Ta có: \(BM=\frac13BC\)
=>\(S_{ABM}=\frac13\cdot S_{ABC}\)
Ta có: \(CN=\frac14BC\)
=>\(S_{ANC}=\frac14\cdot S_{ABC}\)
Ta có: \(S_{ABM}+S_{AMN}+S_{ANC}=S_{ABC}\)
=>\(S_{AMN}=S_{ABC}\left(1-\frac13-\frac14\right)=S_{ABC}\cdot\frac{5}{12}\)
=>\(S_{ABC}=50:\frac{5}{12}=120\left(\operatorname{cm}^2\right)\)