(D) 8 cm
Given, AD = 34 cm and AB = 30 cm
Consider the figure given below,
Draw OL ⊥ AB
Since the perpendicular from the centre of a circle to a chord bisects the chord,
∴ AL=LB= 12 AB=15cm
In right angled Δ OLA,
OA2=OL2+AL2
∴ (17)2=OL2+(15)2
⇒ 289=OL2+225
⇒ 289=OL2+225
⇒ OL2=289−225=64
∴ OL=8cm
[taking positive square root, because lentgh is always positive]
Hence, the distance of the chord from the centre is 8cm.