of any two chords of a circle, show that one which is larger is near to circle.with nice xplanation.answer fastly
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Given: AB and CD are two chords of a circle C(O, r).
OP < OQ, where OP perpendicular to CD and OQ perpendicular to AB
To prove: CD > AB
Proof:
AQ = ½ AB
CP = ½ CD
In right ΔAOQ,
⇒ AO2 = OQ2 + AQ2
⇒ AQ2 = AO2 - OQ2 ....(1)
In triangle COP,
⇒ 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))
⇒ 1/4 CD2 > 1/4 AB2
⇒ CD2 > AB2
⇒ CD > AB
OP < OQ, where OP perpendicular to CD and OQ perpendicular to AB
To prove: CD > AB
Proof:
AQ = ½ AB
CP = ½ CD
In right ΔAOQ,
⇒ AO2 = OQ2 + AQ2
⇒ AQ2 = AO2 - OQ2 ....(1)
In triangle COP,
⇒ 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))
⇒ 1/4 CD2 > 1/4 AB2
⇒ CD2 > AB2
⇒ CD > AB
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