How this instrument works
A regular polygon with n sides splits, without exception, into n congruent isosceles triangles fanned out from its centre, one triangle sitting on each side. A full turn around that centre is 2π radians, so each triangle claims a central angle of 2π ⁄ n; bisecting that wedge gives a right triangle with a base of s ⁄ 2 and an angle of π ⁄ n, and solving for height puts a tangent of π ⁄ n directly under the base — multiply by n triangles and the whole shape's area collapses to A = ns² ⁄ (4·tan(π ⁄ n)). Feed the formula n = 5 or n = 6 and it reduces to the exact pentagon and hexagon formulas this site already lists on their own pages; the difference here is that n is a field you fill in yourself, so a nonagon, a 15-gon, or a 40-gon gets identical treatment without a dedicated page for each one.
Push n upward and the polygon closes in on a circle without ever quite becoming one — a fact Archimedes used more than two thousand years ago to trap π between two numbers, bracketing a circle with a 96-sided polygon inscribed just inside it and another just outside. The algebra shows why the approach works: for large n, tan(π ⁄ n) sits close to π ⁄ n itself, so A ≈ ns² ⁄ (4·π ⁄ n) = n²s² ⁄ (4π), and as n keeps climbing that expression tracks the circumscribed circle ever more tightly. No finite n ever reaches the circle exactly, the same reason no polygon, however many sides it has, can be assigned a value of π that terminates.
At the small end, n = 3 is the floor — three points define the fewest sides that still enclose an area, and the formula reduces cleanly to the familiar equilateral-triangle area, s²√3 ⁄ 4, since tan 60° is exactly √3. The one mistake this general version invites that the fixed-shape pages sidestep is a units slip inside the tangent: π ⁄ n is a radian measure, not a degree count, so a calculator left in degree mode reads tan(π ⁄ 9) as the tangent of roughly a third of one degree rather than 20°, and returns a figure many hundreds of times too small without a single wrong digit typed anywhere else.
- Enter the polygon's side count into Number of sides, n — any whole number from 3 upward, however many sides your shape has.
- Enter the length of one edge into Side length; every side is assumed equal, which is what makes the shape regular in the first place.
- Area updates immediately, computed straight from A = ns² ⁄ (4·tan(π ⁄ n)) with no rounding hidden between steps.
- Checking a pentagon or hexagon by hand? Set Number of sides, n to 5 or 6 and the result should match this site's dedicated pentagon and hexagon pages exactly.
- Raise Number of sides, n while holding Side length fixed to watch Area grow toward, but never reach, the circle the polygon sits inside.
Worked example — a nine-sided tile with 3-unit sides
A nonagon-shaped floor tile is cut with 9 equal sides, each 3 units long, so n = 9 and s = 3. The central angle per wedge is 360° ⁄ 9 = 40°, giving each of the nine triangles a 20° half-angle, and tan 20° works out to 0.363970. Substituting into A = ns² ⁄ (4·tan(π ⁄ n)) gives A = 9 × 3² ⁄ (4 × 0.363970) = 81 ⁄ 1.455881 = 55.6364177439561 square units, the exact figure Area reports for n = 9 and s = 3.
The same total is reachable a second way, through the apothem — the centre-to-edge distance, a = s ⁄ (2·tan 20°) ≈ 4.121216. Half the perimeter times that apothem, ½ × 27 × 4.121216 ≈ 55.6364, lands on the same figure, since A = ½Pa is just the nine-triangle picture regrouped: one route sums triangle areas directly, the other multiplies the whole perimeter by a shared height and halves it.
Questions
What is the area formula for a regular polygon with n sides?
A = ns² ⁄ (4·tan(π ⁄ n)), where n is the number of sides and s is one side's length. It comes from splitting the polygon into n congruent isosceles triangles meeting at the centre and summing their areas; for n = 9 and s = 3 it returns 55.6364177439561, this sheet's own reference value.
How is this different from the pentagon, hexagon, or decagon calculators on this site?
Those pages fix n at 5, 6, or 10 and only ask for a side length, because their formulas already bake the specific angle in. This sheet takes n as a direct input instead, so it covers every regular polygon those dedicated pages don't reach — a heptagon, a 15-gon, a 40-gon — using the identical underlying formula with n left open.
What happens to the area as the number of sides gets very large?
It climbs toward, but never reaches, the area of the circle the polygon sits inside. For large n, tan(π ⁄ n) sits close to π ⁄ n, so the formula behaves like n²s² ⁄ (4π); Archimedes used this same closing-in idea with 96-sided polygons to bracket π between two numbers, roughly two thousand years before decimal approximations existed.
Why does the formula use radians instead of degrees for π ⁄ n?
Because π ⁄ n is the half-angle expressed in radians, and the tangent function needs that argument in the unit the calculation is set up for. Switching a calculator to degree mode and evaluating tan(π ⁄ 9) as if that were roughly a third of a degree, rather than 20°, is the single most common way to land on an area wrong by a wide margin with no typo anywhere else.
Is there a smallest number of sides this formula accepts?
Yes — n = 3, the equilateral triangle, since fewer than three sides cannot enclose an area at all. At n = 3 the formula reduces to s²√3 ⁄ 4, the familiar triangle-area identity, because tan 60° equals exactly √3, one of the few angles in this formula with a clean closed form.
Does this formula work for an irregular polygon with n sides?
No — A = ns² ⁄ (4·tan(π ⁄ n)) assumes every side and every interior angle match, which is what lets one side length and one side count fix the whole shape. An irregular polygon needs its vertices plotted and its area found with the shoelace formula instead, since no single side length describes it.