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, A C 2 = A D 2 + D C 2 x 2 = 8 2 + 16 - x 2 ? x = 10 BC = 32-2x = 32-20 = 12 cm Hence, required area = 1 2 * B C * A D = 1 2 * 12 * 10 = 60 sq cm
EXAMIANS