a) Ta có: $S_{ABCD}=\dfrac12AC.BD$
$\to S_{ABCD}=\dfrac12\cdot a\cdot 2a = a^2$
b) Từ $A$ kẻ đường cao $AH$
Ta có: $S_{ABCD}=AH.CD$
$\quad S_{BCE}=\dfrac12AH.CE =\dfrac12AH.CD$
$\to S_{BCE}=\dfrac12S_{ABCD}$
$\to S_{ABED}=S_{ABCD}+S_{BCE}=\dfrac32S_{ABCD}$
$\to S_{ABED}=\dfrac32a^2$