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Re: Within rectangle ACDF, both ABGH and BCDE are squares, and 3 [#permalink]
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Here's my work.
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x.PNG [ 220.46 KiB | Viewed 8155 times ]


Originally posted by 7jdjones7 on 31 May 2020, 17:39.
Last edited by 7jdjones7 on 31 May 2020, 20:00, edited 2 times in total.
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Re: Within rectangle ACDF, both ABGH and BCDE are squares, and 3 [#permalink]
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7jdjones7 wrote:
Here's my work


Sir, recheck your calculations
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Re: Within rectangle ACDF, both ABGH and BCDE are squares, and 3 [#permalink]
vndnjn wrote:
7jdjones7 wrote:
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.
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Re: Within rectangle ACDF, both ABGH and BCDE are squares, and 3 [#permalink]
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vndnjn wrote:
Area of ACDF: AF*DF = (x+y)*(2x+y)


how did you know that AF =x+y
and DF = 2x+y
where in the question this info is stated ??????????
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Re: Within rectangle ACDF, both ABGH and BCDE are squares, and 3 [#permalink]
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mustafaaldory079 wrote:
vndnjn wrote:
Area of ACDF: AF*DF = (x+y)*(2x+y)


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
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Re: Within rectangle ACDF, both ABGH and BCDE are squares, and 3 [#permalink]
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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.
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Re: Within rectangle ACDF, both ABGH and BCDE are squares, and 3 [#permalink]
7jdjones7 wrote:
Here's my work.

can anyone please explain how DE is x+y?
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Re: Within rectangle ACDF, both ABGH and BCDE are squares, and 3 [#permalink]
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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
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Re: Within rectangle ACDF, both ABGH and BCDE are squares, and 3 [#permalink]
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Re: Within rectangle ACDF, both ABGH and BCDE are squares, and 3 [#permalink]
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