A regular octagon is inscribed in a circle with a radius of 12 inches. Find the perimeter of the octagon. Round to the nearest tenth.

Answer 1

Consider any vertex of the regular octagon. It can be connected to the center of the circle. Two consecutive such lines form the sides of an isoceles triangle. There are 8 such triangles inside the circle.

Each side of these triangles is equal to the radius of the circle: 12 inches in this case. If we drop a perpendicular from the vertex of any of these triangles to its base, we form 2 congruent right triangles. The base of these right triangles is equal to the sine of one-half of the vertex angle times its hypotenuse (sine = opposite/hypotenuse, or sine * hypotenuse = opposite).

The circle's radius is equal to the hypotenuse of these right triangles. Since this is an inscribed octagon, each interior isoceles triangle has a vertex angle of 360 degrees divided by 8, or 45 degrees. Each right triangle measures half of this at its vertex which coincides with the center of the circle. So 12 times sine of 22.5 degrees equals the length of half of a side of the octagon.

There are 16 such sides. So 16 * 12 inches * sin(22.5 degrees) ≈ 73.4752190141 inches, or 73.5 inches to the nearest tenth of an inch.