Heron's Formula Calculator

Heron's formula in two steps

Heron's formula turns three side lengths into an area without needing a height or an angle. The semiperimeter comes first, then the square root of a product of four terms:

s = (a + b + c) / 2
Area = the square root of s x (s - a) x (s - b) x (s - c)

The semiperimeter is half the perimeter, which is why the letter s is used. Every bracket subtracts one side from that half sum, and if one side is longer than the other two together, one bracket turns negative and the square root has no real answer. The formula is reporting a triangle that cannot exist, and the calculator above says so instead of printing a number.

A 3-4-5 triangle shows the whole method. The semiperimeter is (3 + 4 + 5) / 2 = 6, the four factors are 6, 3, 2 and 1, and their product is 6 x 3 x 2 x 1 = 36. The square root of 36 is 6, so the area is 6 square units, which matches the base times height answer of 4 x 3 / 2 = 6.

Why the formula works

The result is attributed to Heron of Alexandria, who recorded it in the Metrica around 60 AD, though the idea is older and Archimedes is credited with an earlier route to the same quantity. The proof that matters for daily use is short: drop a height onto the longest side, write the two right triangles that appear, and eliminate the height with the Pythagorean theorem. The algebra collapses to the four-factor product above.

Two familiar formulas fall out of it as special cases. For a right triangle with legs a and b, the product reduces until the area is a x b / 2. For an equilateral triangle with side a, the brackets all equal a/2 and the whole expression becomes the square root of 3 divided by 4, times a squared, which is the standard equilateral area formula.

Because no angle appears anywhere, the formula is useful exactly where angles are hard to measure: a plot of land with three straight fences, a triangular gable, a bracket cut to fit a corner. Measuring three sides with a tape is easier than sighting an angle with a protractor, and the area follows from the tape alone.

The price is numerical. For a triangle that is nearly flat, one bracket is close to zero and the square root amplifies any rounding in the sides, so a base and a height measured directly gives a steadier answer. For ordinary triangles the difference is far below the precision of a tape measure.

The triangle inequality check

Three lengths make a triangle only when each pair adds up to more than the third side. Written out, all three of these must hold: a + b is greater than c, a + c is greater than b, and b + c is greater than a. The calculator checks all three before it computes anything.

The 1-2-3 case is the boundary. The two short sides just reach the long one, so all three points sit on a straight line, the semiperimeter equals the longest side, one bracket is zero, and the area is zero. Anything that fails the test by a wider margin produces a negative bracket and no real area at all.

This check also catches typos. A plot recorded as 12, 15 and 30 metres is not a triangle, and the failure is more useful than an error message about square roots, because it points at the measurement rather than the arithmetic.

Inradius, circumradius and angles from the same three sides

The area is the gateway to the rest of the triangle. Three more quantities need nothing beyond the three sides.

Inradius: the circle that touches all three sides from inside has radius r = area / s. For 3-4-5 that is 6 / 6 = 1, and for 5-5-5 it is 10.8253 / 7.5 = 1.4434.

Circumradius: the circle through all three corners has radius R = a x b x c / (4 x area). For 3-4-5 that is 3 x 4 x 5 / (4 x 6) = 2.5, and the diameter of 5 matches the hypotenuse, as it must for a right triangle.

Angles: the law of cosines gives each one from the sides. Angle A is the inverse cosine of (b squared + c squared - a squared) / (2 x b x c), and the other two follow by rotating the letters. The three angles always add to 180 degrees, which is a free check on the arithmetic.

The inradius and the area move together, since r is just the area divided by the semiperimeter. The circumradius is the one that jumps around: a long thin triangle pushes it far outside the shape, while an equilateral triangle keeps it close in, at twice the inradius.

Worked examples with the arithmetic shown

Three complete runs through the formula, at three different shapes.

A right triangle, sides 3, 4 and 5. The semiperimeter is (3 + 4 + 5) / 2 = 6. The brackets are 3, 2 and 1, and 6 x 3 x 2 x 1 = 36. The square root of 36 is 6, and the base times height route gives 4 x 3 / 2 = 6 as well.

An equilateral triangle, sides 5, 5 and 5. The semiperimeter is 15 / 2 = 7.5. Each bracket is 2.5, and 7.5 x 2.5 x 2.5 x 2.5 = 117.1875. The square root is 10.8253, close to the 10.8253 that the equilateral shortcut of 0.433 x 25 produces.

