Triangle area by Heron formula:
formula and worked example

When all you have are the sides

The familiar triangle area, half base times height, needs a perpendicular height, and that is often awkward to measure. On a plot of land, a roof gable or a sail panel, you can usually measure the three edges with a tape but have no easy way of dropping a clean perpendicular. Heron's formula takes exactly those three edge lengths and nothing else.

It applies to any triangle, right-angled or not, acute or obtuse. The only requirement is that the three lengths can form a triangle in the first place, which the formula itself helps you check.

Historically the formula is attributed to Heron of Alexandria and is used in surveying because chaining three side lengths on the ground is far easier than constructing a perpendicular across uneven land.

The formula and the semi-perimeter

Define s as the semi-perimeter, half of a + b + c. Then the area is the square root of s x (s - a) x (s - b) x (s - c). The expression looks heavy when written out in full, but it is just four numbers multiplied together under one root.

Each bracket s minus a side tells you how much of the half-perimeter is left after that side. If any side is as long as the sum of the other two, that bracket is zero and the area collapses to nothing, which is how the formula signals a flat, impossible triangle.

It also pays to be consistent about order. The formula is symmetric in the three sides, so it does not matter which one you call a, b or c, and rearranging them leaves the product under the root unchanged.

A triangle with sides 5, 6 and 7

Three edges measure 5, 6 and 7 units.

1. 1Semi-perimeter: (5 + 6 + 7) / 2 = 18 / 2 = 9.
2. 2Subtract each side: 9 - 5 = 4, 9 - 6 = 3, 9 - 7 = 2.
3. 3Multiply the four numbers: 9 x 4 x 3 x 2 = 216.
4. 4Take the square root: root of 216 = 14.697 square units.

A rough check: the triangle has a longest base of 7 and, being fairly balanced, a height near 4.2. Half of 7 x 4.2 is 14.7, which agrees. Another quick test is against an equilateral triangle with the same perimeter of 18, which has side 6 and area about 15.59; a lopsided triangle with the same perimeter must be a bit smaller, as 14.70 is.

Interpreting the answer

Heron's result is the true flat surface of the triangle, so it can be used directly for material takeoff or land area. If you need the height as well, rearrange half base times height: height on side 7 equals 2 x 14.697 / 7 = 4.20. That is a handy by-product when a drawing needs a perpendicular.

Keep an eye on sensitivity when the triangle is thin. For nearly flat triangles two of the brackets become very small, and a one-millimetre error in a side measurement can swing the area by a large proportion. Triangles with decent proportions are forgiving; slivers are not.

For a large triangle the formula can produce a figure that is hard to compare with a plan, so cross-check with a second method when stakes are high: half base times height from a drawing, or the area from coordinates, should land within measurement error of the Heron result.

Lengthening one side

Moving a from 5 to 5.5 changes s to 9.25 and the area to 15.926, up about 8.4%. It is not 10% because the triangle must reshape: stretching one edge with the other two fixed swings the opposite corner and changes the height as well. The relationship is not linear in a single side, which is a key difference from the simple formulas for squares or circles.

There is also a ceiling. With sides b = 6 and c = 7 fixed, side a can only range from just above 1 to just below 13. Past those limits, no triangle exists. Within them, the area peaks when a is such that the angle between b and c is 90 degrees, at a = root of 85, about 9.22, where the area is 21.

Common mistakes with Heron's method

• Using the full perimeter in place of the semi-perimeter, which inflates the answer by a factor of 4.
• Forgetting to subtract each side from s, or subtracting the same side twice.
• Rounding s early. Keep at least four decimals in intermediate steps when sides have decimals.
• Entering lengths that violate the triangle inequality, such as 3, 4 and 8, which makes a bracket negative and the root undefined.

Scenario: a triangular plot

A plot has boundary sides of 30 m, 40 m and 50 m measured with a tape. The semi-perimeter is 60, and the product is 60 x 30 x 20 x 10 = 360,000, whose root is 600 square metres. That equals half of 30 x 40, which confirms the plot is a right triangle. A 1 m slip on the 30 m side, reading 31 m, lifts the area to about 620 square metres, a 3% change. The same slip on the 50 m side, reading 51 m, leaves it near 599.5, because a right angle is already close to the maximum area for those two shorter sides. Seeing which measurement matters before staking out boundaries is the point of running the numbers twice.