Example 5 A regular tetrahedron and a regular square pyramid have congruent lateral faces, so their lateral faces can coincide with each other. How many lateral faces does the polyhedron formed by combining them have?
(19th All-Russian Competition Problem)
As shown in Figure 23-12, the given quadrilateral pyramid is $P-ABCD$, and $l$ is the intersection line of faces $PAD$ and $PBC$. Point $E$ is on $l$ such that $EP=AD=BC$, then $PAEB$ is a tetrahedron with all faces congruent to the side faces of the quadrilateral pyramid.

In fact, $EP \parallel BC$ and $EP$ is equal to $BC$, so $PEBC$ is a parallelogram. Since $PC=BC$, $PEBC$ is a rhombus. Similarly, $PEAD$ is also a rhombus, thus all edges of the tetrahedron $PAEB$ are equal. Therefore, $EPDABC$ is a polyhedron that meets the given conditions, with 5 side faces.