Proving the area formula from the three side lengths

For a triangle with side lengths $a,b,c$, let $s=\tfrac{a+b+c}{2}$. Show that its area equals $\sqrt{s(s-a)(s-b)(s-c)}$.

Use $\text{Area}=\tfrac12ab\sin C$ with the law of cosines. Then $16\,\text{Area}^2=4a^2b^2-(a^2+b^2-c^2)^2$. Factor this difference of squares twice into $(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=16\,s(s-a)(s-b)(s-c)$, and take square roots.