A circle of radius 2.5 cm has a chord of length 4.8 cm. Find the distance of the chord from the centre of the circle.

From figure,

Given: Length of the chord AC = 4.8 cm.

Radius of the circle (r) = 2.5 cm

Distance of the chord from the center of the circle = OB.

B is the midpoint of AC, as OB is perpendicular to the chord AC.

AB = $\dfrac{1}{2}$ AC

= $\dfrac{1}{2} \times (4.8)$

= 2.4 cm.

In Δ OAB, ∠B = 90°

Using Pythagoras theorem,

∴ OA² = OB² + AB²

⇒ (2.5)² = OB² + (2.4)²

⇒ 6.25 = OB² + 5.76

⇒ OB² = 6.25 - 5.76

⇒ OB² = 0.49

⇒ OB = $\sqrt{0.49}$

⇒ OB = 0.7 cm.

Hence, the distance of the chord from the center of the circle is 0.7 cm.