Fig. 142. Fig. 143. Fig. 144. Fig. 145. PROBLEM. 242. Through a given point A (fig. 142) in a given angle BCD, to draw a line BD in such a manner that the parts AB, AD, comprehended between the point A and the two sides of the angle shall be equal. Solution. Through the point A draw AE parallel to CD, take BE = CE, and through the points B and A draw BAD, which will be the line required. For AE being parallel = = to CD, BE: EC :: BA: AD, but AD. PROBLEM. 243. To make a square equivalent to a given parallelogram, or to a given triangle. Solution. 1. Let ABCD (fig. 143) be the given parallelogram, AB its base, and DE its altitude; between AB and DE find a mean proportional XY (239); the square described upon XY will be equivalent to the parallelogram ABCD. For, by construction, AB: XY:: XY: DE; hence --2 XY = AB × DE ; -2 but AB × DE is the measure of the parallelogram, and XY is that of the square, therefore they are equivalent. 2. Let ABC (fig. 144) be the given triangle, BC its base, and AD its altitude; find a mean proportional between BC and half of AD, and let XY be this mean proportional; the square described upon XY will be equivalent to the triangle ABC. For, since BC: XY:: XY: fore the square described upon ABC. -2 AD, XY = BC × 1 AD; there- PROBLEM.. 244. Upon a given line AD (fig. 145) to construct a rectangle ADEX equivalent to a given rectangle ABFC. Solution. Find a fourth proportional to the three lines AD, AB, AC (137), and let AX be this fourth proportional; the rectangle contained by AD and AX will be equivalent to the rectangle ABFC. For, since AD: AB:: AC: AX, AD × AX = ABX AC; therefore the rectangle ADEX is equivalent to the rectangle ABFC. PROBLEM. 245. To find in lines the ratio of the rectangle of two given lines A and B (fig. 148) to the rectangle of two given lines C and D. Solution. Let X be a fourth proportional to the three given lines B, C, D; the ratio of the two lines A and X will be equal to that of the two rectangles Ax B, C × D. For, since B: C:: D: X, CX D=BxX; therefore A B C D :: A× B: B x X :: A : X. : 246. Corollary. In order to obtain the ratio of the squares described upon two lines A and C, find a third proportional Xto the lines A and C, so that we may have the proportion A: C: C: X; Fig. 148. PROBLEM. 247. To find in lines the ratio of the product of three given lines A, B, C (fig. 149), to the product of three given lines P, Q, R. Solution. Find a fourth proportional X to the three given lincs P, A, B; and a fourth proportional Y to the three given lines C, Q, R. The two lines X and Y will be to each other as the products A×B× C, P× Q× R. For, since P: A:: B: X, A× B=Px X; and, by multiplying each of these by C, we shall have AxBX C=CxPx X. In like manner, since C:QRY, QxR=Cx Y; and, by multiplying each of these by P, we shall have PXQXR=PxCxY; therefore the product AX BX C: the product P× Q× R:: C× P×X: P× CxY:: X: Y. Fig. 149. PROBLEM. 248. To make a triangle equivalent to a given polygon. Solution. Let ABCDE (fig. 146) be the given polygon. Fig. 146. Draw the diagonal CE, which cuts off the triangle CDE; through the point D draw DF parallel to CE to meet AE pro Geom. 10 duced; join CF, and the polygon ABCDE will be equivalent to the polygon ABCF, which has one side less. For the triangles CDE, CFE, have the common base CE, they are also of the same altitude, for their vertices D, F, are in a line DF parallel to the base; therefore the triangles are equivalent. Adding to each of these the figure ABCE and we shall have the polygon ABCDE equivalent to the polygon ABCF. We can in like manner cut off the angle B by substituting for the triangle ABC the equivalent triangle AGC, and then the pentagon ABCDE will be transformed into an equivalent trian gle GCF. The same process may be applied to any other figure; for, by making the number of sides one less at each step, we shall at length arrive at an equivalent triangle. 249. Scholium. As we have already seen, that a triangle may be transformed into an equivalent square (243), we may accordingly find a square equivalent to any given rectilineal figure; this is called squaring the rectilineal figure, or finding the quadrature of it. The problem of the quadrature of the circle consists in finding a square equivalent to a circle whose diameter is given. Fig. 147. PROBLEM. 