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Question 1162644: the length of a rectangle exceeds its width by 4 feet. if the width is doubled and the length is decreased by 2 feet, a new rectangle is formed whose perimeter is 8 feet more than the perimeter of hte original rectangle. find the dimensions of the original rectangle.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! the length of a rectangle exceeds its width by 4 feet.
P = 2L + 2W perimeter of the original rectangle
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if the width is doubled and the length is decreased by 2 feet, a new rectangle is formed whose perimeter is 8 feet more than the perimeter of the original rectangle.
2(2w) + 2(L-2) - 8 = 2L + 2w
4w + 2L - 4 - 8 = 2L + 2W
4w - 2w + 2L - 2L - 12 = 0
2w = 12
w = 12/2
w = 6 ft is the original width'
then
6 + 4 = 10 ft is the original length
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Check, subtract the origninal perimeter from the new perimeter
P2: 2(8) + 2(12) = 40
P1: 2(10) + 2(6) = 32
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perimeter differs: 8 ft
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