16. (6 points) As shown in the figure, the area of rectangle $A B C D$ is 60. If $E B=2 A E, A F=F D$, then $S_{\text {quadrilateral } A E O F}=$
【Solution】Solution: Connect $A O, B D$. Since $A F=F D$, we have $S_{\triangle A F O}=S_{\triangle D F O}$.
Since $E B=2 A E$, the area of rectangle $A B C D$ is 60,
$$
\begin{array}{l}
S_{\triangle A B F}=\frac{1}{2} S_{\triangle A B D} \\
=\frac{1}{2} \times \frac{1}{2} \times S_{\text {rectangle } A B C D} \\
=\frac{1}{2} \times \frac{1}{2} \times 60 \\
=15
\end{array}
$$
$$
\begin{array}{l}
E B=2 A E \\
S_{\triangle A D E}=\frac{1}{3} S_{\triangle A B D} \\
=\frac{1}{3} \times \frac{1}{2} \times 60 \\
=10 \\
S_{\triangle E B O}=S_{\triangle D F O}+(15-10)=S_{\triangle D F O}+5=S_{\triangle A F O}+5
\end{array}
$$
$S_{\triangle A E O}=\frac{1}{2} S_{\triangle E B O}$
$$
\begin{array}{c}
S_{\triangle A F O}+S_{\triangle A E O}+S_{\triangle E B O}=15 \\
S_{\triangle A F O}+\frac{1}{2} S_{\triangle E B O}+S_{\triangle E B O}=15 \\
S_{\triangle A F O}+\frac{1}{2} \times\left(S_{\triangle A F O}+5\right)+S_{\triangle A F O}+5=15 \\
S_{\triangle A F O}+\frac{1}{2} S_{\triangle A F O}+2.5+S_{\triangle A F O}+5=15 \\
\frac{5}{2} S_{\triangle A F O}=7.5 \\
S_{\triangle A F O}=3
\end{array}
$$
$S_{\text {quadrilateral } A E O F}=S_{\triangle A D E}-S_{\triangle A F O}=10-3=7$
Answer: The area of $S_{\text {quadrilateral } A E O F}$ is 7.
The answer is: 7.