[tex]AB=\sqrt{HB.BD}=\sqrt{HB.(HB+HD)}=\sqrt{8.(8+18)}=4\sqrt{13}[/tex]
[tex]AD=\sqrt{DH.DB}=\sqrt{DH.(DH+HB)}=\sqrt{18.(18+8)}=4\sqrt{11}[/tex]
Ta có:[tex]\frac{AB}{CD}=\frac{BH}{HD}[/tex] =>[tex]CD=\frac{AB.HD}{HB}=\frac{4\sqrt{13}.18}{8}=9\sqrt{13}[/tex]
[tex]S_{ABCD}=\frac{(AB+CD).AD}{2}=\frac{(9\sqrt{13}+4\sqrt{13}).4\sqrt{11}}{2}=338\sqrt{11}[/tex]