It appears that if you join each short rhombus diagonal with a segment, then erase all the original segments (boards), the result is one of the less symmetric Goldberg polyhedra, < under Class III, frequency (1,2)
This Goldberg polyhedron has 72 Faces, 140 Vertices, 210 Edges. Each edge comes from one rhombus (might not be perfect rhombuses) in the original structure, so that has 210 diamond shapes. Should be 420 boards. Impressive.
I think it will let me load an image, but not as png format...
![enter image description here](
It's correct, I was able to outline two pentagons, with two neighboring hexagons each, the hexagons with edges in common as well. ![enter image description here](