[tex]\displaystyle\it\\\textnormal{In}~\Delta\textnormal{ABC} :\\\cos (\measuredangle B)=\frac{\sqrt{6}}{{3}} \Longleftrightarrow \frac{\textnormal{AB}}{\textnormal{BC}}=\frac{\sqrt{6}}{3} \Longleftrightarrow \textnormal{BC}=\frac{3\textnormal{AB}}{\sqrt{6}}=\frac{\textnormal{AB}\sqrt{6}}{2}=3\sqrt{6}\textnormal{cm}.\\\measuredangle \textnormal{A}=90^{\circ} \stackrel{T.P}{\Longrightarrow} AB^2+AC^2=BC^2 \Longleftrightarrow AC=\sqrt{BC^2-AB^2}=3\sqrt{2}\textnormal{cm}.[/tex]
[tex]\displaystyle\it\\\boxed{\it\textnormal{P}_{\textnormal{ABC}}=\textnormal{AB+BC+CA}=(6+3\sqrt{6}+3\sqrt{2})\textnormal{cm}}.[/tex]
