Heron's Formula

Heron's Formula (sometimes called Hero's formula) is a formula for finding the area of a triangle given only the three side lengths.

Theorem

For any triangle with side lengths ${a}, {b}, {c}$ , the area ${A}$ can be found using the following formula:

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

where the semi-perimeter $s=\frac{a+b+c}{2}$ .

Proof

$[ABC]=\frac{ab}{2}\sin C$

$=\frac{ab}{2}\sqrt{1-\cos^2 C}$

$=\frac{ab}{3}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}$

$=\sqrt{\frac{a^2b^2}{4}\left[1-\frac{(a^2+b^2-c^2)^2}{4a^2b^2}\right]}$

$=\sqrt{\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}$

$=\sqrt{\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16}}$

$=\sqrt{\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}$

$=\sqrt{\frac{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}{16}}$

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

Proof 2

Imagine a triangle with an altitude dropped from vertex A. Now the side of "a" is split into two sections. Label the right section "x" and the left section "a-x". Now we can prove Heron's formula.

$b^2 - (a-x)^2 = h^2$ $b^2 - a^2 + 2ax - x^2 = h^2$

$c^2 - h^2 = h^2$

$b^2 - a^2 + 2ax - x^2 = c^2 - x^2$

$2ax = c^2 - b^2 + a^2$ $x = \frac{c^2 - b^2 + a^2}{2a}$

$h^2 = c^2 - x^2 = (c-x) (c+x)$ $= (c - \frac{c^2 - b^2 + a^2}{2a}) (c + \frac{c^2 - b^2 + a^2}{2a})$ $= \frac{-(2ac + c^2 + a^2) + b^2}{2a} * \frac{(2ac + c^2 + a^2) - b^2}{2a}$ $= \frac{b^2 - (a^2 - 2ac + c^2)}{2a}$ $= \frac{b^2 - (a - c)^2}{2a} * \frac{(a + c)^2 - b^2}{2a}$ $= \frac{(b + a - c)(b - a + c)}{2a} * \frac{(a + c -b)(a + b + c)}{2a}$ $= \frac{(2s - 2c)(2s - 2a)(2s - 2b)(2s)}{4a^2}$

$h = \frac{4\sqrt(s(s - a)(s - b)(s - c))}{2a}$ $= \frac{2\sqrt(s(s - a)(s - b)(s - c))}{a}$

$[ABC] = \frac{1}{2} * a * h$

$[ABC] = \sqrt(s(s - a)(s - b)(s - c))$

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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Heron's Formula

Contents

Theorem

Proof

Proof 2

See Also

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