NTSE Stage 2 SAT 2018 Exam Question Paper

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Question : 54 of 100

Marks: +1, -0

In the adjoining figure, ABC is a triangle in which,

∠B=90º

and its incircle

C_1

has radius 3. A circle

C_2

of radius 1 touches sides AC, BC and the circle

C_1

.Then length AB is equal to

$3+6√3$
$10+3√2$
$10+2√3$
$9+3√3$

Solution:

In

ΔC_1C_2Z

\cosθ=\;\;{2}/{4}=\;\;{1}/{2}

⇒θ=60°

Similarly

∠{X}'{C}_{1}{C}_{2}=60^{∘}

So

∠{ZC}_{1}{X}'=360-(60+60+90)=150

∴∠{BAC}=180-150=30

∠{ZAC}_{1}=15^{∘}

In

ΔZAC_1=15°

\tan15^{∘}=\;\;{3}/{{AZ}}

\;\;{√{3}-1}/{√{3}+1}=\;\;{3}/{{AZ}}

{AZ}=\;\;{√{3}+1}/{√{3}-1}×3=(\;\;{4+2√{3}}/{2})×3

=(2+√{3})×3=6+3√{3}

So,

AB=AZ+BZ=6+3√{3}+3=9+3√{3}

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