10. As shown in Figure 5, in trapezoid $A B C D$, $A D / / B C, C E$ is the bisector of $\angle B C D$, and $C E \perp A B$, with $E$ being the foot of the perpendicular, $B E=2 A E$. If the area of quadrilateral $A E C D$ is 1, then the area of trapezoid $A B C D$ is $\qquad$ .
$10.2 \frac{1}{7}$.
As shown in Figure 9, extend $B A$ and $C D$ to intersect at point $P$, then $C E \perp B P$. Therefore, $\triangle B C P$ is an isosceles triangle. Thus, $E P = B E = 2 A E, A P = A E$.
Hence, $A P: B P = 1: 4$.
Let $S_{\triangle B C P} = S$, then
$$
S_{\triangle P R D} = \left(\frac{1}{4}\right)^{2} S = \frac{S}{16}.
$$

Also, $S_{\triangle B C P} = S_{\triangle A D} + S_{\text{quadrilateral } E C C D}$, which means $\frac{S}{2} = 1 + \frac{S}{16}$: solving for $S$ gives $S = \frac{16}{7}, S_{\triangle P N D} = \frac{S}{16} = \frac{1}{7}$.
Therefore, $S_{\text{trapezoid } A B C D} = S - S_{\triangle P A D} = \frac{15}{7} = 2 \frac{1}{7}$.