The maximum area of rectangle, inscribed in a circle of radius 'r', is :

Options

A. `pi r^(2)`
B. `r^(2)`
C. `2r^(2)`
D. `1/4 pi r^(2)`

Answer

To find the maximum area of a rectangle inscribed in a circle of radius \( r \), we can follow these steps:

### Step 1: Understand the Geometry
Let the rectangle be \( ABCD \) inscribed in a circle with center \( O \) and radius \( r \). The length of the rectangle is \( L \) and the breadth is \( B \). The diagonal \( AC \) of the rectangle is equal to the diameter of the circle, which is \( 2r \).

### Step 2: Use the Pythagorean Theorem
Since \( AC \) is the diagonal of the rectangle, we can apply the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
This gives us:
\[ (2r)^2 = L^2 + B^2 \]
\[ 4r^2 = L^2 + B^2 \]

### Step 3: Express Breadth in Terms of Length
From the equation \( 4r^2 = L^2 + B^2 \), we can express \( B \) in terms of \( L \):
\[ B^2 = 4r^2 - L^2 \]
\[ B = \sqrt{4r^2 - L^2} \]

### Step 4: Write the Area of the Rectangle
The area \( A \) of the rectangle can be expressed as:
\[ A = L \cdot B = L \cdot \sqrt{4r^2 - L^2} \]

### Step 5: Differentiate the Area Function
To find the maximum area, we need to differentiate \( A \) with respect to \( L \):
\[ A = L \sqrt{4r^2 - L^2} \]
Using the product rule:
\[ \frac{dA}{dL} = \sqrt{4r^2 - L^2} + L \cdot \frac{d}{dL}(\sqrt{4r^2 - L^2}) \]
The derivative of \( \sqrt{4r^2 - L^2} \) is:
\[ \frac{d}{dL}(\sqrt{4r^2 - L^2}) = \frac{-L}{\sqrt{4r^2 - L^2}} \]
So,
\[ \frac{dA}{dL} = \sqrt{4r^2 - L^2} - \frac{L^2}{\sqrt{4r^2 - L^2}} \]
Setting this equal to zero for maximum area:
\[ \sqrt{4r^2 - L^2} - \frac{L^2}{\sqrt{4r^2 - L^2}} = 0 \]

### Step 6: Solve for Length \( L \)
Multiplying through by \( \sqrt{4r^2 - L^2} \):
\[ (4r^2 - L^2) - L^2 = 0 \]
\[ 4r^2 - 2L^2 = 0 \]
\[ 2L^2 = 4r^2 \implies L^2 = 2r^2 \implies L = r\sqrt{2} \]

### Step 7: Find the Corresponding Breadth \( B \)
Substituting \( L = r\sqrt{2} \) back into the equation for \( B \):
\[ B = \sqrt{4r^2 - (r\sqrt{2})^2} = \sqrt{4r^2 - 2r^2} = \sqrt{2r^2} = r\sqrt{2} \]

### Step 8: Calculate the Maximum Area
Now, substituting \( L \) and \( B \) into the area formula:
\[ A = L \cdot B = (r\sqrt{2}) \cdot (r\sqrt{2}) = 2r^2 \]

### Conclusion
The maximum area of the rectangle inscribed in a circle of radius \( r \) is:
\[ \boxed{2r^2} \]