If the circumference of a circle and the perimeter of a square are equal, then

Options

A. Area of the circle = Area of the square
B. Area of the circle `gt` Area of the square
C. Area of the circle` lt `Area of the square
D. None of these

Answer

To solve the problem where the circumference of a circle is equal to the perimeter of a square, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Circumference of the Circle:**
The formula for the circumference \( C \) of a circle with radius \( r \) is given by:
\[ C = 2\pi r \]

2. **Define the Perimeter of the Square:**
The formula for the perimeter \( P \) of a square with side length \( a \) is:
\[ P = 4a \]

3. **Set the Circumference Equal to the Perimeter:**
According to the problem, the circumference of the circle is equal to the perimeter of the square:
\[ 2\pi r = 4a \]

4. **Solve for the Relationship Between \( a \) and \( r \):**
To find the relationship between the side length \( a \) and the radius \( r \), we can simplify the equation:
\[ \pi r = 2a \]
Dividing both sides by 2 gives:
\[ a = \frac{\pi r}{2} \]

5. **Calculate the Area of the Circle:**
The area \( A_c \) of the circle is given by:
\[ A_c = \pi r^2 \]

6. **Calculate the Area of the Square:**
The area \( A_s \) of the square is given by the square of its side length:
\[ A_s = a^2 \]
Substituting \( a = \frac{\pi r}{2} \) into the area formula for the square:
\[ A_s = \left(\frac{\pi r}{2}\right)^2 = \frac{\pi^2 r^2}{4} \]

7. **Compare the Areas:**
Now we have:
- Area of the circle: \( A_c = \pi r^2 \)
- Area of the square: \( A_s = \frac{\pi^2 r^2}{4} \)

To compare the two areas, we can express them in terms of \( r^2 \):
\[ A_c = \pi r^2 \]
\[ A_s = \frac{\pi^2 r^2}{4} \]

To compare \( A_c \) and \( A_s \), we can divide \( A_c \) by \( A_s \):
\[ \frac{A_c}{A_s} = \frac{\pi r^2}{\frac{\pi^2 r^2}{4}} = \frac{4}{\pi} \]
Since \( \pi \) is approximately 3.14, \( \frac{4}{\pi} \) is less than 1. Therefore:
\[ A_c < A_s \]

### Conclusion:
The area of the circle is less than the area of the square.