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Heron's Formula Calculator

Result

6.0000units²

Result: 6.0000 units²

Heron's formula gives the area from the three sides alone — no height, no angles. Halve the perimeter to get s, then take √(s(s − a)(s − b)(s − c)). Sides of 3, 4 and 5 give s = 6 and an area of √36 = 6 square units.

The numbers at a glance

Held fixed: Side a 3.000, Side b 4.000.

Side cResult
2.0002.9047
4.0005.5621
5.000Your value6.0000
6.0005.3327
8.0000.0000
10.0000.0000

Worked examples

How it's calculated

A = √(s(s − a)(s − b)(s − c)), s = (a + b + c) ÷ 2

  1. StepMeasure the three sides and enter them in one length unit.
  2. StepCheck the longest side is shorter than the other two combined.
  3. ResultRead the area in square units; the perimeter is a + b + c.

Reference table

Sides a, b, cExactArea
0.5, 0.5, 0.8√0.01440.1200
3, 4, 5√366.0000
5, 5, 525√3 ÷ 410.8253
5, 5, 8√14412.0000
6, 8, 10√57624.0000
7, 8, 912√526.8328
13, 14, 15√705684.0000

Questions

What is Heron's formula?

Heron's formula gives the area of a triangle from its three side lengths alone, without the height or any angle. Find the semi-perimeter s = (a + b + c) ÷ 2, then the area is √(s(s − a)(s − b)(s − c)). It is named after Heron of Alexandria, who described it in the first century AD.

How do I calculate a triangle's area from three sides?

Add the three sides and halve the total to get s. Multiply s by (s − a), (s − b) and (s − c), then take the square root of that product. For a 3-4-5 triangle s = 6, so the area is √(6 × 3 × 2 × 1) = √36 = 6 square units.

Why do some sides give an area of zero?

The three lengths must satisfy the triangle inequality: no side may be as long as, or longer than, the other two combined. Sides of 1, 1 and 5 cannot close into a triangle, so they enclose nothing and the area is zero. Check that the longest side is shorter than the sum of the other two.

When should I use Heron instead of ½ × base × height?

Use Heron's formula when you know all three side lengths but not the height — common in surveying, construction and any field triangle where a perpendicular is awkward to measure. If you already have a base and its perpendicular height, ½ × base × height is simpler and just as exact.

Does it work for right and obtuse triangles?

Yes, for every valid triangle: right-angled, acute, obtuse, scalene, isosceles or equilateral. For a right triangle it agrees with ½ × base × height, and the 3-4-5 triangle gives 6 either way.

Sources and last check

  1. mathworld.wolfram.com

Information, not professional advice.