Answer
$\approx14 \text{ } ft$
Work Step by Step
Using $a^2+b^2=c^2$ or the Pythagorean Theorem, the conditions of the problem translate to
\begin{array}{l}\require{cancel}
x^2+(x+8)^2=36^2
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x^2+[(x)^2+2(x)(8)+(8)^2]=1296
\\
x^2+[x^2+16x+64]=1296
\\
2x^2+16x+64-1296=0
\\
2x^2+16x-1232=0
\\
\dfrac{2x^2+16x-1232}{2}=\dfrac{0}{2}
\\
x^2+8x-616=0
.\end{array}
In the equation above, $a=
1
,$ $b=
8
,$ and $c=
-616
.$ Using the Quadratic Formula which is given by $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
,$ then
\begin{array}{l}\require{cancel}
x=\dfrac{-8\pm\sqrt{8^2-4(1)(-616)}}{2(1)}
\\
x=\dfrac{-8\pm\sqrt{64+2464}}{2}
\\
x=\dfrac{-8\pm\sqrt{2528}}{2}
\\
x=\dfrac{-8\pm\sqrt{16\cdot158}}{2}
\\
x=\dfrac{-8\pm\sqrt{4^2\cdot158}}{2}
\\
x=\dfrac{-8\pm4\sqrt{158}}{2}
\\
x=\dfrac{\cancel2\cdot(-4)\pm\cancel2(2)\sqrt{158}}{\cancel2}
\\
x=-4\pm2\sqrt{158}
.\end{array}
Since $x$ is a measurement (and hence should be nonnegtive), then $
x=-4+2\sqrt{158}
.$
Walking around the lot measures $x+x+8.$ Hence,
\begin{array}{l}\require{cancel}
(-4+2\sqrt{158})+(-4+2\sqrt{158}+8)
\\=
-4+2\sqrt{158}-4+2\sqrt{158}+8
\\=
(-4-4+8)+(2\sqrt{158}+2\sqrt{158})
\\=
4\sqrt{158}
.\end{array}
Cutting across the lot would save a walking distance of
\begin{array}{l}\require{cancel}
4\sqrt{158}-36
\\
\approx14 \text{ } ft
.\end{array}
