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a: Ta có: \(AN=2\times NC\)
=>\(S_{PNA}=2\times S_{PNC};S_{MNA}=2\times S_{MNC}\)
=>\(S_{PNA}-S_{MNA}=2\times\left(S_{PNC}-S_{MNC}\right)\)
=>\(S_{PMA}=2\times S_{PMC}\)
Ta có: \(CM=3\times MB\)
=>\(S_{PMC}=3\times S_{PMB}\)
=>\(S_{PMA}=2\times3\times S_{PMB}=6\times S_{PMB}\)
=>PA=6BP
PB+BA=PA
=>BA=PA-PB=6BP-BP=5BP
=>5BP=6
=>BP=1,2(cm)
PA=PB+BA=1,2+6=7,2(cm)
b: Ta có: \(BM+MC=BC\)
=>BC=3MB+MB=4MB
=>\(S_{ABC}=4\times S_{AMB}\)
TA có: AN+NC=AC
=>AC=2NC+NC=3NC
=>\(AN=\frac23\times AC\)
=>\(S_{AMN}=\frac23\times S_{AMC}\)
Ta có: CM=3/4CB
=>\(S_{AMC}=\frac34\times S_{ABC}\)
=>\(S_{AMN}=\frac34\times\frac23\times S_{ABC}=\frac12\times S_{ABC}\)
PA=1,2AB
=>\(S_{PMA}=\frac65\times S_{BAM}=\frac65\times\frac14\times S_{ABC}=\frac{3}{10}\times S_{ABC}\)
=>\(\frac{S_{PMA}}{S_{AMN}}=\frac{3}{10}:\frac12=\frac{3}{10}\times2=\frac35\)
=>MP=3/5MN
a: Ta có: \(NA=2\times NC\)
=>\(S_{PNA}=2\times S_{PNC};S_{MNA}=2\times S_{MNC}\)
=>\(S_{PNA}-S_{MNA}=2\times\left(S_{PNC}-S_{MNC}\right)\)
=>\(S_{PMA}=2\times S_{PMC}\)
Ta có: CM=3xMB
=>\(S_{PMC}=3\times S_{PBM}\)
=>\(S_{PMA}=2\times3\times S_{PBM}=6\times S_{PBM}\)
=>PA=6BP
Ta có; BP+BA=AP
=>BA=AP-PB=6BP-BP=5BP
=>5BP=6
=>BP=1,2(cm)
AP=AB+BP=6+1,2=7,2(cm)
b: Vì AB/AP=6/7,2=5/6
nên \(S_{ABM}=\frac56\times S_{AMP}\)
BM+MC=BC
=>BC=3MB+MB=4MB
=>\(S_{ABC}=4\times S_{ABM}\)
Vì BM=1/3MC
nên \(S_{ABM}=\frac13\times S_{AMC}\)
Ta có: AN+NC=AC
=>AN=2/3AC
=>\(S_{AMN}=\frac23\times S_{AMC}\)
=>\(\frac{S_{AMN}}{S_{AMB}}=\frac23:\frac13=2\)
=>\(S_{AMN}=2\times S_{AMB}=2\times\frac56\times S_{AMP}=\frac53\times S_{AMP}\)
=>MN=5/3MP
a: Ta có: \(AN=2\times NC\)
=>\(S_{PNA}=2\times S_{PNC};S_{MNA}=2\times S_{MNC}\)
=>\(S_{PNA}-S_{MNA}=2\times\left(S_{PNC}-S_{MNC}\right)\)
=>\(S_{PMA}=2\times S_{PMC}\)
Ta có: \(CM=3\times MB\)
=>\(S_{PMC}=3\times S_{PMB}\)
=>\(S_{PMA}=2\times3\times S_{PMB}=6\times S_{PMB}\)
=>PA=6BP
PB+BA=PA
=>BA=PA-PB=6BP-BP=5BP
=>5BP=6
