# What is the perimeter of a regular octagon with a radius of length 20?

Sep 12, 2015

It depends:

If the inner radius is $20$, then the perimeter is:

$320 \left(\sqrt{2} - 1\right) \approx 132.55$

If the outer radius is $20$, then the perimeter is:

$160 \sqrt{2 - \sqrt{2}} \approx 122.46$

#### Explanation:

Here the red circle circumscribes the outer radius and the green circle the inner one.

Let $r$ be the outer radius - that is the radius of the red circle.

Then the vertices of the octagon centred at $\left(0 , 0\right)$ are at:

$\left(\pm r , 0\right)$, $\left(0 , \pm r\right)$, $\left(\pm \frac{r}{\sqrt{2}} , \pm \frac{r}{\sqrt{2}}\right)$

The length of one side is the distance between $\left(r , 0\right)$ and $\left(\frac{r}{\sqrt{2}} , \frac{r}{\sqrt{2}}\right)$:

$\sqrt{{\left(r - \frac{r}{\sqrt{2}}\right)}^{2} + {\left(\frac{r}{\sqrt{2}}\right)}^{2}}$

$= r \sqrt{{\left(1 - \frac{1}{\sqrt{2}}\right)}^{2} + \frac{1}{2}}$

$= r \sqrt{1 - \frac{2}{\sqrt{2}} + \frac{1}{2} + \frac{1}{2}}$

$= r \sqrt{2 - \sqrt{2}}$

So the total perimeter is:

$\textcolor{red}{8 r \sqrt{2 - \sqrt{2}}}$

So if the outer radius is $20$, then the perimeter is:

$8 \cdot 20 \sqrt{2 - \sqrt{2}} = 160 \sqrt{2 - \sqrt{2}} \approx 122.46$

$\textcolor{w h i t e}{}$
The inner radius will be ${r}_{1} = r \cos \left(\frac{\pi}{8}\right) = \frac{r}{2} \left(\sqrt{2 + \sqrt{2}}\right)$

So $r = \frac{2 {r}_{1}}{\sqrt{2 + \sqrt{2}}}$

Then the total perimeter is

$8 r \sqrt{2 - \sqrt{2}} = 8 \frac{2 {r}_{1}}{\sqrt{2 + \sqrt{2}}} \sqrt{2 - \sqrt{2}}$

$= 16 {r}_{1} \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2 + \sqrt{2}}}$

$= 16 {r}_{1} \frac{\sqrt{2 - \sqrt{2}} \sqrt{2 + \sqrt{2}}}{2 + \sqrt{2}}$

$= 16 {r}_{1} \frac{\sqrt{\left(2 - \sqrt{2}\right) \left(2 + \sqrt{2}\right)}}{2 + \sqrt{2}}$

$= 16 {r}_{1} \frac{\sqrt{2}}{2 + \sqrt{2}}$

$= 16 {r}_{1} \frac{\sqrt{2} \left(2 - \sqrt{2}\right)}{\left(2 + \sqrt{2}\right) \left(2 - \sqrt{2}\right)}$

$= 8 {r}_{1} \left(2 \sqrt{2} - 2\right)$

$= \textcolor{g r e e n}{16 {r}_{1} \left(\sqrt{2} - 1\right)}$

So if the inner radius is $20$, then the perimeter is:

$16 \cdot 20 \left(\sqrt{2} - 1\right) = 320 \left(\sqrt{2} - 1\right) \approx 132.55$

$\textcolor{w h i t e}{}$
How good an approximation for $\pi$ does this give us?

While we're here, what approximation for $\pi$ do we get by averaging the inner and outer radii?

$\pi \approx 2 \left(2 \left(\sqrt{2} - 1\right) + \sqrt{2 - \sqrt{2}}\right) \approx 3.1876$

...so not great.

To get as good an approximation as $\frac{355}{113} \approx 3.1415929$, the Chinese mathematician Zu Chongzhi used a $24576$ ($= {2}^{13} \times 3$) sided polygon and counting rods.

https://en.wikipedia.org/wiki/Zu_Chongzhi