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In a circle of radius 21 cm, an arc subtends an angle of $60^o$ at the centre. Find area of the segment formed by the corresponding chord.
Given:
Radius of the circle $r=21 \mathrm{~cm}$.
Angle subtended by the arc $=60^{\circ}$
To do:
We have to find the area of the segment formed by the corresponding chord.
Solution:
We know that,
Area of the sector $=\pi r^{2} \times \frac{\theta}{360^{\circ}}$
Therefore,
Area of the sector formed by the arc$=\frac{22}{7}(21)^{2} \times \frac{60^{\circ}}{360^{\circ}}$
$=\frac{22}{7} \times 21 \times 21 \times \frac{1}{6}$
$=231 \mathrm{~cm}^{2}$
Therefore,
Area of the segment formed by the corresponding chord $=$ Area of the sector $-$ Area of the triangle formed between the chord and the radius of the circle
$=231-(\frac{1}{2} r^{2} \sin \theta)$
$=231-\frac{1}{2} \times(21)^{2} \times \sin 60^{\circ}$
$=231-\frac{1}{2} \times 441 \times \frac{\sqrt{3}}{2}$
$=231-\frac{441 \sqrt{3}}{4}$
$=231-190.95$
$=40.05 \mathrm{~cm}^{2}$
The area of the segment formed by the corresponding chord is $40.05 \mathrm{~cm}^{2}$.
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