If the perimeter of a rectangular is equal to the perimeter of a parallelogram , then the area of the rectangle is more than that of the parallelogram .

Answer

To solve the problem, we need to analyze the relationship between the perimeter and area of a rectangle and a parallelogram.

### Step-by-Step Solution:

1. **Understanding Perimeter**:
- The perimeter of a rectangle is given by the formula:
\[ P_{rectangle} = 2(l + b) \]
where \( l \) is the length and \( b \) is the breadth.
- The perimeter of a parallelogram is given by the formula:
\[ P_{parallelogram} = 2(a + b) \]
where \( a \) and \( b \) are the lengths of the adjacent sides.

2. **Setting Perimeters Equal**:
- According to the problem, the perimeters of the rectangle and the parallelogram are equal:
\[ 2(l + b) = 2(a + b) \]
- Dividing both sides by 2, we have:
\[ l + b = a + b \]

3. **Simplifying the Equation**:
- Subtract \( b \) from both sides:
\[ l = a \]
- This means that the length of the rectangle is equal to one side of the parallelogram.

4. **Calculating Areas**:
- The area of the rectangle is given by:
\[ A_{rectangle} = l \times b \]
- The area of the parallelogram is given by:
\[ A_{parallelogram} = base \times height = a \times h \]
- Since \( l = a \), we can rewrite the area of the parallelogram as:
\[ A_{parallelogram} = l \times h \]

5. **Comparing Areas**:
- To determine if the area of the rectangle is greater than that of the parallelogram, we compare:
\[ A_{rectangle} = l \times b \quad \text{and} \quad A_{parallelogram} = l \times h \]
- For the rectangle's area to be greater than the parallelogram's area, we need:
\[ l \times b > l \times h \]
- Dividing both sides by \( l \) (assuming \( l > 0 \)):
\[ b > h \]

6. **Conclusion**:
- The statement that the area of the rectangle is more than that of the parallelogram is not universally true. It depends on the relationship between the breadth \( b \) of the rectangle and the height \( h \) of the parallelogram. If \( b > h \), then the rectangle's area is greater; otherwise, it is not.