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If a chord of a circle of radius $10$ cm subtends an angle of $60^\circ$ at the centre of the circle, find the area of the corresponding minor segment of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
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Area of minor segment $= \frac{3.14 \times(10)^2\times 60^\circ}{360^\circ} - \frac{1}{2} \times (10)^2 \times \sqrt{3}$
$= \frac{314}{6} - \frac{173}{4}$
$= 9\frac{1}{12}$ or $9.08$
Hence, area of minor segment is $9.08$ cm$^2$.
$= \frac{314}{6} - \frac{173}{4}$
$= 9\frac{1}{12}$ or $9.08$
Hence, area of minor segment is $9.08$ cm$^2$.