nor PROBLEMS OF CONSTRUCTION. PROPOSITION XI.-PROBLEM. 28. To construct a triangle equivalent to a given polygon. Let ABCDEF be the given polygon. Take any three consecutive vertices, as A, B, C, and draw the diagonal AC. Through B draw BP parallel to AC meeting DC produced in P; join AP. P F E The triangles APC, ABC, have the same base AC; and since their vertices, P and B, lie on the same straight line BP parallel to AC, they also have the same altitude; therefore they are equivalent. Therefore, the pentagon APDEF is equivalent to the hexagon ABCDEF. Now, taking any three consecutive vertices of this pentagon, we shall, by a precisely similar construction, find a quadrilateral of the same area; and, finally, by a similar operation upon the quadrilateral, we shall find a triangle of the same area. Thus, whatever the number of the sides of the given polygon, a series of successive steps, each step reducing the number of sides by one, will give a series of polygons of equal areas, terminating in a triangle. PROPOSITION XII.-PROBLEM. 29. To construct a square equivalent to a given parallelogram or to a given triangle. 1st. Let AC be a given parallelogram; k its base, and h its altitude. Find a mean proportional x between h and k, by (III. 72). The square constructed upon x will be equivalent to the parallelogram, since x2=hXk. 2d. Let ABC be a given triangle; a its base and h its altitude. Find a mean proportional x between ɑ and h; the square constructed upon x will be equivalent to the triangle, since x2 a Xh 13 ah. = B A Α h B a 30. Scholium. By means of this problem and the preceding, a square can be found equivalent to any given polygon PROPOSITION XIII.-PROBLEM. m nt 31. To construct a square equivalent to the sum of two or more given squares, or to the difference of two given squares. 1st. Let m, n, p, q, be the sides of given squares. Draw AB=m, and BC= n, perpendicular to each other at B; join AC. Then (25), AC2 m2 + n2. 2 2 = Draw CD = p, perpendicular to AC, and join AD. Then AD = AC' + p2= m2 + n2 + p2. Draw DE = q perpendicular to AD, and join AE. Then, AE' = AD' + q2 = m2 + n2 + p'q'; therefore, the square constructed upon AE will be equivalent to the sum of the squares 2 constructed upon m, n, p, q. 2 201 C p D B In this manner may the areas of any number of given squares be added. =m, 2d. Construct a right angle ABC, and lay off BA = n. With the centre A and a radius describe an arc cutting BC in C. Then BC2 = AC2 — AB2 n'; therefore, the square con A m n B structed upon BC will be equivalent to the difference of the squares constructed upon m and n. 32. Scholium I. By means of this problem, together with the preceding ones, a square can be found equivalent to the sum of any number of given polygons; or to the difference of any two given polygons. 33. Scholium II. If m, n, p, q, in the preceding problem are homologous sides of given similar polygons, the line AE in the first figure is the homologous side of a similar polygon equivalent to the sum of the given polygons (27). And the line BC, in the second figure, is the homologous side of a similar polygon, equivalent to the difference of two given similar polygons. One side of a polygon, similar to a given polygon, being known, the polygon may be constructed by (III. 80). PROPOSITION XIV.-PROBLEM. 34. Upon a given straight line to construct a rectangle equivalent to a given rectangle. Let k' be the given straight line, and AC the given rectangle whose base is k and altitude h. Find a fourth proportional h', to k', k and h, by (III. 70). Then, the rectangle constructed upon the base k' with the altitude h' is equivalent to AC; for, by the construction, k': k = h: h', whence, k' h' k× h (7). = h B A h' C PROPOSITION XV.-PROBLEM. 35. To construct a rectangle, having given its area and the sum of two adjacent sides. Let MN be equal to the given sum of the adjacent sides of the required rectangle; and let the given area be that of the square whose side is AB. = M B RN Upon MN as a diameter describe a semicircle. At M erect MP AB perpendicular to MN, and draw PQ parallel to MN, intersecting the circumference in Q. From let fall QR perpendicular to MN; then, MR and RN are the base and altitude of the required rectangle. For, hy (III. 47), MR X RN = QR2 = PM == AB2 PROPOSITION XVI.-PROBLEM. 36. To construct a rectangle, having given its area and the difference of two adjacent sides. Let MN be equal to the given difference of the adjacent sides of the required rectangle; and let the given area be that of the square described on AB. Upon MN as a diameter describe a circle. R At M draw the tangent MP P, draw the secant PQR through the centre of the circle; then, PR and PQ are the base and altitude of the required rectangle. For, by (III. 59), PR × PQ : PM' = AB', and the difference of PR and PQ is QR MN. N PROPOSITION XVII.-PROBLEM. 37. To find two straight lines in the ratio of the areas of two given polygons. Let squares be found equal in area to the given polygons, respectively (30). Upon the sides of the right angle A CB, take CA and CB equal to the sides of these squares, join AB and A D B let fall CD perpendicular to AB. Then, by (III. 46), we have AD: DB CA': CB'; therefore, AD, DB, are in the ratio of the areas of the given polygons. PROPOSITION XVIII.-PROBLEM. 38. To find a square which shall be to a given square in the ratio of two given straight lines. Let AB' be the given square, and M : N the given ratio. Upon an indefinite straight line CL, lay M, DE N; upon CE as a G H off CD the perpendicular DF cutting the circum ference in F; join FC, FE; lay off FH = AB, and through H draw HG parallel to EC; then, FG is the side of the required square. For, by (III. 15), we have But FH = AB, therefore the square constructed upon FG is to the square upon AB in the ratio M: N. PROPOSITION XIX.-PROBLEM. 39. To construct a polygon similar to a given polygon and whose area shall be in a given ratio to that of the given polygon. Let P be the given polygon, and let a be one of its sides; let M: N be the given ratio. Find, by the preceding problem, the side a' of a square which shall be to a' in the ratio M: N; upon a', as a homologous side to a, construct the polygon P' similar to P (III. 80); this will be the polygon required. For, the polygons being similar, their areas are in the ratio a': a2, or M: N, as required. M N P a PI a PROPOSITION XX.-PROBLEM. 40. To construct a polygon similar to a given polygon P and equiva lent to a given polygon Q. Find M and N, the sides of squares respectively equal in area to P and Q, (30). Let a be any side of P, and find a fourth proportional a' to M, N and a: upon a', as a homologous side to a, construct the polygon P' similar to P; this will be the required polygon. For, by construction, P a N M P a' |