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

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