## Chord and Diameter

The given semicircle has a diameter of 8 inches (segment $\overline{CD}$). Chord $\overline{AB}$ is half the length and parallel to diameter $\overline{CD}$. Find the distance from the midpoint of chord $\overline{AB}$ to the midpoint of diameter $\overline{CD}$.
Source: mathcontest.olemiss.edu 1/21/2008

SOLUTION

Let $M$ be the midpoint of chord $\overline{AB} \textup{ and }O$ the midpoint of diameter $\overline{CD}$. Since $O$ is the midpoint of the diameter, $O$ is the center of semicircle.

$\triangle OAB$ is an equilateral triangle because $OA=OB=AB=4$ ($OA \textup{ and }OB$ are radii of the semicircle).

$\overline {OM}$ is perpendicular to $\overline{AB}$ because $O$ is equidistant from the endpoints of $\overline{AB}$.

Apply the Pythagorean theorem to the right $\triangle OMB$:
$OM^2+MB^2=OB^2$
$OM^2+2^2=4^2$
$OM^2+4=16$
$OM^2=12$
$OM=\pm \sqrt{12}$
$OM=2\sqrt{3}$

Answer: $2\sqrt{3}$.