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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
=>\(BE=\frac14\times BA\)
=>\(S_{BEI}=\frac14\times S_{BIA}=\frac14\times\frac13\times S_{ABC}=\frac{1}{12}\times S_{ABC}=\frac{360}{12}=30\)
\(AD=\frac{1}{3}\times CD\Rightarrow S_{ABF}=\frac{1}{3}\times S_{BFC}\)
\(BE=\frac{1}{3}\times AB\Rightarrow S_{BEF}=\frac{1}{3}\times S_{ABF}\)
\(\Rightarrow S_{BEF}=\frac{1}{3}\times\frac{1}{3}\times S_{BFC}=\frac{1}{9}\times S_{BFC}\Rightarrow S_{BEF}=\frac{1}{10}\times S_{BEC}\)
\(BE=\frac{1}{3}\times AB\Rightarrow S_{BEC}=\frac{1}{3}\times S_{ABC}\)
\(\Rightarrow S_{BEF}=\frac{1}{10}\times\frac{1}{3}\times S_{ABC}=\frac{1}{30}\times S_{ABC}\)
\(\Rightarrow S_{BAC}=30\times S_{BEF}=5400\left(cm^2\right)\)
Ta có: AE+BE=AB
=>AB=BE+2xBE=3xBE
=>\(S_{AIB}=3\times S_{BEI}=3\times100=300\left(\operatorname{cm}^2\right)\)
Vì AE=2BE
nên \(S_{CEA}=2\times S_{CEB};S_{IEA}=2\times S_{IEB}\)
=>\(S_{CEA}-S_{IEA}=2\times\left(S_{CEB}-S_{IEB}\right)\)
=>\(S_{CIA}=2\times S_{CIB}\)
Ta có: \(CD=3\times AD\)
=>\(S_{BDC}=3\times S_{BDA};S_{IDC}=3\times S_{IDA}\)
=>\(S_{BDC}-S_{IDC}=3\times\left(S_{BDA}-S_{IDA}\right)\)
=>\(S_{BIC}=3\times S_{BIA}\)
=>\(S_{CIA}=2\times3\times S_{BIA}=6\times S_{BIA}\)
\(S_{ABC}=S_{BIA}+S_{AIC}+S_{BIC}\)
\(=S_{BIA}+3\times S_{BIA}+6\times S_{BIA}=10\times S_{BIA}=3000\left(\operatorname{cm}^2\right)\)
a) Ta có:
`S_(ABF)/S_(ABC)=(1/2 xx h xx AF)/(1/2 xx h xx AC) = (AF)/(AC)`
Ta có: `AC=2AF=>(AF)/(AC)=1/2`
`=>S_(ABF)/S_(ABC)=1/2=>S_(ABF)=1/2 xx S_(ABC)=1/2 xx 240=120(cm^2)`
`S_(AEF)/S_(ABF)=(1/2 xx h' xx AE)/(1/2 xx h' xx AB)=(AE)/(AB)`
`AB=3AE=>(AE)/(AB)=1/3`
`=>S_(AEF)/S_(ABF)=1/3=>S_(AEF)=1/3 xx S_(ABF)=1/3 xx 120=40(cm^2)`
`S_(ABC)=S_(EFCB)+S_(AEF)=>S_(EFCB)=S_(ABC)-S_(AEF)=240-40=200(cm^2)`
ko bt:) tick mk đi mong các bn đó
Có phải Giang 5D không