College Algebra, Chapter 1, 1.2, Section 1.2, Problem 38

Suppose that a half acre building lot is five times as long as it is wide.

Let $L$ and $W$ be the dimensions length and width of the building so..


$
\begin{equation}
\begin{aligned}

A =& LW
&& \text{Model (recall that $1$ Acre } = 43,560 \text{ ft }^2 )
\\
\\
\frac{1}{2} (43560) =& LW
&&
\\
\\
21780 \text{ ft }^2 =& LW
&& \text{Perform the condition } (L = 5W)
\\
\\
21780 =& (5W)W
\\
\\
21780 =& 5W^2
&& \text{Solve for } w
\\
\\
W =& \sqrt{\frac{21780}{5}}
&&
\\
\\
W =& 66 \text{ ft}
&& \text{The width of the building}

\end{aligned}
\end{equation}
$


Since, $L = 5W = 5(66) = 330 $ ft

Therefore, the dimensions of the building are $L = 330 $ ft and $W = 66 $ ft.