Question

The maximum area of the rectangle that can be inscribed in a circle of radius $r$ is

• A. $2r^2\ \text{sq. units}$
• B. $\frac{\pi^2}{4}\ \text{sq. units}$
• C. $\pi r^2\ \text{units}$
• D. $r^3\ \text{sq. units}$

Hint:

A highly useful geometric principle to memorize is that for any given closed bounding curve, the inscribed rectangle with the absolute maximum area is always a perfectly symmetric square. For a circle of radius $r$, the diagonal of this square is the diameter $2r$, making its area equal to $\frac{1}{2} \times d^2 = \frac{1}{2}(2r)^2 = 2r^2$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires finding the maximum possible area of a rectangle that can be completely inscribed inside a circle of a fixed radius $r$.

Step 2: Key Formula or Approach:
Let the center of the circle be at the origin $(0,0)$. A vertex of the rectangle in the first quadrant can be represented in parametric coordinates as $(r\cos\theta, r\sin\theta)$, where $\theta$ is the angle made with the positive horizontal axis.
The dimensions of this inscribed rectangle will be: $$\text{Length } = 2r\cos\theta \quad \text{and} \quad \text{Width } = 2r\sin\theta$$ The area $A$ as a function of $\theta$ is: $$A(\theta) = \text{Length} \times \text{Width} = (2r\cos\theta)(2r\sin\theta) = 2r^2(2\sin\theta\cos\theta) = 2r^2\sin(2\theta)$$

Step 3: Detailed Explanation:
To maximize the area function $A(\theta) = 2r^2\sin(2\theta)$, we examine the trigonometric component $\sin(2\theta)$.
The maximum possible value that a sine function can ever achieve is exactly $1$: $$\max(\sin(2\theta)) = 1$$ This maximum occurs when the angle argument equals $90^\circ$: $$2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4}\ (45^\circ)$$ Substituting $\sin(2\theta) = 1$ back into our area equation yields the maximum area: $$A_{\max} = 2r^2(1) = 2r^2\ \text{sq. units}$$ Note that when $\theta = \frac{\pi}{4}$, the length becomes $2r\cos(45^\circ) = \sqrt{2}r$ and the width becomes $2r\sin(45^\circ) = \sqrt{2}r$. Since the length equals the width, the rectangle that maximizes the area is specifically a square.

Step 4: Final Answer:
The maximum area of the inscribed rectangle is $2r^2\ \text{sq. units}$, which corresponds to option (A).