# How to calculate the height of the pyramid?

I have a $n$-sided regular pyramid based on a regular polygon, the length of the side of regular polygon, $s$. Also I know the dihedral angle between the face and the base, $\alpha$.

Question. How to calculate the height of the pyramid, $h$?

My attemp is:

I have found the Thales' method. Thales measured the height of the pyramids by their shadows at the moment when his own shadow was equal to his height.

Let's say $n=4$, $s=1$ unit and $\alpha=60$ then I can find the $R=\frac{s}{2 \cdot sin \frac{180}{n}}$ and $r=R \cdot cos\frac{180}{n}$.

• What side has length $s$? One of the sides in the regular polygon? In that case we need more information to finde the height. – Martin Oct 24 '17 at 6:17
• Why would Thales' method be any different for a pyramid with a differently shaped base? It should work for any object (as long as the base isn't so wide as to not allow you to measure the shadow). – Joppy Oct 24 '17 at 6:47
• @Joppy, yes the Thales' method should be work for any object. But I'd like to estimate the height without sun and shadow. – Nick Oct 24 '17 at 7:19
• You give zero information about the shape of the pyramid. Hence the lower bound is 0, and the upper bound ∞. Artificial examples are possible: Imagine a building like a tower. For each 100m in height, you move the stone bricks 1 μmeter to the center. Eventually you will hit the center. What is your estimate now? Of course, there are some natural regularities, e.g. where you build that pyramid (earth, moon, ..) and what material you use. So there will be a bound smaller infinity. Still, that is the best you can do, given the information in the question. – P. Siehr Oct 24 '17 at 7:41
• What do you mean by "the angle between the side edge and base"? Is the "side edge" an edge from a vertex of the $n$-sided polygon to the top of the pyramid (in which case $\alpha$ is the angle between a line and a plane) or is it an entire triangular face of the pyramid (in which case $\alpha$ is the dihedral angle between that face and the base)? – David K Oct 24 '17 at 12:04

$$r = \frac{s}{2\tan(180/n)}$$
Then you have a vertical right triangle of unknown height $h$, known base $r$, and known angle $\alpha$.
$$\tan(\alpha) = \frac{h}{r}\\ \implies h = r\tan(\alpha)$$