52
A chord of a circle of radius 14 cm subtends an angle of $60^\circ$ 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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Area of minor segment = $\frac{22}{7}\times14\times14\times\frac{60}{360} - \frac{1}{2}\times14\times14\times \frac{\sqrt{3}}{2}$
$=(\frac{308}{3}-49\sqrt{3}) \text{cm}^2$ or $17.9\text{cm}^2$
Area of major segment = $\frac{22}{7} \times 14 \times 14 - (\frac{308}{3}-49\sqrt{3})$
$=616-\frac{308}{3}+49\sqrt{3}$
$=(\frac{1540}{3}+49\sqrt{3}) \text{cm}^2$ or $598.1\text{cm}^2$
$=(\frac{308}{3}-49\sqrt{3}) \text{cm}^2$ or $17.9\text{cm}^2$
Area of major segment = $\frac{22}{7} \times 14 \times 14 - (\frac{308}{3}-49\sqrt{3})$
$=616-\frac{308}{3}+49\sqrt{3}$
$=(\frac{1540}{3}+49\sqrt{3}) \text{cm}^2$ or $598.1\text{cm}^2$