in an isosceles triangle ABC, AB=AC and the altitude from A to BC is 20. If the perimeter of the triangle is 80, find the area of the triangle.
Accepted Solution
A:
Answer: 300 square unitsStep-by-step explanation:Let M be the midpoint of BC. Then AM =20 is the altitude. Let x represent the length BM=MC, and let y represent the length AB=AC. Then the perimeter is ... 2x +2y = 80 x +y = 40 . . . . divide by 2 . . . . . [eq A]The Pythagorean theorem tells us ... x^2 + 20^2 = y^2 . . . . . . . y is the hypotenuse of right triangle AMCRearranging, we have ... y^2 -x^2 = 400 (y -x)(y +x) = 400 (y -x)·40 = 400 y -x = 10. . . . . . . . . [eq B]Subtracting [eq B] from [eq A], we find ... (x +y) -(y -x) = (40) -(10) 2x = 30 x = 15The area of interest is 20x, so is ... A = 20·x = 20·15 = 300 . . . . square units