Question

Pqrs is a square. Sr is a tangent (at point s) to the circle with centre o. T is a point on the circle at which the line joining o and r intersects. If tr = os, then the ratio of area of the circle to the area of the square is

Answer 1

Dr is tangent to the circle ....line perpendicular to sr at s coincide with radius but PS is perpendicular to sr....

Therefore os lies on PS but is not the mid point other calculations are attached in the draft

Answer 2

Answer:

$\pi:3$

Step-by-step explanation:

We are given that PQRS is a square and a circle with centre O

We are given that SR is a tangent to the circle and TR=OR

Let OR =x

OR is a radius of a given circle

Then TR=x

OR=OT+TR=x+x=2x

All radius of a circle are equal

We know that a radius is perpendicular to tangent line

In right-angles triangle OSR

$(OS)^2+(SR)^2=(OR)^2$ (By pythogorous theorem)

$x^2+(SR)^2=(2x)^2$

$x^2+(SR)^2=4x^2$

$(SR)^2=4x^2-x^2=3x^2$

$SR=\sqrt{3x^2}$

$SR=\sqrt{3} x$

PQRS is a square then its all sides are equal

Area of square=$side\times side=(\sqrt{3}x)^2=3x^2$

Area of circle=$\pi(r^2)=\pi x^2$

$\frac{Area\;of\;circle}{Area\;of\;square \;PQRS}=\frac{\pi x^2}{3x^2}=\pi:3$

Hence, the ratio of area of circle to the area of the square=$\pi:3$