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

**Jammix** · February 5th 2013, 06:50 PM

Please don't use sine, cosine, or tangent when solving. Solve it by introducing segments, etc.

**Prove It** · February 5th 2013, 07:24 PM

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

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

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

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.

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

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

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