A chord of length 16 cm is drawn in a circle of diameter 20 cm. Calculate its distance from the centre of the circle.

Given: Length of the chord AC = 16 cm.

Diameter of the circle = 20 cm.

Radius of the circle r = $\Big(\dfrac{20}{2}\Big)$ = 10 cm.

To prove: Distance of the chord from the center of the circle = OB.

Construction: Draw OB ⊥ AC, where O is the center of the circle. Join OA.

Proof:

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

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

= $\dfrac{1}{2}$ x 16 cm

= 8 cm

In Δ OAB, ∠B = 90°

Using Pythagoras theorem,

∴ OA² = OB² + AB²

⇒ (10)² = OB² + (8)²

⇒ 100 = OB² + 64

⇒ OB² = 100 - 64

⇒ OB² = 36

⇒ OB = $\sqrt{36}$

⇒ OB = 6 cm

Hence, the distance of the chord from the center of the circle is 6 cm.