The length of a rectangle is reduced by 30%. By what percent would the width have to be increased to maintain the original area?

Options

A. 42.85%
B. 44.50%
C. 42.50%
D. 44.50%

Answer

To solve the problem, we need to determine how much the width of a rectangle must be increased in order to keep the area constant after the length has been reduced by 30%.

Let's denote:
- Original length of the rectangle = L
- Original width of the rectangle = W
- Original area of the rectangle = A = L × W

1. **Calculate the new length after a 30% reduction:**
- The new length (L') can be calculated as:
\[ L' = L - (30\% \text{ of } L) = L - 0.3L = 0.7L \]

2. **Set up the equation for the area:**
- The original area (A) is:
\[ A = L \times W \]
- The new area (A') after the length is reduced and width is increased (let's denote the new width as W') is:
\[ A' = L' \times W' = (0.7L) \times W' \]
- To maintain the original area, we set A = A':
\[ L \times W = (0.7L) \times W' \]

3. **Simplify the equation:**
- Dividing both sides by L (assuming L ≠ 0):
\[ W = 0.7 \times W' \]

4. **Solve for the new width (W'):**
- Rearranging the equation gives:
\[ W' = \frac{W}{0.7} = \frac{W}{\frac{7}{10}} = W \times \frac{10}{7} \]

5. **Calculate the percentage increase in width:**
- The increase in width is:
\[ \text{Increase} = W' - W = W \times \frac{10}{7} - W = W \left(\frac{10}{7} - 1\right) = W \left(\frac{10 - 7}{7}\right) = W \times \frac{3}{7} \]
- The percentage increase is given by:
\[ \text{Percentage Increase} = \left(\frac{\text{Increase}}{W}\right) \times 100 = \left(\frac{W \times \frac{3}{7}}{W}\right) \times 100 = \frac{3}{7} \times 100 \approx 42.86\% \]

Thus, the width must be increased by approximately **42.86%** to maintain the original area.