Find the area of an isosceles triangle having perimeter 32cm and
length of the unequal side is 12 cm.
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Let △ABC be an isoceles △, such that AB=BC perimeter =a+12+a=2a+12
Also, 2a+12=32cm⇒a=232−12=10cm
∴ar△ABC=s(s−a)(s−b)(s−c),s=2a+b+c=232=16cm
=16(16−10)(16−10)(16−12)=16×6×6×4=4×6×2
=48cm2
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