If the length of a rectangle is increased by \(66 \frac{2}{3} \%\), then by what percent should the width of the rectangle be decreased in order to maintain the same area ?
40%
Finding Percentage Decrease in Width to Maintain Rectangle Area
This problem asks us to find the percentage by which the width of a rectangle must be decreased to keep its area the same, given that its length is increased by a specific percentage. We need to work with the concept of area of a rectangle and percentage changes.
Understanding the Problem Setup
Let the original length of the rectangle be \(L\) and the original width be \(W\). The original area, \(A\), is given by the formula:
\(A = L \times W\)
The problem states that the length is increased by \(66 \frac{2}{3} \% \). Let the new length be \(L'\) and the new width be \(W'\). The requirement is that the new area, \(A'\), remains the same as the original area, \(A\).
\(A' = L' \times W'\)
We are given that \(A' = A\), so:
\(L' \times W' = L \times W\)
Step-by-Step Calculation
Step 1: Convert the percentage increase to a fraction.
The percentage increase in length is \(66 \frac{2}{3} \% \). First, convert the mixed number to an improper fraction:
\(66 \frac{2}{3} = \frac{66 \times 3 + 2}{3} = \frac{198 + 2}{3} = \frac{200}{3}\)
Now, convert the percentage to a fraction by dividing by 100:
\(\frac{200}{3} \% = \frac{\frac{200}{3}}{100} = \frac{200}{3 \times 100} = \frac{200}{300} = \frac{2}{3}\)
So, the length is increased by a factor equivalent to \(\frac{2}{3}\) of the original length.
Step 2: Calculate the new length \(L'\).
The new length \(L'\) is the original length \(L\) plus the increase:
\(L' = L + \text{Increase}\)
\(L' = L + \left(\frac{2}{3} \times L\right)\)
\(L' = L \left(1 + \frac{2}{3}\right)\)
\(L' = L \left(\frac{3}{3} + \frac{2}{3}\right)\)
\(L' = L \left(\frac{5}{3}\right)\)
So, the new length is \(\frac{5}{3}\) times the original length.
Step 3: Use the condition that the area remains the same.
We know that \(L' \times W' = L \times W\). Substitute the expression for \(L'\) into this equation:
\(\left(\frac{5}{3}L\right) \times W' = L \times W\)
Step 4: Solve for the new width \(W'\) in terms of \(W\).
Divide both sides of the equation by \(\frac{5}{3}L\) (assuming \(L \neq 0\)):
\(W' = \frac{L \times W}{\frac{5}{3}L}\)
\(W' = \frac{W}{\frac{5}{3}}\)
\(W' = W \times \frac{3}{5}\)
\(W' = \frac{3}{5}W\)
The new width is \(\frac{3}{5}\) times the original width. Since \(\frac{3}{5} < 1\), the width has indeed decreased.
Step 5: Calculate the percentage decrease in width.
The decrease in width is the original width minus the new width:
\(\text{Decrease} = W - W'\)
\(\text{Decrease} = W - \frac{3}{5}W\)
\(\text{Decrease} = W \left(1 - \frac{3}{5}\right)\)
\(\text{Decrease} = W \left(\frac{5}{5} - \frac{3}{5}\right)\)
\(\text{Decrease} = W \left(\frac{2}{5}\right)\)
The percentage decrease is calculated as:
\(\text{Percentage Decrease} = \frac{\text{Decrease}}{\text{Original Width}} \times 100\%\)
\(\text{Percentage Decrease} = \frac{\frac{2}{5}W}{W} \times 100\%\)
\(\text{Percentage Decrease} = \frac{2}{5} \times 100\%\)
\(\text{Percentage Decrease} = \frac{200}{5}\%\)
\(\text{Percentage Decrease} = 40\%\)
Therefore, the width of the rectangle should be decreased by 40% to maintain the same area after the length is increased by \(66 \frac{2}{3}\%\).
Summary of Calculation
Original Area = \(L \times W\)
Increase in Length = \(66 \frac{2}{3} \% = \frac{2}{3}\)
New Length \(L' = L + \frac{2}{3}L = \frac{5}{3}L\)
Maintain Area: \(L' \times W' = L \times W\)
\(\left(\frac{5}{3}L\right) \times W' = L \times W\)
\(W' = \frac{L \times W}{\frac{5}{3}L} = \frac{3}{5}W\)
Decrease in width = \(W - W' = W - \frac{3}{5}W = \frac{2}{5}W\)
Percentage Decrease = \(\frac{\frac{2}{5}W}{W} \times 100\% = \frac{2}{5} \times 100\% = 40\%\)
| Concept | Formula / Value |
|---|---|
| Original Area | \(A = L \times W\) |
| Length Increase Percentage | \(66 \frac{2}{3} \% = \frac{2}{3}\) |
| New Length \(L'\) | \(L' = L \left(1 + \frac{2}{3}\right) = \frac{5}{3}L\) |
| Area Condition | \(L' \times W' = L \times W\) |
| New Width \(W'\) | \(W' = \frac{L \times W}{L'} = \frac{L \times W}{\frac{5}{3}L} = \frac{3}{5}W\) |
| Width Decrease | \(W - W' = W - \frac{3}{5}W = \frac{2}{5}W\) |
| Percentage Decrease | \(\frac{\text{Decrease}}{W} \times 100 \% = \frac{\frac{2}{5}W}{W} \times 100 \% = 40 \%\) |
Revision Table: Rectangle Area and Percentage Change
| Term | Definition/Concept | Relevance to Problem |
|---|---|---|
| Rectangle Area | Product of its length and width (\(L \times W\)). | Must remain constant. |
| Percentage Increase | Relative change expressed as a fraction of 100. | Applied to the original length. |
| Percentage Decrease | Relative reduction expressed as a fraction of 100. | Needs to be calculated for the width. |
| Fraction Conversion | Converting percentages or mixed numbers into simple fractions. | Essential for easier calculation of new dimensions. |
Additional Information: Maintaining Area with Proportional Changes
When the area of a rectangle needs to be maintained (\(L \times W = \text{constant}\)), if one dimension changes by a certain factor, the other dimension must change by the reciprocal of that factor. For example, if the length becomes \(k \times L\), then the new width \(W'\) must satisfy \((kL) \times W' = L \times W\), which means \(W' = \frac{L \times W}{kL} = \frac{1}{k}W\). The width becomes \(\frac{1}{k}\) times the original width.
In our problem, the length increased by \(\frac{2}{3}\) of its original value, resulting in a new length of \(L \left(1 + \frac{2}{3}\right) = \frac{5}{3}L\). Here, the factor \(k\) is \(\frac{5}{3}\). To maintain the area, the new width must be \(W' = \frac{1}{5/3}W = \frac{3}{5}W\).
The change in width is \(W - \frac{3}{5}W = \frac{2}{5}W\). The percentage decrease is \(\frac{\text{Change}}{W} \times 100\% = \frac{\frac{2}{5}W}{W} \times 100\% = \frac{2}{5} \times 100\% = 40\%\). This confirms our step-by-step calculation.
This reciprocal relationship is a useful shortcut for problems involving maintaining a constant product when two quantities are inversely related, such as the dimensions of a rectangle with constant area, or speed and time for a constant distance.
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