Answer
$\approx16 \text{ } ft$
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+10)^2=40^2
.\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)(10)+(10)^2]=1600
\\
x^2+[x^2+20x+100]=1600
\\
2x^2+20x+100-1600=0
\\
2x^2+20x-1500=0
\\
\dfrac{2x^2+20x-1500}{2}=\dfrac{0}{2}
\\
x^2+10x-750=0
.\end{array}
In the equation above, $a=
1
,$ $b=
10
,$ and $c=
-750
.$ 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{-10\pm\sqrt{10^2-4(1)(-750)}}{2(1)}
\\
x=\dfrac{-10\pm\sqrt{100+3000}}{2}
\\
x=\dfrac{-10\pm\sqrt{3100}}{2}
\\
x=\dfrac{-10\pm\sqrt{100\cdot31}}{2}
\\
x=\dfrac{-10\pm\sqrt{10^2\cdot31}}{2}
\\
x=\dfrac{-10\pm10\sqrt{31}}{2}
\\
x=\dfrac{\cancel2\cdot(-5)\pm\cancel2\cdot5\sqrt{31}}{\cancel2}
\\
x=-5\pm5\sqrt{31}
.\end{array}
Since $x$ is a measurement (and hence should be nonnegtive), then $
x=-5+5\sqrt{31}
.$
Walking around the lot measures $x+x+10.$ Hence,
\begin{array}{l}\require{cancel}
(-5+5\sqrt{31})+(-5+5\sqrt{31}+10)
\\=
(-5-5+10)+(5\sqrt{31}+5\sqrt{31})
\\=
10\sqrt{31}
.\end{array}
Cutting across the lot would save a walking distance of
\begin{array}{l}\require{cancel}
10\sqrt{31}-40
\\
\approx16 \text{ } ft
.\end{array}
