Đáp án: a.$S_{ABCD}=12$ $b.S_{ADM}=3$ $d.S_{AMN}=1$
Giải thích các bước giải:
a.Gọi $AE\perp CD=E$
Ta có:
$S_{ABCD}=2S_{ACD}=2.\dfrac12AE.CD=AE.CD=4.3=12$
b.Vì M là trung điểm AB
$\to S_{ADM}=\dfrac12S_{ADB}=\dfrac12.\dfrac12S_{ABCD}=\dfrac14S_{ABCD}=3$
c.Ta có : $AB//CD\to \dfrac{DN}{MN}=\dfrac{DC}{AM}=\dfrac{AB}{AM}=\dfrac{2AM}{AM}=2$
$\to DN=2NM$
d.Ta có :$DN=2NM\to \dfrac{MN}{DN}=\dfrac12\to\dfrac{MN}{MN+DN}=\dfrac1{1+2}$
$\to\dfrac{MN}{DM}=\dfrac13$
$\to \dfrac{S_{AMN}}{S_{ADM}}=\dfrac13$
$\to S_{AMN}=\dfrac13S_{ADM}=1$