2)
[tex]\it a)\ Fie\ rombul\ ABCD,\ cu\ m(\widehat{BAC})=30^o \Rightarrow m(\widehat{BAD}) = 60^o \Rightarrow\\ \\ \Rightarrow \Delta ABD -echilateral.\\ \\ Fie\ AC\cap BD =\{O\},\ \ AO-\ \^in\breve{a}l\c{\it t}ime\ \^in\ \Delta ABD-\ echilateral\ \Rightarrow[/tex]
[tex]\it \Rightarrow AO = 6\sqrt3 cm\Rightarrow AC = 12\sqrt3cm\\ \\ b)\ \mathcal{A}_{ABCD}=2\cdot\mathcal{A}_{ABD} =2\cdot\dfrac{\ell^2\sqrt3}{4}=\dfrac{12\cdot12\sqrt3}{2}=72\sqrt3\ cm^2[/tex]
[tex]\it c)\ BM=BN=6\sqrt3cm (\^in\breve{a}l\c{\it t}imi\ \^in\ triunghiurile\ echilaterale\ congruente\ ABD,\ BCD)[/tex]
[tex]\it m(\widehat{MBN}) =60^o \Rightarrow \Delta BNM-echilateral \Rightarrow\mathcal{A}_{BNM}=\dfrac{(6\sqrt3)^2\sqrt3}{4}=27\sqrt3\ cm^2\\ \\ \dfrac{\mathcal{A}_{BNM}}{\mathcal{A}_{ABCD}}=\dfrac{27\sqrt3}{72\sqrt3}=\dfrac{3}{8}[/tex]
