7. As shown in Figure 2, in trapezoid
$A B C D$, $A B / / D C$,
$D C=2 A B=2 A D$. If
$B D=6, B C=4$, then
$S_{A B C D}=\ldots \quad\left(S_{A B C D}\right.$ represents
the area of quadrilateral $A B C D$, the same below.
7. 18

As shown in Figure 6, take the midpoint $M$ of $CD$, and connect $BM$, $AM$.

From the given conditions, it is easy to see that quadrilateral $ADMB$ is a rhombus, and quadrilateral $AMCB$ is a parallelogram. Therefore, $AM = BC = 4$.
Noting that $\triangle ADM \cong \triangle MAB \cong \triangle BMC$, we have $S_{ABCD} = \frac{3}{2} S_{\triangle DMB} = \frac{3}{2} \times \frac{1}{2} \times 6 \times 4 = 18$