# Heron's Formula

## Theorem

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

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

$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

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 $z$.

From Pythagoras's Theorem:

$\paren 1: \quad h^2 + \paren {a - z}^2 = b^2$

and:

$\paren 2: \quad h^2 + z^2 = c^2$

By subtracting $\paren 1$ from $\paren 2$:

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

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

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

Substituting for $z$ in $\paren 2$ and simplifying yields:

$h = \sqrt {c^2 - \paren {\dfrac {a^2 + c^2 - b^2} {2a} }^2}$

and so:

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

$\blacksquare$

## Proof 2

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

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

$\sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d}}$

where $s$ is the semiperimeter:

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

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

$\blacksquare$

## Proof 3

Let $\AA$ 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:

$\Area \paren {\triangle AMB} = \dfrac {r c} 2$
$\Area \paren {\triangle BMC} = \dfrac {r a} 2$
$\Area \paren {\triangle CMA} = \dfrac {r b} 2$

Thus:

$(1): \quad \AA = \dfrac {r \paren {c + a + b} } 2 = r s$

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:

 $\ds \dfrac R r$ $=$ $\ds \dfrac s {s - c}$ $\ds \dfrac R {s - b}$ $=$ $\ds \dfrac {s - a} r$ $\ds \leadsto \ \$ $\ds R$ $=$ $\ds \dfrac {r s} {s - c}$ substituting for $R$ $\ds$ $=$ $\ds \dfrac {\paren {s - a} \paren {s - b} } r$ $\ds \leadsto \ \$ $\ds r^2$ $=$ $\ds \dfrac {\paren {s - a} \paren {s - b} \paren {s - c} } s$ rearranging $\ds \leadsto \ \$ $\ds \AA$ $=$ $\ds s \sqrt {\dfrac {\paren {s - a} \paren {s - b} \paren {s - c} } s}$ from $(1)$ $\ds$ $=$ $\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ simplifying

$\blacksquare$

## Proof 4

 $\ds \AA$ $=$ $\ds \dfrac {a b \sin C} 2$ Area of Triangle in Terms of Two Sides and Angle $\ds$ $=$ $\ds \dfrac {a b} 2 \cdot 2 \sin \dfrac C 2 \cos \dfrac C 2$ Double Angle Formula for Sine $\ds$ $=$ $\ds a b \sqrt {\dfrac {\paren {s - a} \paren {s - b} } {a b} } \sqrt {\dfrac {s \paren {s - c} } {a b} }$ Sine of Half Angle in Triangle, Cosine of Half Angle in Triangle $\ds$ $=$ $\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ simplifying

## Also known as

Heron's Formula is also known as Hero's formula.

## 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.