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CAT 2017 Slot 2 — QA Question 20

Geometric CentersEasy

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(√2 - 1) m, then the area, in sq cm, of the triangle ABC is

Answer & solution

Correct answer: 16

Answer: 16

Solution

Easy

Equal perpendicular distances from PP to all three sides mean PP is the incentre, so that distance is the inradius rr. Express the triangle's area two ways — as 12x2\tfrac12 x^2 and as r⋅sr\cdot s — and solve for the leg xx.

1

Set the leg. Let each equal leg be xx; the hypotenuse is 2 x\sqrt2\,x. The inradius equals the given distance.

r=4(2−1)Area =12x2\begin{aligned} &r = 4(\sqrt2 - 1)\\ &\text{Area } = \tfrac12 x^2 \end{aligned}
2

Semiperimeter. Sum the three sides and halve.

s=x+x+2 x2=x+x2=x ⁣(1+12)\begin{aligned} &s = \frac{x + x + \sqrt2\,x}{2} = x + \frac{x}{\sqrt2} = x\!\left(1 + \tfrac{1}{\sqrt2}\right) \end{aligned}
3

Equate the two area forms. Use Area =r s=12x2= r\,s = \tfrac12 x^2 and simplify r ⁣(1+12)r\!\left(1+\tfrac{1}{\sqrt2}\right).

r ⁣(1+12)=4(2−1)⋅2+12⇒ =4(2−1)(2+1)2=4(2−1)2=42=22\begin{aligned} &r\!\left(1 + \tfrac{1}{\sqrt2}\right) = 4(\sqrt2 - 1)\cdot\frac{\sqrt2 + 1}{\sqrt2}\\ &\Rightarrow\ = \frac{4(\sqrt2-1)(\sqrt2+1)}{\sqrt2} = \frac{4(2-1)}{\sqrt2} = \frac{4}{\sqrt2} = 2\sqrt2 \end{aligned}
4

Solve for x. So r s=22 xr\,s = 2\sqrt2\,x, and this equals 12x2\tfrac12 x^2.

22 x=12x2(from step 3)⇒ x=42\begin{aligned} &2\sqrt2\,x = \tfrac12 x^2 \quad\text{(from step 3)}\\ &\Rightarrow\ x = 4\sqrt2 \end{aligned}
5

Area. Plug x=42x = 4\sqrt2 into 12x2\tfrac12 x^2.

Area=12 (42)2=12 (32)=16\begin{aligned} &\text{Area} = \tfrac12\,(4\sqrt2)^2 = \tfrac12\,(32) = 16 \end{aligned}
Area=16 sq units\text{Area} = 16 \text{ sq units}

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CAT 2017 Slot 2 QA Q20: Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB. If the perpendicular d — Solution | TheCATExam