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I presume by diameter of an octagon you mean the length of the diagonal between opposite vertices of a regular octagon.

By completing all the diagonals between opposite vertices of a regular octagon, it divides it up into 8 equal isosceles triangles, with their apexes at the centre of the octagon. The two sides of these isosceles triangles which go from a vertex of the octagon to the centre has a length half that of the diameter.

If the centre point of each side of the octagon is joined to the centre of the octagon, each of the 8 isosceles triangles above are split into two (congruent) right angled triangles. The angle at the centre of the octagon of each of these triangles is (360° ÷ 8) ÷ 2 = 22.5°.

The side opposite this angle is half the length of the side of the octagon and so the perimeter is 16 times its length.

The hypotenuse of the right angled triangles is half the diameter. Thus the trigonometric sine function can be used to find half the octagons side length, and thus its perimeter:

sin = opp/hyp

→ opp = sin x hyp

Perimeter = 16 x opp = 16 x (sin 22.5° x diameter/2) = diameter x 8 x sin 22.5° ≈ diameter x 3.061

This can be generalised to any regular even sided polygon if diameter is defined as the length of the diagonal between opposite vertices (which is also the maximum straight line distance between two points on the perimeter):

perimeter = diameter x number_of_sides x sin(180° ÷ number_of_sides)

As the number of sides increases, "number_of_sides x sin(180° ÷ number_of_sides)" gets closer and closer to π.

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Q: How do you find the perimeter of an octagon if you know the diameter?
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