We are given a circle with center \(O\).
The chord length is
\[
6\text{ cm}.
\]
The perpendicular distance from the center to the chord is
\[
4\text{ cm}.
\]
We know that the perpendicular from the center of a circle to a chord bisects the chord.
Therefore, half of the chord is
\[
\frac{6}{2}=3\text{ cm}.
\]
Now consider the right triangle formed by:
\[
\text{radius of circle},
\]
\[
\text{distance from center to chord}=4\text{ cm},
\]
and
\[
\text{half chord}=3\text{ cm}.
\]
Let the radius be \(r\).
Using Pythagoras theorem:
\[
r^2=3^2+4^2.
\]
\[
r^2=9+16.
\]
\[
r^2=25.
\]
\[
r=5\text{ cm}.
\]
The diameter is twice the radius:
\[
\text{Diameter}=2r.
\]
\[
=2(5).
\]
\[
=10\text{ cm}.
\]
Therefore, the length of the diameter is
\[
10\text{ cm}.
\]