Question

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.

Answer 1

300 square units

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 AMC

Rearranging, 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 = 15

The area of interest is 20x, so is ...

A = 20·x = 20·15 = 300 . . . . square units