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Find the area of an isosceles triangle having the base x cm and one side y cm.

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In ∆ABC, AB = x, BC = x, AC = y………………….Given 

Since , and are the sides of an isosceles triangle and s is 

the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a(s-b)(s-c)}\)where s = \(\frac{a+b+c}2\)[Heron’s Formula]

s = \(\frac{x+y+y}2\) = \(\frac{x+2y}2\)

A = \(\sqrt{\frac{x+2y}{2}({\frac{{x}+{2y}}{2}-{x})}({\frac{{x}+{2y}}{2}-{y}})({\frac{{x}+{2y}}{2}-{y}})}\)

A = \(\sqrt{\frac{x+2y}{2}({\frac{{2y}-{x}}{2})}(\frac{x}{2})(\frac{x}{2})}\) = \(\frac{x}{2}\sqrt{{y}^2-\frac{{x}^2}{4}}\) cm2

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