GATE ECE 2020 | GA | Question: 8

Question

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 _______.                                                              

• A. $\pi a^{2}-a^{2}$
• B. $\pi a^{2}-\sqrt{2}a^{2}$
• C. $\pi a^{2}-2a^{2}$
• D. $\pi a^{2}-3a^{2}$

Comments

maximum possible rectangle inside circle ==>square.

square ==> rombus

area of rombus = (diagonal d1 * diagonal d2)/2 = 2a^2           (d1 =d2 = 2a)

area(circle) - area(square)  == ( pie * a^2 ) - (2a^2)

Answer 1

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. 

Comments

An important fact :- For drawing rectangle of maximum area in a circle then the rectangle will be a square .

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you can also find the area of rectangle like this also : 

Answer 2

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

 

 

Comments

it is rectangle not square so taking l ,l length and breadth is wrong

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@Sanjeev kumar Sen

Every Square is a rectangle but not every rectangle is a square

But some rectangles can be square

Consider PQRS as the rectangle initially

Let $l$ and $b$ be the length and breadth of rectangle PQRS, then its area is $l*b$

Given that $l*b$ is maximum,hence differnce between $l $ and $b$ is minimum 

As $l,b$ are non-negative, the minimum difference will be 0. i..e $l-b=0$ or $l=b$

since $l=b$, the given rectangle PQRS become a square.

 

 

Answer 3

To solve this question, we should know the fact that maximum area here can only be when the rectangle is a square.

So, diagonals (of length 2a here) intersect at 90 degree, 

Therefore total area of rectangle = $4 * (1/2 * base * height)$ = $4 * (1/2 * a * a)$ = $2a^2$

Area of remaining portion =  $\pi a^{2} - 2a^2$

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Derivation of the fact 

We should know $AM \ge GM$ for non-negative numbers (link),

if length is $L$, and width is $W$, then $\frac{L^2 + W^2}{2} \ge \sqrt{L^2 W^2}$

Since we know diagonal is $2a$, using pythagoras theorem: $L^2 + W^2 = (2a)^2 = 4a^2$

Putting it in the equality: $\frac{4a^2}{2} \ge L W$ (L & W has to be positive)

As area of rectangle is $L*W$, the above equation becomes $2a^2 \ge A$

Therefore, maximum value of area is $2a^2 $, which is possible if the rectangle has sides $a\sqrt{2}$

Answer 4

maximum possible rectangle inside circle ==>square.

square ==> rombus

area of rombus = (diagonal d1 * diagonal d2)/2 = 2a^2           (d1 =d2 = 2a)

area(circle) - area(square)  == ( pie * a^2 ) - (2a^2)