Kẻ $BH \perp CD$
$ABHD$ có $\widehat{A}=\widehat{D}=\widehat{BHD}=90^o$
$=>ABHD$ là hình chữ nhật
$=>DH=AB=2(cm);BH=AD=\sqrt{3}(cm);\widehat{ABH}=90^o\\ \widehat{HBC}=\widehat{ABC}-\widehat{ABH}=150^o-90^o=60^o\\ CH=BH\tan 60^o=3(cm)\\ =>CD=CH+HD=5(cm)\\ S_{ABCD}=\dfrac{(2+5)\sqrt{3}}{2}=\dfrac{7\sqrt{3}}{2}$