SSC CHSL 5 Aug 2021 Shift 3 Paper

Section: Quantitative Aptitude

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Question : 64 of 100

Marks: +1, -0

In an isosceles triangle

A B C

,

A B=A C

and

A D

is perpendicular to

B C

. If

A D=12 {cm}

and the perimeter of

△ A B C

is

36 {cm}

, then the length of

B C

(in

{cm}

) is

5
12
10
13

Solution:

Given,

A B=A C=x

(let)
and

A B+B C+C A=36

⇒ \;\; x+x+B C=36

⇒ \;\; B C=36-2 x

. . . (i)
Since,

A D

is perpendicular to

B C

.

∴ \;\; B C=2 × √{x^{2}-(12)^{2}}

. . . (ii)
From Eq. (i) and Eq. (ii),

2 × √{x^{2}-144} \;\; =36-2 x

⇒ \;\; √{x^{2}-144} \;\; =\;{36-2 x}/{2}=18-x

Square both sides,
Square both sides,

\;x^{2}-144\;=324+x^{2}-36 x

⇒\;36 x\;=324+144=468

⇒\;x\;=\;{468}/{36}=13 {cm}

\; ∴ \;\text " From Eq. (i), "\; B C\;=36-2 × 13

\;=36-26=10 {cm}

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