Question

o is the centre of the circle, and lenth of the chood is 8 cm seg op & chord Am If I Cop) -3.cm then find radius of the circle​

Answer 1

Answer:

GIVEN: A circle with centre O, Radius AO = 13 cm, Chord AB = 24 cm

TO FIND : The perpendicular distance of the chord from O. Let it be called OM.

OM is perpendicular to the chord AB.

Perpendicular from the centre O to a chord bisects the chord. So AM = AB/2 = 12 cm

In right triangle AMO , Using Pythagoras’s Theorem,

AO² = AM² + OM²

=> 13² = 12² + OM²

=> 169 = 144 + OM²

=> OM² = 169 - 144 = 25

=> OM = √25 = 5 cm.

Answer 2

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Given: AB and AC are two equal chords of a circle with centre O.

OP⊥AB and OQ⊥AC.

To prove: PB=QC

Proof: OP⊥AB

⇒AM=MB .... (perpendicular from centre bisects the chord)....(i)

Similarly, AN=NC....(ii)

But, AB=AC

⇒ 2AB = 2AC

⇒MB=NC ...(iii) ( From (i) and (ii) )

Also, OP=OQ (Radii of the circle)

and OM=ON (Equal chords are equidistant from the centre)

⇒OP−OM=OQ−ON

⇒MP=NQ ....(iv) (From figure)

In ΔMPB and ΔNQC, we have

∠PMB=∠QNC (Each =90∘ )

MB=NC ( From (iii) )

MP=NQ ( From (iv) )

∴ΔPMB≅ΔQNC (SAS)

⇒PB=QC (CPCT)

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