Given that the figure.
Let the radius of a circle be $r$ cm.
DIAGRAM
The triangle $ \triangle \; \text{OSR}$ is right-angle triangle. so we can apply the Pythagorean theorem.
$\text{(Hypotenuse)}^{2} = \text{(Perpendicular)}^{2} + \text{(Base)}^{2}$
$\Rightarrow \text{(OR)}^{2} = \text{(SR)}^{2} + \text{(SO)}^{2}$
$\Rightarrow \text{(2r)}^{2} = \text{(SR)}^{2} + r^{2}$
$\Rightarrow 4r^{2} – r^{2} = \text{(SR)}^{2} $
$\Rightarrow \text{(SR)}^{2} = 3r^{2} $
$\Rightarrow \boxed{\text{SR} = \sqrt{3} \; r \; cm }$
Now, we can calculate the area of circle and square.
- Area of circle $ = \pi \; r^{2} \; cm^{2}$
- Area of square $ = 3r^{2} \; cm^{2}$
$\therefore$ The required ratio $ = \pi \; r^{2} : 3r^{2} = \pi : 3$
Correct Answer $: \text{A}$