If the length of a rectangle is increased by 80%, what would be the percentage decrease (correct to one place of decimal) in the width to maintain the same area?

Options

A. 0.444
B. 0.425
C. 0.402
D. 0.435

Answer

To solve the problem, we need to determine the percentage decrease in the width of a rectangle when its length is increased by 80%, while keeping the area constant.

### Step-by-Step Solution:

1. **Define the Original Dimensions**:
Let the original length of the rectangle be \( L \) and the original width be \( W \).

2. **Calculate the Original Area**:
The area \( A \) of the rectangle is given by:
\[ A = L \times W \]

3. **Increase the Length by 80%**:
The new length \( L' \) after an 80% increase is:
\[ L' = L + 0.8L = 1.8L \]

4. **Set the New Area Equal to the Original Area**:
To maintain the same area after increasing the length, we set the new area equal to the original area:
\[ A' = L' \times W' = A \]
Substituting the expressions we have:
\[ 1.8L \times W' = L \times W \]

5. **Solve for the New Width \( W' \)**:
Dividing both sides by \( L \) (assuming \( L \neq 0 \)):
\[ 1.8W' = W \]
Now, solving for \( W' \):
\[ W' = \frac{W}{1.8} \]

6. **Calculate the Decrease in Width**:
The decrease in width \( D \) is given by:
\[ D = W - W' = W - \frac{W}{1.8} \]
Simplifying this:
\[ D = W \left(1 - \frac{1}{1.8}\right) = W \left(\frac{1.8 - 1}{1.8}\right) = W \left(\frac{0.8}{1.8}\right) = \frac{0.8W}{1.8} \]

7. **Calculate the Percentage Decrease**:
The percentage decrease in width is given by:
\[ \text{Percentage Decrease} = \left(\frac{D}{W} \times 100\right) = \left(\frac{\frac{0.8W}{1.8}}{W} \times 100\right) \]
Simplifying this:
\[ \text{Percentage Decrease} = \left(\frac{0.8}{1.8} \times 100\right) \approx 44.44\% \]

8. **Round to One Decimal Place**:
Rounding \( 44.44\% \) to one decimal place gives:
\[ \text{Percentage Decrease} \approx 44.4\% \]

### Final Answer:
The percentage decrease in the width to maintain the same area is approximately **44.4%**.