A chord of a circle of radius 14 cm subtends an angle of $60°$ at the centre. Find the area of the corresponding minor segment of the circle. Also find the area of the major segment of the circle.

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Accepted Answer

Step 1: Area of minor segment $= \frac{22}{7} 	imes 14 	imes 14 	imes \frac{60}{360} - \frac{1}{2} 	imes 14 	imes 14 	imes \frac{\sqrt{3}}{2}$
Step 2: $= \frac{308}{3} - 49\sqrt{3}$ cm$^2$ or $17.9$ cm$^2$
Step 3: Area of major segment $= \frac{22}{7} 	imes 14 	imes 14 - (\frac{308}{3} - 49\sqrt{3})$
$= 616 - \frac{308}{3} + 49\sqrt{3}$
$= \left(\frac{1540}{3} + 49\sqrt{3}ight)$ cm$^2$ or $598.1$ cm$^2$

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