represents 30 superficial units, or 30 of those squares, the side of each of which is equal to unity. This is illustrated by figure 102. In geometry, the product of two lines often signifies the same thing as their rectangle, and this expression is introduced into arithmetic to denote the product of two unequal numbers, as that of square is used to express the product of a number by itself. The squares of the numbers 1, 2, 3, &c., are 1, 4, 9, &c. Thus a double line gives a quadruple square (fig. 103), a triple Fig. 103. line a square nine times as great, and so on. THEOREM. 174. The area of any parallelogram is equal to the product of its base by its altitude. Demonstration. The parallelogram ABCD (fig. 97) is equiva- Fig. 97. lent to the rectangle ABEF, which has the same base AB and the same altitude BE (167); but this last has for its measure AB × BE (173); therefore AB × BE is equal to the area of the parallelogram ABCD. 175. Corollary. Parallelograms of the same base are to each other as their altitudes, and parallelograms of the same altitude are to each other as their bases; for, A, B, C, being any three magnitudes whatever, we have generally A× C: Bx C::A: B. THEOREM. 176. The area of a triangle is equal to the product of its base by half of its altitude. Demonstration. The triangle ABC (fig. 104) is half of the Fig. 164. parallelogram ABCE, which has the same base BC and the same altitude AD (16%); now the area of the parallelogram = BC× AD (174); therefore the area of the triangle = BC × AD, or BC × 1 AD. 177. Corollary. Two triangles of the same altitude are to each other as their bases, and two triangles of the same base. are to each other as their altitudes. THEOREM. 178. The area of a trapezoid ABCD (fig. 105) is equal to the Fig. 105. product of its altitude EF by half of the sum of its parallel sides AB, CD. Demonstration. Through the point I, the middle of the side CB, draw KL parallel to the opposite side AD, and produce DC till it meet KL in K. = In the triangles IBL, ICK, the side IB IC, by construction; the angle LIB = CIK, and the angle IBL = ICK, since CK and BL are parallel (67); therefore these triangles are equal (38), and the trapezoid ABCD is equivalent to the parallelogram ADKL, and has for its measure EF × AL. But AL = DK; and, since the triangle IBL is equal to the triangle KCI, the side BLCK; therefore AB+ CD=AL + DK = 2AL; thus AL is half the sum of the sides AB, CD; and consequently the area of the trapezoid ABCD is equal to the product of the altitude EF by half the sum of the sides AB, CD, which may be expressed in this manner; ABCD = EFX ᎯᏴ (AB+ CD). 2 179. Scholium. If through the point I, the middle of BC, IH be drawn parallel to the base AB, the point H will also be the middle of AD; for the figure AHIL is a parallelogram, as well as DHIK, since the opposite sides are parallel; we have therefore AH IL, and DH = IK; but IL = IK, because the triangles BIL, CIK, are equal; therefore AH = DH. = It may be remarked, that the line HI AL=" AB+ CD ; 2 therefore the area of the trapezoid may be expressed also by EF × H1; that is, it is equal to the product of the altitude of the trapezoid by the line joining the middle points of the sides which are not parallel. THEOREM. Fig. 106. the upon 180. If a line AC (fig. 106) is divided into two parts AB, BC, described square the whole line AC will contain the square described upon the part AB, plus the square described upon the other part BC, plus twice the rectangle contained by the two parts AB, BC; which may be thus expressed, AC or (AB+ BC) = AB + BC + 2 AB × BC. Demonstration. Construct the square ACDE, take AF = AB, draw FG parallel to AC, and BH parallel to AE. = The square ACDE is divided into four parts; the first ABIF is the square described upon AB, since AF was taken equal to AB; the second IGDH is the square described upon BC; for, since ACAE, and AB AF, the difference AC-ABAE - AF, which gives BC = EF; but, on account of the parallels, IG =BC, and DG EF, therefore HIGD is equal to the square described upon BC. These two parts being taken from the whole square, there remain the two rectangles BCGI, EFIH, which have each for their measure AB × BC; therefore the square described upon AC, &c. = 181. Scholium. This proposition corresponds to that given in algebra for the formation of the square of a binomial, which is thus expressed, (a + b)2 = a2 +2ab+b2. THEOREM. 