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Re: The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]
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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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]
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]
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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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 \(= (1/2)*XY^2 = (1/2)*(1/9)*(WZ)^2\)

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

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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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 [#permalink]
Sides of the smaller square = 1/3
Area of the smaller square = 1/3 x 1/3 = 1/9
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Re: The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]
VeritasKarishma wrote:
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.


Hi VeritasKarishma

Though the points appear to be vertex of the square, they are nowhere stated.
Is there any logic that the given points are vertex, or is it just based on what we see on the figure, that the points appear near the vertices?
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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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shanks2020 wrote:
VeritasKarishma wrote:
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.


Hi VeritasKarishma

Though the points appear to be vertex of the square, they are nowhere stated.
Is there any logic that the given points are vertex, or is it just based on what we see on the figure, that the points appear near the vertices?


From the figure it seems that the points are the vertices and since this is a PS question and we need to find the answer, it is best to assume they are at the vertices and proceed. Had this been a DS question, we needed to worry that the points may not be at the vertices.
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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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\(d = s * \sqrt{2}\)
thus, \(s = \frac{d}{\sqrt{2}}\)

I. Area of smaller square
say diagonal of smaller square is \(2\sqrt{2}\)
then side of smaller square = \(\frac{2\sqrt{2}}{\sqrt{2}}\) = 2
area of smaller square = \(2^2\) = 4

II. Area of larger square
given diagonal of larger square is 3 times the diagonal of smaller square
if d of smaller square is \(2\sqrt{2}\) then d of larger square is \(3*2\sqrt{2}\) = \(6\sqrt{2}\)
thus, side of larger square = \(\frac{6\sqrt{2}}{\sqrt{2}}\) = 6
area of larger square = \(6^2\) = 36


Area of smaller square : Area of larger square
= \(\frac{4}{36} = \frac{1}{9}\)
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Re: The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]
Hi KarishmaB how do we know that the bigger square does not have a hole in the middle which is the area of the smaller square, in that case the answer would be 1/8. How do we know that?
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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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IN2MBB2PE wrote:
Hi KarishmaB how do we know that the bigger square does not have a hole in the middle which is the area of the smaller square, in that case the answer would be 1/8. How do we know that?


Even if you bend a wire to make a square, its area will remain s^2.
Whether there is actually any material inside it or not is irrelevant. The point is the area it covers from edge to edge. What is or is not there inside it doesn't matter.
When we want to consider only the area outside the area of the smaller square, we ask for the area of the square "frame" and the question wording clarifies it properly.
e.g. ... on a square sheet, a square picture is pasted in the middle... How much area of the sheet is visible now?
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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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Re: The figure shows two squares, and the points W, X, Y, and Z lie on a [#permalink]
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