Fewpal
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he lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?

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Let PQ and RS be two parallel chords of a circle with centre O. 

We have, PQ = 8 cm and RS = 6 cm. 

Draw perpendicular bisector OL of RS which meets PQ in M. Since, 

PQ || RS, therefore, OM is also perpendicular bisector of PQ. 

Also, OL = 4 cm and RL = 1/ 2 RS ⇒ RL = 3 cm 

and PM = 1/ 2 PQ ⇒ PM = 4 cm 

In ∆ORL, we have 

OR2 = RL2 + OL2 [Pythagoras theorem]

⇒ OR2 = 32 + 42 = 9 + 16 

⇒ OR2 = 25 ⇒ OR = 25 

⇒ OR = 5 cm 

∴ OR = OP [Radii of the circle] 

⇒ OP = 5 cm 

Now, in ∆OPM 

OM2 = OP2 – PM2 [Pythagoras theorem] 

⇒ OM2 = 52 – 42 = 25 – 16 = 9 

OM = 9 = 3 cm 

Hence, the distance of the other chord from the centre is 3 cm. Ans.

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