Here's my work.

Sir, recheck your calculations

Did it right when I solved it the first time around, rewrote it wrong for answer I posted. It's been revised.

how did you know that AF =x+y
and DF = 2x+y
where in the question this info is stated ??????????

you can tell if you look at the bottom of the photo. y is one piece and x is the other and since the figure is a square that's how you know

Area of HGEF = xy
Since y is between 3x and 2x, we get
3x^2 > Area of HGEF > 2x^2

Now let's look at area of ACDF. It helps to break this into 3 pieces.
Area of the tiny square = x^2
Area of the tiny rectangle HGEF = xy
Area of the large square = (x+y)^2 = x^2 + 2xy + y^2
Total area of ACDF = 2x^2 + 3xy + y^2
Now we plug in our possible values of y, and we get: 20x^2 > area ACDF > 12x^2

Finally, we can look at the ratios:
20x^2 / 3X^2 > ACDF / HGEF < 12x^2 / 2x^2
6.67 > ACDF / HGEF < 6 --> This range is all greater than 5, so A is greater.

can anyone please explain how DE is x+y?

The figure looks like above

Now , in this question is just to follow the wire until the solution. It is just a consequence or chain

We do know that ACDF is a rectangle so the side AC > CD

We do know that square ABGH have all sides =

we do know that 3x>y>2x

Suppose x= 2 then y =5

From this All sides of ABGH are 2

The parallel side CD is 2+5=7

because ACDF is a rectangle side AF =2+5=7 and CD as well =7

Moreover, we have a rectangle so the side AC must be AT LEAST 8 because it is longer than side CD which is 7

So AC is 8 and because AB is 2, the side portion labeled Z must be 6

The area or the entire rectangle is 8*7=56

HGEF must be also a rectangle because we do know that x = 2 BUT y=5

The area is 2*5=10

So the ratio ACDF to HGEF is = 56/10 and this must be greater than 5

The answer is A

i did all the transformation and it came down to (x-y)^2 + x^2 vs 0 so A gotta be greater