A square-based pyramid has all edges 2 units long. We attach a pyramid to one of the side faces of the pyramid, where the side edges of the new pyramid are of equal length. The total length of the edges of the resulting solid is 18 units. What is the volume of the solid?
The base of the pyramid $A B C D E$ is a square, and every edge is 2 units long, so the side faces are equilateral triangles. The edges of the added pyramid $B C E F$ are equal, making $B C E F$ a regular pyramid. The side edges of this pyramid are longer than the distance from the center of the equilateral triangle $B C E$ to its vertices, so the total length of the side edges is at least $3 \cdot \frac{2 \sqrt{3}}{2} \cdot \frac{2}{3}=2 \sqrt{3}>2$. Therefore, the total length of the edges of the resulting solid can only be 18 if some edges of the original pyramid coincide with the faces of the new solid. The vertex $F$ is located perpendicularly above the centroid of the triangle $E B C$, so $F$ cannot lie in the plane of $A B C D$. If $F$ lies in the plane of $A B E$, then, since the perpendicular bisector plane of the segment $B C$ is a symmetry plane for both the original pyramid and the pyramid $B C E F$, $F$ must also lie in the plane of $C D E$. Therefore, the edges $B E$ and $C E$ of the original pyramid must coincide with the faces of the new solid. The total length of the edges of the new solid will be exactly 18 units if $B F=C F=E F=2$.

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According to this, the resulting solid consists of a square pyramid and a regular tetrahedron. The height of the square pyramid can be easily determined using the Pythagorean theorem in the triangle $A C E$: $\sqrt{2}$, so its volume is $\frac{4 \sqrt{2}}{3}$. The volume of the regular tetrahedron with an edge length of 2 units is $\frac{2 \sqrt{2}}{3}$. Therefore, the volume of the resulting solid is $2 \sqrt{2}$.