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A circle with centre $\text{O}$ is shown in the figure. A rectangle $\text{PQRS}$ of maximum possible area is inscribed in the circle. If the radius of the circle is $a$, then the area of the shaded portion is _______.                                                              

  1. $\pi a^{2}-a^{2}$
  2. $\pi a^{2}-\sqrt{2}a^{2}$
  3. $\pi a^{2}-2a^{2}$
  4. $\pi a^{2}-3a^{2}$
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Migrated from GO Electronics 2 years ago by Arjun

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Theorem: “A perpendicular from the Centre of a circle to a chord, bisects the chord.” (Refer: Proof)

In this case, if a perpendicular is drawn from the center of the circle, it will bisect the base side of the rectangle.

The half-length of this chord would be:  $a\times \cos  45^{\circ}=\frac{a}{\sqrt{2}}$.

Hence the length of the rectangle would be  $2\times \frac{a}{\sqrt{2}}=a\sqrt{2}$

Area of circle $: \pi a^{2}$  and the area of rectangle $: 2a^{2}$

$\therefore$  Area of remaining portion would be $:\pi a^{2}-2a^{2}$ 

Option C is correct. 

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Area of a rectangle of length $l$ and breadth $b$ is $lb$

The maximum value of $lb$ is $l^2$  when $l \geq b$

                                                  $b^2$ when $l \leq b$

$\implies$ Area of the rectangle $PQRS$ is maximum only when it becomes a square

 

 

Answer:

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