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. $5\sqrt{6}$ 

You could've used Heron's formula straight away, actually.
$$\begin{align}
T & = \tfrac{1}{4} \sqrt{(a+b-c)(a-b+c)(-a+b+c)(a+b+c)} \\
  & = \tfrac{1}{4} \sqrt{(2+5-\sqrt{33})(2-5+\sqrt{33})(-2+5+\sqrt{33})(2+5+\sqrt{33})} \\
  & = \tfrac{1}{4} \sqrt{(7-\sqrt{33})(-3+\sqrt{33})(3+\sqrt{33})(7+\sqrt{33})} \\
  & = \tfrac{1}{4} \sqrt{(7+\sqrt{33})(7-\sqrt{33})(\sqrt{33}+3)(\sqrt{33}-3)} \\
  & = \tfrac{1}{4} \sqrt{(49-33)(33-9)} \\
  & = \tfrac{1}{4} \sqrt{16 \cdot 24} \\
  & = \tfrac{1}{4} \sqrt{64 \cdot 6} \\
  & = \tfrac{8}{4} \sqrt{6} \\
  & = 2 \sqrt{6}
\end{align}$$
