[tex]\it \widehat{DAB}=120^o\ (suplementul\ lui\ 60^o)\\ \\ Din\ \Delta DAB-isoscel, AB=AD \Rightarrow \widehat{ABD}= \widehat{BDA} = (180^o-120^o)=30^o\\ \\ Teorema\ sinusurilor\ \hat{\imath}n\ \Delta ABD \Rightarrow \dfrac{AB}{sin30^o}=\dfrac{BD}{sin120^o} \Rightarrow \dfrac{AB}{\dfrac{1}{2}}=\dfrac{12\sqrt3}{\dfrac{\sqrt3}{2}} \Rightarrow \\ \\ \\ \Rightarrow AB=\dfrac{\dfrac{1}{2}\cdot12\sqrt3}{\dfrac{\sqrt3}{2}}=6\sqrt3\cdot\dfrac{2}{\sqrt3}=12\ cm[/tex]
[tex]\it \mathcal{A}_{ABCD}=AB\cdot BC\cdot sin60^o=12\cdot12\cdot\dfrac{\sqrt3}{2}=72\sqrt3\ cm^2[/tex]
