11.4. It is known that the lengths of the sides of a triangle are consecutive natural numbers, and the radius of its inscribed circle is 4. Find the radius of the circumscribed circle of this triangle.
Answer: $\frac{65}{8}$.

Solution: Let the lengths of the sides be $n-1, n, n+1$ for some natural number $n$. By Heron's formula, the area of the triangle $S=\sqrt{\frac{3 n^{2}}{4}\left(\frac{n^{2}}{4}-1\right)}=\frac{1}{2} P \times r=\frac{3}{2} n \times 4=6 n$, from which $n=14, S=84$. Substituting this into the formula $R=\frac{a b c}{4 \mathrm{~S}}=\frac{13 \cdot 14 \cdot 15}{4 \cdot 84}=\frac{65}{8}$.