In the adjoining figure, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.

In the figure, chords AB ∥ CD and O is the centre of the circle.

Radius of the circle = 15 cm

Length of AB = 24 cm and CD = 18 cm.

Join OA and OC.

AB = 24 cm and OM ⊥ AB.

∴ AM = MB = $\dfrac{24}{2}$ = 12 cm (As perpendicular to a chord from the center of the circle bisects it)

In right angle triangle OAM,

⇒ OA² = OM² + AM² (By pythagoras theorem)

⇒ OM² = OA² - AM²

⇒ OM² = 15² - 12²

⇒ OM² = 225 - 144

⇒ OM² = 81

⇒ OM = $\sqrt{81}$ = 9 cm.

Similarly ON ⊥ CD

CN = ND = $\dfrac{18}{2}$ = 9 cm

Similarly In right ∆CNO,

⇒ OC² = ON² + CN² (By pythagoras theorem)

⇒ ON² = OC² - CN²

⇒ ON² = 15² - 9²

⇒ ON² = 225 - 81

⇒ ON² = 144

⇒ ON = $\sqrt{144}$ = 12 cm.

MN = OM + ON = 9 + 12 = 21 cm.

Hence, MN = 21 cm.