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

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Intern
Joined: 07 Aug 2015
Posts: 23
The figure shows two squares, and the points W, X, Y, and Z lie on a  [#permalink]

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Updated on: 02 Sep 2016, 07:43
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00:00

Difficulty:

35% (medium)

Question Stats:

68% (01:15) correct 32% (01:05) wrong based on 190 sessions

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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

Attachment:

Bildschirmfoto 2016-09-02 um 16.33.26.png [ 82.05 KiB | Viewed 3226 times ]
Attachment:

2016-09-02_1842.png [ 8.1 KiB | Viewed 3767 times ]

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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Joined: 02 Sep 2009
Posts: 48067
Re: The figure shows two squares, and the points W, X, Y, and Z lie on a  [#permalink]

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02 Sep 2016, 07:44
Finance0402 wrote:

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

Attachment:
Bildschirmfoto 2016-09-02 um 16.33.26.png
Attachment:
2016-09-02_1842.png

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Joined: 23 Apr 2015
Posts: 321
Location: United States
WE: Engineering (Consulting)
Re: How to solve this geometry question? I don't get it  [#permalink]

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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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Joined: 07 Aug 2015
Posts: 23
Re: The figure shows two squares, and the points W, X, Y, and Z lie on a  [#permalink]

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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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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
Intern
Joined: 05 Sep 2015
Posts: 45
Re: The figure shows two squares, and the points W, X, Y, and Z lie on a  [#permalink]

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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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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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15 Feb 2018, 06:00
Finance0402 wrote:

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

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 $$= (1/2)*XY^2 = (1/2)*(1/9)*(WZ)^2$$

*Edited
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Intern
Joined: 09 Jun 2016
Posts: 15
Re: The figure shows two squares, and the points W, X, Y, and Z lie on a  [#permalink]

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15 Feb 2018, 07:04
1
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

Sent from my SM-G935F using GMAT Club Forum mobile app
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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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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

Sent from my SM-G935F using GMAT Club Forum mobile app

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 &nbs [#permalink] 16 Feb 2018, 04:02
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