[tex]\it \mathcal{P}_{ABC}=BC+AB+AC=BC+4+8=BC+12\ \ \ \ \ \ (1)\\ \\ \mathcal{P}_{ABC}=4(\sqrt3+3)=4\sqrt3+12\ \ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow BC+12=4\sqrt3+12|_{-12} \Rightarrow BC=4\sqrt3\ cm[/tex]
[tex]\it AB^2=4^2=16\\ \\ BC^2=(4\sqrt3)^2=16\cdot3=48\\ \\ AC^2=8^2=64\\ \\ AB^2+BC^2=16+48=64=AB^2\ \stackrel{R.T.P.}{\Longrightarrow}\ \Delta ABC-dr, \widehat B=90^o.[/tex]
[tex]\it \mathcal{A}_{ABC}=\dfrac{c_1\cdot c_2}{2}=\dfrac{AB \cdot BC}{2}=\dfrac{4\cdot4\sqrt3}{2}=\dfrac{16\sqrt3}{2}=8\sqrt3\ cm^2\\ \\ AM-median\breve a \Rightarrow \mathcal{A}_{ACM}=\dfrac{\mathcal{A}_{ABC}}{2}=\dfrac{8\sqrt3}{2}=4\sqrt3\ cm^2[/tex]
