Is there an equidissection of a unit square involving irrational coordinates?

Solution 1:

It turns out the answer is yes. An example is given as follows:

where the green triangles are identical up to rotation and have orthogonal sides of length $\frac{1}{2}+\frac{\sqrt{3}}{6}$ and $\frac{1}{2}-\frac{\sqrt{3}}{6}$. Thus they each have area $\frac{1}{12}$, and the middle white square is then just equidissected into $8$ triangles.

On a related note, I found the following generalization of my question in a paper of Kasimatis and Stein from 1990:

In any equidissection of the square with vertices $(0, 0), (1, 0), (1, 1), (0,1)$ are all the coordinates of all the vertices algebraic?

Presumably, this question is open.