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### Perimeter of Octagon

```
Date: 04/02/97 at 21:59:09
From: Anonymous
Subject: Perimeter of Octagon

Dr.Math,

The person wants to make a vegetable garden in an octagonal shape with
a six foot diameter.  She plans to use special laminated coated logs
for the perimeter of the garden.  She wanted to know how long each
side of the octagon would have to be to give a diamter of 6 feet.  It
is a regular polygon.

We decided to use the formula of circumference of a circle formula and
figured out that the circumference of a circle with a six foot diamter
would be 18.84 and divided by 8 would be 2.355 feet for each part.
We know that the octagon would be smaller than this but wondered if
there was a formula for finding the perimeter of an octagon.  Also,
please note that the diamter used was not from one vertex to another.
It was from the midpoint of one side to the midpoint of the opposite
side.

I figured that maybe the Pythagorean formula was necessary here and
came out with an answer of approximately 2.1213 feet for the
lengths of each section of the octagon.  Was this correct or is there
another way to find the perimeter?

Thanks for your help.

Anonymous
Sewell, New Jersey
```

```
Date: 04/03/97 at 11:20:38
From: Doctor Wilkinson
Subject: Re: Perimeter of Octagon

Your idea of using the circle as an approximation was a good one.

However, if the diameter of the octagon is measured from one side
to the other and not from vertex to vertex, you are making a mistake
in saying that the octagon will be smaller.  The circle you get will
be inside the octagon, not outside, so its circumference will actually
be smaller.

Again you are right in thinking of using the Pythagorean Theorem, but
something seems to have gone wrong, because you are not getting the
right answer.  This problem is somewhat tricky, however, so don't feel

If you take two consecutive vertices P and Q of the octagon and look
at the angle POQ, where O is the center of the octagon, you can see
that POQ measures 45 degrees, because it has to be 1/8 of 360 degrees.

Now if you let X be the midpoint of PQ, you have a right triangle
POX, where OX is 3 (half the diameter of the octagon), and angle POX
is half of 45 degrees.  So the problem is to find PX, which is half
the side of the octagon.  (I suggest you draw a picture of this.)

I don't quite know what to do with the angle of 22 1/2 degrees, so
let's extend the line segment XP to a point Y so that YOX is 45
degrees.  So now we have a new right triangle YOX, and it's an
isosceles right triangle because of that 45-degree angle.  Let's
let x be PX (the length we're trying to find), and let y be YP.
We know that x + y = OX = 3.

Can we get another equation?  We need to add one more point to the
picture.  Draw a perpendicular from P to OY, meeting OY at Z.

Now we have something really useful.  We have congruent triangles
OPZ and OPX, because they share a side, they're both right triangles,
and the angles ZOP and POX are both 22 1/2 degrees.  So PZ is the
same as PX, which is x.  We also have similar triangles YZP and YOX,
because they're both right triangles and they share the angle OYX.
So the ratio of y to x is the same as the ratio of OY to OX.  But
that ratio is just the square root of 2, by the Pythagorean Theorem.

So we get y/x = the square root of 2, or 1.414 approximately.

y = sqrt(2) x

x + y = 3

and we can now solve for x, getting approximately x = 1.242 or
2.484 for the side of the octagon.

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

```
Date: 04/03/97 at 21:40:02
From: Anonymous
Subject: Re: Perimeter of Polygon

Doctor Wilkinson,
Thanks a lot.  I understand your explanation completely.  You did a
nice job with your explanation.
:)

Anonymous
```

```
Date: 10/02/2002 at 09:58:00
From: David J. Dainelis
Subject: Perimeter of an Octagon

Dr. Math,

While following along the explanation I could not visualize "extend
a the line segment xp to a point Y so that YOX is 45 degrees" and
"we need to add one more point to the picture. Draw a perpendicular
from P to OY, meeting at Z."

Could you create a small graphic with the points on it?
Thank you in advance.

David J. Dainelis
Grand Rapids Community College.
```

```
Date: 10/02/2002 at 11:08:02
From: Doctor Jerry
Subject: Re: Perimeter of an Octagon

Hi David,

Here is a simpler solution, involving trig. We don't need any extra
points. We just notice that angle XOP is 45/2 = 22.5 degrees. So

tan(22.5) = XP/OX = XP/3.

So,

XP = 3*tan(22.5) = 1.2426...

I'll describe the construction in a slightly different way. Draw
a set of axes, and draw any circle with center at the origin. Draw
the 45-degree line through the origin, and draw the 135-degree line
through the origin. Let O be the center of the circle, P the point
on the positive x-axis where the circle crosses the x-axis. Let Q
be the point in the first quadrant where the 45 deg line intersects
the circle. Connect P and Q. The line PQ is a side of the octagon.
Let X be the midpoint of PQ. Extend QP downward and to the right
until you find point Y on this extension. Choose Y so that angle YOX
is 45 degrees. The point Z is found by drawing from P a line that
meets YO at a right angle. The meeting point is Z.

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

```
Date: 10/02/2002 at 11:22:20
From: David J. Dainelis
Subject: Thank you (Perimeter of an Octagon)

Thanks that helps a lot. I knew there would be a simpler description.

djd
```

```
Date: 10/02/2002 at 11:37:13
From: Doctor Greenie
Subject: Re: Perimeter of an Octagon

Hello, David -

I read with interest the referenced page in the Dr. Math archives and
Dr. Jerry's response to your message.

The method described for finding the perimeter of the octagon is far
more complex than it needs to be.  Below is a derivation which uses
nothing more complex than the ratios of the lengths of the sides of
an isosceles right triangle.

We will think of the octagon as having been cut from a square 6 feet
on a side.  So to get started, we will sketch a square with side 6
feet and draw lines at 45-degree angles in each corner indicating
where we make the cuts to form the hexagon.

If the length of a side of the octagon is x, then the hypotenuse of
each of the triangles is x, and so the length of each leg of the
triangles is x/sqrt(2).  And the length of each side of the original
square is composed of one side of the octagon and two legs of the
triangles. So we have

6 = x + 2(x/sqrt(2))

from which it follows that

x = 6/(1+sqrt(2)) = 2.485281374239

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

```
Associated Topics:
High School Geometry
High School Practical Geometry
High School Triangles and Other Polygons

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