1. As shown in Figure 2, quadrilateral $ABCD$ is a rectangle, and $AB = 2BC$. Points $M$ and $N$ are the midpoints of sides $BC$ and $CD$, respectively. $AM$ intersects $BN$ at point $E$. If the area of the shaded region is $a$, then the area of rectangle $ABCD$ is $\qquad$
2.1. $5 a$.
As shown in Figure 3, draw $E F \perp B C$ at point $F$.
Let $A B=4 x, E F =h$. Then
$$
C N=2 x, B M=x \text {. }
$$

From $B C=C N$, we know $\angle N B C=45^{\circ}$.
Thus, $B F=E F=h$.
So $M F=x-h$.
By $\triangle M F E \backsim \triangle M B A$ we get
$$
\begin{array}{l}
\frac{E F}{A B}=\frac{M F}{M B} \Rightarrow \frac{h}{4 x}=\frac{x-h}{x} \\
\Rightarrow h=4 x-4 h \Rightarrow h=\frac{4}{5} x .
\end{array}
$$

Therefore, $S_{\triangle B M E}=\frac{1}{2} x \cdot \frac{4}{5} x=\frac{2}{5} x^{2}$.
But $S_{\triangle B C V}=\frac{1}{2} \cdot 2 x \cdot 2 x=2 x^{2}$, so
$$
S_{\text {quadrilateral CMEN }}=S_{\triangle B C N}-S_{\triangle B M E}=\frac{8}{5} x^{2}=a \text {. }
$$

Thus, $x^{2}=\frac{5}{8} a$.
Therefore, $S_{\text {trapezoid } A B C D}=4 x \cdot 2 x=8 x^{2}=5 a$.