[tex]\it \mathcal{A}=\dfrac{\ell^2\sqrt3}{4}=36\sqrt3\ \Big|_{:\sqrt3} \Rightarrow \dfrac{\ell^2}{4}=36\ \Big|_{\cdot4} \Rightarrow \ell^2=144=12^2 \Rightarrow \ell=12\ cm\\ \\ \\ \mathcal{P}=3\cdot \ell =3\cdot12=36\ cm\\ \\ h=\dfrac{\ell\sqrt3}{2}=\dfrac{12\sqrt3}{2}=6\sqrt3\ cm[/tex]