Solution:

Since the hyperbola is symmetric, without the loss of generality, we can have $(1,1)$ as our vertex. Then, since we have the centroid of an equilateral triangle, the angle at the centroid with any two points is $120^ \circ .$ The branch of the hyperbola with negative coordinates can make an angle of at most $90^\circ .$ This means that we can't have two points on the negative branch.

Since the hyperbola is symmetric over $y=x$ and it always decreases, the two points are reflected over $y=x.$ Also, the altitude is on $y=x,$ making the other point also on $y=x.$ This makes the other point $(-1,-1).$ Thus, the circumradius is $2\sqrt 2$ since it is the distance between the two points. This means we have $3$ isosceles triangles with side lengths $2 \sqrt 2$ and angle $120^\circ.$

Therefore, the combined area is $$3\cdot \dfrac{(2\sqrt 2)^2 \cdot \sin(120^ \circ)}2$$$$= 12 \sin(120^\circ)$$$$= 12 \cdot \dfrac{\sqrt{3}}2$$$$=\sqrt{108} .$$ This makes the square $108.$

Thus, the correct answer is C .