ABCD là hình thang có AB//CD
nên \(\dfrac{OA}{OC}=\dfrac{OB}{OD}=\dfrac{AB}{CD}=\dfrac{1}{3}\)
Vì OA/OC=1/3 nên \(\dfrac{S_{AOD}}{S_{DOC}}=\dfrac{1}{3}\)
=>\(S_{AOD}=\dfrac{30}{3}=10\left(cm^2\right)\)
Vì OB/OD=1/3 nên \(\dfrac{S_{AOB}}{S_{AOD}}=\dfrac{1}{3}\)
=>\(S_{AOB}=\dfrac{10}{3}\left(cm^2\right)\)
Vì OA/OC=1/3 nên \(\dfrac{S_{AOB}}{S_{BOC}}=\dfrac{1}{3}\)
=>\(S_{BOC}=10\left(cm^2\right)\)
\(S_{ABCD}=S_{AOB}+S_{BOC}+S_{COD}+S_{AOD}\)
\(=10+10+\dfrac{10}{3}+30=50+\dfrac{10}{3}=\dfrac{160}{3}\left(cm^2\right)\)
Đúng 1
Bình luận (0)
