# When do you use Heron's formula to find area?

Oct 23, 2014

You can use it whenever you know the lengths of all three sides of a triangle.

I hope that this was helpful.

May 20, 2018

Heron's Formula is almost always the wrong formula to use; try Archimedes' Theorem for a triangle with area $A$ and sides $a , b , c$:

$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {\left({a}^{2} + {b}^{2} - {c}^{2}\right)}^{2}$

$\quad = {\left({a}^{2} + {b}^{2} + {c}^{2}\right)}^{2} - 2 \left({a}^{4} + {b}^{4} + {c}^{4}\right)$

$\quad = \left(a + b + c\right) \left(- a + b + c\right) \left(a - b + c\right) \left(a + b - c\right)$

$\quad = 16 s \left(s - a\right) \left(s - b\right) \left(s - c\right)$ where $s = \frac{1}{2} \left(a + b + c\right)$

This last is thinly veiled Heron.

#### Explanation:

Hero of Alexandria wrote in the first century AD. Why we continue to torture students with his result when there are much nicer modern equivalents I have no idea.

Heron's formula for the area $A$ of a triangle with sides $a , b , c$ is

$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ where $s = \frac{1}{2} \left(a + b + c\right)$ is the semiperimeter.

There's no doubt this formula is awesome. But it's awkward to use because of the fraction and, if we start from coordinates, the four square roots.

Let's just do the math. We square and eliminate $s$ which mostly serves to hide a $16$ and an important factorization. You might want to try it yourself first.

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

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

$16 {A}^{2} = \left(a + b + c\right) \left(- a + b + c\right) \left(a - b + c\right) \left(a + b - c\right)$

That's already much better than Heron's form. We save the fraction to the end and there's no more wondering about the meaning of the semiperimeter.

The degenerate case is telling. When one of those factors with a minus sign is zero, that's when two sides add up to exactly the other side. Those are distances between three collinear points, the degenerate triangle, and we get zero area. Makes sense.

The $a + b + c$ factor is interesting. What it tells us is this formula still works if we use displacements, signed lengths, instead of all positive.

The formula is still awkward to use given coordinates. Let's multiply it out; you might want to try it yourself;

$16 {A}^{2} = \left(a + b + c\right) \left(- a + b + c\right) \left(a - b + c\right) \left(a + b - c\right)$

$= \left(- {a}^{2} - a b - a c + a b + {b}^{2} + b c + a c + b c + {c}^{2}\right) \left({a}^{2} - a b + a c + a b - {b}^{2} + b c - a c + b c - {c}^{2}\right)$

$= \left(- {a}^{2} + {b}^{2} + {c}^{2} + 2 b c\right) \left({a}^{2} - {b}^{2} - {c}^{2} + 2 b c\right)$

$= \left(- {a}^{2} + {b}^{2} + {c}^{2} + 2 b c\right) \left({a}^{2} - {b}^{2} - {c}^{2} + 2 b c\right)$

$16 {A}^{2} = 2 \left({a}^{2} {b}^{2} + {a}^{2} {c}^{2} + {b}^{2} {c}^{2}\right) - \left({a}^{4} + {b}^{4} + {c}^{4}\right)$

That form depends only on the squares of the lengths. It's clearly fully symmetrical. We can go beyond Heron now and say if the squared lengths are rational, so is the squared area.

But we can do better if we note

${\left({a}^{2} + {b}^{2} + {c}^{2}\right)}^{2} = \left({a}^{4} + {b}^{4} + {c}^{4}\right) + 2 \left({a}^{2} {b}^{2} + {a}^{2} {c}^{2} + {b}^{2} {c}^{2}\right)$

Subtracting,

$16 {A}^{2} = {\left({a}^{2} + {b}^{2} + {c}^{2}\right)}^{2} - 2 \left({a}^{4} + {b}^{4} + {c}^{4}\right)$

That's the prettiest form.

There's an asymmetric looking form that's usually the most useful. We note

${\left({a}^{2} + {b}^{2} - {c}^{2}\right)}^{2} = \left({a}^{4} + {b}^{4} + {c}^{4}\right) - 2 \left(- {a}^{2} {b}^{2} + {a}^{2} {c}^{2} + {b}^{2} {c}^{2}\right)$

$16 {A}^{2} = 2 \left({a}^{2} {b}^{2} + {a}^{2} {c}^{2} + {b}^{2} {c}^{2}\right) - \left({a}^{4} + {b}^{4} + {c}^{4}\right)$
$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {\left({a}^{2} + {b}^{2} - {c}^{2}\right)}^{2}$