An isosceles triangle, sides 4, 4 and 6. The semiperimeter is 14 / 2 = 7, the brackets are 3, 3 and 1, and 7 x 3 x 3 x 1 = 63. The square root is 7.9373. Cutting the triangle down the middle gives two right triangles with legs 3 and the square root of 7, and 6 x 2.6458 / 2 lands on the same figure.

A scalene triangle, sides 6, 7 and 8. The semiperimeter is 21 / 2 = 10.5, the brackets are 4.5, 3.5 and 2.5, and the product is 413.4375. The square root is 20.3332. This shape has no symmetry to exploit, which is where the formula earns its keep.

Mistakes that give the wrong area

Five errors show up again and again in homework and in field notes.

• Using the perimeter instead of the semiperimeter. Plugging 12 into the 3-4-5 case gives brackets of 9, 8 and 7, and 12 x 9 x 8 x 7 = 6,048, whose square root is 77.77 instead of 6. The halving is not optional.
• Mixing radius and diameter. The inradius is a radius. A circle of diameter 2 in a 3-4-5 triangle has r = 1, so halving the wrong number doubles or halves the area check by a factor of four.
• Squaring the semiperimeter. The formula multiplies four lengths, it does not square s. Writing s squared gives units of length to the fourth power, and the answer stops being an area.
• Rounding s before multiplying. Keep two extra decimals. A semiperimeter of 10.5 rounded to 10 turns a 20.33 area into 16.73, an error of nearly 18 percent.
• Applying it to four sides. Heron's formula takes three sides. A quadrilateral needs Brahmagupta's formula, which uses the same semiperimeter but four brackets and only works for cyclic shapes.

There is a sixth trap that is not an error but a limitation. For a triangle with sides 1, 1 and 1.9999, one bracket is 0.00005 and the square root is sensitive to the last digit of the input. Measuring the base and the height is the better plan for shapes that flat, and the formula is the better plan for everything else.

Units, precision and where the answer lands

The area comes out in square units of whatever the sides were measured in. Sides in metres give square metres, sides in feet give square feet, and sides in centimetres give square centimetres. NIST SP 811 keeps the square metre as the SI unit of area, and every other area unit is a fixed multiple of it.

One acre is 4,046.8564224 square metres, so a triangular plot measured in metres can be reported in acres by dividing the square metre result by that number. A 30 by 40 by 50 metre triangle has an area of 600 square metres, which is 0.1483 acre, a small building lot.

Precision follows the tape, not the calculator. A side measured to the nearest centimetre supports an area to about three significant figures, and printing eight decimals suggests a survey that did not happen. Round the answer to the precision of the sides, and keep the extra digits only while the arithmetic is still running.

For scale, a 3-4-5 triangle of metres is 6 square metres, about the footprint of a small shed. The same numbers in feet give 6 square feet, a tiles-and-cutting-board size. The formula is unit-blind, so the label on the answer is entirely up to the input.

Checking the result without a second formula

A few quick tests catch most wrong answers before they leave the page.

• Compare with the bounding rectangle. The area of any triangle is half the area of the parallelogram built on two sides, so it can never exceed half of the product of the two longest sides. For 6-7-8, half of 56 is 28, and 20.3332 sits comfortably under it.
• Try the equilateral ceiling. For a given perimeter, the equilateral triangle has the largest area. A triangle with perimeter 15 cannot beat the 10.8253 of the 5-5-5 case.
• Check the angles. Add the three angles from the law of cosines. They should total 180 degrees to within a rounding step, and a total near 179.9 is arithmetic, not geometry.
• Redo one bracket. If the area looks off by a factor, one of the four factors in the product is usually the culprit, and a single recomputed bracket finds it faster than a full rerun.
• Swap the sides. The formula is symmetric in a, b and c, so reordering the inputs must give the same area. A result that changes with the order is a sign that a bracket was assigned to the wrong side.

The unit square is the last check. Take sides of 1, 1 and the square root of 2, which is a right isosceles triangle, and the formula returns 0.5, exactly half of the 1 by 1 square it sits in. MathWorld's entry on Heron's formula lists that case alongside the general proof for anyone who wants the algebra in full.