Let the two equal sides (base and height) = $$x$$ cm and hypotenuse = $$y$$ cm => Perimeter = $$2x + y = 2p$$ => $$y = 2p - 2x$$ ----------Eqn(i) Since, it is a right angled triangle, => $$x^2 + x^2 = y^2$$ => $$2x^2 = y^2$$ From (i), => $$2x^2 = 4 (p - x)^2$$ => $$2x^2 = 4p^2 + 4x^2 - 8px$$ => $$2x^2 - 8px + 4p^2 = 0$$ => $$x^2 - 4px + 2p^2 = 0$$ => $$x = frac{4p pm sqrt{16p^2 - 8p^2}}{2}$$ => $$x = frac{4p pm 2p sqrt{2}}{2}$$ => $$x = 2p pm sqrt{2} p$$ $$ecause$$ Perimeter = $$2p$$, => $$x
eq (2p + sqrt{2}p)$$ => $$x = p (2 - sqrt{2})$$ $$ herefore$$ Area of triangle = $$frac{1}{2} imes x imes x = frac{1}{2} imes x^2$$ = $$frac{1}{2} imes [p (2 - sqrt{2})]^2$$  = $$frac{1}{2} imes p^2 imes (4 + 2 - 4 sqrt{2})$$ = $$frac{1}{2} imes p^2 imes (6 - 4 sqrt{2}) = p^2 (3 - 2 sqrt{2}) cm^2$$