Heron's Formula

Heron's Formula finds the area of a triangle using only its three side lengths, making it useful when the height is unknown or hard to measure.

A=s(sa)(sb)(sc),s=a+b+c2A = \sqrt{s(s-a)(s-b)(s-c)}, \quad s = \frac{a+b+c}{2}

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What each symbol means

What Heron's Formula takes
AA
aa
bb
cc
ss
Heron's Formula
SymbolMeaning
AAThe area of the triangle, measured in square units; this is the result you compute with the formula, not a measured length.
aaThe length of one side of the triangle in linear units; must be positive, and the formula requires all three sides to be known.
bbThe length of a second side of the triangle in linear units; must be positive like side aa, and all three sides must be used in the calculation.
ccThe length of the third side of the triangle in linear units; must be positive, and all three sides must satisfy the triangle inequality.
ssThe semi-perimeter, equal to (a+b+c)/2(a+b+c)/2; this is half the triangle's perimeter, so forgetting to divide by 2 will give wrong results.

When to use it

Use Heron's Formula when you know all three side lengths of a triangle but not the height.

Level

Usually taught in: Geometry

Worked examples

1. Find the area of an isosceles triangle

Problem

A triangle has sides of 5 cm, 5 cm, and 6 cm. Find its area using Heron's Formula.
  1. s=5+5+62=162=8s = \frac{5+5+6}{2} = \frac{16}{2} = 8

    First, calculate the semi-perimeter by adding all three sides and dividing by 2.

  2. (sa)(sb)(sc)=(85)(85)(86)=332=18(s-a)(s-b)(s-c) = (8-5)(8-5)(8-6) = 3 \cdot 3 \cdot 2 = 18

    Next, compute the product of the three differences: each semi-perimeter minus one side.

  3. A=s(sa)(sb)(sc)=818=144=12A = \sqrt{s \cdot (s-a)(s-b)(s-c)} = \sqrt{8 \cdot 18} = \sqrt{144} = 12

    Finally, substitute into Heron's Formula: the area is the square root of 8×18=1448 \times 18 = 144, which equals 12.

Answer: A=12 cm2A = 12 \text{ cm}^2

This isosceles triangle (two equal sides) has an area of 12 square centimeters. The integer answer confirms our calculation and shows that Heron's Formula is exact. When a triangle has two equal sides, you can verify this using base and height, but the formula works for all triangle types.

2. Calculate area of a scalene triangle with square root simplification

Problem

A triangle has sides 5, 6, and 7 units. Find its area in simplified radical form.
  1. s=5+6+72=182=9s = \frac{5+6+7}{2} = \frac{18}{2} = 9

    Compute the semi-perimeter: one-half of the total perimeter 5+6+7=185 + 6 + 7 = 18.

  2. 9(95)(96)(97)=9432=2169 \cdot (9-5)(9-6)(9-7) = 9 \cdot 4 \cdot 3 \cdot 2 = 216

    Multiply the semi-perimeter and the three differences: 9×4×3×2=2169 \times 4 \times 3 \times 2 = 216 goes under the square root.

  3. A=216=366=66A = \sqrt{216} = \sqrt{36 \cdot 6} = 6\sqrt{6}

    Simplify the square root by factoring out perfect squares: since 216=36×6216 = 36 \times 6, we get 216=366=66\sqrt{216} = \sqrt{36} \cdot \sqrt{6} = 6\sqrt{6}.

Answer: A=66 units2A = 6\sqrt{6} \text{ units}^2

When the number under the square root is not a perfect square, simplify by factoring out perfect squares. This scalene triangle (all sides different) gives an irrational answer that we leave in radical form. The exact answer is 666\sqrt{6} square units, approximately 14.7 square units.

3. Find the area of a triangular plot of land

Problem

A surveyor measures a triangular plot of land with sides of 20 m, 21 m, and 29 m. What is the area of the plot?
  1. s=20+21+292=702=35s = \frac{20+21+29}{2} = \frac{70}{2} = 35

    Start by calculating the semi-perimeter of the triangle: half of the total perimeter 20+21+29=7020 + 21 + 29 = 70 is 35.

  2. 35(3520)(3521)(3529)=3515146=4410035 \cdot (35-20)(35-21)(35-29) = 35 \cdot 15 \cdot 14 \cdot 6 = 44100

    Substitute the semi-perimeter and each difference into s(sa)(sb)(sc)s(s-a)(s-b)(s-c): multiply 35×15×14×635 \times 15 \times 14 \times 6 to get 44100.

  3. A=44100=210A = \sqrt{44100} = 210

    Take the square root of 44100 to find the area: since 210×210=44100210 \times 210 = 44100, the area is exactly 210 square meters.

