How this instrument works
A regular pentagon has five equal sides and five interior angles of exactly 108°. Fan five wedges out from the shape's centre — each meeting the boundary along one side and closing at a 72° angle where 360° ÷ 5 lands — and the pentagon is nothing more than those five identical isosceles wedges laid edge to edge. Bisect a single wedge and its half-angle, 36°, is exactly the tan(π⁄5) sitting inside A = 5s² ⁄ (4·tan 36°); multiply the resulting triangle height by all five bases and that formula is the whole derivation, no step skipped.
Three pentagons meeting at one point add up to 3 × 108° = 324°, thirty-six degrees short of a full turn, so a floor built from only regular pentagons cannot close without a gap — the Cairo pentagonal tiling solves that with an irregular pentagon instead. Constructing one, though, is another matter: a regular pentagon has been drawable with just a compass and straightedge since antiquity, because 5 is a Fermat prime, while a regular heptagon has no such construction at all — seven is prime, yet it never appears on that short Fermat list. And every diagonal inside a regular pentagon is longer than a side by exactly the golden ratio, φ ≈ 1.618034 — draw all five diagonals and their crossings frame a smaller pentagon at the centre, ready to sprout the same five diagonals over again.
Two numbers make the formula concrete without ever touching a tan key: tan 36° reduces to the nested radical √(5 − 2√5) ≈ 0.726543, and threading that through A = 5s² ⁄ (4·tan 36°) leaves A = (√(25 + 10√5) ⁄ 4)·s², an answer built from nothing but integers and square roots. √5 ≈ 2.236068 is the only irrational ingredient, and it appears twice, once inside each nested root. Shrink s toward zero and both area and perimeter shrink toward zero in step, with nothing left over — the only pentagon a side of length zero can describe is a single point.
That closed form describes a mathematically regular pentagon, and not every five-sided object built in the real world obliges. The Pentagon building in Arlington, Virginia was laid out as a genuinely regular pentagon, each of its five outer walls about 921 feet long, so the formula on this page truly matches its footprint. Home plate on a baseball diamond is a pentagon in name only — a 17-inch square with two back corners folded in to a point — and feeding its longest edge into A = 5s² ⁄ (4·tan 36°) would return a number with no connection to the plate's actual surface.
- Give the Side length field your pentagon's edge measurement — this sheet assumes all five sides match, so one number is all it needs.
- Area (regular pentagon) recalculates the moment Side length changes, straight from A = 5s² ⁄ (4·tan 36°).
- Perimeter follows just as fast: five times whatever figure sits in Side length.
- Working from a real five-sided object? Measure a flat edge with a ruler or tape, not the gap between two non-adjacent corners — that distance is a diagonal, not the side this formula expects.
Worked example — a pentagon with 4-unit sides
Cut a garden step as a regular pentagon with every edge exactly 4 units long, so s = 4. Multiplying through, A = 5 × 16 ⁄ (4 × tan 36°) = 80 ⁄ 2.906170 works out to 27.527638409423474 square units — the exact figure Area (regular pentagon) displays, carried past the sixth decimal rather than stopped there. Perimeter comes back as 5 × 4 = 20.0, the trim length needed for the stone's full boundary.
Route two skips the tangent table entirely: √(25 + 10√5) = √47.360680 ≈ 6.881910, one quarter of that is 1.720477, and s² = 16 times that value lands within a rounding error at the fifteenth decimal of Area (regular pentagon) — two independent derivations meeting at the same answer, the cross-check worth running before trusting a hand calculation. Scale the same pentagon to s = 8 and the closed form stays honest too: A = (√(25 + 10√5) ⁄ 4) × 64 = 110.1105536376939, exactly four times 27.527638409423474, because area answers to the square of the side while perimeter only answers to the side itself.
Questions
What formula finds the area of a regular pentagon?
A = 5s² ⁄ (4·tan 36°) — five congruent isosceles triangles fanned out from the centre, each found from its 72° wedge and then summed. Plug in s = 4 and the formula lands on 27.527638409423474, the exact value Area (regular pentagon) reports for that side length.
Why can't a floor be tiled with only regular pentagons?
Three regular pentagons around one point add up to 3 × 108° = 324°, thirty-six degrees short of the full 360° turn a gap-free tiling needs, and a fourth pentagon overshoots to 432°. No whole number of 108° wedges lands on 360° exactly, which is why pentagon tilings such as the Cairo pattern always rely on irregular pentagons to close the gaps.
Is a regular pentagon constructible with a compass and straightedge?
Yes, and it has been since antiquity — Euclid's Elements, Book IV, Proposition 11 sets out the construction directly, because 5 belongs to the short list of Fermat primes (3, 17, 257, 65537 are the others) that govern which regular polygons a compass and straightedge alone can reach. Seven is not on that list, which is the one detail that keeps a regular heptagon off the classical construction menu.
How is the golden ratio hidden inside a regular pentagon?
Every diagonal of a regular pentagon is exactly φ times a side, where φ = (1 + √5) ⁄ 2 ≈ 1.618034. Drawing all five diagonals traces a five-pointed pentagram whose crossings frame a smaller regular pentagon at the centre — one ready to sprout its own diagonals in the same φ ratio, a self-similar pattern that repeats inward without end.
Will this calculator work for home plate or another irregular pentagon?
No — A = 5s² ⁄ (4·tan 36°) only holds when all five sides and all five interior angles match, which fixes the whole shape from a single length. Home plate is a 17-inch square with two of its corners folded in to a point, not a regular pentagon, so its area has to be pieced together from a square and a triangle rather than read off this formula.
How many diagonals does a regular pentagon have?
Five — one connecting every pair of non-adjacent vertices, given by the general polygon count n(n − 3) ⁄ 2 with n = 5. Draw all five inside a regular pentagon and they cross to trace the five-pointed pentagram, the emblem the Pythagoreans reportedly used to recognise one another, precisely because those five crossing diagonals encode the golden ratio in their lengths.