S u

C

to it, when AB arrives at C, A it will coincide with DC, and by its motion it will have described or generated the parallelogram ABCD; let AB consist of suppose 4 m equal parts, each of which B we will call unity, (or 1.) let Bm=one of those parts, and Br, rs, su, &c. each=Bm; now it is plain, that when AB arrives at r, it will by its motion have described the four rectangles between AB and ær, each of which will be the square of (Bm, that is of) unity; in like manner, when AB arrives at s, u, v, z, C, it will have described 8, 12, 16, 20, 24 squares of (Bm, or) unity: whence it appears, that the area ABCD or 24, is found by multiplying the number of equal parts (called units) contained in AB, or 4, by the number of like parts in BC, or 6. In like manner, if AB contain n units, and BC m units, the area ABCD will contain n× m=nm units: if n=m, the figure ABCD will be a square, and nm will become n2 or m2. Hence the area of a rectangle is found by multiplying the two sides about one of its angles into each other, and the area of a square by multiplying the side into itself.

c

147. Prop. 35. From this proposition, and the preceding article, we derive a method of finding the area of any parallelogram whatever: for let ABCD (see Simson's first figure) be supposed to be a right angled parallelogram, its area will be ABX BC, (by Art. 146.) or the perpendicular AB, drawn into (or multiplied by) the base BC; but DBCF=ABCD by the proposition, DBCF=perp. AB× base BC.

148. Hence we have the following practical rule for finding

The terms multiplying and dividing do not occur in geometrical language; thus, in the expression AB × BC=ABCD, AB is said to be drawn into BC, and ABCD is not called the product of AB and BC, but their rectangle; and AB

C

in expressions like the following C is said to be applied to AB. respect, but the moderns are less made use of in their geometrical problems; but this abuse should as much as possible be avoided.

AB is not said to be divided by C, but The old writers are very particular in this so, as we frequently find arithmetical terms

the area of a parallelogram. 1. Let fall a perpendicular on the base from any point in the opposite side. 2. Multiply the base and perpendicular together, and the product will be the area required.

149. Prop. 37. Since every triangle is half of the parallelogram described upon the same base, and between the same parallels, (see also prop. 41.) and the area of the parallelogram is=perp. × base, (by the last article,) ·.· the area of the triangle will be perp. xhase ; that is, half the perpendicular multiplied

2

into the base, or half the base multiplied into the perpendicular, will give the area of the triangle.

150. Prop. 38. Cor. Hence, if the base BC be greater than the base EF, the triangle ABC will be greater than the triangle EDF; and if BC be less than EF, the triangle ABC will be less than the triangle EDF. Also, if ABC be greater than EDF, then is BC greater than EF; and if less, less.

151. In prop. 42. we are taught how "to describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle." In prop. 44. we are to describe a parallelogram with the two former conditions, and also one more: we are "to apply a parallelogram to a given straight line, which parallelogram shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle;" to "apply a parallelogram to a straight line," means to make it on that straight line, or so that the said line may be one of its sides.

66

152. Prop 45. The enunciation of this proposition is general, if by a given rectilineal figure" we are to understand “ any given rectilineal figure:" but the demonstration applies to only a particular case; for it extends no further than to four-sided figures, and does not even hint at any thing beyond; but the defect is easily supplied as follows:

Let ABCOND be any rectilineal figure; join DB, DC, CN, then having made the parallelogram FKML equal to the quadrilateral figure ABCD, as in the proposition, apply the parallelogram LS=DCN to the straight line LM, having an angle LMS: E, then it may be proved as before, that FL and LP are in the same straight line as are KM and MS; also that PS is parallel to FK, and consequently that FKSP is a parallelogram and equal to ABCND; and applying as before a parallelogram PT=NCO, having the angle PST E, to the straight line PS, FKTR may in like manner be proved to be a parallelogram equal to ABCOND, and having an angle FKT=E; and by a similar process a parallelogram may be made equal to any given rectilineal figure whatever, and having an angle equal to any given rectilineal angle. The foregoing illustration being understood, the corollary to this proposition will be evident.

Cor. Hence we have a method of determining the difference of any two rectilineal figures. Thus ABCOND exceeds DCON by the parallelogram FM.

153. Prop. 46. Cor. In a similar manner the rectangle contained by any two given straight lines may be described.

154. The squares of equal straight lines are equal to one another.

Let the straight lines AB and CD be equal, then will the squares ABEF, CDGH

described on them be equal. For since AB= M

CD by hypothesis, and

F

E

H

G

HC CD (Def. 30.) ·.·

HC AB, but FA=AB,

(Def. 30.). HC=FA;

K B

C

D

wherefore if the square FB be applied to the square HD, so that A may be on C, and AB on CD, B shall coincide with D

because AB CD; and AB coinciding with CD, AF shall coincide with CH because the angle BAF=DBH, (Def. 30. and Ax. 11.) also A coinciding with C, and AF with CH, the point F shall coincide with H, because AF=CH; in the same manner it may be shewn, that FE and EB coincide respectively with HG and GD, therefore the two figures coincide, and consequently are equal by Ax. 8. Q. E. D.

Cor. 1. Hence two squares cannot be described on the same straight line and on the same side of it.

Cor. 2. Hence two rectangles which are equilateral to one another will likewise be equal.

155. If two squares be equal, the straight lines on which they stand will also be equal.

Let ABEF=CDGH, (see the preceding figure) then will AB=CD; for if not, let AB be the greater, and from it cut off AK=CD (3.1) and on AK describe the square AKLM, (46.1) then since AK=CD, the square AL the square CG, (Art. 154.) but AE CG by hypothesis, . AL=AE the greater to the less which is impossible, ·.· AK is not equal to CD, and in like manner it may be shewn that no straight line, either greater or less than AB, can be equal to CD, AB=CD. Q. E. D.

156. Prop. 47. This proposition, which is known by the name of the Pythagorean Theorem, because the philosopher Pythagoras was the inventor of it, is of very extensive application; its primary and obvious use is to find the sum and difference of given squares, the sides of right angled triangles, &c. as is shewn in the following articles .

157. To find a square equal to the sum of any number of given squares. Let A, B, C, D, &c. be any number of given straight lines; it is required to find a square equal to the sum of the squares described on A, B, C, D, &c.

Take any straight line EM, and from any point E in it draw EP perpendicular to EM (11.1); take EF=A, EG=B

d This proposition has been proved in a variety of ways by Ozanam, Tacquet, Sturmius, Ludlam, Mole, and others; it supplies the foundation for computing the tables of sines, tangents, &c. on which the practice of Trigonometry chiefly depends, and was considered by Pythagoras of such prime importance, that (as we are told) he offered a hecatomb, or sacrifice of 100 oxen, to the gods for inspiring him with the discovery of so remarkable and useful a property.

(3.1), join FG, make EL=FG, EH=C, join HL, take EN=

C2 = ) A2 + B2 +C2; and because EN=LH, and EM=D, ': MN2 (ENEM)2=LH2 + D2=) Æ2 + B2 + C2+D2, which was to be shewn; and in the same manner any number of squares may be added together, that is, a square may be found equal to their sum.

158. To find a square equal to the difference of the squares of two given unequal straight lines.

Let A and B be two unequal straight lines, whereof A is

distance DC describe the circle CKF, from E draw EF perpendicular to CH (11.1), and join DF; EF will be the side of the square required.

2

Because FD=(DC=) A, DE=B, and DEF is a right angle, *. (47. 1.) FD2=(DE2+EF)2=) B+EF, that is A2=B2+ EF; take B from each of these equals, and A-B2=EF}2 ; that is, EF is the side of the square, which is the difference required.