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A chord of a circle of diameter $20$ 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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Solution: Radius of circle $= 10$ cm $= r$
$\Delta OPQ$ is an equilateral triangle.
Area of segment $= \frac{1}{6} \times (3.14) \times (10)^2 - \frac{\sqrt{3}}{4} \times (10)^2$
$= \frac{109}{12}$ sq. cm or $9.08$ sq. cm
$\Delta OPQ$ is an equilateral triangle.
Area of segment $= \frac{1}{6} \times (3.14) \times (10)^2 - \frac{\sqrt{3}}{4} \times (10)^2$
$= \frac{109}{12}$ sq. cm or $9.08$ sq. cm