The perimeter of an isosceles right angled triangle is 2 p cm. Its area is

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), => $$2x^2 = 4 (p - x)^2$$

=> $$2x^2 = 4p^2 + 4x^2 - 8px$$

=> $$2x^2 - 8px + 4p^2 = 0$$

=> $$x^2 - 4px + 2p^2 = 0$$

=> $$x = \frac{4p \pm \sqrt{16p^2 - 8p^2}}{2}$$

=> $$x = \frac{4p \pm 2p \sqrt{2}}{2}$$

=> $$x = 2p \pm \sqrt{2} p$$

$$\because$$ Perimeter = $$2p$$, => $$x \neq (2p + \sqrt{2}p)$$

=> $$x = p (2 - \sqrt{2})$$

$$\therefore$$ Area of triangle = $$\frac{1}{2} \times x \times x = \frac{1}{2} \times x^2$$

= $$\frac{1}{2} \times [p (2 - \sqrt{2})]^2$$ 

= $$\frac{1}{2} \times p^2 \times (4 + 2 - 4 \sqrt{2})$$

= $$\frac{1}{2} \times p^2 \times (6 - 4 \sqrt{2}) = p^2 (3 - 2 \sqrt{2}) cm^2$$