A chord of a radius $12\ cm$ subtends an angle of $120^o$ at the centre. Find the area of the corresponding segment of the circle.

Given: A chord of a radius $12\ cm$ subtends an angle of $120^o$ at the centre.

To do: To find the area of the corresponding segment of the circle.

Solution:

Radius of circle $=12\ cm$

$\theta = 120^o$

Area of the segment $=[( \frac{\pi}{360}) \times \theta - sin \frac{\theta}{2} cos \frac{\theta}{2}]r^2$

Area of the segment$={(\pi) \times \frac{120^o}{360^o} - sin \frac{120^o}{2} cos \frac{120^o}{2} } 12^2$

$= [(\pi) \times \frac{1}{3} - sin 60^o cos 60^o ] \times144$

$= ( \frac{\pi}{3} - \frac{1}{2} \times \frac{\sqrt{3}}{2}] \times 144$

$= (\frac{\pi}{3} \times 144 - 144 \times \frac{\sqrt{3}}{4})$

$= 48\pi - 36\sqrt{3}$

$= 12(4\pi - 3\sqrt{3})$

$= 12( 4 \times 3.14 - 3 \times 1.73)$  [$\pi = 3.14,\ \sqrt{3}= 1.73$]

$= 12 (12.56 - 5.19)$

$= 12 \times 7.37$

$= 88.44\ cm^2$

Hence, the area of the corresponding segment of the circle is $88.44\ cm^2$.

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Updated on: 10-Oct-2022

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