In the given figure, AB and CD are two parallel chords of the circle whose centre is O. The lengths of AB and CD are 48 cm and 40 cm respectively. If the distance between the chords is 8 cm, then what is the radius of the circle?

Join OA and OC. Draw OM perpendicular to CD, which intersects AB at N.

We know that, the perpendicular from the centre of a circle to the chord bisects the chord.

$$\begin{gathered} \mathrm{CM}=\frac{1}{2}\mathrm{CD}=\frac{1}{2}\times 40=20 \mathrm{cm} \\ \mathrm{AN}=\frac{1}{2}\mathrm{AB}=\frac{1}{2}\times 48=24 \mathrm{cm} \end{gathered}$$

Given, the distance between the parallel chords AB and CD is 8 cm.

∴ MN = 8 cm

Let ON = x cm

∴ OM = ON + MN = (x + 8) cm

Applying Pythagoras theorem to ΔOAN, we have

OA² = ON² + AN²

⇒ OA² = x² + 24² cm²

⇒ OA² = x² + 576 cm²     …(1)

Similarly, 

OC² = OM² + CM²

⇒ OC² = [(x + 8)² + 20²] cm²

⇒ OC² = (x² + 16x + 64 + 400) cm²

⇒ OC² = (x² + 16x + 464) cm²     …(2)

OA and OC are the radii of the same circle.

∴ OA = OC

⇒ OA² = OC²

⇒ x² + 576 = x² + 16x + 464

⇒ 16x = 112

∴ x = 7

Substituting the value of x in (1), we have

OA² = (7² + 576) cm² = (49 + 576) cm² = 625 cm²

∴ OA = 25 cm

Thus, the radius of the circle is 25 cm.