In a circle of radius $21\, cm$,an arc subtends an angle of $60^{\circ}$ 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 [Use $\pi = \frac{22}{7}$]

(N/A) Radius $(r)$ of the circle $= 21\, cm$.
Angle subtended by the given arc at the centre $(\theta) = 60^{\circ}$.
$(i)$ Length of the arc $= \frac{\theta}{360^{\circ}} \times 2 \pi r$
$= \frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21$
$= \frac{1}{6} \times 2 \times 22 \times 3 = 22\, cm$.
$(ii)$ Area of the sector $= \frac{\theta}{360^{\circ}} \times \pi r^{2}$
$= \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 21 \times 21$
$= \frac{1}{6} \times 22 \times 3 \times 21 = 231\, cm^{2}$.
$(iii)$ In $\Delta OAB$,$OA = OB = 21\, cm$ and $\angle AOB = 60^{\circ}$.
Since the two sides are equal,$\angle OAB = \angle OBA$.
In $\Delta OAB$,$\angle OAB + \angle OBA + \angle AOB = 180^{\circ}$.
$2 \angle OAB + 60^{\circ} = 180^{\circ} \implies 2 \angle OAB = 120^{\circ} \implies \angle OAB = 60^{\circ}$.
Thus,$\Delta OAB$ is an equilateral triangle.
Area of $\Delta OAB = \frac{\sqrt{3}}{4} \times (side)^{2} = \frac{\sqrt{3}}{4} \times (21)^{2} = \frac{441\sqrt{3}}{4}\, cm^{2}$.
Area of the segment = Area of sector $OACB - \text{Area of } \Delta OAB$
$= (231 - \frac{441\sqrt{3}}{4})\, cm^{2}$.