Question

6. Two chords AB and CD of a circle are each at distance of 4 cm from the centre and
AB = 6cm. Find the length of the chord CD.​

Answer 1

Answer:

Given:AB and CD are chords of a circle and they are equidistant 4cm from the center O.

OP and OQ are perpendiculars=4cm from O to AB and CD respectively.

∴ they bisect AB and CD

⇒AP=21AB and DQ=21DC

⇒∠OPA=∠OQD=90∘

Now, between the △AOP and △DOQ we have

OP=OQ=4cm (given)

∠OPA=∠OQD=90∘

⇒OB=OD (radii of the same circle)

∴△BOP is identical to △DOQ by RHS test

∴AP=DQ⇒AB=DC

⇒AB−DC=0