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

let ABC be the isosceles triangle, the AD be the altitude 

Let AB = AC = x then BC= 32-2x       [because parameter = 2 (side) + Base] 

since in an isoceles triange the altitude bisects the base so 

BD = DC = 16-x 

In a triangle ADC,  $AC2=AD2+DC2$ 

$x2=82+16-x2$  $\Rightarrow x=10$ 

BC = 32-2x = 32-20 = 12 cm 

Hence, required area =  $\frac{1}{2}*BC*AD$=  $\frac{1}{2}*12*10$ = 60 sq cm