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 $4(\sqrt{2} - 1)	ext{ cm}$, then the area, in sq. cm, of the triangle $ABC$ is

Accepted Answer

16. Since $P$ is equidistant from all three sides, $P$ is the incenter of $	riangle ABC$, and the given perpendicular distance is the inradius $r = 4(\sqrt{2} - 1)	ext{ cm}$. For a right-angled isosceles triangle with legs $a, a$ and hypotenuse $a\sqrt{2}$: $$r = \frac{a + a - a\sqrt{2}}{2} = \frac{a(2 - \sqrt{2})}{2} = a\left(1 - \frac{\sqrt{2}}{2}ight) = \frac{a(\sqrt{2} - 1)}{\sqrt{2}}$$ Equating to given $r$: $$\frac{a(\sqrt{2} - 1)}{\sqrt{2}} = 4(\sqrt{2} - 1) \implies a = 4\sqrt{2}	ext{ cm}$$ Area of $	riangle ABC = \frac{1}{2} a^2 = \frac{1}{2} (4\sqrt{2})^2 = \frac{1}{2} (32) = 16	ext{ sq. cm}$.

Inradius and Area of Right Isosceles Triangle

Explanation

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