Regular Polygon Perimeter

Every side is the same length, so the distance around is $n$ of them: $P = ns$. "Regular" is doing more work in that sentence than it appears to. It means equal sides and equal angles, and for anything past a triangle the second does not follow from the first — a rhombus has four equal sides and is not a square. For perimeter it makes no difference; for everything else about the shape it makes all the difference.

The useful direction is $s = P/n$, which turns a length of stock into a cut list. An octagonal planter built from 4.8 m of 2×8 has eight sides of 600 mm. But the cut length is only half of what you need, and here is the fact that belongs on this page: the mitre angle. Going once around a closed shape turns you through 360° in total, so each of the $n$ corners turns you $360/n$ — the exterior angle. Split between two mitred ends, each cut is $180/n$: 22.5° for an octagon, 30° for a hexagon, 36° for a pentagon, and the familiar 45° for a square. That number and the side length together are the whole job.

Let $n$ grow and the polygon closes in on a circle, which is exactly how Archimedes got at $\pi$: he computed the perimeters of inscribed and circumscribed 96-gons and trapped the circle's circumference between them. The link to a circle is also how you size a polygon from a hole rather than from a side — for a polygon inscribed in a circle of radius $R$, $s = 2R\sin(\pi/n)$.

Two traps. The first is treating the perimeter as the stock required. Mitred pieces are cut long point to long point, so each one is longer than the finished side by twice the material thickness times $\tan(180/n)$, and every cut eats a saw kerf; ordering exactly $P$ of anything guarantees coming up short. The second catches anyone with a wrench: on a hexagon the side length $s$ is not the size stamped on the fastener. Across the corners is $2s$ and across the flats is $s\sqrt{3} \approx 1.732s$, and hex hardware is specified across the flats — so a 19 mm nut has sides of 10.97 mm and a perimeter of 65.8 mm. Feeding 19 into $s$ here answers a different question entirely. Solving for $n$ should also come back a whole number of at least 3; a fractional result means the side length and perimeter do not describe a closed polygon.