8. (6 points) In the figure, $ABCD$ is a right trapezoid, with the upper base $AD=2$, the lower base $BC=6$, and $E$ is a point on $DC$. The area of triangle $ABE$ is 15.6, and the area of triangle $AED$ is 4.8. Then the area of trapezoid $ABCD$ is $\qquad$.
【Analysis】According to the problem, we can first find the height on the base $A D$ of triangle $A D E$, then find the height on the base $B C$ of triangle $B E C$, which allows us to find the area of triangle $E C B$. It is not difficult to find the area of trapezoid $A B C D$.

【Solution】According to the analysis, first find the height on the base $A D$ of triangle $A D E$ $=4.8 \div\left(\frac{1}{2} \times A D\right)=4.8 \div 1$ $=4.8$,

As shown in the figure, draw $E G \perp B C, E F \perp A B$, clearly $E G=E F$, from the area of the trapezoid, we have
$$
\frac{1}{2} \times(A D+B C) \times A B=\frac{1}{2} \times(2+6) \times(A F+F B)=4 \times(4.8+E G),
$$

The area of trapezoid $=S_{\triangle A D E}+S_{\triangle A B E}+S_{\triangle B C E}=15.6+4.8+\frac{1}{2} \times \mathrm{BC} \times \mathrm{EG}=20.4+\frac{1}{2} \times 6 \times \mathrm{EG}=$ $20.4+3 E G$,
$4 \times(4.8+E G)=20.4+3 E G$, solving this gives: $E G=1.2$, thus the area of trapezoid $A B C D$ $=4 \times(4.8+E G)=4 \times(4.8+1.2)=24$. Therefore, the answer is: 24.