Heron's Formula

01The Statement

Take any triangle with side lengths \(a\), \(b\), and \(c\). Define its semiperimeter \(s\) — literally "half the perimeter":

Then its area \(A\) is given entirely by those three lengths:

That's the whole formula. You never need the height, you never need an angle, and you never need to know where the triangle sits in the plane. Three lengths in, one area out.

02A Little History

Heron of Alexandria (also spelled Hero) was a Greek mathematician and engineer who lived in the first century CE — best estimates put him around 10–70 CE. He was a prolific inventor, credited with an early steam device (the aeolipile), vending mechanisms, and automated theatres.

The formula appears in his work Metrica, a treatise on measurement rediscovered only in 1896. Heron gives a fully worked numerical proof for a triangle with sides 13, 14, 15 (area 84). Though it carries his name, historians believe the result was likely known earlier — possibly to Archimedes two centuries before — but Heron's is the earliest surviving written derivation.

What makes it remarkable is its purely metric character: it converts a question about shape and orientation into one of arithmetic on lengths alone, a recurring theme in computational geometry to this day.

03Intuition — Why the Semiperimeter?

The appearance of \(s\) can feel mysterious. Here is one way to make peace with it. The product

is the formula in its "raw" symmetric form (we will derive it in Proof I). Each factor is a perimeter-like combination of the sides. Dividing each side combination through by 2 turns those four factors into \(2s\), \(2(s-a)\), \(2(s-b)\), \(2(s-c)\); the four 2's multiply to 16 and cancel the leading 16, leaving the clean product under the root.

So the semiperimeter isn't magical — it is just the bookkeeping device that makes a naturally symmetric, four-factor product look tidy. Notice the deep symmetry: swap any two sides and the formula is unchanged, exactly as area should be.

04Worked Examples by Hand

Example 1 — the 3·4·5 right triangle

Sides: \(a=3,\ b=4,\ c=5\).   Semiperimeter: \(s=\tfrac{3+4+5}{2}=6\).

Cross-check: a 3–4–5 triangle is right-angled, so \(A=\tfrac12\cdot3\cdot4=6\). ✓

Example 2 — a scalene 7·8·9 triangle

Semiperimeter: \(s=\tfrac{7+8+9}{2}=12\).

There is no nice right angle to fall back on here — which is precisely where Heron earns its keep.

Example 3 — Heron's own 13·14·15 triangle

Semiperimeter: \(s=\tfrac{13+14+15}{2}=21\).

A perfect integer area — the same example Heron worked in Metrica.

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05Proof I — Algebraic (via the Law of Cosines)

This is the quickest rigorous route. It assumes only two standard facts: the area formula \(A=\tfrac12 ab\sin C\) and the Law of Cosines.

06Proof II — Geometric (via the Inscribed Circle)

This route is closer in spirit to Heron's own and reveals a beautiful by-product: an exact formula for the inradius \(r\). It uses three classical facts, each justified below.

07Interactive — Step-by-Step Calculator

Type three side lengths (or pick a preset). The widget validates the triangle inequality, walks through every intermediate value, and draws the triangle to scale.

08Interactive — Drag a Triangle

Grab any vertex and move it. The side lengths, semiperimeter, and area (computed by Heron's formula) update live. As an independent check, the area is recomputed with the unrelated shoelace formula from the coordinates — watch the two agree to the last digit. The dashed circle is the inscribed circle of radius \(r=A/s\) from Proof II.

09GNU Octave — Code You Can Run

Every snippet below was executed in GNU Octave 8.4; the output shown is the real result. In Octave each function normally lives in its own .m file named after the function.

