Gợi ý: Kẻ AH và CK vuông góc với BD

Lời giải:
Vận dụng bổ đề $S_{ABC}=\frac{1}{2}.AB.AC\sin A$ ta có:

$S_{ABCD}=S_{OAB}+S_{OBC}+S_{ODC}+S_{AOD}$

$=\frac{1}{2}.OA.OB.\sin \widehat{AOB}+\frac{1}{2}.OB.OC.\sin \widehat{BOC}+\frac{1}{2}.OD.OC.\sin \widehat{DOC}+\frac{1}{2}.OA.OD.\sin \widehat{AOD}$

$=\frac{1}{2}.OA.OB\sin 60^0+\frac{1}{2}.OB.OC.\sin 120^0+\frac{1}{2}.OD.OC\sin 60^0+\frac{1}{2}.OA.OD.\sin 120^0$

$=\frac{\sqrt{3}}{4}(OA.OB+OB.OC+OC.OD+OD.OA)$

$=\frac{\sqrt{3}}{4}(AC.BD)=\frac{\sqrt{3}}{4}.4.5=5\sqrt{3}$ (cm vuông)

Hình vẽ:

bạn k mình đi mình giải cho

Mik kb nha

\(S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB\cdot sinAOB=\dfrac{1}{2}\cdot OA\cdot OB\cdot sin60=\dfrac{\sqrt{3}}{4}\cdot OA\cdot OB\)

\(S_{OBC}=\dfrac{1}{2}\cdot OB\cdot OC\cdot sinBOC\)

\(=\dfrac{1}{2}\cdot OB\cdot OC\cdot sin120=\dfrac{\sqrt{3}}{4}\cdot OB\cdot OC\)

\(S_{ODC}=\dfrac{1}{2}\cdot OD\cdot OC\cdot sinDOC\)

\(=\dfrac{1}{2}\cdot OD\cdot OC\cdot sin60=\dfrac{\sqrt{3}}{4}\cdot OD\cdot OC\)

\(S_{AOD}=\dfrac{1}{2}\cdot OA\cdot OD\cdot sinAOD\)

\(=\dfrac{1}{2}\cdot OA\cdot OD\cdot sin60=\dfrac{\sqrt{3}}{4}\cdot OA\cdot OD\)

\(S_{ABCD}=S_{AOB}+S_{AOD}+S_{COD}+S_{COB}\)

\(=\dfrac{\sqrt{3}}{4}\left(OA\cdot OB+OB\cdot OC+OD\cdot OC+OD\cdot OA\right)\)

\(=\dfrac{\sqrt{3}}{4}\cdot\left(OB\cdot AC+OD\cdot AC\right)\)

\(=\dfrac{\sqrt{3}}{4}\left(AC\cdot BD\right)=\dfrac{\sqrt{3}}{4}\cdot4\cdot5=5\sqrt{3}\)