If the perimeter of a square and a rectangle are the same, then the area P and Q enclosed by them would satisfy the condition

Options

A. `P lt Q`
B. `P le Q`
C. `P gt Q`
D. `P = Q`

Answer

To solve the problem, we need to establish the relationship between the areas of a square and a rectangle when both have the same perimeter.

### Step-by-Step Solution:

1. **Define the Perimeter of the Square**:
Let the side length of the square be \( s \).
The perimeter \( P_s \) of the square is given by:
\[ P_s = 4s \]

2. **Define the Perimeter of the Rectangle**:
Let the length of the rectangle be \( l \) and the breadth be \( b \).
The perimeter \( P_r \) of the rectangle is given by:
\[ P_r = 2(l + b) \]

3. **Set the Perimeters Equal**:
Since the perimeters are the same, we can set them equal to each other:
\[ 4s = 2(l + b) \]
Simplifying this gives:
\[ 2s = l + b \]

4. **Calculate the Area of the Square**:
The area \( A_s \) of the square is:
\[ A_s = s^2 \]

5. **Calculate the Area of the Rectangle**:
The area \( A_r \) of the rectangle is:
\[ A_r = l \times b \]

6. **Express \( l \) in terms of \( b \)**:
From the equation \( 2s = l + b \), we can express \( l \) as:
\[ l = 2s - b \]

7. **Substitute \( l \) in the Area of the Rectangle**:
Now substitute \( l \) into the area of the rectangle:
\[ A_r = (2s - b) \times b = 2sb - b^2 \]

8. **Compare the Areas**:
We need to show that \( A_s \) is greater than \( A_r \):
\[ s^2 > 2sb - b^2 \]
Rearranging gives:
\[ s^2 + b^2 > 2sb \]
This is equivalent to:
\[ (s - b)^2 > 0 \]
This inequality holds true as long as \( s \neq b \).

### Conclusion:
Thus, we conclude that if the perimeter of a square and a rectangle are the same, the area of the square \( P \) will always be greater than the area of the rectangle \( Q \) when the rectangle is not a square.