Đáp án:
Câu 3:
$+)$
$ΔMNP$ có $MD$ là phân giác $\widehat{NMP}$
$→\dfrac{DP}{DN}=\dfrac{MP}{MN}=\dfrac{8}{6}=\dfrac{3}{4}$
$ΔMDP$ và $ΔMDN$ có chung đường cao
$→\dfrac{S_{MDP}}{S_{MDN}}=\dfrac{3}{4}$
$+)$
$S_{MNP}=\dfrac{6.8}{2}=24(cm^2)$
$→S_{MDP}+S_{MDN}=S_{MNP}=24(cm^2)$
Vì $\dfrac{S_{MDP}}{S_{MDN}}=\dfrac{3}{4}\to S_{MDP}=\dfrac{3S_{MDN}}{4} $
\[\to \dfrac{3S_{MDN}}{4}+S_{MDN}=24(cm^2)\] \[\to \dfrac{7}{4}S_{MDN}=24(cm^2) \] \[\to S_{MDN}=\dfrac{96}{7}(cm^2)\]
