Two parallel chords of lengths 80 cm and 18 cm are drawn on the same side of the centre of a circle of radius 41 cm. Find the distance between the chords.

Let AB = 18 cm and CD = 80 cm be chords on same side of the center of the circle.

OE ⊥ CD and OF ⊥ AB

We know that,

Perpendicular from the center to the chord, bisects it.

∴ AF = $\dfrac{AB}{2} = \dfrac{18}{2}$ = 9 cm.

CE = $\dfrac{CD}{2} = \dfrac{80}{2}$ = 40 cm.

From figure,

OA = OC = radius = 41 cm.

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ 41² = OE² + 40²

⇒ 1681 = OE² + 1600

⇒ OE² = 1681 - 1600

⇒ OE² = 81

⇒ OE = $\sqrt{81}$ = 9 cm.

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ 41² = OF² + 9²

⇒ 1681 = OF² + 81

⇒ OF² = 1681 - 81

⇒ OF² = 1600

⇒ OF = $\sqrt{1600}$ = 40 cm.

From figure,

⇒ EF = OF - OE = 40 - 9 = 31 cm.

Hence, distance between the chords = 31 cm.