Heron's formula when side lengths include radicals I am helping a 7th grade student prepare for a math contest. I have a copy of a previous year's test, and one of the questions has me perplexed! I think perhaps I'm making this problem harder than it needs to be. Here's the problem:
Find the area of a triangle with sides $2$, $\sqrt{2}$, and $\sqrt{3}-1$.
I would think we'd need to use Heron's formula here, but it gets so messy with all the radicals. This is a problem that she would be expected to solve in 3 minutes or less, so the simpler I can explain it to her, the better. This is a middle school contest, with most students currently enrolled in Algebra 1, so I don't want a solution using trigonometry. 
 A: We can use the fourth form of Heron's formula to get an easy answer.

Let $a=2, b=\sqrt{2}, c=\sqrt{3}-1$. We can use the formula $$A=\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$$ We have $a^2 = 4, b^2 = 2, c^2 =4-2\sqrt{3}$Thus, $$A=\frac{1}{4}\sqrt{32-(2+2\sqrt{3})^2}$$ $$=\frac{1}{2}\sqrt{8-(1+\sqrt{3})^2}$$ $$=\frac{1}{2}\sqrt{4-2\sqrt{3}}$$ $$=\frac{\sqrt{3}-1}{2}$$
A: A 7th grader who does not know Heron's formula can still solve this.
Consider the right triangles with two common vertices $A$ and $B$:


*

*triangle $ABC$ with $AB=1$, $AC=1$, $BC=\sqrt{2}$; its area is $1/2$.

*triangle $ABD$ with $AB=1$, $AD=\sqrt{3}$, $BD=2$; its area is $\sqrt{3}/2$.


Drawing a sketch of these triangles, with point $C$ between $A$ and $D$, we see that the sought-for area is the difference of the above areas, $ABD$ minus $ABC$; i.e. it's the area of triangle $BCD$, which is $\sqrt{3}/2-1/2$.
A: The semi-perimeter is $\frac{1}{2}(1+\sqrt{2}+\sqrt{3})$ so Heron's formula gives:
$$
\begin{align}
4 A & = \sqrt{(1+\sqrt{2}+\sqrt{3}) \cdot (-3+\sqrt{2}+\sqrt{3}) \cdot (1-\sqrt{2}+\sqrt{3}) \cdot (3+\sqrt{2}-\sqrt{3})} \\
 & = \sqrt{\big((1+\sqrt{3})^2-2\big) \cdot \big(2-(3-\sqrt{3})^2\big)} \\
 & = \sqrt{(2+2\sqrt{3})(-10+6\sqrt{3})} = \sqrt{16-8\sqrt{3}} = 2 \sqrt{4-2\sqrt{3}}
\end{align}
$$
Denesting the latter radical gives $4A=2(\sqrt{3}-1)$.
A: Another approach not using Heron formula. Let's call $x$ the angle oposite to the side $\sqrt{3}-1$, so using Cosine rule we get:
$$(\sqrt{3}-1)^2=4+2-2\cdot 2\cdot \sqrt{2}\cos x \Rightarrow \cos x=\frac{\sqrt{3}+1}{\sqrt{8}}\Rightarrow \sin x=\frac{\sqrt{3}-1}{\sqrt{8}}$$
and then the area is:
$$A=\frac{1}{2}\cdot 2\cdot \sqrt{2}\cdot \frac{\sqrt{3}-1}{\sqrt{8}}=\frac{\sqrt{3}-1}{2}$$  
