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

Area of segment $\displaystyle A Y B=\operatorname{ar}($ sector $\displaystyle A O B Y)-\operatorname{ar}(\triangle A O B)$ \[\begin{array}{l} =\frac{120}{360} \times \frac{22}{7} \times 21 \times 21-21 \times 21 \times \sin 60^{\circ} \cos 60^{\circ} \\ =462-441 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} \\ =\left(462-441 \frac{\sqrt{3}}{4}\right) \mathrm{cm}^{2} \\ \text { or } \frac{21}{4}(88-21 \sqrt{3}) \mathrm{cm}^{2} \end{array} \]ALTERNATE SOLUTION:Area of sector $\displaystyle \mathrm{AOBY}=\frac{120}{360} \times \frac{22}{7} \times 21 \times 21$ \[=462 \mathrm{~cm}^{2} \]Draw $\displaystyle \mathrm{OM} \perp \mathrm{AB}, \therefore \angle \mathrm{AOM}=60^{\circ}$ \[\begin{array}{l} \cos 60^{\circ}=\frac{1}{2}=\frac{\mathrm{OM}}{21} \\ \Rightarrow \mathrm{OM}=\frac{21}{2} \mathrm{~cm} \\ \sin 60^{\circ}=\frac{\sqrt{3}}{2}=\frac{\mathrm{AM}}{21} \\ \Rightarrow \mathrm{AM}=\frac{21 \sqrt{3}}{2} \mathrm{~cm} \\ \begin{aligned} \Rightarrow \mathrm{AB} & =21 \sqrt{3} \mathrm{~cm} \\ \therefore \operatorname{ar}(\triangle \mathrm{AOB}) & =\frac{1}{2} \times 21 \sqrt{3} \times \frac{21}{2} \\ & =\frac{441}{4} \sqrt{3} \mathrm{~cm}^{2} \end{aligned} \end{array} \]Area of segment $\displaystyle \mathrm{AYB}=\operatorname{ar}$ (sector AOBY$\displaystyle )-\operatorname{ar}(\triangle \mathrm{AOB})$ \[\begin{aligned} = & \left(462-441 \frac{\sqrt{3}}{4}\right) \mathrm{cm}^{2} \\ & \text { or } \frac{21}{4}(88-21 \sqrt{3}) \mathrm{cm}^{2} \end{aligned} \]