Symbiosis National Aptitude Test SNAP 2015 Paper

Area of ΔAOB = $\frac{1}{2}$ × OA × OB

= $\frac{1}{2}\times 20\times 20$ = 200 $\mathit{c}{\mathit{m}}^2$
and area of sector OACBO
= $\frac{\mathit{\pi }{\mathit{r}}^2\mathit{\theta }}{360°}$ = $\frac{3.14\times 20\times 20\times 90°}{360°}$
= $\frac{3.14\times 400}{4}$ = 314 $\mathit{c}{\mathit{m}}^2$
∴ Area of minor segment
= Area of sector ΔACBO – Area of DAOB
= 314 – 200 = 114 $\mathit{c}{\mathit{m}}^2$