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: \(NA=\frac12\times NC\)

=>\(S_{BNA}=\frac12\times S_{BNC};S_{GNA}=\frac12\times S_{GNC}\)

=>\(S_{BNA}-S_{GNA}=\frac12\times\left(S_{BNC}-S_{GNC}\right)\)

=>\(S_{BGA}=\frac12\times S_{BGC}\)

=>\(S_{BGC}=\frac{30}{2}=15\left(\operatorname{cm}^2\right)\)

\(S_{ABC}=S_{AGB}+S_{AGC}+S_{BGC}\)

\(=30+30+15=75\left(\operatorname{cm}^2\right)\)

Nhanh lên mk k cho 2k các bạn ơi!!^^6

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