Tricky Triangle Area Problem This was from a recent math competition that I was in. So, a triangle has sides $2$ , $5$, and $\sqrt{33}$. How can I derive the area? I can't use a calculator, and (the form of) Heron's formula (that I had memorized) is impossible with the$\sqrt{33}$ in it. How could I have done this? The answer was $2\sqrt{6}$ if it helps.

Edited to add that it was a multiple choice question, with possible answers:
a. $2\sqrt{6}$
b. $5$
c. $3\sqrt{6}$
d. $5\sqrt{6}$ 

Use the $\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$ form of Herons' formula.
$$\begin{align}
  & \frac{1}{4}\sqrt{4\cdot4\cdot25-(4+25-33)^2} \\
= & \frac{1}{4}\sqrt{4^2\cdot25-4^2} \\
= & \sqrt{25-1} \\
= & 2\sqrt{6}
\end{align}$$
