Question

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 :

• A. 72
• B. 60
• C. 80
• D. 64

Answer 1

Answer: (A)

72

Answer 2

Consider rectangle $ABCD$ inscribed within rectangle $PQRS$ as shown in the figure. Let $\theta$ be the angle formed between side $AB$ of $ABCD$ and side $PQ$ of $PQRS$.

Using trigonometry, the dimensions of $PQRS$ are expressed as:

$a = 4 \cos \theta + 2 \sin \theta$ $b = 2 \cos \theta + 4 \sin \theta$

The area of $PQRS$ is given by:

$\text{Area} = (4 \cos \theta + 2 \sin \theta)(2 \cos \theta + 4 \sin \theta)$

Expanding this, we get:

$= 8 \cos^2 \theta + 16 \sin \theta \cos \theta + 4 \sin^2 \theta + 8 \sin^2 \theta$ $= 8 + 10 \sin 2\theta$

The area is maximized when $\sin 2\theta = 1$, i.e., $\theta = 45^\circ$.

Thus, the maximum area is:

$8 + 10 = 18$

Now, we calculate $(a + b)^2$:

$(a + b)^2 = (4 \cos \theta + 2 \sin \theta + 2 \cos \theta + 4 \sin \theta)^2$ $= (6 \cos \theta + 6 \sin \theta)^2$ $= 36(\sin \theta + \cos \theta)^2$

Since $\sin \theta + \cos \theta = \sqrt{2}$ at $\theta = 45^\circ$,$= 36(\sqrt{2})^2 = 36 \times 2 = 72$