ABCD is a parallelogram. G is a point on AB such that AG = 2GB and E is point on DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that: 1. ar(ADEG)=ar(GBCE) 2. ar ( )=1 6 ar(ABCD) 3. ar( )=1 2 ar( ) 4. ar( )=ar( ) 5. Find what portion of the parallelogram is the area of .

ABCD is a parallelogram AG = 2GB, CE = 2DE, BF = 2FC To prove: ABCD is ||gm ar(EBG) = ar(EFC) AB || CD (AB = CD)BG=1 3 AB, DE=1 3 CD BG=DE ADEH is ||gm ar(||gm ADEH) = ar(||gm BCIG) ...(i) ar( ) = ar( ) ..(ii) ( diagonal of ||gm divided into 2 equal areas) (i) and (ii) (||gm ADEG)=ar(||gm GBCE) Height, h of ||gm ABCD and is Its same Base of =1 3 AB area of ABCD = h × ABar( )=1 6 1 3 AB=1 6 h ar(EGB)=1 6 Area(ABCD) let dislaver between EH and CB = xar(EBF)=1 2 =1 2 2 3 BC =1 3 ar(EFC)=1 2 =1 2 1 3 BC =1 2 (EBF) (EFC)=1 2 of EBF If g = altitute from AD to BCar(EFC)=1 2 2 3 g 1 3 BC=1 9 ar(ABCD) (EFC)=2 3 ar EBG ar(EFG)=ar(EGB)+ar(FBF)+ar(EFC) =1 6 ABCD+2ar(EFC)+ar(EFC) =1 6 ABCD+3ar(EFC) =1 6 ABCD+ 3 1 9 = (1 6 +1 3 )ABCD = (1+2 6 )ABCD =1 2 ABCD

ABCD is a parallelogram. G is a point on AB such that AG = 2GB an…

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Question

ABCD is a parallelogram. G is a point on AB such that AG = 2GB and E is point on DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

$\text{ar}(\text{ADEG})=\text{ar}(\text{GBCE})$
$\text{ar}{(\triangle\text{EGB}})=\frac{1}{6}\text{ar}(\text{ABCD})$
$\text{ar}(\triangle\text{EFC})=\frac{1}{2}\text{ar}(\triangle\text{EBF})$
$\text{ar}(\triangle\text{EBG})=\text{ar}(\triangle\text{EFC})$
Find what portion of the parallelogram is the area of $\triangle\text{EFG}.$

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Answer

ABCD is a parallelogram AG = 2GB, CE = 2DE, BF = 2FC To prove: ABCD is ||gm ar(EBG) = ar(EFC) AB || CD (AB = CD)$\text{BG}=\frac{1}{3}\text{AB},\ \text{DE}=\frac{1}{3}\text{CD}$
$\therefore\ \text{BG}=\text{DE}$
$\therefore\ \text{ADEH}$ is ||gm
ar(||gm ADEH) = ar(||gm BCIG) ...(i) ar$(\triangle\text{HEG})$ = ar$(\triangle\text{EGI})$ ..(ii) ($\because$ diagonal of ||gm divided into 2 equal areas) (i) and (ii)$\therefore\text{ar}(||\text{gm}\text{ ADEG})=\text{ar}(||\text{gm}\text{ GBCE})$
Height, h of ||gm ABCD and $\triangle\text{EGB}$ is Its same Base of $\triangle\text{EGB}=\frac{1}{3}\text{AB}$ area of ABCD = h × AB$\text{ar}(\triangle\text{EGB})=\frac{1}{6}\times\text{h}\times\frac{1}{3}\text{AB}=\frac{1}{6}\text{h}\times\text{AB}$
$\text{ar}(\text{EGB})=\frac{1}{6}\text{Area}(\text{ABCD})$
let dislaver between EH and CB = x$\text{ar}(\text{EBF})=\frac{1}{2}\times\text{BF}\times\text{x}=\frac{1}{2}\times\frac{2}{3}\text{BC}\times\text{x}=\frac{1}{3}\times\text{BC}\times\text{x}$
$\text{ar}(\text{EFC})=\frac{1}{2}\times\text{CF}\times\text{x}=\frac{1}{2}\times\frac{1}{3}\text{BC}\times\text{x}=\frac{1}{2}\times\text{ar}\times(\text{EBF})$
$\Rightarrow\text{ar}(\text{EFC})=\frac{1}{2}\times\text{area of EBF}$
If g = altitute from AD to BC$\text{ar}(\text{EFC})=\frac{1}{2}\times\frac{2}{3}\text{g}\times\frac{1}{3}\text{BC}=\frac{1}{9}\text{ar}(\text{ABCD})$
$\Rightarrow\text{ar}(\text{EFC})=\frac{2}{3}\text{ar}\text{ EBG}$
$\text{ar}(\text{EFG})=\text{ar}(\text{EGB})+\text{ar}(\text{FBF})+\text{ar}(\text{EFC})$
$=\frac{1}{6}\text{ABCD}+2\text{ar}(\text{EFC})+\text{ar}(\text{EFC})$
$=\frac{1}{6}\text{ABCD}+3\text{ar}(\text{EFC})$
$=\frac{1}{6}\text{ABCD}+\not3\times\frac{1}{\not9}\times\text{ABCD}$
$=\Big(\frac{1}{6}+\frac{1}{3}\Big)\text{ABCD}$
$=\Big(\frac{1+2}{6}\Big)\text{ABCD}$
$=\frac{1}{2}\text{ABCD}$