* The distinguishing mark of this system of logarithms is, that the index or logarithm of 10 is 1 ; that of 100 is 2 ; that of 1000 is 3 ; &c. And, in decimals, the logarithm of •! is — 1 ; that of *01 is — 2 ; thai of "001 is — 3, die. the log. of 1 being in every systesjt Whence it follows, that the logarithm of any number beV" tween 1 and 10, must be and some fractional parts ; and^ that of a number between 10 and 100, will he 1 and some? fractional parts ; and so on, for any other number whatever. And since the integral part of a logarithm, usually called the Index, or Characteristic, is always thus readily found, it is commonly omitted in the tables ; being left to be supplied by the operator himself) as occasion requires. Another Definition of Logarithms is, that the logarithm of any number is the index of that power of some other num- ber, which is equal to the given number. So, if there be »-r", then n is the log. of n ; where n may be either po- sitive or negative, or nothing, and the root or base r any number whatever, according to the different systems of lo- garithms. When n is = 0, then n is = 1, whatever the value of r is ; which shows, that the log. of 1 is always 0, in every system of logarithms. When n is = 1, then n is = r ; 140 to dial the radix r it always that number whose log. is 1, in every system. When the radix r is = 2 •718281826459 &c. the indices n are the hyperbolic or Napier's log. of the num- bers ]t ; so that n is always the hyp. log. of the number if or (2.718 &c.)\ But when the radix r is = 10, then the index n becomes the common or Briggs's log. of the number n : so that the common log. of any number 10* or n, is n the index of that power of 10 which is equal to the said number. Thus 100, being the second power of 10, will have 2 for its logarithm ; and 1000, being the third power of 10, will have 3 for its logarithm : hence also, if GO be = 10"** 7 , then is 1*69807 the common log. of 50. And, in general, the following de- cuple series of terms, to. 1C, 10', 10 9 , Iff, 10°, io-\ 10-», 10- 3 , 10- 4 , or 10000, 1000, 100, 10, 1, -1, -01, 001, -0001, have 4, 3, 2, 1, 0, -1, -2, -3, -4, for their logarithms, respectively. And from this scale of numbers and logarithms, the same properties easily follow, as above mentioned. problem. To compute the Logarithm to any of the Natural Number* 1, 2, 3, 4, 5, 6fc. RULE I*. Take the geometric series, 1, 10, 100, 1000, 10000, die. and apply to it the arithmetic series, 0, 1,2, 3, 4, &c. as logarithms. — Find a geometric mean between 1 and 10, or /between 10 and 100, or any other two adjacent terms of the aeries, between which the number proposed lies. — In like manner, between the mean, thus found, and the nearest ex- treme, find another geometrical mean ; and so on, till you arrive within the proposed limit of the number whose loga- rithm is sought. — Find also as many arithmetical means, in the same order as you found the geometrical ones, and these will be the logarithms answering to the said geometrical means. * The reader who wishes to inform himself more particularly con- cerning the history, nature, and construction of Logarithms, may con- ssK my Mathematical Tracts, vol. 1, lately published, where he wit sod his cariosity amply gratified. 150 LOGARITHMS. EXAMPX.B. Let it be required to find the logarithm of 9. Here the proposed number lies between 1 and 10. First, then, the log. of 10 is I, and the log. of 1 is fr; theref. (1 + 0) -r- 2 = % = *5 is the arithmetical mean, and -v/(10 X 1) = y/10 = 3*1622777 the geom. mean ; hence the log. of 0*1022777 is '5. Secondly, the log. of 10 is 1, and the log. of 3* 1622777 is -5; theref. (1 + -5) 2 = -75 ia the arithmetic a! mean, and ^(10 X3-102277?)=5-6234132 is the geom. mean ; hence the log. of 5*6234132 is »75. Thirdly, the log. of 10 is 1, and the log* of 5*0234132 is -75; theref. (1 + -75} 2 = 875 is the arithmetical mean, and v/(10 X 5*6234132) = 7*4989422 the geom. mean ; hence the log. of 7-4989422 is -875. Fourthly, the log.of 10 is 1, and the log. of 7-4989422 is -875; theref. (1 + -875) -r- 2 =-9375 is the arithmetical mean, and v/(10 X 7-4989422) = 8-6596431 the geom. mean ; hence the log. of 8-6596431 is -9375. Fifthly, the log. of 10 is 1, and the log. of 8-6596431 is -9375 ; theref. (l+-9375)-i-2=-96875 is the arithmetical mean, and v/( 10 X 8-6596431) = 9-3057204 the geom. mean ; hence the log. of 9-3057204 is -96875. Sixthly, the log. oT 8-6596431 is -9375, and the log. of 9-3057204 is -96875. theref. ( -9375+ -96875) -r-2= -953 125 is the arith-mean, and ^(8-6596431 X 9-3057204) = 8*9768713 the geo- metric mean ; hence the log. of 8-9768713 is -953125. And proceeding in this manner, after 25 extractions, it will be found that the logarithm of 8-9999998 is -9542425 ; which may be taken for the logarithm of 9, as it differs so little from it, that it is sufficiently exact for all practical pur- poses. In this manner were the logarithms of almost all the prime numbers at first computed. RUIjE II*. Let b be the number whose logarithm is required to be found ; and a the number next less than b } so that b — a =1, * For the demonstration of this rule, see my Mathematical Tablet, p. 109, &c. and my Tracts, vol. 1 . LOGARITHMS. 151 the logarithm of a being known ; and let * denote the sum of the two numbers a + b. Then 1. Divide the constant decimal -8085889038 &c. by *, and reserve the quotient : divide the reserved quotient by the square of *, and reserve this quotient ; divide this last quotient also by the square of s y and again reserve the quo- tient : and thus proceed, continually dividing the last quotient by the square of s , as long as division can be made. 2. Then write these quotients orderly under one another, the first uppermost, and divide them respectively by the odd numbers, 1, 3, 5, 7, 0, && as long as division can be made ; that is, divide the first reserved quotient by 1, the second by 3, the third by 5, the fourth by 7, and so on. 8. Add all these last quotients together, and the sum will be the logarithm of b a ; thorefore to this logarithm add also the given logarithm of the said next less number a, so will the last sum be the logarithm of the number b proposed. That is, Log. of b. is log. a + i X ( 1 + J- + -L + -L + &c. where n denotes the constant given decimal '8085889638 dec. examples. Ex. 1. Let it be required to find the log. of the number 2. Here the given number b is 2, and the next less number a is 1, whose log. is ; also the sum 2 + 1=3 = *, and its square * 3 = 9. Then the operation will be as follows : 3 9 9 9 9 9 9 O •80858890-1 •289529054 34109902 3571410 397100 44129 4903 515 01 1 3 5 7 9 11 13 15 •289529054 ( 32109902 ( 3574410 ( 397100 ( 44129 ( 4903 ( 545 ( 01 ( log. of ! { - add log. 1 - •289529(554 10723321 714888 50737 4903 440 42 1 •301029995 •000000000 log. of 2 •301029995 LOflARITMMB. Ex. 3. To compute the logarithm of the number 3. Here b = 3, the next less number a = 2, and the sum a + b = 5 = «, whose square «* is 25, to divide by which, always multiply by -04. Then the operation is as follows : 5 25 25 25 25 25 •868588964 173717793 6948712 277948 11118 445 18 1 ) 3 ) 5 ) 7 ) 11 ) 173717793 6948712 277948 11118 445 18 ( [ Si log. off log. of 2 add 173717798 2316*37 55590 1588 50 2 176091260 3U1029995 log. of 3 sought -477121255 Then, because the sum of the logarithms of numbers, gives the logarithm of their product ; and the difference of the logarithms, gives the logarithm of the quotient of the numbers ; from the above two logarithms, and the logarithm of 10, which is 1, we may obtain a great many logarithms, as in the following examples : EXAMPLE 3. Because 2x2= 4, therefore to log. 2 . -301029995| add log. 2 - -301029995} sum is log. 4 -602059991} EXAMPLE 4. Because 2X3 = 6, therefore to log. 2 - -301029995 add log. 3 - -477121255 sum is log. 6 -778151250 KXAMPUB 5. Because 2 s = 8, therefore log. 2 - -301029995 mult, by 3 3 example. 6. Because 3" = 9, therefore log. 3 - -477121254/, mult by 2 2 gives log. 9 -954242509 EXAMPLE 7. Because y = 5, therefore from log. 10 1 -000000000 take log. 2 -301029995} leaves log. 5 -698970004} EXAMPLE 8. Because 3X4 = 12, therefore to log. 3 - -477121255 add log. 4 * -602059991 gives log. 8 -903089987 j gives log. 12 1-079181246 LOGARITHMS. 158 And thus, computing by this general rule, the logarithms to the other prime numbers, 7, 11, 13, 17, 19, 23, dec. and then using composition and division, we may easily find as many logarithms as we please, or may speedily examine any logarithm in the table*. Description a*d Use if the Table of Logarithms. Having explained the manner of forming a table of the logarithms of numbers, greater than unity ; the next thing to be done is, to show how the logarithms of fractional quan- tities may be found. In order to this, it may be observed, that as in the former case a geometric series is supposed to increase towards the left, from unity, so in the latter case it is supposed to decrease towards the right hand, still be- ginning with unit; as exhibited in the general description, page 152, where the indices being made negative, still show the logarithms to which they belong. Whence it appears, that as + 1 is the log. of 10, so — 1 is the log. of T \ or -1 ; and as + 2 is the log. of 100, so — 2 is the log. of T ^ or *01 : and so on. Hence it appears in general, that all numbers which con- sist of the same figures, whether they be integral, or frac- tional, or mixed, will have the decimal parts of their loga- rithms the same, but differing only in the index, which Mill be more or less, and positive or negative, according to the place of the first figure of the number. Thus, the logarithm of 2051 being 3-423410, the log. of tV» or iiv> or T? ! o7> & c - P art °f *t W 'M 06 113 follows : Numbers. Logarithms. 2 6 5 1 2 0-51 2 6 5-1 2-651 •2 6 5 1 •0 2 6 5 1 •0 2 6 5 1 3-4 2 3 4 1 2-4 2 3 4 1 1-4 2 3 4 1 -4 2 3 4 1 -1-4 2 3 4 1 -2-4 2 3 4 1 -3-4 2 3 4 1 * There are, besides these, many other ingenious methods, which later writers have discovered for finding the logarithms of numbers, la a much easier way than by the original inventor ; but, as they cannot lie understood without a kno'wledgc; of some of the higher branches of Uie mathematics, it is thought proper to omit them, and to refer the a-asderto those works which are written e.xpres ly on the subject. It Vrould likewise much exceed the limits of this compendium, Vo Oat all the peculiar artifices that are made use of fur COn*Vru>cftfe&u& Vol. 1. 21 154 LOGARITHMS. Hence it also appears, that the index of any logarithm, is always less by 1 than the number of integer figures which the natural number consists of: or it is equal to the distance of the first figure from the place of units, or first place of in- tegers, whether on the left, or on the right, of it : and this index is constantly to be placed on the left-hand side of the decimal part of the logarithm. When there are integers in the given number, the index is always affirmative ; but when there are no integers, the index is negative, and is to be marked by a short line drawn before it, or else above it. Thus, A number having 1, 2, 3, 4, 5, &c. integer places, the index of its log. is 0, 1, 2, 3, 4, die. or 1 less than those places. And a decimal fraction having its first effective figure in the 1st, 2d, 3d, 4th, &c, place of the decimals, has always — 1,-2, — 3, — 4, &c. for the index of its logarithm. It may also bo observed, that though the indices of frac- tional quantities are negative, yet the decimal parts of their logarithms are always affirmative. And the negative mark ( — ) may be set either before the index or over it. I. TO FIND IN T1IE TABLE, THE LOGARITHM TO ANY NTMHER*. 1. If the given Number be less than 100, or consist of only two figures ; its log. is immediately found by inspection in the first page of the table, which contains all numbers from 1 to 100, with their logs, and the index immediately annexed in the next column. So the log. of 5 is 0-G9S970. The log. of 23 is 1 -301728. The log. of 50. is 1 -008070. And so on. 2. If the Number be more than 100 but less than 10000; that is, consisting of either three or four figures : the decimal part of the logarithm is found by inspection in the other pages of the table, standing against the given number in this manner ; viz. the first three figures of the given number in the first column of the page, and the fourth figure one of those along the top line of it ; then in the angle of meeting are the last lour figures of the logarithm, and the first two figures of the same at the beginning of the same line in the second entire table of these numbers ; but any information of this kind, which the learner may wish to obtain, may he found in my Tables. See alto the article on Logarithms in the 2d volume, p. 340. &c. ' See the table of Logarithms, at the end of this volume. L0GABITHK8. 156 column of the page : to which is to be prefixed the proper index which is always 1 less than the number of integer figures. So the logarithm of 251 is 2 399674, that is, the decimal •399674 found in the table, with the index 2 prefixed, because the given number contains three integers. And the log. of 94-09 is 1-532627, that is, the decimal -532627 found in the table, with the index 1 prefixed, because tho given number contains two integers. 2. But if the given Number contain more than four figures ; take out the logarithm of the first four figures by inspection in the table, as before, as also the next greater logarithm, subtracting the one logarithm from the other, as also their corresponding numbers the one from the other. Then say, As the difference between the two numbers, Is to the difference of their logarithms, So is the remaining part of the given number, To the proportional part of the logarithm. Which part being added to the less logarithm, before taken out, gives the whole logarithm sought very nearly. EXAMPLE. To find the logarithm of the number 34-0926. Tho log. of 340900, as before, is 532627. And log. of 341000 - - is 532754. The ditfs. are 100 and 127. Then, as 100 : 127 : : 26 : 33, the proportional part. This added to - - - 532627, the first log. Gives, with the index, 1-532660 for the log. of 34-0926. 4. If the number consist both of integers and fractions, or is entirely fractional ; find the decimal part of the logarithm the same as if all its figures were integral ; then this, having prefixed to it the proper index, will give the logarithm re- quired. 5. And if the given number be a proper vulgar fraction : subtract the logarithm of the denominator from the loga- rithm of the numerator, and the remainder will be the loga- rithm sought ; which, being that of a decimal fraction, must *" always have a negative index. 6. But if it be a mixed number ; reduce it to an impro- per fraction, and find the difference of the logarithm of the numerator and denominator, in the same manner as before. 156 LOGARITHMS. EXAMPLES. 1. To find the log. of J?. Log, of 87 - 1-568202 Log. of 94 - 1-973128 Dif. log. of» J —1-595074 Where the index 1 is negative 2. To find the log. of 17$f • First, 17j| = Then, Log. of 405 . 2-807455 Log. of 2ft - 1-361728 Difl ]p|[. of 17iJ 1-245727 n. TO FIND THE NATUHAX ITDMBSS TO ANY GIVEN LOGARITHM. This is to be found in the tables by the reverse method to the former, namely, by searching for the proposed loga- rithm among those in the table, and taking out the corre- sponding number by inspection, in which the proper number of integers are to be pointed off, viz. 1 more than the index. For, in finding the number answering to any given logarithm, the index always shows how far the first figure must be removed from the place of units, viz. to the left hand, or in- tegers, when the index is affirmative ; but to the right hand, or decimals, when it is negative. EXAMPLES. So, the number to the log. ]h532S82 is 34-11. And the number of the log. "l'532b82 is -3111. But if the logarithm cannot be exactly found in the table ; take out the next greater and the next less, subtracting the one of these logarithms from the other, as also their natural numbers the one from the other, and the less logarithm from the logarithm proposed. Then say, As the difference of the first or tabular logarithms, Is to the difference of their natural numbers, So is the differ, of the given log. and the least tabular log. To their corresponding numeral difference. Which being annexed to the least natural number above ta- ken, gives the natural number sought, corresponding to the proposed logarithm. EXAMPLE. So, to find the natural number answering to the given logarithm 1*532708. L04AJUTHM8. 157 Here the next greater and next less tabular logarithms, trith their corresponding numbers, are as below : Next greater 532754 its num. 341000 ; given log. 532708 Next less 532627 its num. 340900 ; next less 532627 — 81 learly the numeral differ, i is the number sought, marking off two integers, because the index of th e given logarithm is 1. Had the index been negative, thus 1-532708, its com- ■ number would have been 340964, wholly decimal. MULTIPLICATION BY LOGARITHMS. RULE. Take out the logarithms of the factors from the table, then add them together, and their sum will be the logarithm of the product required. Then, by means of the table, take out the natural number, answering to the sum, for the pro- duct sought. Take care to add what is to be carried from the decimal part of the logarithm to the affirmative index or indices, or else subtract it from the negative. Also, add the indices together when they are of the same kind, both affirmative or both negative ; but subtract the less from the greater, when the one is affirmative and the other negative, and prefix the sign of the greater to the re- mainder. EXAMPLES. 1. To multiply 23-14 by 5062 Numbers. Logs. 23 14 . 1-364363 5 062 . 0-704322 2. To multiply 2-581926 by 8«457291 Numbers. Logs. 2- 581920 . 0-411944 3- 457291 . 0-538786 **rodoct 117-1347 2 068685 Prod. 8-92648 . 0-950680 158 DIVISION BY &OGABXTHK8. 3. To mult. 3-902 and 597 16 and -0314726 all together. Numbers. 3-902 597 16 •0314728 0-59F287 ! 2-776091 -2-497935 Prod. 73-3333 . 1-865313 Here the — 2 cancels the 2, and the 1 to carry from the decimals is set down. 4. To mult. 3-586, and 2-1046, and 0-8372, and 0-0294 all together. Numbers. Logs. 3 586 - 0-554610 2-1046 - 323170 0-8372 , -1-922829 0-0294 -2-468347 Prod. 01857618 - 1 -268956 Here the 2 to carry cancels the -2, and there remains the —1 to set down. DIVISION BY LOGARITHMS. RULE. From the logarithm of the dividend, subtract the loga- rithm of the divisor, and the number answering to the re- mainder will be the quotient required. Change the sign of the index of the divisor, from affirm- ative to negative, or from negative to affirmative ; then take the sum of the indices if they be of the same name, or their difference when of different signs, with the sign of the greater, for the index to the logarithm of the quotient. Also, when 1 is borrowed, in the left-hand place of the decimal part of the logarithm, add it to the index of the divisor when that index is affirmative, but subtract it when negative ; then let the sign of the index arising from hence be changed, and worked with as before. EXAMPLES. 1. To divide 24163 by 4567. Numbers. Logs. Dividend 24163 - 4 383151 Divisor 4567 - 3-659631 2. To divide 37 -1 49 by 523-76. Numbers. Logs. Dividend 37149 - 1-569947 Divisor 523-76 . 2 719132 Quot. 5-29078 0-723520 Quot. -0709275-2-85081* UfTOLUTIOM BY LOGARITHMS. 150 8. Divide -06314 by -007241 Numbers. Logs. Dividend -06314-2-800305 Divisor -007241-3-859799 4. Todivide-7438by 12-9476. Numbers. Logs* Divid. -7438 -1-871456 Divisor 12-9476 1-112169 Quo*. 8-71979 0-940506 Quot. -057447 - 2-750267 . '•' v — «— Here 1 ctjjirfcd from • the Here the 1 taken from the decimals to pfc 8, jnake* it — 1, makes it become —2, to become — % wen from set down, the other —2, leaves re* maining. Note. The RuletaftThree, or Rule of Proportion, is per- formed by adding the logarithms of tho 2d and 3d terms, jnnd subtracting that of the first term from their sum. In- stances will occur in Plain Trigonometry. INVOLUTION BY LOGARITHMS. RULE. Take out the logarithm of the given number from the ta- ble. Multiply the logarithm thus found, by the index of the power proposed. Find the number answering to the product, and it will be the power required. Note. In multiplying a logarithm with a negative index, by an affirmative number, the product will be negative. But what is to bo carried from the decimal part of the logarithm, will always be affirmative. And therefore their difference wilj be the index of the product, and is always to be made of the same kind with the greater. EXAMPLES. 1. To square the number 2-5791. Numb. Log. Root 2-5791 - - 0-411468 The index - - 2 Power 6-65174 0-822936 2. To find the cube of 3-07146. Numb. Log. Root 3 07146 - - 0-487345 The index - - 3 Power 28-9758 1-462035 ■VOLUTION BY LOGARITHMS. 3. To raise •09163 to the 4th power. Numb. Log. Root •09163 —2-962038 The index - - 4 Pow. -000070494— 5-848152 Here 4 times the negative index being — 8, and 3 to carry, the difference — 5 is the index of the product. 4. To raise 1*0045 to the 365th power. Numb. Log. Root 1 '0045 - - 0-001950 Thejndex - . 365 9750 ■ 11700 5850 0-711750 EVOLUTION BY LOGARITHMS. Take the log. of the given number out of the table. Divide the log. thus found by the index of the root. Then the number answering to the quotient will be the root. Note. When the index of the logarithm, to be divided is negative, and does not exactly contain the divisor, without some remainder, increase the index by such a number as will make it exactly divisible by the index, carrying the units borrowed, as so many tens, to the left-hand place of the deci- mal, and then divide as in whole numbers. EXAMPLES. 1. To find the square root of 365. Numb. Log. Power 365 2) 2-562293 Root 19-10496 1-281146£ 2. To find the 3d root of 12345. Numb. Log. Power 12345 3)4 091491 Root 231116 1-363830} 3. To find the 10th root of 2. Numb. Log. Power 2 - 10) 0-301030 Root 1-071773 0-030103 4. To find the 365th root of 1-045. Numb. Log. Power 1 045 365) 019116 Root 1-000121 0-000052| EVOLUTION BY tOGA 161 5- To find </ <m. Numb. Log* Power 093 2) —2-968483 Root -304959 —1-48424!* Here the divisor 2 is con- tained exactly once in the ne- gative index — 2, and there- fore the index of the quotient it — 1. 6* To find the -00048. Numb. Log. Power -00048 3)— 4-681241 Root -0782973 —2-893747 Here the divisor 3, not being exactly contained in —4, it is augmented by 2, to make up 6, in which the divisor is contained juit 2 timet; then the 2, thus borrowed, being carried to the de- ' * aal figure 6, makes 86, which dividV by S, gives 8, &c. 7. To find 3-1416 X 82 X ff- 8. To find 029IG X 751-3 X fl T 9. Afl 7241 : 3-58 : : 20-46 : ? 10, : -v/ff : : 6-927 : ? Vol. I. 22 ALGEBRA. DEFINITIONS AND NOTATION. . 1. Algebra is the science of investigation by means of symbols. It is sometimes also called Analysis ; and is ft general kind of arithmetic, or universal way of computa- tion. 2. In this science, quantities of all kinds are represented by the letters of the alphabet. And the operations to to performed with them, as addition or subtraction, &c. are de- noted by certain simple characters, instead of being express- ed by words at length. 3. In algebraical inquiries, some quantities are known or E'ven, viz. those whose values are known : and others un- iown, or are to be found out, viz. those whose values are not known. The former of these are represented by the leading letters of the alphabet, a, b, c, d, <3uc. ; and the latter, or unknown quantities, by the final letters, *, y, «, tt, &c. 4. The characters used to denote the operations, are chiefly the following : + signifies addition, and is named plus. — signifies subtraction, and is named minus. X or • signifies multiplication, and is named into. signifies division, and is named by. y/ signifies the square root ; the cube root ; (/ the 4th root, dtc. ; and the nth root. : : : signifies proportion. =s signifies equality, and is named equal to* And so on for other operations. Thus a + b denotes that the number represented by b is to be added to that represented by a. <x — 6 denotes that the number represented by b is to be subtracted from that represented by a. a*b denotes the difference of a and 6, when it is not known which is the greater. vzrorrnoiai jam HbTATiozr. 163 Hi, or a X b, or a . b f expresses the product, by multipli- cation of the numbers represented by a and 4. a -f- ft, or denotes, that the number represented by a b is to be divided by that which is expressed by b. a i b : : e : a\ signifies that a is in the same proportion to b> as e is to a\ , .y # ■= a — & 4- * is an equation, expressing that * is equal to the difference of a and b, added to the quantity c. a, or a*, denotes the square root of a.; $/a, or a^, the eubexoot of a ; and or a' the cube root of the square of a ; JL * also ^/a, or a w , is theath root of a; and ^/a n ora m is the nth power of the mth root of a, or it is a to the — power. fn a 1 denotes the square of a ; a 3 the cube of a ; a 4 the fourth power of a ; and a n the nth power of a. a + b X c, or (a + b) c, denotes the product of the com- 6wnd quantity a + b multiplied by the simple quantity c. sing the bar , or the parenthesis ( ) as a vinculum, to connect several simple quantities into one compound. a + b -f- a — b 9 or a - ^" , , expressed like a fraction, means a — o the quotient of a + b divided by a— b. i/ab + cd 9 or ^ai+crf)^, is the square root of the com* pound quantity ab + cd. And e y/ ab + or c (ab + cd)^ 9 denotes the product of c into the square root of the com- pound quantity ab + cd. a + 6 — c 3 , or (a + b — cf denotes the cube, or third power, of the compound quantity a + b — c. 8a denotes that the quantity a is to be taken 3 times, and 4(a + b) is 4 times a + 6. And these numbers, 3 or 4, showing how often the quantities are to be taken, or multi- plied, are called Co-efficients. Also fx denotes that x is multiplied by { ; thus J Xx or Jx. 5. Like quantities, are those which consist of the same letters, and powers. As a and 3a ; or 2ab and 4ab ; or 30*60 and — 5a*ftc. 6. Unlike Quantities, are those which consist of different letters, or different powers. As a and b ; or 2a and a 9 ; or 4aP and 3abc. 7. Simple Quantities are those which consist of one tettft. only. As 3a, or bob, or 6ak?. 1M .* ALGEBRA. " 8. Compound Quantities are thoae which consist of two or more terms. As a+b, or 2a— 3c, or a+2b - 3c. 9. And when the compound quantity consists of two terms, it is called a Binomial, as a+b ; when of three terms, it is a Trinomial, as a+2b — 3c ; when of four terms, a Quad ri no- mi al, as 2a — 3b+c—4d ; and so on. Also a Multinomial or Polynomial, consists of many terms. 10. A Residual Quantity, is a binomial having one of the terms negative. Asa— 2b. 11. Positive or affirmative Quantities, are those which are to be added, or have the sign +. As a or + a, or obi for when a quantity is found without a sign, it is understood to be positive, or have the sign + prefixed. 12. Negative Quantities, are those which are to be sub- tracted* As — a, or —2ab, or — Sab\ * 13. Like Signs, are cither all positive ( + ), or all nega- tive ( - ). 14. Unlike Signs, are when some are positive ( + ), and others negative ( — ). 15. The Co-efficient of any quantity, as shown above, is the number prefixed to it. As 3, in (he quantity Sab. 16. The power of a quantity (a), is its square (a 8 ), or cube (a 3 ), or biquadrate (a 4 ), &c. ; called also, the 2d power, or 3d power, or 4th power, &c. 17. The Index or Exponent, is the number which denotes the power or root of a quantity. So 2 is the exponent of the square or second power a 2 ; and 3 is the index of the cube or 3d power ; and £ is the index of the square root, at or y/a ; and \ is the index of the cube root, a^, or 18. A Rational Quantity, is that which has no radical sign or index annexed to it. As a, or Sab. 19. An Irrational Quantity, or Surd, is that of which the value cannot be accurately expressed in numbers, as the square root of 2, 3^J^> Surds are commonly expressed by means of the radical sign </ : as y/2, or y/a, or -J/a", or a$. 20. The Reciprocal of any quantity, is that quantity in- verted, or unity divided by it. So, the reciprocal of a, or is i, the reciprocal of % is that of — ~ is ^— -. la b a* x + y a 21. The letters by which any simple quantity is expressed, may be ranged according to any order at pleasure. So the product of a and b, may be cither expressed by ab, or ba ; DEFINITIONS AND NOTATION. * 165 and the product of a, 6, and c, by either abc f or acb, or hoc, or bca> or cab, or cba ; as it matters not which quantities are placed or multiplied first. But it will be sometimes found convenient in long operations, to place the several letters according to their order in the alphabet, as a be, which order also occurs most easily or naturally to the mind. 22. Likewise, the several members, or terms, of which a compound quantity is composed, may be disposed in any order at pleasure, without altering the value of the significa- tion of the whole. Thus, 3a — 2a6+4a6c may also be writ- ten 2a+4abc—2ab, or 4abe+oa -2ab 9 or -2ab+3a+4abc, fee. ; for all these represent the same thing, namely, the quantity which remain^, when the quantity or term iab is subtracted from the sum of the terms or quantities 3a and 4abc» But it is most usual and natural, to begin with a po- sitive term, and with the first letters of the alphabet. SOXV EXAMPLES FOR PRACTICE. In finding the numeral values of various expressions, or com- binations, of quantities. Supposing a=6, and b =5, and c=4, and a*=l, and e=0. Then 1. Will a a + 3aft-c 3 =36 + 90-16 = 110. 2. And 2a*-3a 2 b + c 3 = 432-540 + (54 = - 44. 3. And a 3 x(a+6)-2ate = 36 X 11-240=156. a 3 21ft 4. And +c==^- + 10 = 12+ 16 = 28. 5. And y/2ac+cr or (2ac + c 3 )^ = y/64 = 8. A A 1 / _!_ 2lfC « 0_1_ 40 * 6 - An(1 ^ c + 7C^H^) = 2 + ¥ = 7 - * a j < l3 ~"_v/(^ 3 ~~ ac ) 36 — 1 35 7 * And Sa-v/^+^T) " VJZ7 " "5 ~ 7 ' 8. And t/(* 3 -ac)+ v^(2flc+c»)=l + 8=9. 9. And v'^cT^^+f 1 ) = vT 25 - 2 * + 8) = 3. 10. And <r& + c-d = 183. 11. And0a6-106 2 + c=24. 3? 12. And — - X d = 45. c 18. Andl±*xj = 13J. 166 AMKBRA. 14. And a+b a — b = 1|. 15. And— + e = 45. c 16. And — X « = 0. c 17. And (& — c) X (<*— = L ia And (a+ 6) — (c — 4) = 8 19. And (a + b) — c — d ^ Q. 20. And afc X (P= 144. 21. And acd — d = 23. 22. And o% + Ve + d = V. 23. Aod^X^^lSi. <J — e c — d 24. And v^a 8 + ft 2 — — b' = 4-4936249. 25. And 3ac» + i/a 3 — V = 292-497942. 26. And 4a 1 — 3a ^a 8 — \ab = 72. ADDITION. Addition, in Algebra, is the connecting the quantities together by their proper signs, and incorporating or uniting into one term or sum, such as are similar, and can be united. As 3a -f- 26 - 2a = a + 2o, the sum. The rule of addition in algebra, may be divided into three cases : one, when the quantities are like, and their signs like also ; a second, when the quantities are like, but their signs unlike; and the third, when the quantities are unlike. Which was performed jftfelows*. •The reasons on which these operation! are founded, will readily ap- pear, by a little reflection on the nature of the quantities to be added, or collected together. For, with regard to the first example, where the quantities are 3a and 6a, whatever a represents in the one term, it will represent the same thing in the other; so that 3 times any thing and 6 times the same thing, collected together, must needs make 8 times thatching. As if a denote a shilling ; then 3a is 3 shillings, and 6a is 6 shillings, and their sum 8 shillings. In like manner, —2ab and — 1mb 9 or —2 times any thing, and —7 timet the same thing, make — f> times that thing. * i ADDITION. • 167 CASE I. When the Quantities are Like, and have Like Signs. Add the co-efficients together, and set down the sum ; after which set the common letter or letters of the like quan- tities, and prefix the common sign + or — . Thus, 8a added to 6a, makt a 8a. And — 2ab added to — 7ab, makes —dab. And 5a + 7b added to 7a + U t makes 12a + 10*. ; FOR PRACTICE. bxy 9a — 5bx 2bxy 5a — 4bx 5fey 12a — 2bx bxy a — 7bx Zbxy 2a — bx Uxy 32a — 22bx 18bxy As to the second case, in which the quantities are like, but the signs unlike; the reason of its operation will easily appear, by reflecting, that addition means only the uniting of Quantities together by means of the arithmetical operations denoted by their signs-)- and — , or of addi- tion and subtraction ; which being of contrary or opposite natures, the one co-efficient must be subtracted from the other, to obtain the incor- porated or united mass. As to the third case, where the quantities are unlike, it is plain that such quantities cannot be united into one, or otherwise added, than by means of their signs : thus, for example, if a be supposed to represent a crown, and 6 a shilling ; then the sum of a and u can be neither 2a nor 2*, that is, neither 2 crowns nor 2 shillings, but only 1 crown plus 1 •Billing, that is a +6. In this rule, the word addition is not very properly used ; being much too limited to express the operation here performed. The business of this operation is to incorporate into one mass, or algebraic expression, different algebraic quantities, as far as asUStoa) incorporation or union is possible ; and to retain the algebraic sajfkslbr doing it, in cases where the former is not possible. When wtfjlfo several quantities, some affirmative and some negative ; and the relation of these quantities can in the whole or In part be - discovered ; such incorporation of two or more quantities into one, is plainly effected by the foregoing rules. It may seem a paradox, that what is called addition in algebra, should sometimes mean addition, and sometimes subtraction. But the para- dox wholly arises from the scantiness of the name given to the alge- braic process; from employing an old term in a new and more enlarged sense. Instead of addition, call it incorporation, or union, or g&iking a Islands, or give it any name to which a more extensive idea may be annexed, than that which is usually implied by the word addition : and the paradox vanishes. 168 * AMBMU. 3* dx*+5xy 2<i* — 4y 2* **+ xy 4ax — y 4s 2* a +4*y ax — 8y % 5x*+2*y 5ax — 5y 6x 4x*+3*y 7ax — 2y 15* 5*y 14xy 22xy Ylxy l\xy \xy 30- 13**- -3*y 23- 10**- •4xy 14- 14**- •7xy 10- 16**- •5xy 16- 20**- ' x 9 5xy — 3x + 4a& 8xy— 4x + Sab 3xy — 5* + bob xy— 2* + afc 4*y— * + 7ab CASE II. When the Quantities are alike, ha have Unlike Signs. Add all the affinnatif^co-efficients into one sum, and all the negative ones into'aoother, when there are several of a kind. Then subtract the less sum, or the less co-efficient, from the greater, and to the remainder prefix the sign of the greater, and subjoin the common quantity or letters. So + 5fg and — 3a, united, make + 2a. And — 5a and — 3a, united, make — 2a. ADDITION. 169 OTHER EXAMPLES FOE PRACTICE. + Qx* + 3y - 5x* + 4y — 16x 3 + by + 3s 3 — ly + 2X 3 — 2y — 8x* + 3y + 4ab+ 4 — 4ab + l2 + 7ab— 14 + ab+ 3 — bob— 10 — 3ox* + ax* + 5ax* — 6ax* H-lOv/ax — 3v/ax + Ay/ax — Yty/ax + 3y + 4ax* — y — bax^ + 4y + 2ax* — 2y + 6ax* case m. When the Quantities arc Unlike. Having collected together all the like quantities, as in the two foregoing cases, set down those that are unlike, one after another, with their proper signs. : W - EXAMPLES.™ Sxy 6xy — 12x* 4ax — 130 + 3x* 2ax -4X 1 + 3xy Sx 3 + 3ax + 9*" -5xy +4x« - 2xy Ixy - 4x* + 90 fax -3xy + 4r» yx + 40 -Ox 8 — 2xy+8ax 4xy— 8x" 7ax + 8x* + 7xy Vol. I. 23 170 ALGEBRA. 9x Y 1 4ax - 2x* 9 + x — 5y -li*y 5ax + Zry 2x + 7-/*? + 5y + 3oxy Sif — tex 5y + 3^/ax — 4y — 3x a + 26 10— Ay/ax + 4y 2-v/xy + 14x 3x + 2y —9 + 3-v/jy Add a+6 and 3a — 56 together. Add 5a — 8x and 3a — 4x together. Add 6x- 56+a+8 to — 5a — 4r+4fc — 3. Add a+26-3c-10 to 36 -4a + 5c + 10 and 56 -c. Add a +6 and a — 6 together. Add 3a + 6-10 to>c-<*-a and — 4c + 2a- 36 -7. Add 3a a +6 3 -cto 2rt6-3« 2 +6c— -6. Add a 3 +6 3 c-6 2 to a6 3 ~a6c+6 3 . Add 9a-86 -f 10x-6a*-7c + 50 to $r-3a-5c +46 +6d -10. SUBTRACTION. Set down in one line the first quantities from which the subtraction is to be made ; and underneath them place all the other quantities composing the subtrahend ; ranging the like quantities under each sifter, as in Addition. Then change all the signs (+ and — ) of the lower line, or conceive them to be changed ; after which, collect all the terms together as in the cases of Addition*. * This rule is founded on (he consideration, that addition and sub- traction are opposite to each other in their nature and operation, as are the signs -{-and — , by which they are expressed and represented. So that, since to unite a negative quantity with a positive one of the same kind, has the effect of diminishing it, or subducting an equal positive •UBTRAOTIO.N. 1T1 XXAXPLK3. From la % — 36 9x* ~ 4y + 8 8xy — 3 + 6z — y Take 2a a — 86 bV+5y-4 4xy — 7 — 6x — 4y *l*y + 4 + 12x+3y — 20 — 6x— 5xy 3xy — 9x X 8 — 2ay Rem. 7xy— 12 2$f£%fc — 8 -r-28+3x — 8ry+2ay From 8x»y + 6 5 > /xy + 2x x /xy 7x 3 + 2^/z- 18+ 36 Take— 2x=y + 2 7 x /xy + S—2xy Ox 3 — 12 + 56+ ** Rem. 5xy — 30 7x 3 — 2 (a + b) 3xy 7 + 20a + 10) 7xy — 50 2x 3 ~4(a + 6) Ixy + 12a y/(xy + 10) Rem. 4a*+56 3x 2 — 9y+12 From 5xy— i\5^^ — 3y — 4 Take— 2xy + 6 -4Qfgjb+4 From a + 6, take a — b. From 4a + 46, take b + a. From 4a — 46, take 3a + 56. From 8a — 12x, take 4a — 3x* From 2x — 4a — 26 + 5, take 8— 56 + a + 6x. From3a + 6+c — d — 10, takec + Ua — d. From 3a + b + c — d — 10, take 6— 10 + 3a. From 2a6 + 6 3 — 4c + be — 6, take 3a 3 — r + 6 3 . From a 3 + 36"c + 06* — a6c, take 6 2 + a6 a — a6c. From 12x + 6a— 46 +40, take 46 — 3a + 4x + 6d— 10. • From 2x — 3a + 46 + 6c — 50, take 9a + x + 66 — 6c 40. From 6a — 46 — 12c + 12x, take 2a — 8a + 46 — 5c. ^De from it, therefore to subtract a positive (which is the opposite of ^anitiog or adding) is to add the equal negative quantity. In like map. -^«r, to subtract a negative quantity, is the same in effect as to add or ***rfte an equal positive one. So that, changing the sign of a quantity ^*om-fto— , or from — to -)-, changes its nature from a subductive Quantity to an additive one ; and any quantity is in effect subtracted, by <k*raly changing its sign. TO AL6KBKA. MULTIPLICATION. This consists of several cases, according as the factors are simple or compound quantities. cash. When both the Factors are Simple Quantities. Fcbst multiply the co-efficienti .of ijie two terms together, then to the product annex all the letters in those terms, which will give the whole product required. Note*. like signs, in the factors, produce +, and unlike signs — , in the products. EXAMPLES. 10a —2a 7a — 6* 2b + 2b -4c — 4a 20ab —Sab —28ac +24ax * That this rule for the signs is true, may be thus shown. 1. When -f-a is to be multiplied by + e; the meaning is, that + * is to be taken as many times as there are units in c ; and since the sum of any number of positive terms is positive, it follows that -faX+o makes 4- ac 2. When two quantities are to be multiplied together, the result will be exactly the same, in whatever order they are placed ; for a times c is the same as c times «, and therefore, when — a is to be multiplied by + $ t or + e by — a: this is the same thing as taking — a as many times as there are units in 4-c ; and as the sum of any number of negative terms is negative, it follows that — a X + c, or + a X — • « make or pro- duce — ac. 3. When — a is to be multiplied by — c : here — a is to be subtract- ed as often as there are units in c: but subtracting negatives is the same thing as adding affirmatives, by the demonstration of the rule for sub- traction ; consequently the product is c times a, or -f- ac. Otherwise. Since a — a = 0, therefore (a — a) X — c is also = 0, be- cause multiplied by any quantity, is sUll but ; and since the flrst term of the product, or a X — c is = — ac, by the second case ; then* 1 fore the last term of the product, or — a X — c, must be -f- ac, to make the sum = 0, or — tu+ac = 0; that in, , — a X — c = ~|- ac. Other demonstrations upon the principles of proportion, or by menus of geometrical diagrams, have also been given ; but the above may «tf> ice. 4oc -3a6 — 12a 3 6c Sax — ax +3jcy — 5xyz 4x . yi —6c —4 -5ax — >■* ■ " — — CASE II. When ome of ike Factors is a Compound Quantity. Multiply every term of the multiplicand, or compound quantity, separately, by the multiplier, as in the former case ; placing the products one after another, with the proper signs ; and the result will be the whole product re- quired* XULTIFUGATION. 173 9a 3 * — 2x*y —4xy 4x 3*^ — xy 36aV — 6fy +4*V EXAMPLES. 5a — 3c 3ac — 46 2a 3 — 3c + 5 2a 3a 6c lOa^-Goc Qa^ — 12a* 2a'bc-3bc l + 56c 12x-2ac 4a 25c — 76 -2a 4x — 6 + 3a6 2a6 3c 3 + i 4xy 10s* — 3^ — 4X 3 8a 3 — 2x 3 — 66 2a! 3 174 ALGEBRA. CASS III. When both the Factors are Compound Quantities : Multiply every term of the multiplier by every term of the multiplicand separately ; setting down the products one after or under another, with their proper signs ; and add the several lines of products all together for the whole product required. .A v a+b 3x+2y 2&+*y - 2jr» a+b 4x— by 3x— Sy a*+ab lib*+8xy 6x 3 + 3x^-6*^ +ab+b* - 15xy- I0y 2 —6x 2 y—3xy 3 +6y s a*+2ab+b 2 12X 3 — 7xy-l(y 6x 3 - 3x^-9x^+6^ a+b x 2 +y a'+ab+b* a — b x 2 +y a — b a*+ab x x +yx* a 3 +a~b+ab 2 — ab-b 2 +yx 3 +if — a**— <*>*>'— V a* * x 4 +2yx 3 +3T « 3 * * — 6 3 Note. In the multiplication of compound quantities, it is the best way to set them down in order, according to the powers and the letters of the alphabet. And in the actual operation, begin at the left-hand side, and multiply from the left hand towards the right, in the manner that we write, which is contrary to the way of multiplying numbers. But in setting down the several products, as they arise, in the second and following lines, range them under the like terms in the lines above, when there are such like quantities ; which is the easiest way for adding them up together. In many cases, the multiplication of compound quantities is only to be performed by setting them down one after another, each within or under a vinculum, with a sign multiplicat ion be tween them. As {a + b) X (a — b) X or a + b . a — b . 3a&. „ DTVI8ION. 175 EXAMPLES TOR PRACTTCF. 1 . Multiply 10ac by 2a. 2. Multiply 3a a — 26 by 36. 3. Multiply 3a + 26 by 3a — 26. 4. Multiply X 3 — xy + by x + y. Ans. 20aV. Ans. 9a*b — 66 2 . Ans. 9a a — 4o\ Ans. + y\ 5. Multiply a 3 + a f 6 -f a& 2 + 6 3 by a — 6. Ans. a 4 — 6 4 . 6. Multiply a 8 + ab + 6* by a 3 — «6 + 6 3 . 7. Multiply 3x* — 2xy + 5 by x 2 + 2xy — 6. 8. Multiply 3a 2 — 2** + bx 1 by 3a 2 — 4ax — 7x 2 . 9. Multiply 3x 3 + 2xy + 3y 3 by 2x 3 — 3xY + 3y\ 10. Multiply a 3 + a6 + 6 a by a — 26. Division in Algebra, like that in numbers, is the converse of multiplication ; and it is performed like that of numbers also, by beginning at the left-hand side, and dividing all the parts of the dividend by the divisor, when they can be so divided ; or else by setting them down like a fraction, the dividend over the divisor, and then abbreviating the fraction as much as can be done. This may naturally be distin- guished into the following particular cases. When the Divisor and Dividend arc both Simple Quantities : Set the terms both down as in division of numbers, either the divisor before the dividend, or below it, like the denominator of a fraction. Then abbreviate these terms as much as can be done, by cancelling or striking out all the letters that are common to them both, and also dividing the one co-efficient by the other, or abbreviating them after the manner of a fraction, by dividing them by their common measure. Noli* Like signs in the two factors make + in * ne <l uo " tient ; and unlike signs make — ; the same as in multipli- cation*. - * Became the divisor multiplied by the quotient, must product lVv% dfridend. Therefore, DIVISION. CASE I. 170 ALOKBBA. EXAMPLES. 1. To divide Gab by 3a. Here Gab -e- 3a, or 3a ) Gab ( or ^ = 26. ~ c , , , • abx a 2. Also c-t-c = - = 1; and ate wry = 7— = c oxy y 3. Divide lBx 3 by 8x. Ans. 2*. 4. Divide 12a 8 * 2 by — 3a*x. Ans. — 4x. 5. Divide— 15ay* by 3ay. Ans. — 5*> 9sy 6. Divide — 18ax*y by — Saxx. Ans. CASE. II. When the Dividend i* a Compound Quantity, and the Divitor a Simple one. Divide every term of the dividend by the divisor, as in the former case. EXAMFLE8. 1. (ai + ft a )H-26,or^ a =^ = *a + i6. 2. (10*6 + 15a*) -s- 5a, or 10fl6 + 15fl * =26 + 3*. 3. (30az->48z) -f *, or 30g *~ 48 * = 30a-48. 4. Divide 6ao— 8ax + a by 2a. 5. Divide Sx'-lS + 6x + 6a by 3x. 1. When both the terms are +, tbe quotient must be + ; because + in the divisor X + in the quotient, produces -j- in the dividend. 2. When the terms are both — , the quotient is also + ; because — in the divisor X — in the quotient, produces -f- ia the dividend. 3. When one term is-f and the other—, the quotient must be--; because -f in the divisor X — in the quotient produces — in the divi- dend, or — in the divisor X + in the quotient gives — in the dividend. So that the rule is general ; vis. that like signs give +, and unlike **ffn* give — , in the quotient. DIVISION. 6. Divide 6abc + \2abx — 9a*b by Sab. 7. Divide lOa*x - 25* by 5x. 8. Divide I5a 2 be — Ibacx* + bad? by — 5ac. 9. Divide 15a + Say - lSy 8 by 21a. 10. Divide - 2(W 8 6 8 + OOab 3 by - Qab. CASE III* TFAen Divisor and Dividend are both Compound Quantities. 1. Set them down as in common division of numbers, the divisor before the dividend, with a small curved line be- tween them, and range the terms according to the powers of aomo one of the letters in both, the higher powers before the lower. 2. Divide the first term of the dividend by the first term of the divisor, as in the first case, and set the result in the quotient. 3. Multiply the whole divisor by the term thus found, and subtract the result from the dividend. 4. To this remainder bring down as many terms of the dividend as are requisite for the next operation, dividing as before ; and so on to the end, as in common arithmetic. Note. If the divisor be not exactly contained in the divi- dend, the quantity which remains after the operation is % finished may be placed over the divisor, like a vulgar frac- tion, and set down at the end of the quotient as in arith- metic. EXAMPLES. a - b) a a - %tb + b* (a - b a 3 - ab -ab + V - ab + ^ Vol. I. 24 178 ALGEBRA. a - c) a 3 — 4a 9 c + 4^ - c 3 — (a 3 — 3ac + - 3« 3 c + W - 3a*c + 3'^ oc 3 -c 1 ac 3 - c 3 «_ 2) a 3 — 6a 3 + 12a - 8 (a 3 - 4a + 4 a 3 -2d 3 - 4a 2 + 12a - 4a 3 + 8a 4a — 8 4a — 8 « + *) a * - 3x 4 (a 3 - a 3 * + ax 3 — ~ 3 - a 4 + a 3 x a + * — a 3 * - 3x* — a 3 x — aV aV - 3x 4 a 3 x 3 -f • ax 3 — ar 3 - 3x* - 2x 4 EXAMPLES FOB PRACTICE. 1. Divide a 3 + 4ax + 4x 3 by a + 2x. Ans. a + 2*. 2. Divide a 3 - 3a 3 z + 3a* 3 - z 3 by a - z. Ans. a 3 — 2az + z\ 3. Divide 1 by 1 + a. Ans. 1 - a + a 2 — a 3 + &c. 4. Divide 12x 4 — 102 by 3x - 6. Ans. 4x 3 + Sx 2 + 16x + 32. 5. Divide a 5 - 5a 4 6 + 10a 3 6 3 - 10a 3 6 3 -f 5a6 4 — b* by a 3 - 2a& + b\ Ans. a 3 - 3a 3 6 + Sab 2 - 6'. FRACTIONS. 179 6. Divide 48s 3 - 96a* 3 - 64a 2 z + 150a 3 by 2z - 3a. 7. Divide b* — 36 V + 36 2 x 4 — x*by 6 3 — 2b'x + 36s 3 — x 3 . 8. Divide a 1 — x 7 by a — x. ■ 9. Divide a 3 + 5a ? x + 5ax 3 + x 3 by a + x, 10. Divide a 4 + 4a 2 6 3 — 326* by a + 26. Ik Divide 24a 4 — 6< by 3a — 26. ALGEBRAIC FRACTIONS. Algebraic Fractions have the same names and rales of operation, as numeral fractions in common arithmetic ; <as Appears in the following Rules and Cases. To reduce a Mixed Quantity to an Improper Fraction. Multiply the integer by the denominator of the fraction, and to the product add the numerator, or connect it with its proper sign, 4" or — ; then tho denominator being set under this sum, will give the improper fraction required. EXAMPLES. 1. Reduce 3}, and a ^- to improper fractions. OA (3X5) +4 15 + 4 10 , . First, 3| = ~- — = — = — the Answer. And, a — - = ( aXar )~ ^_ = — — - the Answer. xx x a 2 z i a a 2. Reduce a + -p and a to improper fractions. o a «. . , a 2 (a X 6) + a 3 a6 + a 2 , A . 180 ALGEBRA. 3. Reduce 5f to an improper fraction. An*. V • 4. Reduce 1 — ^ to an improper fraction. Ans. 5. Reduce 2a — ** gx a ^ an improper fraction. 4x 4r 18 6. Reduce 12 -| jr- — to an improper fraction. 1 — 3a c 7. Reduce x H — to an improper fraction. 2r* 3a 8. Reduce 4 + 2x : — — to an improper fraction. CA8k II. To reduce an Improper Fraction to a Whole or Mixed Quantity. Divide the numerator by the denominator, for the inte- gral part; and set the remainder, if any, over the denomina- tor, for the fractional part ; the two joined together will be the mixed quantity required. EXAMPLES. 1. To reduce and to mixed quantities. o o First, y = 16 -T- 3 = 5J, the answer required. And, ^^g— = (ab + a 2 ) -r- b = a + Answer. 2ac — 3a 3 ,3ax + 4x f 2. To reduce und , to mixed quanti- c a + x ties. « 2ac — So 2 M n nK 3a 2 _ First, = (2ac — 3a 2 ) c = 2a . Answer. 3ax+4x a x 2 And, Ip_==(3ax+4x 2 Wa+s)=3x+— - Ans. a+x x / \ / a+x _ , 33 _ 2ax — 3x> . , 3. Reduce — and to mixed quantities. 3x* Ans. 6|, and 2x — . 4a a x 2n 2 -f- 26 4. Reduce and — - — - — to whole or mixed quan- dties. FRACTIONS. 181 5. Reduce **** , ^ and ^ ^ t o whole or mixed x+y x—y quantities. a D . 10a 1 — 4a + 6 4 . , o. Reduce — to a mixed quantity. 7. Reduce r- rr~ a « to a mixed quantity. Sa 1 + 2a 9 — 2a — 4 n ' case in. To reduce Fractions to a Common Denominator. Multiply every numerator, separately, by all the deno- minators except its own, for the new numerators ; and all the denominators together, for the common denominator. When the denominators have a common divisor, it will be better, instead of multiplying by the whole denominators, to multiply only by those parts which arise from dividing by the common divisor. Observing also tho several rules and directions, as in Fractions in the Arithmetic. EXAMPLES. 1. Reduce - and - to a common denominator. x z Here - and - = a% and — , by multiplying the terms of x z tz xz the first fraction by z, and the terms of the 2d by x. 2. Reduce -, and — to a common denominator. x o c n a x j & abc ex 7 , b*x . Here - , - , and — = — and 7 — , by multiplying the x b c bcx bcx bcx r J ° terms of the 1st fraction by be, of the 2d by cx, and of the 3d by bx. 3. Reduce — and ^ to a common denominator. x 2c 4ac , Sbx Ans. — - and — . 2cx 2cx 4. Reduce ^ and — - to a common denominator. b 2c k 4ac , 3ab+2b* Ans.^and— b — , 182 ALGEBRA. 5a 36 5. Reduce and and 4 a*, to a common denominator. Sx 2c lOffc ,06a: J 24cox Ans. - — and — and — — • hex ocx ocx 6. Reduce jj and and 26+ ~, to fractions having a com- 4 206 ,18a6 J 486 2 +72a mon denominator. Ans. - — and --r-, and — -rn . 246 246 246 1 2a 2 2tt 3 +6 3 7. Reduce - and — and — r-r~ to a common denomina- 3 4 o+6 tor. 8. Reduce — and and ^- to a common denominator. 4u s 3a 2a CASE IV. Tb find the greatest common Measure of tlie Terms of a Fraction. Divide the greater term by the less, and the last divisor by the last remainder, and so on till nothing remains ; then the divisor last used will be the common measure required ; just the same as in common numbers. But note, that it is proper to range the quantities accord- ing to the dimensions of some letters, as is shown in division. Note also, that all the letters or figures which are common to each term of the divisors, must be thrown out of them, or must divide them, before they are used in the operation. EXAMPLES. 1. To find the greatest common measure of — ^r-r „» ar J +6c J a6 + 6 3 ) or 2 + be 2 or a + 6 ) ac 2 + bc z (c 2 ac 3 + 6c 2 Therefore the greatest common measure is a + 6. a 3 — n6 a 2. To find the greatest common measure of — — -r-j-r,. a-+2a64-6 1 FRACTIONS. 188 a*+2ab+P) a 3 — <io 3 (a — tab— 2ab 2 ) a 2 4- 2*6 + 6 3 or a-f 6 ) <r + 2ab + b* (a + b a m 4- a6 oA + ft 2 ab + b 1 Therefore a-\-b is the greatest common divisor. a 3 — 4 3. To find the greatest common divisor of ^q^- Ans. a — 2» 4. To find the greatest common divisor of "r^gr- 5. Find the greatest com. measure of —-—r.-r.r-r —r-—7-r m • CASK V. To reduce a Fraction to its lowest Terms, Find the greatest common measure, as in the last pro- blem. Then divide both the terms of the fraction by the com- mon measure thus found, and it will reduce it to its lowest terms at once, as was required. Or divide the terms by any quantity which it may appear will divide them both as in arithmetical fractious. EXAMPLES. , n , ab+b* . t 1. Reduce — z~~r~. — :» to its lowest terms. ac 2 +bc* ab+b 2 ) ar'+bc 1 or a +6 ) ucP+bc 2 (c 3 ac-+b(r Here ab-\-b 2 is divided by the common factor b. Therefore a + b is the greatest common measure, and hence a+b) rf ^^", ^r=-? > is the fraction required. acr-\-bcr <r S-b-c . , 2. To reduce -r-r-r-. . ... to its least terms. cr+%bc+b* 184 ALOEBRA. Hero, by a process similar to that of Ex. 2, Case rv. 9 we find c + b is the greatest common measure, and hence e + b) -f ° -;.= C . ^ C is the fraction required. T ' c*+2bc + b 9 c+b M c 3 — ft 3 c*+bc+V 3. Reduce -r — rrs t0 its lowest terms. Ans. . -?-,-» £i J 4. Reduce n to its lowest terms. Ans. a' — o ar-x-or a 1 — o 4 . , 5. Reduce -= — 5-^-7-0—53 — TJ to lts lowest terms. a 3 — Serb +3*o a — o J 6. Reduce - 1 - T - -~ i - r . , , . to its lowest terms. a 3 c +3aV+3ac 3 +c* CASE VI. To aaV2 Fractional Quantities together. If the fractions have a common denominator, add all the numerators together ; then under their sum set the common denominator, and it is done. If they have not a common denominator, reduce them to one, and then add them as before. EXAMPLES. 1. Let ~ and be given, to find their sum. a . a 4a , 3a 7a . Here — + — = ^ + «F is the sum required. 2. Given — , and to find their sum. be a _.. a , 6 c aco* , 66d ,bcc acd4-6W+occ Uere y + 7 + 7 ,= kS + 6^ + 6^ "635" the sum required. 3s 2 2ax * 3. Let a — and 6 + be added together. * In the addition of mixed quantities, it is best to bring the fractional parts only to a common denominator, and to annex their sum to the ram of the integers, with the proper sign. And the same rule may be ob- served for mixed quantities in subtraction also. See, also, the note to Addition of Fractions in the Arithmetic. FRACTIONS. 185 Here a — r- + 6 H = a 7 r + — — b c be be . . , 2abx-3cx* . . = a + H £ 9 the sura required. be , ...4* . . 20ta+6ax 4. Ada jr- and -=r together. Ans. - — — — 7 — . 3a 56 6 15a6 5. Add ~, ~ and together. Ans. £{a. « AJJ 2a— 3 .5a . . 9a -6 6. Add — - — and — together. Ans. — - — . 7. Add 2a -| — to 4a -\ Ans. Ga Hf ^ — . O 4 *U 8. Add 6a, and ^ and -^jp together. o a^ 5 * j 6a ,3a+2 4 9. Add — , and — and — — — together. 10. Add 2a, and ~ and 3 + ^ together. 11. Add 8a + ^ and 2a — ^ together. , CASE VII. To subtract one Fractional Quantity from another. Reduce the fractions to a common denominator, as in addition, if they have not 1 a common denominator. Subtract the numerators from each other, and under their difference set the commoi denominator, and the work is done. EXAMPLES. 3a 4a 1. To find the difference of — andy. „ 3a 4a 21a 16a 5a . . J ir . , Here — — y = — - — — - = ^ is the d ffer. lequired. 2* To find the difference of ^7 -- - and 4c 36 Vol. I. 25 difference of &» ■»* "4* 4. ^ ^ 4 * 7. T«k« — 9 2a from 4a c cask vin. Beto-g-Jt5 40 , ______ to* FRACTIONS. 187 2. Required the product of |, ~, and 5?. TxTxT "^-H^e product required. 3* Required the product of ~ and !^p~* „ 2m x (a + b) 2aa + 2ab. _ _ H6r6 6— (2T+T) = 23T+-£ ^ produCt 4a 6a 4. Required the product of 7 and -r- • O DC 5. Required the product of ^ and —j*. 6. To multiply y, and ~ 9 and ^ together. ao 3a* 7. Required the product of 2a + -g^and o. Required the product of — — and ^ ^ ». 2a 4-1 2a— 1 9. Required the product of 8a, and — - — and 10. Multiply« + ±-^byr-£ + ^. CASE IX. To oroide one Fractional Quantity by another. Divide the numerators by each other, and the deno- minators by each other, if they will exactly divide. But, if •Hot, then invert the terms of the divisor, and multiply by it exactly as in multiplication*. * 1. If the fraction! to be divided have a common denominator, take the numerator of the dividend for a new numerator, and tfae numerator "Of the divisor for the new denominator. ft. When a fraction is to be divided by any quantity, it is the tame thing whether the numerator be divided by it, or the denominator mul- tiplied by it 8. When the two numerators, or the two denominators, can be divi- ded by some common quantity, let that be done, and the ajtttUM&iwA ' Instead ef the fractions first propweiL 188 ALGEBRA. EXAMPLES. a 3a 1. Required to divide - by a 3a a 8 8a 2 . Here i + T = 4 X 3a = I2S " 3 * e qU0Uent ' 2. Required to divide ?^ by w 3a 5c 3a 4d I2ad 6ad _ Here » + 45 = 26 X 5c = TOfc = 56? the qUOUenU « m j 2a+6 _ 3a+26 „ 3. Todmde— - 6 by— Here, 2a+6 w 4a+6 8a*+6ab+b* _ 3a--26 X 3a+26 = Tra*- tbe qUOtMmt n ^ md ^ 4. To divide ^r- 3 by- a a 3 +6 3 J a+&' H — a+6_3a a X(a+6)_ 3a . 6 aM-6> X a (a 3 +6 3 ) Xa a* — ab + 6* 18 quotient required. 5. To divide — by ~. 4 J 12 6. To divide -f by 3x. 5 7. To divide — - — by — « 8. To divide jr^-- Ly^. 4>X. — 1 u 9. To divided by I?. 5 3 56 10. Todividc^by^-. 4ca oa „ _. . . 5a 4 - to f5a6 11. Dmde^-^^by 160 INVOLUTION. Involution is the raising of powers from any proposed root ; such as finding the square, cube, biquadrate, &c. of any given quantity. The method is as follows. * Multiply the root or given quantity by itself, as many times as there are units in the index less one, and the last product will be the power required. Or, in literals, mul- tiply the index of the root by the index of the power, and the result will be the power, the same as before. Note. When the sign of the root is +, all the powers of it will be + ; but when the sign is — , all the even powers will be +, and all the odd powers — ; as is evident from mul- tiplication. examples. a, the root a 3 3= square a 3 = cube a 4 = 4th power a 1 = 5th power 6zc. a 9 , the root a 4 ~ square a 6 = cube a" = 4th power a ,0 = 5th power dec. — * 2a, the root +* 4a a = square — 8a 3 =- cube + 16a 4 = 4th power — 32a 5 = 5ih power — 3ab*, the root + 9c?b* = square — 270*6*= cube + 81a 4 6 f = 4th power — 243a , 6 ie = 5th power 2ar 2 — the root , 4aV + .— = square 8aV — = cube 276* + - Q j- 6 T 4th P ower ^g, the root a a 4^ = square ?L =cube a 3 166 4 = bi( l uadrate * Any power of the product of two or more quantities, is equal to the tame power of each of the factors, multiplied together. And any power of a fraction, is enual to the same noyr*T cA tab trap aerator, divided by the like power of the denominator* 190 ALOBBBA. x — a =root x+ a=root x — a x+a x 2 — ax x 9 + ax — at + a* +ax + a* X s — 2ax + a 2 square x* + 2ax + a 9 x — a x + a x 5 — 2ax a +o a x x 3 + 2a* 8 f a 1 * —ax 2 +2a»x— a 3 + ax 3 + 2a a x + «* x 3 — 3ax* "4* Sa'x-— a 3 x 3 + Sax 2 + So 3 * + a 3 the cubes, or third powers, of x — a and x + a. KXAMJ LES FOR PRACTICE. 1. Required the cube or 3d power or 3a a . 2. Required the 4th power of 2a s b. # 3. Required the 3d power of — 4a a o 3 . Q^X 4. To find the biquadrate of — ~ a . 5. Required the 5th power of a — 2x. 6. To find the 6th power of 2a*. Sir Isaac Nekton's Rule for raising a Binomial to*any Power whatever* : . # 1. To find the Terms without, the Co-efficients. The inder. of the first, or leading quantity, begins with the index of the given power, and in the succeeding terms decreases con- tinually by 1, in every term to the last ; and in the 2d or Alto, powers or roots of the same quantity, are multiplied by one> another, by adding iheir exponents ; or divided, by subtracting their exponents. aa Thus, «3 X«* = aa+a = as. And as -f-aa or — = as— a — a . as * This rule, expressed in general terms, is as follows: <.+x)*=a^». a-ix+n . "^a^a-j-n. 'I" 1 . "^-v ^ <«-^)»^-«.a»-ix4-n . *^ l a*~ z"-n. *~ l . n ^ 2 a --y Ac. /toil. Xb* ma of the co-efficienU, In every power, it equal Co ther- INVOLUTION. 101 following quantity, the indices of the terms are t 1,2, 3, 4, &c. increasing always by 1. That is, the first term will con- tain only the 1st part of the root with the same index, or of the same height as the intended power : and the last term- of the series wUl contain only the 2d part of the given root, when raised also to the same height of the intended power : bat all the other or intermediate terms will contain the pro- ducts of some powers of both the members of the root, in such sort, that the powers or indices of the 1st or leading member will always decrease by 1, while those of the 2d member always increase by 1. 2. To find the Co efficients. The first co-efficient is al- ways 1, and the second is the same as the index of the in- tended power ; to find the 3d co-efficient, multiply that of the 2d term by the index of the leuding letter in the same term, mod divide the product by 2 ; and so on, that is, multiply the co-efficient of the term last found by the index of the leading quantity in that term, and divide the product by the number of terms to that place, and it will give the co-efficient of the term next following ; which rule will find all the co-efficients, one after another. Note. The whole number of terms will be 1 more than the index of the given power : and when both terms of the root are + , all the terms of the power will be + ; but if the second term be — , all the odd terms will be + > and all the even terms — , which causes the terms to be + and — alter- nately. Also the sum of the two indices, in each term, is always the same number, viz. the index of the required power ; and counting from the middle of the series, both ways, or towards the right and left, the indices of the two terms are the same figures at equal distances, but with mutu- ally changed places. Moreover, the co-efficients are the same numbers at equal distances from the middle of the se- ries, towards the right and left ; so by whatever numbers they increase to the middle, by the same in the reverse order they decrease to the end. number 2, when raised to that power. Thus 1— |— 1 = 2 in the first power ; 1+2+1 =4=2- in the square; 1 +3 + 3 + 1 = 8 =2* in the cube, or third power : and so on. A. trinomial or a quadrinomial may be expanded in the same manner. Thua, to raise a — ft + c — d to the 6th power, put a — b = x, e — d—z, mod raise i + z to the 6th power ; after which substitute for the powers of x and y their corresponding values in terms of o— b, aad c— d, and their powers respectively. 192 AXGKBRA. EXAMPLES. 1. Let a + x be involved to the 5th power. The terms without the co-efficients, by the 1st rule, will be a 5 , a 4 *, aV, aV, or 4 , **, and the co-efficients, by the 2d rule, will be 5Xj 10X3 10X2 5X1 it 5, , — 3—, — — , — — ; or, 1, 5, 10, 10, 5, 1 ; % Therefore the 5th power altogether is a 5 + 5a 4 * + lOaV + lOaV + 5a* 4 + X s . But it is best to set down both the co-efficients and the powers of the letters at once, in one line, without the inter* mediate lines in the above example, as in the example here below. The operation is very easily effected by performing the division first. 2. Let a — x be involved to the 6th power. The terms with the co-efficients will be a 8 — 6a 5 * + 15aV-20aV + 15aV — 6a* 5 + a*. 8. Required the 4th power of a — x. Ans. a 4 — 4a 3 x + 6aV- 4ax 3 + ac< And thus any other powers may be set down at once, in the same manner, which is the best way. 4. Involve a — * to the ninth power ; x — y to the tenth power, and a + b — c to the fourth power. EVOLUTION. Evolution is the reverse of Involution, being the method of finding the square root, cube root, 6zc. of any given quan- tity, whether simple or compound. case 1. To find the Boots of Simple Quantities. Extract the root of the co-efficient, for the numeral part * r and divide the index of the letter or letters, by the index oC EVOLUTION. 193 the power, and it will give the root of the literal part ; then annex this to the former, for the whole root sought*. EXAMPLES. 1. The square root of 4a 3 , is 2a. 2. The cube root of 8a 3 , is 2a' or 2a. 3. The square root of -ttt-, or \Z~7Tn~9 is — v" 5. 16a 4 6 8 2ab 2 4. The cube root of — -~--z- 9 is ^ - 2/2a. 27c 3 3c v 5. To find the square root of 2db\ Ans. a6 3 v/2. 6. To find the cube root of - 64a <6 8 . Ans. -4ci6 3 . ~ m ^ i , „8dr6' 2a6 2 7. To find the square root of -57-7-. Ans. — J - . 3c J c 3c a To find the 4th root of 81aW. Ans. 3aV&. 9. To find the 5th root of — 32aW. Ans. — 2ab*/b. CASE II. To find the square root of a Corn-pound Quantity. This is performed like as in numbers, thus : 1. Range the quantities according to the dimensions of one of the letters, and set the root of the first term in the quotient. 2. Subtract the square of the root, thus found, from the first term, and bring down ihe next two terms to the re- mainder for a dividend ; and take double the root for a di- visor. * Any evpn root of nn affirmative quantity, 1. ay be ml'.ier -j- or — : thus the square root of -|- a* is either + a i or — a » because -f- a X + a = -f- «a . and — a < — a = -f- a* aUo. But an odd root of any rj'ianti'y will have the fame s : gn as the quan- tity itsjlt : thus tbe cube root of -f- a* is -J- .i, and tue cub'j root of — a * k — a ; for -r- >l X + « X + * - 4~ a * » n,, d — a * — '* X — a — — « 3 • Any evon root of a negative quantity ia imp' i«ib'e ; for neithe; -|- a X+ «t nor — a v — a can prr.cHce — /?a. Any root of a prod ict is cq.ial to the like root o each of l! 1 f -jcto-s moltipl-pd together. Tor the root of a fraction, *ake the roM of t'ie numerator and the ro-^t of the dciominutor. Vol. 1. / 104 ALGEBRA* 3. Divide the dividend by the divisor, and annex the re- sult both to the quotient and to the divisor. 4. Multiply the divisor, thus increased, by the term last set in the quotient, and subtract the product from the divi- dend. And so on, always the same, as in common arithmetic. EXAMPLES. 1. Extract the square root of a 4 — 4a 5, &+6a s &» — 4ab*+b*. a 4 — 4a*6 + 6a»6« — 4a6» + 6 4 (a" — tab + b* the root. 2f—2ab)—4a*b + 6a 2 6 3 — 4a 3 6 + 4a 2 6* 2a»_4a* + 6 s ) 2a 2 6 a — 4ab> + 6 4 2a*b 2 — 4a6 3 + b l 2. Find the root of a 4 + 4tfb + 10a 2 6 s + 12a6 3 + 96*. a* + 4a 3 * + lOaV + 12a6 3 + 96 4 (a 2 + 2a£ + 3^ 2a 2 + 2a6) 4a?b + lOa 2 * 2 4a 3 6 + 4a a 6 a 2a a + 4a& + 36 2 ) 6a 8 6 2 + 12aft 3 + 96 4 6aW + 12a6 3 + 9b* 3. To find the square root of a 1 + 4a 3 + 6a 2 + 4a + I - Ans. a 2 + 2a + 1 - 4. Extract the square root of a 4 - 2a 3 + 2a 2 — a + J. Ans. a 2 — a + ■£ — 5. It is required to find the square root of a 3 — ab. CASE III. To find the Boats of any Powers in general. This is also done like the same roots in numbers, thus ; Find the root of the first term, and set it in the quotie*"***' —Subtract its power from that term, and bring down tJ^* second term for a dividend. — Involve the root, last found, Jbe next lower power, and multiply it by the index of ***** EVOLUTION* 105 given power, for a Jivisor. — Divide the dividend by the di- visor, and set the quotient as the next term of the root. — Involve now the whole root to the power to be extracted ; then subtract the power thus arising from the given power, and divide the first term of the remainder by the divisor first found ; and so on till the whole is finished *. EXAMPLES. ' 1. To find the square root of a 4 — 2a' , &+3aV— 2aft 3 +tt ««_2a b + 3a*V— 2aP + b* {a 3 —ab + &*, 2a 3 )— 2a b ** 2a b + a fl 6 > = (a 8 — ab)' 2a 3 ) 2a a #» <t—2ab + 3a a 6 a — 2a6 3 + © 4 = (a* - ab + b 3 f. 2. Find the cube root of a 9 - 6a 5 + 21a 4 -44a 3 + 63a* — 54a + 27. a 8 — 6a» + 2 la 4 — 44a 3 + 63a 2 — 54a + 27 (a 3 — 2a+3. a 9 3a 1 ) — 6a 5 a*— 6a 5 + 12a 4 — 6a 3 = (a 3 — 2«) 3 3a 4 ) + 9a 4 a t -fa>+*\a*--44a : ±G3a 3 --54a+2rr= (a 3 — 2a+3)\ * As this method, in high powers, may be thought too laborious, it will not be improper to observe, that the roots of compound quantities may sometimes be easily discovered, thus: Eitract the roots of some of the most simple terms, and connect them together by the sign -f or — , as may be judged most suitable for the purpose.— Involve (he compound root, thus found, to the proper power ; then, if this be the same with the given quantity, it is the root required. — But if it be found to differ only in some of the signs, change them from -f to — , or from — to + , till its power agrees with the given one throughout. Thus, in the 5th example, the root 3a — 26, is the difference of the toots of the first and last terms ; and in the 3d example, the mot ft*— 1 * + x, is the sum of the roots of the 1st, 4th, and 6th terms. The 'Ipi nay also be observed of the 6th example, where the KOQttofo^k '3HM'Umi first and last terms. t 106 ALGEBRA- 3. To find the square root of a' — 2ab + 2ax + P — 2bx + x* Ans. a — b + x. 4. Find the cube root of a 6 — 3a 5 + 9a 4 — - l&r + 18a 3 — 12a + 8. Ans. a 3 — a + 2. 5. Find the 4th root of 81a 4 — 216a 3 6 + 216a 3 6 3 - 90a6. + 166 4 . Ans. 3a - 26. 6. Find the 5th root of a s — 10a 4 + 40a 3 — 80a 3 + 80a — 32. Ans. a — 2. 7. Required tho square root of 1 — **• 8. Required the cube root of 1 — x\ SURDS. Sum>s are such quantities as have no exact root ; and are usually exprcbded by fractional indices, or by means of the radical sign y/. Thus, 3^, or ^3, denotes tho square root of 3 ; and 2 1 , or ^/2 3 , or ^/4. the cube root of the square of 2; where »ht numer'or shows tho power to w! 'zh the quantity is to be raised, and the denominator its root. PROHLFM I. To reduce a Rational Quantify to the Farm of a Surd. R * ise the piver* quantity to the power denoted by the index of the surd ; then over or above the new quantity set the radical sign, an I it will be of the form required. EXAMPLES. « 1. To reduce 4 to the form of the square root. First, 4 2 — 4 X 4 ■= 10 ; then y/ J 6 is the answer. 2. To reduce 3<r to the form of the cube root. First 3<r X 3i 3 =^ X 3/7 = \3a>) ' = 27 a G ; then \y27a* or (27a*)* is the answer. 3. Reduce C to the form of the cube root. Ans. (210)* or ?/216. 4. Reduce lab to the form of the square root. 8VXDI. 197 5. Reduce 2 to the form of the 4th root. Ans. (16)*. 6. Reduce a 4 to the form of the 5th root. 7. Reduce a + x to ihe form of the square root. 8. Reduce a — * to the form of the cube root. PROBLEM II. To reduce Quantities to a Common Index. 1. Reduce the indices of the given quantities to a com- mon denominator, and involve each of them to the power denoted by its numerator; then 1 set over the common de- nominator will form the common index. Or, 2. If the common index be given, divide the indices of the quantities by the given index, and the quotients will he the new indices for those quantities. Then over the said quan- tities, with their new indices, set the given index, and they will make the equivalent quantities Bought. EXAMPLK8. 1. Reduce 3* and 5* to a common index. Here £ and } = J z and fy. Therefore 3^ and 5^ = (3')^ and (5*)^ = V 5 and 1 = ■ J/243 and l {/25. 2. Reduce a* and b* to the same common index |. Here, f •+ $ = J X \ = f the 1st index, and Jr} = iX \ = | the 2d index. Therefore (a 6 )' and (6*)', or yAr 5 and are the quanti- ties. 3. Reduce 4* and 5* to the common index \. Ans. (256*)* and 25*. 4. Reduce afc and r* to the common index j-. Ans. (a 1 )* and (x')*. 5. Reduce d 2 and x 3 to the same radical sign. Ans. y/a* and y/x*. 6. Reduce (a + x)* and (a — x)^ to a common index. 7. Reduce (a + b)^ and (a — t)* to a comumVbfa** 1G8 alg::bra. l'RORI.r.M nr. To reduce. Surds to more. Simple Terms. Divide the s'.ird, if possible, into two factors, cne of which is a power of the kind that accords wiih the root sought ; as a complete square, if it be a square root, a complete cube, if it he a cube root ; and so on. Set the root f this com- plete power before the surd expression which indicates the root of the other factor ; and the quantity is reduced, ad re- quired. If the surd be a fraction, the reduction is effected by mul- tiplying both its numerator and denominator by some number that will transform the denominator into a complete square, cube, &c. its root will be the denominator to a fraction that will stand before the remaining part, or surd. Sec Example 3, below. EXAMPLKS. 1. To reduce v /32 to simpler terms. Here ^32 = ^(10X2) = ^/lO x ^/2~4 X v/2 =4 </2. 2. To reduce ^/3*20 to simpler terms. V320 =y(t)4 x 5) = yoi x y:> = i x y5 =^ 4 ys. 3. Reduce -/J* to simpler terms. 44 44 , 4i 4.11 H _ 2 2 .55 _ IB'* 5 ' 4. Reduce v/75 to its simplest terms. Ans. 5^/3. 5. Reduce ^/189 to its simplest terms. Ans. 3^/7. 0. Reduce '^/\;\f to its simplest terms. Ans. -JJ/10. 7. Reduce ^/loarh to its simplest terms. Ans. f>tf v /36. Note. There are other cases of reducing algebraic surds to simpler forms, that are practised on several occasions ; one of which, on account of its simplicity and usefulness, may be here noticed, viz. in fractional forms having compound surds in the denominator, multiply both numerator and de- nominator by the same terms of the denominator, but having one sign chanced, from to — or from — to +, which will reduce the fraction »■> a rational denominator. Ex. To reduce ^ 2 ?- + -^-~, multiply it bv ^ and it becomes - - — = 8 + 2 v /l. r >. 49 SURDS. 190 * „ 3v/15 — 4*'5 ,. , . _ «/lf> — y5 ■ Also, to reduce - _ — ; multiply it bv -- — — and y/lr +■ v 'o * ,/la— ^ . , 65—7,770 65 -33 v /3 13— 7,/3 it becomes — - - — — = — . 15— o 10 2 And the same method may easily be applied to examples with three or more surds. PROBLK3I IV. To add Suid Quantities together 1. Bring all fractions to a common denominator, and reduce the quantities to their simplest terms, as in the last problem. — 2. Reduce also such quantities as have unlike indices to other equivalent ones, bavin" a common index. — 3. Then if the Mini part be the same in them all, annex it to the sum of the rational pars. \vi;h the sign of multiplica- tion, and it will give the total sum recr-iired. But if the surd pan be not ilu> s«ni<' in all the quantities, they can only be added by the signs + and — . KX AMI* LI-IS. 1. Required to add ^/IS and X /S2 together. FirstV18=V(9X2; = %/2; ™<l — y/{WX2)^4y/2: Then, 3y2 +4^/2= (3+4)^/2=7^2= sum required. 2. It is required to add */375. and yi92 together. First,y375=^/( 1 2f> X 3 ; =5y 3:and y 102=^/(0.1x3) =4^/3: Then, 5^3 + 4^3 = (5+ l)y.*J 9y3 = sum required. 3. Required the sum of\/27 and x /4t\ Ans. 7 y/S, 4. Required the sum of \/f>0 and >/72. Ans. 11^/2. 5. Required the sum of and v'rs- -^ ns - t^V^' G. Required the sum of -y 50 and i[/lbli. Ans. 5|/7. 7. Required the sum of \/ \ and \/ , ! f Ans. -Jy2. 8. Required the sum of 3^/a-ft and SyMbV/j. fruhlfm v. 7'ri t //W the. Diijcrvncc of Surd Quantities. Pre pa it k th^ quantities the same w:«y as in the last rule : then subtract the rational parts, and to the remainder acuiftx. the common-surd, for the difference of the. surtta tec^ivt^* 200 ALGEBRA. But if the quantities have no common surd, they can only be subtracted by means of the sign — -. EXAMPLES. 1. To find the difference between v/320 and ^/80. First, v/320=V(64 X 5) = 8 ✓ 5;and /S0= % /{ 16X5) =4^5. Then, 8^/5 — 4^/5 = 4^/5 the difference sought. 2. To find the difference between yi'ZS and ^/54. Fint,l/128=i/(64*2)=4y2; andy54 = V(27x2)=3y2- Then, 4J/2 - 3^/2 = the difference required. 3. Required the difference of y/75 and ^/48. Ans. ^/3. 4. Required the difference of $/256 and ]/32. Ans 2^4. 5. Required the difference of y/% and Ans. Jv'S. 6. Find the difference of t/| and yf. Ans. ^^/^ 7. Required the difference of y f and (/V* Ans.fj^/75. 8. Find the difference of v/24a*o a and ^/M'* 4 . Ans. ^/(36 3 — 206)^/6. PROBLEM VI. To multiply Surd Quantities together. Reduce the surds to the same index, if necessary ; next multiply the rational quantities together, and the surds to- gether ; then annex the one product to the other for the whole product required ; which may be reduced to more simple terms if necessary. EXAMPLES. 1. Required to find the product of 4y/\2 and 3^2. Here, 4*3X y/ 12x^2 = 12 v /(12v2) = 12 v /24=12 v /(4x6) = 12X2X^/6=24^/6, the product required. 2. Required to multiply \%/} by . Here lXtf/f xyt^iV* V*=rV Xl/«=A X * X ^ 18 =J r ^/18, the product required. 3. Required the product of SyZ and 2^/8. Ans. 24. 4. Required the product of 1^/4 and jyl2. Ans. 5. To find the product of f and rWi* Ans. jtV15. 6. Required the product of 2^/14 and 3y4. Ans. 12^/7. 7. Required the product of 2a* and a 3 . Ans. 2n f . 8. Required the product of (a+A)^ and (a+&)^. Ml 9. Required the product of 2x+^b and 2x - y/b. 10. Required the product of (af 2y/6)*, and 11. Required the product of 2x* and 3x'»- 12. Required the product of 4x* and2y". PSOBLEH VII. 7b divide one Surd Quantity by another. Reduce the surds to the same index, if necessary ; then take the quotient of the rational quantities, and annex it to the quotient of the surds, and it will give the whole quotient required ; which may be reduced to more simple terms if requisite. EXAMPLES. 1. Required to divide 6 1/ 90 by 3^/8. Here6-r- 3.^(96 + 8) = 2^12 =2 /(4X3) =2 X2v/3 = 4v/3, the quotient required. 2. Required to divide 12^/280 by 3$/5. Here 12 -5- 3 = 4, and 280 + 5 = 56 = 8 X 7 = 2 3 . 7 ; Therefore 4x2 XV 7 = 18 tne quotient required. 3. Let 4y/50 be divided by 2^/5. Ann. 2^10. 4. Let 6 \/100 be divided 3^5. Ans. 2^/20. 5 Let fv/jV be divided by j v'f. Ans. fv/5. 6. Let f be divided by | \/%. Ans. 7. Let | y/a, or |a^, be divided by fa*. Ans. f a^. 8. Let be divided by <A. 9. To divide 3a" by 4a". PROBLEM VIII. To involve or raise Surd Quantities to any Power. Raise both the rational part and the surd part Or multi- tiply the index of the quantity by the index of the power to which it is to be raised, and to the result annex the power of die rational parts, which will give the power Tec\uvc«A» Vol. I. 27 202 ALGEBRA. EXAMPLES. 1. Required to find the square of Jo^. Firs', (f ) a = } X {- = A> a « d («*)■ = a 4 X2 = a i =r^ Therefore, Jafy = ^a, is the square required. 2. Required to find the square of J<A. First, i X i = i, and (a*) f = a* = aj/a ; Therefore = Ja ^/a is the square required* 3. Required to find the cube ef }^/6 or f X 6^. First, (*y = § X § X § - and (6*) 3 = 6* = 6^6 ; Theref. (|^tt) 3 = ft X6 V 6= V the cube required. 4. Required the square of 2^/2. Ans. 4{/4» 5. Required the cube of 3^, or </3. Ans. 3 6. Required the 3d power of £ %/3. Ans. J v& 7. Required to find the 4th power of Ans. £. l 8. Required to find the with power of a n . 9. Required to find the square of 2 + v^3. PROBLEM IX. To evolve or extract ihe Roots of Surd Quantities* . Extract both the rational part and the surd part. Or divide the index of the given quantity by the index of the * The square roet of a binomial or residual surd, « -|- 6, or a — K may he found thus : Take Va* — b 2 = c ; then Vo+T= + V*-^ ; and V« — o = V— g V ~2~' Thus, the square root of 4 + 2v3 = 1 + v3 ; and the square root of 6 — 2v5 = V5 — 1. But for the cnbe, or any higher root, no general role is known. For more on the subject of Surds, see BonrycastU's Algebra, the 8rev edition, and the EUmentary Treatise ofAbgcbra, by Mr. J. R. Ynmg. ARI T HME TICAL PROGRESSION. 203 root to be extracted ; then to the result annex the root of the rational part, which will give the root required. RXAHPLE8. 1* Required to find the square root of 16^/6. First, v/16 = 4, and (6^)* = 6* * = 6*; theref. (16 y 6)* = 4 . 6* = 4(/G, » the sq. root required. 2. Required to find the cube root of ^ Tint, ^ = J, and (i/3)* = s' " 5 * 3 = 3* ; theref. ^/3^ = | . 3^ = it/3, is the cube root required. 3. Required the square root of 6 s . Ans. 6^/6. 4. Required the cube root of |a 3 6. Ans. l atyb. 5. Required the 4th root of 16a* . Ans. 2y/a. 6. Required to find tie mth root of aA 7. Required the square root of a 9 — 6a y/b + 96. ARITHMETICAL PROPORTION AND PRO- GRESSION. Arithmetical Proportion is the relation whic'i two quantities, of the same kind, bear to each other, in respect to their difference. Four quantities are said to be in Arithmetical Proportion, when the difference between the first and seconJ is i vju... io the difference between the third and fourth. Thus, 3, 7, 12, 16, and a 9 a + 6, c, c + 6, are arith- metically proportional. Arithmetical Progression is when a series of quantifies V either increase or decrease by the same common difference. Thus, 1, 8, 5, 7, 9, 11, dec. and a, a -f- 6, a + 26, a +36, «+ 46, a + 56, &c. are series in arithmetical progression, whose common differences are 2 and 6. The most useful part of arithmetical proportion and pro. fwssion has been exhibited in the Arithmetic. The same may be given algebraically, thus : 204 ALGEBRA. Let a denote the least term, z the greatest term, d the common difference, n the number of the terms, and s the sum of the series ; then the princip U properties are expressed by these equa- tions, viz. 1. x = a + d. (n — 1) 2. a = z — €*.(»— 1) 3. s = (a 4. a = (z — \d . n — l )n, 5. « = (a + \d . n — l)n. Moreover, when the first term a is or nothing, the theorems become z = d (n — 1) and $ = Jam. EXAMPLES FOR PRACTICE. 1. The first term of an increasing arithmetical series is 1, the common difference 2, and the number of terms 21 ; re- quired the sum of the series ? First, 1 + 2 X 20 = 1 + 40 = 41, is the last term. 1 4- 41 Then —~— x 20 = 21 X 20 = 420, the sum required. A 2. The first term of a decreasing arithmetical series is 199, the common difference 3, and the number of terms 67 ; re- quired the sum of the series ? First, 199 — 3 . 66 = 199 — 198 = 1, is the last term. 199 4- 1 Then X 67 = 100 X 67 = 6700, the sum re- quired. 3. To find the sum of 100 terms of the natural numbers 1, 2, 3, 4, 5, 6, dec. And. 505a 4. * Required the sum of 99 terms of the odd numbers 1, 3, 5, 7, 9, dec. Ans. 9801. • The tarn of any number (w) of terms of the Arithmetical aerie* of odd numbers 1, 3, 5, 7, 9, &c. is equal to the square (*a ) of that nnfi> ber. That is, ' If 1, 3, ft. 7, 9 t &c. be the numbers, 1hpn will 1 2 s , 3 f , 4\ 6», be the sums of 1, 2 3, fcc. terms, Thus»0-4-l= 1 or 1M be sum of 1 term, 14-3= 4 or 2', the sum of 2 terms, 4 -j- 5 = -9 or 3-', the sum of 3 terms, 9 + 7 = 16 or 4» , the sum oC 4 tenai, &c ARITHMETICAL PBOGBESflXON. 90S 5. The first term of a decreasing arithmetical series is 10, the common difference 1, and the number of terms 21 ; re- quired the sum of the series ? Ans. 140. 6. One hundred stones being placed on the ground, in a straight line, at the distance of 2 yards from each other ; how far will a person travel, who shall bring them one by one to a basket, which is placed 2 yards from the first stone ? Ans. 11 miles and 840 yards. APPLICATION OF ARITHMETICAL PROGRES- SION. Qu. i. A Tin angular Battalion * consists of thirty ranks, in which the first rank is formed of one man only, the second of 3 ; the 3d of 5 ; and so on : What is the strength of such * triangular battalion ? Answer, 900 men. Qu. ii. A detachment having 12 successive days to march, with orders to advance the first day only 2 leagues, the second 3£, and so on, increasing l£ league each day's march : What is the length of the whole march, and what is the last day's march ? Answer, the last day's march is 18J leagues, and 123 leagues is the length of the whole march. Qu. in. A brigade of sappers*)*, having carried on 15 yards of sap the first night, the second only 13 yards, and For, by the 1st theorem, 1 + 2 (n - 1) - 1 -f2n - 2 = 2* - 1 » the last term, when the number of term* is n ; to this last term 2* — 1, ■dd the first term 1, gives 2n the sum of the extremes, or n half the sum of the extremes ; then, by the 3d theorem, nXn = n" is the sum of all the terms. Hence it appears, in general, that half the sum of the extremes b always the same as the number of the terms, n ; and that the sum of alt the terms is the same rs the square of the same number, as. See more on Arithmetical Proportion in the Arithmetic. * By triangular battalion, is to be understood, a body of troops ranged In the form of a triangle, in which the ranks exceed each other by an equal number of men: if the first rank consist of one man only, and the difference between the ranks be al«o 1, then it? form is that of an equilateral triangle; and when the difference between the ranks is mora than 1, its form may then be an isosceles or scalene triangle. The practice of forming troops in this order, which is now laid aside, Was formerly held in greater esteem than forming them in a solid square, at admitting of a greater front, especially when the troops were to make simply a stand on all sides. t A brigade of sappers consists generally of 8 men, divided tc\\ut\V] feto two parties. While one of these parties is advancing fat wp, \\v<* •(kerb faro \3hwg the gabions, fa* cines, and other neceuory \m*o\ a tm«D\A * 206 ALGEBRA. so on, decreasing 2 yards every night, till at last they car* r ried on in one night only 3 yards : What is the number of nights they were employed ; and what is the whole length of the sap. ||r Answer, they were employed 7 nights, and the length of the whole sap was 03 yards. Qu. iv. A number of gabions* being given to be placed in six ranks, one above the other, in such a manner as that each rank exceeding one another equally, the first may con- sist of 4 gabions, and the last of 9 : What is the number of gabions in the six ranks; and what is the difference between each rank ? Answer, the difference between the ranks will be 1, and the number of gabions in the six ranks will be 39. Qu. v. Two detachments, distant from each other 37 leagues, and both designing to occupy nn advantageous post equi-distant from each other's camp, set out at different times ; the first detachment increasing every day's march 1 league and a half, and the second detachment increasing each day's march 2 leagues : both the detachments arrive at the same time ; the first after 5 days' march, and the second after 4 days' march : What is the number of leagues marched by each detachment each day ? The progression r 7 T , 2fc, 5^, 6^, answers the con- ditions of the first detachment : and the progression If-, 3}, 5f , 7|, answers the condition of the second detachment. and when the first party is tired, (he second takes its place, and so on, till each man in turn has been at (he head of the nap. A sap is a >ma11 ditch, between 3 and 4 feet in breadth and depth ; and is distinguished from the trench by its breadth only, the trench having between 10 and 15 feet breadth. As an encouragement to tappers, the pay for all the work carried on by the whole brigade is given to the survivors. • Gabions are baskets, open at both ends, made of ozier twigs, and of a cylindrical form ; those made use of at the trenches are 2 fret wide, and about 3 feet high ; which, being filled with earth, serve aia shelter from the enemy's fire: and those made u«e of to construct bat- teries, are generally higherand broader. There is another sort of gabion, made usa of to raise a low parapet : its height is from 1 to 2 feel, and 1 foot wide at top, hut somewhat less at bottom, to give room for placing the muszle of a firelock between them : these gabions serve instead 0? sand bags. A sand bag is generally made to contain about a cubic foot of earth. PILIXO OF BALLS. 