[tex]a)~A_{ABCD}=2 \cdot A_{ABC}=2 \cdot \frac{AB \cdot BC \cdot sin( \angle ABC)}{2}=AB \cdot BC \cdot sin(60\textdegree) = \\ \\ =8 \cdot 5 \sqrt{3} \cdot \frac{\sqrt{3}}{2}=60~(cm^2). [/tex]
[tex]b)~A_{ABCD}= 4 \cdot A_{BOC}=4 \cdot 13 =52 ~(cm^2). \\ \\ c)~m( \angle BAD)=60 \textdegree ~si~AB=AD \Rightarrow \Delta ABD-echilateral \Rightarrow \\ \Rightarrow AB=AD=BD=16~cm. \\ \\ A_{ABCD}= 2 \cdot A_{ABD}=2 \cdot \frac{AB^2 \sqrt{3}}{4}= \frac{16^2 \sqrt{3}}{2}= 128~ \sqrt{3}~(cm^2).[/tex]