Answer: A=210 m2A = 210 \text{ m}^2

This real-world example shows why Heron's Formula is invaluable: a surveyor can measure the three boundary distances but cannot easily find the height. The perfect square result means the area is exact and practical for land records. Without Heron's Formula, you would need to measure the perpendicular height, which is often difficult or impossible in the field.

Common mistakes

Where Heron's Formula usually goes wrong
Answer came out wrong
Using the full perimeter instead of the semi-perimeter: A=(a+b+c)(a+b+ca)(a+b+cb)(a+b+cc)A = \sqrt{(a+b+c)(a+b+c-a)(a+b+c-b)(a+b+c-c)}
Always divide the perimeter by 2 first to get s=a+b+c2s = \frac{a+b+c}{2}, then apply A=s(sa)(sb)(sc)A = \sqrt{s(s-a)(s-b)(s-c)}.
Forgetting to take the square root at the end and leaving the answer as A=s(sa)(sb)(sc)A = s(s-a)(s-b)(s-c) without the radical symbol.
Always include the square root: A=s(sa)(sb)(sc)A = \sqrt{s(s-a)(s-b)(s-c)} with the full radical symbol, and evaluate it to get the final answer.
Using negative or zero side lengths, or applying the formula when the triangle inequality is violated (e.g., a+bca + b \leq c).
Before using the formula, check that all sides are positive and that the sum of any two sides is greater than the third: a+b>ca + b > c.
The mistakeWhy it is wrongThe fix
Using the full perimeter instead of the semi-perimeter: A=(a+b+c)(a+b+ca)(a+b+cb)(a+b+cc)A = \sqrt{(a+b+c)(a+b+c-a)(a+b+c-b)(a+b+c-c)}This breaks the formula structure entirely, as ss must specifically equal (a+b+c)/2(a+b+c)/2 for Heron's Formula to work correctly.Always divide the perimeter by 2 first to get s=a+b+c2s = \frac{a+b+c}{2}, then apply A=s(sa)(sb)(sc)A = \sqrt{s(s-a)(s-b)(s-c)}.
Forgetting to take the square root at the end and leaving the answer as A=s(sa)(sb)(sc)A = s(s-a)(s-b)(s-c) without the radical symbol.The square root is essential; without it, the quantity s(sa)(sb)(sc)s(s-a)(s-b)(s-c) has units of length to the fourth power, not area (length squared).Always include the square root: A=s(sa)(sb)(sc)A = \sqrt{s(s-a)(s-b)(s-c)} with the full radical symbol, and evaluate it to get the final answer.
Using negative or zero side lengths, or applying the formula when the triangle inequality is violated (e.g., a+bca + b \leq c).Side lengths must be positive, and the three sides must satisfy a+b>ca + b > c, a+c>ba + c > b, and b+c>ab + c > a; otherwise, they cannot form a triangle, and the value under the square root becomes zero or negative.Before using the formula, check that all sides are positive and that the sum of any two sides is greater than the third: a+b>ca + b > c.

Tips and when to use something else

  • If you know the base and height, use A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height} instead—it's faster than Heron's Formula.
  • For right triangles, just multiply the two legs and divide by 2: A=12abA = \frac{1}{2}ab, which is quicker than finding the hypotenuse first.
  • When you don't get a perfect square under the radical, simplify by factoring out perfect squares to get a cleaner answer in radical form.
  • The formula works for any triangle—acute, right, or obtuse—as long as you know all three sides and the triangle is valid.

Frequently asked questions

When should I use Heron's Formula instead of the standard area formula with base and height?
Use Heron's Formula when you know all three side lengths but do not know (or cannot easily measure) the height. If you have the height, the formula A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height} is simpler. Heron's Formula is also useful for obtuse triangles, where the height would fall outside the base.
Does Heron's Formula work for right triangles?
Yes, Heron's Formula works for any valid triangle, including right triangles. However, for right triangles it is usually easier to multiply the two legs and divide by 2, since you do not need to find the hypotenuse. Only use Heron's Formula for a right triangle if you know all three sides.
What if my three side lengths don't satisfy the triangle inequality?
If the three lengths do not satisfy the triangle inequality (e.g., a+bca + b \leq c), they cannot form a triangle. Heron's Formula will fail because the expression under the square root becomes negative or zero. Always verify that a+b>ca + b > c, a+c>ba + c > b, and b+c>ab + c > a before applying the formula.
Why does Heron's Formula use the semi-perimeter instead of just the three sides?
The semi-perimeter ss appears in Heron's Formula because of the algebra used to derive it from other area methods. Using ss allows the formula to avoid fractions and simplifies the expression s(sa)(sb)(sc)s(s-a)(s-b)(s-c) into a form that is easy to remember and compute.

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Reviewed 2026-09-18