An arc of a circle of radius 21 cm subtends an angle of at the centre. Find the area of the minor segment of the circle made by the corresponding chord.

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Solution

Q.
An arc of a circle of radius 21 cm subtends an angle of $60^{o}$ at the centre. Find the area of the minor segment of the circle made by the corresponding chord.

Ans.
In $∆$ OAB, OA = OB = 21cm

$\Rightarrow$ $∠$ OAB = OBA = $60^{o}$

( $∵$ $∠$ AOB = $60^{o}$ and angles opposite to equal sides & Angle sum property of triangle)

Thus, $∆$ OAB is equilateral triangle.

Area of the minor segment = Area of sector OAB – Area of triangle of OAB

$= \frac{θ}{360^{0}} \times π r^{2} - \frac{\sqrt{3}}{4} a^{2} = \frac{60^{0}}{360^{0}} \times \frac{22}{7} \times 21 \times 21 - \frac{\sqrt{3}}{4} 21^{2}$

$= 231 - \frac{441 \sqrt{3}}{4} {cm}^{2}$

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