bien tích ABC = 27 cm2

BOM = CON

( Vẽ hình ra sẽ thấy nhé)

bạn ghi các giải giúp tớ nhé

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

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

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

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

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

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

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

27cm2

bằng 27 cm2

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: \(S_{BOC}+S_{AOC}+S_{AOB}=S_{ABC}\)

=>\(S_{ABC}=S_{OAB}+\frac32\times S_{OAB}+\frac13\times S_{OAB}=S_{OAB}\left(\frac43+\frac32\right)=S_{AOB}\cdot\frac{11}{6}\)

=>\(S_{AOB}=\frac{6}{11}\cdot S_{ABC}=\frac{6}{11}\cdot60=\frac{360}{11}\left(\operatorname{cm}^2\right)\)

=>\(\frac{S_{AOB}}{S_{ABN}}=\frac{\frac{6}{11}\cdot S_{ABC}}{\frac34\cdot S_{ABC}}=\frac{6}{11}:\frac34=\frac{6}{11}\times\frac43=\frac{24}{33}=\frac{8}{11}\)

=>\(\frac{AO}{AN}=\frac{8}{11}\)

=>\(\frac{AO}{ON}=\frac83\)

453

tick nha

\(\)Tính tỉ số \(\dfrac{S_{AOB}}{S_{AOC}}\)

\(\dfrac{S_{AOB}}{S_{AOC}}=\dfrac{h\cdot AO\div2}{h\cdot OC\div2}=\dfrac{AC}{OC}\)

S(AKN) = 1/3S(NKC)
=>S(ABK) = 1/3S(BCK)
Mà S(AKM) = S(BKM) = 1/2 S(ABK) = 1/2 x 1/3=1/6S(BKC)
Mặt khác:

S(AKM) = S(BKM)

=> S(AKC) = SBKC) 
=> S(BKC) = 42 :(1+1+6+6) x 6 =18 cm²