Question

In a circle of radius $21cm$ ,an arc subtends an angle of ${60}^{\circ }$ at the centre.Find

$\left(i\right)$ the length of the arc

$\left(ii\right)$ the area of the sector

$\left(iii\right)$ the area of the minor segment and

$\left(iv\right)$ the area of the major segment.

Answer 1

$\theta ={60}^o$

$r=21cm$

(i) Length of arc $=\frac{60}{360}\times 2\pi \times 21$

$=\frac{1}{6}\times 2\times \frac{22}{7}\times 21=22cm$

(ii) Area of sector$=\frac{60}{360}\times \pi \times (21{)}^2$

$=\frac{1}{6}\times \frac{22}{7}\times (21{)}^2=231c{m}^2$

(iii) Minor segment

Area of $\Delta AOB=\frac{1}{2}\times r^2\times \sin \theta =\frac{1}{2}\times (21{)}^2\times \sin 60$

$=\frac{1}{2}\times 21\times 21\times \frac{\sqrt{3}}{2}=190.95c{m}^2$

Area of circle$=\pi r^2=\frac{22}{7}\times (21{)}^2=1386c{m}^2$

Area of minor segment ABC$=$Area of circle$-$ Area of $\Delta$AOB

$=1386-190.95$

$=1195.05c{m}^2$

(iv) Area of major segment$=$Area of circle - Area of minor segment

$=1386-1195.05$cm

$=190.95c{m}^2$.