# Find the length of the side of a regular octagon inscribed in a circle of radius 1

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• Feb 5th 2013, 06:50 PM
Jammix
Find the length of the side of a regular octagon inscribed in a circle of radius 1
Please don't use sine, cosine, or tangent when solving. Solve it by introducing segments, etc.
• Feb 5th 2013, 07:24 PM
Prove It
Re: Find the length of the side of a regular octagon inscribed in a circle of radius
The easiest way is to realise that the octagon is made up of eight congruent isosceles triangles, each with two lengths 1/2 a unit long and with their included angle as 45 degrees.

Then the area of each triangle is

\displaystyle \displaystyle \begin{align*} A &= \frac{1}{2} ab\sin{C} \\ &= \frac{1}{2}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \sin{\left( 45^{\circ} \right)} \\ &= \frac{1}{8}\cdot \frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{2}}{16} \end{align*}

Therefore the entire area of the octagon is \displaystyle \displaystyle \begin{align*} 8 \cdot \frac{\sqrt{2}}{16} = \frac{\sqrt{2}}{2} \end{align*}.

I'm not sure if it can be done without using some trigonometry tbh...

Edit: I misread the question, I thought you wanted the area. The length of each segment is easily found using the Cosine Rule, since you have two sides of each triangle and their included angle.

\displaystyle \displaystyle \begin{align*} c^2 &= a^2 + b^2 - 2ab\cos{(C)} \\ c^2 &= \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 - 2\left( \frac{1}{2} \right)\left( \frac{1}{2} \right) \cos{\left( 45^{\circ} \right)} \\ c^2 &= \frac{1}{4} + \frac{1}{4} - \frac{1}{2} \left( \frac{\sqrt{2}}{2} \right) \\ c^2 &= \frac{2 - \sqrt{2}}{4} \\ c &= \frac{\sqrt{ 2 - \sqrt{2} }}{2} \end{align*}

The length of each side in the octagon is \displaystyle \displaystyle \begin{align*} \frac{\sqrt{2 - \sqrt{2}}}{2} \end{align*} units.