The second member of this equation is the difference of two squares, and hence can be factored; that is, AD2 – [2 ab +(a2 + b2 — c2)] [2 ab − (a2 + b2 — c2)] = = 4 a2 (2 ab + a2 + b2 - c2) (2 ab-a2 b2 + c2) 4 a2 [(a + b)2 − c2] [c2 — (a — b)2] 4 a2 [(a + b −c) (a + b + c)] [c − (a − b)] [c+a − b] 4 a2 (a+b−c) (a + b + c) (c − a + b) (c + a − b). But (by 254), Area of ABC = 1⁄2 BC × AD 1. If the sides of a triangle are 13, 14, 15, find the area. 2. In the above, find the radius of the inscribed circle (see 257, Ex. 3). 3. The area of a rhombus is 24 and its side is 5; find the lengths of the diagonals. 4. If the sides of an isosceles triangle are a, a, and b, show that its OAD = OA × DF, OAB = OA × BE; hence show 6. Show that two quadrilaterals are equal when they have the following parts of the one respectively equal to the corresponding parts in the other: I. Four sides and one diagonal. II. Four sides and one angle. III. Two adjacent sides and three angles. IV. Three sides and the two included angles. 7. Construct a square having a given diagonal. 8. Show that the sum of the squares of the sides of a triangle is equal to double the square of the bisector of the base together with double the square of half the base. PROBLEMS IN CONSTRUCTION. PROPOSITION XIII. PROBLEM. 272. To construct a square equivalent to the sum of two given squares. C M N A P Let M and N be the given squares. To construct a square equivalent to their sum. It is known (from 265) that the square upon the hypotenuse -is equivalent to the sum of the squares on the other two sides. Therefore construct a right triangle whose base will be equal to a side of M, say AB, and whose altitude will be equal to a side of N, say AC, then the hypotenuse, say BC, will be a side of the square required. EXERCISES. 1. To construct a square equivalent to the sum of any number of squares. 2. To construct a square equivalent to the difference of two given squares. PROPOSITION XIV. PROBLEM. 273. To construct a square equivalent to a given parallelogram. Let ABCD be the given parallelogram. To construct a square equivalent to ABCD. It is known (from 251) that the area of the parallelogram ABCD = AB × DE, therefore any square to be equivalent to ABCD must have such a side that its square must be equal to AB × DE. It is known (from 200) that when the square of one quantity is equal to the product of two other quantities the former is said to be a mean proportional to the other two. Hence find (by 238) a mean proportional to AB and DE, say FG, then the square on FG will be the required square equivalent to ABCD. EXERCISES. 1. To construct a square equivalent to a given triangle. 2. To construct a square equivalent to the sum of two given triangles. PROPOSITION XV. PROBLEM. 274. Upon a given straight line, to construct a rectangle equivalent to a given rectangle. Let ABCD be the given rectangle, and EF the given line. To construct upon EF as a base a rectangle equivalent to ABCD. The area of the given rectangle is AB X DA, therefore if EF is the base of the required rectangle, its altitude must be such a value that when multiplied by EF the product will be equal to AB × AD; that is, a fourth proportional to EF, AB, and DA. Therefore find (by 238) a fourth proportional to EF, AB, DA; suppose it is HE, that is 275. To construct a rectangle equivalent to a given square, having the sum of its base and altitude equal to a given line. Let M be the given square and AB the given line. To construct a rectangle equivalent to M, having the sum of its base and altitude equal to AB. It is known (from 228) that the perpendicular let fall from any point in the circumference upon the diameter is a mean proportional between the segments into which it divides the diameter. Hence we take AB as the diameter (183) and find a point on the circumference which is as far from the diameter as the side of the square. To do this, erect at A (by 184, Ex. 1) a perpendicular to AB, say AC, through C, draw a line parallel to AB (by 189), say CF; then where CF intersects the circumference, say D, let fall the perpendicular DE. We know (by 228) but DE2 = AE × EB, DE2 M. = |