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The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle?
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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, (AC)^2=(AD)^2+(DC)^2 x^2=8^2+(16−x)^2 ⇒x=10 BC = 32-2x = 32-20 = 12 cm Hence, required area = (1/2)*BC*AD= (1/2)*12*10 = 60 sq cm
By: MIRZA SADDAM HUSSAIN ProfileResourcesReport error
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