Heron's Formula

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Theorem

Let $\triangle ABC$ be a triangle with sides $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.


Then the area $A$ of $\triangle ABC$ is given by:

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

where $s = \dfrac{a + b + c} 2$ is the semiperimeter of $\triangle ABC$.


Proof 1

Construct the altitude from $A$.

Let the length of the altitude be $h$ and the foot of the altitude be $D$.

Let the distance from $D$ to $B$ be $z$.

Heron1.png

From Pythagoras's Theorem:

$(1): \quad h^2 + \left({a - z}\right)^2 = b^2$

and:

$(2): \quad h^2 + z^2 = c^2$

By subtracting $(1)$ from $(2)$:

$2az - a^2 = c^2 - b^2$

which can be expressed in terms of $z$ as:

$z = \dfrac{a^2 + c^2 - b^2}{2a}$

Substituting for $z$ in $(2)$ and simplifying yields:

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

and so:

\(\displaystyle Q\) \(=\) \(\displaystyle \frac 1 2 a \sqrt{c^2 - \left({\frac{a^2 + c^2 - b^2}{2a} }\right)^2}\) $\quad$ Area of Triangle in Terms of Side and Altitude $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt{\frac{4 c^2 a^2 - \left({a^2 + c^2 - b^2}\right)^2} {16} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt{\frac{\left({2ac - a^2 - c^2 + b^2}\right) \left({2ac + a^2 + c^2 - b^2}\right)} {16} }\) $\quad$ Difference of Two Squares $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt{\frac{\left({b^2 - \left({a - c}\right)^2}\right)\left({\left({a + c}\right)^2 - b^2}\right)} {16} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt{\frac{\left({b - a + c}\right) \left({b + a - c}\right) \left({a + c - b}\right) \left({a + b + c}\right)} {16} }\) $\quad$ Difference of Two Squares $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt{\frac{\left({a + b + c}\right) \left({a + b - c}\right) \left({a - b + c}\right) \left({-a + b + c}\right)} {16} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt{\left({\frac{a + b + c} 2}\right) \left({\frac{a + b + c} 2 - c} \right) \left({\frac{a + b + c} 2 - b}\right) \left({\frac{a + b + c} 2 - a}\right)}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt{s \left({s - c}\right) \left({s - b}\right) \left({s - a}\right)}\) $\quad$ Definition of semiperimeter $\quad$

$\blacksquare$


Proof 2

A triangle can be considered as a cyclic quadrilateral one of whose sides has degenerated to zero.

From Brahmagupta's Formula, the perimeter of a cyclic quadrilateral is given by:

$\sqrt{\left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right)}$

where $s$ is the semiperimeter:

$s = \dfrac{a + b + c + d} 2$

The result follows by letting $d$ tend to zero.

$\blacksquare$


Proof 3

Heron3.png

Let $T$ be the area of $\triangle ABC$.

Construct the incircle of $\triangle ABC$.

Let the incenter of $\triangle ABC$ be $M$.

Let the inradius of $\triangle ABC$ be $r$.

$\triangle ABC$ is made up of three triangles: $\triangle AMB$, $\triangle BMC$ and $\triangle CMA$.

From Area of Triangle in Terms of Side and Altitude, the areas of $\triangle AMB$, $\triangle BMC$ and $\triangle CMA$ are given by:

$\mathcal A \left({\triangle AMB}\right) = \dfrac {rc} 2$
$\mathcal A \left({\triangle BMC}\right) = \dfrac {ra} 2$
$\mathcal A \left({\triangle CMA}\right) = \dfrac {rb} 2$

Thus:

$(1): \quad T = \dfrac {r \left({c + a + b}\right)} 2 = r s$

where $s$ is the semiperimeter of $\triangle ABC$.


Construct the excircle of $\triangle ABC$ with excenter $N$ tangent to $AB$, and to $AC$ and $BC$ produced at $D$ and $E$ respectively.

We note that $s = CD = CE$.

Therefore:

$DA = s - b$
$EB = s - a$

Note that:

$AF + DA = BG + EB$

and:

$AF + BG = C$

Note also that:

$\triangle NDC$ is similar to $\triangle MFC$
$\triangle NDA$ is similar to $\triangle AFM$

from which:

$\dfrac R r = \dfrac s {s - c}$
$\dfrac R {s - b} = \dfrac {s - a} r$

Substituting for $R$:

$R = \dfrac {r s} {s - c} = \dfrac {\left({s - a}\right) \left({s - b}\right)} r$

and so:

$r^2 = \dfrac {\left({s - a}\right) \left({s - b}\right)\left({s - c}\right)} s$

Thus $(1)$ becomes:

$T = s \sqrt {\dfrac {\left({s - a}\right) \left({s - b}\right)\left({s - c}\right)} s} = \sqrt {s \left({s - a}\right) \left({s - b}\right)\left({s - c}\right)}$

$\blacksquare$


Source of Name

This entry was named for Heron of Alexandria.


Historical Note

Arabic sources from the Middle Ages inform us that Heron's Formula was actually due to Archimedes of Syracuse.

However, Heron's is the earliest proof that survives.


Sources