Question

Given \[PQRS\] 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.\[\dfrac{\pi }{{\sqrt 3 }}\]
B.\[\dfrac{\pi }{3}\]
C.\[\dfrac{\pi }{{\sqrt 2 }}\]
D.\[\dfrac{{\sqrt 3 }}{\pi }\]

Answer 1

Hint: We have given that \[PQRS\] is a square. \[RS\] is tangent to the circle at \[(S)\]. We have also given that \[TR = OS\]. We have to find the ratio of the areas of circle and square. Firstly we have to consider the radius of the circle then we calculate the value of \[OR\] in the form of radius. This helps us to find the value of tangent \[RS\]. \[RS\] is also the side of a square. So we will get to the side of the square. After that we can calculate the area of the square and circle and their ratio.

Complete step-by-step answer:

The figure of the statement is given as \[PQRS\] is square and \[RS\] is tangent to a circle.

We have given that \[TR = OS\]

Let us consider that radius of circle is \['R'\]

So the value of \[OS\] and \[OT\] is \['R'\]

As we have given that \[OS\] is equal to \[TR\] so value of \[TR\] is also equal to \['R'\]

As from the figure \[OR = OT + TS\]

So value of \[OR = R + R\]

\[OR = 2R\]

We know that tangent to circle is perpendicular to radius at point of contact so

\[RS\] is perpendicular to \[OS\]

So \[\angle OSR = {90^0}\]

\[\vartriangle OSR = {90^0}\] Is right angled triangle

By Pythagoras theorem

\[{(OR)^2} = {(OS)^2} + {(SR)^2}\]

$\Rightarrow$ \[{(2R)^2} = {(R)^2} + {(SR)^2}\]

Here, In this step \[{(R)^2}\] is transferred from R.H.S to L.H.S and with that positive sign changes to negative.

\[4{R^2} - {R^2} = S{R^2}\]

$\Rightarrow$ \[3{R^2} = S{R^2}\]

Taking square root on both side

$\Rightarrow$ \[SR = \sqrt {3R} \]

Side of square \[ = \sqrt {3R} \]

Area of square is \[ = A{(Side)^2}\]

So area of square \[ = (\sqrt {3R{)^2}} - 3{R^2}\]

Area of circle is \[ = \pi {R^2}\]

Ratio of area of circle to area of square \[ = \dfrac{\pi }{3}\]

So option \[\left( B \right)\]is correct.

Note: Circle is a figure in a plane whose all points are all a constant distance from a fixed point. The constant distance is called radius and the fixed distance is called radius square is a quadrilateral whose all sides are equal and angle between each adjacent side is equal to \[{90^0}\].