Question 243566: The perimeter of a rectangle is 800 feet, and length of it is 50 feet less than half the width. What are it's dimensions?
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! CORRECTION! I read the problem too fast the first time. (sorry)
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The perimeter of a rectangle = 2L + 2W, where L=length and W=width.
We are told L is 50 ft less than HALF the W, so L = .5W -50.
We are told the perimeter = 800.
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Substituting we have:
800 = 2L + 2W
800 = 2(.5W-50) + 2W
Multiplying the factor
800 = W - 100 + 2W
Collecting like terms
800 = 3W - 100
Adding 100 to both sides
900 = 3W
3W = 900
Dividing both sides by 3
W = 900/3 = 300
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Looking back, we know L = .5W-50, so L = 150 - 50 = 100.
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Check the perimeter.
2W + 2L = 2(300) + 2(100) = 800.
This is what it's supposed to be.
Done.
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