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# CAT 2017 [slot 2] Question with solution 20

Question 54:
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
Explanation:

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

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