Four, $E, F$ are
on the sides $B C$ and $C D$
of rectangle $A B C D$,
if the areas of $\triangle C E F$,
$\triangle A B E$, $\triangle A D F$
are 3,
4, 5 respectively. Find the area $S$ of $\triangle A E F$.
Let $AB = a, BC = b$, then $BE = \frac{8}{a}$,
$$
\begin{array}{l}
CE = b - \frac{8}{a}, DF = \frac{10}{b}, FC = a - \frac{10}{b}. \\
\left\{\begin{array}{l}
\frac{1}{2}\left(b - \frac{8}{a}\right) \times \left(a - \frac{10}{b}\right) = 3, \\
ab = 3 + 4 + 5 + S .
\end{array}\right. \\
S = \sqrt{144 - 80} = 8 .
\end{array}
$$