## Precalculus (6th Edition) Blitzer

The amount of cable used is $\sqrt{36+{{x}^{2}}}+\sqrt{{{x}^{2}}-20x+164}$
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 $\sqrt{36+{{x}^{2}}}+\sqrt{{{x}^{2}}-20x+164}$.