AB and CD are two parallel chords of a circle such that AB = 24 cm and CD = 10 cm. If the radius of the circle is 13 cm. find the distance between the two chords.

AB and CD are parallel chords of a circle of radius 13 cm. AB = 24 cm ⇒ half chord = 12 cm. CD = 10 cm ⇒ half chord = 5 cm.

Distance of AB from centre: d₁ = √(13² − 12²) = √25 = 5 cm.

Distance of CD from centre: d₂ = √(13² − 5²) = √144 = 12 cm.

The chords lie on opposite sides of the centre.

Distance between the chords = 5 + 12 = 17 cm.

Answer: 17 cm.

The perpendicular from the centre to a chord bisects the chord. Apply Pythagoras to the two right triangles formed by the radius, half-chord and centre-to-chord distance.