Find the area of the segment $AYB$ shown in the figure if the radius of the circle is $21 \, cm$ and $\angle AOB = 120^{\circ}$. (Use $\pi = \frac{22}{7}$)

(N/A) Area of the segment $AYB = \text{Area of sector } OAYB - \text{Area of } \triangle OAB$ ......$(1)$
Now,area of the sector $OAYB = \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \, cm^2 = 462 \, cm^2$ ......$(2)$
For finding the area of $\triangle OAB$,draw $OM \perp AB$ as shown in the figure.
Note that $OA = OB$. Therefore,by $RHS$ congruence,$\triangle AMO \cong \triangle BMO$.
So,$M$ is the mid-point of $AB$ and $\angle AOM = \angle BOM = \frac{1}{2} \times 120^{\circ} = 60^{\circ}$.
Let $OM = x \, cm$.
So,from $\triangle OMA$,$\frac{OM}{OA} = \cos 60^{\circ}$.
$\frac{x}{21} = \frac{1}{2} \implies x = \frac{21}{2}$.
So,$OM = \frac{21}{2} \, cm$.
Also,$\frac{AM}{OA} = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$.
So,$AM = \frac{21\sqrt{3}}{2} \, cm$.
Therefore,$AB = 2 \times AM = 2 \times \frac{21\sqrt{3}}{2} = 21\sqrt{3} \, cm$.
So,area of $\triangle OAB = \frac{1}{2} \times AB \times OM = \frac{1}{2} \times 21\sqrt{3} \times \frac{21}{2} = \frac{441\sqrt{3}}{4} \, cm^2$ ......$(3)$
Therefore,area of the segment $AYB = \left( 462 - \frac{441\sqrt{3}}{4} \right) cm^2$ [From $(1), (2)$ and $(3)$].
$= \frac{21}{4} (88 - 21\sqrt{3}) \, cm^2$.