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Prove That, of Any Two Chords of a Circle, the Greater Chord is Nearer to the Centre. - ICSE Class 10 - Mathematics

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ConceptArc and Chord Properties - If Two Chords Intersect Internally Or Externally Then the Product of the Lengths of the Segments Are Equal

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 Rt.  ΔAOM,
 `AO^2  = AM^2  + OM^2`
`r^2 = (1/2 AB)^2  + OM^2 `
`r^2 = 1/4 AB^2 + OM^2`              ………..(i)
Again in Rt.  ΔONC,
`OC^2 = NC^2  + ON^2`
`⇒ 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.

  Is there an error in this question or solution?

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Solution Prove That, of Any Two Chords of a Circle, the Greater Chord is Nearer to the Centre. Concept: Arc and Chord Properties - If Two Chords Intersect Internally Or Externally Then the Product of the Lengths of the Segments Are Equal.
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