The perimeters of a square and a rectangle are equal, If their area be 'A' m' and 'B' m' then correct statement is
एक वर्ग और आयत की परिधि बराबर है। यदि उनके क्षेत्रफल 'A' मी०' और 'B' मी० हो, तो सही कथन क्या है?

Options

A. `A lt B`
B. `A le B`
C. `A gt B`
D. `A ge B`

Explanation

To solve the problem, we need to analyze the relationship between the areas of a square and a rectangle when their perimeters are equal.

### Step-by-Step Solution:

1. **Define the Variables:**
- Let the side length of the square be \( a \).
- Let the length and width of the rectangle be \( L \) and \( B \) respectively.

2. **Write the Perimeter Equations:**
- The perimeter of the square is given by \( P_{square} = 4a \).
- The perimeter of the rectangle is given by \( P_{rectangle} = 2L + 2B \).

3. **Set the Perimeters Equal:**
- Since the perimeters are equal, we have:
\[ 4a = 2L + 2B \]
- Simplifying this gives:
\[ 2a = L + B \quad \text{(1)} \]

4. **Write the Area Equations:**
- The area of the square is:
\[ A = a^2 \]
- The area of the rectangle is:
\[ B = L \times B \]

5. **Substitute \( L \) from Equation (1):**
- From equation (1), we can express \( L \) as:
\[ L = 2a - B \quad \text{(2)} \]
- Substitute this expression for \( L \) into the area of the rectangle:
\[ B = (2a - B) \times B \]

6. **Rearranging the Equation:**
- Rearranging gives:
\[ B = 2aB - B^2 \]
- Rearranging further leads to:
\[ B^2 - 2aB + A = 0 \]

7. **Analyzing the Quadratic Equation:**
- The discriminant of this quadratic equation must be non-negative for real solutions:
\[ D = (2a)^2 - 4 \cdot 1 \cdot A = 4a^2 - 4A \]
- For the area of the square \( A \) to be greater than the area of the rectangle \( B \), we need:
\[ 4a^2 - 4A \geq 0 \implies a^2 \geq A \]

8. **Conclusion:**
- Since \( A = a^2 \), we conclude that:
\[ A > B \]
- Therefore, the area of the square \( A \) is always greater than the area of the rectangle \( B \).

### Final Statement:
Thus, the correct statement is \( A > B \).