1. As shown in Figure 1, in trapezoid $ABCD$, $DC \parallel AB$, $DC : AB = 1 : 2$, $MN \parallel BD$ and bisects $AC$. If the area of trapezoid $ABCD$ is $\overline{ab}$, and $S_{\triangle AMN} = \overline{ba}$, then $\bar{a} + \bar{b} + \overline{ba} =$
ニ、1.99.
$\triangle A M N \backsim \triangle A B D$, similarity ratio $=\frac{\frac{1}{2} A C}{\frac{2}{3} A C}=\frac{3}{4}$,
$$
\begin{array}{l}
S_{\triangle A M N}=\frac{9}{16} S_{\triangle A B D}=\frac{9}{16} \times \frac{2}{3} S_{A B C D}=\frac{3}{8} S_{A B C D} \Rightarrow(10 b \\
+a)=\frac{3}{8}(10 a+b) \Rightarrow 77 b=22 a, 7 b=2 a, \text { when } a=7
\end{array}
$$

when $a=7$, $b=2 . \therefore \overline{a b}+\overline{b a}=99$.