kẻ [TEX]BK \bot CD, BE//AC (E \in CD)[/TEX]
ta có [TEX]DK^2=BD^2-BK^2 (pitago) \Rightarrow DK^2=15^2-12^2=81 \Rightarrow DK=9(cm)[/TEX]
tự c/m ABCE là hbh => AB=CE => AB+CD=CD+CE=DE
có [TEX]AC//BE, BD\bot AC \Rightarrow BD\bot BE \Rightarrow \hat{DBE}=90^o[/TEX]
xét [TEX]\Delta DBE (\hat{DBE}=90^o),BL\bot DE[/TEX]
[TEX]\Rightarrow BD^2=DK.DE \Rightarrow DE=25[/TEX]
[TEX]\Rightarrow S_{ABCD}=\frac{(AB+CD).AH}{2}=\frac{DE.AH}{2}= \frac{25.12}{2}=150cm^2[/TEX]