Octagon in a Circle

To find the perimeter $(p)$ : split the octagon in to 8 isosceles triangles with two of the sides being the radius of the circle. The circle is a unit circle so its radius is 1. As we have two sides and the angle in between them we can use the cosine rule to work out the remaining length:

$a^2 =b^2+c^2 -2bcCosA$

Where $a$ is the length of a side of the octagon, $b$ and $c$ are the two radii and $A$ is the angle in the center of the octagon.

As a the angles around a point sum to ${ 360 }^{ \circ }$ the angle $A$ must equal $\frac { 360^{ \circ } }{ 8 } =45^{ \circ }$ .

Therefore the cosine rule becomes: $a^{ 2 }=1^{ 2 }+1^{ 2 }-2\times 1\times 1\times Cos(45)$

as: $Cos(45)=\frac { \sqrt { 2 } }{ 2 }$

we can simplify to: $a^{ 2 }=2-\sqrt { 2 }$

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

The perimeter is 8 times the length so: $p=8a$

Substituting in $a$ : $p=8\sqrt { 2-\sqrt { 2 } }$