Question

The maximum area of a rectangle inscribed in a circle of diameter \( R \) is:

• A. \( R^2 \)
• B. \( \frac{R^2}{2} \)
• C. \( \frac{R^2}{4} \)
• D. \( \frac{R^2}{8} \)

Hint:

For a rectangle inscribed in a circle, the maximum area occurs when the diagonals are equal, and the area is half the product of the diagonals.

Answer 1

The Correct Option is B

Step 1: {Find the maximum area of the rectangle}
The diagonal of the rectangle inscribed in the circle is equal to the diameter \( R \), so the diagonal \( d = R \). The maximum area of the rectangle is given by: \[ {Max Area} = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times R \times R = \frac{R^2}{2}. \]

Answer 2

Step 1: Understanding the problem
We are asked to find the maximum area of a rectangle inscribed in a circle with diameter \( R \). This means the rectangle is fully enclosed within the circle, with its four corners touching the circle.

Step 2: Set up the geometry
The circle has a radius of \( \frac{R}{2} \), and its center is at the origin \( (0, 0) \). The equation of the circle is: \[ x^2 + y^2 = \left( \frac{R}{2} \right)^2 = \frac{R^2}{4} \] We place the rectangle such that its sides are parallel to the coordinate axes. The corners of the rectangle will lie on the circle, and the sides of the rectangle will have lengths \( 2x \) and \( 2y \), where \( (x, y) \) represents one of the rectangle's corners.

Step 3: Area of the rectangle
The area \( A \) of the rectangle is given by: \[ A = 2x \cdot 2y = 4xy \] We are tasked with maximizing \( A = 4xy \) subject to the constraint \( x^2 + y^2 = \frac{R^2}{4} \).

Step 4: Use the constraint
To maximize the area, observe that we can use symmetry. If we set \( x = y \), the rectangle becomes a square. Substituting \( x = y \) into the constraint equation \( x^2 + y^2 = \frac{R^2}{4} \), we get: \[ 2x^2 = \frac{R^2}{4} \quad \Rightarrow \quad x^2 = \frac{R^2}{8} \quad \Rightarrow \quad x = \frac{R}{2\sqrt{2}} \] Thus, the length of each side of the square is \( 2x = \frac{R}{\sqrt{2}} \).

Step 5: Maximum area of the rectangle
The area of the square is: \[ A = 4xy = 4 \cdot \frac{R}{2\sqrt{2}} \cdot \frac{R}{2\sqrt{2}} = \frac{R^2}{2} \] Therefore, the maximum area of the rectangle inscribed in a circle of diameter \( R \) is:

\( \frac{R^2}{2} \)