Answer
The amount of cable used is \[\sqrt{36+{{x}^{2}}}+\sqrt{{{x}^{2}}-20x+164}\]
Work Step by Step
Consider the length of cable from the top of the $6\text{-foot}$ pole to the ground to be ${{h}_{1}}$ and the length of the cable from the top of the $\text{8-foot}$ pole to the ground to be ${{h}_{2}}$
According to the Pythagoras theorem
In a right-angle triangle,
${{h}^{2}}={{p}^{2}}+{{b}^{2}}$ …… (1)
Substitute $6$ for $p$ , $x$ for $b$ and ${{h}_{1}}$ for $h$ in equation (1)
$\begin{align}
& {{h}_{1}}=\sqrt{{{6}^{2}}+{{x}^{2}}} \\
& =\sqrt{36+{{x}^{2}}}
\end{align}$
The length of the cable from the top of the $6\text{-foot}$ pole to the ground
${{h}_{1}}=\sqrt{36+{{x}^{2}}}$
Substitute $8$ for $p$, $10-x$ for $b$ and ${{h}_{2}}$ for $h$ in equation (1)
${{h}_{2}}=\sqrt{{{8}^{2}}+{{\left( 10-x \right)}^{2}}}$
Use formula ${{\left( A-B \right)}^{2}}={{A}^{2}}-2AB+{{B}^{2}}$ in the above equation
$\begin{align}
& {{h}_{2}}=\sqrt{64+100-20x+{{x}^{2}}} \\
& =\sqrt{164-20x+{{x}^{2}}}
\end{align}$
Or
${{h}_{2}}=\sqrt{{{x}^{2}}-20x+164}$
The length of the cable from the top of the $\text{8-foot}$ pole to the ground
${{h}_{2}}=\sqrt{{{x}^{2}}-20x+164}$
Total length of cable is $\sqrt{36+{{x}^{2}}}+\sqrt{{{x}^{2}}-20x+164}$
The required amount of cable used in terms of the distance from the $6\text{-foot}$ pole to the point where the cable touches the ground, $x$ , where the length of the other pole is $\text{8-foot}$ and distance between poles is $\text{10-foot}$, is $\sqrt{36+{{x}^{2}}}+\sqrt{{{x}^{2}}-20x+164}$.
