Volume of pentahedron having all sides of length 1?

What is the volume of pentahedron having all sides of length 1? And what is the height of the side which is above the equilateral triangle of side length 1 in that pentahedron?

Edit. not square pyramid. Pentahedron having one face of equilateral/isosceles triangle of all sides one. I.e, having minimal faces.

• By pentahedron do you mean a pyramid with a square base? Or two tetrahedra joined at a pair of faces. Do you know the formula for the volume of a pyramid? – Ross Millikan Jun 18 '14 at 15:22
• @RossMillikan not square pyramid. Pentahedron having one face of equilateral/isosceles triangle of all sides one. I.e, having minimal faces. – waqar Jun 18 '14 at 15:34

This is easy to see. Suppose there are $F$ faces and $E$ edges. If we count the 3 edges on each face, we count $3F$ edges total. But this counts each edge twice, because each edge belongs to 2 faces. So the correct number of edges is $E = \frac12\cdot 3F$. But $E$ must be an integer, so $F$ must be even.
In particular, $F=5$ gives $E=\frac{15}2$, which is impossible.
The pentahedron that is not a pyramid is a prism with two equilateral triangle and three square faces. The volume is the area of the triangle, $\frac {\sqrt 3}4$ times the height of $1$, for a total of $\frac {\sqrt 3}4$. When resting on a square face, the height is the altitude of a unit equilateral triangle: $\frac {\sqrt 3} 2$