Equilateral Area

It can be easily proved by congruence of triangles that all the small triangles are congruent to each-other.Therefore the triangle $DEF$ is equilateral,although this is not important to our problem.It's not hard to find that $FC=BE=AD=\dfrac{124}{3}$ and $BD=EC=AF=\dfrac{62}{3}$ .Lets take a look at the small triangles which are congruent,these are $AFD$ , $FEC$ and $BDE$ .We know that one of their sides is equal to $\dfrac{124}{3}$ and the other one is $\dfrac{62}{3}$ .We also know the angle between them is $60$ degrees.Therefore there areas are easy to find using trigonometry. Let $S$ represent the area of one of these triangles which are congruent.

$S=\dfrac{1}{2}*\dfrac{124}{3}*\dfrac{62}{3}*\dfrac{\sqrt{3}}{2}$ because $sin60=\dfrac{\sqrt{3}}{2}$ .Multiplying it by $3$ we get the whole area of the small triangles which is $3S=\dfrac{124*62*\sqrt{3}}{12}$ .The area of the larger triangle $ABC$ is equal to $\dfrac{3844*\sqrt{3}}{4}$ .If we substract the whole area of the smaller triangles from the areaa of the large we will get our desired result which is approximately $\boxed{555}$