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### Side Length of a Regular Octagon, Without Trigonometry

```Date: 09/19/2003 at 17:12:28
From: Gail
Subject: Finding the length of the side of an octagon

I teach middle school math.  A friend asked me how to find the length
of one side of a regular octagon that has a (side-to-side) diameter of
16 feet.  I was not sure how to go about it, and would appreciate your
help.
```

```
Date: 09/20/2003 at 02:02:06
From: Doctor Greenie
Subject: Re: Finding the length of the side of an octagon

Hi, Gail --

Think of the octagon as a square, with the four corners cut off at
45-degree angles.  The pieces cut off are 45-45-90 right triangles.

Let's use "x" to denote the length of each of the legs of each of
these triangles.  Then the hypotenuse of each of those triangles is
x*sqrt(2).  But that hypotenuse is one of the sides of the octagon,
so each side of the octagon has length x*sqrt(2)

You are calling the diameter of the octagon the distance from side
to side; this is the same as the length of the side of the original
square.  But if we picture the original square with the corner
triangles marked but not yet cut off, then each side of that square
(i.e., the diameter of the octagon) is made up of three segments in
the ratio 1:sqrt(2):1.  The side of the octagon is the longer of
these three pieces; then the fraction of the whole diameter of the
octagon which is the length of one of the sides of the octagon is
given by the ratio

sqrt(2)
-----------
2 + sqrt(2)

So in general, if the diameter of the octagon is x, then the length
of the side of the octagon is

sqrt(2)
x * -----------
2 + sqrt(2)

And in your particular case, with a given diameter of 16, the length
of the side of the octagon is

sqrt(2)
16 * -----------
2 + sqrt(2)

I hope this helps.  Please write back if you have any further

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 09/26/2003 at 20:36:18
From: Gail
Subject: Thank you

Dr. Greenie,

Thank you for your quick response.  I appreciate this very
much and your explanation was understandable.

Thank you again,
Gail
```
Associated Topics:
High School Euclidean/Plane Geometry

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