Question

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?

Answer 1

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.