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.

  300 square units

Step-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 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