In a circle with center $O$, a chord $PQ$ is such that $OM\pm PQ$ meeting $PQ$ at $M$. Then
- ${ OQ }^{ 2 }={ OM }^{ 2 }+\dfrac { 1 }{ 2 } { PQ }^{ 2 }$
- ${ OQ }^{ 2 }={ OM }^{ 2 }+\dfrac { 1 }{ 4 } { PQ }^{ 2 }$
- ${ MQ }^{ 2 }={ OM }^{ 2 }- { OQ }^{ 2 }$
- ${ OM }^{ 2 }={ MQ }^{ 2 }- { OQ }^{ 2 }$
Reveal answer
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B
Correct answer
Explanation
In right triangle OMQ, OQ^2 = OM^2 + MQ^2. Since M is the midpoint of chord PQ, MQ = PQ/2. Thus MQ^2 = (PQ/2)^2 = PQ^2/4. Substituting gives OQ^2 = OM^2 + PQ^2/4.
AI explanation
Since the perpendicular from the center bisects the chord, the length of MQ is half of PQ, meaning MQ equals PQ divided by 2. In the right triangle OMQ, we use the Pythagorean theorem to relate the hypotenuse OQ and the legs OM and MQ. The theorem states that OQ squared equals OM squared plus MQ squared. Substituting MQ with one half of PQ gives the final identity OQ squared equals OM squared plus one fourth of PQ squared.