Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then is equal to:

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Solution

Q.
Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let $a$ and $b$ be the sides of the rectangle PQRS when its area is maximum. Then ${(a + b)}^{2}$ is equal to: [2024]

1 80

2 60

3 64

4 72

Ans.
(4)

In $∆ C D S$ , we have

$\cos θ = \frac{C S}{C D}$

$\Rightarrow \cos θ = \frac{C S}{4}$

$\Rightarrow C S = 4 \cos θ$

Similarly, in $∆ B C R$ we have

$\sin θ = \frac{C R}{2}$

$\Rightarrow C R = 2 \sin θ$

$∴$ $S R = C S + C R = 4 \cos θ + 2 \sin θ$

Similarly, $P S = 4 \sin θ + 2 \cos θ$

$∴$ Area of $P Q R S = P S \times S R$

$= 16 \cos θ \sin θ + 8 \cos^{2} θ + 8 \sin^{2} θ + 4 \sin θ \cos θ$

$= 8 (\cos^{2} θ + \sin^{2} θ) + 10 (2 \sin θ \cos θ)$

$= 8 + 10 \sin (2 θ)$

Area is maximum, when $\sin 2 θ = 1 \Rightarrow θ = 45 °$

$∴$ $Side a = 4 \sin θ + 2 \cos θ$ $[∵ θ = 45^{o}]$

$= \frac{4}{\sqrt{2}} + \frac{2}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3 \sqrt{2}$

and $b = 4 \cos θ + 2 \sin θ = 3 \sqrt{2}$

Now, ${(a + b)}^{2} = {(6 \sqrt{2})}^{2} = 72$

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