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Ai trả lời giúp mk đi , cả lời giải và phép tính mai mk fai nộp rồi
Ta có: AE=3EB
=>\(S_{CEA}=3\times S_{CEB};S_{IEA}=3\times S_{IEB}\)
=>\(S_{CEA}-S_{IEA}=3\times\left(S_{CIB}-S_{EIB}\right)\)
=>\(S_{CIA}=3\times S_{CIB}\)
Ta có: AF=2FC
=>\(S_{BFA}=2\times S_{BFC};S_{IFA}=2\times S_{IFC}\)
=>\(S_{BFA}-S_{IFA}=2\times\left(S_{BFC}-S_{IFC}\right)\)
=>\(S_{BIA}=2\times S_{BIC}\)
Ta có: \(S_{CIA}+S_{BIA}+S_{BIC}=S_{ABC}\)
=>\(S_{ABC}=S_{BIC}+2\times S_{BIC}+3\times S_{BIC}=6\times S_{BIC}\)
=>\(\frac{S_{AIB}}{S_{ABC}}=\frac26=\frac13\)
Ta có: BE+EA=BA
=>BA=3BE+BE=4BE
=>\(BE=\frac14\times BA\)
=>\(S_{BEI}=\frac14\times S_{BIA}=\frac14\times\frac13\times S_{ABC}=\frac{1}{12}\times S_{ABC}=\frac{360}{12}=30\)
Sửa đề: AC=3AN
TA có: \(AM=\frac13\times AB\)
=>\(S_{AMC}=\frac13\times S_{ABC}=\frac13\times180=60\left(\operatorname{cm}^2\right)\)
TA có: \(AN=\frac13\times AC\)
=>\(S_{AMN}=\frac13\times S_{AMC}=\frac13\times60=20\left(\operatorname{cm}^2\right)\)
Ta có: \(S_{AMN}+S_{MNCB}=S_{ABC}\)
=>\(S_{MNCB}=180-20=160\left(\operatorname{cm}^2\right)\)
TA có: AN+NC=AC
=>NC=AC-AN=3AN-AN=2AN
=>\(S_{BNC}=2\times S_{BNA};S_{ENC}=2\times S_{ENA}\)
=>\(S_{BNC}-S_{ENC}=2\times\left(S_{BNA}-S_{ENA}\right)\)
=>\(S_{BEC}=2\times S_{BEA}\)
=>\(S_{BEA}=\frac12\times S_{BEC}\)
Ta có: AM+MB=AB
=>MB=AB-AM=3AM-AM=2AM
=>\(S_{CMB}=2\times S_{CMA};S_{EMB}=2\times S_{EMA}\)
=>\(S_{CMB}-S_{EMB}=2\times\left(S_{CMA}-S_{EMA}\right)\)
=>\(S_{CEB}=2\times S_{CEA}\)
=>\(S_{CEA}=\frac12\times S_{CEB}\)
Ta có: \(S_{AEB}+S_{AEC}+S_{BEC}=S_{ABC}\)
=>\(S_{BEC}+\frac12\times S_{BEC}+\frac12\times S_{BEC}=S_{ABC}\)
=>\(2\times S_{BEC}=S_{ABC}=180\)
=>\(S_{BEC}=\frac{180}{2}=90\left(\operatorname{cm}^2\right)\)
=>\(S_{AEB}=S_{AEC}=\frac12\times90=45\left(\operatorname{cm}^2\right)\)
Ta có: \(\frac{S_{AEC}}{S_{AMC}}=\frac{CE}{CM}\)
=>\(\frac{CE}{CM}=\frac{45}{60}=\frac34\)
=>\(\frac{CE}{EM}=3\)
=>\(\frac{EM}{EC}=\frac13\)
Ta có: AC=2AF
=>F là trung điểm của AC
=>FA=FC
=>\(S_{BFA}=S_{BFC};S_{KFA}=S_{KFC}\)
=>\(S_{BFA}-S_{KFA}=S_{BFC}-S_{KFC}\)
=>\(S_{BKA}=S_{BKC}=200\left(\operatorname{cm}^2\right)\)
Ta có: AE+EB=AB
=>EB=AB-AE=3AE-AE=2AE
=>\(S_{CEB}=2\times S_{CEA};S_{KEB}=2\times S_{KEA}\)
=>\(S_{CEB}-S_{KEB}=2\times\left(S_{CKA}-S_{EKA}\right)\)
=>\(S_{CKB}=2\times S_{CKA}\)
=>\(S_{CKA}=\frac{200}{2}=100\left(cm^2\right)\)
\(S_{ABC}=S_{AKB}+S_{AKC}+S_{BKC}\)
\(=200+200+100=500\left(\operatorname{cm}^2\right)\)
Ta có: \(AE=\frac13\times AB\)
=>\(S_{AKE}=\frac13\times S_{AKB}=\frac{200}{3}\left(\operatorname{cm}^2\right)\)
=>\(\frac{S_{AKE}}{S_{AKC}}=\frac{200}{3}:100=\frac{200}{300}=\frac23\)
=>\(\frac{KE}{KC}=\frac23\)