Question

In in the given figure, $△ABC$ is a right angle isosceles triangle with $AB=BC=12\text{cm}$. An arc is drawn considering $AR$ as the radius, if the area of region $CPQ=BRQ$, then the area of circle with radius $AR$ is
(correct answer + 1, wrong answer - 0.25)

• A. $476{\text{cm}}^2$
• B. $196\pi {\text{cm}}^2$
• C. $169\pi {\text{cm}}^2$
• D. $576{\text{cm}}^2$

Answer 1

The correct option is D $576{\text{cm}}^2$

Let $Ar\left(CPQ\right)=m,Ar\left(BQR\right)=n$


$Ar\left(CPQ\right)=Ar\left(BQR\right)\Rightarrow m=n$

Now, we know that
$Ar\left(ARQP\right)=Ar\left(△ABC\right)+n-m\Rightarrow Ar\left(ARQP\right)=Ar\left(△ABC\right)\Rightarrow \frac{\pi }{4\times 2\pi }\times \pi (AR{)}^2=\frac{1}{2}\times 12\times 12(\because \theta =\frac{\pi }{4})\therefore \pi (AR{)}^2=576{\text{cm}}^2$