Prove that the chord near to the centre is larger than the chord away from the centre of a circle .
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Prove that the chord near to the centre is larger than the chord away from the centre of a circle .
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No spams
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PROOF: We know that perpendicular drawn from the centre to the chord , bisect the chord.
Since, OL is perpendicular AB, so
AL = AB/2
and
CM = CD/2
In triangle OAL and OCM.
By pythagoras theorem,
OA^2 = OL^2 + AL^2
Similarly,
OC^2 = OM^2 + CM^2
Now, OA = OC ( radius)
=> OA^2 = OC^2.
=> OL^2 + AL^2 = OM^2 + CM^2 ------(1)
Now, AB > CD
Then, AB/2 > CD/2
=>AL > CM
=> AL^2 > CM^2
=> OL^2 + AL^2 > OL^2 + CM^2
=> OM^2 + CM^2 > OL^2 + CM^2
From equation 1.
=> OL^2 = OM^2
=> OL > OM.
HENCE PROVED.
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