Two parallel chords are drawn in a circle of diameter 30.0 cm. The length of one chord is 24.0 cm and the distance between the two chords is 21.0 cm; find the length of the other chord.

Let AB and CD be the two parallel chords.

Given,

Length of one chord is 24.0 cm. Let AB = 24 cm.

Draw OE ⊥ CD and OF ⊥ AB.

We know that,

Perpendicular from the center to the chord, bisects it.

∴ AF = $\dfrac{AB}{2} = \dfrac{24}{2}$ = 12 cm.

From figure,

OA = OC = radius = $\dfrac{\text{Diameter}}{2} = \dfrac{30}{2}$ = 15 cm.

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ 15² = OF² + 12²

⇒ 225 = OF² + 144

⇒ OF² = 225 - 144

⇒ OF² = 81

⇒ OF = $\sqrt{81}$ = 9 cm.

Given,

Distance between two chords = 21 cm

∴ FE = 21 cm

⇒ OE = FE - OF = 21 - 9 = 12 cm.

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ 15² = 12² + CE²

⇒ 225 = 144 + CE²

⇒ CE² = 225 - 144

⇒ CE² = 81

⇒ CE = $\sqrt{81}$ = 9 cm.

As, perpendicular from center to the chord, bisects it.

∴ CD = 2 × CE = 2 × 9 = 18 cm.

Hence, length of other chord = 18 cm.