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:
$\text{(A) } \frac{3r^2}{16}\quad \text{(B) } \frac{\pi r^2}{16}\quad \text{(C) } \frac{\pi r^2\sqrt{2}}{8}\quad \text{(D) } \frac{r^2\sqrt{3}}{32}\quad \text{(E) } \frac{r^2\sqrt{6}}{48}$
Because $AO = OB = AB$, $\triangle AOB$ is an [equilateral triangle](https://artofproblemsolving.com/wiki/index.php/Equilateral_triangle), and $\angle OAB = 60^\circ$.  Using 30-60-90 triangles, $AM = \tfrac{r}{2}$, $AD = \tfrac{r}{4}$, and $DM = \tfrac{r\sqrt{3}}{4}$.  Thus, the area of $\triangle ADM$ is $\tfrac{1}{2} \cdot \tfrac{r}{4} \cdot \tfrac{r\sqrt{3}}{4} = \boxed{\textbf{(D) } \tfrac{r^2 \sqrt{3}}{32}}$.