Heron’s Formula: Find Triangle Area From Three Sides

If you know a triangle’s three side lengths but not its height, the usual area rule, one-half base times height, seems to leave you stuck. Heron’s formula solves exactly that problem. It turns the three sides into an area directly, so it is especially useful for scalene triangles, surveying problems, and geometry questions in which drawing an altitude would create extra work.

The method is named for Heron of Alexandria, a Greek mathematician and engineer who described it in Metrica. The formula is compact, but using it well requires three habits: check that the side lengths can form a triangle, calculate the semiperimeter carefully, and keep enough precision until the final step.

The formula and what each part means

Let the side lengths be a, b, and c. First find the semiperimeter, written s:

s = (a + b + c) / 2

Then calculate the area A:

A = √[s(s – a)(s – b)(s – c)]

The semiperimeter is simply half the distance around the triangle. Each expression inside the square root compares that half-perimeter with one side. If the side lengths form a real triangle, all four factors are nonnegative. Their product has units to the fourth power, and taking the square root returns square units, exactly what an area should have.

Check the triangle inequality first

Before substituting numbers, make sure the longest side is shorter than the sum of the other two sides. Lengths 7, 8, and 9 pass because 9 is less than 7 + 8. Lengths 2, 3, and 6 fail because 6 is greater than 2 + 3, so they cannot close to form a triangle. If the longest side equals the sum of the other two, the shape collapses into a straight line and its area is zero.

Worked example: sides 7, 8, and 9

Consider a triangle with side lengths 7, 8, and 9 units. Since 9 < 7 + 8, the triangle inequality is satisfied. Start by adding the sides and dividing by two:

s = (7 + 8 + 9) / 2 = 24 / 2 = 12

Now subtract each side from the semiperimeter. The four factors in Heron’s formula are 12, 12 – 7 = 5, 12 – 8 = 4, and 12 – 9 = 3. Substitute them without rounding:

A = √(12 × 5 × 4 × 3) = √720

The radical can be simplified because 720 = 144 × 5:

A = 12√5 ≈ 26.83 square units

This answer is reasonable. If 9 is treated as the base, the triangle’s height must be less than 7, so its area must be less than one-half times 9 times 7, or 31.5 square units. The calculated area of about 26.83 falls below that upper bound. A quick reasonableness check like this can expose a calculator entry error before it becomes a final answer.

Why Heron’s formula works

The formula connects the familiar area rule to the law of cosines. Choose angle C between sides a and b. Using trigonometry, the triangle’s area is A = (1/2)ab sin C. The law of cosines gives cos C = (a² + b² – c²) / (2ab).

Because sin² C = 1 – cos² C, the angle can be removed from the area expression. After squaring and simplifying, the result becomes:

16A² = (a + b + c)(-a + b + c)(a – b + c)(a + b – c)

Each factor has a simple relationship to the semiperimeter: a + b + c = 2s, -a + b + c = 2(s – a), and so on. Replacing the four factors produces 16A² = 16s(s – a)(s – b)(s – c). Divide by 16 and take the positive square root because area cannot be negative. Heron’s formula follows.

This derivation also explains the triangle inequality. If one side is too long, one of the factors becomes negative. The algebra is signaling the same geometric fact: those three segments cannot enclose an area.

Common mistakes and how to catch them

• Using the full perimeter for s: add all three sides, then divide by two. For sides 7, 8, and 9, s is 12, not 24.
• Skipping the triangle check: Heron’s formula assumes the side lengths make a triangle. Test the longest side against the other two before calculating.
• Rounding too early: keep the product or radical exact as long as possible. Round only the final decimal unless a problem says otherwise.
• Losing a factor: the expression has four factors: s, s – a, s – b, and s – c. Parentheses help prevent calculator errors.
• Forgetting square units: if the sides are measured in centimeters, the area is measured in square centimeters, written cm².

A useful calculator routine is to record the semiperimeter and the three differences on separate lines before multiplying. This small pause makes the arithmetic easier to inspect. When an answer looks surprising, estimate a maximum possible area from one-half base times a plausible height and compare it with the result.

Where the formula is useful

Heron’s formula is valuable whenever three side lengths are easier to obtain than a perpendicular height. A surveyor can divide an irregular plot into triangles, measure their sides, and add their areas. Builders and engineers can use the same reasoning for triangular panels, roof sections, trusses, or cross-sections. Mapmakers and computer programs also use triangle-area calculations as building blocks for larger shapes.

The formula is not limited to scalene triangles, though that is where its advantage is easiest to see. It also works for isosceles and equilateral triangles. For an equilateral triangle with side length x, substituting s = 3x/2 simplifies to the familiar result A = (√3/4)x².

Heron’s formula is most useful as a bridge between measurement and geometry: three distances are enough to recover an area. Check that the sides can form a triangle, find the semiperimeter, substitute all four factors, and keep the units squared. With that sequence in place, a problem that seems to be missing its height becomes straightforward.