Starting to use "cut and fold" to explore ratios - thanks Baravaille! . . . which leads to a simple extrapolation from a hexagon! Which is, of course beguiling nonsense! Simply rotate the yellow triangles 1/3 to the position of the blue triangles and you have your Enneagon! Base = 3 overlapping Hexagons Enneagon Beetle, wings closed. Enneagon Beetle, in flight. Three Overlapping Hexagons with Natural Trisections . . . . . . . . provide everything needed after all! There is a slight difference between the edge of an equilateral triangle and the length of the edge of an isosceles triangle with the same height but 1/3 base that I find quite intriguing . . . . Student work - all the essential elements!
When AB = x, OC = √[x²-(x/2)²]
= √[x²-(x²/4)]
= √(3/4x²)
= x(√3)/2
When
AB = x, OD = √[x²-(x/2)²+(x/6)²]
= √[x²-(x²/4)+(x²/36)]
= √[(3/4x²)+(1/36x²)]
= √[(27/36x²)+(1/36x²)]
= √(28/36x²)
= √(7/9x²)
= x(√7)/3
Now I'm off to attempt pentagons by dividing squares into fifths! 90/5=18 18*4=72 18*6=108 | 
