Question 445109: the length of a rectangle is 2 cm less than twice its width. If the perimeter is 56 cm, find the dimensions Found 2 solutions by chriswen, stanbon:Answer by chriswen(106) (Show Source):
You can put this solution on YOUR website! Let x be the width.
Let 2x-2 be the length.
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P=2(l+w)
56=2(2x-2+x)
56=2(3x-2)
56=6x-4
6x=56+4
6x=60
x=60/6
x=10
2x-2=18
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Therefore, the dimensions of the rectangle is 10cm by 18cm.
You can put this solution on YOUR website! the length of a rectangle is 2 cm less than twice its width.
Let width be "W".
Then length is "2W-2"
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If the perimeter is 56 cm, find the dimensions
Perimeter = 2(width + length)
56 = 2(W + 2W-2)
28 = 3W-2
3W = 30
W = 10 cm (width)
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2W-2 = 2*10-2 = 18 cm (length)
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Cheers,
Stan H.
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