What is the area of a triangle with sides $\sqrt{5}$, $\sqrt{10}$, $\sqrt{13}$? I found a "fun algebra problem" that asks you to find the area of a triangle whose sides are $\sqrt{5}$, $\sqrt{10}$, $\sqrt{13}$.  After some algebra hell trying to work with Heron's formula, I plugged the question into Wolfram and it spit out 3.5.
Is there some elegant way to reach this?  My algebra kungfu has so far been too weak.

For another interesting approach, consider the Law of Cosines, $a^2 = b^2 + c^2 - 2bc\cos(\alpha)$. If we let $a = \sqrt{13}$, $b = \sqrt{5}$, and $c = \sqrt{10}$, then we find that $13 = 15 - 10\sqrt{2}\cos(\alpha)$, and thus that $\cos(\alpha) = \sqrt{2}/10$. Using $\cos(\alpha)$, we can calculate $\sin(\alpha)$ through some basic trigonometric manipulation to find that $\sin(\alpha) = \sqrt{98}/10$. Using the area formula for triangles $A =\frac{1}{2}bc\sin(\alpha)$, we find that 
$$A = \frac{1}{2}bc\sin(\alpha) = \frac{1}{2} \cdot\sqrt{50}\cdot\frac{\sqrt{98}}{10}=\frac{7}{2}$$
