# Heron's Formula

## Definition

**Heron's formula** is a formula that can be used to find the area of a triangle, when given its three side lengths. It can be applied to any shape of triangle, as long as we know its three side lengths. The formula is as follows:

The area of a triangle whose side lengths are $a, b,$ and $c$ is given by

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

where $s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}$, semi-perimeter of the triangle.

Other useful forms are

$\begin{aligned} A&=\frac 1 4\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}\\ \\ A&=\frac 1 4\sqrt{ \big[(a+b+c)(a+b-c) \big] \times \Big[\big(+(a-b)+c\big)\big(-(a-b)+c\big) \Big]}\\ A&=\frac 1 4\sqrt{\Big[(a+b)^2-c^2\Big] \times \ \Big[c^2-(a-b)^2\Big] }\\ \\ A&=\frac{1}{4}\sqrt{4a^2b^2-\big(a^2+b^2-c^2\big)^2}\\ A&=\frac{1}{4}\sqrt{2\left(a^2 b^2+a^2c^2+b^2c^2\right)-\left(a^4+b^4+c^4\right)} \\ A&=\frac{1}{4}\sqrt{\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)}. \end{aligned}$

Although this seems to be a bit tricky (in fact, it is), it might come in handy when we have to find the area of a triangle, and we have no other information other than its three side lengths.

## Proof of Heron's Formula

This formula follows from the area formula $A=\frac{1}{2}ab\sin C$.

By the law of cosines, $\cos C=\frac{a^2+b^2-c^2}{2ab}$.

Substituting into the Pythagorean identity $\sin C=\sqrt{1-\cos^2 C}$ yields Heron's formula (after a series of algebraic manipulations). $_\square$

## Examples

Find the area of the triangle below.

Since the three side lengths are all equal to 6, the semiperimeter is $s=\frac{6+6+6}{2}=9$. Therefore the area of the triangle is

$A=\sqrt{9\times(9-6)\times(9-6)\times(9-6)}=9\sqrt{3}.\ _\square$

Find the area of the triangle below.

Since the three side lengths are 4, 5, and 7, the semiperimeter is $s=\frac{4+5+7}{2}=8$. Therefore the area of the triangle is

$A=\sqrt{8\times(8-4)\times(8-5)\times(8-7)}=4\sqrt{6}.\ _\square$

What is the area of a triangle with side lengths 13, 14, and 15?

Since the three side lengths are 13, 14, and 15, the semiperimeter is $s=\frac{13+14+15}{2}=21$. Therefore the area of the triangle is

$A=\sqrt{21\times(21-13)\times(21-14)\times(21-15)}=84.\ _\square$

Find the area of the triangle below.

Since the three side lengths are 6, 8, and 10, the semiperimeter is $s=\frac{6+8+10}{2}=12$. Therefore the area of the triangle is

$A=\sqrt{12\times(12-6)\times(12-8)\times(12-10)}=24.\ _\square$

## Find the area of a triangle with side lengths $4,13$ and $15$.

We have $a=4, b=13, c=15$ and $s=\frac{4+13+15}{2}=16$. Hence,

$A = \sqrt{16(16-4)(16-13)(16-15)} = 24. \ _\square$

## Find the area of the triangle outlined in black.

We can use the Pythagorean theorem to find that the side lengths are $5, \sqrt{ 29}, 2 \sqrt{10}$.

If we used the direct form of $A = \sqrt{ s (s-a)(s-b)(s-c) }$, we will quickly get into a huge mess because these lengths are not integers.Instead, we will use an alternate form of Heron's formula:

$\begin{aligned} A & = \frac{1}{4}\sqrt{2\big(a^2 b^2+a^2c^2+b^2c^2\big)-\big(a^4+b^4+c^4\big)} \\ & = \frac{1}{4} \sqrt{ 2 ( 25 \times 29 + 25 \times 40 + 29 \times 40) - 25^2 - 29^2 - 40^2 } \\ & = \frac{1}{4} \sqrt{ 2704 } \\ & = 13. \ _\square \end{aligned}$

Note: This triangle appears in Composite Figures, which is an easier approach.

## Find the area of the triangle below.

Since the three side lengths are all equal to 6, the semiperimeter is $s=\frac{6+6+6}{2}=9$. Therefore the area of the triangle is

$A=\sqrt{9\times(9-6)\times(9-6)\times(9-6)}=9\sqrt{3}.\ _\square$

## Find the area of the triangle below.

Since the three side lengths are 4, 5, and 7, the semiperimeter is $s=\frac{4+5+7}{2}=8$. Therefore the area of the triangle is

$A=\sqrt{8\times(8-4)\times(8-5)\times(8-7)}=4\sqrt{6}.\ _\square$

## What is the area of a triangle with side lengths 13, 14, and 15?

Since the three side lengths are 13, 14, and 15, the semiperimeter is $s=\frac{13+14+15}{2}=21$. Therefore the area of the triangle is

$A=\sqrt{21\times(21-13)\times(21-14)\times(21-15)}=84.\ _\square$

## Find the area of the triangle below.

Since the three side lengths are 6, 8, and 10, the semiperimeter is $s=\frac{6+8+10}{2}=12$. Therefore the area of the triangle is

$A=\sqrt{12\times(12-6)\times(12-8)\times(12-10)}=24.\ _\square$

## Additional Problems