Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$. 
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.

What I Tried: Here is a picture :-

I know the centroid divides each of the medians in the ratio $2:1$ . So $AD = 3\sqrt{3}$ , $BE = 3\sqrt{2}$ , $CF = 3$ . 
From this site :- https://mathworld.wolfram.com/TriangleMedian.html, I find that the area of the triangle will be :- $$\frac{4}{3}\sqrt{s_m(s_m - m_1)(s_m - m_2)(s_m - m_3)}$$
Where $m_1,m_2,m_3$ are the medians of the triangle and $s_m = \frac{m_1 + m_2 + m_3}{2}$ .
After putting the respective values for the medians I get that $[\Delta ABC]$ is :-
$$\frac{4}{3}\sqrt{\Bigg(\frac{3(\sqrt{3} + \sqrt{2} + 1)}{2}\Bigg)\Bigg(\frac{3(\sqrt{2} + 1 - \sqrt{3})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + 1 - \sqrt{2})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + \sqrt{2} - 1)}{2}\Bigg)}$$
$$\rightarrow \frac{4}{3}\sqrt{\frac{81(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3})(\sqrt{3} + 1 - \sqrt{2})(\sqrt{3} + \sqrt{2} - 1)}{16}}$$
I am almost to the answer (assuming I made no mistake), but I think this simplification is getting complicated. How do I proceed next?
Can anyone help me?
 A: (This is not likely to be what you're looking for.)
I think in this problem you can use a simpler solution.
Construct point $H$ outside $\overline{AC}$ such that $AGCH$ forms a Parallelogram. We have

*

*$\overline{AH}=\overline{GC}=2$

*$\overline{AG}=2\sqrt3$

*$\overline{GE}=\overline{EH}\Longrightarrow \overline{GH}=\overline{GB}=2\sqrt2$
Since $\overline{AG}^2=\overline{AH}^2+\overline{GH}^2$, we know that $\angle AHG=90^{\circ}$.
Note that $\triangle AGE=\frac{1}2\triangle AGH=\frac{1}2\cdot\frac{1}2\cdot2\cdot2\sqrt2=\sqrt2$.
Therefore $\triangle ABC=6\triangle AGE=6\sqrt2$.

A: From where you left,
$A =  \displaystyle 3 \sqrt{{(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3})(\sqrt{3} + 1 - \sqrt{2})(\sqrt{3} + \sqrt{2} - 1)}}$
Take the first two terms, it is of the form $(a-b)(a+b)$ so we have,
$(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3}) = 2\sqrt2$
Next two terms can be taken as (a+b-c)(a-b+c)
$((\sqrt{3} + 1) - \sqrt{2})((\sqrt{3} -1) + \sqrt{2}) = 3 - 1 - 2 + \sqrt 2 (\sqrt3 + 1) - \sqrt2 (\sqrt3 - 1) = 2 \sqrt2$
So $A = 6 \sqrt2$
A: I think, it's better to use $$S_{\Delta ABC}=\sqrt{p(p-a)(p-b)(p-c)}=$$
$$=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}=\frac{1}{4}\sqrt{\sum_{cyc}(2a^2b^2-a^4)}$$ because from the given easy to get $a^2$, $b^2$ and $c^2$.
Indeed, $$\frac{1}{3}\sqrt{2b^2+2c^2-a^2}=2\sqrt3,$$
$$\frac{1}{3}\sqrt{2a^2+2c^2-b^2}=2\sqrt2$$ and
$$\frac{1}{3}\sqrt{2a^2+2b^2-c^2}=2,$$
which gives $$\frac{1}{3}(a^2+b^2+c^2)=4(3+2+1)$$ or
$$a^2+b^2+c^2=72,$$ which gives
$$2(72-a^2)-a^2=108$$ or $$a^2=12.$$
By the similar way we obtain: $b^2=24$ and $c^2=36$, which gives $$S_{\Delta ABC}=6\sqrt2.$$
A: It is indeed a theorem that can be generalized, although Mathworlddoes n't say so explicitly.
The area of a triangle formed by medians ( computed for example from Brahmagupta/Heron formula ) is three-fourths the area formed by the corresponding sides of the given triangle.
It can also be proved by projective geometry.
Linear scale $k=\sin \frac{\pi}{3}$ can be established from relative proportions of the simplest equilateral triangle.
In our case  hypothetical medians after scaling up full sides from centroid by $\text{50%} :  3(\sqrt 3, \sqrt 2,1)$ calculates to $\dfrac{9}{\sqrt 2};$
So the circumscribed triangle area would be:
$$ \dfrac{{\dfrac{9}{\sqrt 2}}} {\sin^2\dfrac{\pi}{3}} = 6 \sqrt 2.$$
