In a circle, if the distance of chord \(PQ\) from the centre is twice the distance of another chord \(RS\) from the centre, then can we decide that \(RS = 2 PQ\)? Give interpretation for your answer.
Verification:
Let the perpendicular bisector from the centre \(O\) of the circle meet the midpoint of the chords \(RS\) and \(PQ\) at \(B\) and \(A\) respectively.
Here, \(OB\) \(=\) [].
By the theorem, the distance between the chord from the centre is proportional to its length.
\(PQ\) \(RS\)
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, \(RS\) \(2 PQ\).