=>BP=1,2(cm)
PA=PB+BA=1,2+6=7,2(cm)
b: Ta có: \(BM+MC=BC\)
=>BC=3MB+MB=4MB
=>\(S_{ABC}=4\times S_{AMB}\)
TA có: AN+NC=AC
=>AC=2NC+NC=3NC
=>\(AN=\frac23\times AC\)
=>\(S_{AMN}=\frac23\times S_{AMC}\)
Ta có: CM=3/4CB
=>\(S_{AMC}=\frac34\times S_{ABC}\)
=>\(S_{AMN}=\frac34\times\frac23\times S_{ABC}=\frac12\times S_{ABC}\)
PA=1,2AB
=>\(S_{PMA}=\frac65\times S_{BAM}=\frac65\times\frac14\times S_{ABC}=\frac{3}{10}\times S_{ABC}\)
=>\(\frac{S_{PMA}}{S_{AMN}}=\frac{3}{10}:\frac12=\frac{3}{10}\times2=\frac35\)
=>MP=3/5MN
Ta có SAMN = SCMN (AN =NC và chung đường cao)
Diện tích tam giác AMC: 7 x 2 = 14 (cm2)
Diện tích tam giác BMC: 14 x 2 = 28 (cm2) (BM gấp đôi AM cung đường cao kẻ từ C)
Diện tích hình tứ giác BCNM: 28 + 7 = 35 (cm2) (SBCNM=SNMC+SMBC)
Đáp số: 35 cm2.
LINK:
https://olm.vn/hoi-dap/tim-kiem?q=cho+tam+gi%C3%A1c+ABC+,+tr%C3%AAn+AB++l%E1%BA%A5y+%C4%91i%E1%BB%83m+M+sao+cho+AM+=+MB+.tr%C3%AAn+AC+l%E1%BA%A5y+%C4%91i%E1%BB%83m+N+sao+cho+AN+=+2+NC+.+%C4%90%C6%B0%E1%BB%9Dng+th%E1%BA%B3ng+MN+c%E1%BA%AFt+BC+k%C3%A9o+d%C3%A0i+t%E1%BA%A1i+D.+ch%E1%BB%A9ng+t%E1%BB%8F+r%E1%BA%B1ng+BC+=CD&id=248698
a: Ta có: \(AN=2\times NC\)
=>\(S_{PNA}=2\times S_{PNC};S_{MNA}=2\times S_{MNC}\)
=>\(S_{PNA}-S_{MNA}=2\times\left(S_{PNC}-S_{MNC}\right)\)
=>\(S_{PMA}=2\times S_{PMC}\)
Ta có: \(CM=3\times MB\)
=>\(S_{PMC}=3\times S_{PMB}\)
=>\(S_{PMA}=2\times3\times S_{PMB}=6\times S_{PMB}\)
=>PA=6BP
PB+BA=PA
=>BA=PA-PB=6BP-BP=5BP
=>5BP=6
=>BP=1,2(cm)
PA=PB+BA=1,2+6=7,2(cm)
b: Ta có: \(BM+MC=BC\)
=>BC=3MB+MB=4MB
=>\(S_{ABC}=4\times S_{AMB}\)
TA có: AN+NC=AC
=>AC=2NC+NC=3NC
=>\(AN=\frac23\times AC\)
=>\(S_{AMN}=\frac23\times S_{AMC}\)
Ta có: CM=3/4CB
=>\(S_{AMC}=\frac34\times S_{ABC}\)
=>\(S_{AMN}=\frac34\times\frac23\times S_{ABC}=\frac12\times S_{ABC}\)
PA=1,2AB
=>\(S_{PMA}=\frac65\times S_{BAM}=\frac65\times\frac14\times S_{ABC}=\frac{3}{10}\times S_{ABC}\)
=>\(\frac{S_{PMA}}{S_{AMN}}=\frac{3}{10}:\frac12=\frac{3}{10}\times2=\frac35\)
=>MP=3/5MN