SOLUTION: The length of a rectangle is four times its width. Its perimeer is 90 m. Find its dimentions. Use P= 2l+2w.

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: The length of a rectangle is four times its width. Its perimeer is 90 m. Find its dimentions. Use P= 2l+2w.      Log On

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Question 109105: The length of a rectangle is four times its width. Its perimeer is 90 m. Find its dimentions. Use P= 2l+2w.
Answer by HyperBrain(694) About Me  (Show Source):
You can put this solution on YOUR website!
Let l=length
____w=width

IF length of a rectangle is four times its width, then l=4w

The perimeter,
P=2l%2B2w
becomes
P=2%284w%29%2B2w=%282%2A4%29w%2B2w=8w%2B2w=10w

But, the perimeter is 90 m . SO,
10w=90
w=10
Thus, the width measures 10 m long

again,
l=4w=4%2A10=40

Thus, the length measures 40 m long

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