[tex]m(\ \textless \ C)=30^0 ;
in tr ADC cu T\ \textless \ 30^0, AC=12[/tex]
cu T>P avem [tex]DC=6\sqrt{3}[/tex]
In tr ABD
[tex]sin 60^0=\frac{AD}{AB}=\frac{\sqrt{3}}{2}=\frac{6}{AB}\\
AB=4\sqrt{3}\\
BC=8\sqrt{3} (T\ \textless \ 30^0)[/tex]
[tex]A=\frac{AB\cdot AC}{2}=24\sqrt{3}\\
P=12\sqrt{3}+12=12(\sqrt{3}+1)\\
[/tex]
Tr CED este asemenea cu trCAB raportul ariilor fiind egal cu patratul raportului de asemanare care este DC/BC=3/4
[tex]\frac{A_{CED}}{A_{abc}}=\frac{9}{16}\\
\frac{A_{ABC}-A_{CED}}{24\sqrt{3}}=\frac{16-9}{16}\\
\frac{A_{AEDB}}{24\sqrt{3}}=\frac{7}{16}\\
A_{AEDB}=\frac{21\sqrt{3}}{2}[/tex]
