# Heron's Formula

#### Contents

## 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{align} 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[(+(a-b)+c)(-(a+b)+c) \big]}\\

&=\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}. \end{align}\]

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.

## 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:

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

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

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

## Proof of Heron's Formula

This formula follows from the area formula \(A=\dfrac{1}{2}ab\sin C\). By the law of cosine we have \(\cos C=\dfrac{a^2+b^2-c^2}{2ab}\). Substituting this in \(\sin C=\sqrt{1-\cos^2 C}\) and a lot of manipulations give Heron's formula. \(_\square\)

Heron's Formula can also be written as

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