Question

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 ___

Answer 1

Formula for in-radius of a right angled triangle = $\frac{a+b-c}{2}$
Since the triangle is an isosceles triangle, both the legs are equal.
So, $\frac{a+a-a\sqrt{2}}{2}=4(\sqrt{2}-1)\Rightarrow \sqrt{2}a(\sqrt{2}-1)=8(\sqrt{2}-1)\Rightarrow a=4\sqrt{2}$
Therefore, area of the triangle ABC = $\frac{1}{2}\times a^2=\frac{1}{2}\times (4\sqrt{2}{)}^2=16$