VIII.3. The legs of a right triangle are $3 \mathrm{~cm}$ and 4 cm. Determine the distance between the centers of the inscribed and circumscribed circles of the triangle.
Solution: $\overline{\mathrm{AB}}=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}=5 ; \quad P=\frac{a \cdot b}{2}=\frac{12}{2}=6 \mathrm{~cm}, \quad$ and $s=\frac{1}{2}(\mathrm{a}+\mathrm{b}+\mathrm{c})=6 \mathrm{~cm}$.

It is known that the radius of the circumscribed circle $\mathrm{R}=\frac{1}{2} \overline{\mathrm{AB}}=2.5 \mathrm{~cm}$, and $\mathrm{P}=\mathrm{s} \cdot \mathrm{r}$, i.e., $r=\frac{P}{s}=\frac{6}{6}=1 \mathrm{~cm}$

$\overline{\mathrm{AD}_{1}}=\overline{\mathrm{AD}}=4-1=3 \mathrm{~cm}$.

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$\overline{\mathrm{DS}}=\overline{\mathrm{AD}}-\overline{\mathrm{AS}} ; \quad \overline{\mathrm{DS}}=3-2.5=0.5 \mathrm{~cm} . \mathrm{O}$ in $\triangle \mathrm{DOS}$ it follows: $\overline{\mathrm{OS}}^{2}=\overline{\mathrm{OD}}^{2}+\overline{\mathrm{DS}}^{2}$ : $\overline{\mathrm{OS}}^{2}=1^{2}+0.5^{2}=1+0.25=1.25 . \overline{\mathrm{OS}}=\sqrt{1.25}=\frac{\sqrt{5}}{2} \mathrm{~cm}$.