Jarris the triangle is playing in the $(x, y)$ plane. Let his maximum $y$ coordinate be $k$. Given that he has side lengths 6,8 , and 10 and that no part of him is below the $x$-axis, find the minimum possible value of $k$.
By playing around, we find that Jarris should have his hypotenuse flat on the $x$ axis. The desired minimum value of $k$ is then the length of the altitude to the hypotenuse. Thus, by computing the area of the triangle in two ways, $\frac{1}{2} \cdot 10 \cdot k=\frac{1}{2} \cdot 6 \cdot 8$ and so $k=\frac{24}{5}$. $\fbox{\frac{24}{5}}$.