Heron's Formula

Theorem

Given a triangle $\triangle ABC$ with sides $a$, $b$, and $c$ opposite points $A$, $B$, and $C$, respectively.

Let $s$ be the semiperimeter, so $s = \dfrac{a + b + c} 2$.

Then the area $A$ of the triangle is given by the formula $A = \sqrt{s \left({s - a}\right) \left({s - b}\right) \left({s - c}\right)}$.

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 $z$.

Then $h^2 + (a - z)^2 = b^2$ and $h^2 + z^2 = c^2$ from Pythagoras's Theorem.

By subtracting these two eqns, we get $2az - a^2 = c^2 - b^2$, which simplifies to $z = \dfrac{a^2 + c^2 - b^2}{2a}$.

Plugging back in and simplifying yields $h = \sqrt{c^2 - \left(\dfrac{a^2 + c^2 - b^2}{2a}\right)^2}$, and so:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \text{Area}\)	\(=\)	\(\displaystyle \frac 1 2 a \sqrt{c^2 - \left(\frac{a^2 + c^2 - b^2}{2a}\right)^2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	From Area of a Triangle in Terms of Side and Altitude
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt{\frac{4 c^2 a^2 - \left({a^2 + c^2 - b^2}\right)^2} {16} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt{\frac{\left({2ac - a^2 - c^2 + b^2}\right) \left({2ac + a^2 + c^2 - b^2}\right)} {16} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	From Difference of Two Squares
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt{\frac{\left({b^2 - \left({a - c}\right)^2}\right)\left({\left({a + c}\right)^2 - b^2}\right)} {16} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt{\frac{\left({b - a + c}\right) \left({b + a - c}\right) \left({a + c - b}\right) \left({a + b + c}\right)} {16} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	From Difference of Two Squares
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt{\frac{\left({a + b + c}\right) \left({a + b - c}\right) \left({a - b + c}\right) \left({-a + b + c}\right)} {16} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt{\left({\frac{a + b + c} 2}\right) \left({\frac{a + b + c} 2 - c} \right) \left({\frac{a + b + c} 2 - b}\right) \left({\frac{a + b + c} 2 - a}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt{s \left({s - c}\right) \left({s - b}\right) \left({s - a}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	From the 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 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.

$\blacksquare$

Source of Name

This entry was named for Heron of Alexandria.

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $4.5$
• George F. Simmons: Calculus Gems (1992), Chapter $\text {B}.1$: Appendix