Cho tứ giác ABCD có AC=8, BD=10, \(\widehat{AOB}=3...

A

- Hướng dẫn giải

Ta có: \(AH\perp BD,CK\perp BD\) và \(CK=OC.sin30^{\circ}, AH=BD.sin30^{\circ}\)

\(S_{ABCD}=S_{ABD}+S_{BCD}=\frac{1}{2}.AH.BD+\frac{1}{2}.CK.BD=\frac{1}{2}.BD.(AH+CK)\)

mà \(AH+CK=(AO+CO).sin30^{\circ}=AC.sin30^{\circ}\)

\(\Rightarrow S_{ABCD}=\frac{1}{2}.BD.AC.sin30^{\circ}=\frac{1}{2}.BD.AC.\frac{1}{2}=20\)