207 OF COMPUTING SHOT OR SHELLS IN A FINISHED PILE. Shot and Shells are generally piled in three different forms, called triangular, square, or oblong piles, according as their base is cither a triangle, a square, or u rectangle. Fig. 1. C G Fig. 2 abcdef, fig. 3, is an oblong pile. A. triangular pile is formed by the continual laying of tri- angular horizontal courses of shot one above another, in such a manner, as that the sides of these courses, called rows, decrease by unity from the bottom row to the top row, which ends always in 1 shot. A square pile is formed by the continual laying of square horizontal courses of shot one above another, in such a man. ner, as that the sides of these courses decrease by unity from the bottom to the top row, which ends also in 1 shot. In the triangular and the square piles, the sides or faces being equilateral triangft&, the shot contained in those faces form an arithmetical progression, having for first term unity, and for last term and number of terms, the shot CAmV&va&& •7 208 AXGKBRA. in the bottom row ; for the number of horizontal rows, or the number counted on one of the angles from the bottom to the top, is always equal to those counted on one side in the bottom : the sides or faces in either the triangular or square piles, are called arithmetical triangles ; and the numbers contained in these, arc called triangular numbers : abc, fig. 1, kfg, fig. 2, are arithmetical triangles. The oblong pile may be conceived as formed from the square pile abcd ; to one side or face of which, as ad, a number of arithmetical triangles equal to the face have been added : and the number of arithmetical triangles added to the square pile, by means of which the oblong pile is formed, is always one less than the shot in the top row ; or which is the same, equal to the difference between the bottom row of the greater side and that of the lesser. Qu. vi. To find the shot in the triangular pile jlbod, fig. 1, the bottom row ab consisting of 8 shot. Solution. The proposed pile consisting of 8 horizontal courses, each of which forms an equilateral triangle ; that is, the shot contained in these being in an arithmetical progres- sion, of which the first and last term, as also the number of terms, ore known ; it follows, that the sum of these particu- lar courses, or of the 8 progressions, will be the shot con- tained in the proposed pile ; then The shot of the first or lower ) triangular course will be $ (8 + l)X4— 36 the second - - - - (7 + 1) X 3J- = 28 the third - - - - (6 + 1) X 3 = 21 the fourth - - - - (5 + 1) X 2i = 15 the fifth - - - - (4 + 1) X 2 = 10 the sixth - - - - (3 + 1) X 1 J = 6 the seventh - - - . (2 + 1) X 1 = 3 the eighth - - - - (1 + 1) X J = 1 Total 120 shot in the pile proposed. Qu. vn. To find the shot of the square pile efgh, fig. 2, the bottom row sf consisting of 8 shot. Solution. The bottom row containing 8 shot, and the second only 7 ; that is, the rows farming the progression, 8, 7, 6, 5, 4, 3, 2, 1, in which each of the terms being the square root of the shot contained in each separate square FILING OF BALLS. 209 ; course employed in forming the square pile ; it follows, that the sum of the squares of these roots will be the shot requir- ed ; and the sum of the squares divided by 8, 7, 6, 5, 4, 3, 2, 1, being 204, expresses the shot in the proposed pile. Qu. viii. To find the shot of the oblong pile abcdep, fig. 3 ; in which bf = 16, and bc = 7. Solution. The oblong pile proposed, consisting of the square pile abcd, whose bottom row is 7 shot ; besides 9 arithmetical triangles or progressions, in which the first and last term, as also the number of terms, are known ; it follows, that, if to the contents of the square pile - - 140 we add the sum of the 9th progression - 252 &\, r their total gives the contents required - - 392 shot. REMARK I. The shot in the triangular and the square piles, as also the shot in each horizontal course, may at once bo ascer- tained by the following table : tho vertical column a con- tains the shot in the bottom row, from 1 to 40 inclusive ; the column b contains the triangular numbers, or number of each course ; the column c contains the sum of the tri- angular numbers, that is, the shot contained in a triangular pile, commonly called pyramidal numbers ; the column n contains the square of tho numbers of the column a, that is, the shot contained in each square horizontal course ; and the column £ contains the sum of these squares or shot in a square pile. Vol. I 28 210 ALQMBEAm c B A D E Pyramidal Triangular Natural Square of the natural Sum of (beta square numbers. number* d umbers. numbers* numbers. 1 1 1 1 1 4 3 2 A 4 c O 10 6 3 1 A 14 20 10 4 lo OA 4U 35 15 5 2d 00 56 21 6 Qfi OD 84 28 7 a n 49 1 Afk 140 120 36 64 204 165 45 9 81 AO C 285 220 55 10 100 385 286 66 11 121 506 364 78 12 144 690 455 91 13 169 819 560 105 14 196 1015 680 120 15 225 1240 816 136 16 256 1496 969 153 17 289 1785 1140 171 18 324 2109 1330 190 • 19 361 2470 1540 210 20 400 2870 1771 231 21 441 3311 2024 253 22 484 3795 9 MOO 976 23 529 4324 2600 300 24 576 4900 2925 325 25 625 5525 3276 351 26 676 6201 3654 378 27 729 6930 4060 406 28 784 7714 4495 435 29 841 8555 4960 465 30 900 9455 5456 496 31 961 10416 5984 528 32 1024 11440 6545 561 33 1089 12529 7140 595 ! 34 1156 13685 7770 630 35 1225 14910 8436 666 36 1296 16206 9139 703 37 1369 17575 9880 741 38 1444 19019 10660 780 39 1521 20540 11480 820 40 1600 22140 Thus, tho bottom row in a triangular pile, consisting of shot, the contents will be 1330 ; and when of 19 in the squfl»-i* FILING 09 BALLS. 211 t m . pile, 3470.— In die same manner, the contents either of a 71 square or triangular pile being given, the shot in the bottom row may be easily ascertained. The contents of any oblong pile by the preceding table may be also with little trouble ascertained, the less side not exceeding 40 shot, nor the difference between the less and the greater side 40. Thus, to find the shot in an oblong pile, the less side being 15, and the greater 35, we are first to find the contents of the square pile, by means of which the oblong pile may be conceived to be formed ; that is, we arc to find the contents of a square pile, whose bottom row is 15 shot : which being 0B4O, we are, secondly, to add these 1240 to the product 2400 of the triangular number 120, answering to 15, the number expressing the bottom row of the arithmetical triangle, multiplied by 20, the number of jtito. triangles* and their sum, being 3640, expresses the "^ppUber of shot V-ihn proposed oblong pile. • SEHARK II. The following algebraical expressions, deduced from the investigations of the sums of the powers of numbers in arithmetical progression, which are men upon many gunners' callipers' 11 , serve to compute with ease and expedition the shot or shells in any pile. That serving to compute any triangular > (n+2)X(n+l)Xn pile, is represented by $ 6 That serving to compute any square ) (n+ l)X( 2n+l)xn pile, is represented by $ (5 In each of these, the letter n represents the number in the bottom row : hence, in a triangular pile, the number in the bottom row being 30 ; then this pile will be (30+2) X (30+1) X V = 4960 shot or shells. In a square pile, the number in the bottom row being also 30 ; then this pile will be (30 + 1) X (60 + 1) X V = W5 5 shot or shells. • Callipers are large compasses, with bowed shanks, serving to tnke the diameters of convex and concave bodies. The gunners' callipers consist of two thin rules or plates, which are moveable quite round a Joint, by the plates folding one over the other : the length of each rule or plate is 6 inches, the breadth about 1 inch. It is usual to represent, -*>tt the plates, ■ variety of scales, tables, proportions, &c. such as are esteemed useful to be known by persons employed about artillery ; hut, «xeept the measuring of the caliber of shot and cannon, and the measur- ing of salient and re-enterine angles, none of the articles, with YiVfa>i the callipers are usually filled, an essential to thai taitroxneal 212 ALGEBRA. That serving to compute any oblong pile, is represented by (2n + 1 + 3m) X (n + 1) X n . «» « ■ « . , . ^ ! ' — - — ' — - , in which the letter n denotes o the number of courses, and the letter m the number of shot, less one, in the top row ; hence, in an oblong pile the num- ber of courses being 30, and the top row 31 ; this pile will be COTT+90 X 30 + 1 X y = 23405 shot or shells. REMARK III. One practical rule, of easy reelection, will include the three cases of the triangular, square^ and rectangular, com- plete piles. Thus, recurring to the diagrams 1, 2, and 3, we shall have, balls in (bd + a + c) X £bdo = triar* liar pile. (ef + ef + g) X £gfh = aquare pile. (bf + bf +ae) X £abc = rectangular pita Hence, for a general rule : add to the number of balls or shells in one side of the base, the numbers in its two paral- lels at bottom and top (whether row or ball), the sum being multiplied by a third of the slant end or face, gives the number in the pile. % GEOMETRICAL PROPORTION, AND PRO- GRESSION. Geometrical Proportion contemplates the relation of quantities considered as to what part or what multiple ono is of another, or how often one contains, or is contained in, another. — Of two quantities compared together, the first is called the Antecedent, and the second the Consequent, Their ratio is tho quotient which arises from dividing the one by the other. Four Quantities are proportional, 'when the two couplets have equal ratios, or when the first is the same part or mul- tiple of the second, as the third is of the fourth. Hius, 3, 6, 4, 8, and a, ar, &, fer, are geometrical proportional*. or hr For f = f =2, and -r = -y = r. And they are staled thus, 3 : 6 : : 4 : 8, &c. See the Arithmetic. Geometrical Progression is one in which the terms have GEOMETRICAL PROGRESSION. 213 all successively the same ratio ; as 1, 2, 4, 8, 16, dec. where the common ratio is 2. The general and common property of a geometrical pro- gression is, that the product of any two terms, or the square of any one single term, is equal to the product of every other two terms that are taken at an equal distance on both sides from the former. So of these terras, 1, 2, 4, 8, 16, 82, 64, czc. 1X04 = 2X 32 = 4X 16 = 8X 8=64. In any geometrical progression, if a denote the least term, x the greatest term, r the common ratio, n the number of the terms, 8 the sum of the series, or all the terms ; any of these quantities may be found from the others, by means of these general values or equations, viz. 2. z = a X r*~ l . 1 — 4. n = a = l°g» r + log, g — log * a log. r log.r 5. * = X a = r- X— r = -. r — 1 r — 1 r*~ l r - 1 When the series is infinite, then the least term a is nothing, and the sum s = r— 1 In any increasing geometrical progression, or series be- ginning with 1, the 3d, 5th, 7th, czc. terms will be squares ; the 4th, 7th, 10th, czc. cubes ; and the 7th will be both a square and a cube. Thus, in the series 1, r, r", r 3 , r* 9 r 5 , r*, r 7 , r 1 , r*, &c. r 2 , r 4 , r 1 , r*, are squares ; r*, r", r 9 , cubes ; and r° both a square and a cube. In a decreasing geometrical progression, the ratio, r, is a 1— r» fraction, and then s = - -a. If n be infinite, this becomes * = 1 — r ' a ^ >e * n 8 ^ **** term r t 214 ALGSUU* When four quantities, a, or, b, br, or 2, 6, 4, 12, are pro- portional ; then any of the following forms of those quantities are also proportional, viz. 1. Directly, a : or : : b : br ; or 2 : 6 : : 4 : 12. 2. Inversely, «r : a : : fcr : b ; or 6 : 2 : : 12 : 4. 3. Alternately, a : b ; : or : br ; or 2 : 4 : : 6 : 12. 4. Compoundedly, a : a+ar ::b: b+br ; or 2 : 8 : : 4 : 10. 5. Dividedly, a : or— a : : b : ftr—ft ; or 2 : 4 : : 4 : 8. 6. Mixed, ar-fa: or — a : : br+b : ftr— 6 ; or 8 : 4 : : 16 : 8. 7. Multiplication, ae : arc : : be : ftrc ; or 2.3 : 6.3 : : 4 : 12, 8. Division, — : — : : b : br ; or 1 : 3 : : 4 : 12. c c 9. The numbers a, 6, c, J, are in harmonical proportion, when a : d : : a b : c d ; or when their reciprocals -j-, -i, are in arithmetical proportion. EXAMPLES. 1. Given the first term of a geometrical series 1, the ratio 2, and the number of terms 12 ; to find the sum of the series 7 First, 1 X 2 11 = 1 X §048, is the last term. 2048 X 2-1 4096 — 1 Ant% - A . . , Then : = = 4095, the sum required. <£ — 1 1 „ 2. Given the first term of a geometric series |, the ratio and the number of terms 8 ; to find the sum of the series 1 First, | X (i) 7 = }X T } T = sly , is the last term. Then (i - *U X i) + (W) = (*- T * T ) -7- i = f ff Xf = a{ { , the sum required. 3. Required the sum of 12 terms of the series 1, 3, 9,27, 81, &c. Ans. 265720. 4. Required the sum of 12 terms of the series, 1, £, -fr. A. &c. Ans. fflfff 5. Required the sum of 100 terms of the series, 1, 2, 4, 8, 16, 32, &c. Ans. 1267650600228220401496703205375. See more of Geometrical Proportion in the Arithmetic. INFINITE SERIES. An Infinite Series is formed either from division, dividing by a compound divisor, or by extracting the root of a com* pound surd quantity, or by other general processes ; and is - ntrnuTB series. 215 such at, being continued, would run on infinitely, in the manner of a continued decimal fraction*. But, by obtaining a few of the first terms, the law of the progression will be manifest ; so that the series may thence be continued, without actually performing the whole operation. problem I. To reduce Fractional Quantities into Infinite Series by Division. Divide the numerator b)- the denominator, as in common division ; then the operation, continued as far as may be thought necessary, will give the infinite series required. EXAMPLE. 2ab 1. To change — p-r into an infinite scries. j7 ^ a + b 2oft..(2&- 2ab + 2b 2 2b 2 2&' 2b* * + b)2ab..{2b- — + -^ — ^ +&c. a or or — 2b 2 2b* — 2b 2 — — a 2b 3 a 2P W a + a 2 2b* a 2 2b* 2b 9 a 2 a 3 * The doctrine of infinite series was commenced by Dr. Wallis ; who, fn his arithmetical works published in 1657, first reduced the frac- tal |-~ by a perpetual division into the infinite series a + ox + at* + , if s + ar4 + &c. 316 AL0BBJU. 2. Let c ^ an 8 e ^ * nto 811 infinite series. 1 — a) l....(l + a + a , + o» + a«+ &c. a a — a» a* -a' 3. Expand — into an infinite series. Ans.^X(l-f + ^-| + &c.) o a or or 4. Expand ^3-^ into an infinite series. a a a + a 1 x 5. Expand ^ ^ into an infinite series. Ans. 1 — 2x + 2* 1 — 2x 3 + 2x\ &c. a a 6. Expand - — r-rr, into an infinite series. r (* + 6) a Ans. 1 f- -3- 3, &c a <r a 3 7. Expand ^ ^ ^ = J, into an infinite series. PROBLEM II. 2b reduce a Compound Surd into an Infinite Series. Extract the root as in common arithmetic ; then the operation, continued as far as may be thought necessary, will give the series required. But this method is chiefly of use in extracting the square root, the operation being too tedious for the higher powers. ZHFUTITB SERIES. 217 EXAMPLES. 1. Extract the root of a 3 — x 2 in an infinite series. „._*»(«____ _ — - _ T &c. 2« — — i 2a) X 4 — x a + — ^ 4a* X s X 4 x 4 X 4 X 9 X* 4^ §0*" + 64a« X« ** . V 8a 4 04a° x 8 X s ■ -4 &c. 8a 4 T 16a° dec. 04a« 2. Expand ^/(l + 1) = \/2, into on infinite series. Ans. l + £-J- + -,V-r5F &c. 3. Expand ^/(l — 1) into an infinite scries. Ans. i-yV— t{? &c. 4. Expand </{c? + x) into an infinite series. 5. Expand ^/(o 3 — 2bx — x 3 ) to an infinite series. PROBLEM III. To extract any Root of a Binomial: or to reduce a Binomial St . d into an infinite Series. This will be done by substituting the particular letters of the binomial, with their proper signs, in the following gene- ral theorem or formula, viz. , _ v ^ ^ _ m in — 7i , m — 2ji (* + Pft) w = P n + — aq + ~2^" " Ba "* f*n~~ Ctt + &c- Vol. I. 29 218 ALCKBSA. and it will give the root required : observing that p denotes m the first term, a the second term divided by the first, « the index of the power or root ; and a, b, c, d, dec. denote the several foregoing terms with their proper signs. EXAMPLES. . 1. To extract the sq. root of a 1 + V 9 in an infinite series* Here p « a 1 , a = and — = g 2 therefore m m p w = (a a )*= (a*) = a = a, the 1st term of the series. — Aa = i XaX^=^ = b, the 2d term. m-n 1—2 v i» v 6» ft* _ . . -s — sa = —j— x — X — = — ^r— 1 ■= c, the 3d term. . 2n 4 2a a* 2.4<r w _2n 1-4 i« V 36* t . ... Hence a + ^ - g — + j-g- - &c. or a + 2«""8^ + 16^ j2^ 7 «fec.u.the senw required. 2. To find the value of ; — ^—7^, or its equal (a — x)~~* in an (a — x)* n N ' infinite series*. Here p=o, q=— =-o- , «, and — = ^? = -2 ; theref. a n 1 * ATofe. To facilitate the application of the rule to fractional eiem- pies, it is proper to observe, that any surd may be taken from the de- nominator of a fraction and placed in the numerator, and vice versa, by only changing the sign of its indei. Thus, L = 1 x ar-a or only x-*; and = 1 X (« + or (a + *)-»; (o» + x») J x («* - **H ; &c. The theorem above given is only the Binomial Theorem so expreav ed as to facilitate its application to roots and series. INFIlCtTB SXStES. 219 = [a)- 9 = a~* = ^- ■= a, the first term of the series. — aq = -2X~X— = ~=2a-^r = b, the 2d term. * tr a or jr — ft 2jf —x 3x* — Tjia =-| x^X-— X = ^~- = 3«-V = c, the 3d. 2» ■ a 3 a a* «— 2» 4 ^Si 1 ^ — x 4x» . g . aft 1 a 4 a a 5 Hence or* + 2a- 3 * + 3a- 4 *" + 4a-** 3 + dec. or 1 , 2x , 3*» 4r* 5x* . . . . , — r + — H — — + — — + — - dec is the series required. Qi Or ft* ft* fl J a 9 8. To find the value of , in an infinite series. a— a? X* X 9 X 4 Ans.a+*+- + -+-&e. 4. To eqpand ✓ „ — in a series. . 1 at* , 3x« 5x* . 5. To expand t r-ri in an infinite series. («-*)" a i . 26 , 3i» 4i» 56 4 . Ans. 1 H h — r + — ^- H — - <fcc. o a* a*. a 4 6. To expand ^/o 1 — x a or (a 3 — x*)^ in a series. x* x 4 x* 5x f Ans. «__-__ I6^~i28?^. f . 7. To find the value of ?/ (a 3 - 6 s ) or (tf 3 — 6 3 )* in a series. 6 3 5b 9 m 8. To find the value of V (<* + «*) or (ci'+x 5 )* in a series. X s 2c 10 6x l 8 o — 6 9. To find the square root of ^ in an infinite series. 220 ALGEBSA. a* 10. Find the cube root of -tttt i° a series. o» . 2&« 3a 3 9a 8 81a» Ans.l__+--.- — &c INFINITE SERIES : PART THE SECOND. PROBLEM I *• A series being given, to find the several orders of dif- ferences of the successive terms. Rule i. Subtract the first term from the second, the second from the third, the third From the fourth, and so on ; the several remainders, will constitute a new series, called the first order of differences. ii. In this new series, take the first terra from the second, the second from the third, dec. as before, and the remainders will form another new series, called the second order of dif. ferences. in. Proceed in the same manner for the third, fourth, fifth, <3fc. orders, until either the differences become 0, or the work will be carried as far as is thought necessaryf . EXAMPLES. 1. Given the series 1, 4, 8, 13, 19, 26, &c. to find the several orders of differences. * The study of this second part of Infinite Series may be conve- niently postponed till Simple and Quadratic Equations have been learnt. t Let a, 6, e, d, e, &c. be the terms of a given series, then if d = the first term of the nth order of differences, the following theorem will exhibit the value of d : vis. ;fc a =P *6 + n . n "7* . c + n . *T"^ - . n— 2 , . 11— 1 n— 2 is— 3 . . , . —3— • d ± * • — y- • —3— • —5— • « Hhi fcc- (to n + 1 terms) = d, where the upper signs must be taken when n is an even number, and the lower signs when it is odd. If the differences be very great, the logarithms of the quantities may be used, the differences of which will be much smaller than those of ^ the cjuantities themselves ; and at the close of the operation the natural number answering to the logarithmical result will be the answer. See Emsnan's ZHfnimtal Mcihod; prop. 1. INFIN I TE SSBIE8. 391 Thus 1, 4, 8, 13, 19, 26, dec. the given series. Then, 3, 4, 5, 6, 7, dec. the first differences. And 1, 1, 1, 1, dec. the second differences. Also 0, 0, 0, dec. the third differences, where the work evidently must terminate. 2. Given the series 1, 4, 8, 16, 32, 64, 128, dec. to find the several orders of differences. Here 1, 4, 8, 16, 32, 64, 128, dec. given series. And 3, 4, 8, 16, 32, 64, dec. 1st diff. 1, 4, 8, 16, .32, &c 2nd diff. 3, 4, 8, 16, &c. 3rd diff. 1, 4, 8, dec. 4th diff. 3, 4, fcc. 5th diff. 1, &c. 6th diff. 3. Find the several orders of differences in the series 1, 2, 3, 4, &c. Ans. First diff. 1, 1, 1, 1, &c. Second diff. 0, 0, 0, &c. 4. To find the several orders of differences in the series 1, 4, 9, 16, 25, &c. of squares. Ans. First differences 3, 5, 7, 9, dec. Second, 2, 2, 2, &c. Third, 0, 0, dec. 5. Required the orders of differences in the series 1, 8, 27, 64, 125, &c. being cubes. 6. Given 1, 6, 20, 50, 105, dec. to find the several orders of differences* PROBLEM, n. To Find any term of a given series. Rule i. Let a, ft, c, d, e, dec. be the given series ; d r , d", d m , <Z* V , dec. respectively, the first term of the first, second, third, fourth, dec. order of differences, as found by the ore- ceding article; n = the number denoting the place or the term required. n. Then will a + ^1. d' + ^=1 . ^ . d> + 1 1 A 1 w— 2 n— 3 n— 1 n-2 n-3 n — i ■ to the nth term required. 388 ALGEBRA* XX AMPLE 8* 1. To find the 10th term of the series 2, 5, 9, 14, 20, dec Here 2, 5, 9, 14, 20, dec. series. 3, 4, 5, 6, dec. 1st diff. 1, 1, 1, dec. 2nd diff. 0, 0, dec 3rd diff. Where d' = 3, d' = 1, d'" = 0, also a = 2, n = 10 ; wherefore a H j— . d' H — ---g— . d f = (2 -| j — X3 + 12pIxi^?Xl=)2+27 + 86 = e6 = the 10th term required. 2. To find the 20th term of the series 2, 6, 12, 20, 30, dee. Here a = 2, n = 20 ; and Art. 12. 2, 6, 12, 20, 30, dec. series. 4, 6, 8, 10, dec. 1st diff. 2, 2, 2, dec. 2nd diff. or d' = 4, d' = 2 ; whence a -\ j— . a' + — p- . — ^— . d" = (2 + — X 4+ ^ X ^X2 = )2 + 76 + 342 = 420 = the 20th term required. 3. Required the 5th term of the series, 1, 3, 6, 10, dec. Ans. 16. 4. To find the 10th term of the series, 1, 4, 8, 13, 19, dee. Ans. 64. 5. Required the 20th term of the series, 1, 8, 27, 64, 125, dec. Ans. 8000. PROBLEM HI. If the succeeding terms of a given series be at an unit's distance from each other, to find any intermediate term by interpolation. Rule 1. Let y be the term to be interpolated, x its distance from the beginning of the series, d', d% d h \ d ir , dec* the tint terms of the several orders of differences. INFINITE SEBIES. 223 2. Then willa + *d' + * . ^ . d' +*. ^ . <T"+ * . . ^-jp • ^-j^ . <# v + &c. = y, the term required. EXAMPLES. 1. Given the logarithmic sines of 3° 4', 3° 5', 3° 6', 8° 7', and 3° 8', to find the sine of 3° 6' 16*. Series. Logarithm*. Istdiff. 2nddiff. 3rddif. 8*4' 8-7283366 23516 1QA 3 5 8-7306882 23390 ~ifl 1 3 6 8-7330272 23263 ~\ZL — * 3 7 8-7353535 23140 ~ KM 3 8 8-7876675 Here x "= (3*6' 15'- 3° 4' = 2' 15' = ) f = the distance of the term y, to be interpolated; a = 8-7283366, d' = 23516, d" = - 126, d'" = 1, and y = a + <rd' + *. 2=1' * + x ?=1. . = (a + K + Hd' + M*" = ) 8-7283366 + 0052911 — -00001771875 + •0000000117 = 8-73360999296, the log. sine required. 2. Given the series -fa, fa, iV iV> t0 find the tenn which stands in the middle, between fa and fa. Ans. y| T . 3. Given the logarithmic sines of 1° 0', 1° \', 1 Q 2', and 1° 3', to find the logarithmic sine of 1° 1' 40". Ans. 8-2537533. PROBLEM IV. 2b find any intermediate Term by Interpolation, when the first Differences of a Series of equidifferent Terms are small. Rule 1. Let a, b, c, d, e, <kc. represent the given series, and n = the number of terms given. 2. Then will a — nb + n . n ~ 1 . c — n*. . n ~ 2 . d + n . n ^-^ . . . e + dec. = 0, from whence 55 o 4 by transportation, dec. any required term may be obtained*. • For the investigation of these roles, see Emenon't Diffemdiel Msthod. ▲L0JBBJU. EXAJCPLBS. 1. Given the square root of 10, 11, 12, 13, and 15, to find the square root of 14. Here n = 5, and e is the term required. a = (-/10 = ) 31622776 b = (^11 = ) 3-3166248 c = (v/12=) 8-4641016 d = L/12 = ) 3-6055512 / = (^15=) 8-8729833 And since n = 5, the series must be continued to 6 terms. Therefore a — nft + n.— . c — n . — . .d + n — 1 n — 2 n — 3 n — 1 n — 2 n — 3 n — 4 ./=0. Whence, by transposition, in order to find c we shall have n — 1 n — 2 n— 8 , . n — 1 , n. g . 3 . —j— .e = — a + nft — n. — .c + n n— 1 9i — 2 , . n -1 n — 2 n — 3 n — 4 r .- 2 -.- 3 ~,<*+».-2-.-3 j-.-g-./; *" in numbers becomes 5c = — 3-1622776 + (5 X 3-3166248) — (10 x 3-4641016) + (10 X 3-6055512) + 3-8729833 =* 56-5116193— 37-8032936=18-7088257, and e= 18 7083257 5 ss 3-74166514 = the root, nearly. 2. Given the square roots of 37, 38, 39, 41, and 42, to find the square root of 40. Ans. 6-32455532. 3. Given the cube roots of 45, 46, 47, 48, and 49, to find the cube root of 50. Ans. 3-684033. PROBLEM V. 7b revert a given Series. When the powers of an unknown quantity are contained in the terms of a aeries, the finding the value of the unknown quantity in another series, which involves the powers of the nornriTE series. 225 quantity to which the given series is equal, and known quan- tities only, is called reverting the series*. Rule 1. Assume a series for tho value of the unknown quantity, of the same form with the series which is required to be reverted. 2. Substitute this series and its powers, for the unknown quantity and its powers, in the given series. 3. Make the resulting terms equal to the corresponding terms of the given series, whence the values of the assumed co-efficients will be obtained. examples. 1. Let ax + bx* + cr* + dx* + &c. = z be given, to find the value of x in terms of z and known quantities. Let z*= x, then it is plain that if z n and its powers be sub- stituted in the given series for x and its powers, the indices of z will be n, 2n, 3n, 4n, &c. and 1 ; whence n= 1, and the differences of these indices are 0, 1, 2, 3, 4, &c. Where- fore the indices of the series to be assumed, must have the same differences ; let therefore this series be az + bz 3 + cz 3 + dz 4 + ©zc. = x. And if this series be involved, and sub- stituted for the several powers of x, in the given series, it will become axz + obz 2 + acz 3 + avz* + dec. * + 6aV + 26ab* 3 + 2b ac z K + dec. * * * + £hV + dec. * * + CA 3 Z 3 + 3cA a HZ l +&C * * * + dA 4 Z 4 . + &C Whence, by equating the terms which contain like powers of *, we obtain (oaz = z, or) a = -i- ; {aitz 2 + 6a V = 0, whence) b = ( — = ) - ~, (acz*+ 2 W+ caV=0, t 2oab+ca 3 K W-ac whence) c = ( = ) - — 5 - — ; J> = ( — - &bAC+bi?+3cA a B+dA* . 5abc-5b 3 -a>d _ . 1 x= ) , &c. and conse- * Other methods of reversion are given by different mathemalic;axv%» Tltie above is selected for it* simplicity. Vol. I. 30 326 ALGEBRA. quently x=(\z+Bz > +cz i + &c.=)-^ ^ + — s*— = . 2* + sc. the senes required. a This conclusion forms a general theorem for every similar series, involving the like powers of the unknown quantity. 2. Let the series x — x 2 + x 3 — x* + dec. = *, be propos- ed l'>r reversion. Here a = 1, b = — 1, c = 1, d = — 1, &c. these values being substituted in the theorem derived from the preceding example, we thence obtain x = z + z 2 + z 3 +z 4 + &c. the answer required. ^2 x 3 jgi 3. Let x — + - — — + = y, be given for rever- sion. Substituting as before, we have a = l,6= — |, c =4, and d =— £, &c. These values being substituted, we shall y2 y3 yA have x = y + + ~r + &c. from which if y be given, 2 24 and sufficiently small for the series to approximate, the value of x will be known. PROBLEM VI. To find the Sum of n Terms of an Infinite Series. Rule 1. Let a, b, c, d, c, &c. be the given scries, s = the sum of n terms, and d', d% d"\ d w , &c. respectively the first terms of the several orders of differences, found by prob. 1. 2. Then will na + n . ^ . d' + n . 5=1 . ^ . d*+ n-l n-2 n-3 n— 1 n— 2 n— 3 n— 4 . d lv + &c. = s, the sum of n terms of the series, as was re- quired. Case 1. To find the sum of n terms of the series 1, 2, 3, 4, 5, dec. First, 1, 2, 3, 4, 5, &c. the given series. 1, 1, 1, 1, &c, first differences. 0, 0, 0, &c. second differences. Here o=l, d'=l, <T=0 ; then will na + n . ^ . cT INFINITE SERIES. 22ff ^ ^""^ ' — , which, (since a and d' each = 1) = 2n+«* — * v » . a + 1 A , . , = ) ~ = 8, the sum required. EXAMPLES. 1. Let the sum of 20 terms of the above series be re- quired. „ on j n^n+T 20X21 01A . Here » = 20, and * = — ^— = — - — = 210, the ans. 2. Let the sum of 1000 terms be required. Ans. 500500. 3. Let the sum of 12345 terms be required. Case 2. To find the sura of n terms of the series, 1, 3, 5,7,9, <&c. Here 1, 3, 5, 7, 9, &c. the given series. 2, 2, 2, 2, &c. . . . first difference. 0, 0, 0, &c. . . . second difference. Wherefore a = 1, d r = 2, <T = 0, and na + n . <*' ft 3 — ft == (na H — . d' = (since a = 1 and a* = 2) n + n 3 — n =)** = *, the sum required. EXAMPLE. To find the sum of 10 terms of the above series. Here » = 10, and s =- (ft 3 =) 100, the answer. Case 3. To find the sum of it terms of the series of squares 1, 4, 9, 16, 25, &c. Here 1, 4, 9, 16, 25, ©zc. the series. 3, 5, 7, 9, &c 1st difF. 2,2,2, <fcc 2nddi(F. 0, 0, &c 3rd diff. Whence a =* 1, d' = 3, d" = 2, d'" = 0, and na + n . _.^ + ._._.^ = ( w+ .3^.. — . + 2n. n— 2 _ 3a* -n n*-3n ? -f-2n n .Ai T l.-'n 1 1 ~' "3 T~ + 3 " ' ft * 4hs sum required. 838 ALGEBRA. EXAMPLE. Let the sum of 30 terms of the above series be required. Here „= 30; wherefore^l^±i)=?^«- 9455, the answer. See the table, pa. 210, PROBLEM VII. To find the Sums of Series, by the Method of Subtraction. This method will be rendered evident by two or three simple examples. EXAMPLE 1. Let 1 + i + ± + } + ozc. in inf. = s then \ + J + } + I + ozc. in inf. = * — 1. by sub. ig + i- 4 + i- + & c. mm/. = 1. EXAMPLE 2. Let 1 + | + * + &c. = * then i + i + i + i + <&c. = s = |. 2 2 2 2 * 8ub - O + 274. + 375 + 4T6 + &C ' = * °'- 5 - 2 'i!3 + ^4 + 3 1 5 + i + &C - = *- EXAMPLE 3. 1 1.2 + 1 2.3 + 1 3.4 1 2.3 + 1 3.4 + 1 4.5 b y 8Ub -r|-3 + 2li + ro + &c - e= * INFINITE SBBXE8. EXAMPLE 4. Find the s»m of the series ^ + ji-g + — + &c. Take away the last factor out of each denominator, and 1,1 , 1 , - " rane 23 + + 678 + * C - S!! * * >ub OS + SXft + £§X0 + * C ' - * KXAKFLK 4. Find the sum of the infinite series _1 , 1 1 , I 8.4.6.8 4.6.8.10 6.8.10.12 8.10.12.14 Ans. ^t- EXAMPLE 5. Find the sum of the infinite series 1 , 1 , 1_ , . *" * q ll i^ q n ia l* ■ n ia inr oa ' flcc « 3.5.8.11 4 5.8.11.14 4 8.11.14.17 r 11.14.17.20 Ans. PROBLEM vni. To sum an infinite series by supposing it to arise from the expansion of some fractional expression. Rule. Assume the series equal to a fraction, whose de- nominator is such, that when the series is multiplied by it, the product may be finite ; this product being equal to the numerator of the assumed fraction, determines its ^jue. EXAMPLES. 1. Required the sum of the infinite series x a? * 280 ALGEBRA. Assume the series = -r^— 1 — x then x + x 2 + x 3 + d&c. into 1 — x x + x 3 + x 3 + d&c. — x* — x 3 — dec. 2 = X .\ x + x* + x 3 + dec* = j-^-. Thus, if x = J, then i + * + i + &c = | -^ j = 1 ; if x = J, then i + i + ft + &c. = $ -r | = J. 2. Required the sum of the infinite series x + 2x a + 8r* + dec. Assume the series = (l—x) a 1— 2x+x a ' then x + 2x* + 3X 3 + dec. into 1 — 2x x a x + 2r a + 3r» + dtc. _2x 3 — 4i 3 — dec. + V + d:c. V x + 2i a + 3x 3 + dec. = - a . (1— x) If x = J, then ^ + f + | + T V + dec. = i -T- -J = 2. If x = J, then J + J + ft + ? * r + dtc. = | -j- J = f. And so on, in other cases *. 3. Find the sum of the infinite series x + 4x 2 + 9x* + 16x 4 + dcc. . *(l+x) * TbvPpfrfeceding is only a sketch of an inexhaustible subject. For the algebraical investigation of infinite series, consult Dodson's Math* malical Repository, and Mr. J. R. Younu's Mnebra. The subject, how- ever, is much more extensively treated by means of the flaiional •BjJjsift 231 SIMPLE EQUATIONS. An Equation is the expression of two equal quantities with the sign of equality (=) placed between them. Thus 10 — 4 = 6 is an equation, denoting the equality of the quantities 10 — 4 and 6. Equations are either simple or compound. A Simple Equation, is that which contains only one power of the un- known quantity, without including different powers. Thus, x — a = b + c, or ax a = b 9 is a simple equation, containing only one power of the unknown quantity x. But x 2 — 2ax ■= b 2 is a compound one. GENERAL RULE. Reduction of Equations, is the finding the value of the unknown quantity. And this consists in disengaging that quantity from the known ones; or in ordering the equa- tion so, that the unknown letter or quantity may stand alone on one side of the equation, or of the mark of equality, without a co-efficient ; and all the rest, or the known quan- ties, on the other side. — In general, the unknown quantity is disengaged from the known ones, by performing always the reverse operations. So, if the known quantities arc con- nected with it by + or addition, they must be subtracted ; if by minus ( — ), or subtraction, they must be added ; if by multiplication, we must divide by them ; if by division, we must multiply ; when it is in any power, wo must extract the root ; and when in any radical, we must raise it to the power. As in the following particular rules ; which are founded on the general principle, that when equal operations are performed on equal quantities, the results must still be equal ; whether by equal additions, or subtractions, or mul- tiplications, or divisions, or roots, or powers. PAB^ftcULAR BULK T. When known quantities arc connected with the unknown hy + or — ; transpose them to the other side of the equa- tion, and change their signs. Which is only adding or sab- tracting the same quantities on both sides, vu ottat to 232 all the unknown terms on one side of the question, and all the known ones on the other side*. Thus, if x + 5 = 8 ; then transposing 5 gives x=8 — 5=3. And if x— 3 + 7 = 9; then transposing the 3 and 7, gives x=9 + 3 — 7 = 5. Also, if* — a + b^cd 9 then by transpc sing a and ft, it is x =» a — b + cd. In like manner, if 5x — 6 = 4x + 10, then by transposing 6 and 4x, it is 5x — 4x — 10 + 6, or x = 1G. RULE II. When the unknown term is multiplied by any quantity ; divide all the terms of the equation by it. Thus, if ax=ab — 4a; then dividing by a, gives x=&— 4. And, if 3x + 5 = 20 ; then first transposing 5 gives 3x = 15 ; and then by dividing by 3, it is x = 5. In like manner, if ax+3a6=4c a ; then by dividing by a, it 4e* 4c 2 is x+36= — ; and then transposing 36, gives x = 36. RULE III. WiiEif the unknown term is divided by any quantity ; we must then multiply all the terms of the equation by that divisor ; which takes it away. Thus, if | = 3 + 2: then mult, by 4, gives x = 12 + 8=20. And, if- = 3b + 2c — d: a then mult, by a, it gives x = Sab + 2ac — ad. * Here it is earnestly recommended that the pupil be accustomed, at every line or step in the reduction of the equations, to name the par- ticular operation to be performed in the equation in the last line, in or- der to produce the next form or state of the equation, in applying each of these rules, according as the particular form of the equation may require ; applying them according to the order in which they are here placed : and beginning every line with the words Then by, as in the following specimens of Examples; which two words will always bring to his recollection, that he is to pronounce what particular operation he is to perform on the last line, in order to give the next ; allotting always a single line for each operation, and ranennq the equations neat- ly just under each other, in the several line*, as they are successively prodaced. SIMPLE EQUATIONS. 238 Then by transposing 3, it is * r And multiplying by 5, it is 3r Lastly, dividing by 3, gives x 10. 50. RULE IV. When the unknown quantity is included in any root or surd : transpose the rest of the terms, if there he any, by Rule 1 ; then raise each side to such n power as is denoted by the index of the surd ; viz. square each hide when it is the square root ; cube each side when it is the cube loot ; &c. which clears that radical. Thus, if y/x — 3=4; then transposing 3, gives y/x = 7 ; And squaring both sides gives x = 40. And, if y/(2r+ 10) = 8: Then by squaring it, it becomes 2r + 10 = 04 ; And by transposing 10, it is 2r = 54 ; Lastly, dividing bv 2. gives x 27. Also, if 3/(3r + 4/+ 3 = ; Then by transposing 3, it is y(3r + 4) = 3 ; And by cubing, it is 3x -f- 4 — 'SI : Also, by transp> ; :ig4, it h !>r = 23 ; Lastly, dividing by 3, gives x = 7J. When that side of the equation which contains the un- known o/jantity is a complete power, or can easily be reduced to one, by rule 1, 2, or 3 ; then extract the runt of the said power on both sides of the equation ; lhat is, extract the square root when it is a square power, or the cube root when it is a cube, &c. Thus, if x 2 + 8t + 1G = 30, or (x 4) a = 30 ; Then by extracting the root, it is x + 4 = ; And by transposing 4, it is x = — 4 = 2. And if 3x : — 10 = 21 + 35. Then, by transposing If), it is 3< 2 = 75 ; And dividing by 3, gives x 1 ~ 25 ; And extracting the root, gtv«:s i = 5. Also, if Jx 3 — fl = 24. Then transposing 0, jives J.r* = 30 ; KT.LK v. Vol. L 31 284 ALGEBRA. And multiplying by 4, gives 3s 3 = 120 ; Then dividing by 3, gives x* = 40 ; Lastly, extracting the root, gives x = ^40 = 6*824556* RULE VI. Whe.v there is any analogy or proportion, it is to be changed into an equation, by multiplying the two extreme terms together, and the two means together, and making the one product equal to the other. Thus, if 2x : 9 : : 3 : 5. Then mult, the extremes and means, gives 10* = 27 ; And dividing by 10, gives x = 2 T 7 T . And if Jx : a : : 56 : 2c. Then mult. c.\!rcmes and means, gives jcx = bob ; And multiplying by 2, gives 3cx = 10ad; t i r r i o l0ab Lastly, dividing by 3c, gives x = Also, if 10— x : lx : : 3 : 1. Then mult, extremes and means, gives 10 — x = 2x ; And transposing x, gives 10 =■= 3x ; Lustly, dividing by 3, gives 3£ = x. RULE VII. When the same quantity is found on both sides of an equa- tion, with the same sign, cither plus or minus, it may be left out of both : and when every term in an equation is either multiplied or divided by the same quantity, it may be struck out of them all. Thus, if 3x + 2a = 2a + b : Then, by taking away 2a, it is 3x = b. And, dividing by 8, it is x = ±b. Also, if there be 4ax + 6ab = 7ac. Then striking out or dividing by a, gives 4x + Gb = 7c. Then by transposing 66, it becomes 4x = 7c — 6b; And then dividing by 4, gives x = jc — $6. Again, if fx — f = V — f - Then, taking away the £, it becomes fx = f s ° ; And taking away the 3's, it is 2x = 10 ; Lastly, dividing by 2, gives x = 5. SIMPLE EQUATIONS. 286 MISCELLANEOUS EXAMPLES. 1. Given 7* — 18 = 4x + 6 ; to find the value of x. Firsts transposing 18 and 4x gives 3x = 24 ; Then dividing by 3, gives x = 8. 2. Given 20 — 4* — 12 = 02 — 10* ; to find x. First, transposing 20 and 12 and lOx, gives 6x = 84 ; Then dividing by 6, gives x = 14. 3. Let 4ax — 5b = Sdx + 2c be given : to find x. First, by trans. 56 and 9dx 9 it is 4ax — 3cfx = 55 + 2c : Then dividing by 4a — 3d, gives x = |~~~^» 4. Let 5x" — 12x = 9x + 2X 3 be given ; to find x. First, by dividing by x, it is 5x — 12 = 9 -f- 2x ; Then transposing 12 and 2x, gives 3x=21 ; Lastly, dividing by 3, gives x =■• 7. 5. Given 9ax 3 — 15abx* = Gar* + 12ax 3 ; to find x. First, dividing by Sax 3 , gives 3x — 56 = 2x + 4 ; Then transposing 56 and 2x, gives x=56 -fi. — XXX 6* Let - — - + ^ =: 2 be given, to find x. First, multiplying by 3, gives x — f r-f- 3 ! x - 6 ; Then, multiplying by 4, gives x + V*x =24. Also multiplying by 5, gives 17x = 120 : Lastly, dividing by 17, gives x = 7y T . ~ ^- x — 5 , x , _ x — 10 „ , 7. Given — — f- - = 12 — ; to find x. First, mult, by 3, gives x — 5 + £r=3G — x + 10 ; Then transposing 5 and x, gives *2» , 4-- , ;r=-51 ; And multiplying by 2, gives 7x = 102 ; Lastly, dividing by 7, gives x=l 1$. 3x 8. Let v^-j- + 7 = 10, be given ; to find x. First, transposing 7, give3 ^^=3 ; Then squaring the equation, gives J x —9 236 ALGEBRA* Then dividing by 3, gives \x = 3 ; Lastly, multiplying by 4, gives x = 12. 6a 3 9. Let 2x + 2v/(a 2 + x 3 ) = , be given, to find*. First, mult, by ^/(a 2 +x t ) 9 gives 2s v /(a 2 +x a )+2a , +2x> = 5a 3 . Thentransp. 2a 3 and 2X 3 , gives 2x v ^(a a +r , )=3a 1 — 2x»; Then by squaring, it is 4x* X (a* + x 3 ) = (3a 1 — 2*") 1 ; That is, 4aV -f- 4x 4 = 9a 4 — 12aV + 4x 4 ; By taking 4x 4 from both sides, it is 4a 3 x 3 =9a 4 — 12aV; Then transposing 12a a x 3 , gives 16a 3 x 3 =9a 4 ; Dividing by a 3 gives lCx 3 = 9a 3 ; And dividing by 16, gives x 2 =f 9 a 9 ; Lastly, extracting the root, gives x=Ja. EXAMPLES FOR PIIACTICK. 1. Given 2r — 5 + 16 = 21 ; to find x. Ans. x=5. 2. Given Ox — 15 = x + 6 ; to find x. Ans. *=4i. 5. Given 8— 3x r 12 = 30-5x+4 ; to find x. Ans. x=7. 4. Given x + Jx — Jx=rl3 ; to find x. Ans. x— 12. 5. Given 3x+ Jx+2=5x— 4; to find x. Ans, *=4. 6. Given 4ax+Ja — 2=ax — bx ; to find x. AnS ' X=: 9a+3S- 7. Given -}x — -f \x = J ; to find x. Ans. x — ff. 8. Given v /(4-fx)=4— v/x ; to find x. Ans. x = 2J. X" 9. Given 4a + x = ; to deter, x. Ans. x = — 2a. 4(7 -f- x 10. Given x /{la l + x 3 )= «/(4& 4 + x 4 ) ; to find x. Ans. x = \/— sT' 2a" 11. Given y/x + y/(2a+x) = — ~— , ; to find x. </{Za-\-x) Ans. x = |a. 12. Given —V + ----- = 2b ; to find x. l+2x 1 — 2x Ans. x=£ \/"^" # 13. Given a+x= v/(a 3 -fx ^/^P+a*)) ; to find x. Ans. x = — — <*• a SIMPLE EQUATIONS. 287 OF REDUCING DOUBLE, TRIPLE, &C. EQUATIONS, CONTAINING TWO, THREE, OR MORE UNKNOWN QUANTITIES. PROBLEM I. lb exterminate two Unknown Quantities; Or, to reduce the two Simple Equations containing them, to a Single one. RULE I. Find the value of one of the unknown letters, in terms of the other quantities, in each of the equations, by the methods already explained. Then put those two values equal to each other for a new equation, with only one unknown quan- tity in it, whose value is to be found as before. Note. It is evident that we must first begin to find the values of that letter which is easiest to be found in the two proposed equations. EXAXPLE8. 1. Given ^ 5^ 2y = 14 | ; t0 find x and y * In the 1st equat. trans. 3y and div. by 2, gives x = ^ ^ ; 14+2y In the 2d trans. 2y and div. by 5, gives x = — — £ ; o . 1. . 1 • 14 + 2 y l?-3y Putting these two values equal, gives — — ~ = — ^ i Then mult, by 2 and 5, gives 28 + Ay = 85 — 15y ; Transposing 28 and 15y, gives 1% = 57 ; And dividing by 19, gives y = 3. And hence x = 4. Or, effect the same by finding two values of y, thus 17-2x In the 1st equat. tr. 2x and div. by 3, gives y = — g — ; 5x— 14 In the 2d tr. 2y and 14, and div. by 2, gives y = — — ; 5a; -14 17— 2x Putting these two values equal, gives — — — = — g — \ 288 ALGEBRA. Mult, by 2 and by 3, gives 15x — 42 = 34 — 4x ; Transp. 42 and 4x 9 gives 19x = 76 ; Dividing by 19, gives x = 4. , Henoe y = 8, as before. 2. Given { jjlSfa* j 5 tofindxandy. Ans. x = a + b 9 and y = — Ji. 3. Given 3* + y = 22, and 3y + x = 18; tofindxandy. Ans. x =s 6, and y « 4. 4. Given + jy = 3J | ? to ^ n< * x y# Ans. x = 6, and y = 3. , 2a: . 3y 22 _ 3a? . 2y . 67 „ _ 5. Given T + ^= y and- + ^=-; to find. and y. Ans. a? =3, and y = 4. 6. Given x + 2y = s y and x* — 4y* = <P ; to find x and y. An»- * = — ^— , and jf = — ^— . 7. Given x — 2y = d> and x : y : : a : b ; to find x and y. ^•^^^^^ RULE II. Find the value of one of the unknown letters, in only one of the equations, as in the former rule; and substitute this value instead of that unknown quantity in the other equation, and there will arise a new equation, with only one unknown quantity, whose value is to be found as before. Note. It is evident that it is best to begin first with that letter whose value is easiest found in the given equations. EXAMF E8. 1. Given |*J + 1* | ; to find x and y. This will admit of four ways of solution ; thus ; First, 17 3 tf in the 1st eq. trans. 3y and div. by 2, gives x = — — - . This val. subs, for x in the 2d, gives 85 — 2y = 14; Hub. by 2, this becomes 86 — 15y — 4y = 28; snmx equations. 289 Transp. 15y and 4y and 28, gives 57 = 19y ; And dividing by 19, gives 3 = y. l*en, = ^ = 4. 14 + 2v Hly, in the 3d trans. 2y and div. by 5, gives x = — - — - ; * This subst. for * in the 1st, gives ^jtl? + 3y =s 17 ; Mult, by 5, gives 28 + 4y + 15y = 85 ; Transpo. 2*5, gives 19y = 57 ; And dividing by 19, gives y = 3. 14 + 2y Then * = — - s= 4, as before. o 8d1y, in the 1st trans. 2x and div. by 3, gives y = ? 34 4 X This subst for y in the 2d, gives Ex ^ = 14 ; Multiplying by 3, gives 15x — 34 + 4x = 42 ; Transposing 34, gives 19x = 76 ; And dividing by 19, gives x = 4. 17 2x Hence y = — = 3, as before. 3 5x — 14 4thly, in the 2d tr. 2y and 14 and div. by 2, gives y = — - — . 15x 42 This substituted in the 1st, gives 2x H - = 17 ; Multiplying by 2, gives 19x — 42 = 34 ; Transposing 42, gives 19x = 70 ; And dividing by 19, gives x = 4 ; 5 X 14 Hence y = — = 3, as before. 2. Given 2x + 3y = 29, and 3x — 2y = 11 : to find x and y. Ans. x = 7, and y = 5. ,3. Given | * * ^Z. l \ ^ ; to find x and y. Ans. x atv^^^^ MO ALGEBRA. 4. Given | ^^20 | ;*<> find x and y. Ans. x = 6, and y = 4. 5. Given g + 3y = 21, and | + 3x = 29 ; to find r and y. Ans. x = 9, and y = 6L 6. Given 10 - | = | + 4, and + f - 2 = ' 3y — x -~r 1 ; to find x and y. Ans. x = 8, and y =■ 6. 7. Given x : y : : 4 : 3, and x 3 — y 1 = 37 ; to find % and y. Ans. x = 4, and y = 8.^ RULE III. Lbt the given equations be so multiplied, or divided, &c. and by such numbers or quantities, as will make the terms which contain one of the unknown quantities the same in both equations ; if they are not the same when fir3t pro- posed. Then by adding or subtracting the equations, according as the signs may require, there will result a now equation, with only one unknown quantity, as before. That is, add the two equations when the signs are unlike, but subtract them when the signs are alike, to cancel that common term. Note. To make two unequal terms become equal, as above, multiply each term by the co-efficient of the other. EXAMPLES. r g,. __. 9 ^ Given i 2x + 5y = 16 \ ' t0 find x and y ' Here we may either make the two first terms, containing rr , equal, or the two 2d terms, containing y, equal. To mak.^ the two first terms equal, we must multiply the 1st equation by 2, and the 2d by 5 ; but to make the two 2d terms equaT, we must multiply the 1st equation by 5, and the 2d by 3 ; Ma follows. SIMPLE EQUATIONS. 241 1. By making the two first terms equal : Mult the 1st equ. by 2, gives lOx — 6y = 18 And mult, the 2d by 5, gives lOx + 25y = 80 Subtr. the upper from the under, gives Sly = 62 And dividing by 31, gives y = 2. Hence, from the 1st given equ. * = — -— ? = 3. 2. By making the two 2d terms equal : Mult, the 1st equat. by 5, gives 25x — 15y = 45 ; And mult, the 2d by 3, gives Ox + 15y = 48 ; Adding these two, gives 31x = 93 ; And dividing by 31, gives x = 3. c y 9 3 Hence, from the 1st equ. y = — - — = 2.' MISCELLANEOUS EXAMPLES. 1. Given + 6y = 21, and ^ + 5x = 23 ; to find x and y. Ans. x = 4, and y = 3. 2. Given + 10 = 13, and + 5= 12 ; to find x and y. Ans. a: = 5, and y = 3. 3 . Give „?^ + *=10,and^ + ? = 14 ; tofind 5 4 o o x and y. Ans. x = 8, and y = 4. 4. Given 3x+4y=38, and 4r — 3y ~ 9 ; to find x and y. Ans. x = 6, and y = 5. problem: hi. y To exterminate three or more Unknotcn Quantities ; Or, to reduce the simple Equations, containing them, to a Single one* RULE. This may be done by any of the three methods in the last problem : viz. 1. After the manner of the first rule in the last problem, find the value of one of the unknown letters in each of the given equations ; next put two of these values equal to each other, and then one of these and a third value equal, and so on for all the values of it ; which gives a new set of c^ftvcycA, Vol. 1. 32 243 AXGEBSJU with which the same process is to be repeated, and so on till there is only one equation, to be reduced by the rules for a single equation. 2. Or, as in the 2d rule of the same problem, find the value of one of the unknown quantities in oue of the equations only; then substitute this value instead of it in the other equations ; which gives a new set of equations to be resolved as before, by repeating the operation. 3. Or, as in the 3d rule, reduce the equations, by multi- plying or dividing them, so as to make some of the terms to agree : then, by adding or subtracting them, as the signs may require, one of the letters may be exterminated, dec. as before. EXAMPLES. Cx + y + z = 91 I. Given < x 4- 2y +3z = 16 > ; to find x 9 y, and z. (x + 3y +4* = 21) 1. By the 1st method : Transp. the terms containing y and z, in each cqua. gives x = 9 — y — 2, *= 16 — 2y — 3z, x = 21 — 3y — 4z; Then putting the 1st and 2d values equal, and the 2d and 3d values equal, give 9— y— z = 16— 2?/ — 32, 16 _2y —32 =21 - 3y — 4z ; In the 1st trans. 9, 2, and 2y, gives y = 7 — 2z ; In the 2d trans. 16, 3z, and 3y, gives y = 5 — z ; Putting these two equal, gives 5 — z = 7 — 22. Trans. 5 and 2z, gives 2 = 2. Hence y = 5 — 2 = 3, and x = 9 — y — 2 — 4- 2dly. By the 2d method : From the 1st equa. x =• 9 — y — 2 ; This value of x substit. in the 2d and 3d, gives 9 + y + 22 = 16, 9 + 2y + 32 = 21 ; lii the 1st trans. 9 and 22, gives y = 7 — 2z ; This substit. in the last, gives 23 —2 = 21; Trans, z and 21, gives 2 = 2. Hence again y = 7 — 22 = 3, and a: = 9 — y — * — SIMPLE EQUATIONS. 248 8d1y. By the 3d method : subtracting the 1st equ. from the 2d, and the 2d from the 3d, gives y + 2z = 7, y+ * == 5 ; Subtr. the latter from the former, gives z = 2. Hence y = 5 — z = 3, and x = 9 — y — : = 4. (*+ y+ * = 18) 2. Given ^x + 3y + 2* = 38V; + + i* = i<0 Ans. x = + 4y + i* = 27) + Jy + i* = 20 } ; to + *y + = 16 J to find x, y, and z. Ads. x = 4, y = 6, z = 8. > + 4y + i* = 27, 3. Given < * + |y + |z = 20 J. ; to find x, y, and z, Ans. x = 1, y = 12, z = 60. 4. Given * — y = 2, x — z = 3, and y + z = 9 • to find x, y, and z. Ans. x = 7, y = 5, z = 4. 5. Given < 3x + 4y + 5z = 46 ) ; to find x, y, and z. ( 2x + 6y + 8z = 68 ) ix(x + y + z)= 4jL^ 6. Given J y (x + y + z) = ; to find x, y, and z. + * + = 10?J J A COLLECTION OF QUESTIONS PRODUCING SIMPLE EQUATIONS. Quest. 1. To find two numbers, such, that their sum shall be 10, and their difference 6. Let x denote the.greater number, and y the less"" Then, by the 1st condition x + y = 10, And by the 2d - - . x — y = 6, Transp. y in each, gives x =■= 10 — y, and x = 6 + y ; Put these two values equal, gives 6 + y= 10 — y; Transpos. 6 and — y, gives - 2y = 4 ; Dividing by 2, gives - - y = 2. And hence - - • - x = 6 -f- y = 8. * la these solutions, as many unknown letters are always used as there are' unknown numbers to be found, purposely for exercise in the nodes of reducing the equations : avoiding the short ways of notation, which, though they may give neater solutions, afford less eiertV6«\u practising the several rules in reducing equations, 244 AL€SBBA. Quest. 2. Divide 100Z among a, b, c, so that ▲ may hate 20Z more than b, and b 10/ more than c. Let x = a's share, y = b's, and z = c's. Then x + y + % — 100, x = y + 20, y = z + 10. In the 1st suhstit. y + 20 for x, gives 2y + z + 20 = 100 ; ' In this substituting z + 10 for y, gives 3* + 40 = 100 ; By transposing 40, gives - - 3z = 60 ; And dividing by 3, gives - - z = 20. Hence y = z + 10 = 30, and x = y + 20 = 50. Quest. 3. A prize of 500Z is to be divided between two persons, so as their shares may be in proportion as 7 to 8 ; required the share of each. Put x and y for the two shares ; then by the question, 7 : 8 : : x : y, or mult, the extremes, and the means, 7y = 8x, and x + y = 500 ; Transposing y, gives x = 500 — y ; This substituted in the St, gives 7y = 4000 — Sy ; By transposing 8y, it is*f 5y = 4000 ; By dividing by 15, it gives y = 266f ; And hence x = 500^- y = 233£. Quest. 4. What fraction is that, to the numerator of which if 1 be added, the value will be \ ; but if 1 be added to the denominator, its value will be £ ? x Let — denote the fraction. y Then by the quest. * ^ 1 = 4, and — = 4. The 1st mult, by 2 and y, gives 2x + 2 = y ; The 2d mult, by 3 and y + 1, is 3x = y + 1 ; The upper taken from the under leaves x — 2 = 1; By transpos. 2, it gives x = 3. And hence y = 2x + 2 = 8 ; and the fraction is f . Quest. 5. A labourer engaged to serve for 30 days oflK these conditions : that for every day he worked, he was receive 20d, but for every day he played, or was absent, he» was to forfeit lOd. Now at the end of the time he had t» receive just 20 shillings, or 240 pence. It is required ts» SIMPLE EQUATIONS. 2*5 find how many days he worked, and how many he was idle? Let x he the days worked, and y the days idled. Then 20x is the pence earned, and lOy the forfeits ; Hence, by the question - x + y = 30, and 20x - lOy = 240 ; The 1st mult, by 10, gives lOx + lOy = 300 ; These two added, give - 30x =■ 540 ; This div. by 30, gives . x = 18, the days worked ; Hence - y = 30 — x = 12, the days idled. Quest. 6. Out of a cask of wine which had leaked away 30 gallons were drawn ; and then, being guaged, it appear- ed to be half full ; how much did it hold? Let it he supposed to have held x gallons, Then it would have leaked }x gallons, Conseq. there had been taken away }x + 30 gallons. Hence Jx = \x + 30 by the question. Then mult, by 4, gives 2x = x + 120 ; And transposing x, gives x = 120 the gallons it held. Quest. 7. To divide 20 into rVo such parts, that 3 times tlie one part added to 5 times the other may make 76. Let x and y denote the two parts. Then by the question - - x + y = 20, and 3x + by = 76. Mult, the 1st by 3, gives - 3x + 3y = 60 ; Subtr. the latter from the former, gives 2y = 16 ; And dividing by 2, gives - - y = 8. Hence, from the 1st, - x = 20 — y = 12. Quest. 8. A market woman bought in a certain number of eggs at 2 a penny, and as many more at 3 a penny, and •old them all out again at the rate of 5 for two-pence, and so doing, contrary to expectation, found she lost 3d ; what number of eggs had she ? Let x = number of eggs of each sort, Then will -£x = cost of the first sort, And \x = cost of the second sort ; But 5:2 : : 2x (the whole number of eggs) : |* ; Hence jx = price of both sorts, at 5 for 2 pence ; Then by the question |r + J* — f x = 3 ; JVfult. by 2, gives - x + \x — -fx = 6 ; And mult, by 3, gives 5x — *£x — IB; Also mult, by 5, gives x = 90, the number of tfjgt *t each sort. 246 ALGEBRA* Quest. 9. Two persons, a and b, engage at play. Be- fore they begin, a has 80 guineas, and b has 60. After a Certain number of games won and lost between them, a rises with three times as many guineas as b. Query, how many guineas did a win of b ? Let x denote the number of guineas a won. Then a rises with 80 + x> And n rises with 60 — x ; Theref. by the quest. 80 + x = 180 — Sx ; Transp. 80 and 3x, gives Ix = 100 ; And dividing by 1, gives x = 25, the guineas won. QUESTIONS FOB PRACTICE. 1. To determine two numbers such, that their difference may be 4, and the difference of their squares 64. Ans. 6 and 10. 2. To find two numbers with these conditions, viz. that half the first with a third part of the second may make 9, and that a 4lh part of the first with a 5th part of the second may make 5. Ans. 8 and 15. 3. To divide the number 20 into two such parts, that a 3d of the one part added to a 5th of the other, may make 6. Ans. 15 and 5. 4. To find three numbers such, that the sum of the 1st and 2d shall be ?, the sum of the 1st and 3d 8, and the sum of the 2tl and 3d 0. Ans. 3, 4, 5. 5. A father, dying, bequeathed his fortune, which was 2800/, to his son and daughter, in this manner ; that for eve- ry half crown the son might have, the daughter was to have a shilling. What then were their two shares ? Ans. The son 2000/ and the daughter 800L 6. Three persons, a, b, c, make a joint contribution, which in the whole amounts to 400/ : of which sum b contributes twice as much as a and 20/ more ; and c as much as a and b together. What sum did each contribute ? Ans. a 60/, b 140/, and c 2001. 7. A person paid a bill of 100/ with half guineas and crowns, using in all 202 pieces ; how many pieces were there of each sort ? Ans. 180 half guineas, and 22 crowns. SIMPLE EQUATIONS. 247 8. Says a to u, if you give me 10 guineas of your money, , I , Bhall then have twice ?.s much us you will have left : but says n to a, give me 10 of your guineas, and then I shall have 3 times as many as you. How many had each ? A us. a 22, n 2G. 9. A person goes to a tavern with a curtain quantity of money in his poc!:et, where ho spends 2 .shillings ; he then borrows as much money as h: j had left, and going to another tavern, he there spends '2 shillings also ; then borrowing again as much money as was left, he went to a third tavern, where likewise he spent 2 shillings ; and thus repeating tho same at a fourth tavern, he then had nothing remaining. What sum had he at first ! A us. 3*. Orf. 10. A man with his wife and child dine together at an inn. The landlord charged 1 shilling for the child ; and for the woman he charged as much as for the child and \ as much as for the man ; and for the man he charged as much as for the woman and child together. How much was that for each ? Ans. The woman 20c/ and the man 32</. 11. A cask, which held (50 gallons, was tilled with a mixture of brandy, wine, and cyder, in this manner, viz. the cyder was gallons more than tho brandy, and the wine was as much as the cyder and \ of the brandy. How much was there of each ? Ans. Brandy 15, cyder 21, wine 24. 12. A general, disposing his armv into a square form, finds that he has 2 SI men more than a perfect square : but increasing the side by 1 niau, he men wants 25 men to be a complete square. How iininv men had lie under his com- mand ? " Ans. '24000. 13. What number is that, to which if 3. 5, and 8, bo severally added, the three Minis shall be in geometrical pro- gression ? Ans. 1. 14. The stock of three traders amounted to 7*50/ : the shares of the first and second exceeded that of the third by 240 : and the sum of the 2d and 3d exceeded the first by 360. What was the share of each .' Ans. The 1st 200, the 2d 300, the 3d 200. 15. What two numbers are those, which, !;cing in the ratio of 3 tu 1, their product is equal to 12 times their sum 248 ALGEBRA. 16. A certain company at a tavern, when they came to settle their reckoning, found that had there been 4 more in company, they might have paid a shilling each less than they did ; but that if there had been 3 fewer in company, they must have paid a shilling each more than they did. What then was the number of persons in company, what each paid, and what was the whole reckoning ? Ans. 24 persons, each paid 7s, and the whole reckoning 8 guineas. 17. A jockey has two horses : and also two saddles, the one valued at 18/. the other at 3/. Now when he sets the - better saddle on the 1st horse, and the worse on the 2d, it makes the first horse worth double the 2d ; but when he places the better saddle on the 2d horse, and the worse on the first, it make* the 2d horse worth three times the 1st. What then were the values of the two horses ? Ans. The 1st 0/, and the 2d 9/. 18. What two numbers arc as 2 to 3, to each of which if 6 be added, the sums will be as 4 to 5 ? Ans. and O. 19. What arc those two numbers, of which the greater is to the less as their sum is to 20, and as their difference is to 10? Ans. 15 and 45. 20. What two numbers are those, uhosc difference, sum, and product, are to each other, as the three numbers 2, 3, 5 ? Ans. 2 and 10. 21. To find three numbers in arithmetical progression, of which the first is to the third as 5 to 0, and the sum of all three is 03. Ann. 15, 21, 27. 22. It is required to divide the number 2i into two such parts, that the quotient of the greater part divided by the less, may be to the quotient of the less part divided by the greater, as 4 to 1. Ans. 10 and 8. 23. A gentleman being asked the age of his two sons, answered, that if to the sum of their ages 18 be added, the result will be double the age of the elder ; but if be taken from the difference of their ages, the remainder will be equal to the age of the younger. What then were their ages ? Ans. 30 and 12. 24. To find four numbers such, that the sum of the 1st, 2d, and 3d shall be 13 ; the sum of the 1st, 2d, and 4th, 15 ; the sura of the 1st, 3d, and 4th, 18 ; and lastly, the sum of the 2d, 3d, and 4th, 20. Ans. 2, 4, 7, 9. 4/ / r*6 i^tk M 6x> y » r f& i** to ic~/r J tiro *-¥ 'ft* -11 i+t.s.i+tjfo ° isL^ri* -fa 2-* It* •V v «*r iiZ.C-?> r**y~i u-*y> v . 1 v --vr lor>fr|i T ■»-*-/f xaej. QUADRATIC EQUATIONS. 249 • 25. To divide 48 into 4 such parts, that the first increased by 3, the second diminished by 3, the third multiplied by 3, and the 4th divided by 3, may be all equal to each other. Ans. 6, 12, 3, 27. QUADRATIC EQUATIONS. Quadratic Equations are either simple or compound. A simple quadratic equation, is that which involves the square only of the unknown quantity. As ax 3 = b. The solution of such quadratics has been already given in simple equations. A compound quadratic equation, is that which contains the square of the unknown quantity in one term, and the first power in another term. As or 1 + bx = c. All compound quadratic equations, after being properly reduced, fall under the three following forms, to which they must always be reduced by preparing them for solution. 1. x 3 + ar = 6 2. x a — ax = b 3. x* — ax = —b The general method of solving quadratic equations, is by what is called completing the square, which is as follows : 1. Reduce the proposed equation to a proper simple form, as usual, such as the forms above ; namely, by transposing all the terms which contain the unknown quantity to one aide of the equation, and the known terms to the other ; placing the square term firs., and the single power scr md ; dividing the equation by the co-efficient of the square or first term, if it has one, and changing the signs of all the terms, when that term happens to be negative, as that term must always be made positive before the solution. Then the proper solution is by completing the square as follows, viz. 2. Complete the unknown side to a square, in this man. ner, viz. Take half the co-efficient of the socond term, «r»d square it ; which square add to both sides of the equation, then that side which contains the unknown quantity will be a complete square. Vol. L 33 250 ALGEBRA. 3. Then extract the square root on both sides of the equation*, and the value of the unknown quantity will be determined, making the root of the known side either + or — , which will give two roots of the equation, or two values of the unknown quantity. * As the square root of any quantity may be cither + or—; there- fore all quadratic equations admit of two solutions. Thus, the square root of -j- n * 's either -j- n or — n; for -j- n X + « and — n X — • are each equal to -|- n a . But the square root of — n 8 , or V — is imaginary or impossible, as neither -j- n nor — n, when squared, gives — i* So. in the first form, x*~±-az - ft, where x -j- fa is found = V(b + the root may be either -J- V(b + fa 2 ), or — V(b -j- fa 2 \ since either of them being multiplied by itself produces b -\- fa\ And this ambi- guity is expressed by writing the uncertain or double sign + before V(b 4- fa*) ; thus x = ± V(b + fa 2 ) — fa. In this form, where i - Hh V(b -\- fa 2 ) — fa, the first value of a:, via. x = -j- V(6 + d« v ) — 4» is always affirmative ; for since J« y -f b is greater than fa-, the greater square must necessarily have the greater root; therefore v(b fa 7 ), will always be greater than Vjir 3 , or its equal fa ; and consequently -j- V(b-\- fa*) — fa will always be affirm- ative. The second value, viz. x — — \/(b -)- fa 2 ) — fa will always \te nega- tive, because it is composed of two negative terms. Therefore when x 9 -}- ax = b, we shall have x — -j- \'{b -f- ja 3 ) — for the affirmative value of x % and x = — -f- Ja 2 ) — for the negative value of x. In the second form, where x = ± V(6 -f- -f- fa tne fi^t value, viz. i — + H" ~r ?l a is always affirmative, since it is composed of two affirmative terms. But the second value, viz. x = — V(6-f fa 7 ) •4- £a, will always be negative; for since b-\- fa 2 is greater than 4«* t therefore v(6 -{- fa") will be greater than vfa 2 , or its equal ^cr ; and consequently — V(n -|- fa' : ) \- fa is always a negative quantity. Therefore, when x 2 — ar — 0, we shall havez — -j- \ '(b -j- fa 2 ) fa for the affirmative value of x ; and x = — - v(& -)- fa'<) -j- j« for toe negative value of x; so that in both the first arid second forms, the un- known quantity has always two value>, one of which is positive, and the other negative. But, in the third form, whero .t - -± \ '(fa 2 — b) -' r fa, both the values of x will be positive, when fa 2 is greater than b. For the first value, viz. x =z + V(\a l — b) -f- fa will then be affirmative, being com- posed of two affirmative terms. The second value, viz. s - — V(fa- — b) -i- fa is affirmative also ; for since fa 2 is greater than fa 2 — b, therefore Vfa- or fa is greater than V(fa 2 — b) ; and consequently — v'(i«- — 6) -f- A<i will always be an affirmative quantity. So that, when x* — ax — — b, wc shall have z — + v(fa l — M 4a. and also x — \ {fa 2 — b) -f- fa, for the values of x, both positive. But in this third form, if b be greater than fa 1 , the solution of the pro- posed question will be i n possible. For bince the square of any quan- tity (whether that quantity be affirmative or negative) is always affirma- tive, the square root of a negative quantity is impossible, and cannot be assigned. But when b is greater than fa\ then fa* — b is a nega- tive quuntity ; and therefore its root V(fa J — b) is impossible, or ima- ginary ; consequently, in that case, z = fa + V(fa~ — b), or the two rood or values of x, are both impo^'ible^ or ioiaginary quantities. QUADRATIC KQUATIONS. 251 Note, 1. The root of the first side of the equation, is always equal to the root of the first term, with half the co- efficient of the second term joined to it, wi;ii its sign, whether + or — . 2. All equations, in which there are two terms including the unknown quantity, and which have the index of the one just double that of the other, are resolved like quadratics, by completing the square, as above. Thus, x* + ax 2 6, or x** + ax n = b, or x + ax* = 6, or (x 2 ± ax) 2 ±. m{x 2 :£ ax) = 6, are analsgous to quadra- tics, and the value of the unknown quantity may be deter, mined accordingly. 3. For the construction of Quadratics, see vol. ii. EXAMPLES. 1. Given x* + 4x = 60 ; to find x. First, by completing the square, x' 2 + 4r + 4 = 64 ; Then, by extracting the root, x + 2 = ±. 8 ; Then, transpos. 2, gives x = 6 or — 10, the two roots* 9 2. Given x a — 6x + 10 = 65 ; to find x. First, trans. 10, gives x 2 — 6x = 55 ; Then by complet. the sq. it is x 2 — 6x + 9 = 64 ; And by extr. the root, gives x — 3 = ± 8 ; Then trans. 3, gives x = 1 1 or — 5. 3. Given 3x 2 — 3* + 9 *= 8| ; to find x. First div. by 3, gives x 2 — x + 3 = 2 J ; Then transpos. 3, gives x 2 — x = — J ; And compl. the sq. gives x 2 — x + J = ^ ; Then extr. the root gives x — ^ = i ; And transp. J, gives x = | or 4. Given i* 2 — £r + 30J = 52£ ; to find x. First by transpos. 30J, it is \x 2 — \x = 22$ ; Then mult, by 2 gives x 2 — §x=44£ ; And by compl. the sq. it is x 2 — §x4-J=44J. Then extr. the root gives x — £ = tt 6} ; And trans. £, gives x = 7 or <— 6£ ; 5. Given ax 2 — bx = c ; to find x. First by div. by a, it is x 1 — - s = — ; 9B& AXGBBRA. Then compl. the sq. gives a*—- a: +^ f =- b 4ac+ b* And extrnc. the root, gives * — ^ = :£ — ^ — ; Then transp. gives * = ± + 6. Given a* — 20**= 6 ; to find a?. First by compl. the sq. gives a?— 2ox+o a ==a a 4-o ; And extract, the root, gives a*— a = ± ^/(a 2 — b) ; Then transpos. a, gives a* = ± ^/(a 1 + 6) + a ; And extract, the root, gives a? = + ^/[a + ^/(a'+i)]. SXAMFLB8 FOR PRACTICE*. J. Given a?— 62-7 = 33 ; to find 2. Ans. 2 = 10 or-4. * 1. Cubic equations when occurring in pairs, may usually be reduced to quadratic*, by extermination. Thus, Suppose 4z 4- 3x* -r- 6z = 150 > and 3z*-f 2x a + 2l = 106 5 Then molt. 1st equa. by 3, and 2d by 4, 1223 4- 9xa 4- 15z = 450 12x 3 4-8** + gz =420 By aobtr. z* 4- 7x = 30 Compl. the sq. z» 4- 7x + = 30 + ±f. = xfi. Extr. the root z + f = ± V z = £=3 or — 10. 2. Sometines, when the unknown square has a co-efficient, the fol- lowing method may be advantageously adopted : viz. Having transposed the known terms to one side and the unknown terms to the other, multiply each side by 4 times the co-efficient of the unknown square. Add the square of the co-efficient of the simple power of the un- known quantity, to both sides; the first side will then be a complete square. Extract the root, and the value of the unknown quantity will be ob- tained. Thus, if 5x a + 4z = 28. Then mult, by 4 X 5 t lOOx* 4- 80x = 660 Add 4 2 . . 100x2 4- 80x4-16* = 576 Eitr. the root, lOz -f 4 = ± 24 Transposing . lOz = 20 or — 28 Dividing by 10, x = 2, or — 2 8. The principal advantage of this method, which is due to the Indians, Is that it dots not introduce fractions into the operation. It will have the same advantage in cases where the square has 00 co-efficient, if that of the simple fwwer be an odd number. QUADRATIC EQUATIONS. 208 2. Given X 1 — 5x— 10 = 14 ; to find x. Ans. x = 8 or — 3. 3. Given 5x* + Ax — 90 = 114 ; to find x. Ans. a? = 6 or — ftf • 4. Given Jx 1 — $x + 2 =» 9 ; to find x. Ana. x = 4 or - 3J. 5. Given 3x* — 2x" = 40 ; to find a?. Ans. x = 2 or —2. 6. Given Jx— J^/x = 1J ; to find x. An*, x =9 or 2}. 7. Given ix* + fx = J ; to find x. * Ans. x = — | ± | */70 — -7277668 or —2-0611000. 8. Given x 8 + 4x* = 12 ; to find x. Ans. x = = 1-259921, or y — 6 = - 1-817121. 9. Given X 2 + 4x = a* + 2 ; to find x. Ans. x — v'(« 2 +6)^2. questions pboduoing quadratic equations. 1. To find two numbers whose difference is 2, and pro- duct 80. Let x and y denote the two required numbers. Then the first condition gives x — y = 2, And the second gives xy •= 80. The n transp. y in the 1st gives x = y + 2 ; This value of x suhstitut. in the 2d, is y 3 + 2y = 80 ; Then comp. the square gives y 3 + 2y + 1 = 81 ; And extrac. the root gives y + 1 = 9 ; And transpos. 1 gives y = 8 ; And therefore x = y + 2 = 10. 2. To divide the number 14 into two such parts, that their product may be 48. Let x and y donote the two parts. Then the 1st condition gives x + y *= 14, And the 2d gives xy = 48. Then transp. y in the first gives x =-* 14 — y ; This value subst. for x in the 2d, is 14y — j/ 5 = 48 ; Changing all the signs, to make the square positive, gives y 2 — 14y = — 48 ; Then com pi. the square gives y 3 — 14y + 49 = 1 ; And extrac. the root gives y — 7 = 4-1; Then transpos. 7, gives y = 8 or 6, the two parts. 3. What two numbers are those, whose sum, product, and difference of their squares, are all equal to each other ? Let x and y denote the two numbers. Then the 1st and 2d expression give x + y =. xy* And the 1st and 3d give x + y = a?— . 254 ALGEBRA. Then the last equa. div. by x + y, gives 1 = x — y ; And transpos. y, gives y + 1 = x ; This val. substit. in the 1st gives 2y + 1 = y' + y ; And transpos. 2y, gives 1 = y 2 — y ; Then complet. the sq. gives f = y* — y + j ; And extracting the root gives J ^5 = y — £ ; ■ And transposing \ gives i v/5 + J = y ; And therefcre x = y + 1 = J + And if these expressions be turned into numbers, by ex- tracting the root of 5, &c. they give x = 2-6180 +, and y= 1-6180 +. 4. There are four numbers in arithmetical progression, of wtth the product of the two extremes is 22, and that of the means 40 ; what are the numbers ? Let x = the less extreme, and y = the common difference ; Then r, x+y, x-f 2y, x-fc3y, will be the four numbers. Hence, by the 1st condition .t 2 + 3.ry = 22, And by the 2d x 2 + Sry + 2y a = 40. Then subtracting the first from the 2d gives 2y* = 18 ; And dividing by 2 gives y 1 = 9 ; And extracting the root gives y = 3. Then substit. 3 for y in the 1st, gives x 2 r|n 9x = 22 ; And completing the square gives x 3 "+ 9x + V = "f 9 ; Then extracting the root gives x -f- J- = y ; And transposing g- gives a: = 2 the least number. Hence the four numbers are 2, 5, 8, 11. 5. To find 3 numbers in geometrical progression, whose sum shall be 7, and the sum of their squares 21. Let x, y, and z denote the Ihree numbers sought. * Then by the 1st condition xz = y 2 , And by the 2d x + y + z = 7, And by the 3d x a + + z* = 21. Transposing y in the 2d gives x + z = 7 — y ; Sq. this equa. gives x 2 ■+ 2xx + 2 a = 49 — 14y + y"; Substi. 2y 2 for 2xs, gives x 3 + 2y= + z 2 = 49 — 14y f ; Subtr. y 2 from each side, leaves x 3 + y 2 + z 2 = 49 — 14y ; Pulling the two values of x 2 + y 3 + z 3 > 21 = 49 _ - m equal to each other, gives $ y * Then transposing 21 and 14y, gives 14 v = 28 ; And dividing by 14, gives y = 2. Then substit. 2 for // in the 1st equa. gives xz = 4, And in the 4th, it gives x + z = 5 ; Transposing z in the last, gives x = 5 — z ; This subst. in the next above, gives bz — z* = 4 ; QUADRATIC EQUATIONS. 255 Changing all the signs, gives r 1 — hz = — 4 ; Then completing the square, gives z 2 — 5z + V = { ? And extracting the root gives z — \ = ±: \\ Then transposing £, gives * and i = 4 and 1, the two other numbers* ; So that the three numbers are 1, 2, 1. QUESTIONS KOR FRACTICK. 1. What number is that which added to its square makes 42? Ans. C, or — 7. 2. To find two numbers such, that the less may be to \jfie greater as the greater is to 12, and that the sum of Jhf ir squares may be 45. j\ns. 3 anR>. 3. What two numbers are those, whose difference is 2, and the difference oi' their cubes 98? Ans. 3 and 5. 4. W T hat two numbers arc those, whose sum is 0, and the sum of their cubes 72 ? * Ans. 2 and 4. 5. What two numbcis are those, whose product is 20, and the difference of their cubes (51 Ans. 4 and 5. 6. To divide the number 11 into two such parts, that the product of their squares may be 784. Ans. 4 and 7. 7. To divide the number 5 into two such parts, that the sum of their alternate quotients may b«; that is of the two quotients of each part divided by the other. Ans. i r s nd 4. 8. To divide 12 into two such parts, Ihsit their product may be equal to 8 limes their dilfei'Micc. Ans. 1 and 8. 1). To divide the number 10 into two such parts, that the square of 4 times the less part, may be 112 more than the square of 2 times the greater. Ans. 4 and 6. 10. To find two numbers such, that the sum of their squares may be 80, and their sum multiplied by the greater may produce 104. J^ns. 5 and 8. 11. What number is that, which bcin<* divided by the product of its two dibits, the quotient is V ; ; but when 9 is subtracted from it, there remains a uurnher Laving the same digits inverted ? A ns. 32. 12. To divide 20 into three !>;• its suc : i. the continual product of all throe may b* 2"0, ;nnl th.it th« difference of the first and second may b^ 2 h <..* than thi* diflc re nee of the second and third. Ans. 5, 6, 9. 13. To find three numbers in arithmetical progression^ such that the sum of their squares may he wuixYifc tori 256 ALGEBRA. arising by adding together 3 times the first and 2 times the second and 3 times the third, may amount to 32. Ans. 2, 4, 6. 14. To divide the number 13 into three such parts, that their squares may have equal differences, and that the sum of those squares may be 75. Ans. 1, 5, 7. 15. To find three numbers having equal differences, so mat their sum may be 12, and the sum of their fourth powers 062. Ans. 3, 4, 5.. 16. To find three numbers having equal differences, and such that the square of the least added to the product of the two greater may make 28, but the square of the greatest adflsd to the product of the two less may make 44. W Ans. 2, 4, 0. 17. Three merchants, a, r, c, on comparing their gains find, that among them all they have gained 1444/ ; and that b's gained added to the square root of a's made 920/ ; but if added to the square root of c*s it made 912/. What were their several gains ? Ans. a 400, b 900, c 144. .18. To find three numbers in arithmetical progression, so that the sum of their squares shall be 93 ; also if il e first be multiplied by 3, the second by 4, and the third by" 5, the sum of the products may be 66." Ans. 2, 5, 8. 19. To find two numbers such, that their product added to their sum may make 47, and their sum taken from the sum of their squares mr.y leave 02. Ans. 5 and 7* RESOLUTION OF CUBIC AND HIGHER EQUATIONS. A Cubic Equation, or Equation of the 3d degree or power, is one that contains the third power, of the unknown quantity. As i 3 - ar- + bx = c. A Biquadratic, or Double Quadratic, is an equation that contains the 4th power of the unknown quantity : As x* — ax 3 -f bx J — cx = d. An Equation of the 5th Power or Decree, is one that contains the 5th power of the unknown quantity. As x 3 — ux A + W — cx 2 + clx =-= e. And so on, for all other higher powers, \fhere it is to be noted, however, that all the powers, or terms, in the =4 y=J" ne < — **- \ . «*- •••• CUBIC, X4UAT10X8. equation, are supposed to be freed from surds or fractional exponents. There are many particular and prolix rules usually given for the solution of some of the above-mentioned powers or equations. But they may be all readily solved by the following easy rule of Double Position, sometimes called Trial -and-E rror *. i RULE. 1. Find, by trial, two numbers, as near the true root as you can, and substitute them separately in the given equa- tion, instead of the unknown quantity ; and find how much the terms collected together, according to their signs 7- or — , differ from the absolute known term of the equation, marking whether these errors are in excess or defect. 2. Multiply the difference of the two numbers, found or taken by trial, by either of the errors, and divide the pro- duct by the difference of tho errors, when they are alike, but by their sum when they arc unlike. Or say, As the difference or sum of the errors, is to the difference of the two numbers, so is either error to the correction of its sup- posed number. 3. Add the quotiont, last found, to the number belonging to that error, when its supposed number is too little, but subtract it when too great, and the result will give tho true root nearly. 4. Take this root and the nearest of the two former, or any other that may be found nearer : and, by proceeding in like manner as above, a root will be had still nearer than before. And so on, to any degree of exactness required. Note 1. It is best to employ always two assumed num- bers that shall differ from each other only by unity in the last figure on the right hand ; because then the difference, or multiplier, is only 1. It is also best to use always the least error in the above operation. NoU 2. It will be convenient also to begin with a single * See, farther, that portion of vol. ii. which relates to equations, their construction, be. A new and ingenious general method of solving equations has been recently discovered by Messrs. //. Atkinson, Holared, and Horner, inde- pendently of each other. For the best pratical view of this new <Ki«\ta&. and its applications, consult the Elementary Treatise of Algebra, TAx, J. R, Young; a work which deserves our cordial reuimoiefedtttoft* Vol. L 34 258 -"ALGEBRA. figure at first, trying several single figures till thero be found the two nearest the truth, the one too little, and the other too great ; and in working witli them, find only one more figure. Then substitute this corrected result in the equation, for the unknown letter, and if the result prove too little, substitute also the number next greater for the second sup. position ; but contrarywise, if the former prove too great, then take the next less number for the second supposition ; and in working with the second pair of errors, continue the quotient only so far as to have the corrected number to four places of figures. Then repeat the same process again with this last corrected number, and the next greater or less, as the case may require, carrying the third corrected number to tight figures ; because each new operation commonly doubles the number of true figures. And thus proceed to any extent that may be wanted. EXAMPLES. Ex. 1. To find the root of the cubic equation ar 1 + jr + x= 100, or the value of x in it. Here it is soon found that x lies between i and 5. As- sume therefore these two num- bers, and the operation will be as follows : k 1st Sup. 2d Sup. 4 - x 5 Hi - X s - 25 61 - - 125 84 100 —10 sums - but should be - errors - +55 the sum of which is 71. Then as 71 : 1 : : 1G : -2 Hence x = 4 '2 nearly. Again, suppose 4*2 and 4-3, and repeat the work aa fol- lows : 155 100 1st Sup. 42 1704 74 088 05-928 100 x x 2 X 3 sums 2d Sup. 4-3 . 18-49 . 79-507 102-297 100 072 errors +2-297 the sum of which is &l ! As 0-309 :-l : : 2-297: 0-036 I This taken from - 4-300 loaves x nearly = 4-264 CUBIC, Ac. EQUATIONS. 250 Again, suppose 4*264, and 4*265, and work as follows: 4*264 X 4*265 18181696 X* 18*100225 77-526752 x 1 77*581310 99-972448 sums 100036535 100 100 -0027552 errors - +0 036535 tbe sum of which is -064087. Then as -064087 : -001 : : 027552 : 0004299 To this adding - 4*264 gives x very nearly = 4-2644209 The work of the example above might have been much shortened, by the use of the Table of Powers in the Arith- metic, which would have given two or three figures by in- spection. But the example has been worked out so particu- larly as it is, the better to show the method. Ex. 2. To find the root of the equation z 3 — 15x 2 + 63x = 50, or the value of x in it. Here it soon appears that x is very little above 1. Suppose therefore 1 *0and 1*1, and work as follows : 10 1*1 63 - 63x - 69*3 —15 — 15* 1 —1815 1 . X s - 1-331 49 50 ~ sums • 52*481 50 —l - errors +2*481 3*481 sum of the errors. As 8*481 : 1 : : *1 : -03correct. 