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### The Diameter of an Octagon

```
Date: 7/25/96 at 11:59:16
From: Anonymous
Subject: The Diameter of an Octagon

I'm trying to figure out the diameter of an Octagon. I have looked
thing. If you could help, that would be great.  All I know is
that each side of the octagon is 2 in.

Thank you,
Michael
```

```
Date: 7/25/96 at 13:17:55
From: Doctor Paul
Subject: Re: The Diameter of an Octagon

Unless you say otherwise I'm assuming you're referring to a regular
octagon (like a stop sign).

In theory, an octagon is any eight-sided polygon, so without a diagram
it's impossible to measure the diameter. I will show you the diameter
for a regular polygon.  There are two different diameters that exist.
I will show you how to get both.

Diameter 1:
Short diameter -

__
/  \      The short diameter measures from one side to the
A |    | B   one directly opposite it... like from the center of
\__/      segment A to the center of segment B.

Here's what we're solving for:
__
/  \
|    |
------     we want to know the distance CD (this is NOT the
C      D    distance from vertex to vertex).

The distance is    2                2
------   + 2  +  -------
sqrt(2)          sqrt(2)

which is approximately equal to 4.828.

In the general case where a is the length of a side, the short
diameter is equal to:

2 * a
------ + a
sqrt(2)

Diameter 2:
The long diameter -
this one measures from vertex to opposite vertex.

Divide the octagon up into eight different triangles.  The angle at
the top (closest to the center of the octagon) is 45 degrees.  The
other two angles are 67.5 degrees each.  Let's use the law of Sines to
solve for the radius and then double it to get the long diameter:

sin 45    sin 67.5
------ = ---------
2          x

cross multiply and solve yields:  x = 2.613

We double this to get the long diameter...it is:  5.226

In the general case where the side equals 'a'
the long diameter is equal to:

2 * a * sin 67.5
----------------
sin 45

I hope this helps.

Regards,

-Doctor Paul,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 7/25/96 at 13:43:27
From: Doctor Pete
Subject: Re: The Diameter of an Octagon

There are several methods one could use to solve this, but I will give
the most elementary, which only requires knowledge of the Pythagorean
Theorem, which states that in a right triangle, the sum of the squares
on the two legs is equal to the square on the hypotenuse.

To begin, here's a picture to give you an idea of what to look for:

B ____________________ C
/|                  |\
/  |                  |  \
/    |                  |    \
/      |                  |      \
/        |                  |        \
/          |_                 |_         \
A /____________|_|________________|_|__________\ D
R                  P

The above trapezoid shows three sides of the octagon;
AB = BC = CD = 2.  P and R are the feet of the perpendiculars dropped
from C and B, respectively. Thus BCPR is a rectangle, and in
particular, BC = PR.  Then note that AR = BR = CP = DP, because angles
ABR = BAR = DCP = CDP = 45 degrees.  Thus by the Pythagorean Theorem,
[AR]^2 + [BR]^2 = [AB]^2 = 2^2 = 4, from which it follows that
2[AR]^2 = 4 or [AR]^2 = 2.  Hence AR = Sqrt[2], and
AD = AR +RP + PD = Sqrt[2] + 2 + Sqrt[2] = 2 Sqrt[2] + 2 =
2(Sqrt[2] + 1).

Now, AD is not the diameter of the octagon.  Consider point E, the
next vertex of the octagon going clockwise from point D:

A _____________________________________________ D
- _            2(Sqrt[2]+1)              |_|
- _                                    |
- _                                |
- _                            |
- _                        |
- _                  2 |
- _                |
- _            |
- _        |
- _    |
- _| E

(In the above diagram, points B, C, P, R were omitted.)  Then AE is
the diameter of the octagon.  Again by the Pythagorean Theorem,
[AD]^2 + [DE]^2 = [AE]^2.  Hence

AE = Sqrt[(2(Sqrt[2] + 1))^2 + 2^2]
= Sqrt[4(2 + 2 Sqrt[2] + 1) + 4]
= 2 Sqrt[2 + 2 Sqrt[2] + 1 + 1]
= 2 Sqrt[4 + 2 Sqrt[2]]
= 2 Sqrt[2] Sqrt[1 + Sqrt[2]] ,

which is the result you requested.

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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