a)
[tex]\it AB=\dfrac{\mathcal{P}}{4}=\dfrac{48}{4}=12\ cm\\ \\ \\ \left.\begin{aligned} \it ABCD-romb \Rightarrow AB=AD \Rightarrow \Delta ABD-isoscel \\ \\ \it \widehat{A}=60^o\end{aligned} \right\} \Rightarrow \Delta ABD-echilateral \Rightarrow \\ \\ \\ \Rightarrow BD=AB=12\ cm[/tex]
b)
[tex]\it AB||CD\ (laturi\ opuse\ ale\ rombului);\ \ M\in AB;\ \ N\in CD \Rightarrow AM||ND\ \ \ (1)\\ \\ M=mijlocul\ lui \ AB \Rightarrow AM=12:2=6\ cm\\ \\ N=mijlocul\ lui \ CD \Rightarrow ND=12:2=6\ cm\\ \\ deci,\ \ AM=ND=6\ cm\ \ \ \ (2)\\ \\ (1),\ (2)\ \Rightarrow AMND-paralelogram[/tex]