1*00 Hence xa 1-03 nearly. Again, suppose the two num. bers 1*03 and 1-02, &c. as follows : 1*03 . i - 1-02 64*89 - 63* 64*26 —15*9135- IS* 3 — 15*6060 1-092727 x 3 1 -061208 50 069227 sums 49-715208 50 50 + •0(S9227er rors — -284792 •284792* As -354019 : -01 : : -069227: •0019555 This taken from 1 -03 leaves x nearly = 1*02804 960 Note 3. Every equation has as many roots as it contains dimensions, or as there are units in the index of its highest power. That is, a simple equation has only one value of the root ; but a quadratic equation has two values or roots, a cubic equation has three roots, a biquadratic equation has four roots, and so on. When one of the roots of an equation has been found by approximation, as above, the rest may be found as follows. Take, for a dividend, the given equation, with the known term transposed, with its sign changed, to the unknown side of the equation ; and, for a divisor, take x minus the root just found. Divide the said dividend by the divisor, and the quotient will be the equation depressed a degree lower than the given one. Find a root of this new equation by approximation, as be- fore, or otherwise, and it will be a second root of the origin- al equation. Then, by means of this root, depress the se- cond equation one degree lower, and from thence find a third root ; and so on, till the equation be reduced to a quadratic ; then the two roots of this being found, by the method of com- pleting the square, they will make up the remainder of the roots. Thus, in the foregoing equation, having found one root to be 1 '02804, connect it by minus with x for a divisor, and the equation for a dividend, &c. as follows : x — 1-02804 ) s 3 - 15r» + 63* — 50 ( x* - 13-97196* + 48-63627 = 0. Then the two roots of this quadratic equation, or - - - X s — 13-97196* = — 48-63627, by completing the square, are 6-57653 and 7*39543, which are also the other two roots of the given cubic equation. So that all the three roots of that equation, viz. r 1 - Ibx 2 + 63j? = 50, and the sum of all the roots is found to be 15, being equal to the co-efficient, of the 2d term of the equation, which the sum of the roots always ought to be, when they are right. Note 4. It is also a particular advantage of the foregoing rule, that it is not necessary to prepare the equation, as for other rules, by reducing it to the usual final form and state of equations. Because the rule may be applied at once to an unreduced equation, though it be ever so much embarrassed arc 1 02804 and 6-57653 and 7-39543 sum 15-00000 CUBIC. mvATiont. 9tl by mid and compound quantities. As in the following ex- ample : Ex. 3. Let k be required to find the root x of the equation y(144** — (jr» + 20)») + + 24) 1 ) » 114, or the value of x in it. By a few trials it is soon found that the value of x is but little above 7. Suppose therefore first that x is=7, and then x = 8. First, when x = 7, Second, when * = 8. 47-906 . -/[144* 1 — («» + 20)n - 46-476 65-384 . v/flOO* 1 — (x a + 24)«] 113- 290 114- 000 —0-710 +1759 - the sums of these - the true number • the two errors 115-750 114-000 +1-750 As 2-469 : 1 0-710 : 0-2 nearly. 70 Therefore x = 7-2 nearly. Suppose again x = 7-2, and then, because it turns out too great, suppose x also = 7-1, dec. as follows : Supp. x = 7-2, 47-990 . v /[144x a — (s 8 + 20) 2 ] 66-402 . ^[196**— (x* + 24) 2 ] 414*392 - the sums of these 114-000 - the true number +0-392 0123 the two errors Supp. * = 7-l, - 47-973 . 65-904 113- 877 114- 000 —0128 As -515 : 1 *123 : -024 the correction, 7-100 add Therefore x = 7*124 nearly the root required. Nate 5. The same rule also, among other more difficult forms of equations, succeeds very well in what are caiUd exponential ones, or those which have an *^a*a&« an ALGEBRA. ty in the exponent of the power ; as in the following ex- ample : Ex. 4. To find the value of x in the exponential (equation For more easily resolving such kinds of equations, it is convenient to take the logarithms of them, and then com- pute the terms by means of a table of logarithms. Thus, the logarithms of the two sides of the present equation are x X log. of x = 2, the log. of 100. Then, by a few trials, it is -soon perceived that the value of x is somewhere be- tween the two numbers 3 and 4, and indeed nearly in the middle between them, but rather nearer the latter than the former. Taking therefore first x = 3 5, and then =■ 3*6, and working with the logarithms, the operation will be as follows : First Supp. x = 3-5. . Log. of 3-5 = 0-544068 then 3-5 Xlog.3 5=1 -904238 Second Supp. x = 3*6. Log. of 3 6 = 0*556303 then 3-6 X log. 3 6=2-002689 the true number 2-000000 I the true number 2*000000 error, too little, — -095762 •002689 error, too great, + -002689 •098451 sum of the errors. Then. As -098451 : -1 : : -002689 : 00273 the correction taken from 3-60000 leaves - 3*59727 = x nearly. By repeating the operation with a larger table of loga- rithms, a nearer value of x may be found 3*597285. This method, indeed, may be a little improved in practice : for since x* = a, we have by logarithms x X log. x = log. a ; and again, log. x + log. log. x = log. log. a. We have therefore only to find a number, which, added to its log. will will be equal to the log. of the log. of the given number ; and the natural number answering to this number, is the va- lue of x required. In illustration of the above, take the 12th example : — «• = 123456789. First, log. 123456789 = 8 -0915143, and log. 8*0915148 = -9080298. Searching in a table of loga- liSaoB, we find the nearest number -93651 ; which added to CUBIC, AiCrn EQUATIONS. 263 its logarithm — 1-0715124 = -9080224. The next higher number '93652 + its log. = -9080371. Hence -9080371 -9080298 •9080224 -9080224 74 — 147 = -503 147 74 Therefore, the number sought is '93651503, the natural number answering to which is 8-640026 the value of x, which is true to the last figure, the value given by Dr. Hut* ton being 8-6400268. The common logarithmic solution fails when a is less than unity, its log. being then negative. In this case, assume x = 1 -f- y, and a =- 1 e, which transforms the given equa. x* = a f to & = y. Taking the logs, twice, we get y log. t = log. y, and log. y + log. of log. e = log. of log. y ; or, putting log. y = », and log. of log. e = *, we have v + * = log. v, an equation easy to solve. Ex. 5. To find the value of x in the equation x 3 + lOx 3 + 5* = 260. Ans. x = 4-1 179857. Ex. 6. To find the value of x in the equation x 3 — 2x=50. Ans. 3-8648854. Ex. 7. To find the value of x in the equation x 3 + 2x 2 — 23x = 70. Ans. x = 5-13457. Ex. 8. To find the value of x in the equation x 3 — 17x 2 + 54x = 350. Ans. x = 14-95407. Ex. 9. To find the value of x in the equation x 4 — 3x* — 75x = 10000. Ans. x = 10-2609. Ex. 10. To find the value of x in the equation 2x 4 — 16x 3 + 40X 3 — 30x = — 1. Ans. x = 1-284724. Ex. 11. To find the value of x in the equation X s + 2x 4 + Sx 3 + 4X 8 + 5x = 54321 . Ans. x = 8*414455. Ex. 12. To find the value of x in the equation x* = 123456789. Ans. x = 8 6400268. Ex. 13. Given 2x 4 — 7x 3 + llx 3 — 3x = 11, to find *. Ex. 14. To find the value of x in the equation. (3x* — 2y/x + 1)? — (x 2 — 4* v /x + 3 v 'x)* = 56. Ans. x = 18-360877. 90* To resolve Cubic Equations bp Cardan's Rule. Though the foregoing general method, by the application of Double Position, be the readiest way, in reul practice, of finding the roots in numbera of cubic equations, as well as of all the higher equations universally, we may here add the particular method commonly called Cardan's Rule, for re- solving cubic equations, in case any person should choose occasionally to employ that method ; although it is only ap- plicable when two of the roots are impossible. The form that a cubic equation ihust necessarily have, to be resolved by this rule, is this, viz. z 1 + az = 6, that is, wanting the second term, or the term of the 2d power z*. Therefore, after any cubic equation hns been reduced down to its final usual form, x 3 + px* + qx = r, freed from the co-efficient of its first term, it will then be necessary to take away the 2d term px 2 ; which is to be done in this manner ; Take Jp, or £ of the co-efficient of the second term, and annex it, with the contrary sign, to another unknown letter z, thus z — Jp ; then substitute this for x, the unknown letter in the original equation x 3 + px 2 + qx = r, and there will result this reduced equation z' J ±az b, of the form proper for applying the following, or Cardan's rule. Or take c = J a, and d = J6, by which the reduced equation takes this form, z 3 + 3cz = 2d. Then substitute the values of c and d in this form, % = V[d + + J)] + V[d - y/{* + O], > or , = „[* + + O] - w ^ w+ ^ y \ and the value of the root z, of the reduced equation z 3 + az = b, will be obtained. Lastly, take x = z — |p, which will give the value of r, the required root of the original equation x 3 + px 3 + qx = r, first proposed. One root of this equation being thus obtained, then de- pressing the original equation one degree lower, after the manner described, p. 260, the other two roots of that equa- tion will be obtained by means of the resulting quadratic equation. Note. When the co-efficient a, or c, is negative, and c 3 is greater than this is called the irreducible case, because then the solution cannot be generally obtained by this rule*. * Suppose a root to consist of the two parts z and y, so that (x + y) as jr; which sum substituted for z, in the given equation *3 + a* = cubic, 4ce» MVAirom. 985 Ex. To find the roots of the equations 3 — 6ac* +10* = 8. First to take away the 2d term, its co-efficient being — 6, its 3d part is — 2 ; put therefore x = * + 2 ; then = z 3 + 6s 9 + 12z + 8 - 6**= -6* 1 — 24* — 24 + 10* = + 10* + 20 theref. the sum z 3 # - 2* + 4 = 8 or z 3 * — 2* = 4 Here then a = — 2, 6 = 4, c = — |, d = 2. Theref. V[rf+^(W)]-l/[2+^(4-A)]«l/(2+^ W)^ and Vt^-^+c 3 )] ^/[2-v/(4-A)]-l/(2-^ W)» 3/(2- y v /3)=0-42265 ^ then the sum of these two is the value of * = 2. Hence x = * + 2 = 4, one root of * in the eq. a 3 — Ox* + lOx = 8. To find the two other roots, perform the division, Sec. as in p. 261, thus: x _ 4 ) x 3 — 6x a + lOx — 8 ( X s — 2x + 2 =s x 3 — 4x" — 2**+ lOx — 2r» + 8x 2r — 8 2x — 8 it becomes x 3 + y 3 + 3xy (x+ v) +a (x + y) = 6. Again, suppose 3sjf = — a ; which substituted, the last equation becomes x 3 -|- = 6. Now, from the square of this equation subtract four times the equation xy = — Ja, and there results x 8 — 2xy + 3^ = * 9 + aS* 3 * tbe square root of which is x 3 — y 3 = V (6* + A fli )- This being added to and taken from the equation x 3 + y 3 = 6, gives • C 2xs = 6 + V (6 a + A *) = 6 4- 2 V [(J6) 3 4- (J*) 3 !, i 2y* = 6 - V (6* — aS « ) = * — 2 V + (J*) 3 ] ; or { S! = 5S ± » V [S J ^ } • Hence ' di * d '"* * * — extracting the cube roots, we have x = l/d -f- v (d 2 -f c 1 ), and y =s %/d — V(d 2 + c 3 ) ; the sum of these two gives the first form of the root s above stated. And that the 2d form is equal to the first will be evident by reducing the two 2d quantities to the same denominator. When c is negative, and c 3 greater than o* f the root *pp*axiY*w& imaginary form. Vox. I. 35 906 of BiKPix umnunr. Hence x 3 — 2* = — 2, or ** — 2*+ 1 «= — 1, and *— 1 = rhv' — 1 = lor=l — — l,thetwo other roots sought. Ex. 2. Given X s — 6a* + 36x = 44, to find *. Ans. * = 2*32748. Ex. 3. To find the roots of x> — 7x* + 14x*= 20. Ans. x = 5, or = 1 + ^/ — 3, or = 1 — ^/ — 3. Ex. 4. Find the three roots of x 3 + 6x = 20. OF SIMPLE INTEREST. As the interest of any sum, for any time, is directly pro- portional to the principal sum, and to the time ; therefore the interest of 1 pound, for 1 year, being multiplied by any given principal sum, and by the time of its forbearance, in years and parts, will give its interest for that time. That is, if there be put r = the rate of interest of 1 pound per annum, p = any principal sum lent, t = the time it is lent for, and a = the amount or sum of principal and interest ; then is pfi = the interest of the sum p, for the time t, and conseq. p + prt or p X (1 + rt) = dy the amount for that time. From this expression, other theorems can easily be de- duced, for finding any of the quantities above mentioned : which theorems, collected together, will be as follows : 1st, a = p + prt the amount ; 2d, p = ■ J* ■ the principal ; 3d, r = the rate ; 4th, t = the time. pt pr For Example. Required to find in what time any princi- pal sum will double itself, at any rate of simple interest. In this case, we must use the first theorem, o =-p + prt, in which the amount a must be made = 2p, or double the principal, that is, p + prt = 2p, or prt = p, or rt = 1 ; and hence t = -. r Hence r being the interest of 11 for 1 year, it follows, that the doubling at simple intern^ ia to the quotient of COMPOUND INTEREST. 907' any sum divided by its interest for 1 year. • So, if the rate of interest be 5 per cent, then 100 -s- 5 = 20, is the time of doubling at that rate. Or the 4th theorem gives at once # a—p 2p—p 2-1 1 . ' r t = = — — — = = the same as before. pr pr r r COMPOUND INTEREST. Besides the quantities concerned in Simple Interest, namely, p = the principal sum, r = the rate of interest of 11 for 1 year, a = the whole amount of the principal and interest, t = the time* there is another quantity employed in Compound Interest, viz. the ratio of the rate of interest, which is the amount of II for 1 time of payment, and which here let be denoted by B, viz. K = 1 + r, the amount of 11 for I time. Then the particular amounts for the several times may be thus computed, viz. As II is to its amount for any time, so is any proposed principal sum, to its amount for the same time ; that is, as }l : R : : p : pa, the 1st year's amount, 1Z : R : : pn : pR a , the 2d year's amount, 11 : R : : pR 2 : J>R 3 , the 3d year's amount, and so on. Therefore, in general, pn 1 = a is the amount for the t year, or t time of payment. Whence the following genera! theorems are deduced : 1st, a = j>r« the amount ; 2d, p = ~ the principal ; Jd,K = y-theratio; 4tM = ^ ^4°---^^ ' v p loe. of r * - ■ OOXPOTCTD From which, any one of the quantities may bo found, when die rest are given. As to the whole interest, it is found by barely subtracting die principal/) from the amount a. Example. Suppose it be required to find, in how many years any principal sum will double itself, at any proposed rate of compound interest. In this case the 4th theorem must be employed, making m a* 2p ; and then it is I log* a~log.jp log, 2 p — log. p _ log. 2 leg. n log. n log. r" So, if the rate of interest be 5 per cent per annum ; then B = 1 + -06 -» 1*05 ; and henee log. 2 '301030 log. 1-05 * -021189 14-2067 nearly ; that is, any sum doubles itself in 14£ years nearly, at the rate of 5 per cent, per annum compound interest. Hence, and from the like question in simple interest, above given, are deduced the times in which any sum doubles itself, at several rates of interest, both simple and compound ; viz. At" 3 f 4 4* 5 6 7 8 9 10 J per cent. -per annum interest, 1/. or any other sum, will double itself in the following years. AtSimp.Int. At Comp.Int in 50 40 33i 28| 25 22J Kj 20 2 16| ? 14| 12* Hi I 10 in 35 0028 28-0701 23-4498 201488 17-6730 15-7473 h- 14-2067 1 11-8957? 10-2448 9 0065 8 0432 7-2725 The following Table will very much facilitate calculations of compound interest on any sum, for any number of years, at various rates of interest. cokpouhd iivtibxst. 968 The Amounts of 12 in any Number of Tears. Yrs. 3 3} 4 5 6 1 1*0300 1-0350 1-0400 1-0450 1-0500 1-0600 2 1-0609 10712 1-0816 1-0920 1-1025 11236 3 1-0927 11087 1-1249 11412 1-1576 11910 4 1-1255 1-1475 1-1699 1-1925 1-2155 1-2625 * 1-1593 1-1877 1-2167 1-2462 1-2763 1-3382 6 1-1948 1-2293 1-2653 1-3023 1-3401 1-4185 7 1-2299 1-2723 1-3159 1-3609 1-4071 1-5036 8 1-2668 1-3168 1-3686 1«4221 1-4775 1-5939 9 1-3048 1-3629 1-4233 1-4861 1-5513 1-6895 10 1-3439 1-4106 1-4802 1-5530 1-6289 1-7909 11 1-3842 1-4600 1-5895 1-6229 1-7103 1-8983 12 1-4258 1-5111 1-6010 1-6959 1.7959 2-0122 13 1-4685 1-5640 1-6651 1-7722 1-8856 21329 14 1-5126 1-6187 1-7317 1-8519 1-9799 2-2609 15 1-5580 1-6753 1-8009 1-9353 20789 2-3966 16 1-6047 1-7340 1-8730 2 0224 21829 2-5404 17 1-6528 1-7947 1-9479 21134 2-2920 2-6928 18 1-7024 1-8575 2 0258 2-2085 2-4066 2-8543 19 1-7535 1-9225 2-1068 2-3079 2-5270 30256 20 1-8061 1-9828 21911 2-4117 2-6533 3-2071 The use of this Table, which contains all the powers, r', to the 20th power, or the amounts of 1Z, is chiefly to calcu- late the interest, or the amount of any principal sum, for any time, not more than 20 years. For example, let it be required to find, to how much 5231 will amount in 15 years, at the rate of 5 per cent, per annum compound interest. In the table, on the line 15, and in the column 5 per cent is the amount of 1/, viz. . - 2 0789 this multiplied by the principal - 523 gives the amount - - 1087-2647 or .... 1087Z5*3J<*. and therefore the interest 564Z 5* 3\d. Note 1. When the rate of interest is to be determined to any other time than a year ; as suppose to £ a year, or J a year, dec. : the rules are still the same ; but then t will ex- press that time, and k must be taken the amount foe tihsX time also. 270 ANNUITIES. Nole 2, When the compound interest, or amount, of any sum, is required for the parts of a year ; it may be determin- ed in the following manner : 1st, For any time which is some aliquot part of a year : — Find the amount of 11 for 1 year, as before ; then that root of it which is denoted by the aliquot part, will be the amount of 11. This amount being multiplied by the principal sum, will produce the amount of the given sum as required. 2d, When the time is not an aliquot part of a year : — Reduce the time into days, and take the 365th root of the amount of 11 for 1 year, which will give the amount of the same for 1 day. Then raise this amount to that power whose index is equal to the number of days, and it will be the amount for that time. Which amount being multiplied by the principal sum, will produce the amount of that sum as before. — And in these calculations, the operation by loga- rithms will be very useful. OF ANNUITIES. Annuity is a term used for any periodical income, arising from money lent, or from houses, lands, salaries, pensions, &c. payable from time to time, but mostly by annual pay- ments. Annuities are divided into those that are in Possession, and those in Reversion : the former meaning such as have commenced ; and the latter such as will not begin till some particular event has happened, or till after some certain time has elapsed. When an annuity is forborn for some years, or the pay- ments not made for that time, the annuity is said to be in Arrears. An annuity may also be for a certain number of years ; or it may be without any limit, and then it is called a Per- petuity. The Amount of an annuity, forborn for any number of years, is the sum arising from the addition of all the annui- ties for that number of years, together with the interest due upon each after it becomes due. ANNUITIES. 271 The Present Worth or Value of an annuity, is the price or sum which ought to be given for it, supposing it to be bought off, or paid all at once. Let a = the annuity, pension, or yearly rent ; n = the number of years forborn, or lent for ; R = the amount of 11 for 1 year ; m =s the amount of the annuity ; v = its value, or its present worth. Now, 1 being the present value of the sum b, by propor- tion the present value of any other sum a, is thus found : as r : 1 : : a : ~ the present value of a due 1 year hence. In like manner ~ is the present value of a due 2 years hence ; for r : 1 : : - : So also ~, — , —, &c. will R R a R 3 R* R 4 be the present values of a, due at the end of 3, 4, 5, &c. years respectively. Consequently the sum of all these, or a i a i a i a ic ,1.1.1.1. X ^ a continued to n terms, will be the present value of all the n years' annuities. And the value of the perpetuity, is the sum of the series to infinity. But this series, it is evident, is a geometrical progression, having ~ but for its first term and common ratio, and the number of its terms n ; therefore the sum v of all the terms, or the present value of all the annual payments, will be „ = * * R " X a, or — J *—^-^- X ^ 1 R 1 R n R When the annuity is a perpetuity ; n being infinite, r* is also infinite, and therefore the quantity ~ becomes = 0, therefore — ^-j- X ~ also = ; consequently the expression becomes barely v = — — _ ; that is, any annuity divided by R "™ ■ ~ •* the interest of 11 for 1 year, gives the value of tto \*%T^etoai- ty. So, if the rate of Interest be 5 per cent. 372 ANNUITIES. Then 100a -f- 5 = 20a is the value of the perpetuity at 5 per cent : Also 100a -f- 4 = 25a is the value of the per- petuity at 4 per cent. : And 100a -r- 3 = 33£a is the value of the perpetuity at 3 per cent. : and so on. Again, because the amount of 11 in n years, is R n , its increase in that time will be R n — 1 ; but its interest for one single year, or the annuity answering to that increase, is r — 1 ; therefore, as r — 1 is to r a — 1, so is a to m ; that n n — 1 is, m = — X a. Hence, the several cases relating to r — 1 Annuities in Arrear, will be resolved by the following equations : m = v = n = Rn— 1 R— 1 X a = CR« ; R» — 1 X- = m R— 1 X R" ? ? R — 1 R n 1 X m = R— 1 R n — 1 log. m - -log. V 'log. a = — ^ X m = — — ^ x vr* ; 77iR — m + a log. R log. R Log & — * ^* m — v Rp R n ' R 1 In this last theorem, r denotes the present value of an annuity in reversion, after p years, or not commencing till after the first p years, being found by taking the difference between the two values ——4 X — and = — , for n R— 1 R n R— 1 RP years and p years. But the amount and present value of any annuity for any number of years, up to 21, will be most readily found by the two following tables. ANNUITIES. 273 TABLE I. The Amount of an Annuity of 1/ at Compound Interest. YrB. at 3 per c. 31 perc. 4 per c. 4 J perc. 5 per c. 6 per c. 1 10000 10000 1 0000 1 0000 10000 10000 2 20300 2 0350 20400 2 0450 20500 20600 3 3 0909 3*1062 3*1216 3- 1370 3 1525 31836 4 41836 4-2149 42465 42782 43101 43746 5 5 3091 53625 5 4163 5-4707 55256 56371 6 64684 65502 66330 67169 68019 69753 7 76625 77794 78983 80192 8 1420 83938 8 88923 90517 92142 9 3800 95491 98975 9 10 1591 103685 10-5828 10-8021 110266 11 4913 10 114639 117314 120061 122882 12 5779 131808 11 12-8078 131420 13 4864 138412 14-2068 149716 12 14 1920 146020 150258 15-4640 159171 16-8699 13 156178 16 1130 16 6268 171599 17-7130 18*8821 14 170863 17 6770 182919 189321 19 5986 21 0151 15 18-5989 19 2957 203236 20 7841 21 5786 232760 16 20 1569 209710 21 8245 22 7193 236575 25-6725 17 21 7616 22-7050 236975 24-7417i 25 8404 282129 IS 234144 244997 25 6454 268551 28 1324 309057 19 25 1169 26 3572 276712 290636 305390 3376C0 20 268704 282797 29*7781 31 3714 33 0660 367856 21 286765 302695 31 9692 33 7831 35 7193 399927 table ii. The Present Value of un Annuit}' of 1/. Yrs. at 3 perc. 3£ per c. 4 per c. 1 09709 09662 09615 2 1 9135 1-8997 1*8861 3 28286 2*8016 27751 4 37171 36731 36299 5 4-5797 4*5151 44518 6 5-4172 5*3286 52421 7 62303 61145 60020 8 70197 6*8740 77327 9 77861 7-6077 74353 10 8*5302 8*3166 8 1109 11 9-5256 90016 8-7605 12 9-9540 96633 93851 13 10-6350 10 3027 9-9857 14 11 2961 10*9205 105631 15 11 9379 11 5174 11 1181! 16 12 5611! 120941 11-65231 17 131661 126513 12 1657 18 13-7535 13 1S97 12-659?! 19 143238 13*7098 13 1339 20 148775 142124 135903 21 154150/ 14 69801 14 0292 4} por c. I 5 per c. 09569 1 8727 27490 3*5875 4- 3^C0 51579 5- 8927 65959 7-2688i 7*9127| 8*5289' 91 186! 9*6S29| 10 2228: 10-7396| 11 2340. 11 7072: 12 16001 1259331 13 0079 13 4047' 09524 1-8594 27233 35460 4 :)23b 50757 57864 64632 7* 1078 7 7217 83G54 88633 9-3S3S 989S6 10-3797 6 per c. 95. 4 2 (>7oC 34651 42124 49173 55824 62095- 68017 73601 78869 8-3838 88527 92950 971*3 10 8378: 10 1069 11-2741 11 6S96 120853 10-4773 10*8276 11 158V Vol. I. 36 274 ANNUITIES. To find the Amount of any Annuity forlorn a certain number of years. Take out the amount of 12 from the first table, for the proposed rate and time ; then multiply it by the given annuity ; and the product will be the amount, for the same number of years, and rate of interest. And the converse to find the rate of time. Exam. To find how much an annuity of 507 will amount to in 20 years, at 3£ per cent, compound interest. On the line of 20 years, and in the column of 3 J per cent, stands 28*2797, which is the amount of an annuity of 12 for the 20 years. Then 28-2797 X 50, gives 1413-9851 = 1413/ 19* 8d for the answer required. To find the Present Value of any Annuity for any number of years. — Proceed here by the 2d table, in the same manner us above for the 1st table, and the present worth required will be found. Exam. 1. To find the present value of an annuity of 502, which is to continue 20 years, at 3J per cent. — By the table, the present value of 11 for the given rate and time, is 14-2124; therefore 14-2124 X 50 = 710-022 or 7102 12* 4d is the present value required. Exam 2. To find the present value of an annuity of 202, to commence 10 years hence, and then to continue for 11 years longer, or to terminate 21 years hence, at 4 per cent, interest. — In such cases as this, we have to find the difference between the present values of two equal annuities, for the two given times ; which therefore will be done by subtracting the tabular value of the one period from that of the other, and then multiplying by the given annuity. Thus, tabular value for 21 years 14*0292 ditto for 10 years 8*1109 the difference 5-9183 multiplied by 20 gives - 118*3602 or - - 1 1 8Z 7* 3 £c2 the answer. END OF THE ALGEBRA. 275 GEOMETRY, DEFINITIONS. 1. A Point is that which has position, bat no magnitude, nor dimensions ; neither length, breadth, nor thickness. 2. A Line is length, without breadth or thickness. S. A Surface or Superficies, is an extension or a figure of two dimensions, length and breadth ; but without thickness. 4. A Body or Solid, is a figure of three di- mensions, namely, length, breadth, and depth, or thickness. 5. Lines are either Right, or Curved, or Mixed of these two. 6. A Right Line, or Straight Line, lies all iu the same direction, between its extremities ; and is the shortest distance between two points. When a Line is mentioned simply, it means a Right Line. 7. A Curve continually changes its direction between its extreme points. 8. Lines are either Parallel, Oblique, Per- pendicular, or Tangential. . 9. Parallel Lines are always at the same perpendicular distance ; and they never meet, though ever so far produced. 10. Oblique lines change their distance, and would meet, if produced on the side of the least distance. 11. One line is Perpendicular to another, when it inclines not more on the one side 276 GEOMETRY. than the other, or when the angles on both sides of it are equal. 12. A line or circle is Tangential, or is a Tangent to a circle, or other curve, when it touches it, without cutting, when bath are produced. 13. An Angle is the inclination or open- ing of two lines, having different directions, and meeting in a point. 14. Angles are Right or Oblique, Acute or Obtuse. 15. A Right Angle is that which is made by one line perpendicular to another. Or when the angles on each side are equal to one another, they are right angles. 16. An Oblique Angle is that which is made by two oblique lines ; and is either less or greater than a right angle. 17. An Acute Angle is less than a right angle. 18. An Obtuse Anglo is greater than a right angle. 19. Superfices are either Plane or Curved. 20. A Plane Superficies, or a Plane, is that with which a right line may, every way, coincide. Or, if the line touch the plane in two points, it will touch it in every point. But, if not, it is curved. 21. Plane Figures are bounded either by right lines or curves. 22. Plane figures that are bounded by right lines have names according to the number of their sides, or of their angles ; for they have as many sides as angles ; the least number being three. 83. A figure of three sides and angles is called a Triangle. And it receives particular denominations from the relations of its sides and angles. 24. An Equilateral Triangle is that whose three sides are all equal. ' 25. An Isosceles Triangle is that which has two sides equal. DEFINITIONS. 277 26. A Scalene Triangle is that whose three sides are all unequal. 27. A Right-angled Triangle is that which has one right angle. ^ 28. Other triangles are Oblique-angled, and are either obtuse or acute. 20. An Obtuse-angled Triangle has one ob- tuse angle. 30. An Acute-angled Triangle has all its three angles acute. 31. A figure of Four sides and angles is call- ed a Quadrangfe, or a Quadrilateral. 32. A Parallelogram is a quadrilateral which has both its pairs of opposite sides parallel. And it takes the following particular names, viz. Rectangle, Square, Rhombus, Rhomboid. 33. A Rectangle is a parallelogram, having a right angle. 34. A Square is an equilateral rectangle ; having its length and breadth equal. 35. A Rhomboid is an oblique-angled paral- lelogram. 36. A Rhombus is an equilateral rhomboid ; having all its sides equal, but its angles ob- lique. 37. A Trapezium is a quadrilateral which hath not its opposite sides parallel. 88. A Trapezoid has only one pair of oppo- site sides parallel. 39. A Diagonal is a line joining any two op* poaite angles of a quadrilateral. □ /J 40. Plane figures that have more than four sides are, in general, called Polygons : and they receive other particular names, according to the number of their sides or angles. Thus, 41. A Pentagon is a polygon of five sides ; a Hexagon, of six sides; a Heptagon, seven; an Octagon, eight; %ttocu agon, nine ; a Decagon, ten ; an Undecagon, tawa \ vodeoagon, twelve sides. 278 GEOMETRY. 42. A Regular Polygon has all its sides and all its angles equal. — If they are not both equal, the polygon is Irregular. 43. An Equilateral Triangle is also a Regular Figure of three sides, and the Square is one of four : the former being also called a Trigon, and the latter ^tetragon. 44. Any figure is equilateral, when all its sides are equal : and it is equiangular when all its angles arc equal. When both these arc equal, it is a regular figure. 45. A Circle is a plane figure bounded by a curve line, called the Circumference, which is every whore equidistant from a certain point within, called its Centre. The circumference itself is often called a circle, and also the Periphery. 46. The Radius of a circle is a line drawn from the centre to the circumference. 47. The Diameter of a cirle is a line drawn through the centre, and terminating at the cireumfcrencc on both sides. 48. An Arc of a circle is any part of the circumference. 49. A Chord is a right line joining the ex. tremitics of an arc. 50. A Segment is any part of a circle bounded by an arc and its chord. 51. A Semicircle is half the circle, or a segment cut off by a diameter. The half circumference is sometimes called the Semicircle. 52. A Sector is any part of a circle which is bounded by an arc, and two radii drawn to its extremities. • 53. A Quadrant, or Quarter of a circle is a sector having a quarter of the circumference for its arc, and its two radii are perpendicular to each other. A quarter of the circumference is sometimes called a Quadrant. DEFINITIONS. 279 B A d 54. Tho Height or Altitude of a figure is a perpendicular let fall from an angle, or its vertex, to the opposite side, called the base. 55. In a right-angled triangle, the side op- posite the right angle ailed the Hypothc- nuse ; and the other two sides are called the Legs, and sometimes the Base and Perpen- dicular. 56. When an angle is denoted by three letters, of which one stands at the angular point, and the other two on the two sides, that which stands at the angular point is read in the middle. 57. The circumference of every circle is supposed to be divided into 360 equal parts called degrees ; and each degree into 60 Mi- nutes, each Minute into 60 Seconds, and so on. Hence a semicircle contains 160 degrees, and a quadrant 90 degrees. 58. The Measure of an angle, is an arc of any circle contained between the two lines which form that angle, the angular point being the centre ; and it is estimated by the number of degrees contained in that arc. 59. Lines, or chords, are said lo be Equi- distant from the centre of a circle, when per- pendiculars drawn Lo them from the centre are equal. 69. And the right line on which the Great- er Perpendicular falls, is said to be farther from the centre. 61. An Angle In a Segment is that which is contained by two lines, drawn from any point in the arc of the segment, to the two extremities of that arc. 62. An Angle On a segment, or an are, is that which is contained by two lines, drawn from anv point in the opposite or supplemental part of the circumference, to tho extremities of the arc, and containing the arc between (hem. 63. An Angle at the circumference, is that S^/V^ whose angular point or summit is any where ( f\ in the circumference. And an angle at the ' / ^ v centre, is that whose angular point is at the centre. 280 GEOMETRY. 64. A right-lined figure is Inscribed in a circle, or the circle Circumscribes it, when all the angular points of the figure are in the circumference of the circle. 65. A right-lined figure Circumscribes a circle, or the circle is Inscribed in it, when all the sides of the figure touch the circumference of the circle. 66* One right-lined figure is Inscribed in another, or the latter circumscribes the former, when all the angular points of the former are placed in the sides of the latter. 67. A Secant is a line that cuts a circle, lying partly within, and partly without it. 66. Two triangles, or other right-lined figures, are said to be mutually equilateral, when all the sides of the one are equal to the corresponding sides of the other, each to each ; and they are said to be mutually equiangular, when the angles of the one are respectively equal to those of the other. 68. Identical figures are such as are both mutually equi- lateral and equiangular ; or that have all the sides and all the angles of tho one, respectively equal to all the sides and all the angles of the other, each to each ; so that if the one figure were applied to, or laid upon the other, all the sides of the one would exactly fall upon and cover all the sides of the other ; the two becoming as it were but one and the same figure. 70. Similar figures, are those that have all the angles of the one equal to all the angles of the other, each to each, and the sides about the equal angles proportional. 71. The Perimeter of a figure, is the sum of all its sides taken together. 72. A Proposition, is something which is either proposed to be done, or to be demonstrated, and is either a problem or a theorem. 73. A Problem, is something proposed to be done. 74. A Theorem, is something proposed to be demonstrated. 75. A Lemma, is something which is premised, or demon- trated, in order to render what follows more easy. 76. A Corollary, is a consequent truth, gained immediate- ly from some preceding truth, or demonstration. 77. A Scholium, is a remark or observation made upon something going before it. 281 , AXIOMS. 1. Things which are equal to the aame thing arc equal to each other. 2. When equals are added to equals, the wholes are equal. 3. When equals are taken from equals, the remainders are equal. 4. When equals are added to uncquals, the wholes are un- equal. 5. When equals are taken from unequal*, the remainders are unequal, 6. Things which are double of the same thing, or equal things, are equal to each other. ^ 7. Things which are halves of the same thing, are equal. . .8. Every whole is equal to all its parts taken together. 0. Things which coincide, or fill the same space, are iden- tical, or mutually equal in all their parts. 20. All right angles are equal to ono another. 21. Angles that have equal measures, or arcs, are equal. TIIEOREX T. in all If two triangles have two sides and the included angle in the one, equal to two sides and the included angle the other, the triangles will be identical, or equal in respects. In the two triangles jlbc, def, if the side ac be equal to the side dp, and the side bc equal to the side ef, and the angle c equal to the angle f ; then will the two triangles be identical, or equal in all respects. ^ For conceive the triangle abc to be applied to, ot ^Yas&& on, the triangle def, in such a manner that the pov&l c m*$ Vol 1 37 \ 482 •EOXETftT. coincide with the point r, and the side ac with the side sv 9 which is equal to it. Then, since the angle f is oqunl to the angle c (by hyp.), the side bc w«ill fall on the side r.r. Also, because ac is equal to or, and nc equal to kf (by hypO, the point a will co»»cido witn the point i>, and ihe jx inr a with the point k ; consequently the side An will coincide with the side dk. Therefore the two triangles are ilt-nlical, and have all their other corresponding parts equal (ax. ?)), namely, the side ab equal to the side de, the angle a to the angle d, and the angle b to the angle e. q. e. d. THEOREM II. When two triangles have two angles and the included side in the one, equal to two angles and the included side in the other, the triangles are identical, or have their other sides and angle equal. Let the two triangles abc, def, q -g have the angle^i equal to the anglo D, the angle b equal to t L - 1 — and the side ab equal to 1 then these two triangles 1 tical. For, conceive the triangle abc to be placed on the triangle def, in such manner that the side ab muy lull exactly on the equal side dk. Then, since the angle a is equal to the angle i> (by hyp.) } the side ac must fall on the side df ; and, in like manner, because the angle h is equal to the angle >:, the side bc must fall on the side kf. Thus the three sides of the triangle a»c will be exactly placed on the, three sides of the triangle dkf : consequently the two triangles are identical (ax. 9), having the other two sides ac, bu, equal to the two df, rf, and the remaining angle c equal to the remaining angle f. q. e. d. theorem III. In an isosceles triangle, the angles at the base are equal. . Or, if a triangle have two sides equal, their opposite angles will also bc equal. If the triangle abc have the side ac equal to the side bc : then will the angle b be equal to the angle a. For, conceive the angle c to be bisected, or divided into two equal parts, by the line cd, making the angle acd equal to the angle — jj- BCD. I to the anglo A the angle e, / I > the side df. ; / I \ will be iden- * t, THEOREM. Then, the two triangles, acd, bci>, have two sides and the contained angle of tho one, equal to two sides and the contained angle of tho other, viz. the side ac equal to Br, the angle acd equal to bcd, and the side cd common ; there- fore these two triangles arc identical, or equal in all respocts (th. 1) ; and consequently tho angle a equal to the angle b. q. E. 1). Carol. 1. Hence the lino which hisects the vertical angle of an isosceles triangle, bisects the base, and is also perpen- dicular to it. Carol . 2. Hence too it appears, that every equilateral tri- angle, is also equiangular, or has all its angles equal. THEOREM IT, When a triangle has two of its angles equal, the sides opposite to them are also equal. If the triangle abc, have the angle cab equal to the angle cba, it will also have the side ca equal to the side ch. For, if ca and cb he not equal, let ca be the greater of tho two, and let da be equal t3 en, and join db. Then, because da, ab, arc equal to cb, ba, each to each, and the angle dab to cba (hyp.), the triangles dab, cba, are equal in all respects (th. 1), a part to the whole, which is absurJ ; therefore ca is not greater than cb. fn the same way it may be proved, that cb is not greater than ca. They are therefore equal. f. d. Cord. Hence every equiangular triangle is also cquu lateral. THEOREM V. Whex two triangles have all the three sides in the one, equal to all the three sides in the other, the triangles are identical, or have also their three triangles equal, each to each* Let the two triangles abc, abd, have their three sides respectively, equal, viz. the side ab equal to ab, ac to ad, and nc to bd ; then shall the two triangles be identical, or have their angles equal, viz. those anglos •K0ML1KV. that are opposite to the equal sides ; g namely, the angle bac to the angle bad, the angle abc to the angle abd, and the angle c to the angle d. A For, conceive the two triangles to be joined together by their longest j) equal sides, and draw the line cd. Then, in the triangle acd, because the side ac is equal to ad (by hyp.), the angle acd is equal to the angle adc (th. 3). In like manner, in the triangle rcd, the angle BCD is equal to the angle bdc, because the side bc is equal to bd. Hence then, the angle acd being equal to the angle adc, and the angle bcd to the angle bdc, by equal additions the sum of the two angles acd, bcd, is equal to the sum of the two adc, bdc, (ax. 2), that is, the whole angle acb equal to the whole angle adb. Since then, the two sides ac, cb, are equal to the two sides ad, db, each to each, (by hyp.), and their contained angles acb, adb, also equal, the two triangles abc, abd, are identical (th. 1), and have the other angles equal, viz. the angle bac to the angle bad, and the angle abc to the angle akd. u. l. d. TilKOKJ m VI. Win:* one line meets another, the angles which it makes on the same side of the otli^r, are together equal to two right angles. Let the line ah meet the line cd : then will the two angles abc, abd, taken to- gether, ho equal to two right angles. For, first, when the two angles abc, abd, are equal to each other. tlic\ are both of them right angles (def. 15.) But when the angles are unequal, suppose bb drawn per- pendicular to cd. Then, since the two angles ebc, fbd, are right angles (def. 15), and the an^le ebd is equal to the two angles eba, add, together (ax. 8), the three angles, ebc, kba, and abd, arc equal to two right angles. But the two angles ebc, eba, arc together equal to the angle abc (ax. 8). Consequently the two angles abc, abd, are also equal to two right angles, q. e. d. Corol. 1. Hence also, conversely, if the two angles abc, abd, on both sides of the line ab, make up together two right angles, then cb and bd form one continued right line cd. THEOREMS. 385 Corol. 2. Hence, alt the angles which can be made, at any point b, by any number of linos, on the same side of the right line cd, are, when taken all together, equal to two right angles. Cord. 3. And, ns all the angles that can be made on the other side of the line cn are also equal to two right angles ; therefore all the angles that can be made quite round a point b, by any number of lines, are equal to four right angles. Coral. 4. Hence also the whole circumfer- ence of a circle, being the sum of the mea- sures of all the angles that can be made about the centre f (def. 57), is the measure of four right angles. Consequently, a semicircle, or 180 degrees, is the measure of two right an- gles ; and a quadrant, or 90 degrees, the measure of one right angle. THEOREM VII. When two lines intersect each other, the opposite angles are equal. Let the two lines ad, vu, intersect in the point e ; then will the angle aec be /C equal to the angle bed, and the angle > i aed equal to the angle ceb. a } /TQ For, since the lino cc meets the line / ab, the two angles aec, bec, taken to- D gethcr, arc equal to two ricjht angles (lli. 6). In like manner, the line he, meeting the line cd, makes the two angles dkc, red, equal to two right angles. Therefore the sum of the two angles aec, bec, is equal to the sum of the two bec, ki d (ax. 1 ;■. And if the angle nr.c, which is common, be taken away from both these, the remaining angle aec will be equal to the remaining angle bed (ax. 3). And in like manner it may be shown, that the angle aid is equal to the opposite angle bec. theorem viii. When one side of a triangle is produced, the outward angle is greater than either of the two vkw&x& c^c*\\» angles. 980 ozomnr. Let abo be a triangle, having the aide ab produced to i> ; then will the outward angle cno be greater than cither of the inward opposite angles a or c. For, conceive the side bc to be bi- sected in the point e, and draw the line ak, producing it till ef bc equal to ak ; and join»BF. Then, since the two triangles aec, hf.f, have the side ae — the side kf, and the side ce = the side be (by suppos.) and the included or opposite angles at e also equal (th. 7}, therefore those two triangles arc equal in all respects (th. 1), and have the angle c = the corresponding angle ebf. But the angle crd is greater than the angle ebf ; consequently the said outward angle cbd is afso greater than the angle <:. In like manner, if cb bo produced to g, and ab b? 1 i- sected, it may bc shown that the outward angle abc, or it* equal cbd, is greater than the other angle a. THEOREM IX. TnE greater side, of cvrry triangle, is oiprs'tc to the greater ungle ; and the greater angle opposite to the greater aide. Lot abc be a triangle, having the side q ab greater than the side ac ; then will the angle acb, opposite the greater side ab, be greater than the angle b, opposite the less aide ac. For, on the greater side ab, take the part ad equal to the less side ac, and join cd. Then, since bcd is a triangle, the outward angle Anc if. greater than the inward opposite angle b (th. 8). But the angle acd is equal to the said outward angle abc, because ad is eqal to ac (th. 3). Consequently the angle acd also is greater than the angle b. And since the angle acd is only a part of acb, much more must the whole angle acb be greater than the angle n. a. k. d. Again, conversely, if the angle c he greater than the angle b, then will the side ab, opposite the former, be greater than the side ac, opposite the latter. For, if ab bc not greater than ac, it must bo cither eguai to it, or leas than k. But it cannot be equal, for TRSOBEKB- then the angle c would be equal to the angle b (th- 3), which it is not, by the supposition. Neither can it bo less, for then the angle c would be less than the angle n, by the former part of this ; which is also contrary to the supposition. Tho aide ab, then, being neither equal to ac, nor less than it, must necessarily be greater, a. u. d. tiieobem x. The sum of any two sidtes of a triangle is greater than the third side. Let abc be a triangle ; then will the sum of any two of its sides be greater than the third side, as for instance, ac + cb greater than ab. For, produce ac till cn be equal to CB, or ad equal to the sum of the two AC + cb; and join bi> Then, because CD is equal to en (by constr.), the angle i> is equal to the angle cbd(iIi. 3). But the angle ahd is greater than the angle cbd, consequently it must also be greater than the angle in And, since the greater side of any triangle is op. posite to the greater nngle (th. 9), the side ai> (of the tri- angle ahd) is greater tbun the side au. But ad is equal to ac and cd, or ac and cb, taken together (by constr.) ; there- fore ac + cb is also greater than ab. q. e. d. Carol. The shortest distance between two points, is a single right line drawn from the one point to the other. THE0BE3I XI. The difference of any two sides of a triangle, is less than the third side. Let abc be a triangle ; then will the D difference of any two sides, as ab — ac, be less than the third side sc. For, produce the less side ac to d, till ad be equal to the greater side ab, . so that <"D may be the difference of tl.o -A. B two sides ah — ac ; and join bd. Then, because ad is equal to ab (by constr.), the opposite angels D and abd are equal (th. 3). J kit the angle cbd is less than the angle abd, and consequently also less than the eaual angle d. And since the greater side of any \xv»&$& OEOatCTKT. opposite to the greater angle (th. 9), the side cd (of the tri- anglo bcd) is less than the side bc. a. e. d. Otherwise* Set off upon ah a distance ai equal to ao. Then (th. |0) ac + cb is greater than ab, that is, greater than ai + ib. From these, take away the equal parts AC, Ai, respectively ; and there remains cb greater than ic. Con* sequently, ic is less than cb. a. e>- d. THEOREM XII. When a lino intersects two parallel lines, it makes the alternate angles equal to each other. Let the line ef cut the two parallel line ab, cd ; then will the angle aef be equal to the alternate angle efd. For if they are not equal, one of them must be greater than the other; let it be efd for instance, which is the greater, if possible ; and conceive the line fh to bc drawn, cutting off the part or angle efb equal to the angle AEF, and meeting the line ab in the point n. Then, since the outward angle aef, of the triangle bef, is greater than the inward opposite angle efb (th. 8) ; and since these two angles also are equal (by the constr.) it fol- lows, that those angles are both equal and unequal at the same time : which is impossible. Therefore the angle efd is not unequal to the alternate angle aef, that is, they are equal to each other, q. e. d. Corol. Right lines which are perpendicular to one, of two parallel lines, are also perpendicular to the other. THEOREM XIII. When a lino, cutting two othrr lines, makes the alter- nate angles equal to each oilier, those two lines are paral- lel. Let the lino F.r, cutting the two lines ab, cn, make the alternate angles aef, dfe, equal to each other ; then will ab be parallel to cd. For if they be not parallel, let some other line, as fg, be parallel to ab. Then, because of these parallels, the angle aef is equal to the alternate angle efg (th. 12). But the angle aef is equal to the angle efd (by hyp.) There- fore the anglo efd is equal to the angle efg "(ax. 1) ; that is, a part is equal to the whole, which is impossible. Therefore oo line but cd can be parallel to ab. q. e. d. THEOREMS. 280 Cord. Those lines which are perpendicular to the same lines, are parallel to each other. THEOREM XIV. When a line cuts two parallel lines, the outward angle is equal to the inward opposite one, on the same side ; and the two inward angles, on the sume side, equal to two right angles. Let the line ef cut the two parallel lines ab, cd ; then will the outward an- gle eob be equal to the inward oppo- site angle ghd, on the same side of the line ef ; and the two inward angles bgh, ghd, taken together, will be equal to two right angles. For since the two lines ab, cd, are parallel, the angle agh is equal to the alternate angle cud, (th. 12.) But the angle agh is equal to the opposite angle egb (th. 7). Therefore the angle egb is also equal to the angle ghd (ax. 1). q. e. d. Again, because the two adjacent angles egb, bgh, are to- gether equal to two right angles (th. (5; ; of which the angle xgb has been shown to be equal to the angle ghd ; therefore the two angles bgh, gud, taken together, are also equal to two right angles. Carol. 1. And, conversely, if one line meeting two other lines, make the angles on the same side of it equal, those two lines are parallels. Corol. 2. If a line, cutting two other lines, make the sum of the two inward angles on the same side, less than two right angles, those two lines will not be parallel, but will meet each other when produced. THEOREM XV. Those lines which arc parallel to the same line, are parallel to each other. Let the lines ab, cd, be each of them G - p parallel to the line ef ; then shall the lines A iS ab, cd, be parallel to each other. C 35 For, let the line Gibe perpendicular jr """I p to ef. Then will this line be also per- I pendicular to both the lines ab, cn (corol. th. 12), and con- sequently the two lines ab, cd, are parallels (corol. th. 13\. a. i>. Vol. I. 38 390 OKOKETAY. THEOREM XVI. Whex one sido of a triangle is produced, the outward angle is equal to both the inward opposite angles together. Let the side ab, of the triangle ahc, be produced to d ; then will the outward angle cbd be equal to the sum of the two inward opposite angles a and c. For, conceive be to be drawn pa- rallel to the side ac of the triangle. Then iic, meeting the two parallels ac, be, makes the alter- nate angles c and cbe equal (th. 12). And ab, cutting the same two parallels ac, be, makes the inward and outward angles on the same side, a and ebd, equal to each other (th. 14). Therefore, by equal additions, the sum of the two angles a and c, is equal to the sum of the two cbe and : that is, to the whole angle cbd (by ax. 2), <i» E. D. THEOREM XVII. In any triangle, the sum of all the three angles is equal to two right angles. Let abc be any plane triangle ; then the sum of the three angles a + b + c is enuul to two right angles. For, let the side ab he produced to D. Then the outward angle cbd is equal to the sum of the two inward opposite angles a + c (th. 16). To each of these equals add the in- ward angle b, then will the sum of the three inward angles a + b + c be equal to the sum of the two adjacent angles abc + cbd (ax. 2). But the sum of these two last adjacent angles is equal to two right angles (th. 6). Therefore aba the sum of the three angles of the triangle a + b + c ii equal to two right angles (ax. 1). q. e. d. Carol. 1. If two angles in one triangle, be equal to two angles in another triangle, the third angles will also be equal (ax* 3), and the two triangles equiangular. Cord. 2. If one angle in one triangle, be equal to mm angle in another, the sums of the remaining angles will afcj» be equal (ax. 3). THEOREMS. 291 CoroZ. 3. If one angle of n triangle be right, the sum of the other two will also be equal to a right angle, and each of them singly will be acute, or less thun a right angle. Cmxi. 4. The two least angles of every triangle are acute, or each less thaa a right angle. THEOREM XVIII. Iff any quadrangle, the sum of all the four inward angles, is equal to four right angles. Let a bcd be a quadrangle ; then the sum of the four inward angles, a + b + c + d is equal to four right angles. Let the diagonal ac be drawn, dividing the quadrangle into two triangles, abc, adc. Then, because the sum of the three angles of each of these triangles is equal to two right angles (th. 17) ; it follows, that the sum of all (he angles of both triangles, which make; up the lour angles of the quadrangle, must be equal to four ri*» In angles (ax. 2). ci. K. D. Cord. 1. Hence, if three of the angles be right ones, tl.e fourth will also be a right angle. Carol. 2. And if the sum of two of the four angles be equal to two right angles, the sum of the remaining two will also be equal to two right angles. THEOREM XIX. In any figure whatever, the sum of all the inward angles, taken together, is equul to twice as many right angles, wanting four, as the figure has sides. Let abcde be any figure ; then the eum of all its inward angles, a + h + c + D + E, is equal to twice as many right angles, wanting four, as the figure lias sides. For, from any point r, within it, draw lines, pa, i»b, re, &c. to all the. angles, dividing the polygon into as m iny tri- angles as it has sides. Now the sum of the three angles of each of these triangles, is equal to two riijht angles (h. 17) ; therefore the sum of the angles of alt the triangles is equal to twice as many right angles as the figure has sides. But the sum of all the angles about the point nYikh m w> 292 GEOMETRY. many of f lie angles of the triangles, but no part or the to- ward angles of the polygon, is equal to four right angles (corol. 3, th. 6), and must he deducted out of the former sum. Hence it follows that the sum of all the inward angles of the polygon alone, a + b + c + i> + e» is equal to twice as many right angles as the figure has sides, wanting the said four right angles, u. e. d. TI1EOREX XX. When every side of any figure is produced out, the sum of all the outward angles thereby made, is equal to four right angles. Let a, b, c, dtc. be the outward angles of any polygon, made by pro- ducing all the sides ; then will the sum a + b + c + i) + e, of all those outward angles, be equal to four right angles. For every one of these outward an- gles, together with its adjacent inward angle, make up two right angles, as A+a equul to two right angles, being the two angles made by one line meeting another (th. 6). And there being as many outward, or inward angles, as the figure has sides ; therefore the sum of all the inward and outward angles, is equal to twice as many right angles as the figure has sides. But the sum of all the inward angles with four right angles, is equal to twice as many right angles as the figure has sides (th. 19). Therefore the sum of all the inward and all the outward angles, is equal to the sum of all the inward angles and four right angles (by ax. 1). From each of these take away all the inward angles, and there remains all the outward angles equal to four right angles (by ax. 3j. THEOREM XXI. A perpendicular is the shortest line that can be drawn from a given point to an indefinite line. And, of any other linos drawn from the same point, those that are nearest the perpendicular are less than those more remote. If ab, ac, ad, &c. be lines drawn from the given point a, to the indefinite line de, of which ab is perpendicular ; then shall the perpendicular ab be less than ac, and ac less than ad, &c. For, the angle b being a t\gV\t quo, the TREOBEKI. 298 angle c is acute (by cor. 3, th. 17), and therefore less than the angle b. But the less angle of a triangle is subtended by the leas side (th. 9). Therefore the side ab is less than the side ac. Again, the angle acb being acute, as before, the adjacent angle acd will be obtuse (by th. tf) ; consequently the angle d is acute (corol. 3, th. 17), and therefore is less than the angle c. And since the less side is opposite to the less angle, therefore the side ac is less than the side ad* q. s. d. Carol. A perpendicular is the least distance of a given point from a line. THEOREM XXII. Thb opposite sides and angles of any parallelogram are equal to each other ; and the diagonal divides it into two equal triangles. Let abcd be a parallelogram, of which the diagonal is bd ; then will its opposite sides and angles be equal to each other, and the diagonal bd will divide it into two equal parts, or triangles. For, since the sides ab and dc are pa- rallel, as also the sides ad and bc (defin. 32), and the line bd meets them ; therefore the alternate angles are equal (th. 12), namely, the angle abd to the angle cdb, and the angle adb to the angle cbd. Hence the two triangles, having two angles in the one equal to two angles in the other, have also their third angles equal (cor. 1, th. 17), namely, the angle a equal to the angle c, which are two of the opposite angles of the parallelogram. Also, if to the equal angles abd, cdb, be added the equal angles cbd, abd, the wholes will be equal (ax. 2), namely, the whole angle abc to the whole ado, which are the other two opposite angles of the parellelogram. ft. e. d. Again, since the two triangles are mutually equiangular and have a side in each equal, viz. the common side bd ; therefore the two triangles are identical (th. 2), or equal in all respects, namely, the side ab equal to the opposite side dc, and ad equal to the opposite side bc, andlta^fata triangle abd equal to the whole triangle bcd. <fc- fe« t>% 294 GEOMETRY. Corol. 1. Hence, if one angle of a parallelogram be a right angle, ail the other three will also be right angles, and the parallelogram a rectangle. Corol. 2. Henc; also, the sum of any two adjacent angles of a parallelogram is equal to two right angles. THEOREM XXIII. Every quadrilateral, whose opposite sides are equal, is a parallelogram, or has its opposite sides parallel. Let a bcd be a quadrangle, having the opposite sides equal, namely, the side ab equal to nc, and ad equal to hc ; then shall these equal sides be also parallel, and the figure a parallelogram. For, let the diagonal bd he drawn. Then, the triangles, abd, cud, being mutually equilateral (by hyp.), they are also mutually equiangular (th. 5), or have their corresponding angles equal ; consequently the opposite sides are parallel (th. 13) ; viz. the side ah parallel to nc, and ad parallel to bc, and the figure is a parallelogram, a. k. d. THEORKM XXIV. Those lines which join the corresponding extremes of two equal and parallel lines, are themselves equal and parallel. Let ab, nc, be two equal and parallel lines ; then will the lines ad, bc, which join their extremes, be also equal and parallel. [Sec the fig. above.] For, draw the diagonal bd. Then, because ab and dc are parallel (by hyp.), the angle abd is equal to the alternate angle bdc (th. 12). Hence then, the two triangles having two sides and the contained alines equal, viz. the side ab equal to the aide dc, and the side bo common, and the con- tained angle abd equal to the contained angle bdc, they have the remaining sides and angles also respectively equal (th. 1) ; consequently ad is equal to bc, and also parallel to it (th. 12). u. £. d. THEORKM XXV. Parallelograms, as also triangles, standing on the same base, and between the samo parallels, are equal to each other. THEOREMS. 295 Let abcd, abep, be two parallelograms, J> C T E and abc, abf, two triangles, standing on V A A 7 the same base ab, and between the same \ / V ; / parallels ab, de ; then will the parallelo- \ / gram abcd be equal to the parallelogram y/ V/ abbp, and the triangle abc equal to the Js^ JJ triangle abf* For, since the line de cuts the two parallels af, be, and the two ad, bc, it makes the angle e equal to the angle afd, and the angle d equal to the angle bce (th. 14) ; the two triangles adf, bce, are therefore equiangular (cor. 1, th. 17) ; and having the two corresponding sides ad, bc, equal (th. 22), being opposite sides of a parallelogram, these two triangles are identical, or equal in all respects (th. 2). If each of these equal triangles then be taken from the whole space abed, there will remain the parallelogram abef in the one case, equal to the parallelograms abcd in the other (by ax. 3). Also the triangles abc, abf, on the same base ah, and between the same parallels, are equal, being the halves of the said equal parallelograms (th. 22). q. r. d. CoroL 1. Parallelograms, or triangles, having the same base and altitude, are equal. For tho altitude is the same as the perpendicular or distance between the two parallels, which is every where equal, by the definition of parallels. Corel. 2. Parallelograms, or triangles, having equal bases and altitudes, are equal. For, if the one figure be applied with its base on the other, the bases will coincide or be the same, because they are equal : and so the two figures, having the same base and altitude, are equal. theorem xxvi. If a parallelogram and a triangle, stand on the same base, and between the same parallels, the parallelogram will be double the triangle, or the triangle half the paral- lelogram. Let abcd be the parallelogram, and abe a triangle, on the same base ab, and between the same parallels ab, de ; then will the parallelogram abcd he double the triangle abe, or the triangle half the parallelo- gram. For, draw the diagonal ac of the parallelogram, divid\\i% it into two equal parts (th. 22). Then bec&uafe lY\& XxwoqgtK* 296 GEOMETRY. abc, abe, on the same base, and between the same parallels, are equal (th. 25) ; and because the one triangle abc is half the parallelogram abcd (th. 22), the other equal triangle abe is also equal to half the same parallelogram abcd. q. e. d. Card. 1. A triangle is equal to half a parallelogram of the same base and altitude, because the altitude is the perpendi- cular distance between the parallels, which is every where equal, by the definition of parallels. Carol. 2. If the base of a parallelogram be half that of a triangle, of the same altitude, or the base of the triangle be double that of the parallelogram, the two figures will be equal to each other. THEOREM XXVn. Rectangles that are contained by equal lines, are equal to each other. Let bd, fh, be two rectangles, having jj C H G the sides ab, bc, equal to the sides ef, "7 fo, each to each ; then will the rectangle / bd be equal to the rectangle fh. / For, draw the two diagonals ac, eg, A. B £ F dividing the two parallelograms each into two equal parts. Then the two triangles abc, efg, are equal to each other (th. 1), because they have the two sides ab, bc, and the contained angle b, equal to the two sides ef, fg, and the contained angle f (by hyp.}. But these equal triangles are the halves of the respective rectangles. And because the halves, or the triangles, are equal, the wholes, or the rectangles db, hf, are also equal (by ax. 6). a. e. d. Carol. The squares on equal lines are also equal ; for every square is a species of rectangle. THEOREM XXVIII. The complements of the parallelograms, which are about the diagonal of any parallelogram, are equal to each other. Let ac be a parallelogram, bd a dia- gonal, eif parallel to abof dc, and <;iu J) G parallel to "ad or bc, making ai, ic, com- K j plements to the parallelograms eg, iif, E which are about the diagonal db : then / ~j will the complement ai be equal to the h complement xc i THEOREMS. 207 For, since the diagonal db bisects the three parallelograms ac, eg, hp (th. 22) ; therefore, the whole triangle dab being equal to the whole triangle dcb, and the parts dei, ihb, re* sportively equal to the parts dgi,ifb, the remaining parts Ar, ic, must also be equal (by ax. 3). q. e. d. THEOREM XXIX. A trapezoid, or trapezium haying two sides parallel, is equal to half a parallelogram, whose base is the sum of those two sides, and its altitude the perpendicular distance between them. Let abcd be the trapezoid, having its two sides ab, dc, parallel ; and in ah produced take be equal to dc, so that ae may be the sum of the two parallel sides ; produce dc also, and let kf, gc, G B E bh, be all three parallel to ad. Then is af a parallelogram of the same altitude with the trapezoid abcd, having its base ae equal to the sum of the parallel sides of the trapezoid ; and it is to be proved that the trape- zoid abcd is equal to half the parallelogram af. Now, since triangles, or parallelograms, of equal bases and altitude, are equal (corol. 2, th. 25), the parallelogram no is equal to the parallelogram he, and the triangle cob equal to the triangle chb ; consequently the line bc bisects, or equal- ly divides, the parallelogram af, and abcd is the half of it. Q. E. D. THEOREM XXX. J3 G H C The sum of all the rectangles contained under one whole line, and the several parts of another line, any way divid- ed, is equal to the rectangle contained under the two whole lines. Let ad be the one line, and ab the other, divided into the parts ae, ef, fb ; then will the rectangle contained by ad and ab, be equal to tlio sum of the rectangles ot ad and ae, and ad and ef, and ad and fb : thus expressed, AD • AB = AD . AE + AD . EF + AD . FB. For, make the rectangle ac of the two whole lines ad, ab ; and draw eg, fh, perpendicular to ab, or parallel to ad, to which they are equal (th. 22). Then \ta ^iWAa rectangle ac is made up of all the other TectaaeW Vol. I. 39 A. E f B 208 GEOMETBY. G H C fc. Hut these rectangles arc contain- ed by ad and ae, eg and ef, i n and fb ; which aro equal to the rectangles of ad and ae, ad and ef, ad and fb, because ad is equal to each of the two eg, fh. Therefore the rectangle \d. ab is equal to the sum of all l he other rectangles ad • AE, AD . EF, AD . FB. Q. E. D. Corol. If a right line be divided into any two parts, the square on the whole line, is equal to both the rectangles of the whole line and each of the parts. X FB THEOREM XXXI. The square of the sum of two lines, is greater ti.an the sum of their squares, by twice the rectangle of the said lines. Or, the square of a whole line, is equal to the squares of its two parts, together with twice the rectangle of those parts. Let the line ab be the sum of any two lines ac, ch ; then will the square of ab js- H ij be equal to the squares of .u?, cb, together with twice the rectangle of ac . cb. That 1S, AB a = AC 2 + cb 3 + 2ac . cn. I For, lot \ni)E be the square on the sum * C B or whole line ab, and acfg the square on the part ac. Produce cf and ar to the other sides at H and i. From the lines ch, ci. which are equal, being each equal to the biiVs of the square ab or bd (th. 22), take the parts CF, gf, which arc also equal, being the sides of the square Ai'. am 1 there remains *u equal to fx, which are also equal to mi. or, being the opposite sides of the parallelogram, fleucc the figure hi is equilateral : and it has all its angles riiiht onus teorol. 1, th. 22) : it is therefore a square on the line fi, or the square of its equal cn. Also the figures ef, fb, arc equal to two rectangles under ac and cb, because c;r is equal to ac. and fh or fi equal to cb. But the whole square ad is made up of the four figures, viz. the two tiquarch af, id, and the two equal rectangles ef, fb. That i? ? the .-square of aji is equal to the squares of ac, cb, toge- ihcr with twice th" rectangle of ac, cb. a. u. d. 0>r. /. Hence, if a line be divided into two equal parts ; the .square of the whole line will be equal to four times the square of half the hue. I THEOREMS. 299 E ID THEOREM XXXII. The square of the difference of two lines, is less than the sum of their squares, by twice the rectangle of the said lines. Let ac, bc, be any two lines, and ab their difference : then will the square of ab & be less than the squares of ac, bc, by twice the rectangle of ac and bc. Or, AB* s* AC 1 + BC 1 — 2AC . BC. For, let abde be the square on the dif- ference ab, and acfo the square on the A. B line ac. Produce ed to h ; also produce K I db and hc, and draw ki, making bi the square of the other line bc. Now it is visible that the square ad is less than the two squares af, bi, by the two rectangles ef, di. But gf is equal to the one line ac, and ue or fii is equal to .the other line bc ; consequently the rectangle kf, contained under i:« and of, is equal to the rectangle of ac and bc. Again, fh being equal to ci or bc or Dir, by adding the common part hc, the whole hi will bo equal to the whole rc, or equal to ac ; and consequently the figure di is equal to the rectangle contained by ac and bc. Hence the two figures ef, di, arc two rectangles of the two lines ac, bc ; and consequently the square of vn is less than the squares of ac, bc, by twice the rectangle AC . BC. Q. E. B. THEOREM XXXUI. The rectangle under the sum and difference of two lines, is equal to the difference of the squares of those lines*. Let ab, ac, be any two unequal lines ; then will the difference of the squares of ab, ac, be equal to a rectangle under their sum and difference That is, AB 1 — AC 3 = AB -f- AC . AB — AC. For, let abde be the square of ab, and acfo the square of ac. Produce db till bh be equal to ac ; draw hi parallel to ab or ed, and produce fc both ways to i and k. E G C F * This and the two preceding theorems, arc evinced algebraically, by the three expression! ( a + 6 ) J = a ' + %ib + 6» a* + 6» -f- 2ab (« — 6 )3 = a> — 2a6 4 fta = a> + 63 — 2a6 (a + 6)(ft-6):=«*— 6*. 800 GEOMETRY. Then the difference of the two squares ad, af, is evi- dently the two rectangles ef, kb. But the rectangles ef, bi are equal, being contained under equal lines ; for ex and bh are each equal to ac, and ge is equal to cb, being each equal to the difference between ab and ac, or their equals ae and ag. Therefore the two ef, kb, are equal to the two xb 9 bi, or to the whole kh ; and consequently kh is equal to the difference of the squares ad, af. But kh is a rect- angle contained by dh, or the sum of ab and ac, and by kd, or the difference of ab and ac. Therefore the difference of the squares of ab, ac, is equal to the rectangle under their sum and difference. Q. e. d. theorem xxxiv. In any right angled triangle, the square of the hypo- thenuse, is equal to the sum of the squares of the other two sides* Let abc be a right-angled triangle, Q having the right angle c ; then will the square of the hypothenuse ab, be equal to the sum of the squares of the other -° two sides ac, cb. Or ab 3 = ac* + BC a . For, on ab describe the square ae, and on ac, cb, the squares ag, bh ; then draw ck parallel to ad or be ; and join ai, bf, cd, ck. Now, because the line ac meets the two cg, cb, so as to make two right angles, these two form one straight line gb (corol. 1, th. 6). And because the angle fac is equal to the angle dab, being each a right angle, or the angle of a square ; to each of these equals add the common angle bac, so will the whole angle or sum fab, be equal to the whole angle or sum cad. But the line fa is equal to the line ac, and the line ab to the line ad, being sides of the same square ; so that the two sides fa, ab, and their included angle fab. are equal to the two sides ca, ad, and the contained angle cad, each to each : therefore the whole triangle afb is equal to the whole triangle acd (th. 1). But the square ag is double the triangle afb, on the same base fa, and between the- same parallels fa, ob (th. 26) ; in like manner the parallelogram ak is double the triangle acd, on the same base ad, and between the same parallels ad, ck. And since the doubles of equal things, are equal (by ax. 6) ; therefore the square ag is equal to the parallelogram ak. THEOREMS. 301 la like manner, the other square bh is proved equal to the other parallelogram bk. Consequently the two squares ag and bh together, are equal to the two parallelograms ak and bk together, or to the whole square ak. That is, the sum of the two squares on the two less sides, is equal to the square on the greatest side. a. e. d. Cord. 1. Hence, the square of either of the two less sides, is equal to the difference of the squares of the hypothenuse and the other side (ax. 3) ; or, equal to the rectangle con- tained by the sum and difference of the said hypothenuse and other side (th. 33). Carol. 2. Hence also, if two right-angled triangles have two sides of the one equal to two corresponding sides of the other ; their third sides will also be equal, and the triangles identical. THEOREM XXXV. In any triangle, the difference of the squares of the two sides, is equal to the difference of the squares of the segments of the base, or of the two lines, or distances, included between the extremes of the base and the perpen- dicular. Let abc be any triangle, having cd perpendicular to ab ; then will the difference of the squares of ac, bc, be equal to the difference of the squares of ad, bd ; that is, _ _ A ^ ^ AC* BC 3 = AD 2 BD 2 . A BD A DB For, since ac 2 is equal to ad 2 + cd 2 > , „. N and bc 2 is equal to bd 2 + c d 2 \ W ihm M > ; Theref. the difference between ac 2 and bc 2 , is equal to the difference between ad 2 + cd 2 and bd 2 + cd 2 , or equal to the difference between ad 2 and bd 3 , by taking away the common square cd 3 . q. e. d. Carol. The rectangle of the sum and difference of the two sides of any triangle, is equal to the rectangle of the sum and difference of the distances between the perpendicular and the two extremes of the base, or equal to the rectangle of the base and the difference or sum of the segments, according as the perpendicular falls within or without the triangle. 302 GEOMETRY. That is, (ac+bc) . (ac — bc) = (ad+bd) . (ad — bd) Or, (ac+bc) . (ac — bc) = ab . (ad — bd) in the 2d fig. And (ac+bc) . (ac— bc) = ab . (ad+bd) in the lrt fig. THEOREM XXXVI. In any obtuse-angled triangle, the square of the side sub- tending the obtUBe angle, is greater than the sum of the squares of the other two sides, by twice the rectangle of the base and the distance of the perpendicular from the ob- tuse angle. Let abc be a triangle, obtuse angled at b, and cd perpen- dicular to ab ; then will the square of ac be greater than the squares of ab, bc, by twice the rectangle of ab, bd. That is, ac 3 = ab 3 + bc 8 + 2a n . bd. See the 1st fig. above, or below. For, ad 2 = ab 2 + bd 3 + 2ab . bd (th. 31). And ad 1 + en 8 = ab 3 + bd* + cd 2 + 2ab . bd (ax. 2). But ad 2 + cd 2 = ac 3 , and bd 2 + cd* = bc 2 (th. 34). Therefore ac 2 = ab 2 + bc 2 + 2a b . bd. q. e. d. THEOREM XX XVI I. In any triangle, the square of the side subtending an acute angle, is less than the squares of the base and the other aide, by twice the rectangle of the base and the distance of the perpendicular from the acute angle. Let abc be a triangle, having ^ „ the angle a acute, and cd perpen- ^ dicular to ab ; then will the square ''■ of bc, be less than the squares of ab, ac, by twice the rectangle of ab, ad. That is, bc = == ab 2 + W .B D A. J)B ac* — 2ad . AB. For BD 3 = AD 3 + AB 2 — 2ad . AB (th. 32). And BD 3 + DC 2 = AD 2 + DC 2 + AB 2 — 2 AD . AB (aX. 2), Therefore bc 3 = ac 2 + ab 2 — 2ad . ab (th. 34). q. s. d. THEOREM XXXVIII. Ik any triangle, the double of the square of a line drawn from the vertex to the middle of the base, together with THEOREMS. 303 double the square of the half base, is equal to the sum of the squares of the other two sides. Let abc be a triangle, and <;d the line drawn from the vertex to the middle of 9 the base ab, bisecting it into the two equal parts ad, db ; then will the sum of the squares of ac, ch, be equal to twice the sum of the squares of <;«, ad ; or ac 3 + cb 3 = 2cd 3 + 2\D a . / ' j A DEB For AC 8 = CD 3 + ad 3 + 2ad • de (th. 36). And bc 3 = CD* + bd 2 — 2ad . db (th. 37). Therefore ac 3 + bc 3 = 2cd 3 + ad 3 + bd 3 = 2cd 3 -f- 2ad 3 (ax. 2). <j. e. d. tiikork;i xxxix. In an isosceles triangle, the square of a line drawn from the vertex to any point in the base, together with the rect- angle of the segments of the base, is equal to the square of one of the equal sides of the trim.gle. Let abc be the isosceles triangle, and cd c a line drawn from the vertex to any point D in the base : then will the square of ac, be equal to the square of ci>, together with the rectangle of ad and on. That is, AC 3 = CD 3 + AD . DB. For AC 3 — CD 3 = AE 3 — DE 3 (th. 35). = AD . DB (th. 33). Therefore, ac 2 = cd 2 -f- ad . db (ax. 2). a. k. d. A 13 THEOREM XL. In any parallelogram, the two diagonals bisect each other ; and the sum of their squares is equal to the sum of the squares of all the four sides of the parallelogram. Let abcd be a parallelogram, whost; "JJ C diagonals intersect each other in e : then will ae be equal to ec, and be to i:i> : and the sum of the squares of ,u\ m>, will be equal to the sum of the square s of ab, ik\ cd, da. That is, ae = ec, and bj: — ed, and ac 3 + bd' = ab s + bc 3 -t ci> r ua • 804 GEOMETRY. For, the triangles aeb, dec, are equiangular, because they have the opposite angles at e equal (th. 7), and the two lines ac 9 bd, meeting the parallels ab, dc, make the angle bae equal to the angle dce, and the angle abe, equal to the angle cde, and the side ab equal to the side dc (th. 22) ; therefore these two triangles are identical, and have their corresponding sides equal (th. 2), viz. ae = ec, and be a ed. Again, since ac is bisected in e, the sum of the squares ad 3 + DC 3 = 2AE 3 + 2de s (th. 38). In like manner, ab' + bc* =■ 2ae s + 2be 3 or 2de 3 . Theref. ab s + bc 3 + CD* + da 1 = 4 a e 3 + 4de 3 (ax. 2). But, because the square of a whole line is equal to 4 times the square of half the line (cor. th. 31), that is, ac* = 4ae\ and bd 3 = 4de 3 : Theref. ab 3 + bc 3 + cd 3 -4 da 3 = ac 2 f bd 3 (ax. 1). Q. E. D. Cor. 1. If ad = dc, or the parallelogram be a rhombus; then ad 3 = ae 3 + ED 3 , CD 3 =■ de* + ce 1 , dsc. Cor. 2. Hence, and by th. 34, the diagonals of a rhom- bus intersect at right angles. THEOREM XLI. If a line, drawn through or from the centre of a circlet bisect a chord, it will bc perpendicular to it; or, if it be perpendicular to the chord, it will bisect both the chord and the arc of the chord. Let ab be any chord in a circle, and ct> a line drawn from the centre <- to the chord. Then, if the chord be bisected in the point d, cd will bc perpendicular to ab. Draw the two radii ca, cb. Then the two triangles acd, bcd, having ca equal to cb (def. 44), and cd common, also ad equal to db (by hyp.) ; they have all the three sides of the one, equal to all the three sides of the other, and so have their angles also equal (th. 5). Hence then, the angle ado being equal to the angle bdc, these angles are right angles, and the line cd is perpendicular to ab (def. 11). Again, if cd bo perpendicular to ah* then will the chord THEOUM*. Mft ab be bisected at the point d, or have ad equal to db ; and the arc abb bisected in the point e, or have ae equal bb. For, having drawn ca, cb, as before : Then, in the tri- angle abc, because the side ca is equal to the side cb, their opposite angles a and b are also equal (th. 3). Hence then, in the two triangles acd, bcd, the angle a is equal to the angle b, and the angles at d are equal (def. 11) ; therefore the third angles are also equal (corol. 1. th. 17). And having the side cd common, they have also the side ad equal to the side db (th. 2). Also, since the angle ace is equal to the angle bob, the arc ae, which measures the former (def. 57), is equal to the arc be, which measures the latter, since equal angles must have equal measures. Corol. Hence a line bisecting any chord at right angles, passes through the centre of the circle. THEOREM XLII. If more than two equal lines oan be drawn from any point within a circle to the circumference, that point will bo the centre. Let abc be a circle, and d a point within it : then if any three lines, da, db, dc, drawn from the point d to the circumference, be equal to each other, the point d will be the centre. Draw the chords ab, bc, which let be bisected in the points e, f, and join de, df. Then, the two triangles, dak, dbe, have the side da equal to the side db by supposition, and the side ae equal to the side eb by hypothesis, also the side db common : therefore these two triangles are identical, and have the angles at e equal to each other (th. 5) ; conse. quently de is perpendicular to the middle of the chord ab . (def. 11), and therefore passes through the centre of the circle (corol. th. 41). In like manner, it may be shown that df passes through the centre. Consequently the point d is the centre of the circle, and the three equal lines da, db, dc, are radii. 0» Vol. 1. 40 306 •KOMETRY. THEOREM XLIII. If two circles placed one within another, touch, the centre* of the circles and the point of contact will be all in the same right line. Let the two circles abc, ade, touch one A. anottier internally in the point a ; then will the point a and the centres of those circles be all in the same right line. Let f be the centre of the circle ahc, through which draw the diameter afc. Then, if the centre of the other circle g can be out of this line ac, let it be sup- posed in some other point as o ; through which draw the line fis, cutting the two circles in b and d. Now in the triangle afg, the sum of the two sides fg t ca, is greater than the third side af (th. 10), or greater than its equal radius fb. From each of these take away the common part fg, and the remainder ga will be greater than the remainder gb. But the point g being supposed the centre of the inner circle, its two radii, ga, gd, are equal to each other ; consequently od will also be greater than gb. But ade being the inner circle, gd is necessarily less than gb. So that up is both greater and less than gb ; which is absurd. Consequently the centre o cannot be out of the lino afc. q. e. n. THEOREM XLIV. If two circles touch one another externally, the centres of the circles and the point of contact will be all in the same right line. Let the two circles ybc, ape, touch one another externally at the point \ : then will / ^ the point of contact a and the centres of the I i two circles be all in the same right line. \w /I Let f be the centre of the circle aim:. through which draw the diameter afc, and / produce it to the other circle at e. Then, if \ F * - - — ..v, ... . the centre of the other circle ade can be out \C ofyhc line fk, let. i(, if possible, be supposed in some other point as g ; and draw the lines a«:, rune, cutting the two circles in b and d. ./ THEOREMS. 507 Then, in the triangle afg, the sum of the two sides af, ao, is greater than the third side fg (th. 10). But, f and o being the centres of the two circles, the two radii ga, gd, •are equal, as are also the two radii af, fb. Hence the sura of ga, af, is equal to the sum of ou, bf ; and therefore this latter sum also, od, bf, is greater than gf, which is absurd. Consequently the centre o cannot be out of the line ef. 4.B.D. THEOREM XLV. Asnc chords in a circle, which are equally distant from the centre, are equal to each other ; or if they be equal to each other, they will be equally distant from the centre. Let ab, cd, be any two chords at equal C distances from the centre g ; then will /7\A\ these two chords ab, cd, be equal to each fc l \ / Vp V other. I / qT \ J Draw the two radii ga, gc, and the V V/ two perpendiculars ge, of, which are the B >s*_^>T3 equal distances from the centre g. Then, the two right-angled triangles, gae, gcf, having the side ga equal the side gc, and the side ge equal the side gf, and the angle at e equal to the angle at f, therefore those two triangles are identical (cor. 2, th. 34), and have the line ae equal to the line cf. But ab is the double of ae, and cd is the double of cf (th. 41) ; therefore ab is equal to cd (by ax. 6). u. e. d. Again, if the chord ab be equal to the chord cd ; then will their distances from the centre, ge, gf, also be equal to each other. For, since ab is [equal cd by supposition, the half ai: is equal the half cf. Also the radii ga, gc, being equal, as well as the right angles e and f, therefore the third sides are equal (cor. 2, th. 34), or the distance ge equal the distance of. a. fi. d. theorem xlvi. A line perpendicular to the extremity of a radius^ va tangent to the circAe. 