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Question
Prove that, of any two chords of a circle, the greater chord is nearer to the centre.
Solution
Given: A circle with centre O and radius r. OM ⊥ AB and ON ⊥ CD Also AB > CD
To prove: OM < ON
Proof: Join OA and OC.
In right ΔAOM,
AO2 = AM2 + OM2
`=> r^2 = (1/2 AB)^2 + OM^2`
`=> r^2 = 1/4 AB^2 + OM^2` ...(i)
Again in right ΔONC,
OC2 = NC2 + ON2
`=> r^2 = (1/2 CD )^2 + ON^2`
`=> r^2 = 1/4 CD^2 + ON^2` ...(ii)
From (i) and (ii)
`1/4 AB^2 + OM^2 = 1/4 CD^2 + ON^2`
But, AB > CD ...(Given)
∴ ON > OM
`=>` OM < ON
Hence, AB is nearer to the centre than CD.
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