SSC CHSL 19 Mar 2020 Shift 2 Paper

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Question : 72 of 100

Marks: +1, -0

The area of the quadrant of a circle whose circumference is

22 {\cm}

, will be

$3.5 {\cm}^{2}$
$10 {\cm}^{2}$
$38.5 {\cm}^{2}$
$9.625 {\cm}^{2}$

Solution:

Let the radius of the circle be

r {\cm}

. Given, circumference of the circle

=22 {\cm}

⇒

2 \πr=22

⇒

2 \×{22}/{7} \×r=22

⇒

r={7}/{2}=3.5 {\cm}

∴ Area of the quadrant of the circle

={1}/{4} \×\πr^{2}

={1}/{4} \×{22}/{7} \×({7}/{2})^{2}

={1}/{4} \×{22}/{7} \×{7}/{2} \×{7}/{2}

=9.625 {\cm}^{2}

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