906 •BOMBTBY. Let the line adb be perpendicular to the radius cd of a circle ; then shall ab touch the circle in the point d only. From any other point s in the line ab draw cfb to the centre, cutting the circle in f. Then, because the angle d, of the trian- 51e cde, is a right angle, the angle at e is acute (cor. 3, th. 7), and consequently less than the angle d. But the greater side is always opposite to the greater angle (th. 9) ; there* fore the side ce is greater than the Bide cd, or greater than its equal cf. Hence the point e is without the circle ; and the same for every other point in the line ab. Consequently the whole line is without the circle, and meets it in the point d only. J THEOREM XLVIZ. Whew a line is a tangent to a circle, a radius drawn to the point of a contact is perpendicular to the tangent. Let the line ab touch the circumference of a circle at the point d ; then will the radius cd be the perpendicular to the tangent ab. [See the last figure.] For the line ab being wholly without the circumference except at the point d, every other line, as ce, drawn from the centre c to the line ab, must pass out of the circle to arrive at this line. The line cd is therefore the shortest that can be drawn from the point c to the line ab, and conse- quently (th. 21) it is perpendicular to that line. Carol. Hence, conversely, a line drawn perpendicular to a tangent, at the point of contact, passes through the centre of the circle. THEOREM XLVIII. The angle formed by a tangent and chord is measured by half the arc of that chord. Let ab be a tangent to a circle, and cd a chord drawn from the point of contact c ; then is the angle bcd measured by half the arc cfd, and the angle acd measured by half the arc con. Draw the radius ec to the point of contact, and the radios bf perpendicular to the chord at h. i THEOREMS. 809 Then the radius kf, being perpendicular to the chord cd, bisects the arc cfd (th. 41 ). Therefore cf is half the arc cfd. In the triangle ceh , the angle h being a right one, the sum of the two remaining angles e and c is equal to a right angle (cor. 3, th. 17), which is equal to the angle bce, because the radius ce is perpendicular to the tangent. From each of these equals take the common part or angle c, and there remains the angle e equal to the angle bcd. But the angle b is measured by the arc cf (def. 57), which is the half of cfd ; therefore the equal angle bcd must also have the same measure, namely, half the arc cfd of the chord cd. Again, the line gef, being perpendicular to the chord cd, bisects the arc cod (th. 41). Therefore co is half the arc cod. Now, since the line ce, meeting fo, makes the sum of the two angles at s equal to two right angles (th. 6), and the line cd makes with ab the sum of the two angles at c equal to two right angles ; if from these two equal sums there be taken away the parts or angles ceh and bch, which have been proved equal, there remains the angle ceo equal to the angle ach. Rut the former of these, cfo, being an angle at the centre, is measured by the are cg (def. 57) ; consequently the equal angle acd must also have the same measure co, which is half the arc cod of the chord CD. Q. E. D. Carol. 1. The sum of two right angles is measured by half the circumference. For the two angles bcd, acd, which make up. two right angles, are measured by the arcs cf, co, which make up half the circumference, fo being a diameter. Carol. 2. Hence also one right angle must have for its measure a quarter of the circumference, or 90 degrees. THE OBEX XLIX. Ah angle at the circumference of a circle is measured by half the arc that subtends it. Let bac be an angle at the circumference ; 3) A 18 it has for its measure, half the arc bc which subtends it. For, suppose the tangent db passing through the point of contact a ; then, the 3) A. IS GEOMETRY. angle dac being measured by half the arc arc, and the angle dab by half the arc ab (th. 48) ; it follows, by equal tub. traction, that the difference, or angle bag, must be measured by half the arc bc, which it stands upon. a. e. d. THEOREM L. All angles in the same segment of a circle, or standing on the same arc, are equal to each other. Let c and d be two angles in the game segment acdb, or, which is the same thing, standing on the supplemental arc aeb ; then will the angle c be equal to the angle d. For each of these angles is measured by half the arc aeb ; and thus, having equal B measures, they are equal to each other (ax. 11). THEOREM LI. An angle at the centre of a circle is double the angle at the circumference, when both stand on the same arc. Let c be an angle at the centre c, and d an angle at the circumference, both standing on the same arc or same chord ab : then will the angle c bo double of the angle i>, or the angle d equal to half the angle c. For, the angle at the centre c is measured by the whole arc aeb (def. 57), and the angle at the circum- ference d is measured by half the same arc aeb (th. 49) ; therefore the angle d is only half the angle c, or the angle c doubles the angle d. THEOREM LII. An angle in a semicircle, is a right angle. If abc or adc be a semicircle ; then any D angle d in that semicircle, is a right angle. For, the angle d, at the circumference, is measured by half the arc abc (th. 49), that is, by a quadrant of the circumference. But a quadrant is the measure of a right angle (cor. 4, th. 6; or cor. 2, th. 48). Therefore the angle d is a rigtit angle. THE0REHS. 311 THEOREM LIU. The angle formed by a tangent to a circle, and a chord drawn from the point of contact, is equal to the angle in the alternate segment. If ab be a tangent, and ac a chord, and D any angle in the alternate segment adc ; then will the angle d be equal to the angle bac made by the tangent and chord of the arc a ec. For the angle d, at the circumference, is measured by half the arc aec (th. 49) ; and the angle bac, made by the tangent and chord, is also nfeasured by the same 1 half arc aec (th. 48) ; therefore these two angles are equal (ax. 11). THEOREM LIV. The sum of any two opposite angles of a Quadrangle in- scribed in a circle, is equal to two right angles. Let abcd be any quadrilateral inscribed in a circle ; then shall the sum of the two opposite angles a and c, or u and d, be equal to two right angles. For the angle a is measured by half the arc dcb, which it stands on, and the angle c by half the arc dab (th. 49) ; therefore the sum of the two angles a and c is measured by half the sum of these two arcr., that is, by half the circumference. But half the circumference is the measure of two right angles (cor* 4, th. G) ; therefore the sum of the two opposite angles a and c is equal to two right angles. In like manner it is dhown, that the sum of the other two opposite angles, d and B, is equal to two right angles, u. k. d. THEOREM LV. Ir any side of a quadrangle, inscribed in a circle, be pro- duced out, the outward angle will be equal to the inward opposite angle. If the side An, of the quadrilateral abcd, inscribed in a circle, be produced to e ; the outward angle dak will bo equal to the inward opposite angle c. ns GEOMETRY. For, the mim of the two adjacent angles dab and dab is equal to two right angles (th. 6) ; and the sum of the two opposite angles c and dab iB also equal to two right angles (th. 54) ; therefore the former sum, of the two angles dab and dab, is equal to the latter Bum, of the two c and dab (ax. 1). From each of these equate taking away the common angle dab, there remains the angle dab equal the angle c. Q. E. D. THEOREM LVI. Awv two parallel chords intercept equal arcs. Let the two chords ab, cd, be parallel : then will the arcs ac, bd, be equal ; or AC = BD. Draw the line bc. Then, because the lines ae, cd, are parallel, the alternate an- gleB b and c are equal (th. 12}. But the angle at the circumference b, is measured by half the arc ac (th. 49) ; and the other equal angle at the circumference c is measured by half the arc bd : therefore the halves of the arcs ac, bd, and consequently the arcs themselves, are also equal, q. e. d. THEOREM LVII. When a tangent and chord are parallel to each other, they intercept equal arcs. Let the tangent abc be parallel to the chord df ; then are the arcs bd, bf, equal ; that is, bd = bf. Draw the chord bd. Then, because the lines ab, df, are parallel, the alternate angles d and b are equal (th. 12). But the angle b, formed by a tangent and chord, is measured by half the arc bd (th. 48) ; and the other angle at the circum- ference d is measured by half the arc bf (th. 49) ; therefore the arcs bd, bf, are equal, a. e. d. THBOBJUC8. 313 THEOREM LVin. Tn angle formed, within a circle, by the intersection of (wo chords, is measured by half the sum of the two inter- cepted arcs. * Let the two chords ab, cd, intersect at the point e : then the angle aec, or deb, is measured by half the sum of the two arcs AC, DB. Draw the chord ap parallel to cd. Then because the lines ap, cd, are parallel, and ab cuts them, the angles on the same side a and deb are equal (th. 14). But the angle at the circumfer- ence a is measured by half the arc bf, or of the sum of fd and db (th. 49) ; therefore the angle £ is also measured by half the sum of fd and db* Again, because the chords af, cd, are parallel, the arcs ac v fd, are equal (th. 56) ; therefore the sum of the two arcs ac, db, is equal to the sum of the two fd, db ; and consequently the angle e, which is measured by half the latter sum, is also measured by half the former, q. e. d. theorem: lix. The angle formed, out of a circle, by two secants, is mea- sured by half the difference of the intercepted arcs. Let the angle e be formed by two se- cants eab and ecd ; this angle is measur- ed by half the difference of the two arcs Ac, db, intercepted by the two secants. Draw the chord af parallel to cd. Then, because the lines af, cd, are parallel, and ab cuts them, the angles on the same side a «nd bed are equal (th. 14). But the angle a, at the circum- ference, is measured by half the arc bf (th. 49), or of the difference of df and db : therefore the equal angle b is also measured by half the difference of df, db. Again, because the chords, af, cd, are parallel, the arcs ac, fd, are equal (th. 56) ; therefore the diffettfic* ot ^* Vol. I. 41 314 GEOMETRY. two arcs AC) db, is equal to the difference of the two »r, db. Consequently the angle e, which is measured by half the latter difference, is also measured by half the former. q. e. d. THEOREM LX. The angle formed by two tangents, is measured by half the difference of the two intercepted arcs. Let eb, ed, be two tangents to a circle at the points a, c; then the angle e is measured by half the difference of the two arcs CFA, CGA. Draw the chord af parallel to ed. Then, because the lines, af, ed, are pa- rallel, and eb meets them, the angles on the same side a and e are equal (th. 14). But the angle a, formed by the chord af and tangent ab, is measured by half the arc af (th. 48) ; therefore the equal angle e is also measured by half the same arc af, or half the difference of the arcs cfa and cf, or cga (th. 57). Carol. In like manner it is proved, that the angle e, formed by a tangent ecd, and a secant eab, is measured by half the difference of the two intercepted arcs ca and cfb. theorem lxi. When two lines, meeting a circle each in two points, cut one another, either within it or without it; the rectangle of the parts of the one, is equal to the rectangle of the parts of the other ; the parts of each being measured from the point of meeting to the two intersections with the cir- cumference. THEOREMS. 815 Let the two lines ab, cd, meet each trther in e ; then the rectangle of ae, eb, will be equal to the rectangle of ce, ed. Or, AE . EB = CE . ED. For, through the point e draw the dia- meter fo ; also, from the centre h draw the radius dh, and draw hi perpendicular to CD. Then, since dbh is a triangle, and the perp. hi bisects the chord cd (th. 41), the line cb is equal to the difference of the segments di, ei, the sum of them being be. Also, because h is the centre of the circle, and the radii dh, fh, oh, are all equal, the line eg is equal to the sum of the sides dh, he ; and ef is equal to their difference. But the rectangle of the sum and difference of the two sides of a triangle is equal to the rectangle of the sum and difference of the segments of the base (th. 35) ; therefore the rectangle of fe, eo, is equal to the rectangle of ce, ed. In like manner it is proved, that the same rectangle of fe, eg, is equal to the rectangle of ae, eb. Consequently the rectangle of ae, eb, is also equal to the rectangle of ce, ed (ax. 1). Q. E. D. Coral. 1. When one of the lines) in the second case, as de, by revolving about the point e, comes into the position of the tan- gent ec or ed, the two points c and d run- ning into one ; then the rectangle of ce, ed, becomes the square of ce, because ce and de are then equal. Consequently the rectangle of the parts of the secant, ae . eb, is equal to the square of the tangent, ce 9 . Carol. 2. Hence both the tangents ec, ef, drawn from the same point e, are equal ; since the square of each is equal tolhe same rectangle or quantity ae • eb. THEOREM LXII. In equiangular triangles, the rectangles of the corresponding or like sides, taken alternately, are equal. 816 GKOVETMT. Let abc, dbf, be two equiangular triangles, having the angle a = the angle d, the angle b = the angle e, and the angle c = (he angle f ; also the like si-les ab, de, and ac, df, be- ing those opposite the equal angles : then will the rectangle of ab, df, be equal to the rectangle of ao, de. In ba produced take ag equal to or ; and through the three points b, c, g, conceive a circle bcoh to be described, meeting ca produced at h, and join oh. Then the angle o is equal to the angle c on the same arc bh, and the angle h equal to the angle b on the same arc co (th. 50) ; also the opposite angles at a are equal (th. 7) : therefore the triangle agh is equiangular to the triangle acb, and consequently to the triangle dfb also. But the two like sides ao, df, are also equal by supposition ; conse- quently the two triangles aoh, dfk, are identical (th. 2), having the two sides ao, ah, equal to the two df, de, each to each. But the rectangle ga . ab is equal to the rectangle ha • ac (th. 61) : consequently the rectangle df . ab is equal to the rectangle de . ac. a. e. d. THEOREM LXI1I. The rectangle of the two sides of any triangle, is equal to the rectangle of the perpendicular on the third side and the diameter of the circumscribing circle. Let cd be the perpendicular, and ce the diameter of the circle about the triangle abc ; then the rectangle ca . cb is = the rectangle cd • ce. For, join be : then in the two triangles acd, ecb, the angles a and e are equal, standing on the same arc bc (th. 50) ; also the right angle d is equal the angle b, which is also a right angle, being in a semicircle (th. 52) : therefore these two triangles have also their third angles equal, and are equian- gular. Hence, ac, ce, and cd, cb, being like sides, nib- tending the equal angles, the rectangle ac . cb. of the first and last of them, is equal to the rectangle ce . cd, of the other two (th. 62). THEOREMS. 317 THEOREM LXIV. The square of a line bisecting any angle of a triangle, together with the rectangle of the two segments of the oppo- site side, is equal to the rectangle of the two other sides in- cluding the bisected angle. Let cd bisect the angle c of the triangle abc ; then the square cd 9 + the rectangle ad . db is = the rectangle ac . cb. For, let cd be produced to meet the cir- cumscribing circle at e, and join ae. Then the two triangles ace, bcd, are equiangular : for the angles at c are equal by supposition, and the angles b and e are equal, standing on the same arc ac (th. 50) ; consequently the third angles at a and Dare equal (cor. 1, th. 17) : also ac, cd, and ce, cb, are like or corresponding sides, being opposite to equal angles : therefore the rectangle ac . cb is = the rectangle cd . ce (th. 62). But the latter rectangle cd . ce is = cd* + the rectangle cd . de (th. 30) ; therefore the former rect- angle ac . cb is also = cd 3 + cd . de, or equal to cd 2 + ad . db, since cd . de is = ad • db (th. 61). a. e. d. theorem lxv. The rectangle of the two diagonals of any quadrangle inscribed in a circle, is equal to the sum of the two rect- angles of the opposite sides. Let abcd be any quadrilateral inscribed in a circle, and ac, bd, its two diagonals : then the rectangle ac . bd is = the rect- angle ab . dc + the rectangle ad . sc. For, let ce be drawn, making the angle bce equal to the angle dca. Then the two triangles acd, bce, are equiangular ; for the angles a and b are equal, standing on the same arc dc ; and the angles dca, bce, are equal by supposition ; conse- quently the third angles adc, bec, are also equal : also, ac, bc, and ad, .be, are like or corresponding sides, being oppo- site to the equal angles : therefore the rectangle ac . be it the rectangle ad . bc (th. 62). 318 GEOMETRY* Again, the two triangles abc, dec, are equiangular : for the angles bac, bdc, are equal, standing on the same arc bc ; and the angle dce is equal to the anglo bca, by adding the common angle ace to the two equal angles dca, bce ; there- fore the third angles e and abc are also equal : but ac, dc, and ab, de, are the like sides : therefore the rectangle AC . de is = the rectangle ab . dc (th. 62). Hence, by equal additions, the sum of the rectangles ac . be + ac . de is = ad . bo + ab . dc. Hut the for- mer sura of the rectangles ac . be + ac . de is = the rect- angle ac . bd (th. 30) : therefore the same rectangle ac • bd is equal to the latter sum, the rect. ad . bc + the rect. ab . dc (ax. 1). Q. E. D. CoroL Hence, if abd be an equilateral triangle, and c any point in the arc bcd of the circumscribing circle, we have ac = bc + dc. For ac . bd being = ad . bc + ab - dc ; dividing by bd == ab = ad, there results ac = bc + dc* OF RATIOS AND PROPORTIONS. DEFINITIONS. Dep. 76. Ratio is the proportion or relation which one magnitude bears to another magnitude of the same kind, with respect to quantity. Note* The measure, or quantity, of a ratio, is conceived, by considering what part or parts the leading quantity, called the Antecedent, is of the other, called the Consequent ; or what part or parts the number expressing the quantity of the former, is of the number denoting in like manner the latter. So, the ratio of a quantity expressed by the number 2, to a like quantity expressed by the number 6, is denoted by 2 divided by 6, or | or £ : the number 2 being 3 times con- tained in 6, or the third part of it. In like manner, the ratio of the quantity 3 to 0, is measured by £ or £ ; the ratio of 4 to 6 is } or |; that of 6 to 4 is $ or £ ; &c. 77. Proportion is an equality of ratios. Thus, 78. Three quantities are said to be proportional, when the ratio of the first to the second is equal to the ratio of the THEOREMS. 319 second to the third. As of the three quantities a (2), b (4), c (8), where $==1 = 1, hoth the same ratio. 79. Four quantities are said to be proportional, when the ratio of the first to the second, is the same as the ratio of the third to the fourth. As of the four, a (4), n (2), c (10), o (5), where a = y> = 2, both the same ratio. Note* To denote that four quantities, a, b, c, d, are pro* portional, they are usually stated or placed thus, a : b : : c : d; and read thus, a is to b as c is to d. But when three quan- tities are proportional, the middle one is repeated, and they are written thus, a : b : : b : c. The proportionality of quantities may also be expressed very generally by the equality of fractions, as at pa. 118. Thus, if - = then a : b : : c : d, also b : a : r c : p, and B D A : o : : b : d, and c : a : : b : d. 80. Of three proportional quantities, the middle one is said to be a Mean Proportional between the other two ; and the last, a Third Proportional to the first and second. 81. Of four proportional quantities, the last is said to be a Fourth Proportional to the other three, taken in order. 82. Quantities are said to be Continually Proportional, or in Continued Proportion, when the ratio is the same between every two adjacent terms, viz. when the first is to the second, as the second to the third, as the third to the fourth, as the fourth to the fifth, and so on, all in the same common ratio. As in the quantities 1, 2, 4, 8, 16, &c. ; where the com- mon ratio is equal to 2. 83. Of any number of quantities, a, b, c, d, the ratio of the first a, to the last d, is said to be Compounded of the ratios of the first to the second, of the second to the third, and so on to the last. 84. Inverse ratio is, when the antecedent is made the consequent, and the consequent the antecedent. — Thus, if 1 : 2 : : 3 : 6 ; then inversely, 2 : 1 : : 6 : 3. 85. Alternate proportion is, when antecedent is compared with antecedent, and consequent with consequent. — As, if 1 : 2 : : 3 : 6 ; then, by alternation, or permutation, it will be 1 : 3 : : 2 : 6. 86. Compound ratio is, when the sum of the antecedent and consequent is compared, either with the to\u&qp£feV> « 820 GEOMETRY* with the antecedent.— Thus, if 1 : 2 : : 3 : 6, then by com- position, l+2:l::3 + 6:3, an( ] 1 -f 2 : 2 : : 3 + 6 : 6. 87. Divided ratio, is when the difference of the antecedent and consequent is compared, either with the antecedent or with the consequent. — Thus, if 1 : 2 : : 3 : 6, then, by di- vision, 2 — 1:1 : : 6— 3 : 3, and 2 — 1 : 2 : : 6 — 3:6. Nate. The term-Divided, or Division, here means subtract, ing, or parting ; being used in the sense opposed to com- pounding, or adding, in def. 86. THEOREM LXVI. Equimultiples of any two quantities have the same ratio as the quantities themselves. Let a and b be any two quantities, and m\, j»b, any equi- multiples of them, m being any number whatever : then will wia and mB have the same ratio as a and b, or a : b : : jra : fllB. mB b . For — = -, the same ratio. HIA A Cord. Hence, like parts of quantities have the same ratio as the wholes ; because the wholes are equimultiples of the like parts, or a and r are like parts of ma and ms. THEOREM LXVII. If four quantities, of the same kind, be proportionals ; they will be in proportion by alternation or permutation, or the antecedents will have the same ratio as the conse- quents*. * The author's object in these propositions was to simplify the doc- trine of ratios and proportions, by imagining that the antecedents and consequences may always be divided into parts that are commensura- ble. But it is known to mathematicians that there are certain quantities or magnitudes, such as the side and the diagonal of a square, which cannot possibly be divided in that manner by means of a common mf* sure. The theorems themselves are true, nevertheless, when applied to these incommensurabUs ; since no two quantities of the same kind can possibly be assigned, whose ratio cannot be expressed by that of two numbers, so near, that the difference shall be less than the least number that can be named. From the greater of two unequal magnitudes we may take, or suppose taken, its Hoif y from the remaining half, its half, THEOREMS. 821 Let a : b : : uiA : »iB ; then will a : wia : : b : jrb. For — = —i and — - = ~> both the same ratio. a 1 b 1 and so on, by continual bisections, until there shall at length be left a magnitude less than the least of two magnitudes ; or. indeed, less than the least magnitude that can be assigued ; and this principle furnishes a ground of reasoning. Or, somewhat differently, let a and b be two constant quantities, a and b two variable quantities, which we can render as small as we please, if we have an equality between i-j-fl, and b -}- 6, or, in other words, if the equation a+o = b-|-6 holds good whatever are the va- lues of a and 6, it may be divided into two others, a = b, between the constant quantities, and, a = b t between the variable quantities, and which latter must obtain for all their states of magnitude. For if, on the contrary, we suppose a = b ^ q, we shall have I — b = 6 — a = q, an absurd result ; since the quantities a and 6 being susceptible of diminishing indefinitely* their difference cannot always be = q. This is the principle which constitutes the method of limits. In general, one magnitude is called a limit of another, tchen we can make this latter ap- proach so near to the former, that their difference shall be less than any given magnitude* and yet so that the two magnitudes shall never become strictly equal. Let us here apply the principle to the demonstration of this proposi- tion, that the ratio of two angles acb, nop, is equal to that of the arcs, o6, np, comprised between their sides, and drawn from their respective summits as centres with equal radii. If the 'arcs pn f ba, are commensurable, their common measure 6m will be contained n times in iro, r times in ba ; so that we shall - have the equal ratios ¥L = -. Through each 6a r point of division, m, n', <fcc. draw the lines mc, n'c, &c. to the summits c, and o, the angles proposed will be divided into n, and r, equal angles, 6cm, men', poq, qor, be. We shall, therefore, have = *. Hence ' r ' boa r is =?—, since each of tbem is equal to the ratio -. bca 6a r If the arcs are incommensurable, divide one of them, 6a, into a num- ber r of equal parts, 6m, mn', be. and set off equal parts pq, qr } &c. upon the other arc pn; and let s be the point of division that falls nearest to n. Draw oss. Then, by the preceding, 6a, ps, being commensurable, we shall have ? — = the angle pos = pon + iros, arc ps = pn + ns. bca ba Therefore, pon ifos pn_ ns bca bca ~ ba ba' Here nos and ns are susceptible of indefinite variation, according as we change the common measure, 6m, of 6a ; they may, therefore, ba Vol. I. 42 832 GEOMETRY. Otherwise. Let a : b : : c : d ; then shall b r a : A C For, let - = - = r ; then a = bt, and c = dt : there- B D fore b A, C TT bI.d 1 -, and d =■ -. Hence - = and - = — , B D ' Whence it is evident that — = — (ax. 1), or b : a : : d : c» A C ' In a similar manner may most of the other theorems he demonstrated. THEOREM LXTUI. If four quantities be proportional ; they will be in proportion by inversion, or inversely. Let a : b : : ira : ms ; then will b : a : : mB : ma. For — = — , both the same ratio. mB b THEOREM LXIX. If four quantities be proportional ; they will be in proportion by composition and division. Let a : b : : mx : mB ; 1 Then will b ± a : a : : ms ± mx : mx, and b ± a : b : : kb dfc mx : mB. _ mx A , f»B B For — =* — — ; and — = — — . «iB rb m\ b±a mB±.m\ b rb a Carol. It appears from hence, that the sum of the greatest and least of four proportional quantities, of the same kind, exceeds the sum of the other two. For, since . - - - a : a + b : : m\ : mx + ms, where a is the least, and fflA + «»b the greatest ; then m + 1 . a + «b, the sum of the greatest and least, exceeds m + 1 . a + b, the sum of the two other quantities. THEOREM JLXX. If, of four proportional quantities, there be taken any equimultiples whatever of the two antecedents, and any equi- rendered as small as we please, while the other quantities remain the same. Consequently, by the nature of limits, as above explained, we have the equal ratios -~ = or for : bac ::pn: ba. bca oa r THEOREMS. 323 multiples whatever of the two consequents ; the quantities resulting will still be proportional. Let a : b : : «a : wib ; also, let pA and pmA be any equimultiples of the two antecedents, and qs and qmB any equimultiples of the two consequents ; then will .... px : qB : : pmA : qmB. For VOIL = 2? both the same ratio. pmA pA ■ t THEOREM LXXI. If there be four proportional quantities, and the two con- sequents be either augmented or diminished by quantities that have the same ratio as the respective antecedents ; the results and the antecedents will still be proportionals. Let a : b : : wa : mn, and n\ and nmA any two quan- tities having the same ratio as the two antecedents ; then will a : b ± nA mA : ms ± nmA. „ mi* ± nmA b ± tia , For = , -both the same ratio. IRA A THEOREM LXXII. If any number of quantities be proportional, then any one of the antecedents will be to its consequent, as the sum of all the antecedents, is to the sum of all the conse- quents. ^ * Let a : b : : mA : wib : : tiA : ub, &c. ; then will - - . a : b : : a + wa + iia : b + mB + iib, &c. ' B+ma+itB (l+ro+n)B b , For : = 77-; ; — r~ = — , the same ratio. A+roA+ftA (l+m+n)A A THEOREM LXXHI. If a whole magnitude be to a whole, as a part taken from the first, is to a part taken from the other ; then the re- mainder will be to the remainder, as the whole to the whole. Let a : b : : then will a b ; ; a a : b b. n % 324 GEOMETRY. m B— — B For as — , both the same ratio. m a A — A theorem: lxxiv. If any quantities be proportional ; their squares, or cubes, or any like powers, or roots, of them, will also be propor- tional. Let a : b : : i?ia : i?ib ; then will a* : b*' : : m n A n : m*B". _ m n B n b» . . . r or — — - = — , both the same ratio. ro*A w a" See also, th. vni. pa, 118. THEOREM LXXV. If there be two sets of proportionals ; then the products or rectangles of the corresponding terms will also be pro- portional. Let a : b : : wa : mB, and c : d : : nc : hd ; then will ac : bd : : mnAC : uitibd. mnBD bd . , . ror = — , both the same ratio. fflttAC AC THEOREM LXXVI. If four quantities be proportional ; the rectangle or pro- duct of the two extremes, will be equal to the rectangle or product of the two means. And the converse. Let a : b : : ota : ms ; then is a X wb = b Xj»a = hiae, as is evident. THEOREM LXXVII. If three quantities be continued proportionals ; the rect- angle or product of the two extremes, will be equal to the square of the mean. And the converse. Let a, «a, m 9 A be three proportionals, or a : tfiA : : mA ; m a A ; then is a X vp?k =* mV, a» \* owtaciSi. THEOREMS. 325 THEOREM LXXVI11. If any number of quantities be continued proportionals the ratio of the first to the third, will be duplicate or the square of the ratio of the first and second ; and the ratio of the first and fourth will be triplicate or the cube of that of the first and second ; and so on. ' Let a, mA, jti'a, jti 3 a, dec. be proportionals ; then is — = — ; but -4- = —7 ; and -4- = \ ; dtc. mA m m a A mr wta mr THEOREM LXXIX. Triangles, and also parallelograms, having equal altitudes, are to each other as their bases. Let the two triangles adc, def, have I CK. the same altitude, or be between the same parallels ae, ce ; then is the surface of the triangle adc, to the surface of the triangle def, as the base ad is to the AB BOH i s base de. Or, ad : de : : the triangle adc : the triangle def. For, let the base ad be to the base de, as any one num- ber m (2), to any other number n (3) ; and divide the respective bases into those parts, ab, bd, do, gm, he, all equal to one another ; and from the points of division draw the lines bc, fg, fh, to the vertices c and f. Then will these lines divide the triangles adc, dkf, into the same number of parts as their bases, each equal to the triangle abc, because those triangular parts have equal bases and altitude (cor. 2, th. 25) ; namely, the triangle abc equal to each of the triangles bdc, dfg, gfh, hfe. So that the tri- angle adc, is to the triangle dfe, as the number of parts m (2) of the former, to the number n (3) of the latter, that is, as the base ad to the base de (def. 79)*. In like manner, the parallelogram adki is to tho parallelo- gram defk, as the base ad is to the base de ; each of these having the same ratio as the number of their parts, m to ft* Q. E. D. * If the bases ad, de, of two triangles that have a common ^%t\VL%, aire incommensurable to each other, the ratio of tto trvi&tjto Ha* wjM^ gUnd'wg, equal to that of their bases. 926 GEOMETRY* THEOREM LXXX. Triangles, and also parallelograms having equal bases, are to each other as their altitudes. Let abc, bep, be two triangles having the equal bases ab, be, and whose altitudes are the perpendicu- lars co, fh ; then will the triangle abc : the triangle bef : : co : fh. For, let bk be perpendicular to ab, and equal to cg ; in which let there be taken bl = fh ; drawing ak and al. Then triangles of equal bases and heights being equal (cor. 2, th. 25), the triangle abk is = abc, and the triangle abl = bef. But, considering now abk, abl, as two tri- angles on the bases bk, bl, and having the same altitude ab, these will be as their bases (th. 79), namely, the triangle abk : the triangle abl : : bk : : bl. But the triangle abk = *abo, and the triangle abl = bef, also bk = cg, and bl = fh. Theref. the triangle abc : triangle bef : : cg : fh. And since parallelograms are the doubles of these triangles, having the same bases and altitudes, they will likewise have to each other the same ratio as their altitudes, q. e. d. Corel. Since, by this theorem, triangles and parallelo- grams, when their bases are equal, arc to each other as their altitudes; and by the foregoing one, when their altitudes are equal, they are to each other as their bases ; therefore uni- For, first, if possible, let the triangle kcd be to the triangle acd, not as ed to ad, but as some other line bd greater than ed, is to AD. Let ah be a part, or measure of ad. less than be, and let di be that multiple frf ah, which least exceeds de, and which by the note B I E .D MA to th. 67, may be made as small as we please. Let cb, ci, be drawn, i evidently falls between e and b, because (by hyp.) ei is less than am. But icd : acd : : id : ad, by th. 79. Also, by hyp. ecd : acd : : bd : ad, greater than the. ratio of id : ad, or of Ico : acd ; and consequently, ecd is greater than icd : which is impoui- - bU, By a like reasoning it may be shown, that ecd cannot be to acd, .as a line feu than ed, is to ad. Consequently, it must be ecd : acd : : *D I ad. Similar reasoning, founded upon vYte \rc*wdvcv^ tvoU, &$oUes alto te the case of parallelograms. THEOREM!. 327 versally, when neither are equal, they are to each other in the compound ratio, or as the rectangle or pibduct of their bases and altitudes. c ~~A Q B r d THEOREM LXXXI. If four lines be proportional ; the rectangle of the ex- tremes will be equal to the rectangle of the means. And, conversely, if the rectangle of the extremes, of four lines, be equal to the rectangle of the means, the four lirttes, taken alternately, will be proportional. Let the four lines a, b, c, d, be A proportionals, or a : b : : c : d ; B then will the rectangle of a and d be ^ equal to the rectangle of b and c ; or the rectangle a . d = b . c. For, let the four lines be placed with their four extremities meeting in a common point, forming at that point four right angles ; and draw lines parallel to them to complete the rectangles p, q, r, where p is the rectangle of a and d, a the rectangle of b and c, and r the rectangle of b and d. " Then the rectangles r and r, being between the same parallels are to each other as their bases a and b (th. 79) ; and the rectangles q and r, being between the same parallels, are to each other as their bases c and d. But the ratio of a to b, is the same as the ratio of c to n, by hypothesis : therefore the ratio of p to a, is the same as the ratio of q to r ; and consequently the rectangles f and q are equal. Q. E. D# Again, if the rectangle of a and d, be equal to the rectangle of b and c ; these lines will be proportional, or a : b : : c : D.j For, the rectangles being placed the same as before : then, because parallelograms between the same parallels, are to one another as their bases, the rectangle r : r : : a : b, and q : r : : c : d. But as p and a are equal, by supposition, they have the same ratio to n, that is, the ratio of a to b is equal to the ratio of c to d, or a : b : : c : d. q. e. d. Carol. 1. When the two means, namely, the second and third terms, are equal, their rectangle becomes a square of the second term, which supplies the place of both the second and third, And hence it follows, that when three lines an 828 GEOMETRY. proportionals, the rectangle of the two extremes is equal to the square of the mean ; and, conversely, if the rectangle of the extremes be equal to the square of the mean, the three lines are proportionals. Corel. 2. Since it appears, by the rules of proportion in Arithmetic and Algebra, that when four quantities are pro* portional, the product of the extremes is equal to the product of the two means ; and, by this theorem, the rectangle of the extremes is equal to the rectangle of the two means ; it fol- lows, that the area or space of a rectangle is represented or expressed by the product' of its length and breadth multiplied together. And, in general, a rectangle in geometry is similar to the product of the measures of its two dimensions of length and breadth, or base and height. Also, a square is similar to, or represented by, the measure of its side multiplied by itself. So that, what is shown of such products, is to be un- derstood of the squares and rectangles. CoroL 3. Since the same reasoning, as in this theorem, holds for any parallelograms whatever, as well as for the rectangles, the same property belongs to all kinds of paral- lelograms, having equal angles, and also to triangles, which are the halves of parallelograms ; namely, that if the sides about the equal angles of parallelograms, or triangles, be reciprocally proportional, the parallelograms or triangles will be equal ; and, conversely, if the parallelograms or triangles be equal, their sides about the equal angles will be recipro- cally proportional. Corol. 4. Parallelograms, or triangles, having an angle in each equal, are in proportion to each other as the rectangles of the sides which arc about these equal angles. THEOREM LXXXII. If a line be drawn in a triangle parallel to one of its sides, it will cut the other two sides proportionally. £ / Let de be parallel to the side bc of the A triangle abc ; then will ad : db : : ae : ec. /\ For, draw be and cd. Then the tri- X)/ \j£ angles dbe, dce, are equal to each other, /x^\A because they have the same base de, and ^\ are between the same parallels de, i:c B C (th. 25). But the two triangles, ade, bde, on the bases ad, db, have the same altitude ; and the two triangles ade, cde, on the bases ae, ec, have also the same THEOREMS. 329 altitude ; and because triangles of the same altitude are to each other as their bases, therefore the triangle ade : bde : : ad : db, and triangle ade : cde : : ae : ec. But bde is ~ cde ; and equals roust have to equals the same ratio ; therefore ad : db : : ae : ec. a* e. d. Carol. Hence, also, the whole lines as, ao, are propor- tional to their corresponding proportional segments (corol. th. 60), viz. ab : ac : : ad : ae, and ab : ac : : bd : ce. theorem lxxxiii. A Line which bisects any angle of a triangle, divides the opposite side into two segments, which are proportional to the two other adjacent sides. Let the angle acb, of the triangle abc, be bisected by the line cd, making the angle r equal to the angle s : then will the segment ad be to the segment db, as the side ac is to the side cb. Or, - - - - ad : db : : ac : cb. For, let be be parallel to cd, meeting AC produced at e. Then, because the line bc cuts the two parallels cd, be, it makes the angle cbe equal to the alter- nate angle s (th. 12), and therefore also equal to the angle r, which is equal to * by the supposition. Again, because the line ae cuts the two parallels uc, be, it makes the angle E equal to the angle r on the same side of it (th. 14). Hence, in the triangle bck, thfjangles b and e, being each equal to the angle r, are equal to each other, and consequently their opposite sides cb, ce, are also equal (th. 3). But now, in the triangle abe, the line cd, being drawn parallel to the side be, cuts the two other sides ab, ae, pro- portionally (th. 82), making ad to db, as is ac to ce or to its equal cb. q. e. d. Vol. L 43 830 GEOXETXT. THEOREM LXXXIV. Euuiangular triangles are similar, or have their like sides proportional- Let abc, def, be two equiangular tri- angles, having the angle a equal to the angle d, the angle b to the angle e, and consequently the angle c to the angle f ; then will ab : ac : : de : df. A. B For, make dg = ab, and dh = ac, and F join gii. Then the two triangles abc, dgh, having the two sides ab, ac, equal to the two dg, dii, and the contained an- gles a and d also equal, are identical, or equal in all respects (th. 1), namely, the angles b and c are equal to the angles g and n. But the angles b and c are equal to the angles e and f by the hypo- thesis ; therefore also the angles g and 11 are equal to the angles e and f (ax. 1), and consequently the line qh ia paral- lel to the sido ef (cor. 1, th. 14). Hence then, in the triangle def, the line gh, being parallel to the side ef, divides the two other sides proportionally, making do : dii : : de : df (cor. th. 82). But dg and dh are equal to ab and ac ; therefore also - - - - • ab : ac : : de : df. a. e. d. THEOREM LXXXV. Triangles which have their sides proportional, arc equi- angular. In the two triangles abc, def, if ab : de : : ac : df : : bc : ef ; the two triangles will have their corresponding /\ angles equal. For, if the triangle abc bc not equian- A B gular with the triangle dkf, suppose some <*• P other triangle, as deg, to be equiangular with abc. But this is impossible : for if the two triangles abc, deg, were equian- gular, their sides would be proportional (th. 84). So that, ab being to de as ac to dg, and ab to de as bc to eg, it follows that do and eg, being fourth proportionals to the same three quantities, as well as the two df, ef, the Conner, dg, eg, would be equal THEOREMS. 881 to the latter, df, bp* Thus, then, the two triangles def, deg, having their three sides equal, would be identical (th. 5) ; which is absurd, since their angles are unequal. THEOREM LXXXVI. Triangles, which have an angle in the one equal to an angle in the other, and the sides about these angles pro- portional, are equiangular. Let ABO, def, be two triangles, having the angle a = the angle d, and the sides ab, ac, proportional to the sides db, dp : then will the triangle abc be equiangular with the triangle dep. For, make do = ab, and dh = ac, and join on. Then, the two triangles abc, doh, having two sides equal, and the contained angles a and d equal, are identical and equiangular (th. 1), having the angles o and h equal to the angles b and c. But, since the sides dg, du, are proportional to the sides dk, dp, tho line air is parallel to ef (th. 82) ; hence the angles e and f are equal to the angles o and h (th. 14), and consequently to their equals b and c. Q. e. d. [See fig. th. lxxxiv.] THEOREM LXXXV1I. In a right-angled triangle, a perpendicular from the right angle, is a mean proportional between the segments of the hypothenuse; and each of the sides, about the right angle, is a mean proportional between the hypothenuse and the adjacent segment. C Let abc be a right-angled triangle, and cd a perpendicular from the right angle c to the hypothenuse ab ; then will A D B cd be a mean proportional between ad and db ; ac a mean proportional between ab and ad ; bc a mean proportional between ab and bd. Or, ad : cd : : cd : db ; and ab : bc : : bc : bd ; and ab : ac : : ac : ad. For, the two triangles, abc, adc, having the right angles at c and d equal, and the angle a common, have their third angles equal, and are equiangular, (cor. 1, th. 17). In, tikfe manner, the two triangles abc, bdc, having xY&TUjgoX wct^e* 382 GEOMETRY. at c and d equal, and the angle b common, hare their third angles equal, and are equiangular. Hence then, all the three triangles, abc, adc, bdc, being equiangular, will have their like sides proportional (th. 84) ; viz. ad : CD : : cd : db; and xr: ac : : ac : ad; and ab : bc : : bc : bd. q. e .d. Carol. 1. Because the angle in a semicircle is a right angle (th. 52) ; it follows, that if, from any point c in the periphery of the semicircle, a perpendicular be drawn to the diameter ab ; and the two chords ca, cb, be drawn to the extremities of the diameter : then are ac, bc, cd, the mean proportionals as in this theorem, or (by th. 77), .... CD* = AD . DB ; AC 3 = AB . AD ; and BC 1 = AB . BD. Corol. 2. Hence ac 3 : bc 9 : : ad : bd. Corol. 3. Hence we have another demonstration of th. 34. For since ac 3 = ab . ad, and bc 3 = ab . bd ; By addition ac 3 + bc 9 = ab (ad + bd) = ab 3 . THEOREM LXXXVIII. Equiangular or similar triangles, are to each other as the squares of their like sides. Let abc, def, be two equiangular triangles, ab and de being two like sides : then will the triangle abc be to the triangle def, as the square of ab is to the square of dk, or as ah 2 to de 2 . For, the triangles being similar, they have theirlikesidesproportional (th.84), and are to each other as the rectangles of the like pairs of their sides (cor. 4, th. 81) ; theref. ab : de : : ac : df (th. 84), and ab : de : : ab : de of equality : theref ab 3 : de 9 : : ab • ac : de . df (th. 75). But A abc : A def : : ab . ac : de . df (cor. 4, th. 81), theref. A abc : A def : : ab 3 : de 9 . a. s. d. a £ THEOREMS • 383 THEOREM LXXX1X. All similar figures are to each other, as the squares of their like sides. Let abcde, fohik, be any two similar figures, the like sides being ab, fg, and bc, oh, and so on in the same order: then will the figure abcde be to the figure fohik, as the square of ab to the square oft or as ab 3 to fo 9 . For, draw be, bd, ok, 01, dividing the figures into an equal number of triangles,* by lines from two equal angles b and g. The two figures being similar (by suppos.), they are equi- angular, and have their like sides proportional (def. 67). Then, since the angle a is = the angle f, and the sides ab, ae, proportional to the sides fo, fk, the triangles abe, fok, are equiangular (th. 88). In like manner, the two triangles bcd ghi, having the angle c = the angle h, and the sides bc, cd, proportional to the sides gh, hi, are also equiangular. Also, if from the equal angles aed, fkj, there be taken the equal angles aeb, fko, there will remain the equals bed, gki ; and if from the equal angles cde, hik, be taken away the equals cdb, hio, there will remain the equals bde, oik ; so that the two triangles bde, gik, having two angles equal, are also equiangular. Hence each trian- gle of the one figure, is equiangular with each corresponding triangle of the other. But equiangular triangles arc similar, and are proportional to the squares of their like sides (th. 88). Therefore the A abb : A fgk : : ab 3 : fg 3 , and A bcd : A ohi : : nc 3 : gii 2 , and A bde : A gik : : de 3 : iK a . But as the two polygons are similar, their like sides are proportional, and consequently their squares also propor- tional ; so that all the ratios ab 2 to fg 3 , and bc 9 to oh 9 , and dm 9 to ik 9 , are equal among themselves, antlxonsequently the corresponding triangles also, abe to fob* and bcd to ohi, and bde to gik, have all the same ratio, viz. that of ab 9 to fg 3 : and hence all the antecedents, or the figure abcde, have to all the consequents, or the figure fghik, the same ratio, viz. that of ab 3 to fo 9 QCh. T&y o.. i>» 384 GEOMETRY* THEOREM XC. Similar figures inscribed in circles, have their like sides, and also their whole perimeters, in the same ratio as the diameters of the circles in which they are inscribed. . Let ABCDE, FGHIK, T> 1 be two similar figures, ^^>J* inscribed in the circles whose diameters are al andFM; then will each side ab, bc, die. of the one figure be to the like side gf, oh, &c. of the other figure, or the whole perimeter ab + bc + die. of the one figure, to the whole perimeter fg + oh + dec. of the other figure, as the diameter al to the diameter fm. For, draw the two corresponding diagonals ac, fh, as also the lines bl, gm. Then, since the polygons are similar, they are equiangular, and their like sides have the same ratio (def. 67) ; therefore the two triangles abc, fgh, have the angle b = the angle g, and the sides ab, bc, proportional to the two sides fg, Gir, consequently these two triangles are equiangular (th. 80), and have the angle acb = fhg. But the angle acb = alb, standing on the same arc ab ; and the angle fhg = fmg, standing on the same arc fg ; therefore the angle alb = fmg (ax. 1). And since the angle abl = fgm, being both right angles, because in a semicircle ; therefore the two triangles abl, fgm, having two angles equal, are equiangular ; and consequently their like sides are proportional (th. 84) ; hence ab : fg : : the diameter al : the diameter fm. In like manner, each side bc, cd, dec. has to each side oh, in, dec. the same ratio of al to fm ; and consequently the sums of them are still in the same ratio, viz. ab + bc + cd, &c. : fg + gh + hi, dec. : : the diam. al : the diam. fm (th. 72). Q. E. D. THEOREM XCI. Similar figures inscribed in circles, are to each other a» the squares of the diameters of those circles. Let abcde, foiiis, be two similar figures, inscribed in the circles whose diameters are al and fm ; then the surface of the polygon abode will be to the surface of the polygon FGHIK, as AL 3 to FM 3 . THEOREMS. 385 For, the figures being similar, are to each other as the squares of their like sides, ab 2 to Ft; 3 (th. 86). But, by the last theorem, the sides ab, fg, are as the diameters al, fm ; and therefore the squares of the sides ab 3 to fg 3 , as the squares of the diameters al* to fx 1 (th. 74). Consequently the polygons abcde, fohik, are also to each other as the squares of the diameters al 3 to fm 3 (ax. 1). u. e. d. [See fig. th. xc] THEOREM XCII. The circumferences of all circles are to each other as their diameters 1 '. * The truth of theorems 92, 93, and 94, may be established more satisfactorily than in the text, upon principles analogous to those of the two last notes. Theorem. The area of any circle abd is equal to the rectangle con- tained by the radius, and a straight line equal to half the circumference. If not, let the rectangle be lest than the circle aid, or equal to the circle fhh : and imagine ed drawn to touch the interior circle in f, and meet the circumference abd in e and d. Join cd, cutting the arc of the interior circle in k. Let fii be a quadrantal arc of the inner circle, and from it take its half, from the remainder Us half, and so on, until an arc fi is obtained, less than fit. Join ci, produce it to cut ed in l, and make fo = fl : so snail lo be the side of a regular polygon circumscribing the circle frh. It is manifest that this polygon is less than the circle abd, because it is contai.ud within it Because the trinngle gcl is half the rectangle of base gl and altitude cf, the whole polygon of which gcl is a constituent triangle, is equal to half the rectangle whose base is the perimeter of that polygon and altitude cf- But that perimeter is less than the circumference abd, because each portion of it, such as gl, Is less than the corresponding arch of circle having radius cl, and there- fore, * fortiori, less than the corresponding arch of circle with radius ca. Also ce is less than ca. Therefore the polygon of which one side is ol, is less than the rectangle whose base is half the circumference abd and altitude ca; that is, (by hyp.) less than the circle fwh, which it conta *s: which is absurd. Therefore, the rectangle under the radius and half the circumference is not less than the circle abd. And by a similar process it may be shown that it is not greater. Consequently, H is equal to that rectangle, q. e. d. Theorem. The circumferences of two circles abd, abd, are as their radii. If possible, let the radius ac, be to the radius ac, as the cir- cumference abd to a circum- ference ihk less than abd. Draw the radius cic, and the straight line/tg- a chord to the circle abd, and a tangent to the circle ihk in i. From eb, a quarter of the circumference ofa&t, take 336 GEOMETRY. Let d, d y denote the diameters of two circles, and c, e 9 their circumferences ; then will d : d : : c : c, or d : c : : d : c. For (by theor. 90), similar polygons inscribed in circles have their perimeters in the same ratio as the diameters of those circles. Now as this property belongs to all polygons, whatever the number of the sides may be ; conceive the number of the sides to be indefinitely great, and the length of each inde- finitely small, till they coincide with the circumference of the circle, and be equal to it, indefinitely near. Then the perimeter of the polygon of no infinite number of sides, is the same thing as the circumference of the circle. Hence it appears that the circumferences of the circles, being the same as the perimeters of such po /gone, are to each other in the same ratio as the diameters of the circles, a- e. d. THEOREM XCIII. The areas or spaces of circles, arc to each other as the squares of their diameters, or of their radii. Let a, a, denote the areas or spaces of two circles, and d, d, their diameters ; then a : a : : d 3 : d 2 . For (by theorem 91) similar polygons inscribed in circles are to each other as the squares of the diameters of the circles. away its half, and then the half of the remainder, and so on, until there be obtained an arc ed less than eg; and from d draw ad parallel to//, it will be the side of a regular polygon inscribed in the circle abd, yet evidently greater than the circle ihk, because each of its constituent tri- angles, as acd contains the corresponding circular sector eno. Let ad be the side of a similar polygon inscribed in the circle adb, and join ac, cd, similarly to ac, ed. The similar triangles acd, acd t give ac : ac : : ad : ad, and : : perim. of polygon in abd : perim. of polygon in abd. But, by the preceding theorem, ac : ac : : circumf. abd : circumf. abd. The perimeters of the polygons are, therefore, as the circumferences of the circles. But, this is impossible ; because, (by hyp.) the perim. of polygon in abd is less than the circumf. ; while, on the contrary, the perim. of polygon in adb is greater than the circumf. ihk. Conse- quently, ac is not to ac, as circumf. adb, to a circumference Ust than adb. And by a similar process it may be shown, that ac is not to ac, as the circumf. dbd, to a circumference less than abd. Therefore ac : ac :: circumf. abd : circumf. abd. e. d. Corol. Since by this theorem, we have c : c : : r : r, or, if c = »R, c = *r ; and, by the former, area (a) : area (a) : : £rc : |rc : we btvo a : a : : fan? : far* : : R* : t* •. d* c» : A THEOREMS. 837 Hence, conceiving the number of the sides of the polygons to be increased more and more, or the length of the sides to become less and less, the polygon approaches nearer and nearer to the circle, till at length, by an infinite approach, they coincide, and become in effect equal ; and then it fol- lows, that the spaces of the circles, which are the same as of the polygons, will be to each other as the squares of the diameters of the circles, q. e. d. Carol. The spaces of circles are also to each other as the squares of the circumferences ; since the circumferences are in the same ratio as the diameters (by theorem 92). THEOREM XCIV. % Thb area of any circle, is equal to the rectangle of half its circumference and half its diameter. Conceive a regular polygon to be in- scribed in the circle ; and radii drawn to all the angular points, dividing it into as many equal triangles as the polygon has sides, one of which is abc, of which the altitude is the perpendicular cd from the centre to the base ab. Then the triangle abc, being equal to a rectangle of half the base and equal altitude (th. 26, cor. 2), is equal to the rectangle of the half base ad and the altitude cd ; con- sequently the whole polygon, or all the triangles added to- gether which compose it, is equal to the rectangle of the common altitude cd, and the halves of all the sides, or the half perimeter of the polygon. Now conceive the number of sides of the polygon to be indefinitely increased ; then will its perimeter coincide with the circumference of the circle, and consequently the altitude cd will become equal to the radius, and the whole polygon equal to the circle. Consequently the space of the circle, or of the polygon in that state, is equal to the rectangle of the radius and half the circumference, a. e. d. Vol. I. 44 338 OF PLANES AND SOLIDS. DEFINITIONS. Def. 88. The Common Section of two Planes, is the line in which they meet, or cut each other. 80. A Line is Perpendicular to a Plane, when it is per* pendicular to every line in that plane which mmka it* 90. One Plane is Perpendicular to Another, when eray line of the one, which is perpendicular to the line of their common section, is perpendicular to the other. 91. The Inclination of one Plane to another, or the angle they form between them, is the angle contained by two lines, drawn from any point in the common section, and at right angles to the same, one of these lines in each plane. 92. Parallel Planes, are such as being produced ever so far both ways, will never meet, or which are every where at an equal perpendicular distance. 93. A Solid Angle, is that which is made by three or more plane angles, meeting each other in the same point. 94. Similar Solids, contained by plane figures, are such as have all their solid angles equal, each to each, and are bounded by the same number of simitar planes, alike placed. 95. A Prism, is a solid whose ends are parallel, equal, and like plane figures, and its sides, connecting those ends, are parallelograms. 96. A Prism takes particular names according to the figure of its base or ends, whether triangular, square, rectangular, pentagonal, hexagonal, &c. 97. A Right or Upright Prism, is that which .has the planes of the sides perpendicular to the planes of the erifa or base. 98. A Parallelopiped, or Parallelopipedon, is a prism bounded by six parallelograms, every opposite two of which are equal, alike, and pa- rallel. DEFINITIONS. 889 A DO. A Rectangular Parallelopidedon, is that whose bound- ins planes are all rectangles, which are perpendicular to each other. 100. A Cube, is a square prism, being bound- ed by six equal square sides or faces, and are perpendicular to each other. 101. A Cylinder is a round prism, having circles for its ends ; and is conceived to be form- ed by the rotation of a right line about the cir- cumferences of two equal and parallel circles, always parallel to the axis. 102. The Axis of a Cylinder, is the right line joining the centres of the two parallel circles, about which the figure is described. 108. A Pyramid, is a solid, whose base is any right-lined plane figure, and its sides triangles, having all their vertices meeting' together in a point above the base, Called the vertex of the pyramid. 104. A pyramid, like the prism, takes particular names from the figure of the base. 105. A Cone, is a round pyramid, having a circular base, and is conceived to be generated by the rotation of a right line about the circum- ference of a circle, one end of which is fixed at a point above the plane of that circle. 106. The Axis of a cone, is the right line, joining the vertex, or fixed point, and the centre of the circle about which the figure is described. 107. Similar Cones and Cylinders, are such as have their altitudes and the diameters of their bases proportional. 108. A Sphere, is a solid bounded by one curve surface, which is every where equally distant from a certain point within, called the Centre. It is conceived to be generated by the rotation of a semicircle about its diameter, which re- mains fixed. 109. The Axis of a Sphere, is the right line about which the semicircle revolves ; and the centre j is the same as that of the revolving semicircle. 110. The Diameter of a Sphere, is any right line passing through the centre, and terminated both ways by the surface. 111. The Altitude of a solid, is the perpendicular drawn from the vertex to the opposite side or base. 840 GEOMETRY. THEOREM XCV. A perpendicular is the shortest line which can be drawn from any point to a plane. Let ab be perpendicular to the plane db ; then any other line, as ac, drawn from the same point a to the plane, will be longer than the line ab. In the plane draw the line bo, joining d thepoints m . Then, because the line ab is perpendi- cular to the plane de, the angle b is a right angle (de£ 90), and consequently greater than the angle c ; therefore the line ab, opposite to the less angle, is less than any other line ac, opposite the greater angle (th. 21). q. e. d. THEOREM ZCVI. A perpendicular measures the distance of any point from a plane. The distance of one point from another is measured by a right line joining them, because this is the shortest line which can be drawn from one point to another. So, also, the dis- tance from a point to a line, is measured by a perpendicu- lar, because this line is the shortest which can be drawn from the point to the line. In like manner, the distance from a point to a plane, must be measured by a perpendicu- lar drawn from that point to the plane, because this is the shortest line which can be drawn from the point to the plane. THEOREM XCVII. The common section of two planes, is a right line. Let acbda, aebfa, be two planes cut- ting each other, and a, b, two points in which the two planes meet ; drawing the line ab, this line will be the common in- tersection of the two planes. For, because the right line ab touches the two planes in the points a ajd b, it THEOREMS. 841 touches them in all other points (def. 20) ; this line is there- fore common to the two planes. That is, the common in- tersection of the two planes is a right line. a. b. d. Cord. From the same point in a plane, there cannot be drawn two perpendiculars to the plane on the same side of it. For, if it were possible, each of these lines would be perpendicular to the straight line which is the common inter- section of the plane and another plane passing through the two perpendiculars, which is impossible. THEOREM XCVIII. If a line be perpendicular to two other lines, at their com- mon point of meeting ; it will be perpendicular to the plane of those lines. Let the line ab make right angles with -q the lines ac, ad ; then will it be per- pendicular to the plane cde which passes through these lines. y^^Ty^ If the line ab were not perpendicular to E^^sT/) the plane cde, another plane might pass through the point a, to which the line ab would be perpendicular. But this is im- possible ; for, since the angles bac, bad, are right angles, this other plane must pass through the points c, d. Hence, this plane passing through the two points a, c, of the line ac, and through the two points a, d, of the line ad, it will pass through both these two lines, and therefore be the same plane with the former, a. e. d. THEOREM XCIX. If two planes cut each other at right angles, and a line be drawn in one of the planes perpendicular to their common intersection, it will be perpendicular to the other plane. Let the two planes acbd, aebf, cut each other at right angles ; and the line co be perpendicular to their common sec- tion ab ; then will co be also perpendicu- lar to the other plane aebf. For, draw eg perpendicular to ab. Then, because the two lines, ac, ox, are perpendicular to the common intersection 53 GEOXETftY. ab, the angle cge is the angle of inclination of the two planes (def. 92). But since the two planes cut each other perpendicularly, the angle of inclination oge is a right angle. And since the line cg is perpendicular to the two Jines oa, oe, in the plane aebf, it is therefore perpendicular to that plane (th. 98). a. e. d. Carol. 1. Every plane, ACB,'passing through a perpendicular <jg to another plane aebf, wiH be perpendicular to that other plane. For, if acb be not perpendicular to the plane aebf, some other plane on the same side of aebf, and passing through ab, will be perpendicular to it. Then, if from the point g a straight line be drawn in this other plane perpendicular to the common intersection, it will be perpendicular to the plane abbf. But (hyp.) co is perpendicular to that plane. There- fore, there will be, from the same point g, two perpendicu- lars to the same plane on the same side of it, which is im- possible (cor. 97). Carol. 2. If from any point g in the common intersection of the two planes acb and aebf perpendicular to each other, a line be drawn perpendicular to either plane, that line will be in the other plane. THEOBEM C. If two lines be perpendicular to the same plane, they will be parallel to each other. Let the two lines ab, cd, be both per- pendicular to the same plane ebdf ; then will ab be parallel to cd. . ■ ■■ For, join b, d, by the line bd in the El B ' h > B plane. The plane abd is perpendicular to I ■■ • J the plane ef (cor. 1, th. 99) ; and therefore the line cd, drawn from a point in the common intersection of the two planes, perpendicular to ef, will be in the plane abd (cor. 2, th. 99). But, because the lines ab, cd, are perpendicular to the plane ef, they are both perpendicular to the line bd in that plane, and they have been proved to be in the same plane abd ; consequently, they are paral- lel to each other (cor. th. 13). a. e. d. Carol. If two lines be parallel, and if one of them be per- pendicular to any plane, the other will also be perpendicular to the same plane. THE0RBMS. 348 THEOREM CI. If one plane meet another plane, it will make angles with that other plane, which are together equal to two right angles. Let the plane acbd meet the plane aebf ; these planes make with each other two angles whose sum is equal to two right angles. For, through any point g, in the common section ab, draw cd, ef, perpendicular to ab. Then, the line cg makes with ef two angles together equal to two right angles. But these two angles are (by def. 92) the angles of inclina* tion of the two planes. Therefore the two planes make an- gles with each other, which are together equal to two right angles. Corel. In like manner, it may be demonstrated, that planes which intersect have their vertical or opposite angles equal ; also, that parallel planes have their alternate angles equal ; and so on, as in parallel lines. • % THEOREM CII. If two planes be parallel to each other ; a line which is perpendicular to one of the planes, will also be perpendicular to the other. Let the two planes cd, ef, be parallel, and let the line ab be perpendicular to the plane cd ; then shall it also be perpendi- cular to the other plane ef. For, from any point g, in the plane ef, draw gh perpendicular to the plane cd, and draw ah, bg. Then, because ba, gh, are both perpendicular to the plane cd, the angles a and h are both right angles. And because the planes cd, ef, are parallel, the perpendiculars ba, gh, are equal (def. 93). Hence it follows that the lines bg, ah, are parallel (def. 9). And the line ab being per- pendicular to the line ah, is also perpendicular to the parallel fine bg (cor. th. 12). In like manner it is proved, that the line ab is perpendicu- lar to all other lines which can be drawn from Ita^wsiX Vto. 844 OEOXBTET. the plane ef. Therefore the line ab is perpendicular to the whole plane ef (def. 00). q. b. d. theobek cm. If two lines be parallel to a third line, though not in the same plane with it ; they will be parallel to each other. Let the lines ab, cd, be each of them parallel to the third line ef, though not in the same plane with it ; then will ab be pa- J) rallel to cd. For, from any point o in the line ef, let oh, 01, be each perpendicular to ef, in the H planes eb, id, of the proposed parallels. Then, since the line ef is perpendicular to the two lines gh, 01, it is perpendicular to the plane ghi of those lines (th. 98). And because ef is perpendicular, to the plane cm, its parallel ab is also per- pendicular to that plane (cor. th. 99). For the same reason, the line cd is perpendicular to the same plane ghi. Hence, because the two lines ab, cd, are perpendicular to the same plane, these two lines are parallel (th. 99). q. e. d. THEOREM CIV. If two lines, that meet each other, be parallel to two other lines that meet each other, though not in the same plane with them ; the angles contained by those lines will be equal. ^*et the two lines ab, bc, be parallel to B the two lines, de, ef ; then will the angle ^ abc beequal to the angle def. ^ For, make . the lins ab, bc, de, ef, all equal to each other, and join ac, df, ad, be, cf. Then, the lines ad, be, joining the equal £> and parallel lines ab, de, are equal and parallel (th. 24). For the same reason, cf, be, are equal and parallel. Therefore ad, cf, are equal and parallel (th. 15) ; and consequently also ac, df (th. 24). Hence, the two triangles abc, def, having all their sides equal, each to each, have their angles also equal, and consequently the angle abc == the angle def. a. e. d. THEOREMS. 845 THEOREM CV. Tax sections made by a plane cutting two other parallel planes, are also parallel to each other. Let the two parallel planes ab, cd, be cut by the third plane efhg, in the lines sr, oh : these two sections kf, oh, will be parallel. Suppose eo, fh, be drawn parallel to each other in the plane efhg ; also let ei, FK, be perpendicular to the plane cd ; and let ig, kh, be joined. Then eg, fh, being parallels, and ei, fk, being both perpendicular to the plane cd, are also parallel to each other (th. 99) ; consequently the angle hfk is .equal to the angle gei (th. 104). But the angle fkh is also equal to the angle big, being both right angles ; therefore the two triangles are equiangular (cor. 1, th. 17) ; and the sides fk, ei, being the equal distances between the parallel planes (def. 93), it fol- lows that the sides fh, eg, are also equal (th. 2). But these two lines are parallel (by suppos.), as well as equal; con" sequently the two lines ef, gh, joining those two equal parallels, are also parallel (th. 24). a. e. d. theorem cvi. If any prism be cut by a plane parallel to its base, the sec- tion will be equal and like to the base. Let ag be any prism, and il a plane parallel to the base, ac ; then will the plane il be equal and like to the base ac, or the two planes will have all their sides and all their ancles equal. For, the two planes ac, il, being parallel by hypothesis ; and two parallel planes, cut by a third plane, having parallel sections (th. 105) ; therefore nt is parallel to ab, and xjl to bc, and lm to cd, and im to ad. But ai and bk are parallels (by def. 95) ; consequently ak is a parallelogram ; and the opposite sides ab, ik, are equal (th. 22). In like manner, it is shown that kx is = bc, and lm = cd, and im s= ad, or the two planes ac, il, are mutually equilateral. But these two planes having their corresponding side* ^an&&, Vol. I. 45 846 GEOMETRY. have the angles contained by them also equal (th. 104), namely, the angle a = the angle i, the angle b « the angle k, the angle c = the angle l, and the angle » = the angle m. So that the two planes ac, il, have all their corresponding sides and angles equal, or they are equal mwA like. a. b. d. THBOBKM CVII* If a cylinder be cut by a plane parallel to its base, the section will be a circle, equal to the base. Let af be a cylinder, and ghi any sec- tion parallel to the base abc ; then will 0111 he a circle, equal to abc. Forf let the planes kb, kf, pass through the .axis of the cylinder mk, and meet the section ghi in the three points h, i, l ; and join the points as in the figure. Then, since kl, ci, are parallel (by def. 102) ; and the plane ki, meeting the two parallel planes abc, ghi, makes the two sections kg, li, pa- rallel (th. 105) ; the figure klic is therefore a parallelogram, and consequently has the opposite sides li, kc, equal, where kc is a radius of the circular base. In like manner it is shown that lh is equal to the radius kb; and that any other lines, drawn from the point L to the circumference of the section ghi, are all equal to radii of the base ; consequently ghi is a circle, and equal to abc. B. D. THEOREM CV1II. All prisms and cylinders, of equal bases and altitudes, ate equal to each other. Let ac, df, be two prisms, and a cylinder, on equal bases, ab, de, and having equal alti- tudes bc, ef ; then will P the solids ac, df, be equal*. For, let pq, bs, be 7 ©J S It E * This, and some other demonstrations relative to solids, are upon the defective principle of Indivisibles, introduced by CaodUriuM in tie year 1635. Unfortunately, demonstrations npon sounder principles would sot accord with the brevity of this Course. 847 may two sections parallel to the bases, and equidistant from them. Then, by the last two theorems, the section pq is equal to. the base, ab, and the section rs equal to the base bb. But the bases, ab, db, are equal, by the hypothesis ; therefore the sections pq, rs, are equal also. In like manner, it may be shown, that any other corresponding sections are equal to one another. Since then every section in the prism ac is equal to its corresponding seetion in the prism or cylinder df, the prisms and cylinder themselves, which are composed of an equal number of all those equal sections, must also be equal. Q. K. D. Corol. Every prism, or cylinder, is equal to a rectangular parallelopipedon, of an equal base and altitude. THEOREM CIZ. Rectangular parallelopipedons, of equal altitudes, are to each other as their bases 3 ". Let ac, eo, be two rectan- gular parallelopipedons, having the equal altitudes ad, eh ; then will the solid ac be to the solid bo, as the base ab is to the base bp. Q R 5 C HO A LIS' For, let the proportion of the base ab to the base ef, be that of any one number m (3) to any other number n (2). And conceive ab to be divided into m equal parts, or rectangles, ai, lk, mb (by dividing an into that number of equal parts, and drawing il, km, paral- lel to bit). And let ef be divided, in like manner, into n equal parts, or rectangles, eo, pf : all of these parts, of both bases, being mutually equal among themselves. And through the lines of division Jet the plane sections lb, ms, pv, pass parallel to aq, et. Then the parallelopipedons ar, ls, mc, ev, pg, are aU equal, having equal bases and altitudes. Therefore the solid ac ia to the solid eg, as the number of parts in the former, to the number of equal parts in the latter ; or as the number * Here, also, the principle of former notes may readily be, the case of ineommeasorablef. 848 GBOBBTRT. of parts in ab to the number of equal parts in Br, that is, as the base ab to the base bf. q. b. d. Carol. From this theorem, and the corollary to the last, it appears that all prisms and cylinders of equ Jl altitudes, are to each other as their bases ; every prism and cylinder being equal to a rectangular parallelopipeaon of an equal base and altitude. THEOREM CZ. V Rectangular parallelopipedons, of equal bases, are to each other as their altitudes. Let ab, cd, be two rectan- n gular parallelopipedons, stand, ing on the equal bases ae, cf ; then will the solid ab be to the solid cd, as the altitude eb is to the altitude fd. For, let ag be a rectangular parallelopipedon on the base ** ^ ab, and its altitude eg equal to the altitude fd of the solid CD. Then ag and cd are equal, being prisms of equal bases and altitudes. But if hb, hg, be considered as bases, the solids ab, ag, of equal altitude ah, will be to each other as those bases hb, hg. But these bases hb, hg, being parallelograms of equal altitude he, are to each other as their bases eb, bo ; therefore the two prisms, ab, ag, are to each other as the lines eb, eg. But ag is equal to cd, and eg equal to fd ; consequently the prisms ab, cd, are*to each other as their al<* titudes, eb, fd ; that is, ab : cd : : eb : fd. q. e. d. Corol. 1. From this theorem, and the corollary to theorem 108, it appears, that all prisms and cylinders, of equal bases, are to one another as their altitudes. * Corol. 2. Because, by corollary 1, prisms and cylinders are as their altitudes, when their bases are equal. And by the corollary to the last theorem, they are as their bases, when their altitudes are equal. Therefore, universally, when neither are equal, they are to one another as the pro- duct of their bases and altitudes. And hence also these products are the proper numeral measures of their quantities or magnitudes. THEOREMS. 849 THEOREM CXI. Sdolaji prisma and cylinders are to each other, at the cubes of their altitudes, or of any other like linear dimensions. Let abcd, efgh, be two similar prisms ; then will the prism cd be to the prism oh, as ab 3 to ef 3 or ad 3 to EH 3 . For the solids are to each other as the product of their bases and altitudes (th. 110, cor. 2), that is, as ac • ad to so • ih. But the bases, being similar planes, are to each other as the squares of their like sides, that is, ac to eg as ab 3 to ef 1 ; therefore the solid cd is to the solid oh, as ab 3 . ad to ef 3 . eh. But bd and fh, being similar planes, hare their like sides pro. portional, that is, ab : ef : : ad : eh, - or ab 3 : ef 3 : : ad 3 : eh 3 : therefore ab 3 . ad : ef 3 . eh : : ab 3 : ef 3 , or : : ad 3 : eh 3 ; conseq. the solid cd : solid oh : : ab 3 : ef 3 : : ad 3 : eh 3 , a. E. D. B THEOREM CXII. In any pyramid, a section parallel to the base is similar to the base ; and these two planes are to each other as the squares of their distances from the vertex. Let abod be a pyramid, and efo a sec. tion parallel to the base bcd, also aih a line perpendicular to the two planes at h and i : then will bd, eg, be two similar planes, and the plane bd will be to the plane eg, as ah 1 to ai 3 . For, join ch, fi. Then because a plane cutting two parallel planes, makes parallel sections (th. 105), therefore the plane abc, meeting the two parallel planes bd, eg, makes the sections bc, ef, parallel : In like manner, the plane acd makes the sections cd, fg parallel. Again, because two pair of parallel lines make equal angles (tb. 104), the two ef, fg, which are parallel to bc, cd, make the angle efo equal the angle bcd. And in like manner it is shown, that each angle in the plane eg is equal to each angle in the plane bd, and con- sequently those two planes are equiangular. OBOXBTBY. Again, the three lines as, ac, ad, making with the parallels bc, bf, and cd, fo, equal angles (th. 14), and * we angles at a being eommon, the two triangles abc, ad, are equiangular, as also the two triangles acd, afg, and .%fm therefore their like sides proportional, namely t * - - - 40 f ajt ; : bo : bf : . 2 cd : re. And in like manner it way be shown, that all the lines in the plane fo, are pro- portional to all the corresponding lines in the base bd. Hence these two planes, haying their angles equal, and their sides proportional, are similar, by def. 68. But, similar planes being to each other as the squares of their like sides, the plane bd : bo : : bc 9 : bf?, or : { ac 9 : AT*, by what is shown above. Also, the two IfAPttfAf* .abc, aif, having the angles h and 1 right ones ph. m\ wad the angle a common, are equiangular, and have there- fore their luce sides proportional, namely, ac : af : : ah : Afc or AC* ; af* : : Ad 9 : ai 9 . Consequently the two planes bd, 30, which are as the former squares ac 9 , af 8 , will be also as the latter squares ah 3 , ai 9 , that is bd : bo : : AH 9 : AI 9 . Q. E. D. THEOREM CX1II. In a cone, any section parallel to the base is a circle ; and this section is to the base, as the squares of their distances from the vertex. Let abod be a cone, and ohi a secti on A. parallel to the base bcd ; then will ghi be a circle, and bcd, obi, will be to each other, as the squares of their distances from the vertex. For, draw alf perpendicular to the two parallel planes ; and let the planes ace, adb, pass through the axis of the cone akb, meeting the section in the three points , H, I, K. 9 Then, since the section ohi is parallel to the base bcd, and the planes ck, dk, meet tbem, hk is parallel to cb, and IK to ps (th. 105). And because the triangles formed by these lines are equiangular, kh : bc : : ax : ab : : ki : bd. But bc is equal to bd, being radii of the same circle ; there- fore xi is also equal to kh. And the same may be shown of driy other lines drawn from the point k to the perimeter of the section ohi, which is therefore a circle (def. 44). Again, bv similar triangles, al : af : : ax : ab, or ; ; XI : bd, hence al 9 : af 3 ; : ki 9 : bd 9 ; but ki 9 : ed 1 : 2 THEOREMS. circle ohi : circle bcd (tb. OS) ; therefore al 9 : af 3 : : circle obi : circle bcd. q. e. d. THEOREM CXIV. All pyramids, and cones, of equal bases and altitudes, are equal to one another. Let abc, def, be any pyramids and cone, of equal bases bc, ef, and equal altitudes ao, dh : then will the pyra- mids and cone abc and def, be equal. For, parallel to the bases and at equal distances \x, do, from the vertices, suppose the planes ik, lm, to be drawn. Then, by the two preceding theorems, - do* : dh 8 : : lm : ef, and an 3 : ao 2 : : ik : bc. But since an-, ao 2 , are equal to do 3 , dh 3 , respectively, therefore ik : bc : : lm : ef. But bc is equal to ef, by hypothesis : therefore ik is also equal to lm. In like manner it is shown, that any other sections, at equal distance from the vertex, are equal to each other. Since then, every section in the cone, is equal to the cor- responding section in the pyramids, and the heights are equal, the solids abc, def, composed of all those sections, must be equal also. q. e. d. THEOREM CXV. Every pyramid is the third part of a prism of the same base and altitude. Let abcdef be a prism, and bdef a pyramid, on the same triangular base def : then will the pyramid bdef be a third part of the prism abcdef. For, in the planes of the three sides of the prism, draw the diagonals bf, bd," cd. Then the two planes bdf, bcd, divide the whole prism into the three pyramids bdef, dabc, dbcf, which are proved to be all equal to one another, as follows. Since the opposite ends of the prism are equal to eu&Vk^fabT, V Mt OBOJCXTBY. the pyramid whoM base is abc and vertex d, if equal to the pyramid whose base is dbf and vertex b (th. 114), being pyramids of equal base and altitude. ' But the latter pyramid, whose base is dbf and vertex B, is the wne solid as the pyramid whose base is bxf and vertex n, and this is equal to the third pyramid whose base is bcf and vertex d, being pyramids of the same altitude and equal bases bef, bcf. Consequently all the three pyramids, which compose the prism, are equal to each other, and each pyramid* is the third part of the prism, or the prism is triple of the pyramid. 4. X. D. Hence also, every pyramid, whatever its figure may be, is the third part of a prism of the same base and attitudes since the base of the prism, whatever be its figure, my be divided into triangles, and the whole solid into triangular prisms and pyramids. Cord. Any cone is the third part of a cylinder, or of a prism, of equal base and altitude ; since it has been proved that a cylinder is equal to a prism, and a cone equal to a pyramid, of equal base and altitude. Scholium. Whatever has been demonstrated of the pro- portionality of prisms, or cylinders, holds equally true of pyramids, or cones ; the former being always triple the latter ; viz} that similar pyramids or cones are as the cubes of their like linear sides, or diameters, or altitudes, fcc. And the same for all similar solids whatever, viz. that they are in proportion to each other, as the cubes of their like linear dimensions, since they are composed of pyramids every way similar. THEOREM CXVI. If a sphere be cut by a plane, the section will be a circlet. Let the sphere abbf be cut by the plane B aub ;.then will the section adb be a circle* If the section pass through the centre of the sphere, then will the distance from the centre to every point in the periphery of that section be equal to the radius of the sphere, and consequently such section is a carafe. Such, in truth, is the circle safb ¥ in the figure. Draw the chord ab, or diameter of the section adb ; per* psndicnhr to which, or to the said section, draw the axis of # THKOSXM8. 