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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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