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Link https://olm.vn/hoi-dap/tim-kiem?q=Cho+h%C3%ACnh+thang+ABCD+c%C3%B3+di%E1%BB%87n+t%C3%ADch+l%C3%A0+360+cm2.Tr%C3%AAn+AB+l%E1%BA%A5y+%C4%91i%E1%BB%83m+M+sao+cho+MB=2MA,tr%C3%AAn+BC+l%E1%BA%A5y+%C4%91i%E1%BB%83m+N+sao+cho+NB=2NC,tr%C3%AAn+CD+l%E1%BA%A5y+%C4%91i%E1%BB%83m+P+sao+cho+PD=2PC+v%C3%A0+tr%C3%AAn+AD+l%E1%BA%A5y+%C4%91i%E1%BB%83m+Q+sao+cho+QD=2QA.N%E1%BB%91i+M,N,P,Q.T%C3%ACm+di%E1%BB%87n+t%C3%ADch+t%E1%BB%A9+gi%C3%A1c+MNPQ.&id=868653
Ta có: BN+NC=BC
=>BC=2NC+NC=3NC
=>\(BN=\frac23\times BC\) ; \(CN=\frac13\times CB\)
Ta có: MB+MA=AB
=>AB=2MA+MA=3MA
=>\(AM=\frac13\times AB;BM=\frac23\times BA\)
TA có: CP+PD=CD
=>CD=2PC+PC=3CP
=>\(CP=\frac13\times CD;DP=\frac23\times CD\)
TA có: DQ+QA=DA
=>DA=2QA+QA=3AQ
=>\(AQ=\frac13\times AD;DQ=\frac23\times DA\)
Ta có: \(BN=\frac23\times BC\)
=>\(S_{ABN}=\frac23\times S_{ABC}\)
Ta có: \(BM=\frac23\times BA\)
=>\(S_{BMN}=\frac23\times S_{ABN}=\frac49\times S_{ABC}\)
TA có: \(DP=\frac23\times DC\)
=>\(S_{DAP}=\frac23\times S_{DAC}\)
Ta có: \(DQ=\frac23\times DA\)
=>\(S_{DQP}=\frac23\times S_{DQC}=\frac23\times\frac23\times S_{DAC}=\frac49\times S_{DAC}\)
TA có: \(AM=\frac13\times AB\)
=>\(S_{DMA}=\frac13\times S_{DAB}\)
\(AQ=\frac13\times AD\)
=>\(S_{AMQ}=\frac13\times S_{AMD}=\frac19\times S_{ABD}\)
TA có:\(CN=\frac13\times CB\)
=>\(S_{DNC}=\frac13\times S_{DBC}\)
Ta có: \(CP=\frac13\times CD\)
=>\(S_{CPN}=\frac13\times S_{DNC}=\frac19\times S_{DBC}\)
Ta có: \(S_{BMN}+S_{DQP}+S_{AQM}+S_{NCP}+S_{MNPQ}=S_{ABCD}\)
=>\(\frac19\times\left(S_{ABD}+S_{BDC}\right)+\frac49\times\left(S_{BAC}+S_{DAC}\right)+S_{MNPQ}=S_{ABCD}\)
=>\(S_{MNPQ}=\frac49\times S_{BADC}=\frac49\times360=160\left(\operatorname{cm}^2\right)\)
Ta có: BN+NC=BC
=>BC=2NC+NC=3NC
=>\(BN=\frac23BC\) và \(CN=\frac13\times BC\)
M là trung điểm của AB
=>\(S_{CBM}=\frac12\times S_{CBA}\)
Vì \(BN=\frac23\times BC\)
nên \(S_{MBN}=\frac23\times S_{MBC}=\frac23\times\frac12\times S_{CBA}=\frac13\times S_{CBA}\)
Ta có: AQ+QD=AD
=>AD=2QA+QA=3QA
=>\(\frac{DQ}{DA}=\frac23\) và \(AQ=\frac13\times AD\)
Ta có: P là trung điểm của CD
=>\(S_{APD}=\frac12\times S_{ADC}\)
\(DQ=\frac23\times DA\) nên \(S_{DQP}=\frac23\times S_{DPA}=\frac23\times\frac12\times S_{ADC}=\frac13\times S_{ADC}\)
=>\(S_{MBN}+S_{QDP}=\frac13\times\left(S_{BAC}+S_{DAC}\right)=\frac13\times S_{ABCD}\)
M là trung điểm của AB
=>\(S_{DMA}=\frac12\times S_{DBA}\)
Vì \(AQ=\frac13\times AD\) nên \(S_{AQM}=\frac13\times S_{AMD}=\frac13\times\frac12\times S_{DBA}=\frac16\times S_{DBA}\)
Vì P là trung điểm của CD
nên \(S_{BPC}=\frac12\times S_{BDC}\)
Vì \(CN=\frac13\times CB\)
nên \(S_{PNC}=\frac13\times S_{BPC}=\frac13\times\frac12\times S_{BDC}=\frac16\times S_{BDC}\)
=>\(S_{AQM}+S_{CNP}=\frac16\times\left(S_{ABD}+S_{BDC}\right)=\frac16\times S_{ABCD}\)
Ta có: \(S_{AQM}+S_{CPN}+S_{BMN}+S_{DQP}+S_{MNPQ}=S_{ABCD}\)
=>\(S_{ABCD}\left(\frac13+\frac16\right)+S_{MNPQ}=S_{ABCD}\)
=>\(S_{MNPQ}=S_{ABCD}-\frac12\times S_{ABCD}=\frac12\times S_{ABCD}=\frac{120}{2}=60\left(\operatorname{cm}^2\right)\)