The lengths of two parallel chords of a circle are $6\text{cm}$ and $8\text{cm}$. If the smaller chord is at a distance of $4\text{cm}$ from the centre, then the distance of the other chord from the centre is:

Let the radius of the circle be $R$.
A line dropped perpendicularly from the center to a chord bisects the chord.

For the smaller chord (length $=6\text{cm}$), the half-length is $\frac{6}{2}=3\text{cm}$. The distance from the center is given as $4\text{cm}$.
Using the Pythagoras theorem inside the right-angled triangle formed with the radius:

$$R^2=3^2+4^2=9+16=25⟹R=5\text{cm}$$

For the larger chord (length $=8\text{cm}$), the half-length is $\frac{8}{2}=4\text{cm}$. Let its distance from the center be $d$.
Using the Pythagoras theorem with the same radius $R=5\text{cm}$:

$$R^2=4^2+d^2$$

$$5^2=4^2+d^2$$

$$25=16+d^2$$

$$d^2=25-16=9⟹d=3\text{cm}$$

Thus, the distance of the other chord from the center is $3\text{cm}$.