Sửa đề: BC=6cm
a, Ta có: \(CN=\dfrac{1}{3}CD=\dfrac{1}{3}.9=3\left(cm\right)\)
\(\Rightarrow S_{ABCN}=\dfrac{\left(CN+AB\right).BC}{2}=\dfrac{\left(3+9\right).6}{2}=\dfrac{12.6}{2}=36\left(cm^2\right)\)
b, Ta có: \(BM=CM=\dfrac{1}{2}CB=\dfrac{1}{2}.6=3\left(cm\right)\)
\(\Rightarrow S_{MCN}=\dfrac{3.3}{2}=\dfrac{9}{2}=4,5\left(cm^2\right)\)
c, Ta có: \(BM=3\left(cm\right)\left(cmt\right)\)
\(\Rightarrow S_{ABM}=\dfrac{9.3}{2}=\dfrac{27}{2}=13,5\left(cm^2\right)\)
Mặt khác: \(S_{ABCD}=9.6=54\left(cm^2\right)\)
\(\Rightarrow S_{AMND}=S_{ABCD}-S_{ABM}-S_{MCN}\)
\(\Rightarrow S_{AMND}=54-13,5-4,5=36\left(cm^2\right)\)
Chúc bạn học tốt!!!