In figure (i) given below, O is the center of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find the :

(i) radius of the circle

(ii) length of chord CD.

(i) Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AM = BM = $\dfrac{24}{2}$ = 12 cm.

In right angle triangle OAM,

⇒ OA² = OM² + AM² (By pythagoras theorem)

⇒ OA² = 5² + 12²

⇒ OA² = 25 + 144

⇒ OA² = 169

⇒ OA = $\sqrt{169}$ = 13 cm.

Hence, radius = 13 cm.

(ii) From figure,

OC = radius = 13 cm.

In right angle triangle OCN,

⇒ OC² = ON² + CN² (By pythagoras theorem)

⇒ CN² = OC² - ON²

⇒ CN² = 13² - 12²

⇒ CN² = 169 - 144 = 25

⇒ CN = $\sqrt{25}$ = 5 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ ND = CN = 5 cm.

CD = CN + ND = 10 cm.

Hence, CD = 10 cm.