Question

In the figure m $\angle$POQ = 30$°$ and radius OP = 12 cm, Find the following (Given $\mathrm{\pi }=3.14$)
(i) Area of sector O-PRQ
(ii) Area of $∆$OPQ
(iii) Area of segment PRQ

Answer 1

(i) Given:
Radius of the circle (OP) = 12 cm
Angle subtended at the centre $\mathrm{\theta }$ = 30$°$
We know:
Area of the sector O-PRQ = $\frac{\mathrm{\theta }}{360}\times \mathrm{\pi }{r}^2$

∴ Area of the sector O-PRQ = $\frac{30}{360}\times 3.14\times {12}^2$

= $3.14\times 12$
= 37.68 cm²


(ii) Area of the triangle = $\frac{1}{2}ab\sin \theta$

$$\begin{gathered} \therefore \mathrm{Ar}\left(△OPM\right)=\frac{1}{2}\times OP\times OM\times \sin 30° \\ =\frac{1}{2}\times 12\times 12\times \frac{1}{2} \\ =36{\mathrm{cm}}^2 \end{gathered}$$

(iii) Area of the segment = Area of sector O-PRQ $-$ Area of $∆$OPQ

= 37.68 $-$36
= 1.68 cm²