Perhaps I'm confused here and/or being too pedantic, but I don't see how you can form a Star of David with two differently sized triangles. The "Star of David" is just another name for a hexagram, which is defined as the star polygon {6/2}. This, in turn, requires the six points used to form the hexagram to be evenly spaced around a circle, which is equivalent to a triangle connecting the 1st, 3rd, and 5th vertices of a regular hexagon and another triangle connecting the 2nd, 4th, and 6th vertices. So, by definition, the two triangles used to create a Star of David must both be equilateral, but more specifically, they must be the same size.
As a consequence of all of this, we can use many known properties of circles and hexagons to make the desired information basically fall out. Suppose we began with a circle of radius r and formed the Star of David by connecting the appropriate equally spaced points. Then, each side the triangles must have length \(\displaystyle r \sqrt{3}\). Furthermore, each of the sides of the triangles are subdivided into three "segments" of equal length. Thus, for the purposes of your problem, you would have \(\displaystyle x = \dfrac{r \sqrt{3}}{3}\), where you can supply the appropriate value of r.
If you're interested, the fact that the length of the sides of the triangle is \(\displaystyle r \sqrt{3}\) can be proven by plugging in two points on the hexagon and cranking it through the distance formula:
\(\displaystyle \sqrt{\left[ r \sin \left( \dfrac{(k+2) \pi}{3} \right ) - r \sin \left( \dfrac{k \pi}{3} \right ) \right]^2 + \left[ r \cos \left( \dfrac{(k+2) \pi}{3} \right ) - r \cos \left( \dfrac{k \pi}{3} \right ) \right]^2}\)
This will involve a lot of trigonometry and algebra, and will likely get messy in places, but if done correctly will lead to the desired result.
