$1 \cdot 102$ There are two triangles with side lengths of $3,4,5$, four triangles with side lengths of 4,5 , $\sqrt{41}$, and six triangles with side lengths of $\frac{5}{6} \sqrt{2}, 4,5$. Using the above triangles as faces, how many tetrahedrons can be formed?
[Solution] Number the three types of triangles as $1, 2, 3$. It is easy to see that triangles $1$ and $2$ are right-angled triangles, while triangle $3$ is an obtuse triangle. By the pigeonhole principle, among the four faces of the tetrahedron, at least two faces are of the same number. Moreover, it is not difficult to prove that in a tetrahedron, if two side triangles are congruent, then the other two side triangles are either congruent triangles or both isosceles triangles. Since there are no isosceles triangles among the given triangles, the four side triangles of the tetrahedron must be two sets of congruent triangles.
(1) Using two $1$-numbered triangles.
Refer to the right figure. Since the edge of length $3$ is the common perpendicular of the two edges of length $4$, we have $x>3$. Also, $\frac{5}{6} \sqrt{2}\sqrt{50}$. Therefore, we have
$$
x \geqslant d>\sqrt{50}>\sqrt{41}>\frac{5}{6} \sqrt{2} .
$$

From this, we know that using four $3$-numbered triangles, or using two $2$-numbered triangles and two $3$-numbered triangles, cannot form a tetrahedron.
(3) Using four $2$-numbered triangles.
If a tetrahedron can be formed, then the two edges of length $4$ would both be the common perpendiculars of the two edges of length $5$, which is impossible.
In summary, using the given triangles as sides, only one tetrahedron can be formed.