SOLUTION: The length of a rectangle is 4 yd longer than its width. If the perimeter of the rectangle is 40 yds, find its area. I am desperate for help
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Question 349720: The length of a rectangle is 4 yd longer than its width. If the perimeter of the rectangle is 40 yds, find its area. I am desperate for help Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! HI,
*Note: the P of a rectangle is the sum of all 4 sides or is generaly written as:
2*l + 2*w = P and area is wrtten as l*w.
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The question states the following to be true:
Let w represent the width
l = (w + 4)
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2*(w + 4) + 2*w = 40yds
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Simplify and solve for w
4w + 8 = 40yds
4w = 32yds
w= 8 yds
l = 12yds
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A = l*w
