SOLUTION: A 50 foot by 60 foot parking lot is torn up to install a sidewalk of uniform width around its perimeter. The new area of the parking lot is three-fourths of the old area. How wide

Question 294418: A 50 foot by 60 foot parking lot is torn up to install a sidewalk of uniform width around its perimeter. The new area of the parking lot is three-fourths of the old area. How wide is the sidewalk?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A 50 foot by 60 foot parking lot is torn up to install a sidewalk of uniform width around its perimeter. The new area of the parking lot is three-fourths of the old area. How wide is the sidewalk?
:
Find the original area of the parking lot: 50 * 60 = 3000 sq/ft
:
Find 3/4 of the area: .75(3000) = 2250 sq/ft
:
Let x = the width of the sidewalk
:
new width*new length = 2250 sq/ft
(50-2x)*(60-2x) = 2250
FOIL
3000 - 100x - 120x + 4x^2 = 2250
Arrange as a quadratic equation on the left
4x^2 - 220x + 3000 - 2250 = 0
:
4x^2 - 220x + 750 = 0
Simplify, divide by 2
2x^2 - 110x + 375 = 0
Solve using the quadratic formula

in this problem a=2, b=-110, c=375

:
I'll let you do the math here, you will get two solutions but only the
smaller value will be a valid solution (x ~ 3.65 ft)
:
Check it in the original equation
(50 - 2(3.65)) *(60 - 2(3.65)) =
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