In the given figure, 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) the radius of the circle

(ii) length of chord CD.

(i) Join OA.

We know that,

Perpendicular from center to chord, bisects the chord.

As, OM ⊥ AB

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

In △ OAM,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OM² + AM²

⇒ OA² = 5² + 12²

⇒ OA² = 25 + 144

⇒ OA² = 169

⇒ OA = $\sqrt{169}$ = 13 cm.

Hence the radius of circle = 13 cm.

(ii) We know that,

Perpendicular from center of the circle to chord, bisects the chord.

As, ON ⊥ CD

∴ CN = $\dfrac{CD}{2}$

⇒ CD = 2CN ………….(1)

Join OC.

In △ OCN,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = ON² + NC²

⇒ 13² = 12² + CN²

⇒ CN² = 13² - 12²

⇒ CN² = 169 - 144

⇒ CN² = 25

⇒ CN = $\sqrt{25}$ = 5 cm.

Substituting value of CN in equation (1), we get :

⇒ CD = 2 × 5 = 10 cm.

Hence, length of chord CD = 10 cm.