Question

$PQRS$ is a square. $SR$ is a tangent (at point $S$) to the circle with centre $O$ and $TR=OS.$ Then find the ratio of area of the circle to the area of the square.

• A. $\frac{\pi }{\sqrt{3}}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{\sqrt{2}}$
• D. $\frac{\sqrt{3}}{\pi }$

Answer 1

The correct option is B $\frac{\pi }{3}$

It is given that $PQRS$ is a square and $SR$ is tangent to circle with centre $O$.

Lets assume the radius of circle be $r$

Now, we know that $TR-OS=r$

Therefore, $OR=OT+TR=r+r=2r$

$\Delta OSR$ is right angled triangle,

Hence, ${OR}^2-{OS}^2={SR}^2$

$\Rightarrow {SR}^2=4r^2-r^2=3r^2$

$\Rightarrow SR=\sqrt{3}r$

Now the side of square is $\sqrt{3}r$

Area of square, $A_s={\sqrt{3}r}^2$

$\Rightarrow A_s=3s^2$

Area of circle: $A_c=\pi r^2$

$\frac{\text{Area of circle}}{\text{Area of square}}=\frac{\pi r^2}{3r^2}=\frac{\pi }{3}$