In triangle $B E N$ shown below with its altitudes intersecting at $X, N A=7, E A=3, A X=4$, and $N S=8$. Find the area of $B E N$
Solution: The idea is to try to find a base and height for the triangle so that we can find the area. By the Pythagorean theorem, $E X=5, N X=\sqrt{65}$, and $S X=1$. Triangles $A X E$ and $B X S$ are similar since they have the same angles. The ratio of their side lengths is $4: 1$, so $B S=3 / 4$ and $B X=5 / 4$. Now using either $N E$ or $N B$ as a base, we get that the area of $B E N$ is $\frac{1}{2} \cdot\left(8+\frac{3}{4}\right) \cdot 6$ or $\frac{1}{2} \cdot\left(4+\frac{5}{4}\right) \cdot 10$, both of which simplify to $\frac{105}{4}$.