Rectangles are inscribed in a circle of radius 5 cm. Consider the fol…

The correct answer is Option (2) → (A) and (B) only

Every diagonal of an inscribed rectangle is a diameter, of length 10 cm. If a diagonal makes an angle θ with one side, \(0 \lt \theta \lt \frac{\pi}{2}\), the sides are \(10\cos\theta\) and \(10\sin\theta\). So

    \( A = 100\sin\theta\cos\theta = 50\sin 2\theta \)
    \( P = 20(\cos\theta + \sin\theta) \)

For the area, \(A' = 100\cos 2\theta = 0\) gives \(\theta = \frac{\pi}{4}\), with \(A'' = -200\sin 2\theta \lt 0\):

    \( \displaystyle A_{\max} = 50\sin\frac{\pi}{2} = 50 \)

For the perimeter, \(P' = 20(\cos\theta - \sin\theta) = 0\) gives \(\theta = \frac{\pi}{4}\), with \(P'' = -20(\sin\theta + \cos\theta) \lt 0\):

    \( \displaystyle P_{\max} = 20\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = 20\sqrt{2} \)

So both maxima occur for the square of side \(5\sqrt{2}\) cm: the maximum area is 50 cm², the maximum perimeter is \(20\sqrt{2}\) cm, and every diagonal is 10 cm.

Checking the statements

(A) The maximum area of such a rectangle is 50 cm². — Correct. \(A = 50\sin 2\theta\) is largest, 50, at \(\theta = \frac{\pi}{4}\).

(B) The rectangle of maximum perimeter is a square. — Correct. At \(\theta = \frac{\pi}{4}\) both sides are \(10 \times \frac{1}{\sqrt{2}} = 5\sqrt{2}\) cm.

(C) The maximum perimeter of such a rectangle is 40 cm. — Incorrect. The maximum is \(20\sqrt{2} \approx 28.3\) cm.

(D) The rectangle of maximum area has a diagonal of length 5 cm. — Incorrect. Its diagonal is a diameter, 10 cm.