Herons Formula for Triangle Area with Examples

Key Points

• Heron's formula is $A = \sqrt{s(s-a)(s-b)(s-c)}$, where $s = \frac{a+b+c}{2}$ is the semi-perimeter.

• The formula needs only the three side lengths, which is why it works when no height is known.

• MacTutor records that Heron proves the formula in Book I of his work Metrica, and that it is his best known mathematical result.

• The semi-perimeter is half the perimeter, so dividing by 2 is part of the formula rather than an optional step.

• If the three lengths cannot form a triangle, one bracket turns negative and the square root has no real value, so the formula detects impossible triangles.

What Is the Herons Formula?

Heron's formula computes the area of any triangle from its three side lengths:

$$A = \sqrt{s(s-a)(s-b)(s-c)}, \qquad s = \frac{a+b+c}{2}$$

No angle and no height is required. The MacTutor History of Mathematics Archive records that Heron of Alexandria proves this result in Book I of Metrica, and describes it as his best known mathematical work.

Why the Semi-Perimeter Appears

The familiar area formula, half base times height, needs a height, and a height is not given when only three sides are known. Dropping a perpendicular to find one requires work that Herons formula skips entirely.

The semi-perimeter is what makes the expression symmetric. Each of the three sides appears once in a bracket of the form $s - a$, so no side is privileged and the answer does not depend on which side is called the base.

Those brackets have a geometric reading too. Each $s - a$ is the distance from a vertex to where the incircle touches an adjacent side, which is why the formula also connects to the triangle's inradius.

Variable Key

Examples of Herons Formula

Example 1

Find the area of a triangle with sides 13, 14 and 15.

$$s = \frac{13+14+15}{2} = 21$$ $$A = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$$

Final answer: $A = 84$ square units

Example 2

Find the area of a triangle with sides 3, 4 and 5, using the full perimeter instead of the semi-perimeter.

The perimeter is $3+4+5 = 12$, and using that directly looks reasonable:

$$\sqrt{12(12-3)(12-4)(12-5)} = \sqrt{12 \times 9 \times 8 \times 7} = \sqrt{6048} \approx 77.8$$

That cannot be right. A triangle with sides 3, 4 and 5 is right-angled, so its area is $\frac{1}{2} \times 3 \times 4 = 6$, and 77.8 is more than twelve times too large.

The formula requires the semi-perimeter, half the total:

$$s = 6, \qquad A = \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6$$

Final answer: $A = 6$ square units, matching the right-angled calculation exactly.

Example 3

Find the area of an equilateral triangle of side 6.

$$s = 9, \qquad A = \sqrt{9 \times 3 \times 3 \times 3} = \sqrt{243} = 9\sqrt{3} \approx 15.59$$

Final answer: $9\sqrt{3} \approx 15.59$ square units

This agrees with the standard equilateral result $\frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4}(36) = 9\sqrt{3}$.

Example 4

A triangular plot has sides 40 m, 24 m and 32 m. Find its area.

$$s = \frac{40+24+32}{2} = 48$$ $$A = \sqrt{48 \times 8 \times 24 \times 16} = \sqrt{147456} = 384$$

Final answer: 384 square metres

Example 5

Test whether sides 2, 3 and 7 form a triangle, using the formula.

$$s = 6, \qquad s - c = 6 - 7 = -1$$ $$A = \sqrt{6 \times 4 \times 3 \times (-1)} = \sqrt{-72}$$

Final answer: No real area exists, so the three lengths cannot form a triangle.

The negative bracket is the formula reporting a violated triangle inequality, since $2 + 3 < 7$. The square root failing is a feature, not an arithmetic accident.

Don't Confuse Herons Formula With These Three

Heron's formula is the one to reach for when only lengths are available. When a height is given, half base times height is faster and involves no square root.

Where Do Students Trip Up on Herons Formula?

Mistake 1: Using the perimeter instead of the semi-perimeter

Where it slips in: The sum of the sides is the natural first computation, and the division by 2 is easy to skip.

Don't do this: $s = a+b+c$

The correct way: $s = \frac{a+b+c}{2}$. The clue is in the name: semi means half.

Students who sanity-check one answer against a triangle they can compute another way, such as the 3, 4, 5 right triangle, catch this on the first attempt. Checking a new formula against a case you already know is the fastest way to find a structural error rather than an arithmetic one.

Mistake 2: Subtracting the wrong quantity in the brackets

Where it slips in: The brackets are written as $s - a$, and it is easy to reverse them to $a - s$.

Don't do this: $\sqrt{s(a-s)(b-s)(c-s)}$

The correct way: Each bracket is $s$ minus a side. Since $s$ is larger than every individual side in a real triangle, all three brackets come out positive.

Mistake 3: Taking the square root too early

Where it slips in: The expression contains four factors, so roots get applied factor by factor.

Don't do this: $\sqrt{s} \times \sqrt{s-a} \times \sqrt{s-b} \times \sqrt{s-c}$ computed with rounded intermediate roots.

The correct way: Multiply all four factors first, then take one square root. Rounding each root separately moves error into the final digits, and in Example 1 the exact product 7056 gives the clean answer 84.

Key Takeaways

• Heron's formula is $A = \sqrt{s(s-a)(s-b)(s-c)}$ with $s$ the semi-perimeter, half the perimeter.

• It needs only the three sides, which makes it the method of choice when no height or angle is available.

• Every bracket is $s$ minus a side, and all three are positive for any genuine triangle.

• A negative value under the root is the formula detecting an impossible triangle, not a mistake.

• To work through triangle geometry with a teacher, explore a geometry tutor or math classes online.

A Practical Next Step

Find the area of a triangle with sides 7, 8 and 9. Then try sides 5, 12 and 13, and check your answer against half base times height, since that triangle is right-angled.

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Read More

• Triangle Formula

• Altitude of a Triangle

• Equilateral Triangle

• Classifying Triangles

• Perimeter Formula