Question

If the area of sector $PQR$ is $18\pi$ square units and $\angle PQR$ is ${45}^0$, then the length of $QR$ is :

• A. $6$
• B. $9$
• C. $9\sqrt{2}$
• D. $12\sqrt{2}$

Answer 1

The correct option is D $12\sqrt{2}$

Area of sector $=\frac{\theta }{360}\times \pi \times r^2$ where $\theta$ is the angle which the arc forms.

$\therefore 18\pi =\frac{45}{360}\times \pi r^2$

$\Rightarrow 18\times \frac{360}{45}=r^2$

$\Rightarrow r^2=18\times 8=144$

$\therefore r=12$ units

So, $PQ=PR=12$ units

Hence, $△PQR$ is isosceles.

Let $PS\perp QR$ such that $P-S-R$.

In $△PSR$,

$\cos 45=\frac{SR}{PR}$

$\frac{1}{\sqrt{2}}=\frac{SR}{12}$

$SR=6\sqrt{2}$ units

Perpendicular drawn from center to the chord bisects the chord.

Hence, $SR=\frac{1}{2}QR$

So, $QR=12\sqrt{2}$ units