The length of a radius of a circle is $13\text{cm}$ and the length of a chord of the circle is $10\text{cm}$, the distance of the chord from the center of the circle is

Let the circle have center $O$ and chord $AB=10\text{cm}$. The radius $OA=13\text{cm}$.
Draw a perpendicular line segment $OD$ from the center $O$ to the chord $AB$.

A perpendicular drawn from the center of a circle to a chord bisects the chord. Therefore:

$$AD=\frac{AB}{2}=\frac{10}{2}=5\text{cm}$$

In right-angled triangle $\Delta ODA$, applying the Pythagorean theorem:

$$O{A}^2=O{D}^2+A{D}^2$$

$$13^2=O{D}^2+5^2$$

$$169=O{D}^2+25$$

$$O{D}^2=169-25=144$$

$$OD=\sqrt{144}=12\text{cm}$$

Thus, the distance of the chord from the center is $12\text{cm}$.