Question

In figure, seg PQ is a diameter of semicircle PNQ. The centre of arc PMQ is O.OP=OQ= 10 cm and m$\angle POQ={60}^2$. Find the area of the shaded portion.(Given $\pi$ = 3.14, $\sqrt{3}$= 1.73)

• A. $30.17c{m}^2$
• B. $32.27c{m}^2$
• C. $35.56c{m}^2$
• D. $40.17c{m}^2$

Answer 1

The correct option is A $30.17c{m}^2$

For a Segment PMQ,
radius $r=10cm$
Measure of are $\theta ={60}^{\circ }$
Area of Segment $PMQ=r^2[\frac{\pi \theta }{360}-\frac{\sin \theta }{2}]$
$={10}^2[\frac{3.14\times 60}{360}-\frac{\sin 60}{2}]$
$=100[\frac{3.14}{6}-\frac{\sqrt{3}}{2}\times \frac{1}{2}]$
$=100[\frac{3.14}{6}-\frac{\sqrt{3}}{4}]$
$=100\left[\frac{6.28-3\left(1.73\right)}{12}\right]$
$=100\left[\frac{6.28-5.19}{12}\right]$
$=\frac{100\times 1.09}{12}=\frac{109}{12}$
$\therefore areaofsegmentPMQ=9.08c{m}^2$
$△OPQsegOP\cong segOQ$
$\therefore \angle OPQ\cong \angle OQP$
Let , $m\angle OPQ=m\angle OQP=x$
$\therefore m\angle OPQ+m\angle OQP+m\angle POQ={180}^{\circ }$
$\therefore x+x+60=180$
$\therefore 2x=180-60$
$\therefore 2x=120$
$\therefore x=\frac{120}{2}$
$\therefore x={60}^{\circ }$
$\therefore m\angle OPQ=m\angle OQP=m\angle POQ={60}^{\circ }$
$\therefore △OPQisanequilateraltriangle$
$\therefore OP=OQ=PQ=10cm$
$DiameterPQ=10cm$
$Radiusr=\frac{10}{2}=5cm$
Area of Semicircle $=\frac{1}{2}\pi r^2$
$=\frac{1}{2}\times 3.14\times 5\times 5=39.28c{m}^2$
Area of the shaded portion = Area of Semi Circle - Area of segment PMQ
$=39.25-9.08=30.17c{m}^2$