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a: Ta có: ABCD là hình chữ nhật
=>AD=BC
Ta có: AE+EB=AB
=>\(EB=AB-\frac14AB=\frac34AB\)
=>\(EB=3\times EA\)
ΔEAD vuông tại A
=>\(S_{AED}=\frac12\times AE\times AD=\frac12\times\frac14\times AB\times AD=\frac18\times AB\times AD=\frac18\times S_{ABCD}\)
ΔEBC vuông tại B
=>\(S_{EBC}=\frac12\times BE\times BC=\frac12\times\frac34\times BA\times BC=\frac38\times BA\times BC=\frac38\times S_{ABCD}\)
=>\(\frac{S_{EBC}}{S_{AED}}=\frac38:\frac18=3\)
=>\(\frac{S_{AED}}{S_{EBC}}=\frac13\)
b: ΔABC vuông tại B
=>\(S_{ABC}=\frac12\times AB\times BC=\frac12\times BC\times CD\) (1)
ΔBCD vuông tại C
=>\(S_{CBD}=\frac12\times CB\times CD\left(2\right)\)
Từ (1),(2) suy ra \(S_{ABC}=S_{BCD}\)
=>\(S_{ABK}+S_{KBC}=S_{KBC}+S_{KCD}\)
=>\(S_{ABK}=S_{KCD}\)
=>\(S_{ABK}=S_{KCD}=3\left(\operatorname{cm}^2\right)\)
AB. Nối ED, EC. Gọi K là giao điểm của EC và DB.
1: AE+EB=AB
=>\(EB=AB-AE=AB-\frac14\times AB=\frac34\times AB\)
=>\(EB=3\times EA\)
ΔADE vuông tại A
=>\(S_{ADE}=\frac12\times AD\times AE\) (1)
ΔBEC vuông tại B
=>\(S_{BEC}=\frac12\times BE\times BC=\frac12\times3\times EA\times AD=3\times\frac12\times AE\times AD\) (2)
Từ (1),(2) suy ra \(\frac{S_{ADE}}{S_{BEC}}=\frac13\)
2: Ta có: \(BE=\frac34\times BA\)
mà BA=CD
nên \(BE=\frac34\times CD\)
BE//CD
=>\(\frac{KE}{KC}=\frac{KB}{KD}=\frac{BE}{CD}=\frac34\)
\(\frac{KE}{KC}=\frac34\)
=>\(\frac{S_{KEB}}{S_{KBC}}=\frac34\)
=>\(S_{KBC}=\frac43\times S_{KEB}\)
Ta có: \(\frac{KB}{KD}=\frac34\)
=>\(\frac{S_{KEB}}{S_{KED}}=\frac34\)
=>\(S_{KED}=\frac43\times S_{KEB}\)
Ta có: \(\frac{EK}{KC}=\frac34\)
=>\(\frac{S_{DKE}}{S_{DKC}}=\frac34\)
=>\(S_{DKC}=\frac43\times S_{DEK}=\frac43\times\frac43\times S_{EKB}=\frac{16}{9}\times S_{EKB}\)
Ta có: \(S_{DKC}-S_{EKB}=3\)
=>\(\frac{16}{9}\times S_{KEB}-S_{KEB}=3\)
=>\(\frac79\times S_{KEB}=3\)
=>\(S_{KEB}=3:\frac79=\frac{27}{7}\left(\operatorname{cm}^2\right)\)
\(S_{BEDC}=S_{BEK}+S_{KBC}+S_{KED}+S_{KDC}\)
\(=S_{KBE}+\frac43\times S_{KBE}+\frac43\times S_{KBE}+\frac{16}{9}\times S_{KBE}\)
\(=\frac{11}{3}\times S_{KBE}+\frac{16}{9}\times S_{KBE}=\frac{49}{9}\times S_{KBE}\)
\(=\frac{49}{9}\times\frac{27}{7}=3\times7=21\left(\operatorname{cm}^2\right)\)
\(S_{BEDC}=\frac12\times\left(BE+DC\right)\times BC\)
\(=\frac12\times BC\times\left(\frac34\times DC+DC\right)=\frac12\times BC\times\frac74\times DC=\frac78\times BC\times DC\)
\(=\frac78\times S_{ABCD}\)
=>\(S_{ABCD}=21:\frac78=24\left(\operatorname{cm}^2\right)\)
