geometry

First of all, the image is below not above 

The hexagon is composed of 6 isoceles triangles. In the sketch above, you just have to draw their apothem and then you get 6x2 right-angled triangles. So you have the same little right-angled triangles composing the whole hexagone. Each isoceles triangle has 6 little right-angled triangles and the total is \(\LaTeX\): \(6\text{isoceles}\triangle\times6\text{little right-angled triangle}=36\). You count \(\LaTeX\): \(4\text{little right-angled}\triangle\times2\text{shaded area}=8\text{shaded little right-angled triangles}\). So the ratio is: \(\frac{36}{8}=\frac{9\times4}{2\times4}=\boxed{\frac92}\). And \(\dfrac92=9:2\).

If we inscribe  the hexagon in a circle, let  the radius  = R

By AAS  we  have 8 congruent  30-60-90  right triangles

The side opposite the 60° angle = R/2

And the side opposite  the   30°  angle =  R / ( 2sqrt (3) )

So.....the area  of  these  8 right triangles  is

8 *(1/2) (R/2) (R  /(2sqrt (3))  =   R^2/sqrt (3)

And  the  side of the  hexagon  = R....so its area =  3sqrt (3)R^2 /2

So   H  /  S    =        3sqrt(3) R^2 / 2               3 sqrt (3) sqrt (3)             9

                              _____________  =          ______________   =     ____

                               R^2 / sqrt (3)                             2                            2

Exactly what   cryptoaops found  !!!!

Good job, cryptoaops  !!!!