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Inradius and Area of Right Isosceles Triangle

CAT 2017 Slot 2 · Quantitative Ability · Medium · Geometry

This is a medium Quantitative Ability question from the CAT 2017 Slot 2 paper. It tests Geometry. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Let PP be an interior point of a right-angled isosceles triangle ABCABC with hypotenuse ABAB. If the perpendicular distance of PP from each of AB,BC,AB, BC, and CACA is 4(2−1) cm4(\sqrt{2} - 1)\text{ cm}, then the area, in sq. cm, of the triangle ABCABC is

Answer

16

Explanation

Since PP is equidistant from all three sides, PP is the incenter of △ABC\triangle ABC, and the given perpendicular distance is the inradius r=4(2−1) cmr = 4(\sqrt{2} - 1)\text{ cm}.

For a right-angled isosceles triangle with legs a,aa, a and hypotenuse a2a\sqrt{2}: r=a+a−a22=a(2−2)2=a(1−22)=a(2−1)2r = \frac{a + a - a\sqrt{2}}{2} = \frac{a(2 - \sqrt{2})}{2} = a\left(1 - \frac{\sqrt{2}}{2}\right) = \frac{a(\sqrt{2} - 1)}{\sqrt{2}}

Equating to given rr: a(2−1)2=4(2−1)  ⟹  a=42 cm\frac{a(\sqrt{2} - 1)}{\sqrt{2}} = 4(\sqrt{2} - 1) \implies a = 4\sqrt{2}\text{ cm}

Area of △ABC=12a2=12(42)2=12(32)=16 sq. cm\triangle ABC = \frac{1}{2} a^2 = \frac{1}{2} (4\sqrt{2})^2 = \frac{1}{2} (32) = 16\text{ sq. cm}.

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