5. As shown in Figure 9, in $\triangle A B C$, $\angle C=$ $90^{\circ}, I$ is the intersection of the angle bisectors $A D$ and $B E$ of $\angle A$ and $\angle B$. Given that the area of $\triangle A B I$ is 12. Then the area of quadrilateral $A B D E$ is $\qquad$
5.24

As shown in Figure 16, construct the symmetric point $F$ of $E$ with respect to $AD$, and the symmetric point $G$ of $D$ with respect to $BE$. Then $F$ and $G$ lie on $AB$, and $AF = AE$, $BG = BD$. Therefore, $\angle AIB = \angle EID = 135^{\circ}$, $\angle DIB = \angle EIA = \angle AIF = \angle BIG = 45^{\circ}$. Thus, $\angle EIF = \angle DIG = 90^{\circ}$, $\angle FIG = 45^{\circ}$. Construct $GK \perp IF$ at $K$, and $DH \perp BI$ at $H$. Since $\mathrm{Rt} \triangle DHI \cong \mathrm{Rt} \triangle GKI$, we have $DH = GK$.

Therefore, we have
$$
\begin{array}{l}
S_{\triangle DIE} = \frac{1}{2} \times EI \times DH \\
S_{\triangle IFG} = \frac{1}{2} \times IF \times GK
\end{array}
$$

Since $EI = IF$ and $DH = GK$, we have
$$
S_{\triangle DIE} = S_{\triangle FFC}.
$$

Therefore, $S_{\text{quadrilateral } ABDE} = 2 \times S_{\triangle ABI} = 24$.