Convex Hexagon divided into 7 Congruent Triangles

Question

So the motivation for this problem comes from the Asian Pacific Math Olympiad 1990, problem 5:

Show that for every integer n ≥ 6, there exists a convex hexagon which can be dissected into exactly n congruent triangles

The interesting part is trying to find a tiling of a convex hexagon into 7 congruent triangles, such as the one below

There are some immediate variations of this one by rearranging some triangles, but I'm curious if there are other tilings with different types of triangles, i.e. non congruent to the one used here.

Thanks

Edit: Some resources
Solutions by Canadian Problems Committee
Solutions from www.vnmath.com, page 7

Answer 1

Take triangles with sides $1$, $3$, and $3$. Line up five of them to form a trapezoid with sides $2$, $3$, $3$, and $3$. Put the sixth one flush on the side of the trapezoid opposite the length $2$ side, and the seventh one flush on the other length $3$ side of the sixth one.