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Mensuration
The length of a rectangle is increased by 34%. By what percentage should the width be reduced to
keep the area of the rectangle same? -
1 Reply
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To solve this problem, let’s assume the original length of the rectangle is L and the original width is W. The original area of the rectangle is given by A = L * W.
According to the problem, the length is increased by 34%. This means the new length is L + 0.34L = 1.34L.
Now, we want to find the percentage reduction in the width, which we’ll represent as x%. The new width would then be (1 – x/100)W, since the width is reduced by x%.
To keep the area of the rectangle the same, the original area (A) should be equal to the new area (A’). So, we have:
A = A’
L * W = (1.34L) * ((1 – x/100)W)
To simplify the equation, we can cancel out the common factors:
1 * W = 1.34 * (1 – x/100) * W
Now, we can cancel out the width (W) from both sides:
1 = 1.34 * (1 – x/100)
Next, we can simplify the equation further:
1 = 1.34 – 1.34x/100
Now, let’s isolate the x term by moving the constants to the other side:
1.34x/100 = 1.34 – 1
Simplifying:
1.34x/100 = 0.34
Now, we can solve for x by cross-multiplying:
1.34x = 0.34 * 100
1.34x = 34
Dividing both sides by 1.34:
x = 34/1.34
x ≈ 25.37
Therefore, the width should be reduced by approximately 25.37% to keep the area of the rectangle the same when the length is increased by 34%.
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