Example of Flattened Irregular 5 Face Pyramid:
I have a series of flattened pyramids which I try to model in 3D. All pyramids are either 4 or 5 faces + base. All use triangles combined from 3 fixed edge dimensions (30,36,42).
How can I find the height of these pyramids? Thanks,
Edit#1: Joseph, Thanks, but I don't think the tetrahedron solution can solve this. I probably wasn't clear enough but I have only the length of the perimeter edges and length to the pyramid vertex..and so don't have enough info to create the intermediate tetrahedrons.
Hopefully this image is clearer:
Vertexes a-e are all on the same plane. Thanks again.
First solve it for a tetrahedron, three lateral faces. Call the three base vertices $a$, $b$, $c$, the apex vertex $v$, and the lengths of the edges from $a$ to $v$: $d_a$, $d_b$, $d_c$. Now solve three equations simultaneously in the unknowns $x,y,z$: \begin{eqnarray} ((x, y, z) - a) \cdot ({x, y, z} - a) &=& d_a^2 \\ ((x, y, z) - b) \cdot ({x, y, z} - b) &=& d_b^2 \\ ((x, y, z) - c) \cdot ({x, y, z} - c) &=& d_c^2 \end{eqnarray} In general you'll find two solutions, $\pm z$; take $z>0$. This gives you $v$ :
Now, in your case, you have either four or five lateral faces, which gives you redundancy. Solve for all combinations of $\binom{4}{3}$ or $\binom{5}{3}$ inscribed tetrahedra and average the solutions for $v$. Of course they should all be identical, but numerical errors will likely creep in.