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UPSC CDS 2 2021 Math Paper

Directions (Q.Nos. 51-53) Consider the following for the next three questions that follow.

A B C

is a triangle with sides

A B=6 {cm}

B C=10 {cm}

and

C A=8 {cm}

. With vertices

A, B

and

C

as centres, three circles are drawn each touching the other two externally.

Question : 53 of 100

Marks: +1, -0

P, Q

R

A, B

C

P=\pi \text{cm}^2

9Q+4R=36\pi \text{cm}^2

Solution: 👈: Video Solution

\because r_1+r_2+r_3=12

(from Q.No. 51)

and

\;\; r_1+r_2=6, \;\; r_2+r_3=10

r_1+r_3=8

\therefore r_1=2 \text{cm}, r_2=4 \text{cm}, r_3=6 \text{cm}

P=\frac{1}{4}\pi r_1^2 = \frac{1}{4}\times \pi \times (2)^2 = \pi \text{cm}^2

9Q+4R=9 \times \frac{\theta_1}{360^{\circ}} \times \pi r_2^2+4

\times \frac{\theta_2}{360^{\circ}} \times \pi r_3^2

=9 \times \frac{\theta_1}{360^{\circ}} \times \pi (4)^2

+4 \times \frac{\theta_2}{360^{\circ}} \times \pi (6)^2

= \frac{144}{360^{\circ}} \pi \theta_1 + \frac{144}{360^{\circ}} \pi \theta_2

= \frac{144}{360^{\circ}} \pi (\theta_1+\theta_2)

= \frac{144}{360^{\circ}} \pi \times 90^{\circ} \; [ \because \theta_1+\theta_2=90^{\circ} ]

=36 \pi \text{cm}^2

Hence, both 1 and 2 are correct.

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