Question

ABCD is a parallelogram, G is the point on AB such that AG = 2GB, E is a point of DC , such that CE = 2DE and F is the point of BC such that BF = 2FC. prove that : 1) ar (quad ADEG) = ar(quad GBCE) 2.) ar( triangle EBG) = 1/6 ar( quad ABCD) 3) ar( triangle EFC) = 1/2 AR(triangle EBF) 4 ) ar( triangle EBG )= ar( triangle EFC) 5 ) find what portion of the area of parallelogram is the area of triangle EFG. "meritnation expert gursheen kaur has answered my question, but the method was incorrect.she supposed that G is the mid-point of AB and F is the midpoint of BC. but it's wrong method. please experts firstly read the question properly before answering it."

Answer 1

1) Given: ABCD is a parallelogram in which AG=2GB,CE=2DE and BF=2FC.

To prove: ar(gmADEG) = ar(gmGBCE)

Proof: Since ABCD is a parallelogram, we have AB  CD and AB =CD.

∴ BG=AB 

and DE= CD=AB 

∴   BG = DE 

∴ ADEH is a gm    [since AH is parallel to DE and AD is parallel to HE]

ar(gmADEH )=ar(gmBCIG )...........(i) 

[since DE=BG and AD=BC parallelogram with corresponding sides equal]

area ( ΔHEG)=area( ΔEGI).............(ii)

[diagonals of a parallelogram divide it into two equal areas]

From (i) and (ii), we get,

ar(gmADEH )+area ( ΔHEG)=ar(gmBCIG )+area( ΔEGI)

∴ ar(gmADEG) = ar(gmGBCE)



2)

Height,  h of gm ABCD and ΔEGB is the same.


Base of ΔEGB =  
area(ABCD) = hAB 
area(EGB) =  


area(EGB) =

 

3)  Let the distance between EH and CB = x


⇒ Area(EFC) =    Area(EBF)

 

4) If g = altitude from AD to BC 
area(EFC) = 
⇒ area(EFC) = area(EBG) 
 

5) area(EFG) = area(EGB)+area(EBF)+area(EFC)