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CAT 2021 Slot 1 — QA Question 15

Tangents & SecantsEasy

A circle of diameter 8 inches is inscribed in a triangle ABC where ∠ABC = 90°. If BC = 10 inches then the area of triangle in square inches is.

Answer & solution

Correct answer: 120

Answer: 120

Solution

Easy

Use the tangent-length property of an inscribed circle: the two tangents drawn from any external vertex are equal. With the incircle radius =4=4 (diameter 88) and the right angle at BB, label the equal tangent segments and use the Pythagorean theorem to find the unknown.

A B C BC = 10
1

Tangent lengths from BB. The incircle has radius r=4r=4. Where it touches BABA and BCBC, the two tangents from BB are equal; since the angle at BB is 90∘90^\circ, those tangent lengths each equal the radius.

BM=BN=4(tangents from B, ∠B=90∘)\begin{aligned} &BM=BN=4\quad\text{(tangents from }B,\ \angle B=90^\circ) \end{aligned}
2

Tangent lengths from CC and AA. From CC: the tangent along BCBC is BC−BM=10−4=6BC-BM=10-4=6, so the tangent from CC on CACA is also 66. Let the tangent length from AA be xx (so AN=xAN=x on ABAB and xx on CACA).

AB=x+4,CA=x+6,BC=10\begin{aligned} &AB=x+4,\qquad CA=x+6,\qquad BC=10 \end{aligned}
3

Pythagoras in right △ABC\triangle ABC.

AB2+BC2=CA2⇒ (x+4)2+102=(x+6)2⇒ x2+8x+16+100=x2+12x+36⇒ 80=4x⇒ x=20\begin{aligned} &AB^2+BC^2=CA^2\\ &\Rightarrow\ (x+4)^2+10^2=(x+6)^2\\ &\Rightarrow\ x^2+8x+16+100=x^2+12x+36\\ &\Rightarrow\ 80=4x\\ &\Rightarrow\ x=20 \end{aligned}
4

Compute the area. The legs are AB=x+4=24AB=x+4=24 and BC=10BC=10.

Area=12⋅AB⋅BC=12⋅24⋅10=120\begin{aligned} &\text{Area}=\tfrac12\cdot AB\cdot BC=\tfrac12\cdot 24\cdot 10=120 \end{aligned}
120 sq inches120\ \text{sq inches}

For a right triangle, the inradius is r=leg1+leg2−hyp2r=\dfrac{\text{leg}_1+\text{leg}_2-\text{hyp}}{2}. With r=4, BC=10r=4,\ BC=10 and legs 10, AB10,\,AB: 4=10+AB−AB2+10024=\tfrac{10+AB-\sqrt{AB^2+100}}{2} gives AB=24AB=24, so area =12⋅24⋅10=120=\tfrac12\cdot24\cdot10=120.

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CAT 2021 Slot 1 QA Q15 — Tangents & Secants solution | TheCATExam