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Find the area of the segment AYB shown in the figure, if the radius of the circle is $21$ cm and $\angle AOB = 120^\circ$. [Use $\pi = \frac{22}{7}$]
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Sol. Area of segment AYB = ar (sector AOBY) – ar ($\triangle AOB$) (1/2 Mark)
$= \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 - \frac{1}{2} \times 21 \times 21 \times \sin 60^\circ$ (1 Mark)
$= 462 - 441 \times \frac{\sqrt{3}}{2} \times \frac{1}{2}$ (1 Mark)
$= (462 - 441\frac{\sqrt{3}}{4})$ cm$^2$ (1/2 Mark)
or
frac{21}{4} (88 – 21
sqrt{3})$ cm$^2$
$= \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 - \frac{1}{2} \times 21 \times 21 \times \sin 60^\circ$ (1 Mark)
$= 462 - 441 \times \frac{\sqrt{3}}{2} \times \frac{1}{2}$ (1 Mark)
$= (462 - 441\frac{\sqrt{3}}{4})$ cm$^2$ (1/2 Mark)
or
frac{21}{4} (88 – 21
sqrt{3})$ cm$^2$