right angles pointed inward, forming a square. The edges of this square are each the length of the base. Given the length of a side a, the span S is The span, then, is equal to the silver ratio times the side, a. The area is then as above: Expressed in terms of the span, the area is Another simple formula for the area is More often the span S is known, and the length of the sides, a, is to be determined, as when cutting a square piece of material into a regular octagon. From the above, The two end lengths e on each side (the leg lengths of the triangles (green in the image) truncated from the square), as well as being may be calculated as Circumradius and inradius The circumradius of the regular octagon in terms of the side length a is and the inradius is (that is one-half the silver ratio times the side, a, or one-half the span, S) Diagonality The regular octagon, in terms of the side length a, has three different types of diagonals: Short diagonal; Medium diagonal (also called span or height), which is twice the length of the inradius; Long diagonal, which is twice the length of the circumradius. The formula for each of them follows from the basic principles of geometry. Here are the formulas for their length: Short diagonal: ; Medium diagonal: ; (silver ratio times a) Long diagonal: . Construction A regular octagon at a given circumcircle may be constructed as follows: Draw a circle and a diameter AOE, where O is the center and A, E are points on the circumcircle. Draw another diameter GOC, perpendicular to AOE. (Note in passing that A,C,E,G are vertices of a square). Draw the bisectors of the right angles GOA and EOG, making two more diameters HOD and FOB. A,B,C,D,E,F,G,H are the vertices of the octagon. A regular octagon can be constructed using a straightedge and a compass, as 8 = 23, a power of two: The regular octagon can be constructed with meccano bars. Twelve bars of size 4, three bars of size 5 and two bars of size 6 are required. Each side of a regular octagon subtends half a right angle at the centre of the circle which connects its vertices. Its area can thus be computed as the sum of eight isosceles triangles, leading to the result: