Question

Find the area of the segment AYB, if radius of the circle is 21 cm and $\angle AOB={120}^o$.

Answer 1

We have;

Area of sector $OAYB=\frac{r^2\theta }{2}=\frac{r^2\theta }{2}=\frac{{\left(21\right)}^2\times 2\pi }{3\times 2}=\frac{21\times 21\times 22}{3\times 7}=462{cm}^2$

Area of $\Delta OAB=\frac{1}{2}\times 21\times 21\times \sin {120}^0=\frac{1}{2}\times 21\times 21\times \frac{\sqrt{3}}{2}=\frac{441\sqrt{3}}{4}{cm}^2$

Therefore area og segment $AYB$=Area of sector $OAYB$-Area of $\Delta OAB$

$=(462-\frac{441}{4}\sqrt{3}){cm}^2$

$\frac{1848-441\sqrt{3}}{4}{cm}^2$