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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: \(AE=\frac13\times AB\)
=>\(S_{AKE}=\frac13\times S_{AKB}=\frac{200}{3}\left(\operatorname{cm}^2\right)\)
=>\(\frac{S_{AKE}}{S_{AKC}}=\frac{200}{3}:100=\frac{200}{300}=\frac23\)
=>\(\frac{KE}{KC}=\frac23\)
Ta có: \(AC=2\times AF\)
=>\(AF=\frac12\times AC\)
=>\(S_{ABF}=\frac12\times S_{ABC}=\frac{240}{2}=120\left(\operatorname{cm}^2\right)\)
Vì AB=3xAE
nên \(AE=\frac13\times AB\)
=>\(S_{AEF}=\frac13\times S_{AFB}=\frac13\times120=40\left(\operatorname{cm}^2\right)\)
Ta có: \(S_{AEF}+S_{BEFC}=S_{ABC}\)
=>\(S_{BEFC}=240-40=200\left(\operatorname{cm}^2\right)\)
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
=>\(BE=\frac14\times BA\)
=>\(S_{BEI}=\frac14\times S_{BIA}=\frac14\times\frac13\times S_{ABC}=\frac{1}{12}\times S_{ABC}=\frac{360}{12}=30\)
ko birt