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.
:
Find the original area; 75 * 60 = 4500 sq/ft
Find 2/3 of that: {{{2/3}}}* 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 = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
In this equation: a=2; b=-135, c=750
{{{x = (-(-135) +- sqrt(-135^2-4*2*750 ))/(2*2) }}} 
:
{{{x = (135 +- sqrt(18225-6000 ))/4 }}} 
:
{{{x = (135 +- sqrt(12225 ))/4 }}}
Two solutions
{{{x = (135 + 110.5667)/4 }}} 
x = {{{245.5667/4}}}
x = 61.4; obviously not a good solution
and
{{{x = (135 - 110.5667)/4 }}} 
x = {{{24.43/4}}}
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