Radius of a circle is 5 cm. Length of chord A B in this circle is 6 cm. What is the distance of this chord from the centre of the circle?

4 cm

Let OB be the radius of the circle and OD be the distance from the center to the chord AB

Now, we know altitude from the center of a circle to its chord bisect the chord

\( {OB }^{ 2} \) = \( {BD }^{ 2} \) + \( {OD }^{2 } \)

= \( {5 }^{ 2} \) = \( {3 }^{ 2} \) + \( {OD }^{2 } \)

= 25 = 9 + \( {OD }^{2 } \)

= \( {OD }^{2 } \) = 16

= OD = 4

Therefore, distance of this chord from the center of the circle is 4 cm.