Question

Find the side of the regular octagon inscribed in a circle of radius 50 mm. Solve in two ways. Also, find the value of each interior and exterior angles of the regular polygon.

Answer 1

$$\alpha=\dfrac{360\degree}{8}=45\degree$$

Consider the isosceles triangle $A_1OA_2.$ The Law of Cosines

$$a^2=R^2+R^2-2R\cdot R\cos\alpha$$

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

$$a=50\sqrt{2-\sqrt{2}}\ mm$$

Let $A_1(\dfrac{50\sqrt{2}}{2}, \dfrac{50\sqrt{2}}{2}), A_2(50,0).$ Then

$$a=A_1A_2=\sqrt{(50-\dfrac{50\sqrt{2}}{2})^2+(0-\dfrac{50\sqrt{2}}{2})^2}$$

$$=50\sqrt{1-\sqrt{2}+\dfrac{1}{2}+\dfrac{1}{2}}=50\sqrt{2-\sqrt{2}}\ (mm)$$

$a=50\sqrt{2-\sqrt{2}}\ mm$

$$\beta=\dfrac{180\degree(8-1)}{8}=157.5\degree$$

$$interior\ angle=157.5\degree$$

The interior and exterior angle add up to 180°

$$exterior\ angle=180\degree-\beta=180\degree-157.5\degree$$

$$exterior\ angle=22.5\degree$$