AHSME 1969 Problem 15

In a circle with center $O$ and radius $r$, chord $AB$ is drawn with length equal to $r$ (units). From $O$, a perpendicular to $AB$ meets $AB$ at $M$. From $M$ a perpendicular to $OA$ meets $OA$ at $D$. In terms of $r$ the area of triangle $MDA$, in appropriate square units, is: $\text{(A) } \frac{3r^2}{16} \text{(B) } \frac{\pi r^2}{16} \text{(C) } \frac{\pi r^2\sqrt{2}}{8} \text{(D) } \frac{r^2\sqrt{3}}{32} \text{(E) } \frac{r^2\sqrt{6}}{48}$