# In a circle of radius 21 cm, an arc subtends an angle of $60^o$ at the centre. Find (i) the length of the arc.(ii) area of the sector formed by the arc.(iii) 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

(i) the length of the arc.

(ii) area of the sector formed by the arc.

(iii) area of the segment formed by the corresponding chord.

Solution:

(i) Let the length of the arc be $l$.

We know that,

Length of arc $=2 \pi r(\frac{\theta}{360^{\circ}})$

Therefore,

Length of the arc $l=2 \times \frac{22}{7} \times 21 \times \frac{60^{\circ}}{360^{\circ}} \mathrm{cm}$

$=132 \times \frac{1}{6} \mathrm{cm}$

$=22 \mathrm{cm}$

The length of the arc is $22 \mathrm{~cm}$.

(ii) We know that,

Area of the sector $=\pi r^{2} \times \frac{\theta}{360^{\circ}}$

Therefore,

Area of the sector formed by the arc$=\frac{22}{7}(21)^{2} \times \frac{60^{\circ}}{360^{\circ}}$

$=\frac{22}{7} \times 21 \times 21 \times \frac{1}{6}$

$=231 \mathrm{~cm}^{2}$

The area of the sector is $231 \mathrm{~cm}^{2}$.

(iii) 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 $40.05 \mathrm{~cm}^{2}$.