Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB. If the perpendicular distance of P from each of AB,BC,and CA is $$4(\sqrt{2}-1)$$ cm,then the area, in sq cm, of the triangle ABC is

Let the length of non-hypotenuse sides of the right angled triangle be $$a$$. Then the hypotenuse h = $$\sqrt{2}a$$
P is equidistant from all the side of the triangle. Hence P is the incenter and the perpendicular distance is the inradius.
In a right angled triangle, inradius = $$\frac{a + b - h}{2}$$
=> $$\frac{a + a - \sqrt{2}a}{2} = 4(\sqrt{2}-1)$$
=> $$ \sqrt{2}a( \sqrt{2} - 1) = 8(\sqrt{2} -1)$$
=> $$ a = 4\sqrt{2}$$
Area of the triangle = $$\frac{1}{2}a^2$$ = 16 sq cm