Question

In the figure, $\angle POQ={30}^o$, $PM\perp OQ$ and radius $OP=12cm$, then find the area of segment $PRQ$ (Given $\pi =3.14$).

• A. $1.68c{m}^2$
• B. $2.46c{m}^2$
• C. $0.68c{m}^2$
• D. none of these

Answer 1

Answer: (A)

$1.68c{m}^2$

Radius $OP=OQ=r=12$ cm

$\theta ={30}^o$

and $PM$ is perpendicular to $OQ$

Area of a sector $OPQR$ $=\pi r^2\left(\frac{\theta }{{360}^o}\right)$

$=3.14\times (12{)}^2\times \left(\frac{30}{{360}^o}\right)$

$=37.68c{m}^2$

In $△OMP,\sin \left({30}^o\right)=\frac{PM}{OP}...........\left(1\right)$

Area of $△OQP=\frac{1}{2}\times OQ\times PM$

$=\frac{1}{2}\times 12\times OP\times \sin {30}^0$ [from $\left(1\right)$]

$=\frac{1}{2}\times 12\times 12\times \frac{1}{2}$

$=36c{m}^2$

Area of segment $PRQ=$ (Area of sector $OPQR$)$-$ (Area of $△OQP)$

$=(37.68-36)c{m}^2$

$=1.68c{m}^2$