Question

In the figure, $PQSO$ is a trapezium in which $PQ\parallel OS,\angle POS={135}^o$ and $\angle OSQ={90}^o$. Points $P,Q$ and $R$ lie on a circle with centre $O$ and radius $12$ cm. The area of the shaded part in $c{m}^2$ is :

• A. $61\frac{2}{7}$
• B. $61\frac{5}{7}$
• C. $73\frac{5}{7}$
• D. $73\frac{2}{7}$

Answer 1

The correct option is B $61\frac{5}{7}$

Given that $PQ||OS$ and angle $POS$ is ${135}^0$ , angle $OSQ$ is ${90}^0$

Given that the radius of circle is $12$ cm

So we have $OP=12$ cm

Let the foot of perpendicular from $O$ to $PQ$ be $N$

So we get $ON=PN=QS=OS=6\sqrt{2}$ cm

The area of trapezium $OSQP$ is $\frac{1}{2}\times 6\sqrt{2}\times 6\sqrt{2}+6\sqrt{2}\times 6\sqrt{2}=36+72=108$

The area of total arc is $\frac{135}{360}\times \pi \times 12\times 12=54\pi$

So, the area of shaded region is $54\pi -108=61\frac{5}{7}$.

Therefor the correct option is B.