Question

# In a circle with center $$O$$, a chord $$PQ$$ is such that $$OM\pm PQ$$ meeting $$PQ$$ at $$M$$. Then

A
OQ2=OM2+12PQ2
B
OQ2=OM2+14PQ2
C
MQ2=OM2OQ2
D
OM2=MQ2OQ2

Solution

## The correct option is A $${ OQ }^{ 2 }={ OM }^{ 2 }+\dfrac { 1 }{ 4 } { PQ }^{ 2 }$$$$OM\pm PQ$$ or OM bisects PQor $$MQ=\frac { 1 }{ 2 } PQ$$or $$MQ^{ 2 }+\frac { 1 }{ 4 } { PQ }^{ 2 }$$By pythagoras theorm,$${ OQ }^{ 2 }={ OM }^{ 2 }+{ MQ }^{ 2 }\\ ={ OM }^{ 2 }+\frac { 1 }{ 4 } { PQ }^{ 2 }$$Mathematics

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