Problem requires hard work!

So the area we want will be maximum, if both the triangle coincide. And as we rotate one of the triangles, then area between them would start getting lesser and lesser. The minimum area would be achieved when the side $FD$ is parallel to $AB$ . So lets start by calculating the areas of the triangles. The areas of both the triangles would be equal.

Now the area of each triangle would be $\frac{\sqrt3(\sqrt3r)^2}{4}=\frac{3\sqrt3r^2}{4}$

So now we have to calculate the area of the hexagon. For that, we calculate the area of the $6$ smaller triangles and subtract them from the area of both the triangles.

The side of each of the smaller triangles would be $\frac13$ of the side of the equilateral triangles, and each of those $6$ smaller triangles would be equilateral too. (Which can be shown easily)

Now, we get that the area of the hexagon is $\frac{3\sqrt3r^2}{4}-3\left(\frac{\sqrt3(\frac{r}{\sqrt3})^2}{4}\right)=\frac{3\sqrt3r^2}{4}-\left(\frac{\sqrt3r^2}{4}\right)=\frac{\sqrt3r^2}2$

So I get the answer as $\frac{\sqrt3}2$

We subdivide the hexagon into 12 right triangles of area $\frac{1}{8}\tan(\pi/6)$ for a total area of $\frac{3}{2}\tan(\pi/6)\approx \boxed{0.87}$