358 the sphere ecgf, through the centre c, which will bisect the chord ab in the point « (th. 41). Also, join ca, cb ; and draw cd, od, to any point d in the perimeter of the sec- tion ADB. Then, because co is perpendicular to the plane adb, it is perpendicular both to ga and od (def. 90). So that coa, cod are two right-angled triangles, having the perpendi- cular co. common, and the two hypothenuses ca, cd, equal, being both radii of the sphere ; therefore the third sides ga, gd, are also equal (cor. 2, th. 34). In like manner it is shown, that any other line, drawn from the centre o to the circumference of the section adb, is equal to ga or gb ; con- sequently that section is a circle. Scholium. The section through the centre, having the same centre and diameter as the sphere, is called a great circle of the sphere ; the other plane sections being little circles. THEOREM. CXVn. Evert sphere is two.thirds of its circumscribing cylinder. Let abcd be a cylinder, circumscribing \ J g the sphere efgh ; then will the sphere sfgh be two-thirds of the cylinder abcd. For, let the plane ac be a section of the sphere and cylinder through the centre i. Join ai, bi. Also, let fih be parallel to ad or bc, and eig and kl parallel to ab or dc, the base of the cylinder ; the latter ■ line kl meeting bi in m, and the circular section of the sphere in w. « - Then, if the whole plane hfbc be conceived to revolve about the line rf as an axis, the square fg will describe a cylinder ag, and the quadrant ifg will describe a hemi- sphere efg, and the triangle ifb will describe a cone lab. Also, in the rotation, the three lines or parts kl, kit, km, as radii, will describe corresponding circular sections of those solids, namely, kl a section of the cylinder, kn a section of the sphere, and km a section of the cone. Now, fb being equal to n or ig, and kl parallel to fb, then by similar triangles ik is equal to km (th. 82^~> And since, in the right-angled triangle ikn, in* is equal to ik 1 + kn 3 (th. 34) ; and because kl is equal to the radius io or in, and km = ik, therefore kl 2 is equal to km 3 + xir* 9 or the square of the longest radius, of tto 8t&~V\t&\^ Vol. I • 46 864 GEOMETRY. sections, is equal to the sum of t^e squares of the two others* And because circles are to each other as the squares of their diameters, or of their radii, therefore the circle described by kl is equal to both the circles described by km and kn ; or the section of the cylinder, is equal to both the corresponding sections of the sphere and cone. And as this is always the. case in every parallel position of kl, it follows, that the cy- linder eb, which is composed of all the former sections, is equal to the hemisphere efg and cone iab, which are com- posed of all the latter sections. But the cone iab is a third part of the cylinder eb (cor. 2, th. 115) ; consequently the hemisphere efg is equal to the remaining two-thirds ; or the whole sphere efgh equal to two-thirds of the whole cylinder abcd. q. e. d. Coral. 1. A cone, hemisphere, and cylinder of the same base and altitude, are to each other as the numbers 1, 2, 3. CoroL 2. All spheres are to each other as the cubes of their diameters ; all these being like parts of their circum- scribing cylinders. CoroL 3. From tho foregoing demonstration it also ap- pears, that the spherical zone or frustum egxf, is equal to the difference between the cylinder eglo and the cone in a, all of the same common height ik. And that the spherical segment pfn, is equal to the difference between the cylinder ablo and the conic frustum aqmb, all of the same common altitude fk. 355 PROBLEMS. PROBLEM I. To bisect a line ab ; that is, to divide it into two equal parts. From the two centres a and b, with any c equal radii, describe arcs of circles, inter, secting each other in c and d ; and draw the line cd, which will bisect the given line A ab in the point e. For, draw the radii ac, bc, ad, bd. Then, because all these four radii are equal, and the side cd common, the two triangles acd, bcd, are mutually equilateral : consequently they are also mutually equiangular (th. 5), and have the angle ace equahto the angle bce. Hence, the two triangles ace, bce, having the two sides ac, ce, equal to the two sides bc, ce, and their contained angles equal, are identical (th. 1), and therefore have the side ae equal to eb. q. e. d. problem n. To bisect an angle bac*. From the centre a, with any radius, de- scribe an arc, cutting off the equal lines ad, ae ; and from the two centres d, e, with the same radius, describe arcs inter, secting in f ; then draw af, which will bisect the angle a as required. * A very ingenious instrument for trisecting an angle, is described in the Mechanic's Magasioe, No. 22, p. 344. A For, join df, if. Then \ the two triangles isr, a»f. baring the two aide* ad 9 df, equal to the two ae, bf (beiag equal radii), and the side af common, they are mutuafiy equilateral ; consequently they are also mutually equiangular (th. 5), and have the angle bat equal to the angle oaf. Scholium. In the same manner is an arc of a circle bi- sected. fboblex m. At a given point o, in a line ab, to erect a perpendicular. From the given point o, with any radius, cut off any equal parts cd, cb, of the given line ; and, from the two centres d and s, with any one radius, describe arcs intersect- _ ing in f ; then join of, which will be per- -4tT"0"TtB pendicular as required* For, draw the two equal radii df, ef. Then the two triangles cdf, cef, having the two sides cd, df, equal to the two ok, bf, and cf common, are mutually equilateral ; consequently they are also mutually equiangular (th. 5), and have the two adjacent angles at c equal to each other ; there- fore the line cf is perpendicular to ab (def. 11). Otherwise* Whew the given point c is near the end of the line* From any point d assumed above the line, as a centre, through the given point c describe a circle, cutting the given line at e ; and through e and the centre d, draw the diameter edf; then join cf, which will be the perpendicular required. For the angle at c, being an angle in a semicircle, is a right angle, and therefore the line cf is a perpendicular (by def. 15). PROBLEM IV. From a given point a, to let fall a perpendicular on a given line bc. From the given point a as a centre, with any convenient radius, describe an arc, cut- ting the given line at the two points d and b ; and from (he two centres d, e, with any radius, describe two arcs, intersecting at f ; then draw aof, which will be perpendicular to bc as required. S07 For, draw the equal radii ad, as, and df, ep. Then the two triangles adf, aef, having the two sides ad df, equal to the two ab, bf, and af common, are mutually equilateral ; consequently they are also mutually equiangular (th. 5), and have the angle dag equal the angle bag. Hence then, the two triangles ado, abg, having the two sides ad, ao, equal to the two ab, ao, and their included angles equal, are there- fore equiangular (th. 1), and have the angles at o equal ; consequently ao is perpendicular to bc (def. 11). Otherwise. WftEN the given point is nearly opposite the end of the line. From any point d, in the given line bc, as a centre, describe the arc of a circle through the given point a, cutting bc in e ; and from the centre e, with the radius ba, describe another arc, cutting the former in f ; then draw agf, which will be perpendicular to bc as required. 3f For, draw the equal radii da, df, and ea, bf. Then the two triangles dak, dfe, will be mutually equilateral ; conse- quently they are also mutually equiangular (th. 5), aqd have the angles at d equal. Hence, the two triangles dag, dfo, having the two sides da, dg, equal to the two df, do, and the included angles at d equal, have also the angles at o equal (th. 1) ; consequently those angles at o are right angles, and the line ag is perpendicular to dg. PROBLEM V. At a given point a, in a line ab, to make an angle equal to a given angle c. From the centres a and c, with any one • jg^ radius, describe the arcs de, fg. Then, with radios de, and centre f, describe an arc, cutting fg in g. Through g draw C D the line ag, and it will form the angle re* ~ quired. \ For, conceive the equal lines or radii, j± jtjj db, fg, to be drawn. Then the two trian- gles CDs, afg, being mutually equilateral, are mutually equi- angular (th. 5), and have the angle at a eqiud to ta* vu^a fe» 868 GEOMETRY* PROBLEM VI. Through a given point a, to draw a line parallel to a given line bo. From the given point a draw a line ad EA to an j point in the given line bc. Then draw the line eaf making the angle at a equal to the angle at d (by prob. 5) ; bo j5~C shall ef be parallel to bc as required. For, the angle d being equal to the alternate angle a, the lines bc, ef, are parallel, by th. 13. problem vii. To divide a line ab into any proposed number of equal parts. Draw any other line ac, forming any angle with the given line ab ; on which EVA set off as many of any equal parts ad, de, \ ef, fc, as tho line ab is to be divided into. 3\V \ \ Join bc ; parallel to which draw the other ^IHCKB lines fg, eh, di : then these will divide ab in the manner as required. — For those parallel lines divide both the sides ab, ac, proportionally, by th. 82. PROBLEM Vin. To find a third proportional to two given lines ab, ac. Place the two given lines ab, ac, forming any angle at a ; and in ab take A B also ad equal to ac. Join bc, and A- ^ draw de parallel to it ; so will ae be the third proportional sought. For, because of the parallels, bc, de, A 2) '0 the two lines ab, ac, are cut propor- tionally (th. 82) ; so that ab : ac : : ad or ac : ae ; there- fore ae is the third proportional to ab, ac. PROBLEM IX. To find a fourth proportional to three lines ab, ac, ad. Place two of the given lines ab, ac, making any angle at A ; also place ad on ab. Join bc ; and parallel to it draw PROBLEMS. 850 de : so shall ae be the fourth propor- tional as required. For, because of the parallels bc, de, the two sides ab, ac, are cut propor- tionally (th. 82) ; so that .... ab : ac : : ad : ae. PROBLEM X. To find a mean proportional between two lines ab, bc. Place ab, bc, joined in one straight line ac : on which, as a diameter, describe the semicircle adc ; to meet which erect the perpendicular bd ; and it will be the mean proportional sought, between ab and bc (by cor. th. 87). . problem XI. To find the centre of a cii Draw any chord ab ; and bisect it p pendicularly with the line cd, which will a diameter (th. 41, cor.). Therefore bisected in o, will give the centre, as reqi ed. PROBLEM XII. To describe the circumference of a circle through three given points a, b, c. From the middle point b draw chords ba, bc, to the two other points, and bi- •"-Hs^JL-JK--* sect these chords perpendicularly by lines ^^T^Si meeting in o, which will be the centre. A<^^y[^5\C Then from the centre o, at the distance I /q\ J of any one of the points, as oa, describe \. J a circle, and it will pass through the two other points b, c, as required. For the two right-angled triangles oad, obd, having the sides ad, db, equal (by constr.), and od common, with the included right angles at d equal, have their third sides oa, ob, also equal (th. 1). And in like manner it is shown that oc is equal to ob or oa. So that all the three oa, ob, go* tid- ing equal, will be radii of the same circta. OROMSTRY. PROBLEM XIII. To draw a tangent to a circle, through a given point When the given point a is in the circum- ference of the circle : Join a and the centre o ; perpendicular to which draw bac, and it will be the tangent, by th. 46. But when the given point a is out of the circle : Draw ao to the centre o ; on which as a diameter describe a semicircle, cutting the given circumference in d ; through R which draw badc, which will be the tangent 1 cs required. For, join do. Then the angle ado, in a semicircle, is a right angle, and consequent- ly ad is perpendicular to the radius do, or is a tangent to the circle (th. 46). PROBLEM xiv. On a given line b to describe a segment of a circle, to contain a given angle c. At the ends of the given line make angles dab, dba, each equal to the given angle c. Then draw ae, be, perpendicular to ad, bd ; and with the centre e, and radius ea or eb, describe a circle ; so shall afb be the segment required, as any angle f made in it will be equal to the given angle c. For, the two lines ad, bd, being perpendicular to the radii ea, eb (by constr.), are tangents to the circle (th. 46) ; and the angle a or b, which is equal to the given angle c by construction, is equal to the angle f in the alternate segment aeb (th. 53). problem xv. To cut off a segment from a circle, that shall contain a given angle c. Draw any tangent ab to the given circle ; and a chord ad to make the E angle dab equal to the given angle c ; then dka will be the segment required, any angle s made in it being equal to the given angle c. PROBLEMS. 861 For the angle a, made by the tangent and chord, whioh ia equal to the given angle c by construction, is also equal to any angle a in the alternate segment (th. 63). problem zvi. To make an equilateral triangle on a given line ab. From the centres a and b, with the distance ab, describe arcs, intersecting in c. Draw ac, bo, and abc will be the equilateral triangle. For the equal radii, ac, bc, are, each of them, equal to ab. problem xvn. To make a triangle with three given lines ab, ac, bc With the centre a, and distance ac, describe an arc. With the centre b, and distance bc, describe another arc, cutting the former in c. Draw ab, bc, and abc will be the triangle required. For the radii, or sides of the triangle, ac, bc, are equal to the given lines ac, ac, by construction. A PROBLEM XVIII. To make a square on a given line ab. Raise ad, bc, each perpendicular and j>_ equal to ab ; and join dc ; so shall abcd be the square sought. For all the three sides ab, ad, bc, are equal, by the construction, and dc is equal and parallel to ab (by th. 24) ; so that all the four sides are equal, and the opposite ones are parallel. Again, the angle a or b, of the parallelogram, being a right angle, the angles are all right ones (cor. 1, th. 22). Hence, then, the figure, having all its sides equal, and alL'towGq&a* right, is a square (def. 34). Vol. I 47 GEOMETRY. problem XIX. To make a rectangle, or a parallelogram, of a given length and breadth, ab, bc. Erect ad, bc, perpendicular to ab, and each equal to bc ; then join dc, and it is done. The demonstration is the same as the _ last problem. B— — C And in the same manner is described any oblique paral- lelogram, only drawing ai> and bc to make the given oblique angle with ab, instead of perpendicular to it. PROBLEM xx* To inscribe a circle in a given triangle abc. Bisect any two angles a and b, with the two lines ad, bd. From the inter- section d, which will be the centre of the circle, draw the perpendiculars db, df, DU, and they will be the radii of the circle required. For, since the angle dae is equal to the angle dag, and the angles at e, o, right angles (by const r.), the two triangles, adb, ado, are equiangular ; and, having also the side ad common, they are identical, and have the sides de, do, equal (th. 2). In like manner it is shown, that df is equal to de or do. Therefore, if with the centre d, and distance de, a circle be described, it will pass through all the three points, e, f, g, in which points also it will touch the three sides of the tri- angle (th. 46), because the radii de, df, do, are perpendicu- lar to them. PROBLEM XXI. To describe a circle about a given triangle abc. Bisect any two sides with two of the perpendiculars de, df, do, and d will be the centre. For, join da, db, dc Then the two right-angled triangles dae, dbe, have the two sides, de, ea, equal to the two de, eb, and the included angles at e equal : those two triangles are therefore P10BLEHS. 383 identical (th. 1% and have the side da. equal to db. In like manner it it shown, that dc is also equal to da or db. So that all the three da, db, dc, being equal, they are radii of a circle passing through a, b, and c. fgOBLBH XXII. To inscribe an equilateral triangle in a given circle. Through the centre c draw any diameter a ab. From the point b as a centre, with the y^/fV^N, radius bc of the given circle, describe an f / |\ \ arc dcb. Join ad, as, db, and ade is the f X-JsA ) equilateral triangle sought x^^l^^ v/ For, join db, dc, bb, bc. Then dcb ^ X^TT^^E is an equilateral triangle, having each side ^jj^ equal to the radius of the given circle. In like manner, bcb is an equilateral triangle. But the angle ade is equal to the angle abb or cbe, standing on the same arc ae ; also the angle akd is equal to the angle cbd, on the same arc ad ; hence the triangle dab has two of its angles, adb, aed, equal to the angles of an equilateral triangle, and therefore the third angle at a is also equal to the same ; so that the triangle is equiangular, and therefore equilateral. PROBLEM XXIII. To inscribe a square in a given circle* Draw two diameters ac, rd, crossing at right angles in the centre e. Then join the four extremities a, b, c, d, with right lines, and these will form the in- scribed square abcd. For the four right-angled triangles aeb, bp.c, ced, dea, are identical be- cause they have the sides ea, eb, ec, ed, all equal, being radii of the circle, and the four included angles at e all equal, being right angles, by the construction. Therefore all their third sides ab, bc, cd, da, are equal to one another, and the figure abcd is equilateral. Also, all its four angles, a, b, c, d, are right ones, being angles in a semicircle. Cooaec^t&Vj the figure is a square. GEOMETRY. PROBLEM XXIV. To describe a square about a given circle. Draw two diameters ac, bd, crossing at right angles in the centre s. Then through their four extremities draw fg, ib, parallel to ac, and fi, oh, parallel to bd, and they will form the square FGHI. For, the opposite sides of parallelo- grams being equal, fg and in are each equal to the diameter ac, and fi and on each equal to the diameter bd ; so that the figure is equilateral. Again, be- cause the opposite angles of parallelograms are equal, all the four angles f, c, h, i, are right angles, being equal to the opposite angles at e. So that the figure fgmi, having its sides equal, and its angles right ones, is a square, and its sides touch the circle at the four points a, b, c, d, being perpen- dicular to the radii drawn to those points. PROBLEM XXV. To inscribe a circle in a given square. Bisect the two sides fg, fi, in the points a and b (last fig.). Then through these two points draw ac parallel to fc or ih, and bd parallel to fi or gh. Then the point of intersection x will be the centre, and the four lines ea, eb, ec, ed, radii of the inscribed circle. For, because the four parallelograms ef, eg, kh, ei, have their opposite sides and angles equal, therefore all the four lines ea, eb, ec, ed, are equal, being each equal to half a side of the square. So that a circle described from the centre E, with the distance ea, will pass through all the points a, b, c, d, and will be inscribed in the square, or will touch its four sides in those points, because the angles there are right ones. PROBLEM XXVI. To describe a circle about a given square. (See fig. Prob. xxiii.) Draw the diagonals ac, bd, and their intersection s will be the centre. For the diagonals of a square bisect each other (th. 40), making ea, eb, ec, ed, all equal, and consequently these are radii of a circle naasin^ vUtou^U the four points a, b, c, p. FftOBUMS. 805 FKOBLEM XXVII. To cut a given line in extreme and mean Let ab be the given line to be divided in ex I re me and mean ratio, that is, so as that the whole line may be to the greater part, as the greater part is to the less part. Draw bc perpendicular to AR/and equal to half ab. Join ac ; and with centre c and distance cb, describe the circle bd ; then with centre a and distance ad, de- scribe the arc dk ; so shall ab be divided in e in extreme and mean ratio, or so that ab : ae : : ab : eb. Produce ac to the circumference at f. Then, auf being a secant, and ab a tangent, because b is a right angle : there- fore the rectangle af . ad is equal to ab 9 (cor. 1, th. 61) ; con* sequently the means and extremes of these are proportional (th. 77), viz. ab : af or ad + df : : ad : ab. But ab is equal to ad by construction, and ab = 2bc = df ; therefore, ab : ae + ab : : ae : ab ; and by division, ab : ae : : ae : eb. PROBLEM XXVIII. To inscribe an isosceles triangle in a given circle, that shall have each of the angles at the base double the angle at the vertex. ^ Draw any diameter ab of the given circle ; and divide the radius cb, in the point d, in extreme and mean ratio, by the last problem. From the point b apply the chords be, bf, each equal to the greater part cd. Then join ae, af, ef ; and akf will be the triangle required. For, the chords be, bf, being equal, their arcs are equal ; therefore the supplemental arcs and chords ae, af, are also equal ; consequently the triangle aef is isosceles, and has the angle e equal to the angle f ; alto the angles at u are right angles. Draw cf and df. Then, bc : cd : : cd : bd, or bc : bf : : bf : bd by constr. And ba : bf : : bf : bo (by th. 87). But bc = £ba ; therefore bo = £bo = go \ therefore the two triangles gbf, ©df, ara VtouvicaX VJ3fc%\V ratio. A t c 806 . C GEOMETRY. £ and each equiangular to arf and agf (th. 87). Therefore their doubles, bfd, afe, are isosceles and equiangular, as well as the triangle hcf ; having the two sides bc, cf, equal, and the angle b common with the triangle bfd. But cd is = df or bf ; therefore the angle c = the angle dpc (th. 4) ; consequently the angle bdf, which is equal to the sum of these two equal angles (th. 16), is double of one of them c ; or the equal angle b or ceb double the angle c. So that cbf is an isosceles triangle, having each of its two equal angles double of the third angle c. Consequently the triangle arf (which it has been shown is equiangular to the triangle c f) has al*o each of its angles at the base double the angle a at the vertex. PROBLEM XXIX. To inscribe a regular pentagon in a given circle. Inscribe the isosceles triangle abc, having each of the angles abc, acr, double the angle bac (prob. 28). Then bisect the two arcs adr, arc, in the points d, e ; and draw the chords ad, dr, ar, ec, so shall adrce be the inscribed equilateral pentagon required. For, because equal angles stand on equal arcs, and double angles on double arcs, also the angles abc, acb, being each double the angle bac, therefore the arcs adr, arc, subtending the two former angles, are each double the arc bc subtending the latter. And since the two former arcs are bisected in d and r it follows that all the five arcs ad, dr, rc, ck, ea, are equal to each other, and con- sequently the chords also which subtend them, or the live sides of the pentagon, are a 1 I equal. Note. In (he constniction, the points d and e are most easily found, by applying bd and ce each equal to bc. problem xxx. To inscribe a regular hexagon in a circle. Apply the radius ao of the given circle as a chord, ar, rc, cd, &c. quite round the circumference, and it will complete the regular hexagon abcdef. Draw the radii ao, ho, co, do, eo, fo, completing six equal triangles ; of which any one, as abo, being equilateral (by const r.) its three angles are all equal (cor. 2, tb. 3), and any one of them, as aob, is one-third of the' FB0BLEMS. 987 whole, or of two right angles (th. 17), or one-sixth of four right angles. But the whole circumference is the measure of four right angles (cor. 4, th. 6). Therefore the arc ar is one.sixth of the circumference of the circle, and consequently its chord ab one aide of an equilateral hexagon inscribed in the circle. And the same of the other chords. Corel. The side of a regular hexagon is equal to the radius of the circumscribing circle, or to the chord of one- sixth part of the circumference*. PROBLEM XXXI. To describe a regular pentagon or hexagon about a circle. Izv the given circle inscribe a regular polygon of the same name or number of sides, as abcdk, by one of the fore- going problems. Then to all its angular points draw tangents (by prob. 13), and these will form the circumscribing poly- gon required. For all the chords, or sides of the inscribed figure, ab, bc, dec. being equal, and all the radii oa, ob, dec. being equal, all the vertical angles about the point o are equal. But the angles oef, oaf, oao, obg, made by the tangents and radii, are right angles ; therefore oef + oaf = two right angles, and oao + obg = two right angles ; consequently, also, aoe + afe = two right an- gles, and oab + a«b « two right angles (cor. 2, th. 18). Hence, then, the angles aoe + afe being = aob + aob, of which aob is = aoe ; consequently the remaining angles r and o are also equal. In the same manner it is shown, that all Hthe angles f, g, h, i, k, are equal. Again, the tangents from the same point fe, fa, are equal, as also the tangents ao, gb, (cor. 2, th. 61) ; and the angles f and g of the isosceles triangles afe, agb, are equal, as well as their opposite sides ae, ab ; consequently those two tri- angles are identical (th. 1), and have their other sides ef, fa, ag, gb, all equal, and fg equal to the double of any one of them. In like manner it is shown, that all the other sides gh, hi, ik, ki, are equal to fg, or double of the tangents gb, bh, &c. * The bent way to describe n polygon of any number of sides, the length of one side being given, is to find the radius of the cATOraMafa» log circle by means of the table, at pa. 413, and \ta t*\» «t \*. 368 GKOKETBY. Hence, then, the circumscribed figure is both equilateral and equiangular, which was to be shown. Cord* The inscribed circle touches the middle of the sides of the polygon. PSOBLEX XXXII. To inscribe a circle in a regular polygon. Bisect any two sides of the polygon by the perpendiculars go, fo, and their intersection o will be the centre of the inscribed circle, and oo or of will be the radius. For the perpendiculars to the tangents af, ao, pass through the centre (cor. r th. 47) ; and the inscribed circle touches C the middle points f, o, by the last corollary. Also, the two sides, ao, ao, of the right-angled triangle aog, being equal to the two sides af, ao, of the right-angled triangle aof, the third sides of, oo, will also be equal (cor. th. 45). There- fore the circle described with the centre o and radius og, will pass through f, and will touch the sides in the points g and f. And the same for all the other sides of the figure. PROBLEM XXXIII. To describe a circle about a regular polygon. Bisect any two of the angles, c and d, with the lines co, do ; then their inter, section o will be the centre of the circum- scribing circle ; and or, or od, will be the radius. For, draw ob, oa, of, &c. to the an- gular points of the given polygon. Then the triangle ocd is isosceles, having the angles at c and D equal, being the halves of the equal angles of the polygon BCD, cdb ; therefore their opposite sides co, do, are equal, (th. 4). But the two triangles oci>, ocb, having the two sides oc, cd, equal to the two oc, cb, and the included angles ocd, Ocb, also equal, will be identical (th. 1), and have their third •ides bo, od, equal. In like manner it is shown, that all the lines oa, ob, oc, od, or, aro equal. Consequently a circle described with the centre o fjnd radius oa, will pass through all the other angular points, b, c, d, dec. and will circum- msribe the polygon. PROBLEMS. PROBLEM XXXIV. To make a square equal to the sum of two or more given squares. Let ib and ac be the sides of two given squares. Draw two indefinite lines ap, aq, at right angles to each other ; in which place the sides ab, ac, of the given squares ; join bc ; then a square described on bc will be equal to the sum of the two squares described on ab and ac (th. 34). In the same manner, a square may be made equal to the sum of three or more given squares. For, if ab, ac, ad, be taken as the sides of the ffiven squares, then, making ae=bc, ad = ad, and drawing he, it is evident that the square on db will be equal to the sum of the three squares on ab, ac, ad. And so on for more squares. PROBLEM XXXV. To make a square equal to the difference of two given squares. Let ab and ac, taken in the same straight line, be equal to the sides of the two given squares. — From the centre a, with the distance ab, describe a circle ; and make cd perpendicular to ab, meet- A C *B ing the circumference in d : so shall a square described on cd be equal to ad' j — ac\ or ab 3 — ac 9 , as required (cor. th. 34). PROBLEM XXXVI. To make a triangle equal to a given quadrangle abcd. Draw the diagonal ac, and parallel j) q to it de, meeting ba produced at e, and /P J9\ join ce ; then will the triangle ceb be st* equal to the given quadrilateral abcd. • / XL For, the two triangles ace, acd, E A B being on the same base ac, and between the same parallels ac, de, are equal (th. 25) ; therefore, if abc be added to each, it will make bce eqaaUo kwct> Ujl^v •Vol. I. 48 870 OBOMBTBY. PROBLEM XXXVII. To make o triangle equal to a given pentagon abcds. Draw da and db, and also ef, co, parallel to them, meeting ab produced at v and o ; then draw dp and do ; so shall the triangle dfo be equal to the given pentagon abcde. For the triangle dfa = dka, and the triangle dob = dcb (th. 25) ; therefore, by adding dab to the equals, the sumw are equal (ax. 2), that is, dab + dap + dbg» dab + dae + dbc, or the triangle dfo = to the pentagon . problem xxxviii. To make a rectangle equal to a given triangle abc. Bisect the base ab in d : then raise de and bf perpendicular to ab, and meeting cf parallel to ab, at e and f : so shall dp be the rectangle equal to the given triangle abc (by cor. 2, th. 26). problem xxxix. To make a square equal to a given rectangle abcd. Produce one side ab, till be be equal to the other side bc. On ae as O . V a diameter describe a circle, meeting 3 ) r J j ^,... bc produced at f : then will bp be the \ \ I side of the square bfoh, equal to the f__ j ! j given rectangle bd, as required ; as AH B E appears by cor. th. 87, and th. 77. 871 APPLICATION OF ALGEBRA TO GEOMETRY. Whbx it is proposed to resolve a geometrical problem algebraically, or by algebra, it is proper, in the first place, to draw a figure that shall represent the several parts or con- ditions of the problem, and to suppose that figure to be the true one. Then having considered attentively the nature of the problem, the figure is next to be prepared for a solu- tion, if necessary, by producing or drawing such lines in it as appear most conducive to that end. This done, the usual symbols or letters, for known and unknown quantities, are employed to denote the several parts of the figure, both the known and unknown parts, or as many of them as necessary, as also such unknown line or lines as may be easiest found, whether required or not. Then proceed to the operation, by observing the relations that the several parts of the figure have to each other ; from which, and the proper theorems in the foregoing elements of geometry, make out as many equations independent of each other, as there are unknown quantities employed in them : the resolution of which equa- tions, in the same manner as in arithmetical problems, will determine the unknown quantities, and resolve the problem proposed. As no general rule can be given for drawing the lines, and selecting the fittest quantities to substitute for, so as always to bring out the most simple conclusions, because different problems require different modes of solution ; the best way to gain experience, is to try the solution of the same problem in different ways, and then apply that which succeeds best, to other cases of the same kind, when they afterwards occur. The following particular directions, however, may be of some use. l«t, In preparing the figure, by drawing lines, let them be either parallel or perpendicular to other lines in the figure, or so as to form similar triangles. And if an angle be given, it will be proper to let the perpendicular be opposite to that -angle, and to fall from one end of a given line, if ^awfttat* 372 APPLICATION or ALGEBRA 2d, In selecting the quantities proper to substitute for, those are to be chosen, whether required or not, which lie nearest the known or given parts of the figure, and by means of which the next adjacent parts may be expressed by addi- tion and subtraction only, without using surds. 3d, When two lines or quantities are alike related to other parts of the figure or problem, the best way is, not to make use of either of them separately, but to substitute for their sum, or difference, or rectangle, or the sum of their alternate quotients, or for some line .or lines, in the figure, to which they have both the same relation. 4th, When the area, or the perimeter, of a figure is given, or such parts of it as have only a remote relation to the parts required : it is sometimes of use to assume another figure similar to the proposed one, having one side equal .to unity, or some other known quantity. For, hence the other parts of the figure may be round, by the known proportions of the like sides, or parts, and so an equation be obtained. For examples, take the following problems. PROBLEM I. In a right-angled triangle, having given the base (8), and the sum of the hypotenuse and perpendicular (9) ; to find both these two sides. Let abc represent the proposed triangle right-angled at b. Put the base ab = 3 = b, and the sum ac + bc of the hypothenuse and perpendicular = 9 = s ; also, let x de- note the hypothenuse ac, and y the perpen- dicular BC. Then by the question - . . x + y = s, and by theorem 34, - - . . x* = y % + b*, By tr»:nspos. y in the 1st equ. gives x = s — y t This value of x substi. in the 2d, gives .... s^-Qsy + y 1 = f + b* 9 Taking away y*on both sides leaves j*... 2sy = b*, By transpos. Usy and 6 j , gives - s^b' = Usy, * 3 — b 2 And dividing by 2s, gives - - — — = y = 4 = bc. Hence x = * — y = 5 = ac. ^ N. B. In this solution, and the following ones, the nota. tkm is made by using as many unknown letters, x and jr, as TO OEOMKBT. 378 there are unknown tides of the triangle, a separate letter for each ; in preference to using only one unknown letter for one side, and expressing the other unknown side in terms of that letter and the given sum or difference of the sides ; though this latter way would render this solution shorter and sooner ; because the former way gives occasion for more and better practice in reducing equations ; which is the very end and reason for which these problems are given at all. PROBLsk'lI. In a right-angled triangle, having given the hypothenuse (5) ; and the sum of the base and perpendicular (7) ; to find both these two sides. Let abc represent the proposed triangle, right-angled at B. Put the given hypothenuse ac = 5 = a, and the sum ab + bc of the base and perpendicular = 7 = s ; also let x denote the base ab, and y the perpendicular bc. Then by the question - - - x + y = s t and by theorem 34 ... x*+ y*= a 2 , By transpos. y in the 1st, gives x = s — y, By substitu. this value for x, gives **— 2sy 4- 2y a = a\ By transposing s*, gives - - 2y* — 2sy — a* — s 7 , By dividing by 2, gives - - y 2 — sy = \a* — ±s* 9 By completing the square, gives y* — sy + Js* = }a* — \s* f By extracting the root, gives - y — \s = %/(i a3 ~" i* 9 ) By transposing is, gives - - y = J* ± y/\\a 2 — \s*) = 4 and 3, the values of x and y. problem in. In a rectangle, having given the diagonal (10\ and the peri- meter, or sum of all the four sides (28) ; to find each of the sides severally* Let abcd be the proposed rectangle ; and put the diagonal ac = 10 = d, and half the perimeter ab + bc or ad + dc — 14 =. a : also put one side ab = ar, and the other side bc = y. Hence, by right-nngled triangles, - - - . - x* + y 7 = d\ And by the question - - - . - x+y — a, Then by transposing y in the 2d, gives x — a — y, This value substituted in the 1st, gives a? — < SUu)V*^=^ 874 APPLICATION OF ALGEBRA Transposing a*, gives . - 2y' — 2ay = d 1 — «*, And dividing by 2, gives - y 1 — ay = \d l — ±a\ By completing the square, it is y* — ay + \a* = \<P — \a\ And extracting the root, gives y — ±a = y/(±d l — Ja 1 ), And transposing \a, gives - y = £a 2: v^(i^ J — l* 1 ) 3 ^ or 6, the values of x and y. problem IV. Having given the base anfoerpendicular of any triangle ; to find the side of a square inscribed in the same. Let abc represent the given triangle, q and EFGit its inscribed square. Put the base ab =■ by the perpendicular cd = a, and the side of the square gf or on = Di x ; then will ci = cd — di = a — x. Then, because the like lines in the similar triangles abc, gpc, are propor- tional (by theor. 84, Geom.), ab : cd : : ge : ci, that is, 6 : a : : x : a — x. Hence, by multiplying extremes and means, qb — bx = ax, and transposing bx, gives ab = ax ab + bx ; then dividing by a + b, gives x = a ^ ^ = gf otch the side of the inscribed square : which therefore is of the same magnitude, whatever the species or the angles of the triangles may be. problem v. In an equilateral triangle, having given the lengths of the three perpendiculars, drawn from a certain point within, on the tlirce sides ; to determine the sides. Let abc represent the equilateral tri- angle, and oe, df, dg, the given per- pendiculars from the point d. Draw the lines da, dh, dc, to the three angular points ; and let fall the perpendicular cu on the base ab. Put the three given per- pendiculars, de = a, df = b, d<; = c, and put x = a 11 or bii, half the side of the equilateral triangle. Then is ac or bc = 2.r, and by right-angled triangles the perpendicular cu = */(ac 2 — ah 1 ) •S3T c ) - TO GEOMETRY. 375 Now, since the area or space of a rectangle, is expressed by the product of the buse arid height (cor. 2, th 81, Geom.), and that a triangle is equal to half a rectangle of equal baso and height (cor. 1, th. ^6), it follows that, the whole triangle abc is = Jab X cn = x X x y/3 = x 2 ^3, the tnungle abd = Jab X do = x X c = cr, the triangle bcd =■ Jbc X dk = x X a = ax, the triangle acd = \ac X df = * X b = bx. But the three last triangles make up, or are equal to, the whole former, or great tnungle ; that is, x* = ax + br + cx ; hence, dividing by *, gives r y/3 = a +6 + c, and dividing by ^/3, gives a-f. & + r x = 6 — , half the side of the triangle sought. Also, since the whole perpendicular cn is =- x y/3, it is therefore = a + b + c. That is, the whole perpendicular cn, is just equal to the sum of all the three smaller perpen- diculars db + nr + do taken together, wherever the point d ii situated. PROBLEM VI. In a right-angled triangle, having given the base (3\ and the difference between the hypoihenuse and perpendicular (1) ; to find both these two sides. PROBLEM VII. In a right-angled triangle, having given the hypothenuse 5), and the difference between the base and perpendicular 1) ; to determine both these two sides. problem viu. Having given the area, or measure of the space, of a rect. angle, inscribed in a given triangle ; to determine the sides of the rectangle. problem u. In a triangle, having given the ratio of the two sides, together with both the segments of the base, made by a per- pendicular from the vertical angle ; to determine the sides of - the triangle. problem x. In a triangle, having given the base, the sum of the other two sides, and the length of a line drawn ftom\Yi« <<i«*ta*\ 376 APPLICATION OF ALGEBRA angle to the middle of the base ; to find the sides of the triangle. PROBLEM XI. In a triangle, having given the two sides about the verti- cal angle, with the line bisecting that angle, and terminating in the base ; to find the base. PROBLEM XII. To determine a right-angled triangle ; having given the lengths of two lines druwn from the acute angles, to the middle of the opposite sides. PROBLEM XIII. To determine a right-angled triangle ; having given the perimeter, and the radius of its inscribed circle. PHOBLEM XIV. To determine a triangle ; having given the base, the per- pendicular, and the ratio of the two sides. PROBLEM xv. To determine a right-angled triangle ; having given the hypothenuse, and the side of the inscribed square. PROBLEM XVI. To determine the radii of three equal circles, described in a given circle, to touch each other and also the circum- ference of the given circle. PROBLEM XVII. In a right-angled triangle, having given the perimeter, or sum of all the sides, and the perpendicular let fall from the right angle on the hypothenuse ; to determine the triangle, that is, its sides. PROBLEM XVIII. To determine a right-angled triangle ; having given the hypothenuse, and the difference of two lines drawn from the two acute angles to the centre of the inscribed circle. -$&~ i-$A t&^h. w-a+yi- t t I "IT If- hZe^dr^ 1 to TO GEOMETRY. 377 PROBLEM XIX. To determine a triangle ; having given the base, the per- pendicular, and the difference of the two other sides. PROBLEM XX. To determine a triangle ; having given the base, the per- pendicular, and the rectangle or product of the two sides* PROBLEM XXL To determine a triangle ; having given the lengths of three lines drawn from the three angles, io the middle of the oppo* site sides. PROBLEM XXXI. In a triangle, having given all the three sides ; to find the radius of the inscribed circle. PROBLEM XXIII. To determine a right-angled triangle ; having given the aide of the inscribed square, and the radius of the inscribed circle. PROBLEM XXIV. To determine a triangle, and the radius of the inscribed circle ; having given the lengths of three lines drawn from the three angles, to the centre of that circle. PROBLEM XXV. To determine a right-angled triangle ; having given the hypothenuse, and the radius of the inscribed circle. PROBLEM XXVI. To determine a triangle ; having given the base, the line bisecting the vertical angle, and the diameter of the circum. scribing circle. Vol. I. 49 97 PLANE TRIGONOMETRY. DEFIMTIOX8. 1. Plank Trigonometry treats of the relations and cal- culations of the sides and angles of plane triangles. 2. The circumference of every circle (as before observed in Geom. Def. 56) is supposed to be divided into 360 equal parts, called Degrees ; also each degree into 60 Minutes, and each minute into 60 Seconds, and so on. Hence a se- micircle contains 180 degrees, and a quadrant 90 degrees. 3. The Measure of an angle (Def. 57, Geom.) is an arc -of any circle contained between the two lines which form that angle, the angular point being the centre ; and it is esti- mated by the number of degrees contained in that arc. Hence, a right angle, being measured by a quadrant, or quarter of the circle, is an angle of 90 degrees ; and the sum of the three angles of every triangle, or two right an* gles, is equal to 180 degrees. Therefore, in a right-angled triangle, taking one of the acute angles from 90 degrees, leaves the other acute angle ; and the sum of the two angles, in any triangle, taken from 180 degrees, leaves the third angle ; or one angle being taken from 180 degrees, leaves the sum of the other two angles. 4. Degrees arc marked at the top of the figure with a small °, minutes with ', seconds with *, and so on. Thus, 57° 30' 12", denote 57 degrees 30 minutes and 12 seconds. 5. The Complement of an arc, is what it wants of a quadrant or 90°. Thus, if ad be a quadrant, then bd is the complement of the arc ab ; and, reciprocally, ab is the complement of bd. So that, if ab be an arc of 50°, & then its complement bd will be 40°. 6. The Supplement of an arc, is what it wants of a semicircle, or 180°. T Thus, if ade be a semicircle, then bde is the supplement of the arc ab ; and, reciprocally, ab DJEFiirrnoifs. 87V is the supplement of the arc bde. So that, if ab be an arc of 50°, then its supplement bob will be 130°. 7. The Sine, or Right Sine, of an arc, is the line drawn from one extremity of the arc, perpendicular to the diameter which passes through the other extremity. Thus, bf is the sine of the arc ab, or of the supplemental arc bde. Hence the sine (bf) is half the chord (bo) of the double arc (bao). 8. The Versed Sine of an arc, is the part of the diameter intercepted between the arc and its sine. So, af is the versed sine of the arc ab, and ef the versed sine of the arc edb. 9. The Tangent of an arc, is a line touching the circle in one extremity of that arc, continued from thence to meet a line drawn from the centre through the other extremity ; which last line is called the Secant of the same arc. Thus, ah is the tangent, and en the secant, of the arc ab. Also, ei is the tangent, and ci the secant, of the supplemental arc bob. And this latter tangent and secant are equal to the former, but are accounted negative, as being drawn in an opposite or contrary direction to the former. 10. The Cosine, Cotangent, and Cosecant, of an arc, are the sine, tangent, and secant of the complement of that arc, the Co being only a contraction of the word comple- ment Thus, the arcs ab, bi>, bein * the complements of each other, the sine, tangent, or secant of the one of these, is the cosine, cotangent, or cosecant of the other. So, bf, the sine of ab, is the cosine of bd ; and bk, the sine of bd, is the cosine of ab : in like manner, ah, the tangent of ab, is the cotangent of bd ; and dl, the tangent of db, is the cotangent of ab ; also, ch, the secant of ab, is the cosecant of bd ; and cl, the secant of bd, is the cosecant of ab. Cord. Hence several important properties easily follow from these definitions ; as, 1st, That an arc and its supplement have the same sine, tangent, and secant ? but the two latter, the tangent und secant, are accounted negative when the arc is greater than a quadrant or 90 degrees. 2d, When the arc is 0, or nothing, the sine and tangent are nothing, but the secant is then the radius oa, the least it can be. As the arc increases from 0, the sines, tangents, and secants, all proceed increasing, till the arc becomes a whole quadrant ad, and then the sine is tkc greatest it can be, being the radius en of the circle ; and both the tangent and secant are infinite. 3d, Of any arc ab, the versed sine af, and cosine bk, or cf, together make up the radius ca of the evecta.— •80 FLAHK TUOONOVXTBT. adius ca, the tangent ah, and the^ secant ch, form a right» angled triangle c % h. So also do the radius, sine, and conn** form another right-angled triangle chf or cbk. As also the radius, cotangent, and cosecant, another right-angled triangle cdl. And all these right-angled triangles are similar to each other. 11. The sine, tangent, or secant of aa angle, is the sine, tangent, or secant of the arc by which the angle is mea- sured, or of tho degrees, dic- ta the same arc or angle. 12. The method of con- structing the scales of chords, sines, tangents, and secants, usually engraven on instru- ments, for practice, is exhi- bited in the annexed figure. 13. A Trigonometrical Canon, is a table showing the length of the sine, tan- Sent, and secant, to everyjg egree and m minute of the quadrant, with respect to the radius, which is expressed by unity or 1, with any number of ciphers. The logarithms ef these sines, tangents, and secants, are also ranged in the tables ; and these are most commonly used, as they perform the calculations by only addition and subtraction, instead of the multiplication and division by the natural sines, etc. ac- cording to the nature of logarithms. Such tables of log. sines and tangents, as well as the logs of common numbers, greatly facilitate trigonometrical computations, and are now very common. Among the most correct are those published by the author of this Course. PROBLEM I. To compute the Natural Sine and Cosine of a Given Arc. Tni9 problem is resolved after various ways. One of these is as follows, viz. by means of the ratio between the diameter PBOBLSK*. 881 and circumference of a circle, together with the known series for the sine and cosine, hereafter demonstrated. Thus, the aemicircumference of the circle, whose radius is 1, being 3-14169^653589703 &c, the proportion will therefore be, as the number of degrees or minutes in the semicircle, is to the degrees or minutes in the proposed arc, so is 3 14169265 dec, to the length of the said arc. This length of the arc being denoted by the letter a ; and its sine and cosine by « and c ; then will these two be ex- pressed by the two following series, viz. _ _ jr r , _a*_ a* _ 9 a 2.3 + 2.3.4.5 2.3.4.5.0.7 = a 3 , a* a 7 , . ° 6 120 5040 2 ^2.3.4 2.3.4.5.6 T C a 9 o* a 8 = l — 2 + 34-720 + &C - Exam. 1. If it be required to find the sine and cosine of 1 minute. Then, the number of minutes in 180° being 10800, it will be first, as 10600 : 1 : : 3 14150205 dec. : •000290888208665 = the length of an arc of one minute. Therefore, in this case, a= -0002908882 and^a 3 = -000000000004 dec. the diflT. \ss= 0002908882 the sine of 1 minute. Also, from 1- take ia* = 0000000423079 dec. leaves c = -9999999577 the cosine of 1 minute. Exam. 2. For the sine end cosine of 5 degrees. Here as 1H)° : 5« : : 3 141 59205 &c. : -08726646 = a the length of 5 degrees. Hence a == -08726K46 — Ja* * - -00011076 + ^ a s = -00000004 these collected give $ = '08715574 the sine of 5\ And, for the cosine, 1 = 1* — ±a 2 = — -00380771 + = -00000241 these collected give c = •99619470 the cosine of 5°. 383 PL ARB TBIGOXOXXTBY. After the same manner, the sine and cosine of any other arc may be computed. Rut the greater the arc is, the slower the series will converge, in which case a greater number of terms must be taken, to bring out the conclusion to the same degree of exactness. Or, having found the sine, the cosine will be found from it, by the property of the right-angled triangle cbf, vis. the cosine cp = y/{cu* — bp 11 ), or c =• y/{l — ?). There are also other methods of constructing the canon of sines and cosines, which, for brevity's sake, are here omitted : some of them, however, are explained under the analytical trigonometry in the second volume of this Course. problem n. To compute the Tangents and Secants. Tub sines and cosines being known, or found by the foregoing problem ; the tangents and secants will be easily found, from the principle of similar triangles, in the follow, ing manner : In the first figure, where, of the arc ab, bf is the sine, cf or bk the cosine, ah the tangent, ch the secant, dl the cotangent, and cl the cosecant, the radius being ca or cb or cd ; the three similar triangles cfb, cah, cdl, give the fol- lowing proportions : 1**, cf : fb : : ca : ah ; whence the tangent is known, being a fourth proportional to the cosine, sine, and radius. 2d, cp : cb : : ca : ch; whence the secant is known, being a third proportional to the cosine and radius : or, being, indeed, the reciprocal of the cosine when the radius is unity. 3d, bf : fc : : cd : dl ; whence the cotangent is known, being a fourth proportional to the sine, cosine, and radius. Or, aii ; ac i i cd : dl ; whence it appears that the co- tangent is a third proportional to the tangent and radius ; or the reciprocal of the tangent to radius 1. 4th bf : bc : : cd : cl ; whence the cosecant is known, being a third proportional to the sine and radius ; or the re- ciprocal of the sine to radius 1. As for the log. sines, tangents, and secants, in the tables, they are only the logarithms of the natural sines, tangents, and secants, calculated as above. Having given an idea of the calculation and use of sines, tangents, and secant*, wt> may uow proceed to resolve the PROBLEMS. 889 several cases of Trigonometry ; previous to which, however, it may be proper to add a few preparatory notes and ob- servations, as below. Note 1. There are three methods of resolving triangles, or the cases of trigonometry ; namely, Geometrical Con- struction, Arithmetical Computation, and Instrumental Opera- tion ; of which the first two will here be treated. In the First Method, The triangle is constructed, by making the parts of the given magnitudes, namely, the sides from a scale of equal parts, and the angles from a scale of chords, or by some other instrument. Then measuring the unknown parts by the same scales or instruments, the solu- tion will be obtained near the truth. In the Second Method, Having stated the terms of the proportion according to the proper rule or theorem, resolve it like any other proportion, in which a fourth term is to fie found from three given terms, by multiplying the second and third together, and dividing the product by the first, in working with the naturul numbers ; or, in working with the logarithms, add the logs, of the second and third terms together, and from the sum take the log. of the first term ; then the natural number answering to the remainder is the fourth term sought. Note 2. Every triangle has six parts, viz. three sides and three angles. And in every triangle proposed, there must be given three of these parts, to find the other three. Also, of the. three parts that are given, one of them at least must be a side ; because, with thf same angles, the sides may be greater or less in any proportion. Note 3. All the cases in trigonometry, may be comprised in three varieties only ; viz. 1**, When a side and its opposite angle are given. ' ■ 2d, When two sides and the contained angle are given. 3d, When the three sides are given. For there cannot possibly be more than these three varie- ties of cases ; for each of which it will therefore be proper to give a separate theorem, as follows : THEOREM I. When a Side and Us Opposite Angle are two of the Give* Parts. Then the unknown parts will be found by this theorem : viz. The sides of the triangle have the same proportion, to each other, as the sines of their opposite ua^tatYm*. 884 PLANE TRIGONOMETRY. That is. As any one aide, la to the sine of its opposite angle ; So is any other side, To the sine of its opposite angle. Demonstr. For, let abc be the pro- C posed triangle, having ab the greatest Y^f\ aide, and bc the least. Take ad = sSv* \ bc, considering it as a radius ; and let |\i \ fall the perpendiculars dk, cf, which £ 3£ Jf B will evidently be the sines of the an- §lea a and a, to the radius ad or bc. Tow the triangles ade, acf, are equiangular ; they therefore have their like sides proportional, namely, ac : cf : : ad or Be : de ; that is, the side ac is to the sine of its opposite an- gle b, as the aide bc is to the sine of its opposite angle a. Note 1. In practice, to find an angle, begin the proportion with a side opposite to a given angle. And to find a aide, begin with an angle opposite to a given side. Note 2. An angle found by this rule is ambiguous, or un- certain whether it be acute or obtuse, unless it be a right angle, or unless its magnitude be such as to prevent the ambiguity ; because the sine answers to two angles, which are supplements to each other ; and accordingly the geome- trical construction forms two triangles with the same parts that are given, as in the example below ; and when there is no restriction or limitation included in the question,- either of them may be taken. The number of degrees in the table, answering to the sine, measurc%he acute angle ; but if the angle be obtuse, subtract those degrees from 180°, and the remainder will be the obtuse angle. When a given angle if obtuse, or a right one, there can be no ambiguity ; for then neither of the other angles can be obtuse, and the geometri- cal construction will form only one triangle. EXAMPLE I. In the plane triangle abc, I ab 345 yards Given { bc 232 yards a 37" 20' Required the other parts. 1. Geometrically. Draw an indefinite line ; on which set off ab = 345, from some convenient scale of equal parts. — Make the angle THEOREM I . 885 a = 37°i. — With a radius of 232, taken from the same scale of equal parts, and centre b, cross ac in the two points, c, c. — Jjastly, join bc, bo, and the figure is construct, ed, which gives two triangles, and shows that the case is am. biguous. Then, the sides ac measured by the scale of equal parts, and the angles b and c measured by the line of chords, or other instrument, will be found to be nearly as below ; viz. ac 174 z.b27« Z.cll5 ft |. or 3741 or 78} or 64 J. 2. Arithmetically. First, to find the angles at c. As side bc 232 . log. 2-3654880 To sin. op. £ a 37° 20' . . 9*78^7958 So side ab 345 - 25378191 To sin. op. c 1 1 5' 36* or 61° 24' 9 9551269 add Z.A 37 20 37 20 the sum 152 56 or 101 44 taken from 180 00 180 00 leaves b 27 04 or 78 16 Then, to find the side ac. As sine Z a 37° 20' To op. side bc 2S2 „ . . i 270 0* So sin. Z b J 78 16 To op. side ac 174 07 or 374*56 - log. 9-7827956 - ' 2-3654*80 9*6580371 9*9908291 '84407*93 2-5735218 EXAMPLE II. In the plane triangle abc, C ab 365 poles Given < £a 57° 12' 24 45 Required the other parts. EXAMPLE III. In the plane triangle abc, ( ac 120 feet Given < bc 112 feet ( Za 57" 27' Required the other parts. Vol. I. 50 Ans. < a c AC BC 98° 3' 154 f £b 64*34' 21' or 115 25 39 £c57 58 39 or 7 7 21 ab 112-65 feet lor 16-47 fa*. 386 PLANE TBIOOH OMETXY. THEOREM II. When too Sides and their Contained Angle are given* F1S8T find the sum and the difference of the given sides. Next subtract the given angle from 180°, and the remainder will be the sum of the two other angles ; then divide that by 2, which will give the half sum of the said unknown an- gles. Then say, As the sum of the two given sides, Is to the difference of the same sides ; So is the tang, of half the sum of their op. angles. To the tang, of half the diff. of the same angles. Add the half difference of the angles, so found, to their half sum, and it will give the greater angle, and subtracting the same will leave the less angle : because the half sum of any two quantities, increased by their half difference, gives the greater, and diminished by it gives the less. All the angles being thus known, the unknown side will be found by the former theorem. Note. Instead of the tangent of the half sum of the un. known angles, in the third term of the proportion, may be used the cotangent of half the given angle, which is the same thing. Demon. Let abc be a plane triangle of which ac, ci, and the included angle c are given : c being acute in the first figure, obtuse in the second. On ac, the longer side, set off cd = cb the shorter ; join bd, and bisect it in e ; also, bisect ad in o, and join », cs, producing the latter to r. Now J(ac + cb) = £(2gd + 2dc) = co and ^(ac — cb) = |(2ag) = ag also {(a + b) = £(cdb + CBD ) x cm and |(b — a) == abc — J sum = abd: C THEOREM II. 397 also, because cs bisects the base of the isosceles triangle cbd, it is perpendicular to it : Therefore ec = tangent of cbd ) . M kp = tangent of abd 5 t0radMI8BB - Lastly, because in the triangle acf, oe is parallel to af (Geom. th. 82) we have co : oa : : ce : ef ; that is, J(ac + cb) : |(ac — cb) : : tan. £(b + a) : tan. J(b — a) ; or, siuce doubling both the antecedent and consequent of the first ratio does not change the mutual relatiou of its terms, we have ac + cb : ac — cb : : tan. J(b + a) : tan. J(b — a), q. e. d. EXAMPLE I. In the plane triangle abc, C ab 345 yards Given < ac 174*07 yards ( £ a 37° 20' Required the other parts. 1. Geometrically. Draw ab = 345 from a scale of equal parts. Make the angle a == 37° 20'. Set off ac = 174 by the scale of equal parts. Join bc, and it is done. Then the other parts being measured, they are found to be nearly as follow ; viz. the side bc 232 yards, the angle b 27°, and the angle c 115°{. 2. Arithmetically. The side ab 345 From 180° 00' the side ac 174*07 take Lk 37 20 their sum 519*07 sum of c and b 142 40 their differ. 170*93 half sum of do. 71 20 As sum of sides ab, ac, - - 519 07 log. 2*7152259 Todiff.ofsides ab.ac, - - 170*93 - 2*2328183 So tang, half num^flc and b 71- 2ff - 10 4712979 To tang, half diff. L s c and b 44 10 - 9-9888903 these added give 115 36 and subtr. give z. b 27 4 888 PLANK TKJOOMOXKTBY. Then, by the former theorem. As sin Z c 1 15" 3ff or 64 24' - log. 0-95518S0 To its ou. side ar 345 - - - 2-53781M Sown. ofZ a37 2ff • - - 07827U58 To its op. side bc 232 - • - 2-3054880 EXAMPLE n. In the plane triangle abc, ( ab 365 poles ( bc 800-86 Given I ac 154-33 Ana. { L b 24« 45' a 57" 12* file 08 3 Required the other parts. EXAMPLE III* In the plane triangle abc, ( ac 120 yards i ab 112-65 Given I bc 112 yards Ans. 1 Z± 57 e 2T 0" ( Zc 57" 58' 39" ( ^b 65 34 21 Required the other parts. TIIEOREM III. When the Three Sides of a Triangle a r give a. First, let full a perpendicular from the greatest angle on the opposite side, or base, dividing it into two segments, and the whole triangle into two right-angled triangles : then the proportion will be, As the base, or sum of the segments, Is to the sum of the other two sides ; So is the difference of those sides, To the diff. of the segments of the base. Then take half this difference of the segments, and add it to the half sum, or the half base, for the greater segment, and subtract the same for the less segment. Ht nee, in each of the two right-angled triangles, there will be known two sides, and the right angle opposite to one of them ; consequently the other angles will be found by the first theorem. Demonsfr. By theor. 35, Geom. the rectangle of the sum and difference of the two sides, is equal to the rectangle of the sum and difference of the two segments. Therefore, by forming the sides of these rectangles into a proportion by THfOREV Iff. 389 tbeor. 76, Geometry, it will appear that the sums and dif- ferences are proportional as in this theorem. N. B. Before you commence a solution of an example to this case, ascertain whether the triangle be rigta-angled or not, by determining whether the square of the rongest side be equal or unequal to the sums of the squares of the other two. If equal, the exircple may be referred to the notes to theorem nr. EXAMPLE i. In the plane triangle abc, Given 4 345 yards *• ~ de8 J bc 174-07 To find the angles. 1. Geometrically. Draw the base ab = 345 by a scale of equal parts. With radius 232, and centre a, describe an arc ; and with radius, 174, and centre b, describe another arc, cutting the former in c. Join ac, bc, and it is done. Then, by measuring the angles, , they will be found to be nearly as follows, vie. Z a 27% L b 37°i, and Lc 11 5° J. 2. Arithmetically* Having let fall the perpendicular cp, it will be, As the base ab : ac + bc : : ac — bc : ap — bp, that is, as 345 : 406 07 : : 57 03 : 68-18 = ap — bp, its half is - 34 09 the half base is 172-50 the sum of these is 206-59 = ap. and their diff. is 138*41 = bp. Then, in the triangle apc, right-angled at p, As the side ac - - 232 - log. 2-3054880 To sin. op. iiP . . 90 s . - 10*0000000 So is the side ap . - 208-59 . 2-3151093 To sin. op. L acp . - 62° 56' . 9-94t6213 which taken from - 90 00 leaves the Lh. 27 04 990 plaice naooHoxmr. Again, in the triangle bpc, right-angled at p 9 As the aide bc . . 174-07 - log. 2-2407230 Toain. op. £p - 00" . . 10-0000000 8oiaaide bp - - 188-41 . . 2-1411675 To ein op. L bcp - - 52° 40' . . 9-0004486 which taken from - 90 00 leaves the Lb 87 20 Also the /acp 62° 66' added to Zbcp 52 40 gives the whole £acb 115 36 So that all the three angles are as follow, viz. the iLA27'4'; the Z.B3T20'; the Lc 115* 36'. The angles a and b may also easily be found by the ex- » AC BO pressions sec. a = — , sec. b = — , or the equivalent logs. EXAMPLE II. In the plane triangle abc, r ( ab 365 poles { L a 57* 12' ss-issa Has - To find the angles exampw in. r; w «n (ab120 (^a 57*27' 0' \ AC 112-65 Ans. < /1b 57 58 39 the sides J BcU2 (^c 64 34 21 To find the angles. The three foregoing theorems include all the eases of plane triangles, both right-angled and oblique. But there are other theorems suited to some particular forms of tri- angles (see vol. ii.), which are sometimes more expeditious in their use than the general ones ; one of which, as the case for which it serves so frequently occurs, may be here ex- plained. TRSOBBM IV. 4 an THEOREM IV. When a Triangle is Right-angled ; any of the unknown part* may be found by the following proportions : viz. As radius Is to either leg of the triangle ; So is tang, of its adjacent angle, To its opposite leg ; And so is secant of the same angle, To the hypothenuse. Demonstr. ab being the given leg, in the * right-angled triangle abc : with the centre a, and any assumed radius ad, describe an arc de, and draw dp perpendicular to ab, or parallel to bc. Then it is evident, from the definitions, that df is the tangent, and af the secant of the arc de, or of the angle a which is measured by that arc, to the radius ad. Then, because of the parallels bc, df, it will be - as ad : ab : : df : bc and : : af : ac, which is the same as the theorem is in words. Note. The radius is equal, either to the sine of 90°, or the tangent of 45 r ; and is expressed by 1, in a table of natural sines, or by 10 in the log. sines. EXAMPLE I. In the right-angled triangle abc, Given \ TAf 48' \ To A0 and B& 1. Geometrically. Make ab = 162 equal parts, and the angle a=53* 7' 48' ; then raise the perpendicular bc, meeting ac in c. So shall ac measure 270, and bc 216. 2. Arithmetically. As radius log. 10*0000000 To leg ab 162 2-2005150 So tang. 53° 7 48" 10-1240871 TolegBc 216 - S-S&M&fllY S03 PLANE raieONOMBTRY. So secant Lk - 53° T 48 7 - 10 2218477 To hyp. ac 270 . 2-4318627 EXAMPLE H. In the right-angled triangle abc, n . 5 the leg ab 180 . (ac 302-0146 UlTcn ) the Z a 62 40' An8, I bc 348-2464 To find the other two side*. Note. There is sometimes given another method for right- angled triangles, which is this : abc being such a triangle, make one leg ab radius ; that is, with centre a, and distance ab, describe an arc bf. Then it is evident that the other leg bc represents the tangent, and the hypo- thenuse ac the secant, of the arc bf, or of the angle a. In like manner, if the leg bc be made radius ; then the other leg ab will re- present the tangent, and the hypothenuse ac the secant, of the arc bo or angle c. But if the hypothenuse be made radius ; then each leg will represent the sine of its opposite angle ; namely, the leg ab the sine of the arc ae or angle c, and the leg bc the sine of the arc cd or angle a. Then the general rule for all these cases is this, namely, that the sides of the triangle bear to each other the same proportion as the parts which they represent. And this is called, Making every side radius. Note 2. When there are given two sides of a right-angled triangle, to find the third side ; this is to be found by the property of the squares of the sides, in theorem 34, Geom. vis* that the square of the hypothenuse, or longest side, is equal to both the squares of the two other sides together. Therefore, to find the longest side, add the squares of the two shorter sides together, and extract the square root of that sum ; but to find one of the shorter sides, subtract the one square from the other, and extract the root of the re- mainder. Or, when the hypothenuse, h, and either the base, b, or the perpendicular, p, are given : then half the sum of log. (a + p) and log. (u — p) — log. b ; and half the sum ofJog. (h -f b) and log. (h — b) = log. p. QflSFUL JPOftMGUB* When b and p are given, the following logarithmic ope- ration may sometimes be advantageously employed ; viz. Find n the number answering to the log. diff., 2 log. p — log. b ; and make b + * = x : then, £ (log. m + log. b) = log. h, the hypothenuse. The truth of this rule is evident : for, from the nature of logarithms. — = if; whence b + n = b + -t- = b 9 +p 9 — - — =* m; and £ (log. n + log.fl) = J log. mb = | log. ( B * + p>) = log. i/(B a + p*) = log. H. Or, stall more simply, find 10 + the diff. (log. p — log. b) in the log. tangents. The corresponding log. secant added to log. b == log. H. Note, also, as many right-angled triangles in integer num- bers as we please may be found by making »* + a 9 5= hypothenuse m 9 — n 9 = perpendicular 2mn ■= base m and n being taken at pleasure, m greater than ft. Before we proceed to the subject of Heights and Distances " we shall give, A CONCISE INVESTIGATION OF SOME OF THE MOST USEFUL TRIGONOMETRICAL FORMULAE. Let ab, ac, ad, be three arches, such that bc = c*, aud o the centre. Join ao, oc, bo. Draw deq and oi per- pendicular, and bdc || to oa. Join Ma and bisect it by the radius on ; and draw ah || to bp. Then is ah = sin. ac oh = cos. ac; alSO DE = EQ = 8in. AD EK = oi = sin. ab ojk = sin. ad + sin. ab dk = sin. ad — sin. ab BI = IM = COS. AB OR => KI = COS. AD MK = COS. AB + COS. AD BK = COS. AB — COS. AD Because the angles at k are right angles : arc bd + arc.Mli = 180*, and arc dc + arc mn = Ifr .% MP = VH = OG = COS. DC = COS. \ ' V Vol, I 51 % V 394 PLANE TRIGONOMETRY. also, because ao = 4(ab + ad) = £baq = angle aoc (at centre) = bdq (at circumf.) = bmq (on same arc) .% triangles aoh, bdk, qmk, are equiangular. Hence — I. oa : ah : : mq : qk ; that is, rad. : sin. ac : : 2 cos. bc : sin. ad + sin. ab II. ao : oh : : bd : dk ; or, rad. : cos. ac : : 2 sin. bc : sin. ad — siii. ab III. ao : oh : : on : kk ; or, rad. : cos. ac : : 2 cos. bc : cos. ab + cos. ad IV. ao : ah : : db : dk ; or, rad. : sin. ac : : 2 sin. bc : cos. ab — cos ad ; also, V. BK . KM = DK . KQ, that is (COS. AB COS. AD) (cos. ab+cos. ad) = (sin. ad— sin. ab) (sin. AD+sin. ab). By reducing the above four proportions into equations, making rad. = 1, we obtain two distinct classes of formula*, thus :— First Class, ac = a, cb = b ; then ad = a + 6, ab = a— b l 1. sin. (a + b) + sin. (a — I) = 2 sin. a cos. b 2. sin. (a + b) — sin. (a — b) = 2 cos. a sin. 6 3. cos. (a — 6) + cos. (a + 6) = 2 cos. a cos. 6 4. cos. (a — b) — cos. (a + b) = 2 sin. a sin. b Second Class, ad = a, ab = b ; then ac = £(a + 6), bc = £(a — 6). 5. sin. a + sin. 6=2 sin. i(a + b) cos. £(a — b) 6. sin. a — sin. 6 = 2 cos.£(a + b) sin. £(a — b) 7. cos. 6 + cos. a ss2 cos.£(a + b) cos. \\a — 6) 8. cos. b — cos. a = 2 sin. i(a + &) sin, £(a — 6) The first class is useful in transforming the products of sines into simple sines, and the contrary. The second facilitates the substitution of sums or. differ- ences of sines for the products, and the contrary. Taking the sum and the difference of equations 1 and 2, also of 3 and 4, remembering that sin. = cos. tan. we obtain the following : Third Class. 9. sin. (a + b) = sin. a cos. b + sin. b cos. a = cos. a co*. h (tan. a + tan. b) USEFUL FOBKiflLfi. 305 10. sin. (a — 6) = sin. a cos. b — sin. b cos. a = cos. £i cos. 6 (tan. a — tan. b) 11. cos. (a + b) = cos. a cos. & — - sin. a sin. b ■s cos. a cos. 6 (1 — tan. a tan. b) 12. cos. (ci — b) = cos. a cos. 6 + sin. a sin. b. = cos. a cos. b (1 +tan. a tan. b). From these, making a = 6, we readily obtain the ex- pressions for sines and cosines of double arcs ; also dividing equation 9 by 11, and equation 10 by 12, we obtain ex- pressions for the tangents of a + b and a — b. Thus we have : — Fourth Class. 13. sin. 2a = 2 sin. a cos. a = 2 ooe.'a tan. a 14. cos. 2a = cos. 2 a — sin. 2 a = cos. 2 a (1 — tan. 2 a) sin. , . . v / • t\ tan. a + tan. 6 15. (a + 6) = tan. (a + 6) = - — — z cos. v ' 1— tan. a tan. 6 sin. , , v /ix tan. o — tan. b (a — b) = tan. (a — ft) = r . cos. v ' ' 1 + tan. a tan. 6 . 16 _ 4 2 tan. a 17. tan. 2a = 18. cot. 2a = 1— tan. 2 a 1— tan. 2 a 2 tan. a Substituting in the second class, for sin. i(a+b), cos. i(a + b) tan. $(a+b), .and for sin. i(a— b), cos. £(a — b) tan. we have : — Fifth Class. 19. cos. &+cos. a=2 cos. £(a+&)cos. 4(a— 6). — Seeequa. 7. 20. cos. 6 — cos.a = tan. $(a+b) tan. £(a— 5) 2 cos. |(a+&) cos. £(a-&) = ton- tan. i(a— &)(cos.&+cos. a) 21. sin. a+sin. 6 = tan. ±(a+&} 2 cos. i( a +&) cos. i(a-5) = tan. £(a+6) (cos. a+cos. b) 22. sin. a -sin. b = tan. ^(a-6) 2 cos. J( a +&) cos. }(*—*) = tan. |(a— 6) (cos. a+cos. 6) ^ sin. a+sin. 5 tan. i{a+b) 23. r — r = - — 77 — 7x - from 21 and 22. sin. a— sm. o tan. J(a— 6) ^ sin. a+sin. 6 w . r rt « 24. r = tan. Ua + 6) : from 21. cos.a+cos. o ^ sin. a— sin. i A , , r M 25. ; r = tan. Ua —6) : from 22. cos.a+cos. b ,v 300 o* nktolm Excunplts for Exercise. 1. Demonstrate that in any right-angled plane triangle tiur following properties obtain : viz. oerp. base (1.) £_£.=£tan* ang. at base. (2.) =tan. ang. at vertex* * 'base * v 'perp. ^ (3.) j^EL'sssin. ang. at base. (4.) — sin. ang. at vertex* (5. ) « sec. ang. at base. (6.) =sec. ang. at vertex* 2. Demonstrate that tan< a + sec. a = tan. (45* + i^)* l*4"tan * a 3. Demonstrate that sec. 2a = - — - — Vi> tnat 1 — tanv * l+tan. 8 A sec. 9 a cosec, 2a = 2 tan. a 2 tan. a 4. Given Axy = ay 8 + ns* ; to find x and y the sine and cosine of an arc. 5. Demonstrate that of any arc, tan. a — sin. 2 = tan. 2 sin. 2 - 6. Demonstrate that if the tan. of an arc be = ^/n, the sine of the same arc is = y/ n n+1" OF HEIGHTS AND DISTANCES, &c. 6y the mensuration and protraction of lines and angles, are determined the lengths, heights, depths, and distances of bodies or objects. Accessible lines are measured by applying to them some certain measure a number of times, as an inch, or a foot, or yard. But inaccessible lines must be measured by taking angles, or by such-like method, drawn from the principles of geometry. When instruments are used for taking the magnitude of the angles in degrees, the lines are then calculated by trigonome- try : in the other methods, the lines are calculated fr#m the principle of similar triangles, or some other geometrical property, without regard to the measure of the angles* AlfD MBTANCEf . 397 Angles of elevation, or of depression, are usually taken either with a theodolite, or with a quadrant, divided into de- grees and minutes, and furnished with a plummet suspended from the centre, and two open sights fixed on one of the radii, or else with telescopic sights. To lake an Angle of Altitude and Depression with tike Quadrant. Let a be any object, as the sun, inoon, or a star, or the top of a tower, or hill, or other eminence : and let it be required to find the measure of the angle abc, which a line drawn from the object makes above the horizontal line bc. Place the centre of the quadrant in the angular point, and move it round there as a oentre, till with one eye at n, the other being shut, you perceive the object a through the sights ; then will the arc gh of the quadrant, cut off by the plumb- line, bii, be the measure of the angle abc as required. The angle abc of depression of any object a, below the horizontal line bc, is taken in the same manner; except that here the eye is applied to the centre, and the measure of the angle is the arc gh, on the other side of the plumb-liner The following examples are to be constructed and calcu- lated by the rules of Trigonometry. B EXAMPLE t. Having measured a distance of 200 feet, in a direct hori- zontal line, from the bottom of a steeple, the angle of eleva- tion of its top, taken at that distance, was found to be 47° 30' ; hence it is required to find the height of the steeple. Construction. Draw an indefinite line ; on which set off ac a 200 equal parts, for the measured distance. Erect the indefinite per- pendicular ab ; and draw cb so as to make the an&la,$ «a OF HEIGHTS 47° SO 7 , tho angle of elevation ; and it is done. Then ab, measured on the scale of equal parts, is nearly 218J. Calculation. As radius - 10-0000000 To ac 200 - - 2-3010300 So tang. L c 47° 80' 10-0379475 To ab 218-26 required 2-3380775 Or, by the nat. tangents, we have ac X tan. bca = 200 X 1-091308 = 218-2616 = ab. EXAMPLE It. What was the perpendicular height of a cloud, or of a balloon, when its angles of elevation were 35° and 64°, as taken by two observers, at the same time, both on the same side of it, and in the same vertical plane ; the distance be* tween them being half a mile or 880 yards ? And what was its distance from the said two observers ? Construction. Draw an indefinite ground line, on which set off the given distance ab = 880 ; then a and b are the places of the observers. Make the angle a = 35°, and the* angle B = 64° ; then the intersection of the lines at c will be the place of the balloon : whence the perpendicular cd, being let fall, will be its perpendicular height. Then, by measure- ment are found the distances and height nearly as follow , viz. ac 1631, bc 1041, dc 936. C Calculation. First, from L b 64' take L a 35 leaves L. acb 29 — ' — / ; A 33 D Then in the triangle abc, As sin. Zacb 29° ... 0-6855712 To op. side ab 880 - 2-9444827 So sin. Lk 35° - 9-7585913 To op. side bc 1041-125 3-0175028 AND DISTANCES. 399 As sin. ^acb 29° - . - 9-6855712 To op. side ab 880 ... 2-9444827 So sin. Z.B 1 16° or 64° - 9-9536602 To op. side ac 1631-442 - - 3-2125717 And in the triangle bcd, ( As sin. L d 90° - - - 10-0000000 To op. side bc 1041-125 - . 3-0175028 So sin. L b 64° - - - 0-9536602 To op. side cd 935-757 - - 2-9711630 EXAMPLE in. Having to find the height of an obelisk standing on the top of a declivity, I first measured from its bottom a distance of 40 feet, and there found the angle, formed by the oblique plane and a line imagined to go to the top of the obelisk, 41 ° ; but after measuring on in the same direction 60 feet farther, the like angle was only 23° 45'. What then was the height of the obelisk ? » Construction. Draw an indefinite line for the sloping plane or declivity, in which assume any point a for the bottom of the obelisk, from which set off the distance ac = 40, and again cd = 60 equal parts. Then make the angle c = 41°, and the angle d = 23° 45' ; and the point b where the two lines meet will be the top of the obelisk. Therefore ab joined, wilr be its height. — Draw also the horizontal line de perp. to ab. Calculation. From the L c 41° 00' take the /d 23 45 loaves the £ dbc 17 15 Then in the triangle dbc, As sin. /.dbc 17° 15' To op. side dc 60 So sin. L d 23 45 To op. side cb 8M88 9-4720856 1-7781513. 9-6050320 i-9\vwn 400 OF 1IBEGHTS And in the triangle abc, As sum of sides cb, ca To diff. of sides cb, ca So tang. }(a + b) To tang. ] (a — b) 121*488 - 20845333 41-488 . 1-6179225 60* 3C - 10-4272623 42 24} - 9*9606516 the diff. of these is £cba 27 5} the sum is L cab 111 54} Lastly, as sin. £cba 27° 5'} . - 9*0582842 To op. side ca 40 . 1*6020600 So sin. Lc - 41° 0' - - 9*8169429 To op. side ab 57-623 - - 1*7607187 Also the L ade ^= bac — 90 r = 21° 54'}. EXAMPLE IV. Wanting to know the distance between two inaccessible trees, or other objects, from tho top of a tower 120 feet high, which lay in the same right line with the two objects, I took the angles formed by the perpendicular wall and lines con* ceived to be drawn from the top of the tower to the bottom of each tree, and found them to be 33° and 64°}. What is the distance between the two objects ? Construcliou. Draw the indefinite ground line bd, and perpendicular to it ba = 120 equal parts. Then draw the two lines ac, ad, making the two angles bac, bad, equal to the given measures 33' and 64°£. So shall c and d be the places of the two objects. Calculation. ^ St - First, in the right-angled triangle abi , As radius - 10-0000000 To ab . 190 ... 4i 2 0791812 Sotaug. /I bac 33 l » - - . 9-8125174 Tob<; . 77-«» - - 1-8916980 ASD DISTANCES. 401 Then in the right-angled triangl* abd, As radius 10 'OOOOOOO To ab • - 120 . 2 0701812 So tang. L bad - 64° 3a - - 10*3215089 To bd • 251-586 - . 2*4006851 From which take bc 77*929 leaves the dist. cd 173*056, as required. Or thus, by the natural tangents, From nat. tan. dab - 64° 30' = 2*0905436 Take nat. tan. cab - . 33 =0 6494076 Difference - • - 1*4471360 If drawu into ab - 120 The result gives cd - = 173-05632 EXAMPLE V. Being on the side of a river, and wanting to know the distance to a house which was seen on the other side, I mea- sured 200 yards in a straight line by the side of the river ; and then, at each end of this line of distance, took the hori- zontal angle formed between the house and the other end of the line ; which angles were, the one of them 68° 2 , and the other 73° 15'. What were the distances from each end to the house ? Construction. Draw the line ab = 200 equal parts. Then draw ac so as to make the angle a = 68 2', and bc to make the angle b=73' 15 . So shall the point c be the place of the house required. The calculation, which i* left for the student's exercise, gives ac = 30619, bc = 296 54. Exam. vi. From the edge of a ditch, of 36 feet wide, surrounding a fort, having taken the angle of elevation of the top of the wall, it was found to be 62*40' : required the height of the wall, and the length of a ladder to reach from my station to the'top of it ? At t height of wall 69-64, Ans " {ladder, 7%Afe*\. Vol. I. 52 Or KXZOHTt Exam, vii. Required the length of a shoar, which being to strut 11 feet from the upright of a building, will support a jamb 23 feet 10 inches trooMhe ground 7. Ans. 26 feet 3 inches. Exam. viii. A ladder, 40 feet long, can be so placed, that it shall reach a window 33 feet from the ground, on one side of the street ; and by turning it over, without moving the foot out of its place, it will do the same by a window 21 feet high, on the other side : required the breadth of the street T Ans. 56-649 feet. Exam. ix. A maypole, whose top was broken off by a blast of wind, struck the ground at 15 feet distance from the foot of the pole : what was the height of the whole maypole, supposing the broken piece to measure 39 feet in length ? Ans. 75 feet Ex ax. x. At 170 feet distance from the bottom of a tower, the angle of its elevation was found to be 52° 30* : required the altitude of the tower ? Ans. 221-55 feet Exam. xi. From the top of a toner, by the sea-side, of 143 feet high, it was observed that the angle of depression of a ship's bottom, then at anchor, measured 35° ; what was the ship's distance from the bottom of the wall ? Ans. 204-22 feet. Exam. xii. What is the perpendicular height of a hifl ; its angle of elevation, taken at the bottom of it, being 46°, and 200 yards farther off, on a level with the bottom, the angle was 31 ° ? Ans. 286-28 yards. Exam. xiii. Wanting to know the height of an inacces- sible tower ; at the least distance from it, on the same hori- zontal plane, I took its angle of elevation equal to 58° ; then going 300 feet directly from it, found the angle there to be only 32° : required its height, and my distance from it at the first station ? k i height 307-53 An8# } distance 198-15 Exam. xiv. Being on a horizontal plane, and wanting to know the height of a tower placed on the top of an inacces- sible hill ; I took the angle of elevation of the top of the hill 40", and of the top of tha tower 51° ; the measuring in a line directly from it to the distance of 200 feet farther, I found the angle to the top of the tower to be 33° 45'. What is the height of the tower ? Ans. 93-33148 feet Exam. xv. From a window near the bottom of a house, which seemed to be on a level with the bottom of a steeple, AND MSTANCM. I took the angle of elevation of the top of the steeple equal 40° ; then from another window, 18 feet directly above the former, the like angle was 37* 10' : required the height and distance of the steeple. A t height 21(1 44 An } distance 250-79 Exam. xvi. Ranting to know the height of, and my distance frofn, an object on the other side of a river, which appeared to be on a level with the place where 1 stood, close by the side of the river ; and not having room to measure backward, in the same line, because of the im. mediate rise of the bank, I placed a mark where I stood, and measured in a direction from the object, up the ascend, ing ground, to the distance of 264 feet, where it was evi. dent that I wan above the level of the top of the object ; there the angles of depression were found to be, viz. of the mark left at the river's side 42°, of the bottom of the object 27°, and of its top 19*. Required the height of the object, and the distance of the mark from its bottom ? a \ height 57-26 Ans " I distance 150-56 Exam. xvii. If the height of the mountain called the Peak of Teneriffe be 2£ miles, as it is very nearly, and the angle taken at the top of it, as formed between a plumb-line and a line conceived to touch the earth in the horizon, or farthest visible point, be 88° 2 ; it is required from these measures to determine the magnitude of the whole earth, and the utmost distance that cpn be seen on its surface from the top of the mountain, supposing the form of the earth to be perfectly globular ? . (dist. 185-943 1 ji in, -Jdiam. 