State true or false.

ABCD
is a parallelogram,

G
is a parallelogram, is the point on

AB
is the point on such that

AG=2GB
such that ,

E
, is a point of is a point of such that

CE=2DE
such that and

F
and is the point of

BC
is the point of such that

BF=2FC
such that . Then

ar(ΔEGB)=
1
6

ar(∥gmABCD)
. Then .

Question

State true or false.

ABCD
  is a parallelogram,

G
  is a parallelogram,  is the point on

AB
  is the point on  such that

AG=2GB
  such that ,

E
 ,  is a point of  is a point of  such that

CE=2DE
  such that  and

F
  and  is the point of

BC
  is the point of  such that

BF=2FC
  such that . Then

ar(ΔEGB)=
    1
    6
   
   ar(∥gmABCD)
 . Then .

Accepted Answer

Given that,

ABCD
  is a parallelogram,

G
  is the point on

AB
  such that

AG=2GB
 .

E
  is a point of  such that

CE=2DE
 .

F
  is the point of

BC
  such that

BF=2FC
 .Draw

EP⊥AB,EQ⊥BC

AG=2GB

⇒AB−GB=2GB

AB=3GB

⇒GB=
      1
      3
     
     AB                  ......(1)

Similarly,

CE=2DE

⇒CD−DE=2DE

CD=3DE

⇒DE=
      1
      3
     
     CD                        ......(2)

And,

BF=2FC

⇒BC−FC=2FC

BC=3FC

FC=
      1
      3
     
     BC                            ......(3)

Area of the parallelogram ABCD = length

×
 breadth

ar(∥gmABCD)=AB×EP
     ……(4)Area of the triangle

=
    1
    2
   
   ×base×height

ar(ΔEGB)=
      1
      2
     
     ×GB×EP

⇒ar(ΔEGB)=
      1
      2
     
     ×
      1
      3
     
     AB×EP                 [by (1)]

=
      1
      6
     
     AB×EP

ar(ΔEGB)=
      1
      6
     
     ar(||gmABCD)    [by (4)]

Hence, the statement is true.Therefore, option 1 is correct.

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