SOLUTION: What is the length of a rectangular parking lot which covers 28,000 square feet and is twice as long as it is wide?

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Question 739588: What is the length of a rectangular parking lot which covers 28,000 square feet and is twice as long as it is wide?
Found 2 solutions by Alan3354, MathLover1:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
What is the length of a rectangular parking lot which covers 28,000 square feet and is twice as long as it is wide?
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Area = L*W = 28000
L = 2W
2W*W = 28000
W^2 = 14000
W = sqrt(14000)
L = 2sqrt(14000)
L =~ 236.64 feet

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
given:
A=+28000ft%5E2
L-length
W-width
if A=+28000ft%5E2, then L%2AW=28000ft%5E2
if L is twice as long as it is W, then
L=2W..substitute in L%2AW=28000ft%5E2 and solve for W
2W%2AW=28000ft%5E2
2W%5E2=28000ft%5E2
W%5E2=28000ft%5E2%2F2
W%5E2=14000ft%5E2
W=sqrt%2814000ft%5E2%29
W=118.3216ft

L=2%2A118.3216ft
highlight%28L=236.6432ft%29