Question

Answer 1

Ta có: \(\frac{DO}{BO}=\frac{8}{16}=\frac12\)

=>\(\frac{S_{AOD}}{S_{AOB}}=\frac12\)

=>\(S_{AOD}=\frac{10}{2}=5\left(\operatorname{cm}^2\right)\)

Ta có: \(\frac{OA}{OC}=\frac{S_{BOA}}{S_{BOC}}\)

=>\(\frac{OA}{OC}=\frac{10}{8}=\frac54\)

=>\(S_{AOD}=\frac54\times S_{DOC}\)

=>\(S_{DOC}=5:\frac54=4\left(\operatorname{cm}^2\right)\)

\(S_{ABCD}=S_{OBA}+S_{OAD}+S_{OBC}+S_{ODC}\)

\(=10+8+5+4=18+9=27\left(\operatorname{cm}^2\right)\)