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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 : 52 of 100

Marks: +1, -0

A

B C

Solution: 👈: Video Solution

\triangle A B C

it is given that

A B = c = 6 \text{ cm}

B C = a = 10 \text{ cm}

and

C A = b = 8 \text{ cm}

s = \;\frac{a+b+c}{2}

where

s

is semi perimeter

\;\; = \;\frac{10+8+6}{2}

\;\; = \;\frac{24}{2}=12 \text{ cm}

\therefore

Area of triangle

\;\; = \sqrt{s(s-a)(s-b)(s-c)}

\Rightarrow\;\; \frac{1}{2} \times B C \times A M

= \;\; \sqrt{12(12-10)(12-8)(12-6)}

\Rightarrow\;\; \frac{1}{2} \times 10 \times A M = \sqrt{12 \times 2 \times 4 \times 6}

\Rightarrow\;5 \times A M = \sqrt{4 \times 3 \times 2 \times 4 \times 2 \times 3}

\Rightarrow\;5 \times A M = 2 \times 3 \times 2 \times 2

\Rightarrow\;A M = \;\frac{24}{5}

\Rightarrow\;A M = 4.8 \text{ cm}

Hence, the length of the required altitude is

4.8 \text{ cm}

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