[tex]\it \mathcal{A}_{ABCD}= \mathcal{A}_{ABD}+ \mathcal{A}_{BCD}\\ \\ \\ \mathcal{A}_{ABD}=\dfrac{\ell^2\sqrt3}{4}=\dfrac{4^2\sqrt3}{4}=\dfrac{4\cdot4\sqrt3}{4}=4\sqrt3\ cm^2\\ \\ \\ \Delta ABD-echilateral \Rightarrow \widehat{ABD}=60^o\\ \\ \\ \widehat{DBC}=\widehat{ABC}-\widehat{ABD}=150^o-60^o=90^o \Rightarrow \Delta DBC-dreptunghic[/tex]
[tex]\it \mathcal{A}_{BCD}=\dfrac{c_1\cdot c_2}{2}=\dfrac{BD\cdot BC}{2}=\dfrac{4\cdot4}{2}=8\ cm^2\\ \\ \\ \mathcal{A}_{ABCD}=4\sqrt3+8\approx4\cdot1,7+8 \Rightarrow \mathcal{A}_{ABCD} \approx 6,8+8 \Rightarrow \mathcal{A}_{ABCD} \approx\ 14,8\ cm^2[/tex]
