SOLUTION: How does a 4th grader find the dimensions of a rectangle when the perimeter is 58 and the area is 54.

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Question 427041: How does a 4th grader find the dimensions of a rectangle when the perimeter is 58 and the area is 54.
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
How does a 4th grader find the dimensions of a rectangle when the perimeter is 58 and the area is 54.
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Perimeter = 2L + 2W = 58 --> L + W = 29
Area = L*W = 54
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L = 54/W
(54/W) + W = 29
54 + W^2 = 29W
W%5E2+-+29W+%2B+54+=+0
(W - 2)*(W - 27) = 0
W = 2
L = 27
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I'm not sure that'll work in the 4th grade, it's been awhile since I was there.
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Try this:
Factor 54 and look for a pair that adds to 29
54 =
1 x 54
2 x 27
3 x 18
6 x 9
Only 2 by 27 fits.
If the solution is not integers, that won't work.