A chord of length 48 cm is at a distance of 10 cm from the centre of a circle. If another chord of length 20 cm is drawn in the same circle, find its distance from the center of the circle.

From figure,

Since, OM ⊥ AB, it bisects it. (As perpendicular from center to chord bisects it)

∴ AM = MB = $\dfrac{48}{2}$ = 24 cm.

In right △AOM,

⇒ AO² = OM² + AM² (By pythagoras theorem)

⇒ AO² = 10² + 24²

⇒ AO² = 100 + 576

⇒ AO² = 676

⇒ AO = $\sqrt{676}$ = 26 cm.

Radius = 26 cm.

Since, ON ⊥ CD, it bisects it. (As perpendicular from center to chord bisects it)

∴ CN = ND = $\dfrac{20}{2}$ = 10 cm.

In right △CNO,

⇒ OC² = ON² + NC² (By pythagoras theorem)

⇒ 26² = ON² + 10²

⇒ ON² = 676 - 100

⇒ ON² = 576

⇒ ON = $\sqrt{576}$ = 24 cm.

Hence, the chord of length 20 cm is at a distance of 24 cm from the centre.