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