Đáp án:
$S_{ABCD}=\dfrac {7\sqrt {3}}{2}$
Lời giải:
Kẻ $DH\bot AB$
$\Rightarrow \sin\widehat {A} = \dfrac {DH}{AD}$
$\Rightarrow \sin60°=\dfrac {DH} {2} \Rightarrow DH =\sqrt {3}$
$\cos A=\dfrac {AH}{AD}$
$\Rightarrow AH = \cos60°. 2=1$
$\Rightarrow DC = AB - 1-1=4,5-2=2,5$
$S_{ABCD} = \dfrac12.\sqrt {3}.(4,5+2,5)$
$=\dfrac {7\sqrt {3}}{2}$