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a: Ta có: AN+NC=AC
=>\(AC=\frac12\times NC+NC=\frac32\times NC\)
=>\(AN=\frac13\times AC\)
=>\(S_{ABN}=\frac13\times S_{ABC}\) (1)
ta có \(BM=\frac12\times BC\)
=>\(S_{ABM}=\frac12\times S_{ABC}\) (2)
Từ (1),(2) suy ra \(\frac{S_{ABN}}{S_{ABM}}=\frac13:\frac12=\frac23\)
b: Ta có: \(AN=\frac12\times NC\)
=>\(S_{AGN}=\frac12\times S_{GNC}\)
=>\(S_{GNC}=10\times2=20\left(\operatorname{cm}^2\right)\)
\(S_{AGC}=10+20=30\left(\operatorname{cm}^2\right)\)
\(BM=\frac12\times BC\)
=>M là trung điểm của BC
Vì MB=MC
nên \(S_{AMB}=S_{AMC};S_{GMB}=S_{GMC}\)
=>\(S_{AMB}-S_{GMB}=S_{AMC}-S_{GMC}\)
=>\(S_{AGB}=S_{AGC}=30\left(\operatorname{cm}^2\right)\)
Ta có: \(NA=\frac12\times NC\)
=>\(S_{BNA}=\frac12\times S_{BNC};S_{GNA}=\frac12\times S_{GNC}\)
=>\(S_{BNA}-S_{GNA}=\frac12\times\left(S_{BNC}-S_{GNC}\right)\)
=>\(S_{BGA}=\frac12\times S_{BGC}\)
=>\(S_{BGC}=\frac{30}{2}=15\left(\operatorname{cm}^2\right)\)
\(S_{ABC}=S_{AGB}+S_{AGC}+S_{BGC}\)
\(=30+30+15=75\left(\operatorname{cm}^2\right)\)
Ta có: BM=2MC
=>\(S_{AMB}=2\times S_{AMC};S_{OMB}=2\times S_{OMC}\)
=>\(S_{AMB}-S_{OMB}=2\times\left(S_{AOC}-S_{MOC}\right)\)
=>\(S_{AOB}=2\times S_{AOC}\)
=>\(S_{AOC}=\frac{40}{2}=20\left(\operatorname{cm}^2\right)\)
Ta có: CN=3NA
=>\(S_{BNC}=3\times S_{BNA};S_{ONC}=3\times S_{ONA}\)
=>\(S_{BNC}-S_{ONC}=3\times\left(S_{BNA}-S_{NOA}\right)\)
=>\(S_{BOC}=3\times S_{BOA}=3\times40=120\left(\operatorname{cm}^2\right)\)
Diện tích tam giác ABC là:
\(S_{ABC}=S_{OAB}+S_{OAC}+S_{OBC}\)
\(=20+40+120=60+120=180\left(\operatorname{cm}^2\right)\)
Bài 2:
\(\dfrac{S_{ABM}}{S_{ABC}}=\dfrac{8}{12}=\dfrac{2}{3}\)
=>\(\dfrac{BM}{BC}=\dfrac{2}{3}\)
=>\(BM=\dfrac{2}{3}\cdot BC=\dfrac{2}{3}\cdot24=16\left(cm\right)\)
Ta có: BM+MC=BC
=>MC+16=24
=>MC=8(cm)
18cm.Mk tính thế nhưng ko bik đúng ko.Vả lại mk ko bik vẽ hình
a: \(AM=\frac23MC\)
=>\(MC=\frac32AM\)
AM+MC=AC
=>\(AC=AM+\frac32AM=\frac52AM\)
=>\(AM=\frac25AC\)
=>\(S_{ABM}=\frac25\times S_{ABC}=\frac25\times60=24\left(\operatorname{cm}^2\right)\)
b: Ta có: BN+NC=BC
=>\(BC=BN+\frac13BN=\frac43BN\)
=>\(CN=\frac14CB;BN=\frac34BC\)
TA có: \(CM=\frac35\times CA\)
=>\(S_{BMC}=\frac35\times S_{ABC}\)
\(CN=\frac14CB\)
=>\(S_{MNC}=\frac14\times S_{BMC}=\frac14\times\frac35\times S_{ABC}=\frac{3}{20}\times S_{ABC}\)
TA có: \(S_{AMNB}+S_{MNC}=S_{ABC}\)
=>\(S_{AMNB}=S_{ABC}-\frac{3}{20}\times S_{ABC}=\frac{17}{20}\times S_{ABC}=\frac{17}{20}\times60=51\left(\operatorname{cm}^2\right)\)
c: Ta có: \(AM=\frac23\times MC\)
=>\(S_{BMA}=\frac23\times S_{BMC};S_{OMA}=\frac23\times S_{OMC}\)
=>\(S_{BMA}-S_{OMA}=\frac23\times\left(S_{BMC}-S_{OMC}\right)\)
=>\(S_{BOA}=\frac23\times S_{BOC}\)
=>\(S_{BOC}=\frac32\times S_{BOA}\)
Ta có: \(NC=\frac13\times NB\)
=>\(S_{ANC}=\frac13\times S_{ANB};S_{ONC}=\frac13\times S_{ONB}\)
=>\(S_{ANC}-S_{ONC}=\frac13\times\left(S_{ANB}-S_{ONB}\right)\)
=>\(S_{AOC}=\frac13\times S_{AOB}=\frac29\times S_{BOC}\)
Ta có: \(S_{BOC}+S_{AOC}+S_{AOB}=S_{ABC}\)
=>\(S_{ABC}=S_{OAB}+\frac32\times S_{OAB}+\frac13\times S_{OAB}=S_{OAB}\left(\frac43+\frac32\right)=S_{AOB}\cdot\frac{11}{6}\)
=>\(S_{AOB}=\frac{6}{11}\cdot S_{ABC}=\frac{6}{11}\cdot60=\frac{360}{11}\left(\operatorname{cm}^2\right)\)
=>\(\frac{S_{AOB}}{S_{ABN}}=\frac{\frac{6}{11}\cdot S_{ABC}}{\frac34\cdot S_{ABC}}=\frac{6}{11}:\frac34=\frac{6}{11}\times\frac43=\frac{24}{33}=\frac{8}{11}\)
=>\(\frac{AO}{AN}=\frac{8}{11}\)
=>\(\frac{AO}{ON}=\frac83\)