A circle with center O inscribed in square ABCD of side 'a' such that it touches the square at points P, Q,R, and S. Area between the circle and square PQRS will be:

Question

Question

A circle with center O inscribed in square ABCD of side 'a' such that it touches the square at points P, Q,R, and S. Area between the circle and square PQRS will be:

\(\frac{a ^2}{2}(\frac{\pi}{2}\ +\ 1)\)
\(\frac{a ^2}{4}(\frac{\pi}{2}\ -\ 1)\)
\(\frac{a ^2}{2}(\frac{\pi}{2}\ -\ 1)\)
\(\frac{a ^2}{4}(\frac{\pi}{2}\ +\ 1)\)

Answer 1

Concept:

Pythagoras theorem: It states that In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

1. Area of Circle = πR²

Where R = radius of the circle

D = 2R

D = Diameter of circle

2. Consider a square of side 'a'.

Area of square = a2

Diagonal of square = √2 a

Calculation:

Let, required area is A.

Given that,

Side of square = a

F1 Sachin Madhuri 07.10.2021 D1

We know that line drawn along the radius is perpendicular to the tangent of the circle.

Therefore,

OP = OQ = OR = OS = a/2 = radius of circle

In a right-angle triangle OPQ, using Pythagoras theorem

OP² + OQ² = PQ²

⇒ (a/2)² + (a/2)² = PQ²

⇒ PQ = a/√2

Therefore,

The side of square PQRS is a/√2

Hence, area b/w circle and square PQRS is given by

A = Area of a circle - Area of square PQRS

⇒ A = \(\pi (\frac{a}{2})^2\ -\ (\frac{a}{\sqrt 2})^2\)

⇒ A = \(\frac{a ^2}{2}(\frac{\pi}{2}\ -\ 1)\)

Hence, option 3 is correct.

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Question

A circle with center O inscribed in square ABCD of side 'a' such that it touches the square at points P, Q,R, and S. Area between the circle and square PQRS will be:

Answer (Detailed Solution Below)

Detailed Solution

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