Question

Problema 3 vă rog.Dau coroană celui mai bun răspuns și puncte!!​

Answer 1

$\it a)\ \mathcal{A}_{ABD}=\dfrac{ \mathcal{A}_{ABCD}}{2}=\dfrac{\ell^2}{2}=\dfrac{6^2}{2}=\dfrac{36}{2}=18\ cm^2\\ \\ \\ BD=\ell\sqrt2=6\sqrt2\ cm \ \ (diagonala\ p\breve atratului\ ABCD)$

$\it \mathcal{A}_ {BFED} = (6\sqrt2)^2=36\cdot2=72\ cm^2\\ \\ \mathcal{A}_ {ABFED} = \mathcal{A}_ {ABD} + \mathcal{A}_ {BFED} ==18+72=90\ cm^2$

$\it b)\ \ AC=BF=6\sqrt2\ cm\ \ \ \ (1)\\ \\ \widehat{CAB}=45^o;\ \ \ \widehat{ABF}=45^o+45^o+45^o=135^o\\ \\ \widehat{CAB}+\widehat{ABF}=180^o \Rightarrow AC||BF\ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow ABFC\ -\ paralelogram$