AB (= 20 cm) is diameter of the given circle and AP (= 16 cm). The distance of chord AP from center O is:

1. 12 cm

2. 18 cm

3. 9 cm

4. 6 cm

Given:

Length of the chord AP = 16 cm.

Diameter of the circle AB = 20 cm.

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

Construction: Draw OC ⊥ AP, where O is the center of the circle.

Since, perpendicular drawn from the center of a circle to a chord bisects it.

∴ OC bisects AP

AC = $\dfrac{1}{2}$ x AP = $\dfrac{1}{2}$ x 16 = 8 cm.

In Δ OAC, ∠C = 90°

Using Pythagoras theorem,

∴ OA² = OC² + AC²

⇒ (10)² = OC² + (8)²

⇒ 100 = OC² + 64

⇒ OC² = 100 - 64

⇒ OC² = 36

⇒ OC = ​$\sqrt{36}$

⇒ OC = 6 cm.

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

Hence, option 4 is the correct option.