SOLUTION: Help! I'm 28 and back in college! A rectangle is twice as long as it is wide. The perimeter is 60 feet. Find the dimensions. Help!

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Question 96227: Help! I'm 28 and back in college! A rectangle is twice as long as it is wide. The perimeter is 60 feet. Find the dimensions. Help!
Found 2 solutions by stanbon, bucky:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A rectangle is twice as long as it is wide. The perimeter is 60 feet. Find the dimensions.
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Draw a picture of a rectangle.
Label the width as "x" ft. and the length must as "2x" ft.
EQUATION:
Perimeter is the distance around the rectangle.
Perimeter = 2*length + 2*width
60 = 2*2x + 2*x
Divde both sides by 2 to get:
30 = 2x + x
Add the like term on the right side to get:
3x=30
Divide both sides by 3 to get:
x=10 feet (width)
2x=20 ft. (length)
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Welcome back.
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Cheers,
Stan H.

Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website!
Back in college? Good for you! Stick with it ...
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Let W represent the width of the rectangle and let L represent the length.
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You are told that the length is twice as long as the width. So 2 widths equals the length. In
equation form you can write this as:
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L = 2W <=== call this equation 1
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Next, you are told that the perimeter (call it P) of the rectangle is 60 feet. The perimeter is
found by adding the dimensions of all 4 sides of the rectangle. Therefore, the equation
for finding the perimeter is:
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P = L + W + L + W <=== call this equation 2
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We now have two equations that apply to this problem. We use one of these equations to
help us solve the other. From equation 1 we know that L equals 2W. So in the second
equation we can replace each L by its equivalent 2W. When we do that replacement equation 2
becomes:
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P = 2W + W + 2W + W
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The terms on the right side all contain W. Therefore, we can add them together (2 + 1 + 2 + 1)
to get 6W.
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This makes the equation become:
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P = 6W
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The problem tells us that P is 60 feet. Substituting 60 for P results in:
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60 = 6W
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Now we can solve for W by dividing both sides of this equation by 6. If you do that division
the equation reduces to:
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10 = W
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So we have found that the width of the rectangle is 10 feet. And equation 1 tells us
that the length of the rectangle is 2 times the width. So the length is 2 times 10 feet
or 20 feet.
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Now check. The perimeter is the distance around the rectangle ... 20 feet + 10 feet + 20 feet
+ 10 feet and sure enough those distances add to 60 feet. Therefore, we have satisfied
the requirements of the problem ... the length is twice as long as the width (20 feet and
10 feet respectively) and the perimeter is 60 feet. Our answers of L = 20 feet and W = 10 feet
are correct.
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Hope this helps you to understand the problem and how to go about solving it.
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