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

1. $\pi /3$
2. $11/7$
3. $3 /\pi$
4. $7/11$

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 $: \text{A}$

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