SOLUTION: The width of a rectangle is 22 centimeters shorter that its length. The perimiter of the rectangular is 160. find its length.

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Question 164311This question is from textbook
: The width of a rectangle is 22 centimeters shorter that its length. The perimiter of the rectangular is 160. find its length. This question is from textbook

Found 2 solutions by algebrapro18, stanbon:
Answer by algebrapro18(249) (Show Source):
You can put this solution on YOUR website!
The width of a rectangle is 22 centimeters shorter that its length. The perimiter of the rectangular is 160. find its length.
let w be the width and L be the length(both in centemeters).
The width of a rectangle is 22 centimeters shorter that its length.
converting this into an equation you get
w = L - 22
The perimiter of the rectangular is 160. find its length.
converting this into an equation you get
2L + 2w(equation of the perimeter of a rectangle) = 160
now we substitute the first equation into the second one and solve for the length.
2L + 2w = 160
2L + 2(L - 22) = 160
2L + 2L -44 = 160
4L -44 = 160
4L = 204
L = 51

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
The width of a rectangle is 22 centimeters shorter that its length. The perimiter of the rectangular is 160. find its length.
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Perimeter = 2(length + width)
160 = 2(L + 22)
80 = L+22
L = 58 cm.
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Cheers,
Stan H.