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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 $88.44\ cm^2$.
Updated on: 10-Oct-2022
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