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=>x2=(82)+(16-x)2
=>32x = 320 => x = 10
BC = (32- 2x) = (32 - 20) cm = 12 cm.
Hence, required area = ((1/2)x*BC * AD) = ((1/2)*12 *10)cm2 = 60 cm2
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