# A Detailed Derivation of the Heron’s Formula

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. 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.