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

Question

An equidissection of a square is a dissection into non-overlapping triangles of equal area. Monsky's theorem from 1970 states that if a square is equidissected into $n$ triangles, then $n$ is even. In 1968, John Thomas proved the following weaker statement: there is no equidissection of a unit square into an odd number of triangles whose vertices are rational numbers with odd denominators.

I wonder to which extent the result of Thomas fails to cover all cases. It is easy to construct an equidissection where the coordinates are rational numbers with even denominators. My question is:

Is there an equidissection of the unit square containing a triangle with at least one irrational coordinate?

Dissecting the square into $6$ or less triangles seems to always yield rational coordinates.

Answer 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.