7918! mLCS - Exam. xvui. Two ships of war, intending to cannonade a fort, are, by the shallowness of the water, kept so fur from it, that they suspect their guns cannot reach it with effect. In order therefore to measure the distance, they separate from each other a quarter of a mile, or 440 yards ; then each ship observes and measuses the angle which the other ship and the fort subtends, which angles are 83" 45' and 85° 15'- What is the distance between each ship and the fort ? . $2292 26 yards. An *' {2296-05 Exam. xix. Wanting to know the breadth of a river, I measured a base of 500 yards in a straight line close by one aide of it ; and at each end of this line I found the anglea Mbtended by||be other end and a tree, close to the haok** 404 OF HEIGHT! AMD DISTAXCBS. the other side of the river, to be 53° and 79* VH* What was the perpendicular breadth of the river t Ana. 520-48 yards. Exam. xx. Wanting to know the extent of a piece of water, or distance between two headlands ; I measured from each of them to a certain point inland, and found^the two distances to be 786 yards and 840 yards ; also the horizontal angle subtended between these two lines was 55° 40. What was the distance required ? Ans. 741 -2 yards. Exam. xxi. A point of land was observed, by a ship at sea, to bear east-by .south ; and after sailing north-east 12 miles, it was found to bear south.east-by-east. It is required to determine the place of that headland, and the ship's dis- tance from it at the last observation ? Ans. 26-0728 miles. Exam. xxii. Wanting to know the distance between a house and a mill, which were seen at a distance on the other side of a river, I measured a base line along the side where I was, of 000 yards, and at each end of it took the angles subtended by the other end and the house and mill, which were as follow, viz. at one end the angles were 58° 2tf and 95° 20', and at the other end the like angles were 53° 3a and 98° 45'. What then was the distance between die house and mill ? Ans. 959-5866 yards. Exam, xxiii. Wanting to know my distance from an in. accessible object o, on the other side of a river ; and having no instrument for taking angles, but only a chain or cord for measuring distances ; from each of two stations, a and b, which were taken at 500 yards asunder, I measured in a di- rect like from the object o 100 yards, viz. ac and bd each equal to 100 yards ; also the diagonal ad measured 550 yards, and the diagonal bc 560. \\ hat was the distance of the object o from each station a and b ? ao 536-81 n ' \ bo 500-47 Exam. xxiv. In a garrison besieged are jjjiree remarkable objects, a, b, c, the distances of which frrtm each other are discovered by means of a map of the place, and are as fol- low, viz. ab 266], ac 530, bc 327 £ yards. Now, having to erect a battery against it, at a certain spot without th(? place, and being desirous to know whether my distances from the three objects be such, as that they may from thence be bat- tered with effect, I took, with an instrument, the horizontal angles subtended by these objects from the station s, and found them to be as follow, viz. the angle asb 13° 30, and the angle B8c28°5tf. Refuted the three distances, a a, MX2ISUBATIOX OF ttAlfBS. 406 ib, sc ; the object b being situated nearest me, and between the two others a and c. i sa 757*14 Ans. < sb 587*10 (sc 655-80 Exam. xxv. Required the same as in the last example, when the object a is the farthest from my station, but still seen between the two others as to angular position, and those an* gles being thus, the angle asb 88 45', and bsc 2*4 80 , also the three distances, ab 600, ac 800, bc 400 vaids ? (•a 710-3 Ans. {sb 1041*85 (sc 93414 Exam. xxvi. If db in the figure at pa. 378 represent a por- tion of the earth's surface, and n the point where the level* ling instrument is placed, then lb will be the difference between the true and the apparent level ; and you nre re- quired to demonstrate that, for distances not exceeding 5 or 6 miles measured on the earth's surface, bl, estimated in feet, is equal to } of the square of bd, taken in miles. MENSURATION OF PLANES. The Area of any plane figure, is the measure of the space contained within its extremes or bounds ; without any re- gard to thickness. This area, or the content of the plane figure, is estimated by the number of little square* that may be contained in it ; the side of those little measuring squares being an inch, or a foot, or a yard, or any other fixed quantity. And hence tl e area or content is said to be so many square inches, or square feet, or square yards, &c. Thus, if the figure to be measured be the rectangle abcd, and the little square b, whose side is one inch, be the mea- suring unit proposed : then as often as the said little square is contained in the rectangle, so many square inches the rectangle is said to contain, which in the present case is 12. D 4 c J A. B m 406 PROBLEM I. 7b find the Area of any Parallelogram ; whether it be a Square, a Rectangle, a Rhombus, or a Rhomboid. Multiply the length by the perpendicular breadth, or height, and the product will be the area*. EXAMPLES. Ex* 1 • To find Ihe area of a parallelogram, the length being 12*25, and breadth or height 8*5. 12*25 length 8*5 breadth C125 0800 104*125 area. Ex. 2. To find the area of a squire, whose side is 35*25 chains. Ans. 124 acres, 1 rood, 1 perch. Ex. 3. To find the area of a rectangular board, whose length is 12} feet, and breadth 9 inches. Ans. 9f feet Ex. 4. To find the content of a piece of land, in form of a rhombus, its length being 6-20 chains, and perpendicular breadth 5*45. Ans. 3 acres, 1 rood, 20 perches. Ex. 5. To find the number of square yards of painting in a rhomboid, whose length is 37 feel, and height 5 feet S inches. Ans. 21^ square yards. * The truth of this rule is proved in the Geom. theor. 81. cor. 9. The same is otherw.e proved ihu*: Let thts foregoing met Angle he the figure piopnsed ; and let the length end lirenrith he dvided inlmeft* nil purls, each equal to lite linear measuring imit t being here 4 forth* length, and H for th»» hreadtli ; and let the opposite points of division bt connected hy right line*.— Then it is evident that three Hues divide tbt rectangle into a n um her of little -squares, each eniial to the sqoait measuring unit i ; and further, that the n urn her of these little squares, or the area of the figure, is eq-ial to the tuimher of linear measuring spin in the length, related at olten as there are linear measuring units in the hrearlth, or height; thai h, equal to the length drawn into the height; whirh here is 4 X 3 or I '2. And it is proved (Geom theor. 25, cor. 2), that any oblique parallelo- gram if equal to a rectangle, of equal length and perjtendirular breadth. Therefore the rale U general for all |*railelograiofl whatever. 0F NAHM. PBOB1 EM II. To find the Area of a Triangle. Rulk im Multiply the base by the perpendicular height, and take half the product for the area*. Or, multiply the one of these dimensions by half the other. EXAMPLES. Ex. 1. To find the area of a triangle, whose J>ase is 625* and perpendicular height 520 links ? Here 6^5 X 260 = 162500 square links, or equal 1 acre, 2 roods, 20 perches, the answer. Ex. 2. How many square yards contains the triangle, whose base is 40, and perpendicular 30 feet ? A ns. 66} square yards. Ex. 3. To find the number of square yards in a triangle, whose base is 49 feet, and height 25J feet. Ans. 684}, or 68*7361. Ex. 4. To find the area of a triangle, whose base is 18 feet 4 inches, and height 11 feet 10 inches ? Ans. 108 feet, 5} inches. Rule n. When two sides and their contained angle are S'ven : Multiply the two given sides together, and take half eir product : Then say, as radius is to the sine of the given angle, so is that half product, to the area of the triangle. Or, multiply that half product by the natural sine of the ■aid angle, for the areaf. * The truth of this rule is evident, because any triangle it the half of a parallelogram of equal base and altitude, by Geom. theor. 26, t For. let ab, ac, be the two given sides, in- Q eluding the given angle a. Now £ ab X cp is the area, by the first rule, cr being the perpendicular. But by trigonometry, as sin. JL r, or radius : ac : : sin. ^ a : cr, which is therefore = ac X era. Z. a, taking radius = 1. Therefore the area £a* X cp is = Iab X ac X tin. £ a, lo radius I ; •r, as radios : sis. 4 a x : Jab x ac : the area. 408 X I2UU RATI05 Ex. 1. What hi the area of a triangle, whose two sides are 80 and 40, and their contained angle 28* 57' ? By Natural Number*. By Lcgariikm. First, J X 40 X 30 = 600, then, 1 : 600 : : -484046 sin. 28- 57' log. 0-684887 600 2 778151 Anawer 290-4276 the area answ. to 2 468088 Ex. 3* How many square yards contains the triangle of which one angle is 45% and its containing sides 25 and 21} feet ? Ans. 20-86047. Ruue in. When the three sides are given : Add all the three sides together, and take half that sum. Next, subtract each aide severally from the said half sum, obtaining three remainders. Then multiply the said half sum and those three remainders all together, and extract the square root of the last product, for the area of the triangle f. t For, let 6 denote the base a a of the triangle abc (see the last fig.), also a the side ac, and e the side ac. Then, by Ibeor. 3, Trigon. as 6 : « + e : : « — ei **~ - = ap — pb the diff. of the seg> ments; - , , oa — ec bb 4- aa — cc 4l _ theref. J6-f — — = — = the segment ap ; hence V^ao 1 — ap*) = the perp. cr, that is, CP. 66 + «« — «» V(as-(— ^ ))= - Wb->~a< +2b€i — »i-f g«y — f i _ V 466 " Bot Jab X cp is the area, that is, 16 X cp = V- J — jg 1 _ vi ma-bb-cc + 2bc^-na + bb + cc + V>c ) (A) . ( « +J±f y ii+ *±! x — *±- e v tt* r •% — v\ 2 2 2 x 2 = V { t X (f — «) X (« — A) X (• — c) J, which is toe role, denotes half the sum of the three sides. The espressioa marked (A), if we put s = 6 + e, and d for 6 — f, to equivalent to J V \ — d — di) j ; which, in most cases, furnishes a satire commodious rale for practice than rule in. here gfren ; espe- Ssatyv t£ the computet have a table of squares at band, OF PLANES. 409 If the aides of the triangle be large, then add the logs, of the half sum, and of the three remainders together, and half their sum will be the log. of the area. Ex. 1. To find the area of the triangle whose three sides are 20, 30, 40. 20 45 45 45 30 20 30 40 40 — — — 25 1st rem. 15 2d rem. 5 3d rem. 2) 00 — — — 45 half sum Then 45 X 25 X 15 X 5 = 84375, The root of which is 290-4737, the area* Ex. 2. How many square yards of plastering are in a triangle, whose sides, are 30, 40, 50 feet ? Ans. 66}. Ex. 3. How many acres, &c. contains the triangle, whose aides are 2569, 4900, 5025 links ? Acs. 61 acres, 1 rood, 39 perches. PROBLEM III. To find the Area of a Trapezoid. Add together the two parallel sides ; then multiply their sum by the perpendicular breadth, or the distance between them ; and take half the product for the area. By Geora. theor. 29. Ex. 1. In a trapezoid, the parallel sides are 750 and 1225, and the perpendicular distance between them 1540 links ; to find the area. 1225 750 1975 X 770 = 152075 square links —-15 arc. 33 perc. Ex. 2. How many square feet are contained in the plank, whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches ? Ans. 13}} feet. Ex. 3. In measuring along one side ab of a quadrangular field, that side, and the two perpendiculars let fall on it from the two opposite corners, measured as foUqw \ content. Vol. I 53 410 WCWITftATTOXf ap = 110 links Ads. 4 acres, 1 rood, 5*792 perches. au= 745 AB = 1110 cp = 352 pa- 595 PROBLEM IT* To find the Area of any Trapezium. Divide the trapezium into two triangles by a diagonal ; then find the areas of these triangles, and add them together. Or thus, let fall two perpendiculars on the diagonal from the other two opposite angles ; then add these two perpen- diculars together, and multiply that sum by the diagonal, taking half the product for the area of the trapezium. Ex. 1. To find the area of the trapezium, whose diagonal is 42, and the two perpendiculars on it 16 and 18. Here 16+18= 34, its half is 17. Then 42 X 17 = 714 the area. Ex. 2. How many square yards of paving are in the tra* pezium, whose diagonal is 65 fe<;t, and the two perpendicu- Ex. 3. In the quadrangular field abcd, on account of ob- structions there could only be taken the following measures, viz. the two sides bc 265 and ad 220 yards, the diagonal ac 378, and the two distances of the perpendiculars from the ends of the diagonal, namely, ae 160, and cf 70 yards. Re- quired the construction of the figure, and the area in acres, when 4840 square yards make an acre ? To find the Area of an Irregular Polygon. Draw diagonals dividing the proposed polygon into tra» peziums and triangles. Then find the areas of all these se- parately, and add them together for the content of the whole polygon. lars let fall on it 28 and 33£ feet ? Ans. 222y 9 yards. Ans. 17 acres, 2 roods, 21 perches. problem v. or PLAinct. abcdefga, in which are given the following diagonals and perpendiculars: namely, To find the Area of a Regular Polygon. Rule i. Multiply the perimeter of the polygon, or sum of its sides, by the perpendicular drawn from its centre on one of its sides, and take half the product for the area 3 ". Ex. 1. To find the area of a regular pentagon, ench side being 25 feet, and the perpendicular from the centre on each side 17-2047737. Here 25 X 5 = 125 is the perimeter. And 17-2047737 X 125 = 2150 5967125. Its half 1075*208356 is the area sought^ Rule n. Square the side . of the polygon ; then multiply that square by the tabular area, or multiplier set against its name in the following table, and the product will be the areaf. • This is only in effect resolving the polygon into as ninny equal tri- angles as it has sides, hy drawing lines from ihe cent re to all the angles; then finding their areas, and adding them all together. t This rule is founded on the property, thitt like polvgon«, being simi- lar figures, are to one another as the Mpiares of their Ifke sides : w»»icb is proved in the Geora. theor. 89. Now, the multipliers iu the table, «re the areas of the respective polygons to the side 1. Whence tLe rule is manifest. ac 55 Ads. 1878}. fd 52 €0 44 em 13 BR 18 oo 12 sp 8 d?23 PROBLEM VI. 413 XXRtCBATIOIf No. of Side*. Names. 3~ Trigon or triangle 4 Tetragon or square 5 Pentagon 6 Hexagon 7 Heptagon 8 Octagon 9 Nonagon 10 Decagon 11 Undccagon 12 Dodecagon Areas, or Radius ot cir- Multipliers. cum. cin !e. 0-4330127 05778503 1-0000000 0-7071068 1-7204774 0-8506508 2-5980762 1-0000000 : -6339124 1 1523824 4-828*271 1-3065628 6-1818242 1-4619022 7-6942088 1-6180340 9-3656399 1-7747324 11 1961524 1-9318517 Exam. Taking here the same example as before, namely, a pentagon, whose side is 25 feet. 9 Then 25 2 being = 625, And the tabular area 1 7204774 ; Theref. 1-7204774 X 625 = 1075-298375, as before. Ex. 2. To find the area of the trigon or equilateral tri- angle, whose side is 20. Ans. 173-20508. Ex. &• To find the area of the hexagon whose side is 20. Ans. 1039-23048. Ex. 4. To find the area of an octagon whose side is 20. Ans. 1931-37084. Ex. 5. To find the area of a decagon whose side is 20. Ans. 3077-68352. Note. If ab * 1, and n the number of sides of the poly- gon, then area of polygon — n times area of the triangle abc = ft ad . dc — n ad tan. c?ad (to rad. ad) = in tan. cad 180° = in cot. ai d = Jn cot. . The ra- ft dius of the circumscribing circle, to side 1, is evidently equal fo J si*c. cad. Mulri-' plying, therefore, the radius of the table by the numeral value of any proposed side, the product is the radius of a circle in which that polygon may be inscribed ; and from which it may readily be constructed. OF PLANES. 41S PROBLEM VII. To find the Diameter and Circumference of any Circle, the one from the other. This may be done nearly, by either of the four following proportion*, viz. As 7 is to 22, so is the diameter to the circumference. Or, As 1 is to 3-1410, so is the diam. to the circumf. Or, As 113 to 355, so is the diam. to the circumf.* And, as 1 : '318309 : : the circumf. : the diameter. * For let abcd benny circle, whose centre is z, and lei ab, bc, he any two equal arcs. Draw the several chords as in the figure, and join be; also draw the diameter da. which produce to r, till bp be equal to the chord bd. Then the two isosceles triangles df.r, dbf, are equiangular, because they have the angle at d common; consequently de : db : : dr : df. But the two triangles afb, dcb, are identical, or equal in all re«pects, because they have the angle f — the an^le biic, being each equal to the angle adb, these being subtended by the equal arcs ab, bc; also the exterior angle fab of the quadrnn- . gle abcd, is equal to the opposite interior angle at c ; and the two triangles have also the side bf = the side bd ; there- fore the side af is also equal in the side dc. Hence the proportion above, viz. de : db : : db : df = da -J- af, becomes dk : db : : db : 2de -|- dc. Then, by taking the rectangles of the extremes and means, it is db-> = 2db : -|-dk . DC. Now, if the radius de be taken = 1, this exprestUafv becomes db 3 = 2-|- i>c. and hence the root db =. V(2-f dc). That is, if the measure of the supplemental chord of any arc be increased by the number 2, the. sq tare root of the sum will be the supplemental chord of half that arc N »w, to apply this to the calculation of the circumference of the cir- cle, let the arc ac be taken equal to \ of the circumference, and be suc- cessively bisected by the above theorem : thus the chord ac of -J- of the 1 circumference, is the side of the inscribed regular hexagon, and is there- , fore equ.il to the radius ae or I : hence, in the right-angled triangle a CD, it « ill be dc ^ V(ad - ac ) = V( ,i — 1) = V3 = 1-7320508U76, the supplemental chord of X of the periphery. Then, by the foregoing theorem, by . always bisecting the aces, end adding 2 to the last square root, there will be found the supplemental chords of the l2th, the 24th, the 48th, the 96th, &c, parts of the peri- phery ; thus. V 3 -7320 108076 : ^/ 3-93 1 85 165*5 : ^3 9828897227 : ^3-9957178165 : v/3-9989291743 : v/3-9997322757 : v/3-9999330678 : V3-9999832669 - 1-9318516525 1-9828097227 1-9957178465 : 1-9939291743 1*9997322757 ; 1-9999330G78 1-9999832669 ° 1 r ^ i US — o T¥ J8 c 3 JX ' *u 414 MBIfSITBATIOlT 1 . . ' V Ex. 1. To.findtfie circumference of the circle whose dia- meter is 20. By the first rale, as 7 : 22 : : 20 : 62j, the answer. Ex. 2. If the circumference of the earth be 24877-4 miles, what is its diameter ? By the 2d rule, as 3-1416 : 1 : : 24877-4 : 7918-7 nearly the diameter. By the 3d rule, as 355 : 113 : : 24877-4 : 7918-7 nearly. i>; u.e4lhrule,asl : -318309 : : 24877 4 : 7918-7 nearly. PROBLEM VIII. To find the Length of any Arc of a Circle. Multiply the decimal -017453 by the degrees in the given arc, and that product by the radius of the circle, for the length of the arc*. Since then it is found that 3*9999832669 is the square of fhe supple- mental chord of the 1536th part of the periphery, let this number be Inker- from 4, which is the square of the diameter, nnd the remainder 0-0000167331 will tie the square of the chord of the said 1 546th part of the periphery, and consequently the root \/O 00O0 167331 = 0-OO40906 U2 is the length of that chord; this number then being multiplied by 1536 gives 6 2S3I7BS for the perimeter of a regular polygon of 1536 sides in* scribed in the circle; which, as the sides of the polygon nearly coin- cide with the circumference of the circle, must also express the length of the circumference itself, very nearly. But now, to show how near this determination is to the truth, let a^p = 0-0040906 1 12 represent one side of such a regular polygon of 1536 side*, and «rt a side of another similar polygon described about the circle; and from the centre e let the perpendicular zqR be drawn, bisecting a p and st in q nud r. Then >ince a<i is = 4 ap - 00i» »4Vi05rt. and ka = I. therefore »q'i — ka'— ai^ 1 = -9991)958167, and consequently its root gives ei = -99991)791)8-1 ; then because of the pa- rallels ap, st, it is p.q : kr : : ap : st : : as the whole in- scribed perimeter: to the circumscribed one. that is, as -9999979084 : 1 : : 6 283176* : 6 28*1920 the perime- ter of the circumscribed polygon. Now, the circum- ference of the circle being greater than the perimeter of the inner poly- gon, but less limn that of the outer, it must consequently b« greater than 6-283178& but less than 6-2831940, and must therefore be nearly equal to £ their sum, <>r 6-2831854, which % in fact is true to the last figure, which fhould be a 3, instead of the 4. Hence the circumference heing 6-2831854 w hen the diameter is 2, it will be the half of that, or 3 1415^27. when the diameter is I, to which the ratio in the rule, viz. 1 to 3- 14 16. is very near. Also the first ratio in the rule, 7 to 22 or 1 to 3f 3- 1428 &c. is another near a pproii na- tion. But the third ratio, 1 13 to 355, — 1 to 3- 14 15929, is the nearest. * It having been found, in the demonstration of the foregoing problem, that when the radius of a cArc\e\t \,\V& ItugLh of the whole circumie- OF FLANKS. 415 Ex. 1. To find the length of an arc of 30 degrees, the radius being feet. Ans. 4-71231. Ex. 2. To find the length of an arc of 12 10', or 12 J, the radius being 10 feet. Ans. 2*1234. PROBLEM IX. To find the Area of a Circle*. Rule i. Multiply half the circumference by half the diameter. Or multiply the whole circumference by the whole diameter, and take \ of the product. Rule ft. Square the diameter, and multiply that square by the decimal a 7854, for the area. Rule in. Square the circumference, and multiply that square by the decimal -07958. Ex. 1. To find the area of a circle whose diameter is 1Q> and its circumference 31 '416. By Rule 1. By Rule 2. By Rule 3. 31-416 -7854 31-416 10 10 3 = 100 31-416 4)314 16 ~* 986-965 78-54 70 34 -07958 78-54 So that the area is 78*54 by all the .three rules. rence is 6 2831854, which consists of 360 degrees ; therefore as 3*0" : 6-2831854: : 1 : 017453, kc. the length of the arc of I degree. Hence the decimal -017453 multiplied hy any number of degrees, will give the length of the arc of those degrees. And because the circumferences and arcs are in proportion as the diameters, or as the radii of the circles, therefore as the radius 1 is to any other radius r, so is the length of the arc above mentioned, to '017453 X degrees in the arc X r> which is the length of that arc, as in the rule. * The first rule is proved in the Geom. theor. 94. And the 9d and 3 1 rules are deduced from the first rule, in this man- ner.— By that rule, de -i- 4 is the area, when d denote* the diameter, and t the circumference. But, by prob. 7, e is = 3- 14 IrW ; therefore ihe said area de -r 4, becomes (tX*14lfi<*-r>4 = -78544', which gives the 2d rale. — Also, by tho snme prob. 7, d is = c -f- 3 1416; therefore again the same first area dr. -f- 4, becomes (c 3*1416) X (c 4) = e- 12-5664, which is = c* X 07958, by taking the reciprocal of 12 5664, or changing ibat divisor into the inuitipler -07958 ; which gives the 3d rule. CoroL Hence the areas of different circles are in proportion to one another, as the square of their diameters or as the square oC Umax ^maNtr fereoces- as before proved in the Geom. Iheor. 416 MEKST7IUTI0N Ex. 2. To find the area of a circle, whose diameter is 7, and circumference 22. Ans. 38$. Ex. 3. How many square yards are in a circle, whose diameter is 3 J feet ? Ans. 1 -069. Ex. 4. To find the area of a circle, whose circumference is 12 feet. Ans. 11-4595. PROBLEM X. To find the Area of a Circular Ring, or of the Space included between the Circumferences of two Circles ; tlie "bite being contained within the other. Take the difference between the areas of the two circles, as found by the last problem, for the area of the ring. — Or, which is the same thing, subtract the square of the less dia- meter from the square of the greater, and multiply their dif- ference by -7854. — Or, lastly, multiply the sum of the dia- meters by the difference of the same, and that product by •7854 ; which is still the same thing, because the product of the sum and difference of any two quantities, is equal to the difference of their squares. Ex. 1. The diameters of two concentric circles being 10 and 6, required the area of the ring contained between their circumferences. Here 10 + 6 = 16 the sum, and 10 — 6 = 4 the diff. Therefore -7854 X 10 X 4 ^= -7854 X 64 - 50 2656, the area. Ex 2. What is the area of the ring, the diameters of whose bounding circles are 10 and 20 ? Ans. 235 62. PROBLEM XI. To find the Area of the Sector of a Circle. Rule i. Multiply the radius, or half the diameter, by half the arc of the sector, for the area. Or, multiply the whole diumeter by the whole arc of the sector, and take £ of the product. The reason of which is the same as for the first rule to problem 9, for the whole circle. Rule ii. Compute the area of the whole circle : then say, as 360 is to the degrees in the «rc of the sector, so is the area of the whole circle, to X\ve <rt toa «&&&t. OP PLAICES. 41? This is evident, because the sector is proportional to the length of the arc, or to the degrees contained in it. Ex. 1. To find the area of a circular sector, whose arc contains 18 degrees ; the diameter being 3 feet. 1. By the first Rule. First, 3-1416 X 3 = 0-4248, the circumference. And 360 : 18 : : 0-4848 : -47124, the length of the arc. Then -47124 X 3 ^4= 1-41372 + 4 « -35343, the area. 2; By the 2d Rule. First, -7854 X 3" = 7-0686, the area pf the whole circle. Then, as 360 : 18 : : 7 0686 : -35343, the area of the sector. Ex. 2. To find the area of a sector, whose radius is 10, and arc 20. Ans. 100. Ex. 3. Required the area of a sector, whose radius is 25, and its arc containing 147° 20'. Ans. 804-3086. PROBLEM XII. To find the Area of a Segment of a Circle. Rule I. Find the area of the sector having the same arc with the segment, by the last problem. Find also the area of the triangle, formed by the chord of the segment and the two radii of the sector. Then add these two together for the answer, when the segment is greater than a semicircle : or subtract them when it is less than a semicircle. — As is evident by inspection. Ex. 1. To find the area of the segment acbda, its chord ▲b being 12, and the radius ae or ce 10. First, As af : sin. L. d 00° : : ad : sin. 86° 52' J = 36*87 degrees, the degrees in the aec or arc ac. Their double, 73*74, are the degrees in the whole arc acb. \ J Now -7854 X 400 = 314-16, the area of VX-/ the whole circle. . / ' Therefore 360° : 73-74 : : 31416 : ^3504, area of the sector acre. Again, ^/(ab» — ad 2 ) = ^(100 — 36) = ^/64 = 8 = de. Theref. ad X de =* 6 X 8 = 48, the area of the triangle AEB. Hence sector acbe — triangle akb = 16-3504, area of seg. acbda. Rule n. Multiply the square of the radius of the circle by either half the difference of the arc acb and its sine (Jrtdrt Vol. I. 54 418 MHIUBJkUOM to the radius 1), or half the sum of the; arc and its sine, Ac* cording as the segment is less or greater than a semicircle ; the product will be the area. The reason of this rule, also* is evident from an mspcetioo of the diagram. Exam, the same as before, in which am « IS, ax a* 10 ; and from the former computation are acb = 78* 44'f- Then, by Button's Mathematical Tables, pp. 840, ccc arc 73° 44'f, to radius 1 =* 1-2870059 sin. 78° 44'}, to radius 1 = -9600010 2) -3270009 •1685084 whence, -1635034 X 10* = 16-35034, the area of thcseg- ment ; very nearly as before. Ex. 2. What is the area of the segment, whose height is 18, and diameter of the circle 50 ? Ans. 636-375. Ex. 3. Required the area of the segment whose chord is 16, the diameter being* 20 ? Ans. 44-728, or 269-432. PROBLEM XIII. To measure long Irregular Figures. Take or measure the breadth at both ends, and at several places, at equal distances. Then add together all these ia* termediate breadths and half the two extremes, which sum multiply by the length, and divide by the number of parts, for the area*. 5 {jlj^t? * This rule is made out as follows: — Let abcd be the irregular piece ; having the several breadths ad, bf, oh, iK y bc, at the equal distances ak, eg, oi, u. Let the several breadths in or- . der be denoted by the corresponding A E & I B letters a, b t c, d. c, and the whole length ab by l\ then compute the areas of the parti into which the Store » divided by the perpendiculars, as so many trapezoids, by prob. 3, aid add them all together. Thus, the sum of the parts is, = x it + tjf x y+ - x ii+Sf • x * = «• + & + c + 4 + ft X V = <JR -V » + « + <*) V. or rums. JVbfe. If the perpendiculars or breadths be not at equal distances, compute all the parts separately, as so many trape- zoids, and add them all together for the whole area. Or else, add all the perpendicular breadths together, and divide their sum by the number of them for the mean breadth, to multiply by the length ; which will give the whole area, not fiur from the truth. Ex. 1. The breadths of an irregular figure, at five equi. distant places, being 8-2, 7-4, 9-2, 10-2, 8-6 ; and the whole length 39 ; required the area. 8*2 35-2 sum. 8-6 39 2 ) 16-8 sum of the extremes. 8168 8-4 mean of the extremes. 1066 7-4 4 ) 1372-8 9-2 343-2 An*. 10-2 36-2 sum. Ex. 2. The length of an irregular figure being 84, and the breadths at six equidistant places 17*4,20-6, 14-2, 16-5, 20*1, 24-4 ; what is the area? Ana. 1550-64. FBOBLBM XIV. To find Ae Area of an Ellipsis or Oval Multiply the longest diameter, or axis, by the shortest ; then multiply the product by the decimal -7854, for the area. As appears from cor. 2, theor. 3, of the Ellipse, in the Conic Sections. Ex. 1. Required the area of an ellipse whose two axes are 70 and 50. Ans. 2748-9. Ex. 2. To find the area of the oval whose two axes are 24 and 18. Ans. 339-2928. which if the whole area, agreeing with the rule : m being the arithmeti- cal memo between the extremes, or half the sum of them both, and 4 the number of the parti. And the same for any other aombet «C ?ute whatever. 480 XBMlUBAnOIC NKOBLXH XV. To find ike Area of an Elliptic Segment. Find the area of a corresponding circular segment, having the same height and the same vertical axis or diameter. Then say, as the said vertical axis is to the other axis, parallel to the segment's base, so is the area of the circular segment before found, to the area of the elliptic segment sought. This rule also comes from cor. 2, theor. 3, of the Ellipse. Ex. 1. To find the area of the elliptic segment, whose height is 20, the vertical axis being 70, and the parallel axis 60. ins. 648 13. Ex. 2. Required the area of an elliptic segment, cut off parallel to the shorter axis ; the height being 10, and the two axes 25 and 35. Ans. 162-03. Ex. 3. To find the area of the elliptic segment, cut off pa- rallel to the longer axis ; the height being 5, and the axes 25 and 35. Ans. 97-8425. PROBLEM XVI. To find the Area of a Parabola, or iU Segment. MultiAy the base by the perpendicular height; then take two-thirds of the product for the area. As is proved in theorem 17 of the Parabola, in the Conic Sections. Ex. 1. To find the area of a parabola ; the height being 2, and the base 12. Here 2 X 12 = 24. Then % of 24 = 16, is the area. Ex. 2. Required the area of the parabola, whose height is 10, and its base 16. Ans. 106f. MENSURATION OF SOLIDS. By the Mensuration of Solids are determined tho spaces included by contiguous surfaces ; and the sum of the met* sures of these including surfaces is the whole surface or su. perficies of the body. The measure of a solid, is called its solidity, capacity, or content. o» solids, 431 Solids are measured by cubes, whose sides are inches, or feet, or yards, dec. And hence the solidity of a body is said to be so many cubic inches, feet, yards, dec. as will fill its capacity or space, or another of an equal magnitude. The least solid measure is the cubic inch, other cubes being taken from it according to the proportion in the fol- lowing table, which is formed by cubing the linear pro* portions. Table of Cubic or Solid Measures. 1728 cubic inches make ' 1 cubic foot 27 cubic feet - 1 cubic yard 166| cubic yards - 1 cubic pole 04000 cubic poles . 1 cubic furlong 512 cubic furlongs 1 cubic mile. PROBLEM 1. To find the Superficies of a Prism or Cylinder. Multiply the perimeter of one end of the prism by the length or height of the solid, and the product will be the sur- face of all its sides. To which add also the area of the two ends of the prism, when required*. Or, compute the areas of all the sides and ends separately, and add them all together. Ex. 1. To find the surface of a cube, the length of each side be ng 20 feet. Ans. 2400 feet. Ex. 2. To find the whole surface of a triangular prism, whose length is , 20 feet, and each side of its end or base 18 inches. Ans. 91*048 feet. Ex. 3. To find the convex surface of a round prism, or cylinder, whose length is 20 feet, and the diameter of its base is 2 feet. Ans. 125-664. Ex. 4. What must be paid for lining a rectangular cistern with lead, at 3d. a pound weight, the thickness of the lead being such as to weigh 71b. for each square foot of surface ; the inside dimensions of the cistern being as follow, viz. the length 3 feet 2 inches, the breadth 2 feet 8 inches, and depth 2 feet 6 inches? Ans. 32. 5s. 9jd. * The truth of this will easily Appear, by considering that the sides of any prism are parallelograms, whose common length is the same as Ibe length of the solid, and their breadths taken all together make op the perimeter of the ends of the same. And the rale is evidently the tame for the tatta* oAiqjYk&ki. MJUOTJRATSOIf PBOBLBMII. To find the Surface of a Pyramid or Cone. Multiply the perimeter of the bate by the riant height, or length of the side, and half the product will evidently be the surface of the sides, or the sum of the areas of all the tri- angles which form it. To which add the area of the end or base, if requisite. Ex. 1. What is the upright surface of a triangular pyra- mid, the slant height being 20 feet, and each side of the base 3 feet? Ana. 90 feet. Ex. 2. Required the convex surface of a cone, or circular pyramid, the slant height being 50 feet, and the diameter of To find the Surface of the Frustum of a Pyramid or Cone, being the lower pari, when the top is cut off by a plane pa- rallel to the base. Add together the perimeters of the two ends, and multiply their sum by the slant height, taking half the product for the answer. — As is evident, because the sides of the solid are trapezoids, having the opposite sides parallel. Ex. 1. How many square feet are in the surface of the frustum of a square pyramid, whose slant height is 10 feet ; also each side of the base or greater end being 3 feet 4 inches, and each side of the less end 2 feet 2 inches ? Ans. 110 feet. Ex. 2. To find the convex surface of the frustum of a cone, the slant height of the frustum being 12± feet, and the circumferences of the two ends 6 and 8*4 feet. To find the Solid Cohtent of any Prism or Cylinder. Find the area of the base, or end, whatever the figure of it may be ; and multiply it by the length of the prism or cylin- der, for the solid content*. Ans. 607-511. PROBLEM m. Ans. 90 feet. PROBLEM IV. • This rale appears from the Geom. tbeor. 110, cor. 2. The fMBe » more particularly shown at follows: Let the annexed rectangular parti OF MUM. Note. For a cube, take the cube of its aide by multiplying this twice by itaelf ; and for a parallelopipedon, multiply the length, breadth, and depth all together, for the content. Ex. 1. To find the solid content of a cube, whose side is 24 inches. Ans. 18834. Ex. 2. How many cubic feet are in a block of marble, its length being 3 feet 2 inches, breadth 2 feet 8 inches, and thickness 2 feet 6 inches ? Ans. 21}. Ex. 3. How many gallons of water will the cistern con- tain, whose dimensions are the same as in the btst example, when 277} cubic inches are contained in one gallon ? Ans. 131-53. Ex. 4. Required the solidity of a triangular prism, whose length is 10 feet, and the three sides of its triangular end or base are 3, 4, 5 feet. Ans. 60. Ex. 5. Required the content of a round pillar, or cylinder, whose length is 20 feet, and circumference 5 feet 6 inches. Ans. 48*1450 feet. PROBLEM V. To find the Content of any Pyramid or Cone. Find the area of the base, and multiply that area by the perpendicular height ; then take £ of the product for the content*. lelopipedon be the solid to be measured, and the cube r the solid measuring unit, its side being 1 inch, or 1 foot, Jtc. ; also, let the length and breadth of the base abcd, and also the height ah, be each divided into spaces equal to the length of the base of the cube p, namely, here 3 in the length and 2 in the breadth, matins; 3 times 2 or 6 squares in the base ac, each equal to the base of the cube p. Hence it is manifest that the parallelopipedon will contain the cube p, as many times as the base ac con- tains the base of the cube, repeated as often as the height ah contains the height of the cube. That is, the contest of any parallelopipedon is found, by multiplying the- are a of the base by the altitude of that solid. And because all prisms and cylinders are equal to parallelopfoedom of equal bases and altitudes, by Geom. theor. 108, it follows that tne rale b general for all such solids, whatever the figure of the base may be. * This rule follows from that of the prism, because any pyramid if -l of a prism of equal base and altitude ; by Geom. theot AvVout. XBSrsVBATIOZf Ex. 1. Required the solidity of a square pyramid, each aide of its base being 30, and its perpendicular height 25. Ads. 7500. Ex. 2. To find the content of a triangular pyramid, whose perpendicular height is 30, and each side of the base 8. Ans. 38-971143. Ex. 3. To find the content of a triangular pyramid, its height being 14 feet 6 inches, and the three sides of its base 5, 6, 7 feet. Aim. 71*0352. Ex. 4. What is the content of a pentagonal pyramid, its height being 12 feet, and each side of its base 2 feet ? Ans. 27-5276. Ex. 5. What is the content of the hexagonal pyramid, whose height is 6*4 feet, and each side of its base 6 inches 1 Ans. 1-38564 feet. Ex. 6. Required the content of a cone, its height being lty feet, and the circumference of its base 9 feet, Ans. 22-56093. PROBLEM VI. To find the Solidity of a Frustum of a Cone or Pyramid. Add into one sum, the areas of the two ends, and the mean proportional between them : and take \ of that sum for a mean area ; which being multiplied by tne perpen- dicular height, or length of the frustum, will give its con- tent*. * Let abcd be any pyramid, of which bcdofe is a frustum. And put a 2 for the area of I he base rcd, b 9 the area of the top, rfo, A the height ih of the frustum, and c the height ai of the top part above it. Then c -\- A = ah is the height of the whole pyramid. Hence, by the last prob. (c-fM is the content of the whole pyramid abcd. and {b'c the content of the top part akfo ; therefore the difference - - - |a*(c + A) — J6 ? c is the content of the frustum bcdgfjc. fiat the quantity c being no dimension of the frustum, it mutt be expelled from this formula, by substituting iti value, found in the following manner. By Geom. theor. 1 12, « J : £» : : (c + A)« : c\ or « : b : : e + A : c, hence (ficom. th. 69) a — b : b : : A : c, 6A cA and« — b : a : : A : c 4- A ; hence therefore c = —j- and c -f- A = — . ; OP SOLIDS. 435 Note. This general rule may be otherwise expressed, as follows, when the ends of the frustum arc circles or regular polygons. In this latter case, square one side of each poly, gon, and also multiply the one side by the other ; add all these three products together ; then multiply their, sum by the tabular area proper to the polygon, and take one-third of the product for the mean area, to be multiplied by the length, to give the solid content. And in the case of the frustum- of a cone, the ends being circles, square the diameter or the circumference of each end, and also multiply the same two dimensions together ; then take the sum of the three pro- ducts, and multiply it by the proper tabular number, viz. by •7854 when the diameters are used, or by '07958 in using the circumferences ; then taking one-third of the product, to multiply by the length, for the content. Ex. 1. To find the number of solid feet in a piece of tim- ber, whose bases are squares, each side of the greater end being 15 inches, and each side of the less end 6 inches ; also, the length or the perpendicular altitude 24 feet. Ans. 19 J. Ex. 2. Required the content of a pentagonal frustum, whose height is 5 feet, each side of the base 18 inches ; and each side of the top or less end 6 inches. Ans. 9-31925 feet. Ex. 3. To find the content of a conic frustum, the altitude being 18, the greatest diameter 6, and the least diameter 4. Ans. 527-788P. Ex. 4. What is the solidity of the frustum of a cone, the ' altitude being 25, also the circumference at the greater end being 20, and at the less end 10 ? Ans. 464-216. Ex. 5. If a cask, which is two equal conic frustums joined together at the bases, have its bung diameter 28 inches, the bead diameter 20 inches, and length 40 inches ; how many gallons of wine will it hold ? Ans. 79-0613. then these values of c and e +h being substituted for them in the ex- pression for the content of the frustum gives that content = *a> - X fl -^j = V* X^=p = lhx(* + ab + b*); which to the rale above given ; ak being the mean between a 9 and a*. Note. If d, d be the corresponding linear dimensions of ihe ends, 6 their difference, m the appropriate multiplier, h the height of the frus- tum, then is the content = JmA (3ixf + 6) ; which is a coavenieot prac- tical expression. Vol- I. 55 ' 426 MKIfttJBATIOIf PROBLEM VII. To find the Surface of a Sphere, or any Segment. Rule i. Multiply the circumference of the sphere by its diameter, and the product will be the whole surface of it*. Rulk u. Square the diameter and multiply that square by 3*1416, for the surface. Rulk hi. Square the circumference ; then either multi- ply that square by the decimal '31 83, or divide it by 8-1416, for the surface. Note. For the surface of a segment or frustum, multiply * These rules come from the following theorems for the surface of a sphere, viz. That the said surface is equal to the curve surface of ^cir- cumscribing cylinder; or Hint it is equal to 4 great circles of the same sphere, nr of l he ?ame diameter; which are thus proved. Let abcd be a cylinder, circumscribing the Sphere eioh ; the former genertitr d by the J± rotation of the rectangle fbch about the axis or dimeter fii ; and the. latter by the rota- tion of the semicircle fgh about the same di- ameter fh. Draw two lines kl, mi», perpen- dicnlar to the axis, intercepting the paits i.n, op, of the cylinder and sphere ; then will the ring or cylindric surface generated by the ro- tation of i n, be equal to the ring or fpherical surface generated by the arc op. For, first, ™ suppose the parallels ki. and ms to be indefi- nitely near together; drawing 10, and also oo. parallel to iji. Then the t\\ o triangles iko. oqp, being equiangular, it is, as op : oq or LR : : so or ki- : ko :: circumference described by kl : circiimf. described by so; therefore the rectangle op.xcirciimf. of ko is equal to the rectangle l*X circumf. of ki.; that is, the ring described by op on the sphere, iseqoal to the ring described by l.n on the rylindt-r. And as this is every where the case, therefore the sums of any corres- ponding number of tln-se are also equal ; that is. the whole surface of the sphere, described by the whole semicircle fgh, is equal to the whole curve surface of the cylinder, described by the height bc ; as well as the surface of any > cement described by fo, equal to the surface of the cor- responding segment described by el. Corol. I. Hence 'he smface of the sphere is equal to 4 of its great cir- cles or equal to the circumference efgh, or of dc, multiplied by tat height bc. or by the diameter fh. Corol. 2. Hence also, the surface of any such part, as a segment or frustum, or zone, is equal to the same circumference of the sphere, mal- tiplied by the height of the said part. And consequently such spheri- cal curve surfaces are to one auotber in the same proportion as their altitudes. OP SOLIDS. 437 the whole circumference of the sphere by the height of the part required. Ex. 1. Required the convex superficies of a sphere, whose diameter is 7, and circumference 22. Ans. 154. Ex. 2. Required the superficies of a globe, whose diameter is 24 inches. Ans. 1809-5616. Ex. 3. Required the area of the whole surface of the earth y its diameter being 71)57 J miles, and its circumferenco 25000 miles. Ans. 198943750 sq. miles. Ex. 4. The axis of a sphere being 42 inches, what is the convex superficies of the segment whose height is 9 inches ? Ans. 1187-5248 inches. Ex. 5. Required the convex surface of a spherical zone, whose breadth or height is 2 feet, and cut from n sphere of 12» feet diameter. Ans. 78-54 feet. PROBLEM Vin. 7b find the Solidity of a Sphere or Globe. Rule i. Multiply the surface by the diameter, and take | of the product for the content*. Or, which is the same thing, multiply the square of the diameter by the circum. ference, and take } of the product. Rulr H. Take the cube of the diameter, and multiply it by the decimal '5236, for the content. Rule in. Cube the circumference, and multiply by •01688. Ex. 1. To find the solid content of the globe of the earth, supposing its circumference to be 25000 miles. Ans. 263750000000 miles. Ex. 2. Supposing that a cubic inch of cast iron weighs •269 of a lb. avoird. what is the weight of an iron bull of 5-04 inches diameter ? * For. nnt d = the diameter, c = the circumference, end t — the surface of the sphere, or of its circumscribing cylinder ; also, a — the number 3-1416. Then, \t is =. the base of the cylinder, or one gre«t circle of the sphere ; and d is the height of the cylinder : then-fore \ds is lite cniifent of the cylinder. But { of the cylinder is the sphere, liy th. 117, Gcom. that it, § of $d$ % or^ds is the sphere ; which is the first rule. Again, because the surface « is = ad*; iherefore ±*h _ \nd* — 5MM\ U the content, as in the 2d rule. Also, d being = c a, therefore t*d* = a* = •01088, the 3d rule for the content* 488 MJBH S UKATfON PROBLEM K. 7b find the Solid Content of a Spherical Segment. * Rule i. From 3 tiroes the diameter of the sphere take double the height of the segment; then multiply the re* mainder by the square of the height, and the product by the decimal -5230, for the content. Rule ii. To 3 tiroes the square of the radius of the segment's base, add the square of its height ; then multiply the sum by the height, and the product by '5236, for the content. Ex. 1 . To find the content of a spherical segment, of 2 feet in height, cut from a sphere of 8 feet diameter. Ans. 41-888. • By corol. 3, of theor. 117, Geom. it ap- pears thnt the spheiic segment prit, is equal to the difference between the cylinder ablo, and the conic frustum abmq. But, putting d = ab or fh the diameter of £ the sphere or cylinder, /* = fk the height of the segment, r = pk the radius of its lm«c, and a = 3*1416; ihen the content of the tone abi is = \ad* X Jfi — : and by the similar cones abi, qmi, as ri 1 : si 1 : : ^ad x : ^ad* X (r-^j—) 1 = the cone <*m ; therefore the cone abi — the cone qsti = &ad* — J^i X (*-""- V = iad* — b*dk* + J** 1 is = the conic frustum of abm<*. And \atl h is — the cylinder ablo. Then the difference of these tw o is jmfti — \ah^ = \ah* X (3rf —21), for the spheric segment pfn ; whirh is ihe first rule. Again, because pk* = fk X kh (cor. to theor. 87, Geom.) or r 1 = k (d- A), therefore = /*, and 2d - <J> = ^+«=^L+i; 3ri -4- A* which being substituted in the former rule, it becomes \ah* X — ^ — = +*A X ( *r'-M ), which is (he 2d rule. Note. By subtracting a segn ent from n half sphere, or from another tegmeiit, the content o( my faiiitam or tone may be found. or MUM. 429 *Ex. 2. What is the solidity of the segment of a sphera, its height being 9, and the diameter of its base 20 ? Ana. 1795*4344. Arte. The general rules for measuring the most useful figu-es having been now delivered, we may proceed to apply them to the several practical uses in life, as follows. [ 430 ] LAND SURVEYING. SECTION I. DESCRIPTION AND USE OF THE INSTRUMENTS. 1. OF THE CHAIN. Land is measured with a chain, called Gunter's Chain, from its inventor, the length of which is 4 poles, or 22 yards, or 66 feet. It consists of 100 equal links ; and the length of each link is therefore j'fo of a yard, or of a foot, or 7 -92 inches. Land is estimated in acres, roods, and perches. An acre is equal to 10 square chains, or as much as 10 chains in length and 1 chain in hreadth. Or, in yards, it is 220 X 22 = 4840 square vards. Or, in poh»s, it is 40 X 4 = 160 square poles. Or, in 'links, it is 1000 X 100 — 100000 square links: these being all the same quantity. Also, an acre is divided into 4 parts called roods, and a rood into 40 parts called perches, which are square poles, or the square of a pole of yards long, or the square of £ of a chain, or of 25 links, which is 625 square links. So that the divisions of land measure will be thus : 625 sq. links — 1 pole or perch 40 perches = 1 rood 4 roods = 1 acre. The lengths of lines measured with a chain, are best set down in links as integers, every chain in length being 100 links ; and not in chains and decimals. Therefore, after the content is found, it will he in square links; then cut off five of the figures on the right h;ind for decimals, and the rest will be acres. These decimals are then multiplied by 4 for roods, and the decimals of these again by 40 for perches. LAUD SUB YE TING. 481 Exam. Suppose tho length of a rectangular piece of ground be ?£2 Lnks, and its breadth 385 ; to fi.ul the area in acres, roods, aud perches. 792 3 04020 385 4 •K680 6336 40 2376 3 (49*0 7-8*200 Ans. 3 aores, roods, 7 perches. 2. OF THE PLAIN TABLE. This instrument consists of a plain rectangular board, of any convenient size : the centre of which, when used, is fixed by means of screws to a three-legged stand, having a hall and socket, or other joint, at the top, by means of which, when the legs are fixed on the ground, the table is inclined in any direction. To the table belong various parts, as follow. 1. A frame of wood, made to fit round its edges, and to be taken off, for the convenience of putting a sheet of paper on the table. One side of this frame is usually divided into equal parts, for drawing lines across the table, parallel or perpendicular to the sides ; and the other side of the frame is divided into 360 degrees, to a centre in the middle of the table ; by means of which the table may be used as a theo- dolite, dec. 2. A magnetic needle arid compass, either screwed into the side of the table, or fixed beneath its centre, to point out the directions, and to be a check on the sights. 3. An index, which is a brass two-foot scale, with either a small telescope, or open sights set perpendicularly on the ends. These sights and one edge of the index are in the same plane, and that is called the fiducial edge of the index. To use this instrument, take a sheet of paper which will cover it, and wet it to make it expand ; then spread it flat on the table, pressing down the frame on the edges, to stretch it and keep it fixed there ; and when the paper is become dry, it will, by contracting again, stretch itself smooth and flat from any cramps and unevenness. On this paper is to be drawn the plan or form of the thing measured. Thus, begin at any proper part of the ground, and make a point on a convenient part of the paper or tabic, to reore. sent that place on the ground ; then fix. vu \tax ^oveft. 482 LAND leg of (he compasses, or a fine steel pin, and apply to it the fiducial edge of the index, moving it round till through the aights you perceive some remarkable object, as the corner of a field, dtc. ; and from the station-point draw a line with the point of the compasses along the fiducial edge of the index, which is called setting or taking the object : than set another object or corner, and draw its line ; do the same by another ; and so on, till as many objects are taken as may be thought fit* Then measure from the station towards as many of toe objects as may be necessary, but not more, taking the requi- site offsets to corners or crooks in the hedges, laying the measures down on their respective lines on the table. Then at any convenient place measured to, fix the table in the same position, and set the objects which appear from that place ; and so on, as before. And thus continue till the work is finished, measuring such lines only as are necessary, and determining as many as may be by intersecting lines of direction drawn from different stations. Of shifting the Paper on the Plain Table. When one paper is full, and there is occasion for more, draw a line in any manner through the farthest point of the last station line, to which the work can be conveniently laid down ; then take the sheet off the table, and fix another on, drawing a line over it, in a part the most convenient for the rest of the work ; then fold or cut the old sheet by the line drawn on it, applying the edge to the line on the new sheet, and, as they lie in that position, continue the last sta- tion line on the new paper, placing on it the rest of the measure, beginning at where thu old sheet left off. And so on from sheet to sheet. When the work is done, and you would fasten all the sheets together into one piece, or rough plan, the aforesaid lines are to be accurately joined together, in the same man* ncr a8 when the lines were transferred from the old sheets to the new ones. But it is to be noted, that if the said join, ing lines, on the old and new sheets, have not the same in- clination to the side of the table, the needle will not point to the original degree when the table is rectified ; and if the needle be required to respect still the same degree of the compass, the easiest way of drawing the line in the same position, is to draw them both parallel to the same sides of the table, by means of the equal divisions marked on the other two sides. SUItVBYlNQ. m 3. OF THE THEODOLITE. The theodolite is a brazen circular ring, divided into 360 degrees, &c. and having an index with sights, or a telescope, placed on the centre, about which the index is moveable ; also a compass fixed to the centre, to point out courses and check the sights ; the whole being fixed by the centre on a stand of a convenient height for use. In using this instrument, an exact account, or field-book, of all measures and things necessary to be remarked in the plan, ntust be kept, from which to make out the plan on re- turning home from the ground. Ilegin at such part of the ground, and measure in such directions as are judged most convenient ; taking angles or directions to objects, and measuring such distances as appear necessary, under the same restrictions as in the use of the plain table. And it is safest to fix the theodolite in the original position at every station, by means of fore and back objects, and the compass, exactly as in using the plain table ; registering the number of degrees cut off by the index when directed to each object ; and, at any station, placing the index at the same degree as when the direction towards that station was taken from the last preceding one, to fix the theodolite there in the original position. The best method of laying down the aforesaid lines of direction, is to describe n pretty large circle ; then quarter it, and lay on it the several numbers of degrees cut off by the index in each direction, and drawing lines from the centre to all these marked points in the circle. Then, by means of a parallel ruler, draw from station to station, lines parallel to the aforesaid lines drawn from the centre to the respective points in the circumference. 4. OF THE CROSS. The cross consists of two pair of sights set at right angles to each other, on a staff having a sharp point at the bottom, to fix in the ground. Tho cross is very useful to measure small and crooked pieces of ground. The method is, to measure a base or chief line, usually in the longest direction of the piece* from corner to corner ; and while measuring it, finding the places where perpendiculars would fall on this line, from the several cor- nere and bends in the boundary of the piece, with the cross, by fixing it, by trials, on such parts of the line, as that through one pair of the sights both ends of the line may appear, and through the other pair the rom«^uKv&%taro&i Vol. I. 66 m urn or corners : and then measuring die lengths of the eni Tp« r pendiculors. BEXAMS* Besides the fore*mentioned instruments, whic't ere mcst commonly used, there are some others ; as, The perambulator, used for measuring roads, and other great distances, level ground, and by the sides of rivers. It has a wheel of 8} feet, or half a pole, in circumference* by the turning of which the machine goes forward ; and the . distance measured is pointed out by an index, which is moved round by clock-work. Levels, with telescopic or other sights, are used to find the' level between place and place, or how much one place is higher or lower than another. And in measuring any sloping or oblique line, either ascending or descending, a small pocket level is useful for showing how many links lor each chain are to be deducted, to reduce the tine to the hori- zontal length. An offset. staff is a very useful instrument, for measuring the offsets and other short distances. It is 10 links in length, being divided and marked at each of the 10 links. 'J en small arrows, or rods of iron or wood, are used to mark the end of every chain length, in measuring lines. And sometimes pickets, or staves with flags, are set up as marks or objects of direction. Various scales are also used in protracting and measuring on the plan or paper ; such as plane scales, line of chords, protractor, compasses, reducing scale, parallel and pc rpen- dicular rules, dec. Of plane scales, there should be several sizes, as a chain in 1 inch, a chain in } of an inch, a chain in £ an inch, &c. And of these, the best for use are those that are laid on the very edges of the ivory scale, to mark off distances, without compasses. SECTION IL THE PRACTICE OF SURVEYING. This part contains the several works proper to be dc n*» in the field, or the ways of measuring by all the instaiaeots, aud in all situations. •UBVBViire, PBO^EH I. 7b measure a Line or Distance. To measure a line on the ground with the chain : Having provided a chain, with 10 small arrows, or rods, to fix one into the ground, as a mark, at the end of every chain ; two persons take hold of the chain, one at each end of it ; nnd all the 10 arrows are taken by one of them, who goes fore- most, and is called the leader ; the other being called the follower, for distinction's sake. A picket, or station-staff, being set up in the direction of the line to be measured, if there do not appear some murks naturally in that direction, they measure straight towards it, 4he leader fixing down an arrow at the end of every chain, which the follower always takes up, as he comes at it, till all the ten arrows are used. 'They are then all returned to the leader, to use over again. And thus the arrows are changed from the one to the other at every 10 chains length, till the whole line is finished ; then the number of changes of the arrows shows the number of tens, to which the fol. lower adds the arrows he holds in his hand, and the number of links of another chain over to the mark or end of the line. So, if there have been 3 changes of the arrows, and the follower hold 6 arrows, and the end of the line cut off 45 links more, the whole length of the line is set down in links thus, 3645. When the ground is not level, but either ascending or de- scending ; at every chain length, lay the offset-staff, or link- staff, down in the slope of the chain, on which lay the small pocket level, to show now many links or parts the slope line is longer than the true level one ; then draw the chain for- ward so many links or parts, which reduces the line to the horizontal direction. problem n. To lake Angles and Bearings. Let b and c be two objects, or two pickets set up perpendicular ; and let it be required to take their bearings, or the angles formed between them at any station a. 1. With the Pimm Table. The table being covered with * paper, and fixed on "ts ataod ; plant it at the station a, and fix a fine pin, or a foot of the compasses, in a proper point of the paper, to repre- sent the place a ; Close by ihe side of this pin lay the fiducial edge of the index, and turn it about, still touching the pin, till one object b can be seen through the sights : then by the fiducial edge of the index draw a lute. In the same manner draw another line in the direction of the other object o» And it is done. 2. With the Theodolite, 4*c Direct the fixed sights along one of tho lines, as ab, by turning the instrument about till the mark b is seen through these sights ; and there screw the instrument fast. Then turn the moveable index round, till through its sights the other mark c is seen. Then the degrees cut by the index, on the graduated limb or ring of the instrument, show the quantity of the angle. 3. With the Magnetic Needle and Compass. Turn the instrument or compass so, that the north end of the needle point to the flower-de-luce. Then direct the sights to one mark ns b, and note the degrees cut by the needle. Next direct the. sights to the other mark c, and note ngnin the degrees cut by the needle. Then their sum or difference, as the case may be, will give the quantity of the angle bac. 4. By Measurement with the Chain, <fc. Measure one chain length, or any other length, along both directions, as to b and c. Then measure the distance 6c, and it is done.— This is easily transferred to paper, by making a triangle Abe with these three lengths, and then measuring the angle a. 0UBV12T1XO. 417 pboblbh in. To survey a Triangular Field abc. 1. By the Chain. ap 794 AB 1321 fc 826 C Having set up marks at the come , which is to be done in all cases where there are not marks naturally ; measure with the chain from a to p, where a perpendicular would fall from the angle c, and set up a mark at p, noting down the distance a p. Then complete the distance ab, by mea- suring from p to b. Having set down this measure, return to p, and measure the perpendicular pc. And thus, having the base and perpendicular, the area from them is easily found. Or having the place r of the perpendicular, the triangle is easily constructed. Or, measure all the three sides with the chain, and note them down. From which the content is easily found, or the figure is constructed. Measure two sides ab, ac, and the angle a between them. Or measure one side ab, and the two adjacent angles a and b. From either of these ways the figure is easily planned ; then by measuring the perpendicular cp on the plan, and multiplying it by half ab, the content is found. problem IV. To Measure a Four-sided Field. 1. By the Chuin. Measure along one of the diagonals, as ao ; and eitfce the two perpendiculars be, bf, as in the last prohWA * 2. By taking some of the Angles. 4*8 LAHP else the tides ab, bo, cd, da. From either of which the figure may be planned and computed as before directed* Otherwise, h C\aU. Measure, on the longest side, the distances at, aq, ab ; and the perpendiculars rc, <u>. 2. J9y tatoig mne ©ffAe itagfat. ap 110 I 852 rc aq 745 505 qd AB 1110 I Measure the diagonal ac (see the last fig. but one), and the angles cab, cad, acb, acd. — Or measure the four sides, and any one of the angles, as bad. Thus. ac 501 cab 37° W CAD 41 15 ub 72 25 acd 54 40 Or thus. ab 486 bc 304 cd 410 da 462 bad 78" 35'. PROBLEM. V, To survey any Field by the Chain only. Having set up marks at the corners, where necessary, of the proposed field abcdefg, walk over the ground, and con* aider how it can best be divided into triangles and trapeziums ; and measure them separately, ns in the last two problems. Thus, the following figure is divided into the two trapeziums abco, gdef, and the triangle ocn. Then, in the first tra- pezium, beginning at a, measure the diagonal ac, and the two perpendiculars cm, iro. Then the base gc and the perpendicular vq. Lastly, the diagonal df, and the two perpendiculars pE, og. All wbich measures write against the corresponding parts of a rough figure drawn to resemble the figure surveyed, or set them down in any other form you choose. «0 Thus. Am 135 130 mo Aft 410 180 TIB AC 550 cq 152 230 q» CG 440 FO 237 120 oo FP 288 80 FD 520 Or thus. Measure all the sides ab, bc, cd, dk, ef, fg, ga ; and the diagonals ac, cg, gd, df. Otherwise. Many pieces of land may be very well surveyed, by mea- suring any base line, either within or without them, with the perpendiculars let fall on it from every corner. For they are by those means divided into several triangles and trape- zoids, all whose parallel sides are perpendicular to the base line ; and the sum of these triangles and trapeziums will be equal to the figure proposed if the base line fall within it ; if not, the sum of the parts which are without being taken from the sum of the whole which are both within and without, will leave the area of the figure proposed. In pieces that are not very large, it will be sufficiently exact to find the points, in the base line, where the several perpendiculars will fall, by means of the crass, or even by judging by the eye only, and from thence measuring to the corners for the lengths of the perpendiculars. — And it will be most convenient to draw the line so as that all the perpen- diculars may fall within the figure. Thus, in the following figure, beginning at a, and mea- suring along the line ao, the distances and perpendiculars on the right and left are as below. Ab AC xd AC A/ AO 315 440 585 610 990 1020 350 6b 70 cc 320 dv 50 470 /f 4t» LA*B PBOBUm TI. To ike Of***. jjtitiwm being a crooked hedge, or brook, dec. Front a measure in a straight direction along the side of it to b. And in measuring along this line ab, observe when you m directly opposite any bends or corners of the boundary, as si c, 4, e, ic.; and from these measure the perpendicular offsets dfc, di, fcc. with the ofiset.staff, if they are not very large, otherwise , with the chain itself ; and the work is done* 'fhe register, or field-book, may be as follows : Oft. left. Baaa line am A c* 62 45 AC di 84 220 Ad ek 70 840 Kt fl 96 510 *f gm 57 034 *g m 91 785 AB Ac d e J 9 3 FROBLEV VH. To survey any PieUjeUh the Plain Table. 1. Prom one Station. PtAirr the table at any angle as c, from which all the other angles, or marks set up, can be seen ; turn the table about till the needle point to the flower-de-luce ; and there screw it fast. Make a point for c on the paper on the table, and lay the edge of the index to c, turning it about c till through the sights ' A. JB you see the mark d : and by the edge of the index draw a dry or obscure line : then measure the distance cd, and lsy that distance down on the line cd. Then turn the index about the point c, till the mark £ be seen through the sights, by which draw a line and measure the distance to b, layiag it on the line from c to s. In like manner determine the positions of ca and cb, V*y \wnxva£the sights auccossircly to •UEVEYIXG. 441 a and b ; and lay the length of those lines down. Then connect the points, hy drawing the black lines cd, de, ka, ab, bo, for the boundaries of the field. From a Station within the Field. When all the other parts cannot be seen from one angle, choose some place O within, or even without, if v more convenient, from which the other part 8 can be seen. Plant the table at O, then fix it with the needle north, and mark the point O on it. Apply the index successively to (), turning it round with the sights to each angle, a, b, c, d, e, drawing dry lines to them hy the edge of the index ; then measuring the distances oa, or, and laying them down on those lines. Lastly, draw the boundaries ab, bc, cd, de, ea. 3. By going round the Figure. When the figure is a wood, or water, or wlien from some other obstruction you cannot measure lines across it ; begin at any point a, and measure around it, either within or without the figure, and draw the directions of all the sides, thus: Plant the table at a ; turn it with the needle to the north or flower-de-luce ; fix it, and mark the point a. Apply the index to a, turning it till you can see the point k, and there draw a line : then the point b, and there draw a line : then measure these lines, and lay them down from a to Kund b. Next move the table to b, lay the index along the line ab, and turn the table about till you can see the mark a, and screw fast the table ; in which position also the needle will again point to the flower-de-luce, as it will do indeed at every station when the table is in the right position. Here turn the index about b till through the sights you seek the mark c ; there draw a line, measure bc, and lay the distance on that line after you have set down the table at c. Turn it then again into its proper position, and in like manner find the next line cd. And so on quite around by e, to a again. Then the proof of the work will be the joining at a : for if the work be all right, the last direction ka on the ground, will pass exactly through the point a on the paper ; and the measured distance will also reach exactly to a. If these do not coincide, or nearly so, some error has been committed, and the work must be examined over again. Vol. I. 57 442 LAND PllOBLEX VIII. To surrey a Field xci'h the Theodolite, «fc. I. From One Point or Station* When all the angles can be seen from one point, as the angle c first fig. to lust prob.), place the instrument at c, and turn it about, till through the fixed sights you see the mark a, and there fix it. Then turn the moveable index about till the mark a be seen through the sights, and note the de- grees cut on the instrument. Next turn the index suc- cessively to k and n, noting the degrees cut off* at each ; which gives all the angles h<;a, bck, bod. Lastly, measure the lines cb, ca, ce, cd ; and enter the measures in a field-book, or rather, against the corresponding parts of a rough figure drawn by guess to resemble the field. 2. From a Point within or without. Plant the instrument at o (last fig.), and turn it about till the fixed sights point to any object, as a ; and there screw it fast. Then turn the moveable index round till the sights point successively to the other points n, c, b, no'ing the degrees cut off at each of them ; which gives all the angles round the point o. Lastly, measure the distances oa, ob, oc, od, oe, noting them down as before, and the work is done. 3. By going round the Field. By measuring round, either % EL-- within or without the field, pro. ceed thus. Having set up marks at b, u, &c. near the corners as usual, plant the instrument at any point a, and turn it till the fixed index be in the direction ab, and there screw it fast : then turn the moveable index to the / direction ac ; and the degrees cut off will be the angle a. Measure the line ab, and plant the instrument at b, and! there in the same manner observe the angle a. Then mea- sure bc, and observe the angle c. Then measure the dis- tance cd, and take the angle d. Then measure de, and take the angle e. Then measure ef, and tak^ the angle r. And lastly, measure the distance fa. To prove the work ; add all the inward angles, a, b, c, Ac. together ; for when the work is right, their sum will be equal to twice as many right angles as the figure has sides, SUtVIYJNG. 443 wanting 4 right angles. But when there is an angle, as f, that bends inwards, and you measure the external angle, which is less than two right angles, subtract it from 4 right angles, or 360 degrees, to give the internal angle greater than ■ sfc-; a semicircle or 180 degrees. Otherwise. < Instead of observing the internal angles, we may take the external angles, formed without the figure by procuring the sides farther out. And in this case, when the work is right, their sum altogether will be equal to 360 degrees. But when one of them, as f, runs inwards, subtract it from the sum of the rest, to leave 360 degrees. problem rx. To survey a Field with crooked Hedges, 4*c. With any of the instruments, measure the lengths and positions of imaginary lines running as near the sides of the field as you can ; and. in going along them, men sure the offsets in the manner before taught ; then you will have the plan on the paper in using the plain table, drawing the crooked hedges through the ends of the offsets ; but in sur- veying with the theodolite, or other instrument, set down the measures properly in a field-book, or memorandum* book, and plan them after returning from the field, by lay. ing down all the lines and angles. So in surveying the piece arcde, set up marks, a, b. r, #2, dividing it so as to have as few sides as may be. Then begin at any station, a, and measure the lines a//, 6c, cd, d>i> taking their positions, or the angles, a, 6, c, d ; and, in going along the lines, measure all the offsets, as at m, n, o, p, &c. along every station. line. And this is done either within the field, or without, as may be most convenient. When there wc* oV^rt^Mrai LAUD within, as wood, water, hills, ozc. then measure without, t» in the next following figure. problem x. 7b Survey a Field, or any other Thing, by ttco Stations. This is performed by choosing two stations from which all the marks and objects can be seen ; then measuring the distance between the stations, and at each station, taking the angles formed by every object fr< m the station line or distance. The two stations may be taken either within the hounds, or in one of the sides, or in the direction of two of the ob- jects, or quite at a distance and without the bounds of the objects or part to be surveyed. In this manner, not only grounds may be surveyed, with- out even entering them, but a map may be taken of t'.ie principal parts of a county, or the chief places nf a town, or any part of a river or coast surveyed, or any other inac- cessible objects ; by taking two stations, on two towers, or two hills, or such-like. B 8T7HVEY1JG. 445 PRO B LEX XI. To survey a large Estate. If the estate be very large, and contain a great number of field*, it cannot welt be done by surveying all the fields singly, and then putting them together ; nor can it be done by taking all the angles and boundaries that enclose it. For in these cases, any small errors will be so much increased, as to render it very much distorted. But proceed a* below. 1. W ilk over the estate two or three times, in order to get a perfect idea of it, or till you can keep the figure of it pretty we I in mind. And to help your memory, draw an eye-draught of it on paper, at least of the principal parts of it, to guide you ; setting the names within the fields in that draught. 2. Choose two or more eminent places in the estate, for station?*, from which all the princip.il parts of it can be seen : selecting th' se stations as far distant from one another as convenient. 3. Take such nng'es, between the station*, as you think necessary, and measure the distances from station to station, always in a ri^ht line : these things must be done, till you get an many angles and lines as are sufficient for determining all the points of station. And in measuring any of these station distance*, mark accurately where these lines meet with any hedges, ditches, roads, lanes, paths, rivulets, dec. ; and where any remarkable object is placed, by measuring its distance from the station- line ; and where a perpendicular from it cuts that line. And thus as you go along unv mnin station. line, take offsets to the ends of all hedges, am/ to any pond, house, mill, bridge, 6zc. noting every thing down that is remarkable. 4 As to the inner parts of the estate, they must be deter- mined, in like manner, by new station lines ; for, after the main stations are determined, and every thing adjoining to them, then the estate must be subdivided into two or three parts by new station lines ; taking inner stations at proper places, where you can have the best view. Measure these Stat ion- lines as you d d tl e first, and all their intersections with hedges, and offsets to such objects as appear. Then proceed to survey the adjoining fields, by taking the angles that the sides make with the station. 'ine, at the intersections, and measuring the distances to each corner, from the inter, sections. For the station. lines will be the bases to \\W \V» future operations ; the situation of aft pun* Ym&% «D&t*Vi 448 LAKD dependent on them ; nnd therefore they should be taken of as great length a* possible ; and it is best for them to run along Home of the hedges or boundaries of one or more fields, or to pass through some of their angles. All things being determined fur these stations, you must take more inner stations, and continue to divide and subdivide till at last you come to single fields : repeating the same work for the inner stations as for the outer ones, till all is done ; and close the work as often as you can, and in as few lines as possible. 5. An estate nmy be so situated that the whole cannot be surveyed together ; because one part of the estate cannot be seen from another. In this case, you may divido it into three or four parts, and survey the parts separately, as if they were lands belonging to different persons ; and at last join them together. 6. As it is necessary to protract or lay down the work as you proceed in it, you must have a scale of a due length to do it by. To get such n scale, measure the whole length of the estate in chains ; then consider how many inches long the map is to he ; and from these will be known how many chains you must have in an inch ; then make the scale ac- cordingly, or choose one already made. PROBLEM XII. To survey a County, or large Trad of Land. 1. Choosk two, three, or four eminent places, for stations; such as the tops of high hills or mountains, towers, or church steeples, which may he seen from one another ; from which most of the towns and other places of note may also be seen ; and so as to be as far dis ant from one another as possible. On these places raise beacons, or long poles, with flags of different colours flying at them, so as to be visible from all the other stations. 2. At all the places which you would set down in the map, plant long poles, with Hags at them of several colours, to distinguish the places from one another ; fixing them on the tops of church steeples, or the tops of houses ; or in the centres of smaller towns and villages. These marks then being set up at a convenient number of places, and such as may be seen from both stations ; go to one of these stations, and, with an instrument to take angles, standing at that station, take ail the angles between the other station nnd each of these marks. Then go to the other station, and take all the angles between the first station and SURVEY:*©. 447 each of the former marks, setting them down with the others, each against its fellow with the same colour. You may, if convenient, also take the angles at some third station, which may serve to prove the work, if the three lines inter* sect in that point where any mark stands. The marks must stand till the observations are finished al both stations ; and then they may be taken down, and set up at new places. The same operations must be performed at both stations, for these new places ; and the like for others. The instrument for taking angles must ba an exceeding good one, made on purpose with telescopic sights, and of a good length of ra- dius. 3. And though it be not absolutely necessary to measure any distance, because, a stationary line being laid down from any scale, all the other lines will be proportional to it ; yet it is better to measure some of the lines, to ascertain the ' distances of places in miles, and to know how many geome- trical miles there are in any length ; as also from thence to make a scale to measure any distance in miles. In measuring any distance, it will not be exact enough to go along the high roads ; which, by reason of their turnings and windings, hardly ever lie in a right line between the stations ; which must cause endless reductions, and require great trouble to make it a right line ; for which reason it can never be exact. But a better way is to measure in a straight line with a chain, between station and station, over hills and dales, or level fields and all obstacles. Only in case of water, woods, towns, rocks, banks, <kc. where we cannot pass, such parts of the line must be measured by the methods of inaccessible distances ; and besides, allowing for ascents and descents, when they are met with. A good compass, that shows the bearing of the two stations, will always direct us to. go straight, when the two stations cannot be seen ; and in the progress, if we can go straight, offsets may be taken to any remarkable places, likewise noting the intersection of the station. line with all roads, rivers, dtc. 4. From all the stations, and in the whole progress, we must be very particular in observing sea-coasts, river-mouths, towns, castles, houses, churches, mills, trees, nicks, sands, roads, bridges, fords, ferries, woods, hills, mountains, rills, brooks, parks, beacons, sluices, floodgates, locks, &c, and in general every thing that is remarkable. 5. After we have done with the first and main station* lines, wbich command the whole county : we must then take inner stations, at some places already determined ; which will divide the whole into several partitions : and from these •tations we must determine the places of as traiw) ^ga land remaining towns as we can. And if any remain in that part y we must take nnre stations, at Home places already determined ; from which we may determine the rent. And thus go through all the parts of *lho county, taking atalkm after station, till we have determined the whole. And in general the station distances must always pass through such remarkable points as have been determined before, by the former stations. rKOBLEM XIII. To survey a Town or City. This may be done with any of the instruments for taking ^ angles, but best of all with the plain table, where every minute part is drawn while in sight. Instead of the common surveying or Gunter's chain, it will he best, for this purpose, to have a chain 50 feet long, divided into 50 links of one foot each, and an offset- staff «;f 10 feet long. Begin at the meeting of two or more of the principal streets, through which you enn have the longest prospect?, to get the longest station lines : there having fixed the in* strument, draw lines jf direction along those streets, using two men as marks, or poles set in wooden pedestals, or per- haps some remarkable places in the houses at the fan her ends, as windows, doors, corners, dtc. Measure those lines with the chain, taking onsets with the staff, at all corners of streets, bendings, or windings, and to all remurkable thing*, as churches, markets, halls, colleges, eminent houses, &c. Then remove the instrument to another station, nlon"r one of these lines ; and there repeat (he same process as before. And so on till the whole is finished. SURVEYING* 449 Thus, fix the instrument at a, and draw lines in the direction of all the streets meeting there ; then measure ab, noting the street on the left at m. At the second station b, draw the directions of the streets meeting there ; and mea- sure from b to c, noting the places of the streets at n and o as you pass by them. At the third station c y take the di- rection of all the streets meeting there, and measure cd. At d do the same, and measure de, noting the place of the cross streets at p. And in this manner go through all the principal streets. This clone, proceed to the smaller and intermediate streets ; and lastly to the lanes, alleys, courts, yards, and every part that it may be thought proper to re- present in the plan. THEOREM XIV. To lay down the Plan of any Survey. Ir the survey was taken with the plain table, we have a rough plan of it already on the paper which covered the table. But if the survey was with any other instrument, a plan of it is to be drawn from the measures that were taken in the survey ; and first of all a rough plan on paper. To do this, you must have a net of proper instruments, for laying down both lines and angles, &c. ; as scales of va- rious sizes, the more of them, and the more accurate, the better, scales of chords, protractors, perpendicular and pa- rallel rulers, <kc. Diagonal scales are best for the lines, because they extend to three figures, or chains, and links, which are 100 parts of chains. But in using the diagonal scale, a pair of compasses must be employed, to take on the lengths of t!io principal lines very accurately. But a scale with a thin edge divided, is much readier for laying down the perpendicular offsets to crooked hedges, and for marking the places of those offsets on the station- line ; which is done at only one application of the cd«;e the scale to that line, and then pricking off all at once the distances along it. Angles are to be laid down, either with a good scale of chords, which is perhaps the most accurate way, or with a large protractor, which is much readier when many angles are to be laid down at one point, as they arc pricked off all at once round the edge of the protractor. In general, all ILies and angles must be laid down on the plan ii> the same order in which they were measured in the field, and in which they are written in the field-book ; lay- ing down first the angles for the position of \vma, wraX. Vol. I. 58 LAJCD lengths of the lines, with the places of the offsets, and then the lengths of the offsets themselve*, all with dry or obscure lines; then a black line drawn through the extremities of nil the offset*, will be the edge or bounding line of the field, ccc. After the principal bounds and lines are laid down, and made to fit or close properly, proceed next to the smaller objects, till you have entered every thing that ought to ap- pear in the plan, as houses, brooks, trees, hills, gates, stiles, roads, lanes, mill*, bridges, woodlands, Ac. dec. The north side of a map or plan is commonly placed uppermost, and a meridian is somewhere drawn, with the compass or flower-de-luce pointing north. Also, in a vacant part, a scale of equal parts or chains is drawn, with the title of the map in conspicuous characters, and embellished with a compartment. Hills are shadowed, to distinguish them in the map. Colour the hedges with different colours ; repre- sent hilly grounds by broken hills and valleys ; draw single dotted lines for foot-paths, and double ones for horse or car- riage roads. Write the name of each field and remarkable place within it, and, if you choose, its contents in acres, roods, and perches. In a very large estate, or a county, draw vertical and ho- rizontal lines through the map, denoting the spaces between them by letters placed at the top, and bottom, and sides, for readily finding any field or other object mentioned in a table. In mapping counties, and estates that have uneven grounds of hills and valleys, reduce all oblique lines, measured up* hill and down. hill, to horizontal straight lines, if that was not done during the survey, before they were entered in the field-book, by making a proper allowance to shorten them. For which purpose there is commonly a small table engraven on some of the instruments for surveying. THE NEW METHOD OF SURVEYING. PROBLEM XV. To survey and plan bp the new Method. Is the former method of measuring a large estate, theac curacy of it depends both on the correctness of the instru- ments, and on the care in taking the angles. To avoid the errors incident to such a multitude of angles, other methods hu\c of late years been used by some few skilful surveyors : the most practical, expeditious, and correct, seems to be the SURVEYING. 