1969 AHSME Problems/Problem 15

Problem

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:

$\text{(A) } \frac{3r^2}{16}\quad \text{(B) } \frac{\pi r^2}{16}\quad \text{(C) } \frac{\pi r^2\sqrt{2}}{8}\quad \text{(D) } \frac{r^2\sqrt{3}}{32}\quad \text{(E) } \frac{r^2\sqrt{6}}{48}$

Solution

$\fbox{D}$

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1969 AHSME Problems/Problem 15

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