SOLUTION: The length of a rectangle is 2 in. more than twice its width. If the
perimeter of the rectangle is 34 in., find the dimensions of the rectangle.
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-> SOLUTION: The length of a rectangle is 2 in. more than twice its width. If the
perimeter of the rectangle is 34 in., find the dimensions of the rectangle.
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Question 102926: The length of a rectangle is 2 in. more than twice its width. If the
perimeter of the rectangle is 34 in., find the dimensions of the rectangle. Found 2 solutions by bucky, MathLover1: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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You are told that L equals 2 inches more than 2W. In equation form this is:
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L = 2W + 2
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Next you are told that the perimeter (P) is 34 inches. The perimeter is found by adding all
the sides. In other words:
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P = L + W + L + W
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By combining like terms on the right side this can be simplified to:
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P = 2L + 2W
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Substitute 34 for the perimeter to get:
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34 = 2L + 2W
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But since L = 2W + 2 we can substitute 2W + 2 for L in the perimeter equation to get:
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34 = 2(2W + 2) + 2W
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Multiply out the right side by finding the product of 2 and each of the terms inside
the parentheses. When you do, the equation becomes:
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34 = 4W + 4 + 2W
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Get rid of the 4 on the right side by subtracting 4 from both sides to get:
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30 = 4W + 2W
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Combine terms on the right side to reduce the equation to:
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30 = 6W
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Solve for W by dividing both sides by 6 to get:
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5 = W
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So we know the width is 5 inches. But our first equation tells us that the Length is
twice the width plus 2 inches ... or 2 times 5 and then add 2. This makes the length 12
inches.
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In summary, the width is 5 inches and the length is 12 inches.
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Check by adding the 4 sides: 5 + 12 + 5 + 12 to get 34 inches for the perimeter, just as
it should be.
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Hope this helps you to understand the problem and how to solve.
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