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The figure shows two squares, and the points W, X, Y, and Z lie on a

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The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]

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The figure shows two squares, and the points W, X, Y, and Z lie on a line. If the distance between W and X is equal to the distance between X and Y and to the distance between Y and Z, then the area of the smaller square is what fraction of the area of the larger square?

A. 1/2
B. 1/3
C. 1/4
D. 1/8
E. 1/9

[Reveal] Spoiler:
Attachment:
Bildschirmfoto 2016-09-02 um 16.33.26.png
Bildschirmfoto 2016-09-02 um 16.33.26.png [ 82.05 KiB | Viewed 2730 times ]
Attachment:
2016-09-02_1842.png
2016-09-02_1842.png [ 8.1 KiB | Viewed 3211 times ]
[Reveal] Spoiler: OA

Originally posted by Finance0402 on 02 Sep 2016, 07:34.
Last edited by Bunuel on 02 Sep 2016, 07:43, edited 1 time in total.
Renamed the topic and edited the question.
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Re: The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]

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New post 02 Sep 2016, 07:44
Finance0402 wrote:
Image
The figure shows two squares, and the points W, X, Y, and Z lie on a line. If the distance between W and X is equal to the distance between X and Y and to the distance between Y and Z, then the area of the smaller square is what fraction of the area of the larger square?

A. 1/2
B. 1/3
C. 1/4
D. 1/8
E. 1/9

[Reveal] Spoiler:
Attachment:
Bildschirmfoto 2016-09-02 um 16.33.26.png
Attachment:
2016-09-02_1842.png


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Re: How to solve this geometry question? I don't get it [#permalink]

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New post 02 Sep 2016, 07:45
Finance0402 wrote:
How to solve this geometry question? I don't get it



Let \(x\) be the diagonal of a square, so the side is \(\frac{x}{\sqrt{2}}\). Area is side * side = \(\frac{x^2}{2}\).
Given diagonal of larger square is 3 times the diagonal of the smaller square.
So Area of small square : Area of larger square = \(\frac{x^2}{2} : \frac{3^2 * x^2}{2}\) = \(\frac{1}{9}\)
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Re: The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]

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New post 02 Sep 2016, 09:00
Senthil1981 wrote:
Finance0402 wrote:
How to solve this geometry question? I don't get it



Let \(x\) be the diagonal of a square, so the side is \(\frac{x}{\sqrt{2}}\). Area is side * side = \(\frac{x^2}{2}\).
Given diagonal of larger square is 3 times the diagonal of the smaller square.
So Area of small square : Area of larger square = \(\frac{x^2}{2} : \frac{3^2 * x^2}{2}\) = \(\frac{1}{9}\)


I dont get to be honest why the side should be \(\frac{x}{\sqrt{2}}\) ?
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The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]

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New post 02 Sep 2016, 09:07
Finance0402 wrote:
Senthil1981 wrote:
Finance0402 wrote:
How to solve this geometry question? I don't get it



Let \(x\) be the diagonal of a square, so the side is \(\frac{x}{\sqrt{2}}\). Area is side * side = \(\frac{x^2}{2}\).
Given diagonal of larger square is 3 times the diagonal of the smaller square.
So Area of small square : Area of larger square = \(\frac{x^2}{2} : \frac{3^2 * x^2}{2}\) = \(\frac{1}{9}\)


I dont get to be honest why the side should be \(\frac{x}{\sqrt{2}}\) ?


It's the property of the square. Diagonal = Side * \(\sqrt{2}\) . Pythogoras theorem
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Re: The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]

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New post 03 Sep 2016, 00:55
formula for side when diagonal is known s^2 = 2 * d^2

area of small square = 2 (xy)^2 ; area of bigger square = 2 (wz)^2
wz=3*xy
area of bigger square = 2 * 9 * xy^2.

ratio of smaller to bigger squares area = 1:9 .
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Re: The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]

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New post 15 Feb 2018, 04:44
WX, XY and YZ as hypotenuse, are equal in lengths. So, this means that a case can be made where the two non-hypotenuse sides will also be equal for each of these lengths.

By this analogy, if the side of the larger triangle is 3, then the side of the smaller triangle will be 1.

So, ratio of areas = 1:9
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The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]

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New post 15 Feb 2018, 06:00
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Finance0402 wrote:
Image
The figure shows two squares, and the points W, X, Y, and Z lie on a line. If the distance between W and X is equal to the distance between X and Y and to the distance between Y and Z, then the area of the smaller square is what fraction of the area of the larger square?

A. 1/2
B. 1/3
C. 1/4
D. 1/8
E. 1/9

[Reveal] Spoiler:
Attachment:
Bildschirmfoto 2016-09-02 um 16.33.26.png
Attachment:
2016-09-02_1842.png


Since \(WX = XY = YZ\), basically
\(XY = (1/3)*WZ\)

Note that a square is a special type of rhombus. Area of rhombus = (1/2)*diagonal1*diagonal2

Area of smaller square \(= (1/2)*(WZ)^2\)

Area of larger square \(= (1/2)*XY^2 = (1/2)*(1/9)*(WZ)^2\)

Answer (E)

*Edited
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Re: The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]

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Dear Karishma. You have mentioned areas of smaller and larger squares wrongly.. the small square area has to be : 1/2*1/9*(wz)2 and for larger square it should be what you mentioned for smaller square

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Re: The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]

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New post 16 Feb 2018, 04:02
kaushiksamrat wrote:
Dear Karishma. You have mentioned areas of smaller and larger squares wrongly.. the small square area has to be : 1/2*1/9*(wz)2 and for larger square it should be what you mentioned for smaller square

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Yeah, I flipped the two. Edited. Thanks.
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Re: The figure shows two squares, and the points W, X, Y, and Z lie on a   [#permalink] 16 Feb 2018, 04:02
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