A chord of length 48 cm is drawn in a circle of radius 25 cm. Calculate its distance from the centre of the circle.

From figure,

AB is the chord and radius = OA = 25 cm.

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

∴ CB = AC = $\dfrac{48}{2}$ = 24 cm.

In right angle triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ 25² = OC² + 24²

⇒ OC² = 25² - 24²

⇒ OC² = 625 - 576 = 49

⇒ OC = $\sqrt{49}$ = 7 cm.

Hence, chord is at a distance of 7 cm from the center.