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

The Correct Option is A

 

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 \(\theta = 45^\circ\),\[ = 36(\sqrt{2})^2 = 36 \times 2 = 72 \]