Area of triangle

Heron's: $\sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semiperimeter and $a,b,c$ are the sides. The semiperimeter can be found by $\frac{A+B+C}{2}=\frac{4+6+8}{2}=\frac{18}{2}=9.$ Now, we have $\sqrt{9(9-4)(9-6)(9-8)}=\sqrt{9*5*3*1}=\sqrt{135}=3\sqrt{15}.$