The figure shows two squares, and the points W, X, Y, and Z lie on a : Problem Solving (PS)

Finance0402

Joined: 07 Aug 2015

Last visit: 09 Apr 2017

Posts: 23

Own Kudos:

46

Given Kudos: 1

Posts: 23

Kudos: 46

02 Sep 2016, 09:00

Kudos

Bookmarks

Senthil1981

Finance0402

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 \(\frac{x}{\sqrt{2}}\) ?

Senthil1981

Joined: 23 Apr 2015

Last visit: 14 Oct 2021

Posts: 230

Own Kudos:

557

[1]

Given Kudos: 36

Location: United States

Concentration: General Management, International Business

Schools: Booth PT '19 Ross Weekend"19

WE:Engineering (Consulting)

Schools: Booth PT '19 Ross Weekend"19

Posts: 230

Kudos: 557

[1]

02 Sep 2016, 09:07

1

Kudos

Bookmarks

Finance0402

Senthil1981

Finance0402

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

Harish Yannam

Joined: 05 Sep 2015

Last visit: 15 May 2020

Posts: 42

Own Kudos:

10

Given Kudos: 12

Posts: 42

Kudos: 10

03 Sep 2016, 00:55

Kudos

Bookmarks

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 .

Nived

Joined: 30 Nov 2017

Last visit: 07 Nov 2019

Posts: 19

Own Kudos:

24

[1]

Given Kudos: 12

Posts: 19

Kudos: 24

[1]

15 Feb 2018, 04:44

Kudos

Bookmarks

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

KarishmaB

Tutor

Joined: 16 Oct 2010

Last visit: 21 Mar 2025

Posts: 15,826

Own Kudos:

72,238

[4]

Given Kudos: 461

Location: Pune, India

Expert

Expert reply

Posts: 15,826

Kudos: 72,238

[4]

15 Feb 2018, 06:00

2

Kudos

2

Bookmarks

Finance0402

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

Show 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

kaushiksamrat

Joined: 09 Jun 2016

Last visit: 17 Apr 2019

Posts: 11

Own Kudos:

4

[1]

Posts: 11

Kudos: 4

[1]

15 Feb 2018, 07:04

1

Kudos

Bookmarks

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

KarishmaB

Tutor

Joined: 16 Oct 2010

Last visit: 21 Mar 2025

Posts: 15,826

Own Kudos:

72,238

Given Kudos: 461

Location: Pune, India

Expert

Expert reply

Posts: 15,826

Kudos: 72,238

16 Feb 2018, 04:02

Kudos

Bookmarks

kaushiksamrat

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.

mozerng

Joined: 08 Feb 2020

Last visit: 08 Oct 2020

Posts: 7

Own Kudos:

31

Given Kudos: 49

Posts: 7

Kudos: 31

18 Jun 2020, 09:16

Kudos

Bookmarks

Sides of the smaller square = 1/3
Area of the smaller square = 1/3 x 1/3 = 1/9

shanks2020

Joined: 02 Dec 2018

Last visit: 21 Mar 2024

Posts: 240

Own Kudos:

36

Given Kudos: 70

Posts: 240

Kudos: 36

30 Sep 2020, 10:22

Kudos

Bookmarks

VeritasKarishma

kaushiksamrat

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?

KarishmaB

Tutor

Joined: 16 Oct 2010

Last visit: 21 Mar 2025

Posts: 15,826

Own Kudos:

72,238

Given Kudos: 461

Location: Pune, India

Expert

Expert reply

Posts: 15,826

Kudos: 72,238

06 Oct 2020, 02:15

Kudos

Bookmarks

shanks2020

VeritasKarishma

kaushiksamrat

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.

sabrinasiregar

Joined: 16 Oct 2020

Last visit: 04 Jun 2023

Posts: 1

Own Kudos:

6

[1]

Posts: 1

Kudos: 6

[1]

25 Mar 2021, 19:32

1

Kudos

Bookmarks

\(d = s * \sqrt{2}\)
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
= \(\frac{4}{36} = \frac{1}{9}\)

IN2MBB2PE

Joined: 20 Aug 2020

Last visit: 17 Feb 2024

Posts: 131

Own Kudos:

31

Given Kudos: 82

Posts: 131

Kudos: 31

08 Apr 2023, 18:01

Kudos

Bookmarks

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?

KarishmaB

Tutor

Joined: 16 Oct 2010

Last visit: 21 Mar 2025

Posts: 15,826

Own Kudos:

72,238

Given Kudos: 461

Location: Pune, India

Expert

Expert reply

Posts: 15,826

Kudos: 72,238

10 Apr 2023, 05:14

Kudos

Bookmarks

IN2MBB2PE

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?

bumpbot

Non-Human User

Joined: 09 Sep 2013

Last visit: 04 Jan 2021

Posts: 36,669

Own Kudos:

963

Posts: 36,669

Kudos: 963

31 May 2024, 10:04

Kudos

Bookmarks

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

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

Prep Toolkit

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

My Rewards