College Algebra, Chapter 1, 1.3, Section 1.3, Problem 90

Suppose that a flagpole is secured on opposite sides by two guy wires, each of which is $5 ft$ longer than the pole. The distance between the points where the wires are fixed to the ground is equal to the length of one guy wire. What is the height of the flagpole to the nearest inch?

Let $h$ be the height of the pole







By Pythagorean Theorem,


$
\begin{equation}
\begin{aligned}

a^2 + b^2 =& c^2
&&
\\
\\
h^2 + \left( \frac{h + 5}{2} \right)^2 =& (h + 5)^2
&& \text{Model}
\\
\\
h^2 + \frac{h^2 + 10 h + 25}{4} =& h^2 + 10 h + 25
&& \text{Expand}
\\
\\
4h^2 + h^2 + 10h + 25 =& 4h^2 + 40h + 100
&& \text{Multiply both sides by 4}
\\
\\
4h^2 - 4h^2 + h^2 - 40h + 10h - 100 + 25 =& 0
&& \text{Combine like terms}
\\
\\
h^2 - 30h - 75 =& 0
&& \text{Simplify}
\\
\\
h^2 - 30h =& 75
&& \text{Add 75}
\\
\\
h^2 - 30h + 225 =& 75 + 225
&& \text{Complete the square: add } \left( \frac{-30}{2} \right)^2 = 225
\\
\\
(h - 15)^2 =& 300
&& \text{Perfect Square}
\\
\\
h - 15 =& \pm \sqrt{300}
&& \text{Take the square root}
\\
\\
h =& 15 \pm 10 \sqrt{3}
&& \text{Add 15}
\\
\\
h =& 15 + 10 \sqrt{3} \text{ and } h = 15 - 10 \sqrt{3}
&& \text{Solve for } h
\\
\\
h =& 15 + 10 \sqrt{3} ft
&& \text{Choose } h > 0
\\
\\
h =& (15 + 10 \sqrt{3}) ft \left( \frac{12 in}{1ft} \right) = 387.85 in = 388 in
&& \text{Convert $ft$ into inches}

\end{aligned}
\end{equation}
$