3. (8 points) In rectangle $A B C D$, $B E=5, E C=4, C F=4, F D=1$, as shown in the figure, then the area of $\triangle A E F$ is $\qquad$ ;
【Solution】According to the analysis, $A D=B E+E C=5+4=9$,
$$
\begin{array}{l}
A B=1+4=5, S_{\triangle E F C}=\frac{1}{2} \times E C \times F C=\frac{1}{2} \times 4 \times 4=8 ; \\
S_{\triangle A B E}=\frac{1}{2} \times A B \times B E=\frac{1}{2} \times 5 \times 5=12.5 ; \\
S_{\triangle A D F}=\frac{1}{2} \times A D \times D F=\frac{1}{2} \times 9 \times 1=4.5 ; \\
S_{\text {rectangle } A B C D}=A B \times A D=5 \times 9=45,
\end{array}
$$

The area of $\triangle A E F$ is equal to the area of the entire rectangle minus the areas of the three triangles.
$$
S_{\triangle A E F}=S_{\text {rectangle } A B C D}-S_{\triangle E F C}-S_{\triangle A B E}-S_{\triangle A D F}=45-8-12.5-4.5=20 \text {. }
$$

Therefore, the answer is: 20.