250. To make a square which shall be equal to the sum or the difference of two given squares. Solution. Let A and B (fig. 147) be the sides of the given squares. 1. If it is proposed to find a square equal to the sum of these squares, draw the two indefinite lines ED, EF, at right angles to each other; take ED = A and EG = B; join DG, and DG will be the side of the square sought. For the triangle DEG being right-angled, the square described upon DG will be equal to the sum of the squares described upon ED and EG. 2. If it is proposed to find a square equal to the difference of the given squares, form in like manner a right angle FEH, take GE equal to the smaller of the sides A and B; from the point G, as a centre, and with a radius GH equal to the other side, describe an arc cutting EH in H; the square described upon EH will be equal to the difference of the squares described upon the lines A and B. For in the right-angled triangle GEH the hypothenuse GHA, and the side GE B; therefore the square described upon EH is equal to the difference of the squares described upon the given sides A and B. 251. Scholium. We can thus find a square equal to the sum of any number of squares; for the construction by which two are reduced to one, may be used to reduce three to two and these two to one, and so of a larger number. Also a similar method may be employed when certain given squares are to be subtracted from others. PROBLEM. 252. To construct a square which shall be to a given square ABCD (fig. 150), as the line M is to the line N. Solution. On the indefinite line EG take EF=M, and FG=N; on EG, as a diameter, describe a semicircle, and at the point F erect upon the diameter the perpendicular FH. From the point H draw the chords HG, HE, which produce indefinitely; on the first take HK equal to the side AB of the given square, and through the point K draw KI parallel to EG; HI will be the side of the square sought. For, on account of the parallels KI, GE, hence HI: HK :: HE: HG; 2 -2 2 HI : HK :: HE: HG (v). HỈ:HK::HẺ:HỔ But, in the right-angled triangle EHG, the square upon HI: the square upon AB:: M: N. Fig. 150 PROBLEM. 253. Upon a side FG (fig. 129), homologous to AB, to describe a Fig. 129. polygon similar to a given polygon ABCDE. Solution. In the given polygon draw the diagonals AC, AD. At the point F make the angle GFH = BAC, and at the point G the angle FGH = ABC; the lines FH, GH, will cut each other in H, and the triangle FGH will be similar to ABC. Likewise upon FH, homologous to AC, construct the triangle FIH similar to ADC, and upon FI, homologous to AD, construct the triangle FIK similar to ADE. The polygon FGHIK will be similar to ABCDE. For these two polygons are composed of the same number of triangles, which are similar to each other and similarly disposed (219). PROBLEM. 254. Two similar figures being given, to construct a similar figure which shall be equal to their sum or their difference. Solution. Let A and B be two homologous sides of the given figures, find a square equal to the sum or the difference of the squares described upon A and B ; let X be the side of this square, X will be, in the figure sought, the side homologous to A and B in the given figures. The figure may then be constructed by the preceding problem. For similar figures are as the square of their homologous sides; but the square of the side X is equal to the sum or the difference of the squares described upon the homologous sides A and B; therefore the figure described upon the side X is equal to the sum or the difference of the similar figures described upon the sides A and B. PROBLEM. 255. To construct a figure similar to a given figure, and which shall be to this figure in the given ratio of M to N. Solution. Let A be a side of the given figure, and X the homologous side of the figure sought; the square of X must be to the square of A, as M is to N (221); X then may be found by art. 252; and, knowing X, we may finish the problem by art. 253. PROBLEM. Fig. 151. 256. To construct a figure similar to the figure P (fig. 151) and equivalent to the figure Q. |
