The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle.

Let ABC be the isosceles triangle and AD be the altitude 

Let AB = AC = x. Then, BC = (32 - 2x). 

Since, in an isosceles triangle, the altitude bisects the base,
so BD = DC = (16 - x).
In triangle ADC, A C 2 = A D 2 + D C 2 

? x 2 = 8 2 + 16 - x 2 ? x = 10 

BC = (32- 2x) = (32 - 20) cm = 12 cm. 

Hence, required area =   1 2 * B C * A D = 1 2 * 12 * 8 = 48 c m 2