Find the area of the minor segment of a circle of radius $14\ cm$, when its central angle is $60^{o}$. Also find the area of the corresponding major segment. [use $\pi =\frac{22}{7}$]

Given: A circle of radius $14\ cm$.

To do: Find the area of the minor segment of a circle and of the corresponding major segment. when its central angle is $60^{o}$

Solution: 

Radius of the circle, $r =14\ cm$

Central Angle$\theta=60^{o}$ ,

Area of the minor segmen$=\frac{\theta }{360^{o}} \times \pi r^{2} -\frac{1}{2} r^{2} sin\theta $

$=\frac{60^{o}}{360^{o}} \times( \frac{22}{7}) \times ( 14)^{2} -\frac{1}{2}( 14)^{2} sin60^{o}$

$=\frac{1\times 22\times 2\times 14}{6} -\frac{1}{2} \times 14\times 14\times \frac{\sqrt{3}}{2}$

$=\frac{22\times 14}{3} -49\sqrt{3}$

$=\frac{308-147\sqrt{3}}{3}$

$\boxed{Area\ of\ the\ minor\ segment\ \ =\frac{308-147\sqrt{3}}{3} \ cm^{2} .\ }$



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Updated on: 10-Oct-2022

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