a: S ADE=1/2*AE*AD
S EBC=1/2*EB*BC=1/2*EB*AD
=>S ADE/S EBC=AE/EB=1/3
b: Xét ΔKDC và ΔKBE có
góc KDC=góc KBE
góc DKC=góc BKE
=>ΔKDC đồng dạng với ΔKBE
=>S KDC/S KBE=(DC/BE)^2=(4/3)^2=16/9
mà S DKC-S EKB=3
nên S DKC=48/7cm2;S EKB=27/7cm2
ΔKDC đồng dạng với ΔKBE
=>KD/KB=DC/BE=4/3
=>S EKD=4/3*S EKB=36/7cm2
=>S BKC=36/7cm2
=>S EBC=36/7+27/7=63/7=9cm2
=>S AED=1/3*9=3cm2
S EBCD=48/7+27/7+36/7+36/7=21cm2
=>S ABCD=24cm2
1: AE+EB=AB
=>\(EB=AB-AE=AB-\frac14\times AB=\frac34\times AB\)
=>\(EB=3\times EA\)
ΔADE vuông tại A
=>\(S_{ADE}=\frac12\times AD\times AE\) (1)
ΔBEC vuông tại B
=>\(S_{BEC}=\frac12\times BE\times BC=\frac12\times3\times EA\times AD=3\times\frac12\times AE\times AD\) (2)
Từ (1),(2) suy ra \(\frac{S_{ADE}}{S_{BEC}}=\frac13\)
2: Ta có: \(BE=\frac34\times BA\)
mà BA=CD
nên \(BE=\frac34\times CD\)
BE//CD
=>\(\frac{KE}{KC}=\frac{KB}{KD}=\frac{BE}{CD}=\frac34\)
\(\frac{KE}{KC}=\frac34\)
=>\(\frac{S_{KEB}}{S_{KBC}}=\frac34\)
=>\(S_{KBC}=\frac43\times S_{KEB}\)
Ta có: \(\frac{KB}{KD}=\frac34\)
=>\(\frac{S_{KEB}}{S_{KED}}=\frac34\)
=>\(S_{KED}=\frac43\times S_{KEB}\)
Ta có: \(\frac{EK}{KC}=\frac34\)
=>\(\frac{S_{DKE}}{S_{DKC}}=\frac34\)
=>\(S_{DKC}=\frac43\times S_{DEK}=\frac43\times\frac43\times S_{EKB}=\frac{16}{9}\times S_{EKB}\)
Ta có: \(S_{DKC}-S_{EKB}=3\)
=>\(\frac{16}{9}\times S_{KEB}-S_{KEB}=3\)
=>\(\frac79\times S_{KEB}=3\)
=>\(S_{KEB}=3:\frac79=\frac{27}{7}\left(\operatorname{cm}^2\right)\)
\(S_{BEDC}=S_{BEK}+S_{KBC}+S_{KED}+S_{KDC}\)
\(=S_{KBE}+\frac43\times S_{KBE}+\frac43\times S_{KBE}+\frac{16}{9}\times S_{KBE}\)
\(=\frac{11}{3}\times S_{KBE}+\frac{16}{9}\times S_{KBE}=\frac{49}{9}\times S_{KBE}\)
\(=\frac{49}{9}\times\frac{27}{7}=3\times7=21\left(\operatorname{cm}^2\right)\)
\(S_{BEDC}=\frac12\times\left(BE+DC\right)\times BC\)
\(=\frac12\times BC\times\left(\frac34\times DC+DC\right)=\frac12\times BC\times\frac74\times DC=\frac78\times BC\times DC\)
\(=\frac78\times S_{ABCD}\)
=>\(S_{ABCD}=21:\frac78=24\left(\operatorname{cm}^2\right)\)
a: AE+EC=AC
=>AC=2EC+EC=3EC
=>\(AE=\frac23\times AC\)
=>\(S_{ABE}=\frac23\times S_{ABC}=\frac23\times\frac12\times S_{ABCD}=\frac13\times S_{ABCD}\)
Ta có: AC=3EC
=>\(S_{ADC}=3\times S_{DEC}\)
=>\(S_{DEC}=\frac13\times S_{ADC}=\frac13\times\frac12\times S_{ABCD}=\frac16\times S_{ABCD}\)
=>\(S_{ABE}=2\times S_{EDC}\)
b: Sửa đề: Tính diện tích hình chữ nhật ABCD
Ta có: \(S_{DEC}=\frac16\times S_{ABCD}\)
=>\(S_{ABCD}=4.5\times6=27\left(\operatorname{cm}^2\right)\)