[tex]\text{Triunghiul este isoscel cu AB=BC} \\
=\ \textgreater \ ~~ \text{Inaltimea BM imparte baza in 2 parti egale.} \\
In ~\Delta ABM,~avem: \\ \ \textless \ BMA = 90^o \\ BM=9\;cm \\ AM=12\;cm \\
AB = \sqrt{BM^2 + AM^2}= \sqrt{9^2 + 12^2}=\sqrt{81 + 144}=\sqrt{225}=\boxed{15\;cm} \\
BC=AB=15\;cm \\
AC=2 \times AM=2 \times 12 = 24\;cm \\
P=BC+AB+AC = 15+15+24 = \boxed{54\;cm}[/tex]