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.

Question

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.

Smart Achievers · Accepted Answer

In $60^{o}$ 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

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

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

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.

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Solution

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

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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.