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

Answer 2

Step 1: Set up geometry and notation
Let the inner rectangle ABCD have side lengths 4 (long side) and 2 (short side). Let it be rotated by an angle θ with respect to the sides of the outer rectangle PQRS (whose sides we denote by a and b). By symmetry, take 0 ≤ θ ≤ 90°. When a tight axis-aligned bounding rectangle contains a rotated rectangle of sides 4 and 2, its side lengths are the sums of component projections:
a = 4 cosθ + 2 sinθ, b = 4 sinθ + 2 cosθ.

Step 2: Express the area of PQRS in terms of θ
Area A(θ) = a·b = (4 cosθ + 2 sinθ)(4 sinθ + 2 cosθ). Expand and simplify using cos²θ + sin²θ = 1 and 2 sinθ cosθ = sin2θ:
A(θ) = 8(cos²θ + sin²θ) + 20 sinθ cosθ = 8 + 10 sin2θ.

Step 3: Maximize the area and find the corresponding a, b
Since −1 ≤ sin2θ ≤ 1, A(θ) is maximized when sin2θ = 1 ⇒ 2θ = 90° ⇒ θ = 45°. Then cosθ = sinθ = √2/2. Substitute into a and b:
a = 4(√2/2) + 2(√2/2) = 3√2, b = 4(√2/2) + 2(√2/2) = 3√2.
So the maximum-area outer rectangle is a square with side 3√2 (area 18).

Step 4: Compute (a + b)²
a + b = 3√2 + 3√2 = 6√2 ⇒ (a + b)² = (6√2)² = 72.

Final answer
72