Answer:
3√3 ≈ 5.20 square units
Explanation:
You want the area of ∆ABC with centroid G such that ∆ABG is an equilateral triangle with side length 2.
Centroid
The centroid of a triangle divides the median into two parts in the ratio 1:2, where the shorter part is the distance to the side being bisected, and the longer part is the distance to the vertex.
Here equilateral triangle ABG with side length 2 will have an altitude of √3, so the full length of CH is 3·√3.
The area of ∆ABC is ...
A = 1/2bh
A = 1/2(2)(3√3) = 3√3
The area of ∆ABC is 3√3 square units.
__
Additional comments
In the attached figure, GH is the median and altitude of ∆ABG. The figure is symmetrical about the line GH, so ∆ABC is isosceles and CH is its altitude and median.
We can find the length GH different ways. Perhaps the easiest is to refer to memory, where we find the side length ratios in ∆GAH are 1 : √3 : 2. This means for GA=2, GH=√3 and AH=1. We can also use trigonometry, which tells us GH/GA = sin(60°) = √3/2. Then GH=GA·√3/2 = √3.