Plz help :(

Thanks! 

wait. in your answer, you used \(2x^2\) as a leg. shouldn't it be \(x^2+(2\sqrt{6})^2=(2x)^2\)?

edit: nvmd read my own question wrong :P

Gavin, I believe you made another error here. For your application of the Pythagorean's theorem, you did the following: 

\(x^2+(2x)^2=(​2\sqrt6)^2\\ x^2+4x^2=4\cdot6\\ 5x^2=24\\\)

However, I do not understand why 2x is the representation of a leg in this equation, but you show in your diagram that it is clearly the representation of the hypotenuse's length. If you wanted to use this method to get the answer, then x and x√​3 should be the lengths. The work would look something like this:
 

\(x^2+\left(\textcolor{red}{x\sqrt{3}}\right)^2=\left(2\sqrt{6}\right)^2\\ x^2+\textcolor{red}{3}x^2=2^2*\sqrt{6}^2\\ \textcolor{red}{4}x^2=4*6\\ \textcolor{red}{4}x^2=24\)

That small error from your work makes the discrepancies highlighted in red. You can, then, solve for x and plug them in for the legs, but you can also use the method below.

The ratio of the sidelengths of the 30-60-90 special right triangle is \(1:\sqrt{3}:2\), so Pythagorean's theorem is not necessary. Of course, you can still use it if you wish. However, it is also possible to generate this answer with your knowledge of proportions.  

\(\frac{2\sqrt{6}}{2}=\frac{\text{leg}_1}{1}\\ \text{leg}_1=\sqrt{6}\)

\(\frac{2\sqrt{6}}{2}=\frac{\text{leg}_2}{\sqrt{3}}\\ \text{leg}_2=\sqrt{6}*\sqrt{3}=\sqrt{18}=3\sqrt{2} \)

Now, add them together. 

\(\left(3\sqrt{2}+\sqrt{6}\right)\text{cm}\approx 6.6921\text{cm}\) is, then, the sum of the length of the sides.  

Yes, I realized my mistake, 

It was a careless one again. :)