5. As shown in Figure 2, in quadrilateral $ABCD$, it is known that $\angle ACB = \angle ADB = 90^{\circ}$, $AC$ intersects $BD$ at point $E$, $AC = BC$, $AD = 4$, $BD = 7$. Then $S_{\triangle AEB}=$ $\qquad$
5. $11 \frac{9}{11}$.

From the problem, we know that points $A$, $B$, $C$, and $D$ are concyclic, and
$$
\angle C B D = \angle C A D \text{, }
$$
$\triangle A B C$ is an isosceles right triangle, with $\angle C B A = 45^{\circ}$.
Rotate $\triangle B C D$ around point $C$ so that point $B$ coincides with point $A$, then $B D$ falls on $A D$, forming $\triangle A C F$.
Thus, $A F = 7$, $D F = 7 - 4 = 3$,
$\angle F D C = \angle A B C = 45^{\circ}$.
Draw $C H \perp D F$ at point $H$. Then
$$
C H = \frac{1}{2} D F = \frac{3}{2} \text{. }
$$

Therefore, $S_{\triangle A C D} = \frac{1}{2} \times 4 \times \frac{3}{2} = 3$.
And $S_{\triangle B C} = \frac{1}{4} \times (4^2 + 7^2) = \frac{65}{4}$, so
$$
\frac{D E}{E B} = \frac{S_{\triangle A C D}}{S_{\triangle A B C}} = \frac{3}{\frac{65}{4}} = \frac{12}{65} \Rightarrow \frac{S_{\triangle A B D}}{S_{\triangle A B E}} = \frac{D B}{E B} = \frac{77}{65} \text{. }
$$

Also, $S_{\triangle A B D} = \frac{1}{2} \times 4 \times 7 = 14$, thus,
$$
S_{\triangle A B E} = \frac{14 \times 65}{77} = \frac{910}{77} = 11 \frac{9}{11} \text{. }
$$