In a right-angled triangle $\Delta ABC$, $\angle B=90^{\circ }$, $AB=6\text{cm}$, and $BC=8\text{cm}$. The radius $r$ of the inscribed circle is:

First, find the hypotenuse $AC$ using the Pythagoras theorem:
$$AC=\sqrt{A{B}^2+B{C}^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{cm}$$

For any right-angled triangle, the inradius $r$ can be computed using the direct formula:

$$r=\frac{\text{Perpendicular}+\text{Base}-\text{Hypotenuse}}{2}$$

$$r=\frac{AB+BC-AC}{2}$$

$$r=\frac{6+8-10}{2}=\frac{4}{2}=2\text{cm}$$