Way to find volume of the solid A solid has a square base of side $s$ . The upper edge is parallel to the base and has length $2s$. All other edges have length $s$ . What is the volume of the solid ? 
NB : The volume of the tetrahedron with all sides length l is $ V = \dfrac{\sqrt2}{12}l^3$
 A: Converting comments to answer.

Especially given your "NB", it seems like you're describing a solid formed by slicing a regular tetrahedron by a plane parallel to (and half-way between) two opposite edges. In that case, the volume of the solid is half the volume of a regular tetrahedron of side $2s$.
Imagine a regular tetrahedron balancing on an edge, with the opposite edge parallel to the table. A plane parallel to the table will be parallel to those opposite edges. If the plane passes halfway between the table and the "upper" edge, then that plane will pass through the midpoints of the four remaining edges; those midpoints determine a square whose edges have half the length of the tetrahedron's edges (since each edge of the square is a "mid-line" of one of the tetrahedron's faces). And clearly the plane cuts the volume of the tetrahedron in half.
Here's a cool animation (though without the "balancing on an edge" thing):

Image credit: http://www.mmmlib.com/anne%20tyng%20maryland.html .
A: We can divide the solid in two tetrahedra and a square pyramid. See fig.1.

Fig.1
The pyramid height $GH$ has measure $x$, and the measure of $DB$ is the double of $x$. 
(Check it out).
Using pythagoras in $\triangle ABD$ we get:
$$x=\frac{s \sqrt{2}}{2}$$
We know that the volume of the solid $V$ is:
$$V=2V_{tetra}+V_{pyr}$$
where $V_{tetra}$ is the volume of each tetrahedron and $V_{pyr}$ is the volume of the square pyramid.
Calculating each term we have:
$$V=2 \frac{\sqrt{2}}{12}s^3+\frac{1}{3}s^2 \frac{s \sqrt{2}}{2} \Rightarrow$$
$$V= \frac{\sqrt{2}}{3}s^3$$
