PDF chapter test TRY NOW
In a circle, if the distance of chord \(AB\) from the centre is twice the distance of another chord \(CD\) from the centre, then can we conclude that \(CD = 2 AB\)? Justify your answer.
Verification:
Let the perpendicular bisector from the centre \(O\) of the circle meet the midpoint of the chords \(CD\) and \(AB\) at \(N\) and \(M\) respectively.
Here, \(ON\) \(=\) [].
By the theorem, the distance between the chord from the centre is proportional to its length.
\(AB\) \(CD\)
The chord length depends on the distance from the centre in a way.
Thus, doubling the distance of the chord from the centre that the length of the chord doubles.
Therefore, \(CD\) \(2 AB\).
Register for free to see more content