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

OR^{2} = RL^{2} + OL^{2} [Pythagoras theorem]

⇒ OR^{2} = 3^{2 }+ 4^{2} = 9 + 16

⇒ OR^{2} = 25 ⇒ OR = 25

⇒ OR = 5 cm

∴ OR = OP [Radii of the circle]

⇒ OP = 5 cm

Now, in ∆OPM

OM^{2} = OP^{2} – PM^{2} [Pythagoras theorem]

⇒ OM^{2} = 5^{2} – 4^{2} = 25 – 16 = 9

OM = 9 = 3 cm

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