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Question 95403: The opposite sides of any rectangle are equal. If a rectangle is twice as long as it is wide and its perimeter is 48 inches, how wide is it? How long?
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Let L represent the length and W represent the width.
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The problem tells you that the length L equals twice the width or 2W. In equation form this
is:
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L = 2W
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From the definition of perimeter, we know that the perimeter P of a rectangle is given by
the equation:
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L + W + L + W = P
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If we add the common terms on the left side we get 2L + 2W so the equation for the perimeter
becomes:
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2L + 2W = P
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But earlier we found that L = 2W so we can substitute 2W for L and make the perimeter
equation become:
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2(2W) + 2W = P
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Multiplying 2 times 2W on the left side results in:
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4W + 2W = P
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Then we can combine the terms on the left side to get 6W and we can substitute 48 inches
for P (which was given in the problem). As a result the perimeter equation becomes:
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6W = 48
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Solve for W by dividing both sides of the equation by 6 which is the multiplier of W.
This division of both sides by 6 results in:
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W = 48/6 = 8
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So we have that the width W is 8 inches. And the problem tells us that the length L
is twice that, so the length is 2 times 8 or 16 inches. Therefore, the rectangle
has 2 sides each 8 inches long, and 2 sides each 16 inches long. That makes the perimeter
(the sum of the 4 sides) equal to 8 + 8 + 16 + 16 = 48 inches. This checks, so our answer
is correct. Width = 8 inches and Length = 16 inches.
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Hope this helps you to understand the problem and to see how it can be worked to find the answer.
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