The given figure shows a circle with centre O. A, B, C, and D are points on the circle such that AB = 16 cm.

If CD = 30 cm, then what is the distance of chord CD from the centre?

Let M be the foot of the perpendicular drawn from centre O to chord CD.

Join OB and OD.

The perpendicular drawn from the centre of a circle to a chord bisects the chord.

$$\begin{gathered} \therefore \mathrm{AL}=\mathrm{LB}=\frac{\mathrm{AB}}{2}=\frac{16}{2}=8 \mathrm{cm} \\ \mathrm{CM}=\mathrm{MD}=\frac{\mathrm{CD}}{2}=\frac{30}{2}=15 \mathrm{cm} \end{gathered}$$

Applying Pythagoras Theorem in right triangle OLB, we obtain

(OB)² = (OL)² + (LB)²

⇒ (OB)² = (15 cm)² + (8 cm)²

⇒ (OB)² = 225 cm² + 64 cm²

⇒ (OB)² = 289 cm²

∴OB = 17 cm

Thus, radius of the circle = 17 cm

∴ OD = 17 cm

Again applying Pythagoras Theorem in right triangle OMD, we obtain

(OD)² = (OM)² + (MD)²

⇒ (17 cm)² = (OM)² + (15 cm)²

⇒ 289 cm² = (OM)² + 225 cm²

⇒ (OM)² = 289 cm² − 225 cm²

⇒ (OM)² = 64 cm²

∴OM = 8 cm

Thus, the distance of chord CD from the centre is 8 cm.