SOLUTION: The local park measures 60 m by 50 m. Part of the park is torn up to install a sidewalk of uniform width about it, reducing the area of the park itself by 321 m2. How wide is the s

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Question 1087632: The local park measures 60 m by 50 m. Part of the park is torn up to install a sidewalk of uniform width about it, reducing the area of the park itself by 321 m2. How wide is the sidewalk?
Found 2 solutions by rothauserc, ikleyn:
Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
let l be the length of the internal rectangle and w be the width of the internal rectangle
:
1) (l+2x) * (w+2x) = 3000
2) 2(l+2x) + 2(w+2x) = 220
3) l * w = 3000 - 321 = 2679
:
expand equation 1 by multiplication of left side of =
:
lw +2xw +2xl +4x^2 = 3000
4) 4x^2 +2x(w+l) = 321
:
use equation 2 to get expression for w+l
:
l +w + 4x = 110
l +w = 110-4x
:
substitute for l+w in equation 4
:
4x^2 +2x(110-4x) = 321
:
4x^2 +220x -8x^2 = 321
4x^2 -220x +321 = 0
:
use quadratic formula to solve for x
:
x = (-(-220) + square root((-220)^2 -4*4*321)) / 2(4) = 53.5
x = (-(-220) - square root((-220)^2 -4*4*321)) / 2(4) = 1.5
:
we accept x = 1.5 m
:
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the sidewalk is 1.5 m wide
******************************
:
Answer by ikleyn(52777)   (Show Source): You can put this solution on YOUR website!
.
The setup is made (and should be made) in 3 lines: 

60*50 - (60-2x)*(50-2x) = 321,

220x - 4x^2 = 321,

4x^2 - 220x + 321 = 0.


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