The area of a triangle a, b, c

This week, I encountered the following problem in my textbook: Show that the area of a triangle ABC is given by \(\sqrt{s(s-a)(s-b)(s-c)}\), where \(s = (a+b+c)/2\) is the semi-perimeter of the triangle. I want to share my solution because it shows how you can avoid time-consuming algebra by factoring instead of expanding.

\[\small\begin{array}{l}
A = \sqrt{A^2}\\
\quad A^2 = \left(\frac{bh}{2}\right)^2 = \frac{b^2h^2}{4}\\
\quad\quad h^2 = \begin{cases} c^2-(c\cos{A})^2 \\ a^2-(b-c\cos{A})^2 \end{cases}\\
\quad\quad c\cos{A} = \frac{c^2+b^2-a^2}{2b}\\
\quad\quad h^2 = c^2 - \left(\frac{c^2+b^2-a^2}{2b}\right)^2 = \Pi\left(c \pm \frac{c^2+b^2-a^2}{2b}\right) = \Pi\frac{2bc \pm (c^2+b^2-a^2)}{2b}\\
\quad \begin{aligned}A^2 &= \cfrac{(2bc + (c^2+b^2-a^2))(2bc - (c^2+b^2-a^2))}{16} \\&= \cfrac{((b+c)^2-a^2)(a^2 - (b-c)^2)}{16} \\&= \cfrac{(b+c+a)(b+c-a)(a-b+c)(a+b-c)}{2\cdot2\cdot2\cdot2} \\&= \cfrac{a+b+c}{2}\cdot\cfrac{a+b+c-2a}{2}\cdot\cfrac{a+b+c-2b}{2}\cdot\cfrac{a+b+c-2c}{2} \\&= s(s-a)(s-b)(s-c)\end{aligned}\\
A = \sqrt{s(s-a)(s-b)(s-c)}
\end{array}\]