Heron's Formula

What if you know the three sides of a triangle but not its height? Heron's formula answers exactly this question. Discovered by Heron of Alexandria around 60 CE, it computes the area of any triangle directly from its three sides , no trigonometry, no construction, no measurement of angles. The formula is: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)},$ where $a, b, c$ are the side lengths and $s = \tfrac{a+b+c}{2}$ is the semi-perimeter (half the perimeter).

Definitions

The semi-perimeter of a triangle with sides $a, b, c$ is $s = \frac{a + b + c}{2}.$

Heron's formula is the area formula: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}.$

(The proof using the law of cosines is beyond this chapter. We will accept the formula and use it.)

The recipe

1. Compute $s = \tfrac{a + b + c}{2}$.
2. Compute the four quantities $s, s - a, s - b, s - c$. (Each should be positive; if not, the triangle inequality fails.)
3. Multiply them all together: $s(s-a)(s-b)(s-c)$.
4. Take the square root.

The arithmetic can be heavy, but the procedure is mechanical.

A worked example. Triangle with sides $a = 13, b = 14, c = 15$.

• $s = \tfrac{13 + 14 + 15}{2} = 21$.
• $s - a = 8, s - b = 7, s - c = 6$.
• $s(s-a)(s-b)(s-c) = 21 \times 8 \times 7 \times 6 = 7056$.
• $\sqrt{7056} = 84$.

Area $= 84$ square units. (This triangle is the famous "$13$-$14$-$15$" with an integer area.)

Verifying with a special case

Right triangle $(3, 4, 5)$. $s = 6$, so $s - 3 = 3, s - 4 = 2, s - 5 = 1$. Product $= 6 \times 3 \times 2 \times 1 = 36$. Area $= 6$. By the direct formula ($\tfrac{1}{2} \times 3 \times 4 = 6$), same answer. Heron's formula passes.

Equilateral triangle of side $a$. $s = \tfrac{3a}{2}$. $s - a = \tfrac{a}{2}$ (three times). Product $= \tfrac{3a}{2} \times \tfrac{a}{2} \times \tfrac{a}{2} \times \tfrac{a}{2} = \tfrac{3 a^4}{16}$. Area $= \sqrt{\tfrac{3 a^4}{16}} = \tfrac{\sqrt{3}}{4} a^2$. This matches the known formula for an equilateral triangle.

Why Heron's formula works (briefly)

The formula can be derived using the law of cosines and the identity $\sin^2 \theta + \cos^2 \theta = 1$. Roughly: starting from $\text{Area} = \tfrac{1}{2} ab \sin C$, express $\cos C$ via the law of cosines, eliminate it using the Pythagorean identity, and simplify. The result is Heron's formula. We will study the law of cosines and trigonometric identities in Class XI; for now, we take Heron's formula on trust.

When to use Heron's formula

• When all three sides are known.
• When no convenient height is available.
• When the triangle is not right, equilateral, or isosceles in an obvious way.

If the triangle is right (and you know which side is the hypotenuse), use $\tfrac{1}{2}$ leg $\times$ leg. If equilateral, use $\tfrac{\sqrt{3}}{4} a^2$. Otherwise, Heron's formula is the universal tool.

Worked examples

Example 1. Find the area of a triangle with sides $5, 6, 7$.

$s = 9$. $s - a = 4, s - b = 3, s - c = 2$. Product $= 9 \times 4 \times 3 \times 2 = 216$. Area $= \sqrt{216} = 6\sqrt{6}$ square units.

Example 2. Find the area of a triangle with sides $13, 14, 15$.

(As computed above.) Area $= 84$ square units.

Example 3. Find the area of a triangle with sides $7, 24, 25$.

$s = 28$. $s - a = 21, s - b = 4, s - c = 3$. Product $= 28 \times 21 \times 4 \times 3 = 7056$. Area $= 84$. (Note: this is a right triangle, since $7^2 + 24^2 = 25^2$. Direct formula: $\tfrac{1}{2} \times 7 \times 24 = 84$. Confirms.)

Example 4. Find the area of an isosceles triangle with sides $5, 5, 6$.

$s = 8$. $s - a = 3, s - b = 3, s - c = 2$. Product $= 8 \times 3 \times 3 \times 2 = 144$. Area $= 12$ square units.

Example 5. Find the area of an equilateral triangle of side $10$ using Heron's formula.

$s = 15$. $s - a = 5$ (three times). Product $= 15 \times 5 \times 5 \times 5 = 1875$. Area $= \sqrt{1875} = 25\sqrt{3}$ square units. Verify: $\tfrac{\sqrt{3}}{4} \times 100 = 25\sqrt{3}$. ✓

Try it yourself

1. Find the area of a triangle with sides $3, 4, 5$.
2. Find the area of a triangle with sides $8, 15, 17$.
3. Find the area of a triangle with sides $7, 8, 9$.
4. Find the area of an equilateral triangle with side $12$ using Heron's formula.
5. Find the area of an isosceles triangle with sides $10, 10, 12$.
6. Find the area of a triangle with sides $10, 12, 16$.
7. Find the area of a triangle with sides $20, 30, 40$.
8. A triangle has sides $9, 12, 15$. Find the area.
9. State the semi-perimeter and check the triangle inequality for sides $5, 6, 11$. Comment on Heron's formula.
10. Find the area of a triangle with sides $11, 60, 61$.

Pitfalls / Insight

• Compute $s$ correctly. $s$ is the semi-perimeter, not the perimeter.
• Don't forget the square root. The product $s(s-a)(s-b)(s-c)$ must be taken under the square root.
• Check the triangle inequality first. If any of $s - a, s - b, s - c$ is non-positive, the three sides do not form a triangle.

Insight. Heron's formula is one of the most elegant results in elementary geometry: a single expression that captures the area of any triangle from its three sides. Two thousand years after Heron, we still use his formula without improvement.