# An Alternative Derivation of Heron’s Formula

In my previous blog titled “An Algebraic Proof of Heron’s Formula“, I algebraically derived the Heron’s formula concerning A, the area of a triangle:

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

where a, b, c are three sides of a triangle and $s=\frac{a+b+c}{2}$.

There is an alternative derivation which requires some basic trigonometry.

Here it is:

Fig. 1

Let $\theta = \angle ABC$ (see Fig. 1),  then

$A=\frac{1}{2}a\;h = \frac{1}{2}a\;(c\sin{\theta})$

i.e.,

$A^2=\frac{1}{4}a^2 c^2 \sin^2{\theta} = \frac{1}{4} a^2 c^2 (1-\cos^2{\theta})\quad\quad\quad\quad\quad(1)$

due to trigonometric identity $\sin^2{\theta} + \cos^2{\theta} =1$.

Moreover, by the law of cosines, $b^2=a^2+c^2-2a c \cos{\theta}$ which implies

$\cos{\theta} = \frac{a^2+c^2-b^2}{2 a c}$,

and (1) becomes

$A^2=\frac{1}{4}a^2 c^2 (1-\frac{(a^2+c^2-b^2)^2}{4 a^2 c^2})$

$=\frac{1}{4}a^2 c^2 \frac{4a^2 c^2-(a^2+c^2-b^2)^2}{4a^2 c^2}$

$=\frac{1}{16}(2a c +a^2+c^2-b^2)(2a c-a^2-c^2+b^2)$

$=\frac{1}{16}((a+c)^2-b^2)(b^2-(a-c)^2)$

$=\frac{1}{16}(a+c+b)(a+c-b)(b+a-c)(b-a+c)$

$=\frac{1}{16}(a+b+c)(b+c-a)(a+c-b)(a+b-c)$

$=\frac{1}{16}\;2^4\;(\frac{a+b+c}{2})(\frac{a+b+c}{2}-a)(\frac{a+b+c}{2}-b)(\frac{a+b+c}{2}-c)$

$= s(s-a)(s-b)(s-c)$

Hence,

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