Let $\alpha$ be the angle between the sides $b$ and $c$ , then we get from the cosines law: $$\cos\alpha =\frac{b^2+c^2-a^2}{2bc}.$$ Using the equation $$\sin\alpha=\sqrt{1-\cos^2\alpha}$$ we get: $$\sin\alpha=\frac{\sqrt{-a^4-b^4-c^4+2b^2c^2+2a^2b^2+2a^2c^2}}{2bc}.$$ Now we know: $$\Delta=\frac{1}{2}bc\sin\alpha.$$ So we get: \begin{eqnarray*} \Delta & = & \frac{1}{4}\sqrt{-a^4-b^4-c^4+2b^2c^2+2a^2b^2+2a^2c^2}\\ & = & \frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}\\ & = & \sqrt{s(s-a)(s-b)(s-c)}. \end{eqnarray*}This is Heron's formula.