Inradius of a right triangle

Let the right angle be at $C$, with $CB=a$ and $CA=b$. The incircle touches the legs at two points. Together with $C$ and the centre $I$, these points form a square of side $r$ (all four angles are right angles and two adjacent sides equal $r$). So the tangent length from $C$ is $r$.

Hence the tangent length from $B$ is $a-r$ and from $A$ is $b-r$. The hypotenuse $AB$ is split by its point of tangency into these two lengths, so
$$(a-r)+(b-r)=c\quad\Longrightarrow\quad r=\frac{a+b-c}{2}.$$

For the $3$-$4$-$5$ triangle, $r=\dfrac{3+4-5}{2}=1$.