Is there an equation to find the lengths of the diagonals of a regular heptagon, given the length of one side?
I've decided to build an object out of building material with straight sides. I have 7 units of the building material, to be arranged on a flat surface, and they are all of the same length. I intend to organize them in a heptagon, which is easy enough when it doesn't need to be a regular heptagon, as I can just connect the end of one to the next until I reach the last which is then connected to the free end of the first. But now that I have a 'ring' of them, and they certainly form a heptagon, I'm struggling to find a way to make sure the heptagon is a regular one, other than just 'eye-balling it". My intuition, and my limited recollection of high-school geometry from a couple decades or so ago, tells me that using the length of the diagonals as guides would serve this purpose perfectly. Unfortunately I don't recall how to determine that length (if, indeed, I ever knew it for heptagons, specifically).
I understand that, for a heptagon, there are 4 diagonals from any one vertex, 2 pairs of 2 different lengths. I suspect finding the length of the shorter of the two would be the most useful to my situation, but a way to find either one, or both, would be appreciated.
$\endgroup$2 Answers
$\begingroup$One can easily find the length of the diagonals of the heptagon using simple trigonometry and a calculator.
Let the side length be x, angle between sides is ${\approx}128.56^{\circ}$Length of shorter diagonal will be $2xsin({128.56\over 2})$
The longer diagonal can also be found similarly. I leave that as a challenge for you to do.
$\endgroup$ $\begingroup$If there isn't we can figure one out.
Consider the heptagon to be centered at $O$ and to have regular vertices at $A_1$ to $A_7$.
The angles of any $A_{n-1}$ to $O$ to $A_{n}$ will be $\frac {360^\circ}{7}$.
The $A_i$ are regularly apart and lie on a circle.
There is a theorem that if $A,B,C$ are three points of a circle and $O$ is the center of the circle then $\angle AOB = 2\angle ACB$.
So consider triangle $A_7A_1A_4$ we figure $\angle A_1A_4A_7= \frac 12 \frac {360}{7} = \frac {180}{7}$ and $\angle A_1A_7A_4=\angle A_7A_1A_4= \frac {180 - \frac {180}{7}}2 = \frac 37*180$.
By law of sines, $\frac {A_4A_1}{\sin \frac 37*180} = \frac {A_4A_7}{\sin \frac 37*180} = \frac {A_1A_7}{\sin \frac {180}{7}}$ so if the side of the heptagon is $s = A_1A_7$ then
$A_4A_7 = A_4A_1 = s*\frac {\sin \frac 37*180}{\sin \frac {180}{7}}$
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