CAT 2017 Slot 2 — QA Question 20

Now since the perpendicular distance of P from each 3 sides is the same, it is the incenter of the triangle ABC.

Let ‘x’ be one of the equal sides of the isosceles right angled triangle ABC.
So the length of the hypotenuse will be √2x.

Area of right triangle ABC = $\frac{1}{2}{\mathrm{x}}^2$

So the Area of triangle ABC can also be expressed as a product of the in radius ‘r’ and semi perimeter ‘s’.

s = $\frac{x+x+\sqrt{2}x}{2}$ = x + $\frac{x}{\sqrt{2}}$

∴ Area = $\left(x+\frac{x}{\sqrt{2}}\right)\left[4\right(\sqrt{2}-1\left)\right]$ = $\frac{x^2}{2}$

⇒ 4√2x - 4x + 4x - 2√2x = $\frac{x^2}{2}$

⇒ 2√2x = $\frac{x^2}{2}$

⇒ x = 4√2

Area of triangle ABC = ½ × 4√2 × 4√2 = 16 sq.units.

Hence, 16.