Area of a regular octagon

A regular polygon with eight sides, each of length $s>0$, has areaJoin its center to all eight vertices of a polygon. The resulting congruent isosceles triangles each have base $s$ and central angle $\pi /4$. Bisecting one gives a right triangle with opposite side $s/2$ and adjacent side $h$, the perpendicular center-to-side distance. HenceAdding the eight triangle areas gives $A=8(sh/2)=4sh$, proving the formula. For side length one, the area is $2+2\sqrt{2}$.