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, $AC2=AD2+DC2$ 

$\Rightarrow x2=82+16-x2\Rightarrow x=10$ 

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

Hence, required area =   $\frac{1}{2}*BC*AD=\frac{1}{2}*12*8=48cm2$