a: CD=2,2-0,4=1,8(m)
Chiều cao là (2,2+1,8):2=2(m)
Diện tích hình thang là \(S_{ABCD}=\dfrac{1}{2}\cdot2\left(2,2+1,8\right)=4\left(m^2\right)\)
b: Vì ABCD là hình thang nên \(\dfrac{S_{ABC}}{S_{ADC}}=\dfrac{AB}{DC}=\dfrac{2.2}{1.8}=\dfrac{11}{9}\)
=>\(S_{ADC}=\dfrac{9}{11}\cdot S_{ABC}\)
Ta có: \(S_{ADC}+S_{ABC}=S_{ABCD}\)
=>\(\dfrac{9}{11}\cdot S_{ABC}+S_{ABC}=4\)
=>\(S_{ABC}=4:\dfrac{20}{11}=4\cdot\dfrac{11}{20}=\dfrac{44}{20}=2,2\left(m^2\right)\)
c: \(S_{ADC}=\dfrac{9}{11}\cdot2,2=1,8\left(m^2\right)\)