What is the circumradius of a triangle $ABC$ with sides of length $6$, $8$, and $10$?
I recognize that this is a right triangle, since $6^2 + 8^2 = 10^2$. That means that the hypotenuse, which is $10$, is also the diameter of the circumcircle. Therefore, the circumradius is half of the diameter, which is $5$. 