[tex]\bf \textit{area of a triangle}\\\\
A=\cfrac{1}{2}bh\qquad
\begin{cases}
b=base\\
h=height\\
----------\\
h=4b\leftarrow
\begin{array}{llll}
\textit{height is 4}\\
\textit{times the base}
\end{array}\\
A=288
\end{cases}\implies 288=\cfrac{1}{2}\cdot b\cdot 2b[/tex]
solve for "b", to see what the base is
what about the height? well, h = 4b
