10.060. The radii of the inscribed and circumscribed circles of a right triangle are 2 and 5 cm, respectively (Fig. 10.59). Find the legs of the triangle.
## Solution.

Given $r=2 \text{cm}, R=5$ cm. In $\triangle A C B, \angle A C B=90^{\circ}, A B=2 R=10$, $A K=A E, K B=M B, A C=A E+C E=A K+C E=A K+r, B C=M B+C M=$ $=K B+C M=r+K B, A C+C B=2 r+A B, A C+C B=2(r+R)$. Let $A C=x$, then $C B=2(r+R)-x, C B=14-x$. Since $A B^{2}=A C^{2}+$ $+B C^{2}$, so $100=x^{2}+196-28 x+x^{2}, x_{1}=8, x_{2}=6$.

Answer: $6 ; 8 \text{ cm}$.