Tứ giác ABCD. S.ADC = 30cm2 S.BDC = 21 cm2 AM = 1/3 AB Tính S.MCD

kẻ AI, MN, BH lần lượt vuông góc với DC
⟹ AI//MN//BH
có: $\displaystyle \frac{S_{ACD}}{S_{BCD}} =\frac{AI}{BH} =\frac{30}{21} =\frac{10}{7}$
trong tam giác KAI có: AI//BH
⟹$\displaystyle \frac{AI}{BH} =\frac{KA}{KB} =\frac{10}{7} \Longrightarrow \frac{AB}{KA} =\frac{3}{10} \Longrightarrow \frac{AM}{KA} =\frac{1}{10}\left( \ vì\ AM=\frac{1}{3} AB\right)$
⟹$\displaystyle \frac{KM}{KA} =\frac{MN}{AI} =\frac{9}{10}$
⟹$\displaystyle \frac{S_{ACD}}{S_{MCD}} =\frac{AI}{MN} =\frac{10}{9} \Longrightarrow S_{MCD} =\frac{9}{10} .30=27cm^{2}$