182. If the line AC (fig. 107) is the difference of two lines AB, Fig. 107. BC, the square described upon AC will contain the square of AB, plus the square of BC, minus twice the rectangle contained by AB = and BC; that is, AC or (AB-BC)= AB+BC — 2 AB × BC. Demonstration. Construct the square ABIF, take AE = AC, draw CG parallel to BI, HK parallel to AB, and finish the square EFLK. The two rectangles CBIG, GLKD, have each for their measure AB × BC; if we subtract them from the whole figure ABILKEA, which has for its value AB + BC it is evident, that there will remain the square ACDE; therefore, if the line AC, &c. 183. Scholium. This proposition answers to the algebraic formula (ab)2 a 2 + b2 — 2 a b. =a 2 +b2 THEOREM. 184. The rectangle contained by the sum and difference of two lines is equal to the difference of their squares; that is, (AB+ BC) × (AB — BC) = AB — BC (fig. 108). Demonstration. Construct upon AB and AC the squares ABIF, ACDE; produce AB making BK = BC, and complete the rectangle AKLE. Fig. 108. Fig. 109. The base AK of the rectangle is the sum of the two lines AB, BC, its altitude AE is the difference of these lines; therefore the rectangle AKLE = (AB + BC) × (AB — BC). But this same rectangle is composed of two parts ABHE + BHLK, and the part BHLK is equal to the rectangle EDGF, for BH = DE, and BK = EF; consequently AKLE=ABHE+EDGF. Now these two parts form the square ABIF, minus the square DHIG which is the square described upon BC; therefore -2 (AB+BC) × (AB — BC) = AB—BC. 185. Scholium. This proposition agrees with the algebraic formula (a + b) x (a—b) = (a2— b2) (Alg. 34). THEOREM. 186. The square described upon the hypothenuse of a right-angled triangle is equal to the sum of the squares described upon the two other sides. Demonstration. Let ABC (fig. 109) be a triangle right-angled at A. Having constructed squares upon the three sides, let fall, from the right angle upon the hypothenuse, the perpendicular AD, which produce to E, and draw the diagonals AF, CH. The angle ABF is composed of the angle ABC plus the right angle CBF; and the angle HBC is composed of the same angle ABC plus the right angle ABH; hence the angle ABF = HBC. But AB = BH, being sides of the same square; and BF = BC, for the same reason; consequently the triangles ABF, HBC, have two sides and the included angle of the one respectively equal to two sides and the included angle of the other; they are therefore equal (36). The triangle ABF is half of the rectangle BEt, which has the same base BF and the same altitude BD (169). Also the triangle HBC is half of the square AH; for, the angle BAC being a right angle as well as BAL, AC and AL are in the same straight line parallel to HB; hence the triangle HBC and the square AH have the same base BH, and the same altitude AB; therefore the triangle is half of the square. † An abridged expression for BDEF. It has already been proved, that the triangle ABF is equal to the triangle HBC; consequently the rectangle BDEF, double of the triangle ABF, is equivalent to the square AH, double of the triangle HBC. It may be demonstrated, in the same manner, that the rectangle CDEG is equivalent to the square AI; but the two rectangles BDEF, CDEG, taken together, make the square BCGF; therefore the square BCGF, described upon the hypothenuse, is equal to the sum of the squares ABHL, ACIK, des BC= cribed the two other sides; or, BC = AB+ AC. -2 the square of the other side; or AB=BC — AC. 188. Corollary 1. Let ABCD (fig. 118) be a square, AC its Fig. 119. diagonal; the triangle ABC being right-angled and isosceles, we 2 -2 = 2 -2 have AC AB+BC=2AB; therefore the square described upon the diagonal AC is double of the square described upon the side AB. This property may be rendered sensible by drawing, through the points A and C, parallels to BD, and through the points B and D, parallels to AC; a new square EFGH is thus formed which is the square of AC. It is manifest that EFGH contains eight triangles, each of which is equal to ABE, and that ABCD contains four of them; therefore the square EFGH is double of ABCD. --2 --2 Since AC: AB::2:1, we have, by extracting the square root, AC: AB::2:1; therefore the diagonal of a square is incommensurable with its side (Alg. 99). This will be more fully developed hereafter. 189. Corollary III. It has been demonstrated, that the square AH (fig. 109) is equivalent to the rectangle BDEF; now, on Fig. 109: account of the common altitude BF, the square BCGF is to the rectangle BDEF as the base BC is to the base BD; therefore or, the square of the hypothenuse is to the square of one of the sides of the right angle, as the hypothenuse is to the segment adjacent to this side. We give the name of segment to that part of the hypothenuse cut off by the perpendicular let fall from the right angle; thus BD is the segment adjacent to the side AB, and DC the segment adjacent to the side AC. We have likewise |