7 prove that of any two chords of a circle then the one which is larger is nearer to the centre

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Prove that, of any two chords of a circle, the greater chord is nearer to the centre.

Solution

Given: AB and CD are two chords of a circle C(O, r). OP < OQ, where OP ⊥ CD and OQ ⊥ AB To prove: CD > AB Proof:

AQ = $\frac{1}{2}$ AB (Perpendicular from the centre of a circle to a chord bisects the cord)

CP = $\frac{1}{2}$ CD (Perpendicular from the centre of a circle to a chord bisects the cord) In right ΔAOQ,

⇒ AO2 = OQ2 + AQ2

⇒ AQ2 = AO2 - OQ2 ....(1) In right ΔCOP,

⇒ CO2 = CP2 + OP2

⇒ CP2 = CO2 - OP2 ....(2) ⇒ OP < OQ

⇒ OP2 < OQ2

⇒ -OP2 < -OQ2
⇒ CO2 - OP2 < AO2 - OQ2 (CO = AO)
⇒ CP2 > AQ2 (from (1) and (2))
⇒$\frac{1}{4}$ CD2 > $\frac{1}{4}$ AB2
⇒ CD2 > AB2
⇒ CD > AB

Mathematics

Concise Mathematics

Standard X

Given: A circle with centre O and radius r.OM ⊥  AB and ON ⊥ CD Also AB > CDTo prove: OM < ONProof: 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.