Fig. 97. Fig. 98. Eig. 99. Hence the triangles DAF, CBE, have the three sides of the one equal to the sides of the other, each to each; they are therefore equal (43). But, if from the quadrilateral ABED the triangle ADF be taken, there will remain the parallelogram ABEF; and, if from the same quadrilateral ABED the triangle CBE, equal to the former, be taken, there will remain the parallelogram ABCD; therefore the two parallelograms ABCD, ABEF, which have the same base and the same altitude, are equivalent. 167. Corollary. Every parallelogram ABCD (fig. 97) is equivalent to a rectangle of the same base and altitude. THEOREM. 168. Every triangle ABC (fig. 98) is half of a parallelogram ABCD of the same base and altitude. Demonstration. The triangles ABC, ACD, are equal (87), therefore each is half of the parallelogram ABCD. 169. Corollary 1. A triangle ABC is half of a rectangle BCEF of the same base BC and the same altitude AO; for the rectangle BCEF is equivalent to the parallelogram ABCD (167). 170. Corollary 11. All triangles, which have equal bases and equal altitudes, are equivalent. THEOREM. 171. Two rectangles which have the same altitude, are to each other as their bases. Demonstration. Let ABCD, AEFD (fig. 99), be two rectangles, which have a common altitude AD; they are to each other as their bases AB, AE. Let us suppose, in the first place, that the bases AB, AE, are commensurable, and that they are to each other, as the numbers 7 and 4, for example; if we divide AB into 7 equal parts, AE will contain four of these parts; erect at each point of division, a perpendicular to the base, we shall thus form seven partial rectangles which will be equal to each other, since they will have the same base and the same altitude (166). The rectangle ABCD will contain seven partial rectangles, while AEFD will contain four; therefore the rectangle ABCD is to the rectangle AEFD, as 7 is to 4, or as AB is to AE. The same rea soning may be applied to any other ratio beside that of 7 to 4; nence, whatever be the ratio, provided it is commensurable, we have ABCD:AEFD::AB: AE. Let us suppose, in the second place, that the bases AB, AE (fig. 100), are incommensurable; we shall have notwithstanding Fig. 100. ABCD:AEFD::AB:AE. For, if this proportion be not true, the three first terms remaining the same, the fourth will be greater or less than AE. Let us suppose that it is greater, and that we have ABCD: AEFD::AB: AO. Divide the line AB into equal parts smaller than EO, and there will be at least one point of division I between E and 0: at this point erect the perpendicular IK; the bases AB, AI, will be commensurable, and we shall have, according to what has just been demonstrated, ABCD: AIKD : : AB : AI. But we have, by hypothesis, ABCD: AEFD::AB: AO. In these two proportions the antecedents are equal, therefore the consequents are proportional (1); that is AIKD : AEFD::AI : AO. Now AO is greater than AI; it is necessary then, in order that this proportion may take place, that the rectangle AEFD should be greater than AIKD; but it is less; therefore the proportion is impossible, and ABCD cannot be to AEFD, as AB is to a line greater than AE. By a process entirely similar it may be shown, that the fourth term of the proportion cannot be smaller than AE; consequently it is equal to AE. Whatever therefore be the ratio of the bases, two rectangles ABCD, AEFD, of the same altitude, are to each other as their bases AB, AE. THEOREM. 172. Any two rectangles ABCD, AEGF (fig. 101), are to each Fig. 101. other, as the products of their bases by their altitudes, that is, ABCD: AEGF :: AB × AD: AE × AF. Fig. 102. Demonstration. Having disposed the two rectangles in such a manner, that the angles at A shall be opposite to each other, produce the sides GE, CD, till they meet in H; the two rectangles ABCD, AEHD, have the same altitude AD; they are consequently to each other as their bases AB, AE. Likewise the two rectangles AEHD, AEGF, have the same altitude AE; these are therefore to each other as their bases AD, AF. We have thus the two proportions ABCD: AEHD::AB: AE, AEHD: AEGF ::AD: AF. Multiplying these proportions in order and observing, that the connecting term AEHD may be omitted, being a multiplier common to the antecedent and consequent, we have ABCD: AEGF :: AB × AD: AE × AF. 173. Scholium. We may take for the measure of a rectangle the product of its base by its altitude, provided that, by this product, we understand that of two numbers which are the number of linear units contained in the base, and the number of linear units contained in the altitude. This measure, however, is not absolute, but relative; it supposes that we estimate, in a similar manner, another rectangle by measuring its sides by the same linear unit; we obtain thus a second product, and the ratio of these two products is equal to that of the rectangles, conformably to the proposition, which has just been demonstrated. If, for example, the base of a rectangle A (fig. 102) be three units and its altitude ten, the rectangle would be represented by the number 3 x 10, or 30, a number which, thus disconnected, has no meaning; but, if we have a second rectangle B, whose base is twelve and altitude seven units, this rectangle will be represented by the number 7 × 12, or 84. Whence the two rectangles A and B are to each other, as 30 to 84. If therefore it is agreed to take the rectangle A, as the unit of measure for surfaces, the rectangle B will have for its absolute measure, that is, it will be equal to superficial units. The more common and simple method is to take the square as the unit of surface; and that square has been preferred, whose side is the unit of length; the measure therefore, which we have regarded as simply relative, becomes absolute. The number 30, for example, by which we have measured the rectangle A, 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 Ax 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. 104. 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 × 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 1B 1C, 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 EFX AL. But AL= DK; and, since the triangle IBL is equal to the triangle K CI, the side BL = CK; 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 = EF × (AB+ CD). 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. AB+ CD ; 2 It may be remarked, that the line HI AL = = therefore the area of the trapezoid may be expressed also by EF × HI; 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. Fig. 106. the THEOREM. 180. If a line AC (fig. 106) is divided into two parts AB, BC, square described upon 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, -2 2 -2 2 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. |