SOLUTION: A parking lot is 75 feet wide by 60 feet long. It is being torn up to install a sidewalk of uniform width all the way around. If the area of the new parking lot is 2/3 of the origi

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Question 357803: A parking lot is 75 feet wide by 60 feet long. It is being torn up to install a sidewalk of uniform width all the way around. If the area of the new parking lot is 2/3 of the original one, find the width of the sidewalk.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A parking lot is 75 feet wide by 60 feet long.
It is being torn up to install a sidewalk of uniform width all the way around.
If the area of the new parking lot is 2/3 of the original one, find the width of the sidewalk.
:
Find the original area; 75 * 60 = 4500 sq/ft
Find 2/3 of that: 2%2F3* 4500 = 3000 sq/ft, is the new parking lot area
:
Let x = the width of the sidewalk
then the dimensions of new parking lot will be: (75-2x) by (60-2x)
:
the area equation
(75-2x)*(60-2x) = 3000
FOIL
4500 - 150x - 120x + 4x^2 = 3000
A quadratic equation
4x^2 - 270x + 4500 - 3000 = 0
4x^2 - 270x + 1500 = 0
simplify, divide by 2
2x^2 - 135x + 750 = 0
Solve for x using the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In this equation: a=2; b=-135, c=750
x+=+%28-%28-135%29+%2B-+sqrt%28-135%5E2-4%2A2%2A750+%29%29%2F%282%2A2%29+
:
x+=+%28135+%2B-+sqrt%2818225-6000+%29%29%2F4+
:
x+=+%28135+%2B-+sqrt%2812225+%29%29%2F4+
Two solutions
x+=+%28135+%2B+110.5667%29%2F4+
x = 245.5667%2F4
x = 61.4; obviously not a good solution
and
x+=+%28135+-+110.5667%29%2F4+
x = 24.43%2F4
x = 6.1 ft is the width of the side walk
:
:
We can check this by finding the area with this value
(75-2(6.1)) * (60-2(6.1)) =
62.8 * 47.8 = 30001.84 ~ 3000, close enough