Heron's Formula

Theorem

In geometry, Heron's formula (or Hero's formula) gives the area of a triangle in terms of three side lengths, $a$ , $b$ , $c$ . Letting $s$ be the semiperimeter of the triangle, $s = \frac{1}{2} (a + b + c)$ , the area $A$ is

$A=\sqrt{s(s-a)(s-b)(s-c)}$

It is named after the first-century engineer Heron of Alexandria (or Hero) who proved it in his work Metrica, though it was probably known centuries earlier.

History

The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book Metrica. Mathematical historian Thomas Heath suggested that Archimedes knew the formula over two centuries earlier, and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that the formula predates the reference given in that work.

A formula equivalent to Heron's was discovered by Chinese mathematician Qin Jiushao:

$A = \frac{1}{2} \sqrt{a^2 c^2 - (\frac{a^2 + c^2 - b^2}{2}^2)}$ published in Mathematical Treatise in Nine Sections (Qin Jiushao, 1247).

Proof

Using basic Trigonometry, we have $[ABC]=\frac{ab}{2}\sin C,$ which simplifies to $[ABC]=\frac{ab}{2}\sqrt{1-\cos^2 C}.$

The Law of Cosines states that in triangle $ABC$ , $c^2 = a^2 + b^2 - 2ab\cos C$ , which can be written as $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ . Thus, $[ABC]=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}.$

Now, we can simplify: \begin{align*} [ABC] &= \sqrt{\frac{a^2b^2}{4} \left( 1 - \frac{(a^2 + b^2 - c^2)^2}{4a^2b^2} \right)} \\ &= \sqrt{\frac{4a^2b^2 - (a^2 + b^2 - c^2)^2}{16}} \\ &= \sqrt{\frac{(2ab + a^2 + b^2 - c^2)(2ab - a^2 - b^2 + c^2)}{16}} \\ &= \sqrt{\frac{((a + b)^2 - c^2)((a - b)^2 - c^2)}{16}} \\ &= \sqrt{\frac{(a + b + c)(a + b - c)(b + c - a)(a + c - b)}{16}} \\ &= \sqrt{s(s - a)(s - b)(s - c)} \end{align*}

Isosceles Triangle Simplification

$A=\sqrt{s(s-a)(s-b)(s-c)}$ for all triangles

$b=c$ for all isosceles triangles

$A=\sqrt{s(s-a)(s-b)(s-b)}$ simplifies to $A=(s-b)\sqrt{s(s-a)}$

Square root simplification/modification

From $A=\sqrt{s(s-a)(s-b)(s-c)}$ We can "take out" the $1/2$ in each $s$ , then we have $A=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$ Using the difference of squares on the first two and last two factors, we get $A=\frac{1}{4}\sqrt{(b^2+2bc+c^2-a^2)(a^2-b^2+2bc-c^2)}$ and using the difference of squares again, we get $A=\frac{1}{4}\sqrt{(2bc)^2-(-a^2+b^2+c^2)^2}$ From this equation (although seemingly not symmetrical), it is much easier to calculate the area of a certain triangle with side lengths with quantities with square roots. One can remember this formula by noticing that when finding the cosine of an angle in a triangle, the formula is $\cos{A}=\frac{-a^2+b^2+c^2}{2bc}$ and the two terms in the formula are just the denominator and numerator of the fraction for $\cos{A}$ , only they're squared. This can also serve as a reason for why the area $A$ is never imaginary. This is equivalent of ending at step $4$ in the proof and distributing.

Note

Replacing $-a^2+b^2+c^2$ as $2bc\cos{A}$ , the area simplifies down to $\frac{1}{4}\sqrt{(2bc\sin{A})^2}$ , or $\frac{1}{4}\cdot2bc\sin{A}$ , or $\frac{1}{2}bc\sin{A}$ , another common area formula for the triangle.

Example

Let $\triangle ABC$ be the triangle with sides $a = 3$ , $b = 4$ , and $c = 5$ . This triangle's semiperimeter is

\begin{align*} s &= \frac{1}{2} (a + b + c) \\ &= \frac{1}{2} (3 + 4 + 5) \\ &= 16 \end{align*}

therefore $s - a = 13$ , $s - b = 12$ , $s - c = 11$ , and the area is

\begin{align*} A &= \sqrt{s(s - a)(s - b)(s - c)} \\ &= \sqrt{13 \times 12 \times 11} \\ &= 24 \end{align*}

In this example, the triangle's side lengths and area are integers, making it a Heronian triangle. However, Heron's formula works equally well when the side lengths are real numbers. As long as they obey the strict triangle inequality, they define a triangle in the Euclidean plane whose area is a positive real number.

External Links

In general, it is a good advice not to use Heron's formula in computer programs whenever we can avoid it. For example, whenever vertex coordinates are known, vector product is a much better alternative. Main reasons:

Computing the square root is much slower than multiplication.
For triangles with area close to zero Heron's formula computed using floating point variables suffers from precision problems.

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