[tex]\it AD=BC=9\ cm\ (laturi\ opuse)\\ \\ \Delta ABD-\ dreptunghic,\ \widehat{BDA}=90^o,\ \widehat{DAB}=60^o,\ \Rightarrow \widehat{ABD}=30^o\\ \\ Th.\ \angle\ 30^o \Rightarrow AB=2\cdot AD=2\cdot9=18\ cm\\ \\ \mathcal{P}=2(AB+BC)=2((+!*)=2\cdot27=54\ cm[/tex]