Tricky Triangle Area Problem

Solution 1:

From the law of cosines ($C^2 = A^2 + B^2 - 2AB\cos \theta$), we get that $(\sqrt{33})^2 = 2^2 + 5^2 - 2 \cdot 2 \cdot 5 \cos \theta$.

Simplifying this, we get $33 = 29 - 20 \cos \theta$, which means that $\displaystyle \cos \theta = -\frac{1}{5}$

Because $\cos^2 \theta + \sin^2 \theta = 1$, we get that $\displaystyle \sin^2 \theta = \frac{24}{25}$. This means that $\displaystyle \sin \theta = \frac{2\sqrt{6}}{5}$ (note that, because $0 \le \theta \le \pi$, $\sin \theta \ge 0$).

The area of the triangle is $\displaystyle \frac{1}{2} A B \sin \theta = \frac{1}{2} \cdot 2 \cdot 5 \cdot \frac{2\sqrt{6}}{5} = 2\sqrt{6}$.

Solution 2:

Use the $\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$ form of Herons' formula.

$$\begin{align} & \frac{1}{4}\sqrt{4\cdot4\cdot25-(4+25-33)^2} \\ = & \frac{1}{4}\sqrt{4^2\cdot25-4^2} \\ = & \sqrt{25-1} \\ = & 2\sqrt{6} \end{align}$$

Solution 3:

You could've used Heron's formula straight away, actually.

$$\begin{align} T & = \tfrac{1}{4} \sqrt{(a+b-c)(a-b+c)(-a+b+c)(a+b+c)} \\ & = \tfrac{1}{4} \sqrt{(2+5-\sqrt{33})(2-5+\sqrt{33})(-2+5+\sqrt{33})(2+5+\sqrt{33})} \\ & = \tfrac{1}{4} \sqrt{(7-\sqrt{33})(-3+\sqrt{33})(3+\sqrt{33})(7+\sqrt{33})} \\ & = \tfrac{1}{4} \sqrt{(7+\sqrt{33})(7-\sqrt{33})(\sqrt{33}+3)(\sqrt{33}-3)} \\ & = \tfrac{1}{4} \sqrt{(49-33)(33-9)} \\ & = \tfrac{1}{4} \sqrt{16 \cdot 24} \\ & = \tfrac{1}{4} \sqrt{64 \cdot 6} \\ & = \tfrac{8}{4} \sqrt{6} \\ & = 2 \sqrt{6} \end{align}$$

Solution 4:

Since this was multiple-choice, I think it's worth noting that you could have guessed the right answer without doing (much) arithmetic: the diagonal of a right triangle with sides $2$ and $5$ has length $\sqrt{2^2+5^2} = \sqrt{29}$; since $\sqrt{33}$ is fairly close to this, the answer should be close to the area of a $2-5-\sqrt{29}$ triangle, which is of course $\frac12(2)(5)=5$. If you imagine how to 'stretch out' the $\sqrt{29}$ diagonal to $\sqrt{33}$, it's clear that the right angle will have to become obtuse, and this in turn means that the area of the $2-5-\sqrt{33}$ triangle will have to be less than $5$; of the provided answers, only $2\sqrt{6}\approx4.472$ is less than $5$ (and of course very close to it).

Solution 5:

Let be $ABC$ the triangle. Consider the altitude $AH$ over the greatest side, the one whose length is $\sqrt{33}$. Now call $x=BH$, so $\sqrt{33}-x=CH$. Apply Pythagoras' theorem to get $$\left\{ \begin{array}{rcl} x^2+h^2&=&4\\ \left(\sqrt{33}-x\right)^2+h^2&=&25 \end{array} \right.$$

Solve for $h$ and you are (almost) done.