2. Let $\triangle A B C$ be a right triangle with legs $A B=6$ and $B C=8$. Let $R$ and $r$ be the circumradius and the inradius of $\triangle A B C$, respectively. Find the sum of $R$ and $r$.
Answer: 7
Solution: Since $\triangle A B C$ is a right triangle, its circumradius is just half of its hypotenuse. By Pythagorean theorem, $A C=\sqrt{6^{2}+8^{2}}=10$ and so $R=\frac{A C}{2}=10 / 2=5$. Now, we can compute the inradius by using the following identity:
$$
\left(r \times \frac{A B+B C+A C}{2}=\operatorname{Area}(\triangle A B C)\right) .
$$

Since $\triangle A B C$ is a right triangle, its area is just $(A B \times B C) / 2=48 / 2=24$. So that
$$
r=\frac{24}{(6+8+10) / 2}=2
$$

Thus, $R+r=5+2=7$.