A Detailed Derivation of the Heron’s Formula

June 9, 2013 GB High School Algebra, High School Mathematics

Heron’s Formula can be used to find the area of a triangle given the lengths of the three sides. A triangle with side lengths $a$ , $b$ , and $c$ , its area $A$ can be calculated using the Heron’s formula

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

where

$s = \displaystyle \frac{a+b+c}{2}$

is the semiperimeter (half the perimeter) of the triangle.

In this post, I will provide a detailed derivation of this formula.

Some Preliminaries

The area of a triangle is half the product of its base and its altitude. In the figure below, $\overline{AM}$ is the altitude of triangle $ABC$ . If the length of the altitude is not given, and an angle measure is given, we can use Trigonometry to calculate the altitude.

Heron's Formula

In the figure above, the altitude $\overline{AM}$ forms right triangle $AMC$ . We know that

$\sin \theta$ = (length of opposite side)/(length of hypotenuse)

so,

$\sin C = \displaystyle\frac{AM }{AC}$ .

Simplifying, we have $AM = AC \sin C$ which is equivalent to

$AM = b \sin C$

Since the base of triangle is $\overline{BM} = a$ and its altitude is $b \sin C$ , its area $A$ is given by the formula

$A = ab \sin C$ . (1)

The Derivation

Now we apply the preceding formula, the Cosine Law and the Pythagorean identity $\sin \theta + \cos \theta = 1$ to derive the Heron’s formula.

Using the Pythagorean identity, and manipulating algebraically

$\sin^2 C + \cos^2 C = 1$

$\sin^2 C = 1 - \cos^2 C$

$\sin^2 C = (1 + \cos C)(1- \cos C)$ (2)

By the Cosine Law, in a triangle $ABC$ with side lengths $a$ , $b$ , and $c$

$c^2 = a^2 + b^2 -2ab \cos C$ .

Calculating for $\cos C$ , we have

$\cos C = \displaystyle\frac{a^2 + b^2 - c^2}{2ab}$ .

Substituting the preceding equation to (2), we have

$\sin^2 C = \left (1 + \displaystyle\frac{a^2 + b^2 - c^2}{2ab} \right ) \left ( 1 - \displaystyle \frac{a^2 + b^2 - c^2} {2ab} \right )$ .

$= \left (\displaystyle\frac{2ab + a^2 + b^2 - c^2}{2ab} \right ) \left ( \displaystyle \frac{2ab - a^2 - b^2 + c^2} {2ab} \right )$ .

$=\displaystyle \frac{ [(a + b)^2 - c^2][c^2 - (a-b)^2 ]}{4a^2b^2}$ .

$\sin^2 C = \displaystyle \frac {( a + b + c)(a + b - c)( c + a - b) (c - a + b)}{4a^2b^2}$ .

Getting the square root of both sides, we have

$\sin C = \displaystyle \frac{ \sqrt{(a + b +c)(a + b - c)(c + a - b) (c -a + b)}}{2ab}$ (3).

Using (1) and (3), we calculate the area of the triangle ,

$A = ab \sin C$

$A = \frac{1}{2} ab \displaystyle \frac{ \sqrt{(a + b +c)(a + b - c)(c + a - b) (c -a + b)}}{2ab}$

$A = \displaystyle \frac{1}{4} \sqrt{(a + b + c)(a + b - c)(c + a - b)(c - a + b)}$ (4)

Now, if we let $s$ be the semiperimeter (half the perimeter) of triangle $ABC$ , then

$a + b + c = 2s$ .

Now, $a + b + c - 2a = b + c - a = 2s -2a = 2(s - a)$ .

Also, $a - b + c = 2(s - b)$ and $a + b - c = 2(s - c)$ .

Substituting the expressions with s to (4), we have

$A = \frac{1}{4} \sqrt{2s [2(s-a)][2(s-b)][2(s-c)]}$

$= \frac{1}{4} \sqrt{16s(s-a)(s-b)(s-a)}$

which is equivalent to

$A = \frac{4}{4} \sqrt{s(s-a)(s-b)(s-a)}$ .

Simplifying, we have $A = \sqrt{s(s-a)(s-b)(s-c)}$ , the Heron’s formula.

2 comments area of triangle, cosine law, heron's formula, heron's formula derivation, heron's formula proof, pythagorean identity

A Detailed Derivation of the Heron’s Formula

Some Preliminaries

The Derivation

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