1 · The basic function

function A = heron(a, b, c)
  % HERON  Area of a triangle from its three side lengths.
  %   A = HERON(a, b, c) returns the area via Heron's formula.
  s = (a + b + c) / 2;            % semiperimeter
  A = sqrt(s * (s-a) * (s-b) * (s-c));
end

2 · Guard against impossible triangles

A robust version refuses inputs that violate the triangle inequality instead of silently returning a complex number.

function A = heron_safe(a, b, c)
  % Validate the triangle inequality before computing.
  sides = sort([a, b, c]);        % ascending: shortest two first
  if sides(1) + sides(2) <= sides(3)
    error("heron_safe: %g, %g, %g cannot form a triangle", a, b, c);
  end
  s = (a + b + c) / 2;
  A = sqrt(s * (s-a) * (s-b) * (s-c));
end

3 · Vectorise over many triangles at once

Octave shines at array work. Swap * for the element-wise .* and you can process whole columns of triangles in one shot — no loop.

a = [3;  5;  8];     % three triangles, one per row
b = [4; 12; 15];
c = [5; 13; 17];

s = (a + b + c) / 2;                       % column of semiperimeters
A = sqrt(s .* (s-a) .* (s-b) .* (s-c))     % element-wise -> column of areas

4 · Cross-check with the shoelace formula

If you happen to have the vertex coordinates, the shoelace (Gauss) formula gives the area directly. It's a great independent test of a Heron implementation.

function A = shoelace(P)
  % P is an n-by-2 matrix of polygon vertices (here, a triangle).
  x = P(:,1);  y = P(:,2);
  n = rows(P);
  idx = [2:n, 1];                 % "next" vertex, wrapping around
  A = 0.5 * abs(sum(x .* y(idx) - x(idx) .* y));
end

5 · Visualising the area surface

Fix one side at \(a=5\) and let the other two roam: the height of the surface is the Heron area, and it vanishes exactly where the triangle inequality fails (the dark border).

a = 5;
b = linspace(0.1, 10, 400);
c = linspace(0.1, 10, 400);
[B, C] = meshgrid(b, c);

s = (a + B + C) / 2;
A = sqrt(s .* (s - a) .* (s - B) .* (s - C));   % complex where invalid

contourf(B, C, real(A), 20, "LineColor", "none");
colorbar; axis equal tight;
xlabel("side b"); ylabel("side c");
title("Heron area of triangles with a = 5");
print("heron_surface.png", "-dpng");

10Numerical Stability — and Kahan's Fix

The textbook formula has a famous flaw on a computer. For a needle-like triangle (one extremely thin sliver), the quantities \(s\) and \(s-a\) are nearly equal large numbers, so subtracting them throws away most of your significant digits — catastrophic cancellation. The result can be badly wrong even in double precision.

William Kahan published a re-grouping that is provably accurate. Sort the sides so that \(a\ge b\ge c\), then evaluate with this exact parenthesisation (the parentheses are not optional):

function A = heron_kahan(a, b, c)
  % Numerically stable area (Kahan). Sort so a >= b >= c first.
  s = sort([a, b, c], "descend");
  a = s(1); b = s(2); c = s(3);
  A = 0.25 * sqrt( (a + (b + c)) * (c - (a - b)) ...
                 * (c + (a - b)) * (a + (b - c)) );
end

Try it yourself — the playground below runs the same IEEE-754 double arithmetic your browser uses, so you'll reproduce the cancellation live.

11Beyond Triangles

Brahmagupta's formula — cyclic quadrilaterals

Around 628 CE the Indian mathematician Brahmagupta generalised Heron to any quadrilateral whose four vertices lie on a circle. With sides \(a,b,c,d\) and \(s=\tfrac{a+b+c+d}{2}\):

Set \(d=0\) and the fourth vertex collapses onto a third — the quadrilateral becomes a triangle and Brahmagupta reduces exactly to Heron. Heron is the \(d\to 0\) shadow of Brahmagupta.

The Cayley–Menger determinant — any dimension

The grand generalisation expresses the squared content of a simplex from its edge lengths alone. For a triangle it is the determinant

which expands precisely to the \(16A^2\) we factored in Proof I. The same determinant, enlarged, gives the volume of a tetrahedron from its six edges, and so on into higher dimensions — Heron's idea, scaled up forever.

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