Question

Let ABCD be a parallelogram of area 10 with AB = 3 and BC = 5. Locate
E, F and G on segments 4B.BC and 4D , respectively, with
AE - BF = AG=2. Let the line through G parallel to EF intersect CD
at H. The area of quadrilateral EFH Gis​

Answer 1

Step-by-step explanation:

Let AF=GD=x & FB=CG=y.

A(AFE)/A (FBE)=x/y.

As A(AFE) =1, A (FBE)= y/x.

FBE is similar to GDE.

Using property of ratio of area of similar triangles,

A(FBE)/A(GDE)=(y/x)^2

So, A(GDE)=x/y.

DGE is similar DCB.

So, A(GDE)/A(CDB)

=(x/x+y)^2-=x/y / 5y+x/y

x^2/(x+y)^2= x/x+5y

x/(x+y)^2 = 1/x+5y

x^2+5xy=x^2+2xy+y^2

3xy=y^2

y=3x

y/x=3.

A(FBE)=3 sq.units

Now, A(FBGC) = 5+3 = 8 sq. units

A(ABCD)= 8*4/3 =-32/3 sq.units