This is a sequel question to Hyperbolic Right-Angled Hexagon
Let $ABCDEF$ be an arbitrary hyperbolic convex right-angled hexagon, let $BE, AD, FC$ be geodesic segments connecting $B$ and $E, A$ and $D, F$ and $C$.
We know that a hyperbolic right-angled hexagon is defined uniquely (up to isometry) by the lengths $l_1, l_2, l_3$ of 3 edges $AB, CD, EF$ respectively.
My question is under which conditions of $l_1, l_2, l_3$, the union of 3 quadrilaterals $ADEF, ABCF$ and $BCDE$ can cover the hexagon? and what is the difference between the hyperbolic area of the union and the hyperbolic area of the hexagon?
ADD: I have an idea that the condition is equivalent to the following: $ADEF+ABCF+BCDE-CDEF-ABCD-ABEF>0$, and then by using Gauss-Bonnet theorem, it is equivalent to $A_1+D_2+C_1+F_2+B_2+E1-C_2-F_1-D_1-A_2-B_1-E_2<0$. Here I think we need to use hyperbolic trigonometry to find the relation between angles and edges of bi-rectangle, but I don't know any fomula in bi-rectangle. I hope someone have the idea.