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Heron’s Formula

Heron's Formula
Hero's Formula

A formula for the area of a triangle used when the lengths of all three sides are known.

 

Triangle with vertices B, C, A and sides labeled a (BC), b (CA), c (AB). Formula: Area = √(s(s−a)(s−b)(s−c)), where s = (a+b+c)/2.

 

 

See also

Semiperimeter

Key Formula

A=s(s−a)(s−b)(s−c)A = \sqrt{s(s-a)(s-b)(s-c)}
Where:
  • AA = Area of the triangle
  • a,b,ca, b, c = Lengths of the three sides of the triangle
  • ss = Semiperimeter of the triangle, defined as s = (a + b + c) / 2

Worked Example

Problem: Find the area of a triangle with sides of length 5, 12, and 13.
Step 1: Find the semiperimeter by adding all three sides and dividing by 2.
s=5+12+132=302=15s = \frac{5 + 12 + 13}{2} = \frac{30}{2} = 15
Step 2: Compute each factor (s − a), (s − b), and (s − c).
s−a=15−5=10,s−b=15−12=3,s−c=15−13=2s - a = 15 - 5 = 10, \quad s - b = 15 - 12 = 3, \quad s - c = 15 - 13 = 2
Step 3: Multiply s by all three factors.
s(s−a)(s−b)(s−c)=15×10×3×2=900s(s-a)(s-b)(s-c) = 15 \times 10 \times 3 \times 2 = 900
Step 4: Take the square root to get the area.
A=900=30A = \sqrt{900} = 30
Answer: The area of the triangle is 30 square units.

Another Example

Unlike the first example (a right triangle with a perfect square result), this triangle is scalene and produces an irrational area, showing how to simplify a radical in the final step.

Problem: Find the area of a triangle with sides of length 7, 8, and 9.
Step 1: Calculate the semiperimeter.
s=7+8+92=242=12s = \frac{7 + 8 + 9}{2} = \frac{24}{2} = 12
Step 2: Compute each factor.
s−a=12−7=5,s−b=12−8=4,s−c=12−9=3s - a = 12 - 7 = 5, \quad s - b = 12 - 8 = 4, \quad s - c = 12 - 9 = 3
Step 3: Multiply all four values together.
s(s−a)(s−b)(s−c)=12×5×4×3=720s(s-a)(s-b)(s-c) = 12 \times 5 \times 4 \times 3 = 720
Step 4: Take the square root. Since 720 is not a perfect square, simplify the radical or give a decimal.
A=720=144×5=125≈26.83A = \sqrt{720} = \sqrt{144 \times 5} = 12\sqrt{5} \approx 26.83
Answer: The area of the triangle is 12√5 ≈ 26.83 square units.

Frequently Asked Questions

When should you use Heron's Formula instead of the standard triangle area formula?
Use Heron's Formula when you know all three side lengths but do not know the height (altitude) or any angle of the triangle. The standard formula A = ½ × base × height requires a height measurement, and the trigonometric formula A = ½ab sin C requires an included angle. Heron's Formula needs only the three sides.
Why is it called Heron's Formula?
The formula is named after Heron of Alexandria, a Greek mathematician and engineer who lived around the 1st century AD. He published a proof of the formula in his work Metrica. Some sources also call it Hero's Formula, since 'Hero' is an alternate spelling of his name.
Does Heron's Formula work for all types of triangles?
Yes. Heron's Formula works for any valid triangle — equilateral, isosceles, scalene, acute, right, or obtuse. The only requirement is that the three side lengths satisfy the triangle inequality (the sum of any two sides must be greater than the third side). If they don't, the expression under the square root will be negative, indicating no real triangle exists.

Heron's Formula vs. Standard Area Formula (½ × base × height)

Heron's FormulaStandard Area Formula (½ × base × height)
FormulaA = √(s(s−a)(s−b)(s−c))A = ½ × b × h
What you needAll three side lengthsA base length and its corresponding height
When to useWhen only side lengths are known (e.g., SSS problems)When the height is known or easy to find
ComplexityRequires computing the semiperimeter and a square rootA single multiplication — simpler arithmetic
Works for all triangles?YesYes, but you must know or be able to calculate the height

Why It Matters

Heron's Formula appears frequently in geometry courses when you face SSS (side-side-side) triangle problems and need to find area without an altitude. It is also essential in coordinate geometry: given three vertices, you can compute side lengths with the distance formula and then apply Heron's Formula directly. Beyond the classroom, surveyors and engineers use it to calculate land areas from measured boundary lengths.

Common Mistakes

Mistake: Using the perimeter instead of the semiperimeter.
Correction: The variable s must equal half the perimeter: s = (a + b + c) / 2. Forgetting to divide by 2 will produce a drastically incorrect area.
Mistake: Subtracting in the wrong order, such as computing (a − s) instead of (s − a).
Correction: Each factor must be (s − side). Since s is always larger than any single side of a valid triangle, each factor should be positive. If you get a negative factor, check your subtraction order.

Related Terms