geometry
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.
Its E. Key step is: $h$ = height, then: $h^2 = 4^2 - \left(\dfrac{d}{2}\right)^2 = 16 - \left(\dfrac{4\sqrt{2}}{2}\right)^2 = 8$. So: $h = 2\sqrt{2}$, and therefore:
$V = \dfrac{Sh}{3} = \dfrac{4^2\cdot 2\sqrt{2}}{3} \approx 15.08$, $d$ = diagonal of the square at base, and $d = 4\sqrt{2}$
I visualized the solution Nick provided:
(I found it because it had used this arrangement to build boats models with a magnet game)
Best Answer
You seem to suggest that you want a pentahedron whose five faces are all triangular. There is no such pentahedron, not with equilateral triangles and not with any other kind of triangles.
Every polyhedron whose faces are all triangular has an even number of sides. For example, the regular tetrahedron (4 sides), the octahedron (8 sides), the snub disphenoid (12 sides), and the rest of these.
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.