[tex]\it In\ triunghiul\ ABC \Rightarrow m(\widehat{BCA}) =180^o-(135^o+30^o) =15^o\\ \\ Th.\ sinusurilor\ \^{i}n\ \Delta ABC \Rightarrow \dfrac{AB}{sin15^o} =\dfrac{BC}{sin30^0} \Rightarrow AB=\dfrac{BC\cdot sin15^o}{sin30^o} \Rightarrow[/tex]
[tex]\it \Rightarrow AB=\dfrac{3\cdot\dfrac{\sqrt6-\sqrt2}{4}}{\dfrac{1}{2}}=\dfrac{3(\sqrt6-\sqrt2)}{4\cdot\dfrac{1}{2}} =\dfrac{3(\sqrt6-\sqrt2)}{2}=\dfrac{3\sqrt2(\sqrt3-1)}{2}[/tex]
[tex]\it \mathcal{A}_{ABCD}=AB\cdot BC\cdot sin135^o =\dfrac{3\sqrt2(\sqrt3-1)}{2}\cdot3\cdot\dfrac{\sqrt2}{2} =\dfrac{9(\sqrt{3}-1)}{2}[/tex]
