Going, going, ‘gon

Surprise, surprise, a Catriona Shearer puzzle.

16.
Going, Going, ‘gon

Six identical squares and a smaller rectangle are fitted into this regular hexagon. What fraction of the hexagon do they cover?

Spoiler!

Is there any way to do this besides the hard way? I looked for a dissection solution but came up with nothing.

Look at just one twelfth of the hexagon, a right triangle, $ABO$ :

$BC$ and $CD$ are two sides of a square. $OA$ and $DE$ are two line segments I added. The central rectangle isn’t shown here because it differs in the different triangles around the hexagon.

If $AC = DE$ is defined to be length 1 and $AB = CE = x$ then $BC = CD = \sqrt{1+x^2}$ and $BD = \sqrt{2+2x^2}$ . But $DO = 2$ and $BD+DO = 2x$ , so

$\sqrt{2+2x^2} = 2x - 2$

$2+2x^2 = 4(x-1)^2$

$x^2-4x+1=0$

$x = 2 + \sqrt{3}$ is the correct solution

Then the height of the triangle $ABO$ is $3+ 2\sqrt{3}$ and its area is $(3+2\sqrt{3})(2+\sqrt{3})/2 = (6+3\sqrt{3}+4\sqrt{3}+6)/2 = 6 + 7\sqrt(3)/2$ , and the hexagon’s full area is $72+42\sqrt{3}$ .