Example 5 Suppose there are two triangles with side lengths $3,4,5$, four triangles with side lengths $4,5, \sqrt{41}$, and six triangles with side lengths $\frac{5}{6} \sqrt{2}, 4,5$. Using these triangles as faces, how many tetrahedra can be formed?
The three given triangles are sequentially numbered as $1, 2, 3$. It is easy to see that triangles 1 and 2 are right triangles, and triangle 3 is an obtuse triangle. Since a tetrahedron has 4 faces, at least $\left[\frac{4-1}{3}\right]+1=2$ faces must be of the same type. Furthermore, it is not difficult to prove that if two side triangles in a tetrahedron are congruent, then the other two side triangles are either two congruent triangles or both isosceles triangles. However, there are no isosceles triangles among the given triangles, so the 4 side faces of the tetrahedron must be two sets of congruent triangles.
(1) Using two 1st type triangles.
Refer to Figure 6-3. Since the edge of length 3 is the common perpendicular to the two edges of length 4, we have $x>3$. Also, $\frac{5}{6} \sqrt{2}\sqrt{50}$, thus
$$
x \geqslant d>\sqrt{50}>\sqrt{41}>\frac{5}{6} \sqrt{2} \text {. }
$$

From this, it is evident that using four 3rd type triangles, or using two 3rd type triangles and two 2nd type triangles, cannot form a tetrahedron.
(3) Using four 2nd type triangles.
If a tetrahedron can be formed, then the two edges of length 4 would both be the common perpendiculars to the two edges of length 5, which is impossible.
In summary, using the given triangles as side faces, only one tetrahedron can be formed.