Show that the perimeter of an octagon is $8(\sqrt{2} -1)$

Question

My question is as follows: The top of a table is made in the shape of a regular octagon by cutting the congruent isosceles triangles from the corners of a $1$ m square piece of wood. Show that the perimeter of the octagon is $8(\sqrt{2} -1)$ metres. Thanks

Answer 1

If $x$ is the octagon side on a side of the square after the cut then

$$\frac{1-x}{2}\sqrt{2}$$

is the other side of the octagon after the cut. Since we want a regular octagon we need: $$ \frac{1-x}{2}\sqrt{2}=x \Rightarrow 1-x=\sqrt{2}x \Rightarrow x=\sqrt{2}-1 $$ and the perimeter is $8x=8(\sqrt{2}-1)$

Answer 2

Might this be the graphical demonstration you had in mind?

The octagon in question is the inner hull of two concentric unit squares. Each side is then evidently equal in length to the side of the little square at upper right, which in turn is equal in length to the excess of the diagonal of the unit square over its unit side (the two dashed segments). Since the diagonal is $\sqrt{2}$, each side of the octagon is $\sqrt{2}-1$, and the perimeter is $8(\sqrt{2}-1)$.