A chord of length 6 cm is drawn in a circle of diameter 10 cm, its distance from the center of the circle is :

1. 6 cm

2. 8 cm

3. 4 cm

4. 10 cm

Let AB be the chord of the circle with center O.

Given,

Diameter = 10 cm

Radius = $\dfrac{\text{Diameter}}{2} = \dfrac{10}{2}$ = 5 cm.

We know that,

Perpendicular from center to chord, bisects the chord.

∴ AD = $\dfrac{AB}{2} = \dfrac{6}{2}$ = 3 cm.

In right angled triangle OAD,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OD² + AD²

⇒ 5² = OD² + 3²

⇒ OD² = 5² - 3²

⇒ OD² = 25 - 9

⇒ OD² = 16

⇒ OD = $\sqrt{16}$ = 4 cm.

Hence, Option 3 is the correct option.