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}.

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About mvtrinh

Retired high school math teacher.
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