3. As shown in Figure 2, in rectangle $A B C D$, $E$ is a point on side $A B$, $A B=14, C E=13, D E$ $=15, C F \perp D E$ at point $F$, connect $A F$ and $B F$. Find the area of $\triangle A B F$.
3. 36.96 .

Let $B E=x$. Then $A E=14-x$.
In Rt $\triangle A D E$ and Rt $\triangle B C E$, applying the Pythagorean theorem respectively, we get
$$
D E^{2}-A E^{2}=A D^{2}=B C^{2}=C E^{2}-B E^{2} \text {, }
$$

which is $15^{2}-(14-x)^{2}=13^{2}-x^{2}$.
Solving for $x$ gives $x=5$.
Then $A D=B C=13^{2}-x^{2}=12$.
Therefore, $S_{\triangle C O E}=\frac{1}{2} A B \cdot B C=84$.
Also, Rt $\triangle C D F \backsim \mathrm{Rt} \triangle D E A$, so
$$
\frac{S_{\triangle C D F}}{S_{\triangle D E 1}}=\frac{C D^{2}}{D E^{2}} \text {. }
$$

Thus, $S_{\triangle C D F}=\frac{14^{2}}{15^{2}} S_{\triangle D E 1}=47.04$.
Since $S_{\triangle C D F}+S_{\triangle A B F}=\frac{1}{2} S_{\text {trapezoid } A B C D}=84$, then
$$
S_{\triangle A B F}=84-S_{\triangle C D F}=36.96 .
$$