Question

In the figure, ‘O’ is the centre of the circle, then

A) EF > CD > AB 

B) EF = CD = AB 

C) AB > CD > EF 

D) None

Answer 1

Correct option is (C) AB > CD > EF

Chord AB is nearest to the centre of circle, chord EF is farest to the centre of circle.

\(\because\) d(AB, 0) < d(CD, 0) < d(EF, 0)

\(\therefore\) \(AB>CD>EF\)

\((\because\) Chord which is more distanced from the centre of the circle is smaller than the other chord)

Answer 2

C) AB > CD > EF