Solution

The inscribed circle has it center coinciding the the cut point of the medians of the equilateral triangle. The height to the vertices are 3:1 to the radius r of the inscribed circle.

 

For the next equilateral triangle r is the height, and the radius of the inscribed circle is again on third of it. The goes on to set up an exhaustion. The area of a circle is πR². So in this case it is an infinite geometric series: πr²Σ3-n sum from zero to infinity. πr²/(1-1/3)=3πr²/2.

 

Adapt that result to the boundary of the edge length 1 of the equilateral triangle 3r=sin(60°)=2/√3. r=2/3√3=0.222675. The area of all the triangles is 0.392699. Compare that to √3s²/4, 0.433013, the area of the equilateral triangle with s=1, the side length.