Lời giải:
Có:
$\frac{S_{MNC}}{S_{BNC}}=\frac{MC}{BC}=\frac{1}{2}$ (do $M$ là trung điểm $BC$)
$\Rightarrow S_{MNC}=\frac{1}{2}\times S_{BNC}$
$\frac{S_{BNC}}{S_{BAC}}=\frac{NC}{AC}=\frac{1}{3}$
$\Rightarrow S_{BNC}=\frac{1}{3}\times S_{BAC}$
Suy ra:
$S_{MNC}=\frac{1}{2}S_{BNC}=\frac{1}{2}\times \frac{1}{3}S_{ABC}=\frac{1}{6}\times S_{ABC}$
$S_{ABC}=6\times S_{MNC}=6\times 4=24$ (cm2)