SSC CHSL 11 Aug 2021 Shift 2 Paper

Section: Quantitative Aptitude

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Question : 65 of 100

Marks: +1, -0

In a

△ A B C

, the length of side

A C

is

4 {cm}

less than five times the length of side

A B

. The length of side

B C

exceeds four times the length of side

A B

by

4 {cm}

. If the perimeter of

△ A B C

is

90 {cm}

, then its area is

$160 {cm}^{2}$
$164 {cm}^{2}$
$180 {cm}^{2}$
$148 {cm}^{2}$

Solution:

Let the length of

A B

be

x {cm}

.

∴ \;\; A C=(5 x-4) {cm}

and

B C=(4 x+4) {cm}

Given, perimeter of

△ A B C=90

⇒ x+(5 x-4)+(4 x+4)=90

⇒ \;\; 10 x=90

⇒ \;\; x=\;{90}/{10}=9 {cm}

∴ A B=9 {cm}, A C=5 × 9-4=41 {cm}

and

B C=4 × 9+4=40 {cm}

Since,

(41)^{2}=(40)^{2}+(9)^{2}

[A C^{2}=B C^{2}+A B^{2}]

∴ △ A B C

is a right angle triangle with hypotenuse of

41 {cm}

.

∴

Area of

△ A B C=\;{1}/{2} × 40 × 9=180 {cm}^{2}

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