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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.

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    $\begingroup$ 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? $\endgroup$ – Ross Millikan Jun 18 '14 at 15:22
  • $\begingroup$ @RossMillikan not square pyramid. Pentahedron having one face of equilateral/isosceles triangle of all sides one. I.e, having minimal faces. $\endgroup$ – waqar Jun 18 '14 at 15:34
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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.

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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: $\frac {\sqrt 3} 2$

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  • $\begingroup$ penta means five-you need five faces. Please provide a sketch of the solid you envision if not the triangular prism and not the pyramid. Note that all triangles will be equilateral if the sides are all 1-you speak of isosceles in an earlier comment. $\endgroup$ – Ross Millikan Jun 18 '14 at 15:56

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