451 following, which is performed, without taking angles, by measuring with the chain only. Choose two or more eminences, as grand stations, and measure a principal base line from one station to another ; noting every hedge, brook, or other remarkable object, as you pass by it ; measuring also such short perpendicular lines to the bends of hedges as may be near at hand. From the ex- tremities of this base line, or from any convenient parts of the same, go off with other lines to some remarkable object situated towards the sides of the estate, without regarding the angles they make with the base line or with one another ; still remembering to note every hedge, brook, or other ob- ject, that you pass by. These lines, when laid down by in- tersections, will, with the base line, form n grand triangle on the estate ; several of which, if need he, being thus mea- sured and laid down, you may proceed to form other smaller triangles and trapezoids on the sides of the former : and so on till you finish with the enclosures individually. By which means a kind of skeleton of the estate may first be obtained, and the chief lines serve as the bases of such triangles and trapezoids as are necessary to fill up all the interior parts. The field-book is ruled into three columns, ns usual. In the middle one are set down the distances on the chain. line, at which any mark, offset, or other observation, is made ; and in the right and left hand columns are entered the off- sets and observations made on the right and left hand re- spectively of the chain-line ; sketching on the sides the shape or resemblance of the fences or boundaries. It is of groat advantage, both for brevity and perspicuity, to begin at the bottom of the leaf, and write upwards ; de- noting the crossing of fences, by lines drawn across the mid. die column, or only a part of such a line on the right and left opposite the figures, to avoid confusion ; and the corners of fields, and other remarkable turns in the fences where off- sets are taken to, by lines joining in the manner the fences do ; as will be best seen by comparing the book with the plan annexed to the field-book following, p. 454. The letter in the left-hand corner at the beginning of every line, is the mark or place measured from; and that at the right-hand corner at the end, is the mark measured to : but when it is not convenient to go exactly from a mark, the place measured from is described such a distance from one mark towards another ; and where a former mark is not mea- sured to, the exact place is ascertained by saying, turn to the right or left hand, such a distance to such a mark, it being always understood that those distances are tuken in the chain line* 4» LAND The characters used are, f for turn to the right handf } for turn to the left hand, and placed over an offset, to show that it is not taken at right angles with the chain- line, but in the direction of some straight fence ; being chiefly used when crossing their directions ; which is a better way of obtaining their true places than by offsets at right angles. When a line is measured whose position is determined, either by former work (as in the case of producing a given line, or measuring from one known place or mark to another) or by itself (as in the third side of the triangle), it is called a fast line, and a double tine across the book is drawn at the conclusion of it ; but if its position is not determined 'as in the second side of the triangle), it is called a loose line, and a single line is drawn across the book. When a line becomes determined in position, and is afterwards continued farther, a double line half through the hook is drawn. When a loose line is measured, it becomes absolutely ne- cessary to measure some other line that will determine its position. Thus, the first line ah or bh, being the base of a triangle, is always determined ; but the position of the second side hj does not become determined, till the third side jb is measured ; then the position of both is determined, and the triangle may be constructed. At the beginning of a line, to fix a loose line to the mark or place measured from, the sign of turning to the right or left hand must be added, as at h in the second, and j in the third line ; otherwise a stranger, when laving down the work, may as easily construct the triangle hjb on the wrong side of the lino ah, us on the right one : hut this error can- not be fallen into, if the sign above named be carefully ob- served. In choosing a line to fix n loose one, care must be taken that it does not make a very acute or obtuse angle ; as in the ^ triangle pur, by the an«rle at b being very obtuse, a small deviation from truth, even the breadth of a point at p or r, would make the error at b, when constructed, very consi. derable ; but by constructing the triangle pnq, such a devia. tion is of no consequence. Where the words leave off are written in the field* book, it signifies that the taking of offsets is from thence discominu. ed ; and of course something is wanting between that and the next offset, to be afterwards determined by measuring some other line. The field. book for this method, and the plan drawn from it, arc contained in the four following pages, engraven on copper plates ; answerable to which the pupil is to draw a - - iaio 004 i4tto -*4 ti 700 •18 . . _ . 400 .10 L 14A«> 1'2J)0 to t 4o loo-» -art m •24 ($* p8o (i 1 •280 - 1.8^ 14(5 4 -** — — X050 02 (1 r r% 050 3* 80 48 307 1 14 _ *I04 2.100 1 \ o 40 7 A 17,*J« 15,10 14*0 L rto L * h 1 5* .! —l 170 — si8o - 1 4404 2000 44 | 18O0 <»O^J — 1840 / 140 4 r 70 1.1*8 1>40 1 ;>l 1 06 1130 8(Jo 44;#o /1 3570 •2 l> 'J. O }} c 2()t() '2 '2 1 O jeofio l(> 40 f J i»lo h , 3 I 7 7 Cili t ft B 049 d i 6 U> i H J ft'+ffo I l lo mill* r 1-1/, \ *** i-| 441 .VI «<j it fkf ^ r 4 * 1 7 < * . si mi i 7 «*& -A 'lit SO - t f. .a , ■UBVJBVIXG* 458 plan from the measures in the field-book, of a larger size, viz. to a scale of a double size will be convenient, such a scale being also found on most instruments. In doing this, begin at the commencement of the field-book, or bottom of the first page, and draw the first line ah in any direction at pleasure, and then the next two sides of the first triangle bhj by sweeping intersected arcs ; and so all the triangles in the same manner, after each other in their order ; and after- wards setting the perpendicular and other offsets at their proper places, and through the ends of them drawing the bounding fences. Note. That the field-book begins at the bottom of the first page, and reads up to the top ; hence it goes to the bottom of the next page, and to the top ; and thence it passes from the bottom of the third page to the top, which is the end of the field-book. The several marks measured to or from, are here denoted by the letters of the alphabet, first the small ones, a, b, c, d y &c, and after them the capitals A, B, C, D 9 dec. But instead of these letters, some surveyors use the numbers in order, 1, 2, 3, 4, dtc. OF THE OLD KIND OF FIELD-BOOK. In surveying with the plain table, a field-book is not used, as every thing is drawn on the table immediately when it is measured. But in surveying with the theodolite, or any other instrument, some kind of a field book must be used to write down in it a register or account of all that is done and occurs relative to the survey in hand. This book every one contrives and rules as he thinks fittest for himself. The following is a specimen of a form which has been formerly used. . It is ruled in three columns, as in the next page. Here Q 1 is the first station, where the angle or bearing is 105° 25'. On the left, at 73 links in the distance or prin- cipal line, is an offset of 92 ; and at 610 an offset of 24 to a cross hedge. On the right, at 0, or the beginning, an offset 25 to the corner of the field ; at 248 Brown's boundary hedge commences ; at 610 an offset 35 ; and at 954, the end of the first line, the denotes its terminating in the hedge. And so on for the other stations. A line is drawn under the work, at the end of every sta- tion line, to prevent confusion. 454 LAKD Form ofthU FiM.Book. Offset* and Remarks on the left. Stations, Bearings, and Distances. Ofiveta and Remarks on the right. 80 OS a cross hedge 34 1 105° 25' 00 73 248 610 054 25 corner Brown's hedge 35 00 house corner 51 34 2 53* 10* 25 120 764 21 20 a tree 40 a stile a brook 30 foot-path 16 cross hedge 18 3 67* 20 61 248 635) 810 973 35 16 a spring 20 a pond Then the plan, on a small scale drawn from the above field- book, will be a* in the following figure. But thn pupil may draw a plan of 3 or 4 time* the size on his paper book. The dotted lines denote the 3 chain or measured linea, and the bluck lines the boundaries on the right and left. But some skilful surveyors now make use of a different method for the field-book, namely, beginning at the bottom 455 of the page and writing upwards ; sketching also a neat boundary on either hand, resembling the parts near tho measured lines as they pas* along ; an example of which was given in the new method of surveying, in the preceding pages. In smaller surveys and measurements, a good way of set- ting down the work, is to draw by the eye, on a piece of paper, a figure resembling that which id to be measured ; and so writing the dimensions, as they are found, against the corresponding parts of the figure. And this method may be practised to a considerable extent, even in the larger surveys. SECTION III. OF COMPUTING AND DIVIDING. PROBLEM XVI. To compute the Contents of Fields. 1. Compute the contents of the figures as divided into triangles, or trapeziums, by the proper rules for these figures laid down in measuring ; multiplying the perpendiculars by the diagonals or bases, both in links, and divide by 2 , the quotient is acres, after having cut off five figures on the right for decimals. Then bring there decimals to roods and perches, by multiplying first by 4, and then by 40. An example of which is ghen in the description of the chain, pag. 480. 2. In small and separate pieces, it is usual to compute their contents from the measures of the lines taken in surveying them, without making a correct plan of them. 3. In pieces bounded by very crooked and winding hedges, measured by offsets, all the parts between the offsets are most accurately measured separately as small trapezoids. 4. Sometimes such pieces as that last mentioned are com* puted by finding a mean breadth, by adding all the offsets together, and dividing the sum by the number of them, ac- counting that for one of them where the boundary meets the station* line (which increases the number of them by I, for the divisor, though it does not increase the sum or quan- tity t > be divided) ; then multiply the length by that mean breadth. 5. But in larger pieces and whole estate** wmm£&%<& 456 LARD many fields, it is the common practice to make a rough plan of the whole, and from it compute the contents, quite inde. pendent of the measures of the lines and angles that wort taken in surveying. For then new lines are drawn in the fields on the plans,' so as to divide them into trapeziums and triangles, the bases and perpendiculars of which are measured on the plan by means of the scale from which it was drawn, and so multiplied together for the contents. In this way, the work is very expeditiously done, and sufficiently correct; for such dimensions are taken as afford the most easy method of calculation ; and among a number of parts, thus taken and applied to a scale, though it be likely that some of the parts will be taken a small matter too little, and others too great, yet they will, on the whole, in all probability, very nearly balance one another, and give a sufficiently accurate result. After all the fields and particular parts are thus computed separately, and added all together into one sum ; calculate the whole estate independent of the fields, by di- viding it into large and arbitrary triangles and trapeziums, and add these also together. Then if this sum be equal to the former, or nearly so. the work is right ; but if the sums have any considerable difference, it is wrong, and they must be examined, and re-computed, till they nearly agree. 6. But the chief art in computing, consists in finding the contents of pieces bounded by curved or very irregular lines, or in reducing such crooked sides of fields or boundaries to straight lines, that shall enclose the same or equal area with those crooked sides, and so obtain the area of the curved figure by means of the right-lined one, which will commonly be a trapezium. Now this reducing the crooked sides to straight ones, is very easily and accurately performed in this manner: — Apply the straight edge of a thin, clear piece of lantern. horn to the crooked line, which is to be reduced, in such a manner, that the small parts cut off from the crooked figure by it, may be equal to those which are taken in : which equality of the parts included and excluded you will presently be able to judge of very nicely by a little prac- tice : then with a pencil, or point of a tracer, draw a line by the straight edge of the horn. Do the same by the other sides of the field or figure. So shall you have a straight- sided figure equal to the curved one ; the content of which, being computed as before directed, will be the content of the crooked figure proposed. Or, instead of the straight edge of the horn, a horse-hair, or fine thread, may be applied across the crooked sides in the same manner ; and the easiest way of using the thread, it to string a amaVV ateufat how with it, either of wire, or cane. IUKVJCYM6. 457 or whale-bone, or such-like slender elastic matter ; for the bow keeping it always stretched, it can be easily and neatly applied with one hand, while the other is at liberty to make two marks by the side of it, to draw the straight line by. EXAMPLE. Thus, let it be required to find thejeontents of the same figure as in Prob. ix, page 443, to a scale of 4 chains to an inch. Draw the 4 dotted straight lines ab, bc, cd, da, cutting off equal quantities on both sides of them, which they do as near as the eye can judge : so is the crooked figure reduced to an equivalent right-lined one of 4 sides, abcd. Then draw the diagonal b», which, by applying a proper scale to it, measures suppose 1256; Also the perpendicular, or nearest distance from a to this diagonal, measures 456 ; and the distance of c from it, is 428. Then, half the sum of 456 and 428, multiplied by the diagonal 1256, gives 555152 square links, or 5 acres, 2 roods, 8 perches, the content of the trapezium, or of the irregular crooked piece. As a general example of this practice, let the contents be computed of all the fields separately in the foregoing plan facing page 453, and, by adding the contents altogether, the whole sum or content of the estate will be found nearly equal to 103} acres. Then, to prove the work, divide the whole plan into two parts, by a pencil line drawn across it any way near the middle, as from the corner I on the right, to the corner near s on the loft ; then, by computing these two large parts separately, their sum must be nearly equal to the former sum, when the work is all right. Vol. I. 59 458 LAND SURVEYING. PROBLEM XVII. To Transfer a Plan to Another Paper, $c. After the rough plan is completed, and a fair one is wanted ; this may be done by any of the following methods* First Method.— Lay the rough plan on the clean paper, keeping them always pressed flat and close together, by weights laid on them? Then with the point of a fine pin or pricker, prick through all the corners of the plan to be copied. Take them asunder, and connect the pricked points on the clean paper, with lines ; and it is done. This method is only to he practised in plans of such figures as are small and tolerably regular, or bounded by right lines. Second Method* — Rub the back of the rough plan over win black-lead powder ; and lay this blacked part on the clean paper on which the plan is to be copied, and in the proper position. Then, with the blunt point of some hard substance, as brass, or such-like, trace over the lines of the whole plan ; pressing the tracer so much, as that the bJack lead under the lines may be transferred to the clean paper : after w hich, take off the rough plan, and trace over the leaden marks with common ink, or with Indian ink. — Or, instead of blacking the rough plan, we may keep constantly a blacked paper to lay between the plans. Third Method. — Another method of copying plans, is by means of squares. This is performed by dividing both ends and sides of the plan which is to be copied into any conve- nient number of equal parts, and connecting the correspond- ing points of division with lines ; which will divide the plan into a number of small squares. Then divide the paper, on which the plan is to be copied, into the same number of squares, each equal to the former when the plan is to be copied of the same size, but greater or less than the others, in the proportion in which the plan is to be increased or diminished, when of a different size. Lastly, copy into the clean squares the parts contained in the corresponding squares of the old plan ; and you will have the copy, either of the same size, or greater or less in any proportion. Fourth Method. — A fourth method is by the instrument called a pentagraph, which also copies the plan in any size required : for this purpose, also, Professor Wallace's eido* graph may be advantageously employed. Fifth method. — A very neat method, at least in copying from u fair plan, is this. Procure a copying frame or glass, made in this manner : namely, a large square of the best AlTOFfCBRS 9 WORK. window glass, set in a broad frame of wood, which can be raised up to any angle, when the lower side of it rests on a table. Set this frame up to any angle before you, facing a strong light ; fix the old plan and clean paper together, with several pins quite around, to keep them together, the clean paper being laid uppermost, and over the face of the plan to be copied. Lay them, with the back of the old plan, on the glass ; namely, that part which you intend to begin at to copy first ; and by means of the light shining through the papers, you will very distinctly perceive every -line of the plan through the clean paper. In this state then trace all the lines on the paper with a pencil. Having drawn that part which covers the glass, slide another part over the glass, and copy it in the same manner. Then another part : and so on, till the whole is copied. Then take them asunder, and trace all the pencil lines over with a fine pen and Indian ink, or with common ink. And thus you may copy the finest plan, without injuring it in the least. OF ARTIFICERS' WORKS, AND TIMBER MEASURING. 1. OF THE CARPENTER'S OR SLIDING RULE. The Carpenter's or Sliding Rule, is an instrument much used in measuring of timber and artificers' works, both for taking the dimensions, and computing the contents. The instrument consists of two equal pieces, each a foot in length, which are connected together by a folding joint. One side or face of the rule is divided into inches, and eighths, or half-quartoM On the same fuce also are several plane scales divided mm twelfth parts by diagonal lines; which are used in plannutg dimensions that are taken in feet and inches. The edge of the rule is commonly divided decimally, or into tenths ; namely, each foot into 400 ARTIFICERS* WORK. parts, and each of these into ten parts again ; so that by means of this last scale, dimensions are taken in feet, tenths, and hundredths, and multiplied as common decimal numbers, which is the best way. On the one part of the other face are four lines, marked b, c, d ; the two middle ones b and c being on a slider, which runs in a groove made in the stock. The same num. berg serve for both these two middle lines, the one being above the numbers, and the other below. These four lines are logarithmic ones, and the three a, s, c, which are all equal to one another, are double lines, as they proceed twice over from 1 to 10. The other or lowest line, d, is a single one, proceeding from 4 to 40. It is also called the girt line, from its use in computing the contents of trees and timber ; and on it are marked wo at 17*15, and AO at 18-05, the wine and ale gage points, to make this in- strument serve the purpose of a gaging rule. On the other part of this face, there is a table of the value of a load, or 50 cubic feet of timber, at all prices, from 6 pence to 2 shillings a foot. When 1 at the beginning of any line is accounted 1, then the 1 in the middle will be 10, and the 10 at the end 100 ; but when 1 at the beginning is counted 10, then the 1 in the middle is 100, and the 10 at the end 1000 ; and so on. And all the smaller divisions are altered proportionally. n. ARTIFICERS' WORK. Artificers compute the contents of their works by several different measures. As, Glazing and masonry, by the foot ; Painting, plastering, paving, 6zc. by the yard, of 9 square feet : Flooring, partitioning, roofing, tiling, 6zc. by the square of 100 square feet : And brickwork, either by the yard of 9 square feet, or by the perch, or square rod or pole, containing 272} square feet, or 30} square yards, being the square of the rod or pole of 16} feet or 5} yards long. As this number 272} is troublesome to divide by, the } is often omitted in practice, and the content in feet divided only by the 272. All works, whether superficial or solid, are computed by the rules proper to the figure of them, whether it be a tri- angle, ot rectang\e, & ^axa\taV)^«d, or any other figure. mUCKLAYSBft' WORK* 401 III- BRICKLAYERS' WORK. Brickwork is estimated at the rate of a brick and a half thick. So that if a wall be more or less than this standard thickness, it must be reduced to it, as follows : Multiply the superficial content of the wall by the number of half bricks in the thickness, and divide the product by 3. The dimensions of a building may be taken by measuring half round on the outside and half round on tho inside ; the sum of these two gives the compass of the wall, to be multi- plied by the height, for the content of the materials. Chimneys are commonly measured as if they were solid, deducting only the vacuity from the hearth to the mantle, on account of the trouble of them. All windows, doors, dec. are to be deducted out of the contents of the walls in which they are placed. The dimensions of a common bare brick are, 8J inches long, 4 inches broad, and 2\ thick ; but including the half inch joint of mortar, when laid in brickwork, every dimen- sion is to be counted half an inch more, making its length 9 inches, its breadth 4), and thickness 3 inches. So that every 4 courses of proper brickwork measures just 1 foot or 12 inches in height. EXAXPLE8. Exam. 1. How many yards and rods of standard brick* work are in a wall whose length or compass is 57 feet 3 inches, and height 24 feet 6 inches ; the wall being 2\ bricks or 5 half bricks thick ? Ans. 8 rods, 17} yards. Exam. 2. Required the content of a wall 62 feet 6 inches long, and 14 feet 8 inches high, and 2| bricks thick ? Ans. 169-753 yards. Exam. 3. A triangular gable is raised 17| feet high, on an end wall whose length is 24 feet 9 inches, the thickness being 2 bricks : required the reduced content ? Ans. 32-08J yards. Ex ax. 4. The end wall of a house is 28 feet 10 inches long, and 55 feet 8 inches high, to the eaves ; 20 feet high is 2£ bricks thick, other 20 feet high is 2 bricks thick, and the remaining 15 feet 8 inches is l£ brick thick ; above which is a triangular gable, of 1 brick thick, which rises 42 courses of bricks, of which every 4 courses make a foot. What is the - whole content in standard measure ? &n&. Y*x&&« 402 carpenters' and joiners' work. IV. MASONS' WORK. To Masonry belong all sorts of stone work ; and the mea- sure made use of is a foot, either superficial or solid. Walls, columns, blocks of stone or marble, &c. are mea- sured by the cubic foot ; and pavements, slabs, chimney- pieces, &c. by the superficial or square foot. Cubic or solid measure is used for the materials, and square measure for the workmanship. In (he solid measure, flic true length, breadth, and thick, ness arc taken and multiplied continually together. In the superficial, there must be taken the length and breadth of every part of the projection which is seen without the general upright face of the building. EXAMPLES* Exam. 1. Required the solid content of a wall, 53 feet 6 inches long, 12 feet 3 inches high, and 2 feel thick ? Ans. 1310J feet. Exam. 2. What is the solid content of a wall, the length being 21 feet 3 inches, height 10 feet 9 inches, and 2 feet thick ? Ans. 521-375 feet. Exam. 3. Required the value of a marble slab, at 8*. per foot ; the length being 5 feet 7 inches, and breadth 1 foot 10 inches ? Ans. 1Z. 1*. lOJd. Exam. 4. In a chimney-piece, suppose the length of the mantle and slab, each 4 feet 6 inches breadth of both together - 3 2 length of each jamb -.44 breadth of both together. 1 9 Required the superficial content ? Ans. 21 feet 10 inches. V. CARPENTERS' AND JOINERS' WORK. To this branch belongs all the wood-work of a house, such as flooring, partitioning, roofing, &c. Large and plain articles are usually measured by the square foot or yard, &c. ; but enriched mouldings, and some other articles, are often estimated by running or lineal mea. sure ; and some things are rated by the piece. In measuring of Joists, take the dimensions of one joist, carpenters' and joiners' work. 463 and multiply its content by the number of them ; consider- ing that each end is let into the wall about & of the thick- ness, as it ought to be. Partitions are measured from wall to wall for one dimen- sion, and from floor to floor, as far as they extend, for the other. The measure of Centering for Cellars is found by making a string pass over the surface of the arch for the breadth, and taking the length of the cellar for the length : but in groin centering, it is usual to allow double measure, on ac- count of their extraordinary trouble. In Roofing, the dimensions, as to length, breadth, and depth, are taken as in flooring joists, and the contents com- puted the same way. In Floor -boarding, take the length of the room for one di- mension, and the breadth for the other, to multiply together for the content. For Stair-cases, take the breadtii of all the steps, by mak- ing a line ply close over ihem, from the top to the bottom, and multiply the length of this line by the length of a step, for the whole area. — By the length of a step is meant the length of the front and the returns nt the two ends ; and by the breadth is to be understood the girts of its two oater sur- faces, or the tread and riser. For the Balustrade, take the whole length of the upper part of the hand-rail, and girt over its end till it meet the top of the newel-post, for the one dimension ; and twice the length of the baluster on the landing, with the girt of the hand-rail, for the other dimension. For Wainscoting, take the compass of the room for the one dimension ; and the height from the floor to the ceiling, making the string ply close into all the mouldings, for the other. For Doors, take the height and the breadth, to multiply them together for the area. — If the door be panneled on both sides, take double its measure for the workmanship ; but if one side only be panneled, take the area and its half for the workmanship. For the Surrounding Architrave, girt it about the uppermost part for its length ; and measure over it, as far as it can be seen when the door is open, for the breadth. Window-shutters, Bases, &c. are measured in like manner. In measuring of Joiners' work, the sttin^ \a roa&fe v& ^ 464 SLATERS AND TILERS' WOBK. close into all mouldings, and to every part of the work over which it passes. EXAMPLES. Ex ax. 1. Required the content of a floor, 48 feet 6 inches long, and 24 feet 3 inches broad? An*. 11 sq. 76} feet. Exam. 2. A floor being 36 feet 3 inches long, and 16 feet 6 inches broad, how many squares are in it ? Ans. 5 sq. 98} feet. Exam. 3. How many squares are there in 173 feet 10 inches in length, and 10 feet 7 inches height, of partitioning ? Ans. 18*3973 squares. Exam. 4. What cost the roofing of a house at 10*. 6d. a square ; the length within the walls being 52 feet 8 inches, and the breadth 30 feet 6 inches ; reckoning the roof £ of the flat? Ans. 121. 12s. ll}d. Exam. 5. To how much, at 0*. per square yard, amounts the wainscoting of a room ; the height, taking in the cornice and mouldings, being 12 feet 6 inches, and the whole com- pass 83 feet 8 inches ; also the three window-shutters are each 7 feet by 8 inches by 3 feet G inches, and the door 7 feet by 3 feet inches ; the door and shutters, being worked on both sides, arc reckoned work and half work ? Ans. 36/. 12*. 2*d. VI. SLATERS' AND TILERS' WORK. In these articles, the content of a roof is found by mul- tiplying the length of the ridge by the girt over from eaves to caves ; making allowance in this girt for the double row of slates at the bottom, or for how much one row of slates or tiles is laid over another. When the roof is of a true pitch, that is, forming a right angle at top; then the breadth of the building, with its half added, is the girt added over both sides nearly. Iu angles formed in a roof, running from the ridge to the eaves, when the angle bends inwnnN. it is called a valley ; but when outwards, it is onllcd n hip. Deductions are made ibr chimney shafts or window holes. PLASTERERS* won. EXAMPLES. Exam. 1. Required the content of a slated roof, the length being 45 feet 9 inches, and the whole girt 34 feet 3 inches? Ans. 174^ yank. Exam. 2. To how much amounts the tiling of a house, at 25*. 6d. per square ; the length being 43 feet 10 inches, and the breadth on the fiat 27 feet 5 inches ; also the caves projecting 16 inches on each side, and the roof of a true pitch ? Ans. 241. 9s. fyd. VH. PLASTERERS' WORK. Plasterers' work is of two kinds ; namely, ceiling, which is plastering on laths ; and rendering, which is plastering on walls : which are measured separately. The contents are estimated either by the foot or the yard, or the square, of 100 feet. Enriched mouldings, dec. are rated by running or lineal measure. Deductions are made for chimneys, doors, windows, &c. examples. Exam. 1. How many yards contains the ceiling which is 43 feet 3 inches long, and 25 feet 6 inches broad ? Ans. 122}. Exam. 2. To how much amounts the ceiling of a room, at 10J. per yard : the length being 21 feet 8 inches, and the breadth 14 feet 10 inches ? Ans. 1/. 9s. 8Jrf. Exam. 3. The length of a room is 18 feet 6 inches, the breadth 12 feet 3 inches, and height 10 feet 6 inches ; to how much amounts the ceiling and rendering, the former at Sd. and the latter at 3d per yard : allowing for the door of 7 feet by 3 feet 8, and a fire-place of 3 feet square ? Ans. 1Z. 13*. 3jd. Exam. 4. Required the quantity of plastering in a room, the length being 14 feet 5 inches, breadth 13 feet 2 inches, and height 9 feet 3 inches to the under side of the cornice, which girts 8} inches, and projects 5 inches from the wall on the upper part next the ceiling ; deducting only for a door 7 feet by 4? Ans. 53 yards 5 feet 3} inches of rendering 18 5 6 of ceiling 39 Of} ofwrarcfe. Vol. I. 60 406 CLAftflUtl' WOfcX. VIII. PAINTERS 9 WORK. Painters' work is computed in square yards. Every part is measured where the colour lies ; and the measuring line is forced into all the mouldings and corners. Windows are done at so much a piece. And it is usual to allow double measure for carved mouldings, dec. EXAMPLES. Exam. 1. How many yards of painting contains the room which is 65 feet 6 inches in compass, and 12 feet 4 inches high ? Ans. 89}£ yards. Exam. 2. The length of a room being 20 feet, its breadth 14 feet 6 inches, and height 10 feet 4 niches ; how many varus of painting are in it, deducting a fire-place of 4 feet by 4 feet 4 inches, and two windows each 6 feet by S feet 2 inches ? Ans. 73^ yards. Exam. 3. What cost the painting of a room, at 6d. per yard ; its length being 24 feet 6 inches, its breadth 16 feet 3 inches, and height 12 feet 9 inches ; also the door is 7 feet by 3 feet 6, and the window-shutters to two windows each 7 feet 9 by 3 feet 8 ; but the breaks of the windows them- selves are 8 feet 6 inches high, and 1 foot 3 inches deep ; in- cluding also the window cills or seats, and the soffits above, the dimensions of which are known from the other dimen. sions : but deducting the fire-place of 5 feet by 5 feet 6 ? Ans. 31 3*. 10}d. IX. GLAZIERS' WORK. Glaziers take their dimensions, either in feet, inches, and parts, or feet, tenths, and hundredths. And they compute their work in square feet. In taking the length and breadth of a window, the cross bars between the squares are included. Also windows of round or oval forms are measured as square, measuring them to their greatest length and breadth, on account of the waste in cutting the glass. examples. Exam. 1. How many square feet contains the window which is 4*25 feet long, and 2-75 feet broad ? Ans. llf PAVEIt'* WORK. 407 Exam. 2. What will the glazing a triangular sky-light tome to, at lOd. per foot ; the base being 12 feet 6 inches, and the perpendicular height 6 feet 9 inches ? Ans. II. 15s. \\d. Exam. 3. There is a house with three tiers of windows, three windows in each tier, their common breadth 3 feet 1 1 inches : now the height of the first tier, is 7 feet 10 inches of the second 6 8 of the third 5 4 Required the expense of glazing at I4d per foot ? Ans. 13*. lis. lO^d. Exam. 4. Required the expense of glazing the windows of a house at 13i. a foot ; there being three stories, and three windows in each story : the height of the lower tier is 7 feet 9 inches of the middle 6 6 of the upper 5 3| and of an oval window over the door 1 10} the common breadth of all the windows being 3 feet 9 inches ? Ans. 121. 5s. 6d. X. PAVERS' WORK. Pavers' work is done by the square yard. And the con* tent is found by multiplying the length by the breadth. EXAMPLS8. Exam. 1. What cost the paving a foot-path, at Ss. 4d. a yard ; the length being 35 feet 4 inches, and breadth 8 feet 3 inches ? Ans. 51. 7s. ll$d. Exam. 2. What cost the paving a court, at 3*. 2d. per yard ; the length being 27 feet 10 inches, and the breadth 14 feet 9 inches ? Ans. 71. 4s. 5}d. Exam. 3. What will be the expense of paving a rectan. gular court-yard, whose length is 63 feet, and breadth 45 feet ; in which there is laid a foot-path of 5 feet 3 inches broad, running the whole length, with broad stones, at 8*. a yard ; the rest being paved with pebbles at 2s. 6d. a yard ; Ans. 401 5s. 10^1. XI. PLUMBERS* WORK. Pmnsfts' work is rated at so much a pound, or else fey the hundred weight of 113 pounds. J Sheet lead, used in roofing, guttering, Ac is from II t» 101b. to the square foot. And a pipe of an inch bore ia cess* feoely l&or 141b, to tbe yard in length, jEgAJt* 1. How much weighs the lead which is 80 feet C inches long, and 3 feet 3 inches broad, at 8|lb. to the a^iarefeot? P Ana. 1091 ^ lb. Exam. 2. What cdst the covering and guttering* reef With lead, at 18s. the cwU ; the length of the roof being 43 feet, and breadth or girt over it 32 feet ; the guttering 57 feet long, and 2 feet wide ; the former 9*831 lb. and the latter 7<37ftlb. to the square foot ? Ana, 1151. 9s. ljd- XII. TIMBER MEASURING. PROBLEM I. 7b find the Area, or Superficial Content of a Board or Plank. Multiply the length by the mean breadth. Note. When the board is tapering, add the breadths at the two ends together, and take half the sum for the mean breadth. 0{ else take the mean breadth in the middle. By the Sliding Rule. Set 12 on b to the breadth in inches on a ; then against the length in feet on b, is the content on a, in feet and fractional parts. bxaxflbs. £xax. 1. What is the value of a plank, at lfd. per feet, whose length is 12 feet 6 inches, and mean breadth 11 iftfbes? Ans. ls.5d. ^ TIMBU MSA1UBIK9. 469 Exam. 2. Requred the content of a board, whose length is 11 feet 2 inches, and breadth 1 foot 10 inches ? Ans. 20 feet 5 inches 8*. Exam. 3. What is the value of a plank, which is 12 feet 9 inches long, and 1 foot 3 inches broad, at 2±d. a foot? Ans. 3* . 3| A Exam. 4. Required the value of 5 oaken planks at 3d. per foot, each of them being 17 1 feet long ; and their several breadths as follows, namely, two of 13£ inches in the middle, one of 14 J inches in the middle, and the two remaining ones, each 18 inches at the broader end, and 11 J at the nar. rower ? Ans. II. 5s. 9±d. PROBLEM II. To find the Solid Content of Squared or Four-tided Timber. Multiply the mean breadth by the mean thickness, and the product again by the length, for the content nearly. By the Sliding Rule. C D D C As length : 12 or 10 : : quarter girt : solidity. That is, as the length in feet on c, is to 12 on d, when the quarter girt is in inches, or to 10 on d, when it is in tenths of feet ; so is the quarter girt on d, to the content on c. Note 1. If the tree taper regularly from the one end to the other ; either take the mean breadth and thickness in the middle, or take the dimensions at the two ends, and half their sum will be the mean dimensions : which multiplied as above, will give the content nearly. 2. If the piece do not taper regularly, but be unequally thick in some parts and small in others ; take several different dimensions, add them all together, and divide their sum by the number of them, for the mean dimensions. EXAMPLES. Exam. 1. The length of a piece of timber is 18 feet 6 inches, the breadths at the greater and less end 1 foot 6 inches and 1 foot 3 inches, and the thickness at the greater and less end 1 foot 3 inches and 1 foot ; required the solid content? Ans. 28 feet 7 inches. Exam. 2. What is the content of the piece of MfhBef, whoee length is 24£ feet, and the mean breadth and thick* ness each 1-04 feet ? Ana. 26} feet. Exam. 3. Required the content of a piece of timber, whose length is 20-38 feet, and its ends unequal squares, the side of the greater being 19 J inches, and the side of the less 9} inches t Ana. 29*7562 feet Exam. 4. Required the content of the piece of timber, whose length is 27*36 feet ; at the greater end the breadth Is 1*78, and thickness 1*23 ; and at the less end the breadth is 1-04, and thickness 0*91 feet ? Ans. 41 -278 feet. PROBLEM m. To find the Solidity of Round or Unfavored Timber. Multiply the square of the quarter girt, or of J of the toean circumference, by the length, for the content. By the Sliding Rule. Ae the length upon c : 12 or 10 upon d :: quarter girt, in 12ths, or lOths, on d : content on c. Note 1. When the tree is tapering take the mean dimen- sions as in the former problems, either by girting it in the fciddle, for the mean girt, or at the two ends, and taking half the sum of the two ; or by girting it in several places, then adding all the girts together, and dividing the sum by the number of them, for the mean girt. But when the tree is very irregular, divide it into several lengths, and find the content of each part separately. 2. This rule, which is commonly used, gives the answer about i less than the true quantity in the tree, or nearly what the quantity would be, after the tree is hewed square in the usual way : so that it seems intended to make an allowance for the squaring of the tree. On this subject, however, Hutton's Mensuration, part v. sect. 4, may be advantageously consulted. EXAMPLES. Exam. 1. A piece of round timber being 9 feet 6 inches long, and its mean quarter girt 42 inches ; what is the content ? Ans. 116$ feet Exam. 2. The length of a tree is 24 feet, its girt at the thicker end 14 feet, and at the smaller end 2 feet ; required the content 7 Ans. 96 feet T1XBXK UASUBHIO. 471 Exam. 3. What is the content of a tree whose mean girt is 3*15 feet, and length 14 feet 6 inches ? Ans. 8-9922 feet. Exam. 4. Required the content of a tree, whose length is IH feet, which girts in five different places as follows, namely, in the first plaoe 9*43 feet, in the second 7*92, in the third 6-15, in the fourth 4*74, and in the fifth 3*16 ? Ans. 42-519525. [ 472 J 1 • CONIC SECTIONS. DEFINITIONS. 1. Conic Sections are the figures made by a plane cut- ting a cone. 2. According to the different positions of the cutting plane there arise five different figures objections, namely, a triangle, a circle, an ellipsis, an hyperboK, and a parabola : the three last of which only are peculiarly called Conic Sec- tions. 3. If the caning plane pass through the vertex of the, cone, and any part of the base, the section will evidently be a triangle ; as vab. 4. If the plane cut the cone parallel to the base, or make no angle with it, the section will be a circle ; as abd. 5. The section dab is an ellipse when the cone is cut obliquely through both sides, or when the plane is inclin- ed to the base in a less angle than the side of the cone is. 6. The section is a parabola, when the cone is cut by a plane parallel to the side, or when the cutting plane and the side of the cone make equal angles with the base. DHflNITlOm. 473 7. The section is an hyperbola, when the cutting plane makes a greater angle with the base than the side of the cone makes. 8. And if all the sides of the cone be continued through the vertex, forming an opposite equal cone, and the plane be also continued to cut the opposite cone, this latter section will be the op- posite hyperbola to the former ; as dac. 9. The Vertices of any section, are the points where the cutting plane meets the sides of that vertical triangular sec- tion which is perpendicular to it ; as a and b. Hence the ellipse and the opposite hyperbolas, have each two vertices ; but the parabola only one ; unless we consider the other as at an infinite distance. 10. The Axis, or Transverse Diameter, of a conic section, is the line or distance ab between the vertices. Hence the axis of a parabola is infinite in length, \b being only a part of it. Ellipse. Hyperbolas. Parabola. 1 1 . The centre c is the middle of the axis. Hence the centre of a parabola is infinitely distant from the vertex. And of an ellipse, the axis and centre lie within the curve ; but of an hyperbola, without. 12. A Diameter is any right line, as ab or db, drawn through the centre, and terminated on each side by the curve ; and the extremities of the diameter, or its intersections with the curve, are its vertices. Hence all the diameters of a parabola are parallel to the axis, and infinite in length. Hence also evei^ ftassfttax <& Vol. I. 61 474 CONIC SECTIONS, the ellipse and hyperbola has two vertices ; but of the pan. bola, only one ; unless we consider the other as at an infinite distance. 13. The Conjugate to any diameter, is the line drawn through the centre, and parallel to the tangent of the curve at the vertex of the diameter. So, fo, parallel to the tangent at i>, is the conjugate to de ; and 111, parallel to the tangent at a," is the conjugate to ab. Hence the conjugate hi, of the axis ab, is perpendicular to it. .. 14. An Ordinate to any diameter, is a line parallel to its conjugate, or to the tangent at its vertex, and terminated by the diameter and curve. So dk, el, are ordinates to the axis ab ; and mn, xo, ordinates to the diameter dk. Hence the ordinates of the axis are perpendicular to it. 15. An Absciss is a part of auy diameter contained between either of its vertices and an ordinate to it ; as ak or bk, or dn or en. Hence, in the ellipse and hyperbola, ever}' ordinate has two determinate abscisses ; but in the parabola only one ; the other vertex of the diameter being infinitely distant. 1G. The Parameter of any diameter, is a third proportional to that diameter and its conjugate, in the ellipse and hyper- bola, and to one absciss and its ordinate in the parabola. 17. The Focus is the point in the axis where the ordinate is equal to half the parameter. As k and l, where dk or el is equal to the semi -parameter. The name focus being given to this point from the peculiar property of it mentioned in the corol. to thcor. 5* in the Ellipse and Hyperbola following, and to theor. in the Parabola. Hence, the ellipse and hyperbola have each two foci ; but the parabola only one. 18. If t)al, fhg, be two opposite hyperbolas, having as for their first or transverse axis, and ab for their second or conjugate axis. And if dm, fbg, be two other opposite hy. pcrbolns having the same axes, hut in the contrary order, namely, ab their fatal txxvs, *iA » tat\t n*»wA\ vWwvlvfiee DEFINXTIOXS* 475 two latter curves d*e, fbg, are called the conjugate hyper- bolas to the two former dae, pro ; and each pair of opposite curves mutually conjugate to the other ; being all for con- venience of investigation referred to one plane, though they are only posited two and two in one plane ; as will appear more evidently from the demonstration of th. 2. Hyperbola. 19. And if tangents be drawn to the four vertices of the ^curves, or extremities of the axes, forming the inscribed rectangle iiikl ; the diagonals hck, icl, of this rectangle, are called the asymptotes of the curves. And if these lfamp. totes intersect at right angles, or the inscribed rectangle be a square, or the two axes ab and ab be equal, then the hy- berbolas are said to be right-angled, or equilateral. SCHOLIUM. The rectangle inscribed between the four conjugate hy- perbolas, is similar to a rectangle circumscribed about an ellipse, by drawing tangents in like manner, to the four ex. tremities of the two axes ; and the asymptotes or diagonals in the hyperbola, arc analogous to those in the ellipse, cut- ting this curve in similar points, and making that pair of conjugate diameters which arc equal to each other. Also, the whole figure formed by the four hyperbolas, is as it were, an ellipse turned inside out, cut open at the extre. mi tics, d, e, f, g, of the said equal conjugate diameters, and those four points drawn out to an infinite distance ; the cur. vature being turned the contrary way, but the axes, and the rectangle passing through their extremities, continuing fixed. And further, if there be four cones cscrf, cop, cmp, cno, having all the same vertex c, and all their axes in the same plane, and their sides touching or coinciding in the common intersecting lines mco, ncv ; then if these four cones be all cut by one plane, parallel to the common plane of their axes, there will be formed the four hyperbolas, gqr, fst, vkl, win, of which each two op- po8itesare equal ; and each pair resembles the conjugates to the other two, as here in the annexed figure ; but they are not accurately the conjugates, except only when the four cones are all equal, and then the four hyperbolic sections are all equal also. [ 476 ] OF THE EIJJPSE. THEOREM X. The Squares of the Ordinntes of the Axis are to each other as the Rectangles of their Abscisses. Let avb be a plane passing through the apju of the cone ; agiii another section of the cone perpendicular to the plane of the former ; ab the axis of this elliptic section ; and fg, hi, or- dimftes perpendicular to it. Then it will be, as fg 3 : hi 3 : : af . eb : ah . iib. For, through the ordinates fg, hi, draw the circular sections xgl, min, parallel^ to the base of the cone, having kl, mn, for their diameters, to which fg, hi, are ordinate, as will as to the axis of the ellipse. Now, by the similar trangles afl, ahn, and bfk, bhh, it is AF l AH * FL : UN, and fb : iib : ; kf : mu ; hence, taking the rectangles of the corresponding terms, it is, the rect. af . fb : ah . hb : : kf . fl : mu . iin. But, by the circle, kf . fl = f.; 8 , and mh . hn = hi 9 ; Therefore the rect. af . fb : ah . hb : : fg 8 : hi*, q. e. d. ^ THEOREM II. As the Square of the Transverse Axis Is to the Square of the Conjugate : So is the Rectangle of the Abscisses To the Square of their Ordinate. ft For, by theor. 1, ac . cb : ad . db : : c« 9 : de* ; But, if c be the centre, then ac.cbs ac 1 , and ca is the semi-conjugate. OF TBS SLUMS. 477 Therefore ac 9 : ad • db : : ac* : db 9 ; or, by permutation, ac 9 : ac 9 : : ad • db : de 9 ; or, by doubling, ab s : ab* : : ad . db : db 1 . a. s. d. Cord. Or. by div. ab : — : : ad . db or ca 9 — cd 9 : db 1 . ' ab that is, ab : p : : ad . db or ca 9 — CD 9 : db 9 ; where p is the parameter^-, by the definition of it. That is, As the transverse, 9 Is to its parameter, So is the rectangle of the abscisses, To the square of their ordinate. THEOREM III. As the Square of the Conjugate Axis Is to the Square of the Transverse Axis, So is the Rectangle of the Abscisses wf the Conjugate, or the difference of the Squares of the Semi-conjugate and Distance of the centre from any Ordinate of that Axis, To the Square of the Ordinate. That is, cb 9 : : ad . db or ca 9 — cd 1 : For, draw the ordinate ed to the transversa ab. Then, by theor. 1, ca 9 : ca 1 : : de 1 : ad .*bb or ca 1 — cd*, or ca 9 : ca 9 : : cd* : ca 9 — cJe 9 , But ca 9 : ca 9 : : ca 9 : ca 9 , theref. by subtr. ca 9 : ca 9 : : ca 9 — cd 1 or ad .db : da 9 . Q. E. D. Carol. 1. If two circles be described on the two axes as diameters, the one inscribed within the ellipse, and the other circumscribed about it ; (hen an ordinate in the circle will be to the corresponding ordinate in the ellipse, as the axis of this ordinate, is to the other axis. That is, ca : ca : : dg : de, and ca : ca dg : dz. For, by the nature of the circle, ad • db = dg* ; theref. by the nature of the ellipse, ca 9 : ca 9 : : ad • db or do* : de 9 , or ca : ca ; : iw \ iv* conic SBCTIOKt. In like manner . ca : ca : : dg : o*e. Also, by equality - dg : de or <:d : : o*e or tc 2 dg. Therefore ego is a continued straight line. Corel. 2. Hence also, as the ellipse and circle are made up of the same number of corresponding ordinates, which are all in the same proportion of the two axes, it follows that the areas of the whole circle and ellipse, as also of any like parts of them, are' in the same proportion of the two axes, or as tfcfe square of the diameter to the rectangle of the two axes ; that is, the areas of the two circles, and of the ellipse, are as the square of each axis and the rectangle of the two ; and therefore the ellipse is a mean proportional between the two circles. THEOREM IV. The Square of the Distance of the Focus from the Centre, is equal to the Difference of the Squares of the Semi- axes. Or, the square of the Distance between the Foci, is equal to the Difference of the Squares of the two Axes. That is, cf 9 = ca 8 - ca* or f/ j = ab 1 — a6 J b For, to the focus f draw the ordinate fe ; which, by the definition, will be the semi parameter. Then, by the nature of the curve - - ca 8 : ca 1 : : ca 2 - of* : Ft* ; and by the def. of the para, ca 8 : ca 2 : : ca 8 : fk 2 ; therefore - . c/i a = ca 2 - cr 2 ; ♦ and by addit. nnd subtr. cf 2 = ca 2 — ca'; or, by doubling, • rf 2 = ab 3 — ab*. q. e. p. Carol. 1. The two semi-axes, nnd the focal distance from the centre, are the sides of a right. angled triangle era ; and the distance Fa from the focus to the extremity of the con- jugate axis, is = ac the semi-transverse. Carol. 2. The conjugate semi-axis ca is a mean proper, ttonal between af, fb, or between a/*, /b, the distances of either focus from the two vertices. For co" = ca? — cf* =■ Vsfc- "V ^ • — w \ = af . fb. Or THE ELLIPSE. THEOREM V. x The Sum of two lines drawn from the two Foci to meet , at any Point in the Curve, is equal to the Transverse Axis. For, draw ag parallel and equal to ca the semi-conjugate ; and join cg meeting the ordinate de in h ; also take ci a 4th proportional to ca, cf, cd. Then by theor. 2, ca 8 : ag 2 : : ca 3 — cd 3 : de 3 ; and, by sim. tri. ca 2 : ag 3 : : ca 3 — cd 3 : ag 8 — dh* ; consequently de 3 = ag 3 — du 2 = ca 3 — dh 3 . Also, fd = cf ^ cd, and fd 2 = cf 2 — 2of . cd + cd 3 ; And, by right-angled triangles, fe 3 = fd 3 + de 2 ; therefore fe 2 = cf 2 + ca 2 — 2cf . cd + cd 2 — dh 3 ; But by theor. 4, cf 3 + ca 3 = ca 2 , % and by supposition, 2cf . cd = 2ca • ci ; theref. fe 3 = ca 3 — 2ca . ci + cd 3 — dh 3 . Again, by supp. ca 3 : cd 3 : : cf 3 or ca 3 — ag 3 : cr ; and, by sim. tri. ca 3 : cd 8 : : ca 3 — ag 3 : cd 3 — dh 3 ; therefore • ci 3 = cd 3 — dh 8 ; consequently fe 2 =■= ca 3 — 2ca . ci + cr*. And the root or side of this square is fe = ca — ci = ai. In the same manner it is found that fz = ca + ci = bi. Conseq. by addit. fe + ft = ai + bi = ab. q. k. d. Cord. 1. Hence ci or ca — fe is a 4th proportional to ca, cr, CD. Coral. 2. And /e — fe = 2ci ; that is, the difference be* tween two lines drawn from the foci, to any point in the curve, is dquble the 4th proportional to ca, cf, cd. CoroL 3. Hence is derived the common method of de- scribing this curve mechanically by points, or with a thread, thus: 480 come accnoxi. In the transverse take the foci f,/, mod any point i. Then with the radii ai, bi, and centres r,/, describe arcs intersecting in b, which will be a point in the curve. In like manner, assuming other points i, as many other points will be found in the curve. Then with a steady hand, the curve line may be drawn through all the points of inter- section E. Or, take a thread of the length ab of the transverse axis, and fix its two ends in the foci f, /, by two pins. Then carry a pen or pencil round by the thread, keeping it always stretched, and its point will trace out the curve line. THEOREM VI. If from any Point i in the Axis produced, a Line il be drawn touching the Curve in one Point l ; and the Or- dinate lh be drawn ; and if c be the Centre or Middle of ab : Then shall cm be to ci as the Square of am to the Square of ai. cm : For, from the point i draw any other line if.h to cut the curve in two points is and h ; from which let fall the perpen- diculars ed and hg ; and bisect do in k. Then, by theor. 1, ad . db : ag . gb : : de 2 : gh 9 , and by sim. triangles, id* : ig 3 : : de 9 : gh 9 ; there f. by equality, ad . db : ag . gb : : id* : ig*. But DB = CB + CD = AC + I'D — AG + DC CG = 2cE + AG, and GB = CB CG = AC CG = AD+ DC — CG = 2cK + AD 5 theref. ad . 2ck + ad . ag : ag . 2c k + ad . ag : : id* : and, by div. dg . 2ck : ig 8 — id 2 or dg . 2ik : : ad . 2ck + AD . AG : ID a , or 2ck : 2ik : : ad . 2ck + ad . ag : id', or ad . 2ck : ad . 2ik : : ad . 2ck + ad . ag : id 9 ; theref. by div. ck : ik : : ad • ag : id' — ad . 2ik, and, by corop. ck : ic : : ad . ag : id 9 — ad . id + ia 9 or - ck : ci : : ad . ag : ai 9 . OF THE ELLIPSE. 481 But, when the line ih, by revolving about the point i, comes into the position of the tangent il, then the points e and ii meet in the point l, and the points d, k, g, coincide with the point m ; and then the last proportion becomes cm : ci : : am 3 : ai s . q. e. d. THEOREM VII. If a Tangent and Ordinate be drawn from any Point in the Curve, meeting the Transverse Axis ; the Semi -transverse will be a Mean Proportional between the Distances of the said two Intersections from the Centre. That is, ca is a mean proportional between cd and or ; or cd, ca, cr, are con- tinued proportionals. For. by tlieor. G, cd : ct : : ad 9 : at 2 that is, cd : ct : : (ca — co) a : (ct — ca) 3 , or - cd : ct : : cd 3 + CA a : ca 9 + ct 1 , and - cd : dt : : cd 3 + ca 9 : ct 9 — cd 9 , or - cd : dt : : cd ! + ca 2 : (ct + cd)dt, or - cd 9 : cd . dt : : cd* + ca 2 : (cd . dt) -f- (ct . dt), hence cd 2 : ca 9 : : cd . dt : ct . dt, and - cd': ca j : : cd : ct. therefore (t^. 78, Geora.) cd : ca : : ca : ct. q. e. d. Cord. 1. Since cr is always a third proportional to cd, ca ; if the points d, a, remain constant, then will the. point T be constant also ; nnd therefore all the tangents will meet in this point t, which are drawn from the point e, of every ellipse described on the same axis ar, where they are cut by the common ordinate def drawn from the point d. Cord. 2. When the outer ellipse, by enlarging, becomes a circle, as at the upper figure at e, then by drawing et perp. to ce, and joining t to the lower e, the tangent to the point e at the ellipse is obtained. theorem viii. If there be any Tangent meeting four Perpendiculars to the Axis drawn from these four Points, namely, the Centre, the two Extremities of the Axis, and the Point of Contact ; those four Perpendiculars will be Proportions. Vol. I. 62 488 COKIC SKCTIOM* That is, AO : de : : ch : bi. For, by theor. 7, tc : ac : : ac : dc, theref. by div. ta : ad : : tc : ac or cb, and by comp. ta : td : : tc : th, and by sim. tri. ao : de : : cn : hi. q. E. D. Cord. 1. Hence ta, td, tc, tb ) ^ . , and TG, TB, TH, TI \ *"> P~pOrtM»^ Fur these are as ao, dr, ch, bi, by similar triangles. Corel. 2. Draw ai to bisect de in p ; then since ta : te : : tc : ti, the triangles tab, tci are similar, as well as the triangles aed, cbi, and adp, abi. Hence - ad : dk : : cb : bi and - ad : dp : : ab : bi .\ de : dp : : ab : cb : : 2 : 1 ; which sug- gests another simple practical method of drawing a tangent to an ellipse. theorem ix. If there be any Tangent, and two Lines drawn from the Foci to the Point of Contact ; these two lines will make equal Angles with the Tangent. That is, the JL vet = L fve. For, draw the ordinate dk and /a parallel to fe. By cor. 1 , theor. 5, ca : cd : : cf : ca — fk, and by theor. 7, ca : cd : : ct : ca ; therefore ct : cf : : ca : ca — fe ; and by add. and sub.TF : rf : : fe : 2ca— fe or /e by th. 5. But by sim. tri. tf : rf : : re :fe ; therefore ft: — fe> and conseq. = /.foe. But because fe is parallel to fe, the k ^fet ; therefore ^.fet = Lfot. q. e. d. Or TUB ILLIP8X 488 Cord. As opticians find that the angle of incidence is equal to the angle of reflection, it appears from this theorem, that rays of Tight issuing from the one focus, and meeting the curve in every point, will be reflected into lines drawn from those points to the other focus. So the ray fn is reflected into fe. And this is the reuson why the points f,/, ure call- ed the foci, or burning points. theorex x. All the Parallelograms circumscribed about an Ellipse are equal to one another, and each equal to the Reclaugle of the two Axes. That is, the parallelogram pqrs the rectangle ab • ab. S Let' kg, eg y be two conjugate diameters parallel to the aides of the parallelogram, and dividing it into four less and equal parallelograms. Also, draw the ordinate* i>k, de, and ck perpendicular to pq; and let the axis c.\ produced meet the sides of the parallelogram, produced if uecessury, in t and t. Then, by theor. 7, ct : ca : : ca : cd, and - - c/ : ca : : ca : erf ; theref. by equality, err : ct : : cd : cd ; but, by sim. triangles, ct : ct : : td : cd, theref. by equality, td : cd : : cd : cd, and the rectangle td . dc = is the square cd\ Again, by theor. 7, cd : ca : : ca : ct, or, by division, cd : ca : : da : at, and by composition, cd : db : : ad : dt ; conseq. the rectangle cd . dt = cd 9 =* ad . dh*. But, by theor. 1, ca*: ca*:: (ad . db or) cd* : de?, therefore - ca : ca . : cd : dk ; or - - - ca i dk : : ca : cd ; By th. 7, - ct : ca : : ca : ca\ * Cord. Bet-nil <« cd 1 = ad . ns — ca^ — «d", therefore ca* = cd 1 ~f- cH* . In like manner, c« J = ss'-f *V. 484 oohic tactions by equality by sim. tri. theref. by equality, But, by trim. tri. theref. by equality, and the rectangle But the reet. theref. the rect. conseq. the rect. ct : ca : : ca : db, ct : ct : : de : de, ct : ca : ; ca : de. ct : ck : : ce : de; cr : ca : : ca : ce, ck • ce ~ ca . ca. ck . ce = the parallelogram cbp*, ca • c<i — the parnllelograra cbic, ab • ab ~ the parallelogram pqrb- q.e*d. THEOREM XX. The Sum of the Squares of every Pair of Conjugate Diame- ters, is equal to the same constant Quantity, namely, the Sum of the Squares of the two Axes. That is, ab* + a&* = eo 8 + eg 9 ; where eo, eg, are any pair of con- jugate diameters. For, draw the ordinates ed, ed y Then, by cor. to Theor. 10, ca 3 = cir + cd\ and - - - co 9 = de' 4- de 1 ; therefore the sum ca 8 + va 1 = cd 2 + dk* + cd? + de*. But, by right-angled as, ce 51 = cn a 4- de', and ce 2 = cd 3 + de 1 ; therefore the sum t e 2 4- cc 2 = cn 2 + de 2 + cd 1 + de 1 . consequently - ca 2 + ca 2 ce 2 + ce* ; or, by doubling, ab 2 + ab 2 = kg 5 + eg 9 . q. b. d. THEOREM XII, The difference between the semi -transverse and a line drawn from the. focus to any point in the curve, is equal to a fourth proportional to the semi -transverse, the dixtnnce from the centre to the focus, and the distance from the centre to the ordinate belonging to that point of the curve. OF trs uunii 485 That is, ac — fe =■ ci, or fe = ai ; and/is — ac ci, or /k = bi. Where ca : cf : : cd : ci the 4th proportional to ca, cf, cd. For, draw ao parallel and equal to ca the semi-conjugate ; and join co meeting the ordinate de in h. Then, by theor. 2, ca 2 : ah 2 : : ca" - cd 2 : dr 2 : and, by sim. tri. ca 9 : ao* : : ca* — cd 2 : ^o a — dh" ; consequently dk 2 »ao j -dh*=oi*— dh*. Also - fd=cf^-cd, end fd^cf 2 — 2cf . cd+cd 2 ; but by right-angled triangles, fd'+dk^fk 2 ; therefore - FK^ci^+ca 2 — 2cf . cd+cd 2 — du 1 . But by theor. 4, ca , +CF a =CA*; and, by supposition, 2cf . cd=2ca . ci ; therefore - - fb 2 =ca 2 — 2c a . ci+cd 2 — do 1 . But by supposition, ca 2 : cd' : : cf 8 or ca 2 — ag 2 : ct a . and, by sim. tri. ca 2 : cn a : therefore - - ci^cd 2 — nn a ; consequently - fe 9 «»ca 2 — 2ca . ci + cr 2 . And the root or side of this square is fe =-= ca — ci = ax. In the same manner is found /e= ca+ci=bi. q. v. d. Carol. 1. Hence ci or ca — fe is a 4th proportional to CA, CF, CD. Carol. 2. And fp. — fe =■ 2ci ; that is, the diffi rence between two lines draw from the foci, to any point in the curve, is double the 4th proportional to ca, cf, cd. : ca 2 — ao 2 : cd 2 — dh 2 ; THEOREM XIII. If a line be drawn from either focus, perpendicular to a tan- gent to any point of the curve ; the distance of their inter- sections from the centre will be equal to the semi-transverse axis. That is, if fp, fp, be perpendi- cular to the tan- gent T17?, then shall cp and cp be each equal to ca or CB. 486 come SECTIONS. For through the point of contact e draw rs, and /* meeting fp produced in o. Then the L gkp = ^ pep, being each equal to the L f&pt and the angles at p being right, and the side pk being common, the two triangles gep, fep are equal in all respects, and woe = fe, and gp = fp. Therefore, since fp = £fg, and fc = \rf, and the angle at f common, the side cp will be -= £/b or J ab, that is cp = ca *©r cb. And in the same manner cp = cx or cb. o> e. v. Card. 1. A circle described on the transverse axis, as a diameter, will pasc through the points p, p ; because all the lines ca, cp, cp 9 cb, being equal, will be radii of the circle. Cord. 2. cp is parallel to /e, and cp parallel to fb. Cord. 8. If at the intersections of any tangent, with the circumscribed circle, perpendiculars to the tangent be drawn, they will meet the transverse axis in the two foci. That is, the perpendiculars pf, pf give the foci r y f, THEOREU XIV. The equal ordi nates, or the ordi nates at equal distances from the centre, on the opposite sides and ends of aa ellipse, have their extremities connected by one right line passing through the centre, and thut line is bisected by the centre. That is, if cd = r«, or the ordinate de = gii ; then shall cr = ch, and k<;ii will be a right line. For when cd = cg, then also is de = on by cor. 2, th. 1. But the L d =r ^ «, being both right angles ; therefore the third side ce = en, and the / dce = ^.gcb, and consequently ech is a right line. Coroh 1. And, conversely, if ecu be a right line passing through the centre ; then shall it he bisected by the centre, or have ce — ch ; also de will be = oh, and cd = co. OF TU SLUMS. 487 Carol. 2. Hence also, if two tangents be drawn to the two ends b, h of any diameter eh ; they will be parallel to each > other, and will cut the axis at equal angles, and at equal distances from the centre. For, the two cd, ca being equal to the two co, cb, the third proportionals <t, cs will be equal also ; then the two sides ce, ct being equal to the two ch, cs, and the included angle ect equal to the included angle bcs, all the other corresponding parts are equal : and^ so the L t = L s, and te parallel to ns. Carol. 3. And hence the four tangents, at the four extre- mities of any two conjugate diameters, form a parallelogram circumscribing the ellipse, and the pairs of opposite sides are each equal to the corresponding parallel conjugate diameters. For, if the diameter eh be drawn parallel to the tangent ts or hs, it will be the conjugate to eh by the definition ; and the tangents to e, k will be parallel to each other, and to the diameter eh for the same reason. THEOREX XV. If two ordinates ed, ed be drawn from the extremities e, e, of two conjugate diameters, and tangents be drawn to the same extremities, and meeting the axis produced in t and r; Then shall cd be a mean proportional between cd, da, and cd a mean proportional between cd, dt. For, by theor. 7, cd : ca : and by the same, cd : ca i theref. by equality, cd : cd : But, by sim. tri. dt : cd theref. by equality, cd : cd In like manner, cd : cd : Corel. 1. Hence cd : cd : : cs : ct. : ca : ct, : <:a : cn ; : cr : ct, : ct : cb; : cd : dt. : cd : da. a. s. p. 488 come lECTiom. « Carol. 2. Hence also cd : erf : : de : de. And the rectangle cd.de = erf . rff , or A cdb = a ede. Corof. 8. Also W J = cd . dt, and cd* ■= cd . rfa, Or erf a mean proportional between cd, dt ; and cd a mean proportional between cd y da. P THEOREM XVI. TTie name figure being constructed as in the last theorem, each ordinate will divide the axis, and the semi-axis added to the external part, in the same ratio. [See the last fig.] That if», da : dt : : dc : db, and dx : rfa : : dc : rfn. For, by Theor. 7, cd : ca : : ca : ct, and by div. en : ca : : ad : at, and by com p. cd : db : : ad : dt, or, - - - da : dt : : dc : db. In like manner, o*a : rfa : : rfc : ds. a. e. D. Corol. \. Hence, and from cor. 3 to the last, it is, cd 1 = CD . cd 3 = cd DT = AD rf*R = Ad , DB = CA a — CD*. ds = CA 8 crf\ Corol. 2. Hence also, ca 2 = cd 2 + cd 2 , and ca 3 = de* + de*. Corol. 3. Further, because ca 2 : ca 1 : : ad . db or erf 2 : de*, therefore ca : ca : : erf : de, likewise ca : ca : : cd : rfe. THEOREM XVII. If from any point in the curve there be drawn an ordinate, and a perpendicular to the curve, or to the tangent at that point : then, the Dist. on the trans, between the centre and ordinate, cd, Will be to the dist. pd, ™ "> As sq. of the trans, axis To sq. of the conjugate. EL ca" : ca' That is, : : dc : dp. D P C For, by theor. % ca? ; ccr *. kt> * to • OF THE ELLIPSE. 489 But, by rt. angled As, the rect. td . dp-=pe s ; V and, by cor. 1, theor. 16, cd • dt=ad . db ; therefore - - - ca 2 : ca' : : td . dc : td . dp, or - - - - - ac 2 : ca 3 : : dc : dp. q. e. d. THEOREM. XVIII. If there be two tangents drawn, the one to the extremity**- of the transverse, and the other to the extremity of any^^ other diameter, each meeting the other's diameter pro- duced ; the two tangential triangles so formed, will be equal. For, draw the ordinate de. Then # By sim. triangles, cd : ca : : ce : cn ; but, by theor. 7, cd : ca : : ca : ct ; the re f. by equal, ca : ct : : ce : ex. The two triangles crt, can, have then the angle c common, and the sides about that angle reciprocally proportional ; those triangles are therefore equal, namely, the a cet = a can. Cord. 1. From each of the equal tri. cet, can, take the common space cape, and there remains the external A pat = apne. Corel. 2. Also from the equal triangles cet, can, take the common triangle ced, and there remains the A ted = trapez. aned. The same being supposed as in the last proposition ; then any lines kq, qg, drawn parallel to the two tangents, shall also cut off equal spaces. That is, That is, the triangle cET=the triangle can. theorem xix. Vol. I. 68 For, draw the ordinate pb. Then the three aim. triangles can, cde, cgh, are to each other as ca*, cd 1 , cg s ; th. by div. the trap, aned : trap, anhg : : ca 2 — cd* : ca j — ce*. But, by theor. 1, de 8 : oq 9 : : ca s — cd* : ca 2 — cc 1 . theref. by equ. trap, aned : trap, anhg : : de* : oq*. But, by8im. As, tri. ted : tri. kqo : : de* : gq\ theref by equality, ankp : ted : : anho : kqc. But, by cor. 2, theor. 18, the trap, aned = A ted; and therefore the trap, anho =" A kqg. In like manner the trap. anA#= AKgg. o. e. d. Carol. 1. The three spaces anhg, tehg, kqg, are aW equal. Carol. 2. From the equals anhg, kqg, take the equals anA^, Kqg, and there remains ghuo =» gqoa. Corol. 3. And from the equals #Ahg, gqao 9 take the common space £?lkg, and there remains the Alqh = At? A. Corol. 4. Again, from the equals kqo, ttshg, take the common space klhg, and there remains tclk = -A Left. Corol. 5. And when by the lines kq, gh, moving with a parallel motion, kq comes into the position ie, where ^ cr is the conjugate to T ca; then the triangle kqg becomes the triangle ntc, and the space anhg becomes the triangle anc ; and therefore the Airc = A anc = tec. Corol. 6. Also when the lines kq and hq, by moviag with a parallel motion, come into the position ce f me, OP THE ELLIPSE. tke triangle lqh becomes the triangle cent, and the space telk becomes the triangle tec ; and theref. the Ac cm = A tec = Aanc = &uic. theorem xx. Any diameter bisects all its double ordinotes, or the lines drawn parallel to the tangent at its vertex, or to its con- jugate diameter. That is, if oq be pa- rallel to the tangent te, or to ce, thenehall La = Mr- For, draw qh, qh perpendicular to the transverse. Then by cor. 3, theor. 19, the A lqh ="= vqh ; but these triangles are also equiangular ; consequently their like sides are equal, or La = 14. Cord* Any diameter divides the ellipse into two equal parts. For, the ordinates on each aide being equal to each other, aud equal in number ; all the ordinates, or the area, on one side of the diameter, is equal to all the ordinates, or the areu, on the other side of iU TBEOKEX XXI. As the square of any diameter Is to the square of its conjugate, • So is the rectangle of any two abscisses To the square of their ordinate. That is, ce 8 : ce 9 :: el • lo or ce 3 — cl 2 : lq*. For, draw the tangent te, and produce the or- dinate ul to the trans- verse at k. Also draw <w» ex perpendicular to the transverse, and meet. ingEG in n and x. Then, similar triangles r \ 403 come iBcnoKt. being as the squares of their like sides, it is, by sim. triangles, Acet : A cut : : ce* : cl 8 ; or, by division, Ackt : trap, tklk : : ck*: ce 1 — cx a . Again, by sim. tri. A : A t<m : : cc* : l<*'. But, by cor. 5thcor. 19, the Ac*m = acet, and, by cor. 4 theor. 10, the Ai<w = trap, tslk ; theref. by equality, ce : ce 1 : : en* — cl* : m*, * or - • ce 3 : ce* : : el . lg : i.q s . q. b. d. Cord. 1. The squares of the ordi nates to any diameter, are to one another as the rectangles of their respective abscisses, or as the difference of the squares of the semi- diameter and of the distance between the ordinate and centre. For they are all in the same ratio of ce 2 to ce\ Corel. 2. The above being a similar property to that be* longing to the two axes, all the other properties before laid down, for the axes, may be understood of any two conjugate diameters whatever, using only the oblique ordinates of these diameters, instead of the perpendicular ordinates of (he axes ; namely, all the properties in theorens 6, 7, 8, 14, 15, 16, 18, and )9. THEOREM XXII. If any two lines, that any where intersect each other, meet the curve each in two points ; then the rectangle of the segments of the one is to the rectangle of the segments of the other, as the square of the diam. parallel to the former to the square of the diam. parallel to the latter. That is, if cr and cr be y' parallel to any two lines ^ phq, pnq then shall ph . hq : pn For, d raw the diameter che, and the tangent te, and its parallels pk, hi, mii, meeting the conjugate of the diameter cr in the points t, k, i, m. Then, because similar triangles are as the squares of their like sides, it is, by sim. triangles cu a : up 2 : : acri : £gpk, and . - - . v:u A \ uiv 1 % \ kvaiw ^hk ; t!*eref. by division, en? *. o# — ^v? *. *. \ *x-vnu OF THE ELLIPSE. Again, by aim. tri. ce 3 : cn 9 : : acts : Ackh; and by division, cr 1 : cr* — cu 9 : : Acte : tehm. But, by cor. 5 theor. 19, the Acte = A cm, and by cor. 1 theor. 19, tf.hg kphg, or tehm = kpbm ; the re 1*. by equ. ce 9 : ce 3 — ch 9 : : cr* : gp 9 —- gh 9 or ph . HQ. In like manner ck* : ce* — ch 9 : : cr 9 : pu . uq. Theref. by equ. cr 3 : cr 3 : : ph . na : pH. Hq. q. e. d. Carol. 1. In like manner, if any other line p'n'q, parallel to cr or to pq, meet phq ; since the rectangles ph'q, p'n'q' are al*o in the same ratio of cr 9 to cr 9 ; therefore reel. phq, : pnq : : pii'q : pu'q'. Also, if another line p ho? be drawn parallel to pq or cr ; * because the rectangles, p Aq', p hq are still in the same ratio, therefore, in general, the rect. phq : puq : : p ha : p hq'. That is, the rectangles of the pans of two parallel lines, are to one another, as the rectangles of the parts of two other parallel linen, any where intersecting the former. Cord. 2. And when any of the lines only touch the curte, instead of cutting it, the rectangles of such become squares, and the general property still attends them. Corel. 3. And hence « : re : : If : /e. [484] OF THE HYPERBOLA. THEOREM I. The Squares of the Ordinate* of the Axis are to each other as the Rectangles of their Abscisses. Let avb be a plane passing through the vertex and axis of the opposite cones; agih an- other section of them perpendi- cular to the plane of the former ; ab the axis of the hyperbolic sections; and fo, hi. ordinatcs perpendicular to it. Then it will be, as fg 2 : hi 3 : : af . fb : ah . hb. For, through the ordinates fg, hi, draw the circular sections kgl, min, parallel to the base of the cone, having ki., m\, for their diameters, to which fg, hi, are ordinates, as well as to the cxis of the hyperbola. Now, by the similar triangles afl, ahn, and bfk, bhm, it is af : ah : : fl : hn, and fh : hr : : kf : mh ; hence, taking the rectangles of the corresponding terms, it is, the rect. af . fb : ah . iiu : : kf . fl : mh . hn. But, by the circle, kf . fl ■» fg 3 , and mh . hn = hi 3 ; Therefore the rect. af . fb : ah . hb : : kg 9 : hi 2 . q. e. D. THEOREM II. As the Square of the Transverse Axis Is to the Square of the Conjugate : So is the Rectangle of the Abscisses To the Square of their Ordinate. That is, ab 3 ac 3 : ac? : ; - OF TflS RTPSRBOLA. For, by theor. 1 , ca . cb : ad . db : : ca* : dr* ; But, if c be the centre, then ac . cb = ac 1 , and ca is the setni.conj. Therefore . ac 1 : ad . db : : cc* : de* ; or, by permutation, ac 9 : oc 9 : : ad . db : dk* ; or, by doubling, ab' : a& 9 : : ad . db : db 1 . <l. e. d. ab 9 Cord. Or, by div. ab : — : : ad • db or cd*— ca 1 : de*, J AB that is, ab i p 1 1 ad • db or cd* — ca* : de* ; ab 3 where p is the parameter — by the definition of it. That is, As the transverse, Is to its parameter, So is the rectangle of the abscisses, To the square of their ordinate. Otherwise, thus ; Let a continued plane, cut from the two opposite cones, the two mutually connected oppo- site hyperbolas hag, hag, whose vertices arc a, a, and bases u<», hg y parallel to each other, fall, ing in the planes of the two pa- rallel circlet* lgk, Jgk. Through c, the middle point of ah, let a plane be drawn parallel to that of lok, it will cut in the cone lvk a circular section whose di- ameter is mn ; to which circu- lar section, let d be a tangent at t. Then, by sim. tri. Acm, AFL and, by sim. tri. acn, an ac : cm AF : I*L ! ac : cn : : of : fk. /, ac • ca i cm • eft 1 1 af • Fa : lf or, ac* : c£* : : af . ra : fg*. FK, In like manner, for the opposite hyperbola ac* \cP n hf .fa :fg*. Hens ct is what.ic usually denominated the semMSonjvgsje to the opposite hyperbolas hak, hak : but it is evidently «sf in the same plane whh them. 406 OOXIC SKCTIOXS. THEOREM III. As the Square of the Conjugate Axis In to the Square of the Transverse Axis, So is the Sum of the Squares of the Semi •conjugate, and distance of the Centre from any Ordinate of the Axis, To the Square of the Ordinate. m ca 9 That i*, ca 1 : : ca 1 + cd 9 dm 9 . For, draw the ordinate ed to the transverse, ab. ca 9 CA 9 ca 9 CA 9 Then, by theor. 1, ca 9 or - - ca* B^ut - ca 3 tfieref. by compos, ca 9 In like manner, ca 9 : ca 9 : : Corol. By the last theor. ca 9 and by this theor. ca 9 : therefore - de* In like manner, de 9 ca 2 CA 9 ca 9 ca 9 I* 9 Df 9 . dE* AD . DB Or CD 1 — CA 9 , C*E 9 — CA 9 . . CA 9 . + erf 9 <*e 9 . + CD' : DC 9 . Q.. E. D. : : cd 9 — ca 9 : de 9 9 :cd 9 +ca 9 : CD 9 — CA 9 : cd 9 — ca 1 :d* 3 , : cd 9 4- CA 9 cd 9 + ca 9 THEOREM IV. The Square of the Distance of the Focus from the Centre, is equal to the Sum of the Squares of the Semi-axes. Or, the Square of the Distance between the Foci, is equal to the Sum of the Squares of the two Axes. That is, f . cf 9 = ca 9 + ca 9 , or Hrt pfi ab 9 + ab*. For, to the focus f draw the ordinate fe ; which, by the definition, will be the semi- parameter. Then, by the nature of the curve . ca 9 : ca 9 : : cf 9 — ca 9 : fe 1 ; and by the def. of the para, ca 9 : ca 9 ; : ca 9 : fe 9 ; therefore - - ca 2 = cf 9 — ca 9 ; and by addition • cf 2 = ca 9 + ca 9 ; or, by doubling, — -V afc* * 4. *• D. Or THE HYPERBOLA. 497 Corol. 1 . The two semi-axes, and the fncal-distance from the centre, are the sides of a right-angled triangle c\a ; and the distance \a is = cf the focal distance. Corol. 2. The conjugate semi-axis ca is a mean propor- tional between af, fb, or between a/*,/b, the distances of either focus from the two vertices. For ca* = cf 3 — ca 2 = (cf + ca) . (cf — ca) = af . fb, THEOREM V. The Difference of two Lines drawn from the two Foci to meet at any Point in the Curve, is equal to the Transverse Axis. For, draw ag parallel and equal to ca the semi. conjugate ; and join cg, meeting the ordinate de produced in h ; also take ci a 4th proportional toe a, cf, cd. Then, by th. 2, ca : ag : : cd 3 — ca 3 : de 1 ; and, by sin. As, ca 2 : ag 3 : : CD i — ca 3 : dh 3 — ag' ; consequently de 3 = dh 3 — ag 3 = dh 3 — ci ? . Also, fd = cf cd, and fd j = cf 3 — 2cf . cd -f • cd 3 ; and, by right. angled triangles, fe 3 = fd 2 + de 3 . therefore fe 3 = cf 3 — ca J — 2cf . cd + cd 2 + dh But, by theor. 4, cf 3 -r~ ca 3 = ca s , and, by supposition, 2c f . cd = 2c a . ci ; theref. fe 3 = ca 3 — 2ca . ci + cd 3 + dh" ; Again, by suppos. ca 2 : cd 2 : : cf* or ca 2 + ag 3 : cr ; and, by sim. tri. ca 3 : cd 3 : ; ca 3 + ag 3 : cd 3 + dh 1 ; therefore - ci 3 = cd 3 + dh 3 ■= en 3 ; consequently fe 3 = ca 3 — 2ca . cr +ci 3 . And the root or side of this square is fe = ci — ca = ai. In the same manner, it is found that/k = ci + ca = bi. Conseq. by subtract. f% — fe = bi — ai = ab. q. b. d, Corol. 1. Hence ch = ci is a 4th proportional to ca, cf, cd. Corol. 2. And /e + fe = 2ch or 2ci ; or fe, ch, fk 9 are in continued arithmetical progression, the common difference being ca the semi-transverse. Vol. I. 64 4fe8 COKIC SECTIONS. \ CoroL 8. Hence is derived the common method of deecrib- *irtg this curve mechanically by points, thus: In the transverse ab, produced, take the foci f,/, nnd anj point i. Then with the radii ai, bi, and centres f,/„ describe arcs intersecting in e, which will be a point in the curve. Id like manner, assuming other points i, as many other points will he found in the curve. Then, with a siendy hand, the curve line may be drawn through all the points of intersection r. In the same manner are constructed the other two or con- jugate hyperbolas, using the axis ab instead of ab. THEOREM VI. If from any Point i in the Axis, a Line il be drawn touching the Curve in one Point l ; and the Ordinate lm be drawn ; : and if c be the Centre or the Middle of ab : Then shall cm be to ci as the Square of am to the Square of ai. **>- That is, cm ; ci : : am 2 : Ai a . For, from the point i draw any line irh to cut the curve in two points e and 11 : from which let fall the perps. ed, hg ; and bisect no in k. Then, by theor. 1, \d . db : ac . gb : : de 3 : gh 9 , and by sim. triangles, ur* : iG a : : de 3 : gh*; theref. by equality, ad . db : ag . gb : : id 9 : ig*. But DB = CH + CD = CA + CD = CG + CD — AG = 2cK AG, and tut = cn + ct: = c\ + cc = cg + cd — ad = 2ck— ad ; thcrcf. a i) . 2c k — ad . ag : ag . 2ck — ad . ag : : id 3 : ig 3 , and. by div. »g . 2ck : ig 3 — id 2 or dg . 2ik : : ad . 2c B — ad . ag : id 2 ; or - 2c k : 2ik : : ad . 2ck — ad . ag : id 3 ; or ad . 2c k : ad . 2ik : : ad . 2c k — ad . ag : id 3 ; theref. by div. ck : ik : : ad . ag : ad . 2ik — id 8 , and, by div. ck : ci : : ad . ag : id 2 — • ad . (id + ia), or - ck : ci : : ad . ag : ai 2 . But, when the line in, by revolving about the point i, comes into the position of the tangent il, then the points e and h meet in the point l, nnd the points d, k, g, coincide with the point >i ; and then the last proportion becomes cm : ci ; : am 3 : ai 3 . E. D. .» / OF THE HTPBfiBOLA. THEOREM VII. If a Tangent and Ordinate be drawn from any Point in the Curve, meeting the Iransverse Axis ; the Semi- 1 runs vewc will he a Mean Proportional between the Distances of the said Two Intersections from the Centre. That is, ca is a mean proportional between cd and ct ; or cd, ca, it, are con- tinued proportionals. For, by th. 6, cd : ct : : ad* : at 3 , that is, - cd : ct : : (cn — ca) s : (ca — ct) 1 , or - - cd : ct : : cd' + ua 3 : ca 2 + cr 3 , and - - cd : or : : cd 3 4- ca 1 : cd 2 — ct 3 , or - - cd : dt ; : cd* 1 + ca 9 : (cd + ct) dt, & or cd* : cd . dt : : cd 8 + ca 3 : cd . dt + ct . td ; hence cd 2 : ca 3 : : cn . dt : ct . td, and co a : ca 3 ; : cd : ct, theref. (th. 78, Geom.) cd : ca : . ca : ct. q. e. d. Carol. Since ct is always a third proportional to cn, ca ; if the points d, a, remain constant, then will the point r be constant also ; and therefore all the tangents will meet in this point t, which are drawn from the point e, of every hyperbo- la described on the same axis ar, where they are cut by the common ordinate deb drawn from the point d. THEOREM VIII. If there be any Tnngent meeting four Perpendiculars to the Axis drawn from these four Points, namely, the Centre, the two Extremities of the Axis, and the Point of Contact ; those four Perpendiculars will be Proportionals. That is, ao : de : : en : bi. ac : nc, tc : ac or cd, TC ' TB, cu : HI. 499 ■ I For, by thcor. 7, tc : ac : . theref. by div. ta : ai> : : and by com p. ta : td : : mad by sim. tri. ao : de : : 600 CONIC 1X011091. Cord. Hence ta, td, tc, tb ) | and TO, TE, Til, ti \ For these are as ag, de, cii, bi, by similar triangles. ore also proportionals. THEOREM. IX. If thore be any Tangent, and two Lines drawn from the Foci to the Point of Contact ; these two Lines will make equal Angles with the Tungent. That is, 1 l^n^^FD the JL fbt = L foe. For, draw the ordinate pe, and fe parallel to fe. By, cor. 1, theor. 