smc

Let the triangle be $△ABC$ with base $\overline{BC}$ and longest parallel segment $\overline{DE}$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$, as in the diagram. By the properties of transversals, we have $\measuredangle ABC=\measuredangle ADE$. Thus, by AA Similarity, we have $△ADE\sim △ABC$ (because they share $\angle BAC$). From the problem, we know that $\frac{AE}{AC}=\frac{9}{10}$, so, by similarity, $\frac{DE}{BC}=\frac{9}{10}$, and so $DE=\frac{9}{10}BC$. Now, let $[△ABC]=A$. Because $\frac{DE}{BE}=\frac{9}{10}$, we know that $\frac{[△ADE]}{[△ABC]}=(\frac{9}{10}{)}^2=\frac{81}{100}$. From the problem, $[BCED]=38$, so $A=[△ADE]+38=\frac{9}{10}A+38$. Solving for $A$ yields $A=\boxed{200}$.