Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: AP+PC=AC
=>\(AC=\frac12PC+PC=\frac32PC\)
=>\(AP=\frac13AC\)
TA có: \(AM=\frac12\times AB\)
=>\(S_{CMA}=\frac12\times S_{CAB}=\frac12\times180=90\left(dm^2\right)\)
Ta có: \(AP=\frac13\times AC\)
=>\(S_{APM}=\frac13\times S_{AMC}=\frac13\times90=30\left(dm^2\right)\)
Ta có: \(CN=\frac12\times CB\)
=>\(S_{ANC}=\frac12\times S_{ACB}=\frac12\times180=90\left(dm^2\right)\)
Ta có: \(CP=\frac23\times CA\)
=>\(S_{CPN}=\frac23\times S_{CNA}=\frac23\times90=60\left(dm^2\right)\)
Ta có: \(BN=\frac12\times BC\)
=>\(S_{ANB}=\frac12\times S_{ABC}=\frac12\times180=90\left(dm^2\right)\)
Ta có: \(BM=\frac12\times BA\)
=>\(S_{NBM}=\frac12\times S_{ANB}=\frac12\times90=45\left(dm^2\right)\)
Ta có: \(S_{APM}+S_{NBM}+S_{MPCN}=S_{ABC}\)
=>\(S_{MPCN}=180-30-45=180-75=105\left(dm^2\right)\)
Sửa đề: AP=1/2PC
Ta có: AP+PC=AC
=>\(AC=\frac12PC+PC=\frac32PC\)
=>\(AP=\frac13AC\)
TA có: \(AM=\frac12\times AB\)
=>\(S_{CMA}=\frac12\times S_{CAB}=\frac12\times180=90\left(dm^2\right)\)
Ta có: \(AP=\frac13\times AC\)
=>\(S_{APM}=\frac13\times S_{AMC}=\frac13\times90=30\left(dm^2\right)\)
Ta có: \(CN=\frac12\times CB\)
=>\(S_{ANC}=\frac12\times S_{ACB}=\frac12\times180=90\left(dm^2\right)\)
Ta có: \(CP=\frac23\times CA\)
=>\(S_{CPN}=\frac23\times S_{CNA}=\frac23\times90=60\left(dm^2\right)\)
Ta có: \(BN=\frac12\times BC\)
=>\(S_{ANB}=\frac12\times S_{ABC}=\frac12\times180=90\left(dm^2\right)\)
Ta có: \(BM=\frac12\times BA\)
=>\(S_{NBM}=\frac12\times S_{ANB}=\frac12\times90=45\left(dm^2\right)\)
Ta có: \(S_{APM}+S_{NBM}+S_{MPCN}=S_{ABC}\)
=>\(S_{MPCN}=180-30-45=180-75=105\left(dm^2\right)\)
diện tích tam giác ABC là
150:4=37,5(cm2)
Đáp số : 37,5cm2
Sửa đề: \(AN=\frac13AC\)
a: MB=MC
=>\(S_{EMB}=S_{EMC}\) và \(S_{NMB}=S_{NMC}\)
=>\(S_{EMB}-S_{NMB}=S_{EMC}-S_{NMC}\)
=>\(S_{ENB}=S_{ENC}\)
b: Ta có: \(AN=\frac13\times AC\)
=>\(S_{EAN}=\frac13\times S_{EAC}\)
=>\(S_{EAC}=3\times250=750\left(\operatorname{cm}^2\right)\)
Ta có: \(S_{EAN}+S_{ENC}=S_{EAC}\)
=>\(S_{ENC}=750-250=500\left(\operatorname{cm}^2\right)\)
=>\(S_{ENB}=500\left(\operatorname{cm}^2\right)\)
Ta có: \(S_{EAN}+S_{BAN}=S_{ENB}\)
=>\(S_{BAN}=500-250=250\left(\operatorname{cm}^2\right)\)
Vì \(AN=\frac13\times AC\)
nên \(S_{BAN}=\frac13\times S_{ABC}\)
=>\(S_{ABC}=250\times3=750\left(\operatorname{cm}^2\right)\)
Ta có: N là trung điểm của BC
=>\(BN=NC=\frac{BC}{2}\)
=>\(S_{ANB}=S_{ANC}=\frac{S_{ABC}}{2}=\frac{180}{2}=90\left(dm^2\right)\)
M là trung điểm của AB
=>\(BM=\frac12\times BA\)
=>\(S_{BMN}=\frac12\times S_{ABN}=\frac12\times90=45\left(dm^2\right)\)
Ta có: AP+PC=AC
=>\(AC=\frac12PC+PC=\frac32PC\)
=>\(AP=\frac13\times AC\)
=>\(S_{BAP}=\frac13\times S_{BAC}=\frac13\times180=60\left(dm^2\right)\)
Ta có: \(AM=\frac12\times AB\)
=>\(S_{AMP}=\frac12\times S_{ABP}=\frac12\times60=30\left(dm^2\right)\)
\(S_{AMP}+S_{BMN}+S_{MPCN}=S_{ABC}\)
=>\(S_{MPCN}=180-45-30=135-30=105\left(dm^2\right)\)