Question

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

Answer 1

Step-by-step explanation:

Two chords AB and CD with centre O Such that AB > CD.

Here,

OA = OC {Radii os same circle}

AE = (1/2)AB andCF = (1/2)CD

(1) From ΔOEA and ΔOFC,we have

OA² = OE² + AE² and OC² = OF² + CF²

∴ OE² + AE² = OF² + CF² {OA = OC and OA² = OC²}

But,

AB > CD

⇒ (1/2)AB > (1/2)CD

⇒ AE > CF

∴ OF² + CF² = OE² + AE² > OE² + CF²   (∵AE > CF)

OF² > OE² and So, OF > OE

Hence, OE < OF i.e, AB is nearer to the centre than CD.

Hope it helps you

#Bebrainly

Answer 2

Heyy mate ❤✌✌❤

Here's your Answer..

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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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