Here, AM=5cm,BM=3cm.
`thereforeAB=AM-BM=5-3=2cm`
It is also given that PQ is the perpendicular bisector of AB.
`therefore AC=CB=(1)/(2)AB=1cm`
and `anglePCA=90^@`
Now, in right triangle PCA,
`PC^2=AP^2-AC^2` (by Pythagoras theorem)
`=(5)^2-(1)^2=24`
`therefore PC =sqrt(24)cm`
`rArrPC=2(sqrt6)cm`
`therefore PQ=2PC` (because perpendicular drawn from the centre to the chord bisects the chord
`=2xx2sqrt(6)cm`
`rArrPQ=4sqrt(6)cm`.