5, ca : cn : : cf : ca + fe, cd : : cr : ca ; cf : : ca : ca + fe ; rf : : fe : 2ca + fe or /e by th. 5. if :: fe :/k; -fe y and conseq. Le = &f$e. fe is parallel to fe, the Le = Z.fkt ; Z. FKT = £fEC. Q. K. D. and by th. 7, ca therefore - ct and by add. and sub. tf But by sini. tri. i f therefore - fv. But, because therefore the Corol. As opticians find that the angle of incidence is equal to the angle of reflection, it appears, from this proposition, that rays of light issuing from the one focus, and meeting the curve in every point, will be reflected into lines drawn from the other focus. So the ray ft: is reflected into fe. And this is the reason why the points f, /, are called foci, or burning points. THEOREM X. All the Parallelograms inscribed between the four Conjugate Hyperbolas are equal to one another, and each equal to the Rectangle of the two Axes. That is, the parallelogram pqrs the rectangle ab . ab. 4 OF THE HYPEBBOLA. Ml Let eg, eg, be two conjugate diameters parallel to the i of the parallelogram, and dividing it into four less and equal parallelograms. Also, draw the ordi nates dk, de, andec perpendicular to pq ; and let the axis produced meet the sides of the parallelograms, produced, if necessary, in T and t. Then, by theor. 7, ct : ca : : ca : cd, and - • cl : ca : : ca : cd ; theref. hy equality, ct : cl : : cd : cd ; but, by sim. triangles, ct : ct : : td : cd, theref. by equality, td : cd : : cd : cd, and the rectangle td • dc is '= the square cd?. Again, hy theor. 7, cd : ca : : ca : ct, or, by division, cd : ca : : da : at, and, by composition, cd : db : : da : dt ; conseq. the rectangle cd . dt : : cd 2 = ad • db*. But, by theor. X*' okr therefore - ca or . ca By theor. 7, . ca By equality - ct By sim. tri. - ct theref. by equality, ct But, by sim. tri. ct theref. by equality, ck and the rectangle ck , But the rect. ck theref. the rect. ca conseq. the rect. ab co 8 : : (ad . db or) aP ; ca de : ct CA CT ; ca : ck : ca : cd : ca cd : ca de : ca ; ce : ca : Ct = CA DE, id. CA. DE. DE ; de. de ; ce. ca. ce = the parallelogram cepc, ce = the parallelogram cere, ab = the paral. pqrs. q. e. d. THEOREM XI. The Difference of the Squares of every Pair of Conjugate Diameters, is equal to the same constant Quantity, namely, the Difference of the Squares of the two Axes. That is. ab" — ab 2 — kg* — eg ; where eg, eg are any conjugate diameters. • Corol. Becnu*e erf 1 = ai> . dk = cd* — ca 9 . therefore ca 9 = cd 1 — cd 2 . la Lice manner ejff = «V* — d* 9 . Mt come SECTIOXS. For, draw the ordirmtcs kd, rd. Then, by cor. to theor. 10, ca 3 = cd 3 — cd* 9 and - - - - ca 1 = de 2 — dk 3 ; theref. the difference ca 3 — ca 3 = t'i) s + DK 3 — cd*—de*. Bui, by right-angled As, CE 7 s= CD* + DK 3 , and - ce 1 = cd* + de 2 ; theref. the difference ce 3 — ce 2 — CD 3 + DK 3 — cd* —de 9 , consequently . ca 3 — Ctl 2 = CK 3 — ck 3 ; or, by doubling, ab 3 — ab 2 = EG 3 -eg 2 . a, e. d. THEOREM XII. All the Parallelograms are equal which are formed between the Asymptotes and Curve, by Lines drawn Parallel to the Asymptotes. That is the lines gk, f.k, ap, aq, being parallel to the asymptotes <;ii, <7 ; then the para I. igek = pural. cpaq. For, let a be the vortex of the curve, or extremity of the semi-transverse axis a<\ perp. to which draw al or a/, which will be equal to the semi-conjugate, by definition 19. Also, draw hkdi/i parallel to l/, Then, by theor. k 2, v\ s : al= : : co 3 — ca 3 : de 3 , and, by parallels, ia 2 : al 2 : : ( n 3 : mi - ; theref. by subtract. ca 2 : al 3 : : <\\ 3 : dii* — dk 3 or rect. ii k . v.h ; conseq. the square al 2 = the rect. in: . f7i. But, by sim. tri. i a : al : : gk ; eh, and, by the same, a a : a/ : : kk : v.h ; theref. by comp. pa . \q . al 3 . : gk . f.k : iik . v.h ; and because al 3 = iik . k/i, theref. pa . aq = <;e . ek. But the parallelograms cgk.k, cpaq, being equiangular, are as the rectangles gk . kk and pa . aq. Therefore the parallelogram gk ~ the paral. pq. That is, all the inscribed parallelograms are equal to one another. q. e. d. CoroL 1. Because the rectangle gkk or cgk is constant, therefore ok is reciprocally as cg, or cg : cp : : pa : gk. And hence the asvmnUiU'. continually approaches towards the curve, bul i\g\*:t vw^^X^ \\ . vvt w^ywaUy C K Q £ h OF THE OYPEBBOLA. as co increases ; and it is always of some magnitude, except when co is supposed to be infinitely great, for then gk is infinitely small, or nothing. So that the asymptote co may be considered us a tangent to the curve at a point infinitely distant from c. Carol. 2. If the abscisses cd, cv, cg, &c. taken on the one asymptote, be in geometrical pro- gression increasing ; then shall the ordi nates uh, ki, ok, &c. parallel to the other asymptote, be a de- creasing geometrical progression, £r having the same ratio. For, all the rectangles cdh, cei, cor, &c. being equal, the ordinates dh, ki, gk, dec, are reciprocally as the abscisses cd, ck, cg, die. which are geometricals. And the reciprocals of geome- trical are also geometricals, and in the same ratio, but de- creasing, or in converse order. THEOREM XIII. The three following Spaces between the Asymptotes and the Curve, arc equal ; namely, the Sector or Trilinear Space contained by an Arc of the Curve and two Radii, or Lines drawn from its Extremities to the Centre ; and each of the two Quadrilaterals, contained by the said Arc, and two Lines drawn from its Extremities parallel to one Asymptote, and the intercepted Part of the other Asymp- tote. That is, The sector cae = paeg = qaek, all standing on the same arc ae. For, bytheor. 12, cpaq = cgkk ; subtract the common space cgiq, there remains the paral. pi = the par. ik ; To each add the trilineal iae, then the sum is the quadr. pa kg = qakk. Again, from the quadrilateral caek take the equal triangles, caq, cek, and there remains the sector cak = qaek. Therefore cae = qaek = faeg. q. e. d. MM cosiic lEcnoifi. SCHOLIUM. In the figure to theorem 12, cor. 2, if c d = 1, and ce, ca, &c. be any numbers, the hyperbolic Bpaces hdei, iegk, &c. are analogous to the logarithms of those numbers. For, whilst the numbers cd, ce, cg, 6zc. proceed in geometries! progression, the correspondent spaces proceed in arithmetical progression ; and therefore, from the nature of logarithms are respectively proportional to the logarithms of those num. hers. If the angle c were a right angle, and cn = dh =1 ; then if ce were "= 10, the space deih would be 2*30258509, &c. ; if co were = 100, then the space dgkh would be 4-60517018 : these being the Napierean logarithms to 10 and 100 respectively. Intermediate a re are corresponding to in- termediate abscissae would be the appropriate logarithms. These are usually called Hyperbolic logarithms ; but I he term is improper : for by drawing other hyperbolic curves between hik and its asymptotes, other systems of logarithms would be obtained. Or, by changing the angle between the asymptotes, the same thing may he effected. Thus, when the angle c is a right angle, or has it:, sine = 1. the hyperbo- lic spaces indicate the Napierean logarithms; hut when the angle is 25 44' 27}", whose sine is = -43420148, A:c. the modulus to the common, or ttriggs's, logaiithms, the spaces deih, &c. measure those logarithms. I*) both eases, if spaces to the right of dh are regarded as posit ice, those Jo the left will be negative; whence it follows that the logarithms of numbers less than 1 arc negative also. The sum or difference of the semi -transverse and a line drawn from the focus to any point in the curve W equal to a fourth proportional to the semi. transverse, the distance from the centre to the focus, and the distance from the centre to the ordinate belonging to that point of the curve. THZOREM XIV. fe+ac=xi, or FK=AI ; and fit — ac=ci, or fK—m. •Where c\ : cf : : cn : ci the 4th propor. to ca, cf, cd. That is, OF TUB HYPERBOLA. fiM For, draw ao parallel and equal to ea the semi. conjugate { and join co meeting the ordinate de produced in h. Then, by theor. 2, CA a : ao* : : cd 2 — ca 2 : de 2 ; and, by aim. As, ca 2 : ao 2 : : co a — ca 8 : dh 3 — AG a ; consequently db 2s5s dh 2 — ag 2 =dh 2 — ca*. Also fd=cf^cd, and fd 3 =cf 3 — 2cf . cd+cd 2 ; but, by right-angled triangles, fd 2 +de 2 =-fr 2 ; therefore fe^cf 3 — ca 7 — 2cf • cD+cD a +DH i . But by theor. 4, cf 2 — ca'=x'A a , and, by supposition, 2cf . cd=2ca . ci ; tlieref. fe 2 =ca 2 — 2ca . ci+cd 2 +dh 2 . But, by supposition, ca 2 : cd 2 : : of 2 or ca 2 +ag 2 : ci 1 ? and, by sim. as, ca 2 : cd 2 : : ca 2 +ao 2 : cd 2 +dh 2 ; c c- therefore - ci^cdM-dh^ch 2 ; consequently - fb^ca 2 — 2ca . ci+ci 3 . And the root or aide of this square is fe=ci — ca=ai. In the same manner is found /e— ci+ca=bi. a. e. b. Coral. 1. Hence ch=ci is a 4th propor. to ca, cf, cd. Cord. 2. And /e+fe=2ch or 2ci ; or fk, ch, /e are in continued arithmetical progression, the commr* difference being ca the semi- transverse. \ Corol. 3. From the demonstration it appears, that de 2 = dh 2 — ao 2 = dm 2 — ca\ Consequently dh is every where greater than de ; and so the asymptote cgh never meets the curve, though they be ever so far produced : but dh and de approach nearer and nearer to a ratio of equality as tlfey recede farther from the vertex, tity at an infinite distance they become equal, and the asymptote is a tangent to the curve at an infinite distance from the vertex. theorem xv. If a line be drawn from either focus, perpendicular to a tan- gent to any point of the curve; the distance of their in- tersection from the centre will be equal to the semi-trans- verse axis. Vol. I. 65 • That is, if ff, fp be per- pendicular to the tangent Trp, then ahall cp ami cp be each equal to ca or ca. For, through the point of contact ■ draw n, aad /a, nenV ing fp produced in e. Then, the £obp« £*bp, being each equal to the l/ap, and the anglee at f being n^at, and the aide pb being common, the two trangtes, obf, i» emeaaal in all respects, and ae «■ » ra, and* or ■* fp. Tho tufa e, aince re^po, and re=^rf p and the eagle, at f cxNaaaocvthe aide cr will be =4/e or 4ab, that ia cp«*ca sccb. And in the same manner op«CA or ca. e> sfeBt Corn/. 1. A circle deecribed on the transverse axis, as' a' diameter, will pass through the points r, p; because all the lines ca, cp, en, being equal, will be radii of the circle. Carol. 2. cr is parallel to /a, and cp parallel to Fa. Coral. 3. If at the intersections of any tangent with the circumscribed circle perpendiculars to the tangent be drawn, -they will meet the transverse axis in the two loci. That is, the perpendiculars ff, pf give the foci f, /. THSOBBM XVI. The equal ordinates, or the ordinatea at equal from the centre, on the opposite sides and ends of aa hyperbola, have their extremities connected by one right line passing through the centre, and that line is by the centre. That is, if en £3 co, or the ordinate dk = oh ; then shall cb = cr, and ech will be a right line. For, when cd = co, then also is ra = on by cor. 9 theor. 1. But the L d = L o, being both right anglee ; therefore the third aide ca =■= cw, and the L MB « L oca, and conaequetrtty ttt^ottafe* Carol. 1. And, conversely, if ech be a right line pairing through the centre ; then shall it be bisected by the centre; or have cr « ch, also de will be = on, and-cn = c«. Carol, 2. Hence also, if two tangents be drawn to the two ends e, H of any diumeier eh ; they will be parallel to each other, and will cut the axis at equal tingles, and at equal dis- tances from the centre. For, the two cd, ca being equal to the two co, cr, the third proportionals or, <:» will be equal also ; then the two sides ce, ct being equal to the two 1.11, < a, and the included angle ect equal to the included angle Res, all the other corresponding parts are equal : and so the L t = JL s, and te parallel to ns. Carol. 3. And hence the four tangents, at the lour ex- tremities of any two conjugate diameters, form a parallelogram inscribed between the hyperbolas, and the pairs of opposite sides are each equal to the corresponding parallel conjugate diameters. — For, if the diameter eh be drawn parallel to the tangent te or hs, it will be the conjugate to eh by the defini- tion ; and the tangents to e, A will be parallel to each other, add to the diameter eh, for the same reason. theobem xvir. If two ordinates ed v ed be drawn from the extremities f, e, of two conjugate diameters, and tangents be drawn to the same extremities, and meeting the axb produced in t and r ; Then shall cd be a mean proportional between cd, da, and cd a mean proportional between cd, dt. For, by theor. 7, cd : ca : : ca : -ct, , and by the same, cd 2 ca : ; ca : cr ; thtref. by equality, cd : cd : 2 ca : ct. But by sim. tri. # nr : cd : : ct : cr ; theref. by equality, (So : cd : : cd : dt. In like manner, cd 2 cd : : cd : d*. a. e. d. Coral. 1. Hence cd : cd : : cr : cr. % Carol. 2. Hence also cn : cd : vdt : de. Aad the rect. cd. djs = cd.de, or A cde ~ &c<fe. JjffiM-S. Akocf~oD.»r,telcaP«e«.dta. 7 Or cd a mean proportional between ca» vr % and cd a mean proportional between cd, dsu The mm igvro befog constructed aetata* lael awea»oeitj»B> pmch ordinate will divide tbe axis, and the ■■— a m added to the external partem the tame ratio. [See the last fig.] That is, da : dt : : dc ; m, and dx :da : :4c: da. For, by theor. 7, cd : ca 8 : ca : cr, and by dir. - cd : oa : : ad t at£ and by eomp. ■ cd : db t : ad : dt, or • • da : or : : dc : db. In like manner, 4a : da : : dc : dm. q. b» d. Carol. 1. Hence, and from cor. 9 to the last prop* it is Cd 2 — CD • DT = AD • DB = CD 5 CA*, and cd* = cd . da = Ad . da = ca" = cd* . Carol. 2. Hence also ca s =cd s — cd', and ca*=de*— db*. Corel. 3. Farther, because ca 9 : ca 2 : : ad . dbotcJ' : db*« therefore ca : ca : : cd : db. likewise ca : ca : : cd : de. THEOBEM XIX. If from any point in the curve there be drawn an ordinate, and a perpendicular to the curve, or to the tangent at that point : then the Dirt, on the trans, between the centre and ordinate, cd, Will be to the dist. pd, As square of trans, axis To square of the conjugate. That is, ca s : co 2 : : dc : dp. For, by theor. 2, ca* : cd* : : ad . db : db 1 , But, by it. angled as, the reel, td . dp = de 1 , and, by cor. 1 theor. 16, cd . dt = ad . db; therefore • - ca* : co 9 : : td . dc : td • dp, or % • • • CA? \ OQ> % \ TO 4»S*B> OF TBS BYMMOLA* an THBOftJBK XX. If there be two tangents drawn, the one to the extremity of the transverse, and the other to the extremity of any other diameter, each meeting the other's diameter produced: the two tangential triangles so formed, will be equal. For, draw the ordinate ox. Then By aim. triangles, cd : ca : : cb : cir ; but, by theor. 7, cd : ca : : ca : ct ; theref. by equal, ca : cb : : cb : ext. The two triangles cet, can have then the angle c com- mon, and the sides about that angle reciprocally proportional ; those triangles are therefore equal, viz. the Acet = A can. q. B. D. CoroL 1. Take each of the equal triangles cbt, can, from the common space cape, and there remains the external A pat = A pub. Cord. 8. Also take the equal triangles cbt, can, from the common triangle cbd, and there remains the A ted = trapez. anbd. The same being supposed as in the last proposition ; then any lines kq, go, drawn parallel to the two tangents, shall also cut off equal spaces. That is, the triangle cbt = the triangle can* THEOREM XXI. That is, * the akqg = trapez. anho. and AKf£ = trapez* ana£. F$ draw the ordinate Djk Then The three mm. trinngl^VAW, cdb, oob 9 are to each other as ca% gd% oe* ; th. by di? . the trap, anbd : trap, akho : : cd*~ca* : co*— CA*. Bor, by thear. 1, ob* t «Q f : : cb*— ca*,: og&— theref. by equ. trap. axbb : trap, akhq : : db" : oq\ Bat, by aim. As, tri. tkd : tri* kqg i : bb* : ao? ; theref. by equal* aned tbd : : a»bg : boo. Bot| by cor. 3 theor. 20, the trap, aitkd .= A tbd ; jind therefore the trap, anho = A kqo. Ill like manner the trap. A*hg — Aiff. q. b. b. Carat. 1. The threo apacea abhg, tkbg, kqo are aD equal. ConL 3. From the equals akbg, kqg, * take the equ*la anA^, Kqg, ' and there remaina gkua =» gfQG. Carol. 8. And from the equals £Ahg, £?qg, take the common space ^i.ho, and there remains the Alqh = A 14*. Cord. 4. Again, from the equals kqg, tbhg, « take the common space klho, and there remaina tklk = alqh. • CoroZ. 5. And when by the lines kq, gh, moving with a parallel motion, kq comes into the position ir, where ca is the conjugato to Ca ; then the triangle kqg becomes the triangle ikc, and the space ax kg becomes the triangle axc ; and therefore the Ainc - aanc~ Atcc. Corol. 6. Also when the lines kq nnd no, by moving with a parallel motion, come into the position re, i*e f the triangle lqh becomes the triangle cm, i nd th ' spac/; tklk l>ec< mes thi triangle tkc ; and theref. the acix a atkc = A axc » Aibc. THKOKEX XXn. Any diameter bisects nil its double ordinate*, or the lines drawn parallel to the tangent at its vertex, or to its 'coojo- gate diameter. or TBI HYtSUOLA. 611 That is, if oq be paral- lel to the tangent te, or to c* , then shall lq = Lq* For, draw qii. qh perpendicular to the transverse. Then by cor. 3 theor. 21, the a lqh =?= &Lqh ; but these triangles are also equimigiilar; conseq. their like sides are oqual, or ui = Lq. Corel.* 1. Any diameter divides the hyperbola into two equaj^parts. For, the ordinates on each side being equal to each other, and equal in number; all the ordinates, or the urea, oa one side of the diameter, is equal to all the ordinates, or the area, on the other side of it. Cord. 2. In like manner, if the ordinate he produced to the conjugate hyperbolas at q', q\ it may be proved that lq' = tq . Or if the tangent te be produced, then kv =ew. Also the diameter gckh bisects all lines 'drawn parallel to tk or oq, and limited either by one hyperbola, or by its two con* jugate hyperbolas. THBOREX XXin. As the square of any diameter Is to the square of its conjugate, So is the rectangle of any two abscisses To the square of their ordinate. That is, ce* : cc* : : el . lg or cx a — cb* : lq*. For, draw the tangent te, and produce the ordi- nate ql to the transverse at k. Also draw qh, ex perpendicular to the trans* . verse, and meeting eg in h and m. Then, similar triangles being as the squares of their like sides, it is, by sim. triangles, £cet : A cut : : ce* : cl* ; i aftay division, AflfT : tap. tub : : on* : €»• — .«■'. Again, by sim. tri. |fli : alqb : : ce a ;LQ 9 . ^ Bat, by oor. 5 theor. 81, the Aoex «* a err, and, by cor. 4 theor. 31, the Alqji trap, raw ; theref. by equality, ©a 8 : or* : : cl 8 en* : aa*, or • • • os* : oe* : : bl • lo : 14 1 . . o» a* ft. Gorrf. 1. The squares of the ordinatee to any diaiajntar, are to one another aa the rectangles of their respective ab» actssee, or aa the difference of the squares of the eeau- dia- meter and of the distance between the ordinate and centre. For they ate all in the same ratio of ca f to ce". Carol. 8. The above being a similar property lo that Be- longing to the two axes, all the other properties before laid down, for the axes, may be understood of any two ronisyfe diameters whatever! using only the oblique onlinatee af these diameters instead of the perpendicular ordinatee of the exes ; namely, all the properties in theorems 6, 7, 8, 16, 17, SO, SI. Carol. 8. Likewise, when the ordinates are continued to the conjugate hyperbolas at a', g\ the same properties Ml obtain, substituting only the sum for the difference of the squares of ce and cl, ../ That is, cB f : ce* : : cl 1 + cb* : W a . 1 And so lq* : LQ* : : cl*— cb* : cl 1 -r Carol. 4. When/ by the motion of La' parallel to haal( that line coincides with ev, the last corollary becomes cb 1 .: ce* : : 2c K* 2 Bv* f or ce* : bv* : : 1 : 2, or ce : bv 2 1 or as the side of a square to its diagonal. That is, in all conjugate hyperbolas, and all their dia- meters, any diameter is to its parallel tangent, in the constant ratio of the aide of a square to ita diagonal. THBOBEM XXIV. If any two lines, that any where intersect each other, mast the curve each in two points ; then The rectangle of the segments of the one ^ Is to the rectangle of the segments of the other, As the square of the diam. parallel to the former To the square of the diam. parallel to the latter. Or THX HYPRRB3LA. 51S That is, if cr and cr be parallel to nny two lines phq, pnq ; then shall cr 8 : cr* : : fh • hq : pu • H£. For, draw the diameter ens, and the tangent tr, and its parallels pk, ri, hii, meeting the conjugate of the diameter cr in the points t, k, i, m. Then, because similar triangles are as the squares of their like side*, it is, by sim. triangles, cr 3 : of* : : Acai : Agpk, and - - cr* : oh 9 : : Acri : A cum ; tlicref. by division, cr* : op 3 — on' : : cki : kpuh. Again, by sim. tri. ck* : cii a : : Actk : A cam ; and by division, cr 1 : c» 3 — ck j : : Acre : teiix. But, by cor. 5 theor. 21, the Acte = Acir, an J by cor. 1 theor. 21, tkiio = kph j, or tehm = kphm ; tlieref. by equ. C* 1 : ch'-ck* : : rR J : gp 1 — oh 1 or ph . hq. In like m inner ce ! : ch j -ck 1 : : cr 1 : pa . Hq. Tberef. byeq'i. cu J : cr 1 : : ph . hh : pH . Hfl. q. r. d. Coral. 1. In like minner, if any other line p'fTf', parallel to cr or to pq % meet phq ; since th > rectangles ph q. p %iq' are also in the same ratio of cr j to cr J ; therefore the rect. phq : pnq : : pii'q : pn'q. Also, if another line p Aq' be drawn parallel to pq or cr ; because the rectangles p Aq , phq are still in the same ratio, therefore, in general, the rectangle phq : pnq : : p'Aq' : phq'. That id, the rectangles of the p irts of two parallel lines, are to one another, as the rectangles of the parts of two other parallel lines, any where intersecting the former. Cord. 2. And when any of the lines only touch the curve, instead of cutting it, the rectangles of audi become squares, and the general property still attends them. Vol. L 66 914 Carol. 3. And hence tb : re : : Is : ft. THEOREM XXV. If a line be drawn through any point of the curves, parallel to either of the axes, and terminated at the asymptotes ; the rectangle of its segments, measured from that point, will 1>e equal to the square of the serai-axis to which it is parallel. That is, the rect. iiek or n*K = ca\ and reel, hsk or hek ca 8 . e For, draw al parallel to ca, and «x to ca. Then hy the parallels. ca 8 : ca'or al* : : cd 8 : dh* ; and, by tlicor. 2, ca 8 : c«* : : cd"— ca* : v*. 9 ; theref. by subtr. ca 2 : ca* : : c%" : dh 8 — de* or HEX* But the antecedents ca 3 , ca 8 are equal, theref, the consequents ca 2 , hex must also be equal* In like manner it is again, by the parallels, ca 3 : ca 2 or al 3 : : cd j : dh> ; and by theor. 3, ca 3 : ca 3 : : cd 3 + ca s : ne* ; there!, by subtr. ca 3 : ca 3 : : ca 3 : ne 3 — dh 3 or hjk. But the antecedents ca 3 , ca 3 are the same, theref the conseq. ca 3 , hck must be equal. In like manner, by changing the axes, is hsk or hek = ca** Corol. 1. Because the rect. hek = the rect. h«k. therefore eh : en : : en : ek. Aud cxma&^iftaxYy \&*ta*Y*J greater than He. OF THE ltVF*XBOLA. MS CSdro/. ft. the rectangle Ark (he rect. hb*, for, by aim. tri. feA : 6k : : e& : EX. 8CHOLIUM. ft is evident that thii proposition is general for any line oblique to the axis aha, namely, that the rectangle (if the seg n mta of any line, cut by ttoe curve, and t originated by the asymptotes, is e.p*l to tfhe squ ire of the semi-diamHter to which the line is parallel. Since the demonstration is drawn from properties that are common to art diameters. THEOREM zxvi. All the rectangles are eqiml which are made of the seg. mints of any parallel lines cut by the curve, and limited by the asymptotes. For, each of the rectangles hex or hsk is equal to the squire of the parallel semi-diamater c? ; and each of the rect- angles huh or hek is oqu il to the sqi ire of the p.inllel semi* diameter oi. and therefore the rectangles of the segments of all parallel 4ines are equal to one another. q. k. d. C)rol. 1. The rectangle hrk being constantly the same, whether the point R is taken on the one side or the other of the point of contact i of the tangent parallel to iik, it follows that the parts uk, kk, of any such line hk, are equal. And because the rectangle HtK is constant, whether the point e is taken in the one or the other of the opposite hy- perbolas, it follows, that the parte He, k«, are also equal. Ctr&L 2. And when hx crimes into the position of the tangent mi, the last corollary becotftds il = id, arid lit — in, and lm = dn. Hence also the diameter cm bisect* all the parallels to dl which are terminate J by the asymprto, a&outj yav — Carol. 8. From the proposition, and the bet corollary, it follows that the constant rectangle hkk or khk hi *= il*. And the equal constant rect. ueK or me = mut or ix 9 — uP. Coral. 4. And hence il = the parallel setni-diameter cs- For, the rect.KHE = il*, and the equal rect. ees = in" — il*, theref. il* =«" — iL* f or n> = 2iL a ; but, by cor. 4 tbeor. 23, is* ■= 2cs> 9 and therefore • il = c*. And an the asymptotes pass through the opposite angles of all the inscribed parallelograms. THEOBB3I XXVn. The rectangle of any two lines drawn from any point ia the curve, parallel to two given lines, and limned by the asymptotes, is a constant quuntity. That i«, if ap, eg, pi be parallels, as also AQi kk, dm p;irallelf, then shall the rect. pap. = rect. gek = red. ira. For, produce ke, md to the other asymptote at u», l. Then, by the parallels, me : <;e : : ld : id ; but - - - ek : ek : : dm : dm ; theref. the fectangle hkk : gek : :ldm : idm. But, by the last thwr. the rect. hkk = ldm ; and therefore the rect. gek = idm = paq. q. a. a. THEOREM XXVIII. Every inscribed triangle, formed by any tangent and the two inter* epted parts of the asymptotes is equal lo a constant quantity ; namely, double the inscribed paral- lelogram. Thai \s, lYte \x\*Ng& ct% — % or «m nraaou. •17 For, since the tangent ts is bisected by the point of contact e, (th. 26, cor. 2), nnd kk is parallel to tc, and ge to ck ; therefore ck, hs, ge, are all equal, as are also co, gt, kk. ^ — , Consequently the triangle urB C JC g = the triangle kes, and each equal to half the constant in- scribed parallelogram gk. And therefore the whole triangle CT8, which is composed of the two smaller triangles and the parallelogram, is equal to double the constant inscribed paral. lelogram gk. q. e. d. TnEOREJf XXIX. If from tho point of contact of any tangent, and the two in- tersections of the curve with a line parallel to the tangent, three parallel linen be drawn m any direction, and ter- minated by either asymptote ; those three lines shall be in continued proportion. That is* if hkm and the tangent 1 1* be parallel, then are the parallels dh, ei, ok in continued propor- tion. C D E L For, by the parallels, ei : il : : dh : nx ; and, by the same, ei : il : : gk : m ; thoref. by compos, ei 9 : il 9 : : dh . gk : hxx ; but, by theor. 26, the rect. hxr = il* ; and theref. the rect. dh . ok s ei 1 , or - • dh : ei : : ei : gk. q. e. d. theorem xxx. Dr.iw the semi-diameters ch, cix, ck ; Tneu shall the sector chi ^ the sector cik. For, because hk and all its parallels are bisected by cis, therefore the triangle c.\h = tri. cxk % ■If and the segment wm » aeg, nm f consequently the sector ao = sec. cik. Carol. If the geometrical proportional* hh, mm 9 w be parallel to the other asymptote, the spaces drib, biko wiB be equal ; for they are equal to the equal sectors chi, cik. So that by taking any geometrical proportionuU cd, cm, go* 4tc. and drawing dm, bi, «k, &c. parallel to the other asymptote, as also the radii uh. ci, ck ; then the sectors chi, cut, dec or the space* dhik, eiku, dec. 1 will be all equal among themselves. Or the sectors chi, chx, &c. or the spaces dhir, dhko, dtc. will be in nrithmeticaj progression. And therefore these sectors, or space*, wilt be analogous to the logarithms of the lines or bases cd, ck, cu, dsc* } namely, cm or dbib the log. of the ratio of co to ck, or of cs to cg, dec. ; or of ei to dii, or of gk to si, Ac ; and chk or phk«; the log. of the ratio of cd to co, die. or of ok to dh, Sic. OF THE PARABOLA. THEOREM I. The Abscisses are proportional to the Squares of their Ordinates. Let avm be a section through the axis of the cone, and agih a parabolic section by n plane per. pendicular to the former, and parallel to the side vm of the cone ; also let afh be the com. mon intersection of the two planes, or the axis of the para- bola, and fg, hi ordinates per- pendicular to it. Then it will be, as ap : ah : : fg* : Hi f . For, through the ordinates fg, hi, draw the circular sec- tions, kgl, min, parallel to the base of the cone, having kl, MR for their diamctera, \o yg^ hi ace ordinates, as well as to the axis of \ta ^itoVhA*. OF THJB PAJUBALA. 619 Then, by similar triangles, af : ah : : fl : hn ; but, because of the parallels, kf = mh ; therefore - - - af : ah : : kf . fl : mh . hn. But, by the circle, kf . fl = FG 2 /and mh . hn — hi 1 ; Therefore - - - af : ah : : fg 9 : hi 9 . o,. E. d. PqS hi' Carol. Hence the third proportional — or — is a con* r r AF AH stant quantity, and is equal to the parameter of the axis, by defin. 16. Or af : fg : : fo : f the parameter. Or tho rectangle p . af = fg 9 . THEOREM II. As the Parameter of the Axis : Is to the Sum of any Two Ordinates : : So is the Difference of those Ordinates : To the Difference of their Abscisses. That is, f : gh + db : : gii — de : dg, Or, p : ki : : ih : ie. For, by cor. theor. 1, p . ag gh 9 , and - • - p . ad -= de 9 ; theref. by subtraction, p . dg = gh 9 — db 9 . Or, - - • p . dg = ki . ih, therefore - - p : Kt : : ih : dg or ei. q. e. d. Cord. Hence, because p . ei = ki . ih* and, by cor. theor. 1, p . ag — gh 9 , therefore - - ag : ei : : gh 9 : ki • ih. So that any diameter ei is as the rectangle of the segments ki, ih of the double ordinate kh. theorem hi. The Distance from the Vertex to the Focus is equal to J of the Parameter, or to Half the Ordinate at the Focus. That is, af - Jfe = {r, where f is the focus. 590 O0XIC tKCTIOXfl* For, the genera! property is af : fb : : rmi f. But, by definition 17, - fe * \r ; therefore also - at = Jfe = Jr. q. b* d» TBKORBM IV. A Line drawn from the Focus to any Point in the Curve, is equal to the Sum of the Focal Distance and the Absciss of the Ordinate to that Point. . 6 That is, FE = FA + AD = GD, taking ao ~ af. For, since fd = ad ^ af, theref. by squaring, But, by cor. theor. 1, theref. by addition, But, by right-ang. tri. therefore and the root or side is or - FD' •=» AF 8 - 2AF . Ad + AD*, DE a = P . AD - 4AF . AD ; FD 5 + DK 3 = AF 1 + 2AP . AD + AD*. FD fl + DE a = FE* J FE 2 = AF a + 2\F . AD + AD*, FE = AF + Al>, fk =t gd, by taking ag = af. u. E. D. Carol 1. If, through the point g, the HHH G» HHH line on be drawn perpendicular to the axis, it in called the directrix of the parabola.* The property of which, from this theorem, it appears, in this : That drawing any lines he parallel to the axis, he is always equal to fe the distance of the forus from the point e. Corol. 2. Hence also the curve is easily described by peinls. Namely, in the axis produced take ag = af the focal dis- tance, and draw a number of lines ee perpendicular to the axis ad; then with the distances gd, gd, gd, dec. as. radii, and the centre f, draw arcs crossing the parallel ordi nates in e, e, f., ©zc. Then draw the curve through all the points E, s, K. l 5 — ! * Rach of the other conic sections has n directrix ; hut the conside- ration of it dotttnoA nccwT \\\ v\w tftttfatat«<tm^taY?«dof tavetUgstinf the general properties ©A \\\« cwv«*. Of TBI PAJUIOLA. «1 THEOREM V. If a Tangent be drawn to any Point of the Parabola, meet- ing the Axis produced; and if an Ordinate to the, Axis be drawn from the point of Contact; then the Absciss of that Ordinate will he equal to the external Part of the Axis, measured from the Vertex. That is, if tc touch the curve at the point c, then is at = ah. Let cc, an indefinitely small portion of a parabolic curve, be produced to meet the prolongation of the axis in t ; and let cm be drawn parallel to ex, and cs parallel to ag the axis. Let, also, p = parameter of the parabola. Then, by aim. tri. cs : sc : : cm : ma -f- at = mt, MT • CS .\ cs = . CM Also, th. 1. cor. p . Am = mc* = ms* + 2ms . sc + sc*, = mc* -f 2mc . sc + sc 1 , and p . am = mc'. Consequently, omitting sc 1 as indefinitely small, and sub* trading the latter equa. from the former, we have p . (aju — am) =" p . cs = 2cs . mc : or, substituting for cs its value above, mt . cs = 2cs . mc ; or p . MT = 2mc* : Consequently, mt • CM :2p . AM (th. 1.) = 2am, and ma = at. Q. E. D. THEOREM VI. If a Tangent to the Curve meet the Axis produced ; then the Line drawn from the Focus to the Point of Contact, will be equal to the Distance of the Focus from the Inter- section of the Tangent and Axis. Vol. I. 07 east coanc sBcnsjs** That i*, tv ss. rr. K <* For, draw the ordinate vc to the point of contact c Then, hy theor. 5, .\r = ad ; therefore • ft ~ af + ad. Kut, by theor. 4. fc = af + ad ; thcref. by equality, fc — ft. Q- b. d. Corol. 1. If ro be drawn perpendicular to th* curve, or to the tangent, at c ; then Khali kg = fc = FT. For, draw rii perpendicular t ) tc, which will also bisect tc, because ft = fc ; and therefore, hy the nature of the pantile!*, fii ulso bisects tg in f. And consequently fg = ft — fc. So tli it f is the centre of a cire'e passing through t, c, g. Corol. 2. The subnormal dg is a constant quantity, and equal to half the parameter, or to 2af, double the focal distance. For, since ice; is a right angle, therefore 1 1> or <Jad : dc : : dc : dg ; but by the def ad : dc : : dc : parameter ; therefore do = half the parameter = 2 aw Corel. 3. The tangent at the vertex an, is a mean propor- tional between AFand ad. For, because fiit is a right angle, therefore - ah is a mean between af, at, or I etween - af, ad, because ad = at. Likewise, - fii is a mean between fa, ft, or between fa, tc. Corol. 4. The tangent tc makes equal angles with fc and the axis ft ; as well as with fc and ci. For, because ft = fc, Therefore the L fct = L ftc. Also, the angle c.cf -= the angle gck, drawing ick parallel to the axis ag. Corol. 5. And because the angle of incidence gck is =* the angle of reflection gcf ; therefore a ray of light falling on the curve in the direction kc, will be reflected to the focus f. That is, all rays parallel to the axis, are reflected totbs focus, or burning uuuvx* Or THE PAS ABOLA. 62* THEOREM VII. If there be any Tangent, and a Do'ihle Ordinate c'riwn from the Point of Contact, and also any Line parallel to the Axis, limited by the Tangent and Double Ordinate : Then ahull the Curve divide that Line in the sum; Ratio as the Line divides the Double Ordinate. That is. ie : ek : ; ck : kl. For, by sim. triangles, cx : ki : : cd : dt or 2da ; but, by the def. the paratn. p : cl : : < o : 2da ; therefore, by equality, p : ck : : «x : ki. • But, by theor. 2, v : ck ::kl:rr; therefore, by equality, cl : kl : : ki . kb ; and, by division, • ck : kl : ; ie : ek. q. e. d. THEOREM VIII. The same being supposed as in theor. 7 ; then shall the External Part of the Line between the Curve and Tun- gen*, be proportioual to the Square of the intercepted Part of the Tangent, or to the Square of the intercepted Part of the Double Ordinate. That is, ie is as ci* or as ck', and IK, TA, ON, PL, C2C. are as ci 3 , ct', co*,'cf 3 , die. or as ck 3 , cd 9 , cm 3 , cl*, &c. For, by theor. 7, ie : ek : : ck : kl, or, by equality, ie : kk : : ck 2 : ck . kl. But, by cor. th. 2, ek isai the rect. ck . kl, therefore - ik id as ck 3 , or as ci 3 . Q. e. d. Cord. As this property is common to every position of the tangent, if the lines ie, ta, o\, due. be appended on the points i, r, o, dtc. and moveable about ihetn, and of such lengths as that their extremities k, a, n, &c. be in the curve of a pa rub da in some one position of the tangent ; Ineu making the tangent revolve about the point c, u ui» v «u.i* 694 CONIC SBCTKKCS* that the extremities e, a, h, dee. will always fern the cum of some parabola, in every position of the tangent. THXORXH u. , The Abscisses of any Diameter, are as the Squares of their Ordinates. That is, cq, cr, csg&c. are as qr 1 , ra*, sn*, &c. Or cq, : cr : : qjb 1 : ra s , &c. For, draw the tangent ct, and the externals, si, at, it o, die. parallel to the axis, or to the diameter, cs. Then, because the ordinates, qe, ka, sn, 6lc. are parallel to the tangent ct, by the definition of them, therefore all the figures iq, tk, os, dtc. are parallelograms, whose op- posite sides arc equal ; namely, - • ^k, ta, on, dec. are equal to - cq, cr, cs, dtc. Therefore, by theor. 8, cq, cr, cs, &c. are as - - \ ci 3 , ct 2 , co a , die. or as their equals - qk 2 , ha*, sn% &c. q. k. d* Cord, Here, like as in theor. 2, life difference of the ah* scisses is as the difference of the squares of their ordinates, or as the rectangles under the sum and difference of the ordinates, the rectangle of the sum and difference of the ord mutes being equal to the rectangle under the difference of the abscisses and the parameter of that diameter, or a# third proportional to any absciss and its ordinate. THEOREM X. If a Line be drawn parallel to any Tangent, and cut the Curve in two Points ; then, if two Ordinates be drawn to the Intersections, and a third to the Point of Contact, those three Ordinates will be in Arithmetical Progression, or the Sum of the Extremes will be equal to Double the Mean. Or Till FABABA&A. 888 That is, BO + HI =• 2CD. For, draw bk parallel to the axis, and produce hi to &• Then, by aim. triangles, kk : hk : : tb or 2ap : cd ; but, bu theor. 2, - bk : hk : : kl : r the param. thereu by equality, 2ad : kl : : cd : p. But, by the defin. 2ad f 2cd : : cd : t» ; theref. the 2d terms are equal, kl = 2cd, that is, . . bo + hi = 2cd. q. b. d. Carol. When the point b is on the other side of ai ; then hi — ob = 2cd. THEOREM XI. Any diameter bisects all its Double Ordinates, or lines parallel to the Tangent at its Vertex. • That is, T H X jj X For, to the axis ai draw the ordinates eg, cd, hi, and M5 parallel to them, which is equal to cd. ♦ Then, by theor. 10, 2mn or 2cd = eg + hi, therefore m is the middle of eh. * And, for the same reason, all its parallels are bisected. Q. b. D. Schol. Hence, ns the abscisses of any diameter and their ordinates huve the same relations as those of the axis, namely, that the ordinates nre bisected by the diameter, and their squares proportional to the abscisses ; ho all the other pro- perties of the axis and ||s ordinates and abscisses, before da* monst rated, will likewise hold good for any diameter and its ordinates and abscisses. And also those of the parameters^ cone ascnoss. understanding the parameter of any diameter, as a third pio- portional to uny absciss and its ordinate. Some of the most materia] of which are demonstrated in the tour following theorems. TUEORXTl III. The Parameter of any Diameter is equal to four Times the Line drawn from the » ocus to the Vertex of that Diane* ter. M N For. draw the ordinate m i parallel to the tangent ct : also cd, perpendicular to the axis a.n, and fh perpendicular to the tangent ct. Then the abscisses ad, cm or at, being equal, by theor. 5, the parameters will be as the. squares of the ordinate;? co, xa or ct, by the definition ; that is, . - !•:/).. cd s . ct*, But by sim. tri. - fii : ft : : co : ct ; therofore - . v : p : : fii* . ft*. But, by cor. 3, th. 6, fii 3 =- fa . ft ; therefore - • i» : p : : fa . ft : ft* ; or, by equality, - v : p : : i a : ft or fc Bui. by theor. 3, p = 4 fa, and therefore - p ~ 4ft or 4fc. q. e. d. Cord. Hence the parameter p of the diameter cm is equal to 4fa + 4ad, or to p + 4ad, that is, the parameter of the axis added to 4a o. THEOREM XIII. If an Ordinate to any Di:i meter pass through the Focun, it will be equal to Half Us Parameter ; and its Absciss equal to One Fourth ot vYv& torm* Vaxvuc&vwu OF TUX PARABOLA. an That is, cm = lp, and me = \p. For, join fc, and draw the tangent ct By (he parallels, cm = ft ; and, hy theor. 6, fc = ft; also, by theor. 12, fc = }p ; therefore - - cm = Jp. Again, by the defin. cm or jp MF. MR £. D. and consequently me = jp = 2cm. Corel. 1. Hence, of any diameter, the double ordinate which passes through the focus, is equal to the parameter, or to quadruple its absciss. C rol. 2. flence, and from cor 1, to theor. 4, and theor. 6 and 12, it appears, that if the directrix on be drawn, and any lines he, he, paral- lel to the axis ; then every parallel he will he equal to ef, or \ of the pa- rameter of the diameter to the point e. THEOBKM XIV. If there be a Tangent, and any Line drawn from the Point of Contact and meeting the Curve in some other Point, as also another Line parallel to the Axis, and limited by the First Line and the Tangent : then shall the Curve divide this Second Line in the same Ratio as the Second Line divides the First Lino. That is, ie : ek : : ck : KL. For, draw lp parallel to ik, or to the axis. Then by theor. 8, ne : fl : : ci* : cf% or, by sini. tri. - ib : fl : : ck ( : cl\ Also, by sim. tri* ik : fl : : ck : cl, or - - - ik ; fl : : ck* : ck • cl ; 9 638 como octioni. * therefore by equality, ir : ik : : ck . cl : cl" ; or - - ie : ik : : ck : cl ; and, by division, ie : bk : : ck : kl. a. r. d % Cord. When ck = kl, then ie = bk = Jik. TneoBBX xv. If from any Point of the Curve there be drawn a Tangent, and also Two Right Line* to cut the Curve ; and Dia- meters be drawn through the Points of Intersection b and i. t meeting those Two Right Lines in two other Prints o and k : then will the Line kg joining these last Twe Points be parallel to the Tangent. For, by theor. 14, ck : kl :: ei : kk ; and by composition, ck : ci. :: ki : ki ; and by the paiallels ck : cl :: oh : lh. Hut, by aim. tri. - ck : cl : ; ki : lh ; theref. by equal. - ki : lh : : gu : lh : consequently - ki = oh, and therefore - kg is parallel and equal to t h. q* b. b. TIIKOREM xvi. If an ordinate be drawn to the point of contact of any tan- gent, and another ordinate produced to cut the tangent ; it will be, as the difference of the ordinates Is to the difference added to the external part, So is double the first ordinate To the sum of the ordinates. That is, kh : ki : : kl : KG. For, by cor. \, faew. \, * \ xvc -» *. dc : da. OF THE PARABOLA. 529 and - . - p : 2dc : : dc : dt or 2da. Hut, by aim. triangles, ki : kc : : dc : dt ; therefore, by equality, j» : 2dc : : ki : kc, or, - . - p : ki : : kl : kc. Again, by theor. 2, p : kh : : kg : kc ; 1 he re fore by equality, kh : ki : : kl : kg. q. e. d. Carol. 1. Hence, by composition and division, it is, kh : ki : : gk : gi, and hi : hk : : hk : kl, also ih : ik : : ik : ig ; that is, ik is a mean proportional between ig and ih. Carol. 2. And from this last property a tangent cun easily be drawn to the curve from any given point i. Namely, draw ihg perpendicular to the axis, and take ik a mean pro- portional between ih, ig ; then dmw kc parallel to the axis, and c will be tbe point of contact, through which and the given point i the tangent ic is to be drawn. THEOREM XVII. If a tangent cut any diameter produced, and if an ordinate to that diameter be drawn from the point of contact ; then the distance in the diameter produced, between the vertex and the intersection of the tangent, will be equal to the absciss of that ordinate. That is, ie = ek. For, by the last th. is : ek : : ck : kl. But, by theor. 11, ck = kl, and therefore ie — ek. Coral. 1. The two tangents ci, li, at the extremities of any double ordinate cl, meet in the same point of the dia- meter of that double ordinate produced. And the diameter drawn through the intersection of two tangents, bisects the line connecting the points of contact. CoroL 2. Hence we have another method of drawing a tangent from any given point i without the curve. Namely, from i draw the diameter ik, in which take ek = ei, and through k draw cl parallel to the tangent at e ; then c and L are the points to which the tangents must be drawn from i. THEOREM XVIII. If a line be drawn from the vertex of any diameter, to cut the curve in some other point, and an ot$voaX« >taoX Vox. /. 68 580 conic tKcnoirt. diameter be drawn to that point, as also another ordinate any where cutting the line, both produced if necessary : 'i he three will be continual proportionals, namely, the two- ord. nates and the part of the latter limited by the said line drawn from the vertex. That is, de, 6H, 01 are crntinual proportionals, or de : gh : : gh : gi. For, by thcor. 9, - - . de 8 : gh* : : ad : ao ; i ml, by sim. tri. - - - de : oi : : ad : ag ; theref. by equality, - - de : gi : : hk* : gh 9 , that is, of the three de, gh, ci, 1st : 3d : : 1st* : 2d*, therefore 1st : 2d : : 2d : 3d, \ 7 y that is, de : Grr : : gh : gi. q. e. d. Cbrol. 1. Or their equals gk, gh, gi, are proportionals ; where ek is parallel to the diameter ad. CoroL 2. Hence it is de : ag : : p : gi, where p is the parameter, or ag : gi : : de : p. For, by the dcfiu. ag : gh : : gh : p. Cord. 3. Hence also the three mn, mi, mo, are propor- tionals, where mo is parallel to the diameter, and am parallel to il.e ordinatcs. For, by theor. 9, - mn, mi, mo, or their equals - ap, ag, ad, are as the squares of pn, gh, de, or of their equals gi, gh, gk, which are proportionals by cor. 1. theorem xix. If a diameter cut any parallel lines terminated by the curve ; the segments of the diameter will be as the rectangle of the segments of those lines. That is, ek : em : : ck . kl : nm . mo. Or, ek is as the rectangle ck . kl. For, draw the diameter rs to which the parallels cl, no are ordinatcs, and the ordinate eq parallel to them. Then ck i» \V\e toffet- cnce, and kx \Y\e sum of the ordinatcs cb \ oVso OF TBS PARABOLA. 531 ttx the difference, and no the sum of the ordinate* eq, ns. And the differences, of the abscisses, are or, as, or kk, km. Then by cor. theor. 9, an : qs : : ck • kl : nm . mo, that in • • kk : km : : ck . kl : nm . mo. Carol 1. The rect. cl . kl = rect, ek and the pa mm. of i>s. For the rect. ck . kl = rect. u« and the param of rs. Carol. 2. If any line cl be cut by two di « meters, kk, csii ; the rectangles of the parts of the line, are as the segments of the diameters. For kk is as the rectangle cz • kl, and ch is as the rectangle cii.iil; therefore kk : oh : : ck . kl : cu . .hl. Carol. 3. If two parallels, cl, no, he cut by two diame- ters, km, gi ; the rectangles of the parts of the parallel* will be as the segments of the respective diameters. For - - - ek : km : : ck . kl : nm . no, „ and . . . ek .: gh : : ck • kl : ch . hl, theref. by equal, km : on : : nm . mo : ch . hl. Carol. 4. When the parallels come info the position of the tangent at p, their two extremities, or points in the curve, unite in rhe point of contact p ; and the rectangle of the parts becomes the square of the tangent, and the same properties still follow them. So that, ev : pv : : pv : • ow : pw : : pw : ev ev ; gw : gh : : pv pv a : p the param. : PW», ; cn. ii l. theorem xx. 'I If two parallels intersect any other two parallels ; the rect. angles of the segments will be respectively proportional. That is, ck . kl : " m - E . 10. For, by cor. 3 theor. $3, pk : at : : ck . kl : gi • in And by the same, ' pk : ut : : dk • KB : ni • io ; therefc byoquaL<iK.KL: ox. « ax . w ;si.io. CoroL When one of the pain of intersecting lines comes k. into the position of their parallel tangents, meeting and limit* ing oach other, the rectangles of their segments become ihe squares of their respective tangents. So that the constant ratio of the rectangles, is that of the square of their parallel tangents, namely, ce • kl: dk • sis : : tang 9 , parallel to cl : tang 9 , parallel to ac THEOBJE* XXI. If there be three tangents intersecting each other ; seek segments will be in the same proportion. That is, gi : in : cu : gd : : dh For, through the points q, i, d, h, draw the diame- ters ok, il, dm, iin ; ns also the lines ci, ei, which are double ordinates to the diameters ok, iin, by cor. 1 theor. 16 ; therefore the diameters gk, dm, iin, bisect the lines cl, ce, le ; hence km = cm — ck = Jce — {cl = )us *e L x r ne, and mn * me — ne = ! ce — £le = Jcl « ck or KL, But, by parallels, gi : in : : kl : ln, and - . cg : cd : : ck : km, also - - dh : he : : mn : ne. But the 3d terms kl, ck, mn are all equal ; as also the 4th terms ln, km ne. Therefore the first and second terms, in all the lines, are proportional, namely, gi : m : : cc : gd : . dii : he. q. e. d» THEOREM XXII. The Area or Space of a Parabola, is equal to Two-Thirds of its Circumscribing Parallelogram. *Lct acb he a semi.pnrubola, cb ihe axis, f the focus, ed the directrix ; then if the line af be supposed to revolve about f as a centre, while the line ae moves along the directrix per- pendicularly to it, the nrea gene- rated by the motion of ae, will always be equal to double the area genera ted hy fa ; and con- sequently the whole external area aegd = double the area or raft PARABOLA- Ml For draw a k parallel, and indefinitely near, to ab ; and draw the diagonals ak' and a'k ; then by th. 6, cor. 4, the angles b'a'a and ka'a are equal, aa' being considered as part of the tangent at a' ; and in the same manner, the angles baa' and faa are also equal to each other ; and since ba = af, and b'a' — a'f ; the triangles ea a' and b'a'a are each equal to the triangle a a'f ; hut the triangle baa' =■ the triangle bb'a, being on the same base and between the same parallels ; therefore the sum of the two triangles bb'a and baa, or the quadrilateral space eaa'b' is double the trilateral space aa'f ; and as this is the case in every position of fa', b'a', it fol- lows that the whole external area baco = double the inter- nal area afc. Hence, Take do = fb, and complete the parallelogram do he, which is double the triangle abf ; therefore the area abc = J the area haco, or 7 of the rectangle aegh, or | of the rectangle abci, because bc = |bo ; that is, the area of a parabola = Jof the circumscribing rectangle. q. b. d.* THBOREM XXIII. The Solid Content of a Paraboloid (or Solid generated by the Rotation of a Parabola about its Axis), is equal to Half its Circumscribing Cylinder. Let ghdd be a cylinder, in which two equal para bo. loids are inscribed ; one bad having its base bcd equal to the lower extremity of the cylinder ; the other gch in- verted with respect to the former, but of equal base and altitude. Let the plane lr parallel to each end of the cylinder, cut all the three so. lids, while a vertical plane may be supposed to cut them so as to define the parabolas shown in the ngure. Then, in the semi-parabola acb, p . ap = pm 1 , also, in the semi-parabola aco, p . cv = pit*, consequently, by addition, p . (ap + cp)=p . ac « pm* + fb*. But, p . ac = cb* = PL*. Therefore pl 3 = pm 3 + ph* : That is, since circles are as the squares of their radii, the p y " e * This Jem »n it ration whs given by Lieut. Drummond of the Roynl Engineers, * hen be wai a gentleman Cadet at the Royal Military Acs- lo a Ctinder »^#«e Hei»j- i« pr. ud ft* ft**e li*lf ike to of the two CirtoUr Kmc* ea. sc. Let * - 3 1410 : B D C Th*n, by th* l**t theor. % *pe Y Air = the ». I id arc, and, by the wifnc \jc X Ar 1 = the solid aec, tl§#?r^f. the difT. £pc X 'ai* 1 — *f ;=the fiust. begc. Rut kit 1 — ai" = ur X ' hi* -r ir% Iheref. Jf* X di X 'ad 4- kt) = the fro* begc. But, by th. I, p X sit — ih. j , and/> X ir = fg ? : the re f. X dj X (in/ 1 + fg*, = the frost, begc. q. e. d. fhoblems, &c for exercise in comc sections. 1. Demonstrate flint if a cylinder be cut obliquely the sec- tion will In; an ellipse. 2. Show how to draw a tangent to an ellipse whose foci are r, f } from a given point i*. 3. Show how to draw a tangent to a given parabola from ii given point v. 4. The diameters of an ellipse are 16 and 12. Required the parameter and the area. 5. The bane and altitude of a parabola are 12 and 0. Re* quired the parameter, and the semi-ordinates corresponding to the abscissa* 2, .'1, and 4. (I. In the actual formation of arches, the voussoirs or arch, gtoauft arc *j cuV a& \o >»tai«3% ^on^adiaUar CONIC SBCTI09CS. 5S5 to the respective points of the curve upon which they stand. By what const ruction* may this be effected for the parabola and the ellipse ? 7. Construct accurately on paper, a parabol.i whose base shall be 12 and altitude 9. 8. A cone, the diameter of whose base is 10 inches, and *hose altitude is 12, is cut obliquely by a plane, which enters at 3 inches from the vertex on one slant slide, and comes out at 3 inches from the base on the opposite slant side* Requir- ed the dimensions of the section ? 9. Suppose the same cone to be cut by a plane parallel to one of the slant sides, entering the other slant side at 4 inches from the vertex, what will be the dimensions of the section ? 10. Let any straight line efb be drawn through f, one of the foci, of an ellipse, and terminated by the curve in b and r ; then it is to be demonstrated that kf . fr = eb.{ pa- rameter. 11. Demonstrate that, in any conic section, a straight line drawn from a focus to the intersection of two tangents makes equal angles with straight lines drawn from the same focus to the points of contact. 12. In every conic section the radius of curvature at any- point is to half the parameter, in the triplicate ratio of the distance of the focus from that point to its distance from the tangent. Also, in every conic section the radius of curvature is pro- portional to the cube of the normal. Also; let fc be the radius of curvature at any point, p, in> an ellipse or hyperbola whose tranverse axis is ab, conju- fpF . rf)i gate ab, and foci f and /: then is pc = ~ -^-t-. ° ' Jab . ab Required demonstrations of these properties* OH Til COMIC NLCTBOH! *S W MW ** AMMftMl *0*A- TIOHC CALLED THS BQ0ATIO]* if Tl| W*TB. • * 1. Fortke gltipm. Let I denote ab, the transverse, or any diameter; * sb|h its conjugate ; * = ak, any absciss, from the extremity of the diam. y = ok. the correspondent ordinate : the two being jointly denominated co-ordinates. Then, theor. 2, ab* : hi* : : ak . kb : dk 3 , that is, 1* : e\ : : x(f — x) : y*, hence iy « c"(<Jr — ar*), or y = -y- y/(tx — x 9 ), the equation of the curve. And from these equations, any one of the four letters or quantities, f, c, x, y, may easily be found, by the reduction of equations, when the other three are given. Or, if p denote the parameter, = c 1 1 by its definition ; then, by cor. th. 2, f.: p : : *' v *-x) : y\ or y*= -y- (fx— x*), which is another form of the equation of the curve. Otherwise. If f = ac the semiaxis ; e «= cr the semiconjugate ; then p =; c* -r t the semi parameter ; x = ck the absciss counted from the centre ; and y = dk the ordinate as before. Then is ak = f— x, and kb = t + x, and ak . kb = {t — x) X (H-*)*** 8 — X s . Then, by th. 2, : c* : : fi— x* : y', and <y = c*(f* — x»), or y = -j- ^/(<* — X s ), the equation of the curve. Or, tip:: <■— x 1 : y\ and y« = (C — x*), another form of the equation to \\ie cwrc*\ Vrom ^\v\tVv <me of the quantities may be found, HiYawi vVv^ rax ZQUATIONS OF THJB CVBVS. 697 2. For tJie Hyperbola. Because the general property of the opposite hyperbolas, with respect to their abscisses and ordinatcs, is the same as that of the ellipse, therefore the process here is the very same as in the former cade for the ellipse ; and the equation to the curve must come out the same also, with sometimes only the change of the sign of a letter or term, from + to — , or from — to +, because here the abscisses lie beyond or without the transverse diameter, whereas they lie between or upon them in the ellipse. Thus, making the same notation for the whole diameter, conjugate, absciss, and ordinate, as at first in the ellipse ; then, the one absciss ak being x, the olher bk will be t + x, which in the ellipse was t — r ; so the sign of x must be changed in the general property and equation, by which it becomes f 3 : c* : : x[t + x) : y 8 ; hence ftp = c* (tx + x*) and y = -j- y/ (tx + jc 2 ), the equation of the curve. Or using p the parameter, as before, it \s,t : p : : x(t+x) : y* or y* = -j- {tx + x* ), another form of the equation to the curve. Otherwise, by using the same letters f, c, p, for the halves of the diameters and parameter, and x for the absciss ck counted from the centre ; then is ak = x — t, and bk = x+t, and the property P : c* : : (< — t) X (x+ 1) : y\ gives *V=r €* (x* — f) 9 or y = T ✓(x*— / a ), where the signs of t* and x % are changed from what they were in the ellipse. Or again, using the semi parameter, tip:: x* — £ : y 3 , and y* = -j- (x 3 — f) the equation of the curve. But for the conjugate hyperbola, as in the figure to theo- rem 3, the signs of both x' J and t % will be positive ; for the property in that theorem being ca 3 : ca 2 : : cd* + ca" : ne 3 , it is f 3 : c* : : x* + f 3 : y > = D« a , or < a y s = c 3 (x a + < > ), and y =* \ y/(*? + 0> tne equation to the conjugate hyperbola. Or, as t : p : : x* + c* : y 8 , and y 3 = (x 3 + f 3 ) also the equation to the same curve. Vol. I. 69 a* x * y*rrvML <r*. jdt imn. vtraa; ». wmi as, >i a9, sa 4r?inacei pan£ei to aansmou*. a/ « ir =«,cisz. Mil a* = «- T"i*&- sr tM«r. 2&l at . iv as t« . ». *>p = rj, :a* to tie */p*rV*a, n«a *.v* 1 Triton sjs4 csm- aac** take* parauet to the asvwaassBflL If lite a;>*r»*4a fc* arx rectangular ar . sr. am. r vfi be tquai to & vfiare. 3. Fir U* P mrmhoU- if x tenrtoi any ataciss beginning mt the vertex, and y ordinate: alvi /> tne :*ararnei*r. Then, oy oor. theorem 1, ak Mi* k f# p. <jt 1 y y p : hence yz = jf 'is the equa- tion f/i t.v; \,htwa'+. Or, if 6i = abscissa and 6 the corre*- fforidiri^ ^rriiordiriat'.-, ilien ~ j = y% is the equation. 1. For the Circle. H':raui«: the circle is only a species of the ellipse, in which two fixoH an; equal to each other ; therefore, m«bmg the (wo fli'irnctorH / and r equal each to d in the foregoing equa* Iioiih to thf! ellip.se, they become y a = cir — -r 1 , when the nhnciMH t h<;giriH at the vertex of the diameter : and y a = |d a x*, when the absciss begins at the centre. Or y = (iJr 4 - .■!■*), and y= ^/(r 2 — s 2 ), respectively, when r is the ruiliiiM. Scholium. lu vvcry onu of theso equations, we perceive that they rise to tlm *Jd or quadratic degree, or to two dimensions ; which in hImu the number of points in which any one of these curves nut) l»<i nit by a right lino. Hence also it is that these four eurvoM are niihI to he linos of the 2d order. And these four urn nil the. lines that aro of that order, every other curve hav- ing muim higher equation, or may be cut in more points by a right lino. Wo nun heir add an important observation with regard to all eurves o\y rwwed by equations : viz. that the origin of iukbxts of noraiMBTKY. 58t when aU the terms ef its equation are affected by one of the variable quantities x or y; and when, on the contrary, there is in the equation one terra entirely known, then the origin of the co-ordinates cannot be on a point of the curve* In proof of this, let the general equation of a curve be asf* + bx*yi =0 ; then, it is evident that if we take »= 0, we shall likewise haveiy* = 0, ory =0; and consequently the origin of the co-ordinates is a point in the curve. So again, if, in the same equation, we take y = 0, it will result thai ax = 0, and * =0, which brings us to the same thing as before. Bat, if the equation of the curve include one known term, as, for example, aw* + fc'y* + cy* — = ; then taking x = a, we shall have cy — =*= 0, or y = '/ which proves that the corresponding point p, of die curve, is distant from the origin of the z'a by the quantity V A simitar troth will flow from making y = 0, when the same equation will give * « ELEMENTS OF ISOPERIMETRY. Dtf. 1. When a variable quantity has it* rimtations re- jpdated by a certain law, or confined within certain limits, it is called a maximum when it has reached the greatest mag- nitude it can possibly attain ; and, on the contrary, when k has arrived at the least possible magnitude, it is called a mi- Isepcrimetors> or f mpeiimeti ic ai Figures, me those which have equal perimeters. Def. 3. The Locus of any point, or intersection, dec. is the right line or curve in which these are always situated. The problem in which it is required to find, among figures of the same or of different kinds, those which, within equal perimeters, shall comprehend the greatest surfaces, has long engaged the attention of mathematicians. Since the admir- able invention of the method of Fluxions, this problem has been elegantly treated by some of the writers on that branch of analysis ; especially by Haclaqrin and Simpson* k wud*> 640 ELEXKm Or fSOPBUMXTKT. more extensive problem was investigated at the time of " the war of problems," between the two brothers John and James Bernoulli : namely, " To find, among all the isoperi- metrical curves between given limits, such a curve, that, con- structing a second curve, the ordinates of which shall be functions of the ordinates or arcs of the former, the area of the second curve shall be a maximum or a minimum." While, however, the attention of mathematicians was drawn to the most abstruse inquiries connected with isoperiraetry, the cfe- menls of the subject were lost sight of. Simpson was the first who called them back to this interesting branch of research, by giving in his neat little book of Geometry a chapter on the maxima and minima of geometrical quantities, and some of w the simplest problems concerning isoperi meters. The next who treated this subject in an elementary manner was Simon Lhuillier, of Geneva, who, in 1782, published his treatise De Relatione mulua Capacitatis el Terminorvm Figurarum, die. His principal object in the composition of that work was to supply the deficiency in this respect which he found in most of the Elementary Courses ; and to determine, with re- garb? to both the most usual surfaces and solids, those which possessed the minimum of contour with the same capacity ; and, reciprocally, the maximum of capacity with the same boundary. M. Legend re has also considered the same sub- ject, in a manner somewhat different from either Simpson or Lhuillier, in his EUments dc Geomttrie. An elegant geo- metrical tract, on the same subject, was also given by Dr. Horsley, in the Philos. Trans, vol. 75, for 1775 ; contained also in the New Abridgement, vol. 13, page 653*. The chief propositions deduced by these four geometers, together with a few additional propositions, are reduced into one system in the following theorems. " Another work on the same general subject, containing mnny valua- ble theorems, has been published since the first edition of this volume, by Dr. CrtsucU of Trinny College, Cambridge. {541 ] SECTION I. SURFACES. THEORXM I. Of all triangles of the same base, and whose vertices fall in a right Tine given in position, the one whose perimeter is a minimum is that whose sides are equally inclined to that line. Let ab be the common base of a series of triangles abc\ abc, &c. whose vertices c', c, fall in the right line lm, jpven in position, then is the triangle of least perimeter that whose sides ac, bc, are inclined to the line lm in equal angles. For, let bm be drawn from b, per. pendicularly to lm, and produced till dm = bm : join ad, and from the point c where ad cuts lm draw bc : also, from any other point c', assumed in lm, draw c'a, c'b, cd. Then the triangles dmc, bmc, having the angle dcm = angle acl (th. 7 Geom.) = mcb (by hyp.), dmc =■ bmc, and dm = bm, and mc common to both, have also dc = bc (th. 1 Geom.). So also, we have c'i> = c'b. Hence ac + cb = ac + cd ~ ad, is less than ac' + c'p (theor. 10 Geom.), or than its equal ac' + c'b. And consequently, ab + bc + ac is less than ab + bc' + ac'. q. e. d. Cor. 1. Of all triangles of the same base and the same al- titude, or of all equal triungles of the same base, the isosceles triangle has the smallest perimeter. For, the locus of the vertices of all triangles of the same altitude will be a right line lm parallel to the base ; and when lm in the above figure becomes parallel to ab, since mcb = acl, mcb =? cba (th. 12 Geom.), acl = cab ; it follows that cab = cba, and consequently ac = cb (th. 4 Geom.) Cor. 2. Of all triangles of the same surface, that which has the minimum perimeter is equilateral. For the triangle of the smallest perimeter, with the same surface, must be isosceles, whichever of the sides be con- sidered as base : therefore, the triaogU of ra^i^^\TMtfust 542 ELEMENTS OF ISOFEllIMETBT. has each two or each pair of its sides equal, and consequently it is equilateral. Cor. 3. Of all rectilinear figures, with a given magnitude and a given number of sides, that which has the smallest perimeter is equilateral. For so long as any two adjacent sides are not equal, we may draw a diagonal to become a base to those two sides, and then draw an isosceles triangle equal lo the triangle so cut off, but of less perimeter : whence the corollary is manifest Sc?iolium. ?■■ To illustrate the second corollary above, the student may proceed thus : assuming an isosceles triangle whoso base is not equal to either of the two sides, and then, taking for anew base one of those sides of that triangle, he may construct an- other isosceles triangle equal to it, but of a smaller perimeter. Afterwards, if the base and sides of this second isosceles tri- angle are not respectively equal, he may construct a third isosceles triangle equal to it, but of a still smaller perimeter ; and so on. In performing these successive operations, he will find that the new triangles will approach nearer and nearer to an equilateral triangle. THEOREM II. Of all triangles of the same base, and of equal perimeters, the isosceles triangle has the greatest surface. Let abc, abo, be two triangles of the same base ab and with equal perimeters, of which the one auc is isosceles, the other is not : then the triangle aim- has a surface (or an altitude) greater than the surface (or than the altitude) of the triangle \bd. Draw c'd through o, parallel to ab, to cut ve (drawn perpendicular to ab) in c : then it is to be demonstrated that ce is greater than c'k. The triangles ac'b, adb, are equal both in base and alti- tude ; but the triangle Ac'n is isosceles, while adb is scalene : therefore the triangle ac'b has a smaller perimeter than the triangle adb (th. 1 cor. 1), or than acb (by hyp.). Conse- quently ac < ac ; and in the right-angled triangles aec, aec, having ae common, we have c'e < ce *. q. e. d. ♦ When two mathtma\\tB\ qpwfl&>»% <&wx»kXm\ <^ SURFACES. 548 Cor. Of all isoperimetrical figures, of which the number of sides is given, that which is the greatest has all its sides equal. And in particular, of all isoperimetrical triangles, that whose surface is a maximum, is equilateral. For, so long as any two adjacent sides are not equal, the surface may be augmented without increasing the perimeter. Remark. Nearly as in this theorem may it be proved that, of all triangles of equal heights, and of which the sum of the two sides is equal, that which is isosceles has the greatest base. And, of all triangles standing on the same base and having equal vertical angles, the isosceles one is the greatest. # THEOREM in. Of all right lines that can be drawn through a given pointy between two right lines given in position, that which is bisected by the given point forms with the other two lines the least triangle. Of all right lines on, ab, cd, that can be drawn through a given point p to cut the right lines ca, cd, given in position, that, ab, wnich is bi- sected by the given point r, forms with ca, co, the least triangle, abc For, let ee be drawn through a parallel to cd, meeting dg (produced if necessary) in e ; then the triangles p jo, pae, are manifestly equiangular ; and, since the corresponding sides pb, pa are equal, the triangles are equal also. Hence pbd will be less or greater than pag, according as cg is greater or less than ca. In the former case, let pacd, which is common, be added to both ; then will bac be less than dgc (ax. 4 Geona.). In the latter case, if pgcb bo added, dco will be greater than bac ; and conse- quently in this case also bac is less than dco. q. e. d. Cor. If ph and pn be drawn parallel to cb and ca re- spectively, the two triangles pam, pbn, will be equal, and these two taken together (since am = pn a mc) will be equal to the parallelogram pmcn : and consequently the parallelogram pmcn is equal to half abc, but less than half doc. From which it follows (consistently with both the al- gebraical and geometrical solution of prob. 8, Application of it denotes that the preceding quantity is less than the succeeding one : when, on the contrary, the separating character is > , it denotes that the preceding quantity is greater th*n the succeeding one. 544 KLXMEXTI OF UOPIUiaTBT. Algebra to Geometry), that a parallelogram is always lew than half a triangle in which it is inscribed, except when the base of the one is half the base of the other, or the height of the former half the height of the latter ; in which case the parallelogram is just half the triangle : this being the maxi- mum parallelogram inscribed in the triangle. Scho* urn. From the preceding corollary it might easily be shown, that the least triangle which can possibly be described about, and the greatest parallelogram which can be inscribed in, any curve concave to its axis, will be whenjL subtangent is equal to half the base of the triangle, or to » whole base of the % parallelogram : and tliat the two figures will be in the ratio of 2 to 1. But this is foreign to the present inquiry. THEOREM IV. Of all triangles in which two side* are given in magnitude, the greatest is that in which the two given sides are perpendicular to each other. For, assuming for base one of the given sides, the surface is proportional to the perpendicular let fall upon that side from the opposite extremity of the other given side : there- fore, the surface is the greatest, when that perpendicular is the greatest ; that is to say, when the other side is not in- clined to that perpendicular, but coincides with it : hence the surface is a maximum when the two given sides are per- pendicular to each other. Otherwise. Since the surface of a triangle, in which two sides are given, is proportional to the sine of the angle in- cluded between those two sides ; it follows, that the triangle is the greatest when that sine is the greatest : but the greatest sine is the sine total, or the sine of a quadrant ; therefore the two sides given make a quadrantal angle, or are perpendicular to each other* q. e. d. THEOREM V. Of all rectilinear figures in which all the sides except one are known, the greatest is that which may be inscribed in a semicircle whose diameter is that unknown side. For, if you suppose the contrary to be the case, then when- ever the figure made with the sides given, and the side un- known, is not inscribable in a semicircle of which this latter SURFACES. 545 i the diameter, viz. wtmever any one of the angles, formed by lines drawn from the extremities of the unknown side to one of the summits of the figure, is not u right angle ; we may make a figure greater than it, in which that angle shall be right, and which shall only differ from it in that respect : therefore, whenever all the angles, formed by right lines drawn from the several vertices of the figure to the extre- mities of the unknown line, are not right angles, or do not fall in the circumference of a semicircle, the figure is not in its maximum state, q. e. d. . THEOREM VI. I Of all figures made £th sides given in number and mag. ' nitude, that which may be inscribed in a circle is the greatest. Let abcdefg be the polygon inscrib- ed, and abcdefg a polygon with equal sides, but not inscri- bable in a circle ; so that ab = <i6, bc = be, die. ; it is affirm, ed that the polygon abcdbfg is greater than the polygon abedefg. Draw the diameter bp ; join ai\ ph ; upon ab = ab ma'ce the triangle abp, equal in all respects to abp ; and join ep. Then, ofthe two figures edebp, pag fe> one at least is not (by hyp.) inscribable, in the semicircle of which ep is the diame- ter. Consequently, one at least of these two figures is smaller than the corresponding part of the figure apbcdrfg (th. 6). Therefore the figure aprcdrfo is greater than the figure apbc 'efg : and. if from these there be taken away the re- spective triangles apb, apb, which are equal by construction, there will remain (ax. 5 Geom.) the polygon abcdefg greater than the polygon abcdefg. q,. e. d. THEOREM VII. The magnitude of the greatest polygon which can be con- tained under any number of unequal sides, does not at all depend on the order in which those lines are connected with each other. For, since the polygon is a maximum under given side*, it is inscribable in a circle (th. 6). And this inscribed polygon is constituted of as many isosceles triangles as it has sides, those sides forming the bases of the respective triangles, the Vol. I. 70 646 ELEMENTS OF ISOmRIMJCTRY. other rides of all the triangle* being radii of the circle, and their common summit the centre of the circle. ConJu»quenujr the magnitude of tho polygon, tbmt is, of the asscmbliige of these triangles, does not at all depend on their d i spos it ion , or arrangement around tho common centre, a- k. d. THEOREM VIU. If n polygon inscribed in a circle have all its sides equal, all its angles are likewise equal, or it is a regular polygon. For, if lines be drawn from the several angles of the poly, gon, to the centre of the circumscribing circle, they will divide the polygon into as many iuuawles triangles as it has! sides ; nnd each of these isosceles triangles will be equal to™ either of the others in all respects, and of course they: will have the angles at their bases all equal : consequently, .the angles of the polygon, which are each made up of two angles at the bases of two contiguous isosceles triangles, will be equal to one another, a. k. d. THEOREM IX. Of all figures having the same number of sides and sequel perimeters, the greatest is regular. For, the greatest figure under the given conditions has all its sides equal (tli. 2. cor.). But since the sum of the sides and the number of them are given, each of them is given : therefore (th. 6\ the figure is inscribable in a circle: nnd consequently (th. 6) all its angles ore equal ; that is, it is regular, a. e. i>. Cor. Hence we see that regular polygons possess the pro- perty of a maximum of surface, when compared with any okher figures of the same name and with equal perimeters. THEOREM X. A regular polygon has a smaller perimeter than an irregular one equal to it in surface, and having the same number cf sides. This is the converse of the preceding theorem, and may be demonstrated thus : Let r and I be two figures equal in surface, and having the same number of sides, of which** is regular, i irregular : let also h' be a regular figure similar to r, and having a perimeter equal to that of i. Then (th. 9) r' > i ; but \ =• a; vYu&tetac* ^ > ^ wuijt are si- 947 stoHar ; o o aja oq ae n fly, perimeter, of b' > perimeter of a ; while per. a' as per. 1 (by hyp.). Hence, per. i > per. a. q. b. d. THBORKW XI* The surfaces of polygons, circumscribed about the same or equal circles, are respectively as their perimeters*. D Let the polygon abcd hi! circumscribed about theeircte vfoh ; and let this polygon be divided into triangles, by lines drawn pfrom its several angles to the centre o of the circle. Then, since each of the tan- gents ab, nc, dec. is perpendicular to its -A. eomspondtnf radius, or, op, &c, drawn to the point of con- tact (th. 46Geom.) ; and since the area of a triangle is equal lb the rectangle of the perpendicular and half the base (Mens, •f Surface*, pr. 3); it follows, that the area of each of the triangles abo, bco, Sic. is equal to the rectangle of the radius of the Circle and half the corresponding side ab, bc, &c. ; and consequently, the area of the polygon abcd, circumscribing the circle, will he equal to the rectangle of the radius of the circle and half the perimeter of the polygon. But, the sur- face of the circle is equal to the rectangle of the radius and half the circumference (th. 04 Geom.). Therefore, the sur- face of the circle, is to that of the polygon, as half the cir- cumference of the former, to half the perimeter of the latter ; or, as the circumference of the former, to the perimeter of the latter. Now, let p and p' be any two polygons circum- scribing a circle c : then, by the foregoing, we htive surf, c : surf, p : : circum. c : perim. p. surf, c : surf, p' :; circum. c : perim. p\ But, since the antecedents of the ratios in both these proper- tions, are equal, the conaequents are proportional : that is, aurf. p : surf, p' : : perim. p : perim. p'« 4. k. d. Cor. 1. Any one of the triangular portions abo, of a po- lygon circumscribing a circle, is to the corresponding circular sector, as the side ab of the polygon, to the arc of the circle included between ao and bo. •This theorem, together with tho analogous onei re«prcling hodips ctaaoiferihMg cylinders nnd spheres, were piveu by Eme.-son in his tfrometry, and their u*e4n tan theory of lso|M»rimetm was just suggest- ed : but the full *ppKc*Uua ef them te that theory is due to Siioee Uuillier. 548 xunanm of hokeikbtet. Car. 2. Every circular arc is greater than Ha chord, aad leas than the sum of the two tangents drawn from its extremi ties and produced till they meet. The first part of this corollary is evident, because a r bt line is the shortest distance between two given points. T <e second part follows at once from this proposition : for ea f ah being to the arch kih, as the quadrangle a boh to the c r cular sector hieo ; and the quadrangle being greater than the sector, because it contains it ; it follows that ea + ah if greater than the arch kih*. Cor. 8. Hence also, any single tangent sa, is greater than its corresponding arc ki. ^ THEOREM XII. If a circle and a polygon, circumscribahle about another circle, are isoperimeterc, the surface of the circle is a geometrical mean proportional between that polygon and a similar polygon (regular or irregular) circumscribed about that circle. Let c be a circle, p a polygon isoperimetrical to that circle, and circumscribahle about some other circle, and p' a polygon similar to p and circumscribahle about the circle c : it is affirmed that p : c : : c : p'. For, p : p' : : peri nr. p : perim a . p' : : circum*. c : perim*. t by tli. 89, geom. and the hypothesis. But (th. 1 1) p ' : c : : per. p' : cir. c ; : per 3 , p' : per. p' x cir.c Therefore p : c : : .... c-ir 9 . c : per. P x cir. c. : : cir. c : per. p' : : c : p'. Q. e. ». THEOREM XIII. If a circle and a polygon, circumscribahle about another circle, are equal in surface, the perimeter of that figure is a geometrical mean proportional between the circum- ference of the tiret circle and the perimeter of a similar polygon circumscribed about it. Let c = p, and let p' be circumscribed about c and similar to c : then it is affirmed that cir. c : per. p : : per. p : per. p'. * This second enrol In r v is introduced, not l>ecfui«* (if its immrdiftle connexion with the subject nndr discission, but because, nottrith- standing its timpVuAly, *v\\Wr% V\wv« employed whole pegetia attempting Us demoi\&Vt*V\w\, SURFACES* For cir. c : per. p' : : c : p' : : p : p' : : per*, p : per 1 . p'. A Up, per. p' : per. p - : : per*, p' : per. p X per. p'. Tbi reforo, cir. c : per. p : : per", p : per. p x per p'. : : per. p : : per. p. q,. b. d. ' THEOREM XIV. The circle is greater than any rectilinear figure of the same perimeter : and it has a perimeter smaller tha » ai»y recti- linear figure of the same surface. For, in the proportion, p : c :: c : p (th. 12), since c < p', therefore p < c. And, in the propor. cir. c . per. p : : per. p : per. p' (th* 13), or, cir. c : per. p' : : cir 3 . c : per*, p, and cir. c < per. p' : therefore, cir*. c < per 1 , p, or cir. c < per. p. q. e. d. Cor. 1. It follows at once, from this and the two pre- ceding theorems, that rectilinear figures which are isoperi* metera, and each circumscribable about a circle, are re- spectively in the inverse ratio of the perimeters, or of the surfaces, of figures similar to them, and both circumscribed about one and the same circle. And that the perimeters of equal rectilineal figures, each circumscribable about a circle, are respectively in the subduplicate ratio of the perimeters, or of the surfaces, of figures similar to them, ana both cir- cumscribed about one and the same circle. Cor. 2. Therefore, the comparison of the perimeters of equal regular figures, having different numbers of sides, and that of the surfaces of regular isoperimetrical figures, is re- duced to the comparison of the perimeters, or of the surfaces of regular figures respectively similar to them, and circum- scribable about one and the same circle. Lemma 1. If an acu*o anglo of a right-angled triangle be divided into any number of equal parts, the side of the triangle opposite to that acute angle is divided into unequal parts, which are greater as they are more remote from the right angles. Let the acute angle c, of the right- Cfev angled triangle acp, be divided into equal V\^V parts, by the lines rc, cd, ce, drawn from \\N\ that angle to the opposite side; then shall \ \ the parts ab, bd, &c. intercepted by the A B J> E 7 650 ELEMENTS OF nOPEBJXETRY. lines drawn from c, be successively longer us they are more lemofe from the right angle a. For, the tingles acd, bce. &c. being bisected by c», CD, d&c. iherefure by thcor. 83 Geom. ac : cd : : ab : bd, and rc : ce : : bu : de, and dc : cf : : de : ef. And by th. 21 Geom. cd > ca, ce > cb, cf > cc, and so on : whence it follows, that db > ab, de > db, and soon. u. e. d. Cor. Hence it is ohviius that, if the part the most remote from the right angle a, be repeated a number of times equal to that into which the acute angle is divided, there will re. suit a quantity greater than the side opposi e to the divided angle. THEOREM XV. If two reg liar figures, circumscribed about the same circle, differ in their number of sides* by unity, that which has the greatest number of sides shall have the smallest peri- meter. Let ca be the radius of a circle, and ab, ad, the haff sides of two regular polygons circumscribed about thai circle., of which the. number of sides differ by unity, being C respectively n + 1 und n. The angles acb, acd, therefore arc respectively the ~^ and the ^ th part of two right angles : consequently these <A- I^D angles ar* as n and n + 1 : and hence, the angle may be conceived divided into n + 1 equal purls, of which bcd is one. Consequently, (cor. to the lemma) (n -f 1) no > ad. Taking, then, unequal quantities from equal quantities, we shall have (n + 1} ad - (n + 1) bd < (n + 1) ad - ad, or(n t!)ab< n . ad. That is, the scmiperimeter of the polygon whose half side is ab, is smaller than the scmiperimeter of the polygon whose half side is ad : whence the proposition is manifest. Cor. Hence, augmenting successively by unity the num. ber of sides, it follows generally, that the perimeters of polygons circumscribed about any proposed circle, become smaller as the number of their sides become greater. THEOREM XVI. The surfaces ef regular isoperimetrical figures are grAriter as the number of their sides is greater : and the peri- meters of equal ve^ulur figures are smaller as the number of their sides is gtuutet. SOLIDS. r,$| For, 1st. Regular isoperimetrical figures are (cor. 1. th. 14) in the inverse ratio of figures similar to tliem circum- scribed about the same circle. And (th. 15) these latter ara smnller when their .number of sides is greater : therefore, on the contrary, the foifaer become greater as they have more sides. 2dly. The perimeters of equal regular figures are (cor. 1 th. 14; in the subduplicute ratio of the perimeters of similar figures otptamscrib$d .about the same circle : and (th. 15) these latter are smaller as they have more side's: therefore the perimtffsrs of the former also arc smaller when the num- ber of then* sides is greater, a. e. d. SECTION II. 4 SOLIDS. THEOREM XV1T. Of all prisms of the same altitude, whose base is given in magnitude and species, or figure, or shape, the right prism has the smallest surface. For, the area of each face of the prism is proportional to its height ; therefore the area of each face is the smallest when its height is the smallest, that is to say, when it is equal to the altitude of the prism itself: and iu that case the prism is evidently a right prism, q. k. d. TIIEOREM xviii. Of all prisms whose base is given in magnitude and species, and whose lateral surface is the same, tho right prism has the greatest altitude, or the greatest capacity. This is the converse of the preceding theorem, and may readily be proved after the manner of theorem 2. TIIEOREM XW. Of all right prisms of the same altitude, whose liases are 4 given in magnitude and of a given number of sides, that whose base is a regular figure has the smallest surface. For, the surface of a right prism of given altitude, and base giveu in magnitude, is evidently proportional to the