10.050. Find the radius of the circle circumscribed around a right-angled triangle if the radius of the circle inscribed in this triangle is $3 \mathrm{~cm}$, and one of the legs is $10 \mathrm{~cm}$.
Solution.

Draw radii $O D, O M$ and $O N$ to the points of tangency (Fig. 10.50). We have $O D=O M=C M=C D$ and, therefore, $B D=10-3=7$ (cm). But $B D=B N$ and $A N=A M$ (tangents drawn from the same point). Let $A N=x$; then $(x+7)^{2}=(x+3)^{2}+10^{2}$, from which $8 x=60$, i.e., $x=7.5$ (cm). Hence, $A B=14.5$ (cm) and we get the answer: $R=0.5 A B=7.25$ (cm).

Answer: $7.25 \text{ cm}$.