a: TA có: BM+MC=BC
=>\(CM=BC-BM=BC-\frac12BC=\frac12BC\)
Ta có: AN+NC=AC
=>\(AN=AC-CN=AC-\frac13AC=\frac23AC\)
Ta có: AP+PB=AB
=>\(PB=AB-AP=AB-\frac14AB=\frac34AB\)
Ta có: \(AN=\frac23AC\)
=>\(S_{ABN}=\frac23\cdot S_{ABC}\)
Ta có: \(AP=\frac14\cdot AB\)
=>\(S_{ANP}=\frac14\cdot S_{ANB}=\frac14\cdot\frac23\cdot S_{ABC}=\frac16\cdot S_{ABC}\)
Ta có: \(BM=\frac12BC\)
=>\(S_{AMB}=\frac12\cdot S_{ABC}\)
Ta có: \(BP=\frac34\cdot BA\)
=>\(S_{BPM}=\frac34\cdot S_{AMB}=\frac34\cdot\frac12\cdot S_{ABC}=\frac38\cdot S_{ABC}\)
Ta có: \(CM=\frac12\cdot CB\)
=>\(S_{AMC}=\frac12\cdot S_{ABC}\)
Ta có: \(CN=\frac13\cdot CA\)
=>\(S_{CMN}=\frac13\cdot S_{CMA}=\frac13\cdot\frac12\cdot S_{ABC}=\frac16\cdot S_{ABC}\)
\(\frac{S_{ANP}}{S_{ABC}}+\frac{S_{BPM}}{S_{ABC}}+\frac{S_{CNM}}{S_{ABC}}\)
\(=\frac16+\frac38+\frac16=\frac38+\frac13=\frac{9+8}{24}=\frac{17}{24}\)
b: \(S_{ANP}+S_{BPM}+S_{CNM}+S_{MNP}=S_{ABC}\)
=>\(S_{MNP}=S_{ABC}\cdot\left(1-\frac{17}{24}\right)=\frac{7}{24}\cdot S_{ABC}\)
=>\(S_{MNP}=\frac{7}{24}\cdot10=\frac{70}{24}=\frac{35}{12}\left(\operatorname{cm}^2\right)\)
