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# Figures X and Y above show how eight identical triangular pieces of ca

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Intern
Joined: 22 Jun 2016
Posts: 47
Re: Figures X and Y above show how eight identical triangular pieces of ca [#permalink]

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09 Oct 2016, 18:55
for square, i dont understand how the side is calculated to be sqaure root 2?
should not that be the amount of the DIAGONAL and not the 4 sides? side = 1. perimeter = 4*1 = 4
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Re: Figures X and Y above show how eight identical triangular pieces of ca [#permalink]

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13 Oct 2016, 02:26
jonmarrow wrote:
for square, i dont understand how the side is calculated to be sqaure root 2?
should not that be the amount of the DIAGONAL and not the 4 sides? side = 1. perimeter = 4*1 = 4

Which solution are you referring to?

We have right isosceles triangles.

If we consider the hypotenuse of the triangle to be 1 (notice that the hypotenuse of the triangle = the side of the square), then the legs of the triangle will be $$\frac{1}{\sqrt{2}}$$ (notice that a leg of the triangle = the width of the rectangle).

If we consider the legs of the triangle to be 1 (notice that the a leg of the triangle = the width of the rectangle), then the hypotenuse will be $$\sqrt{2}$$ (notice that the hypotenuse of the triangle = the side of the square).

In any of the cases the ration of the perimeters comes to be $$2\sqrt{2} :3$$.
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Re: Figures X and Y above show how eight identical triangular pieces of ca [#permalink]

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26 Feb 2017, 01:31
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Guys, you can do it either way. The version with side of square = a is also correct, you only need to do the math when you arrive at 4/3√2!

4/3√2 --> multiply numerator and denominator with √2

4 √2/3√2*√2 = 4√2/3*2 --> cancel out 4 (in numerator)/2 (in denominator) = 2 so what remains is 2√2/3

Makes sense?
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Re: Figures X and Y above show how eight identical triangular pieces of ca [#permalink]

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14 Mar 2017, 04:18
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Bunuel wrote:

Figures X and Y above show how eight identical triangular pieces of cardboard were used to form a square and a rectangle, respectively. What is the ratio of the perimeter of X to the perimeter of Y?

(A) 2:3
(B) $$\sqrt{2}:2$$
(C) $$2\sqrt{2} :3$$
(D) 1:1
(E) $$\sqrt{2}:1$$

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
2015-10-20_1410.png

Responding to a pm:
Quote:
My Query: I am getting A as the answer since I am trying to relate side of square as (a) and perimeter as 4a. Where am I going wrong

You are right about side of square a has perimeter 4a.

But note how the sides of each triangular piece are related. The two legs, which will be identical in all triangles are say of length L each. The triangles are right angled so the hypotenuse will be $$\sqrt{2}L$$.

The square is made up of 4 hypotenuse. The perimeter will be $$4*\sqrt{2}L$$.
The longer sides of the rectangle are made up of 2 legs each and the shorter sides are made up of L each.
Perimeter = 2*2L + 2L = 6L

The ratio of perimeters $$= 4*\sqrt{2} : 6 = 2\sqrt{2} : 3$$

Answer (C)
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Joined: 13 Mar 2017 Posts: 572 Location: India Concentration: General Management, Entrepreneurship GPA: 3.8 WE: Engineering (Energy and Utilities) Re: Figures X and Y above show how eight identical triangular pieces of ca [#permalink] ### Show Tags 16 Aug 2017, 12:51 Bunuel wrote: Figures X and Y above show how eight identical triangular pieces of cardboard were used to form a square and a rectangle, respectively. What is the ratio of the perimeter of X to the perimeter of Y? (A) 2:3 (B) $$\sqrt{2}:2$$ (C) $$2\sqrt{2} :3$$ (D) 1:1 (E) $$\sqrt{2}:1$$ Kudos for a correct solution. [Reveal] Spoiler: Attachment: 2015-10-20_1410.png Let the perpendicular sides of each smaller triangle be a . So hypotenuse = a$$\sqrt{2}$$ Perimeter of square = 4* a$$\sqrt{2}$$ length of rectangle = 2a and breadth = a So, perimeter of rectangle = 2*(2a +a) = 6a So ratio = 4* a$$\sqrt{2}$$/6a = 2$$\sqrt{2}$$/3 Answer C _________________ CAT 99th percentiler : VA 97.27 | DI-LR 96.84 | QA 98.04 | OA 98.95 UPSC Aspirants : Get my app UPSC Important News Reader from Play store. MBA Social Network : WebMaggu Appreciate by Clicking +1 Kudos ( Lets be more generous friends.) What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish". SVP Joined: 12 Sep 2015 Posts: 2140 Location: Canada Figures X and Y above show how eight identical triangular pieces of ca [#permalink] ### Show Tags 11 Sep 2017, 13:25 1 This post received KUDOS Expert's post Top Contributor Bunuel wrote: Figures X and Y above show how eight identical triangular pieces of cardboard were used to form a square and a rectangle, respectively. What is the ratio of the perimeter of X to the perimeter of Y? (A) 2:3 (B) $$\sqrt{2}:2$$ (C) $$2\sqrt{2} :3$$ (D) 1:1 (E) $$\sqrt{2}:1$$ Kudos for a correct solution. [Reveal] Spoiler: Attachment: 2015-10-20_1410.png Let's start by gathering more information about these eight identical triangles. First notice that we have 4 equal angles meeting at a single point. So, each angle must be 90° Now examine the red triangle below. The red triangle is an isosceles triangle since all 4 sides of a square are equal. The two equal angles must add to 90° So, each angle must be 45° Using similar logic, we can conclude that all of the eight triangles are 45-45-90 special right triangles. Now let's examine what happens if we examine one particular 45-45-90 special right triangle, which has sides of length 1, 1 and √2 Use those lengths for all eight identical triangles in the 2 diagrams we get the following: At this point, we can calculate the perimeters. Perimeter of X = √2 + √2 + √2 + √2 = 4√2 Perimeter of Y = 1 + 1 + 1 + 1 + 1 + 1 = 6 So, perimeter of X : perimeter of Y = 4√2 : 6 Divide both sides by 2 to get the equivalent ratio 2√2 : 3 Answer: [Reveal] Spoiler: C ASIDE: Please note that I didn't have to use those specific lengths (1, 1 and √2) for each of the 45-45-90 triangles. I could have used ANY measurements that could be found in a 45-45-90 triangle. For example, I could have used 5, 5 and 5√2, and those measurements still would have yielded the same ratio (after some simplifying) RELATED VIDEOS _________________ Brent Hanneson – Founder of gmatprepnow.com Director Joined: 29 Jun 2017 Posts: 518 GMAT 1: 570 Q49 V19 GPA: 4 WE: Engineering (Transportation) Re: Figures X and Y above show how eight identical triangular pieces of ca [#permalink] ### Show Tags 11 Sep 2017, 13:50 Let a be the side a sqrt 2 x4 = 4a sqrt2 is oerimeter of square And 6a is of rectangle 4a root2 :6a 2 root2:3 option C Sent from my iPhone using GMAT Club Forum _________________ Give Kudos for correct answer and/or if you like the solution. Manager Joined: 07 Jun 2017 Posts: 176 Location: India Concentration: Technology, General Management GMAT 1: 660 Q46 V38 GPA: 3.6 WE: Information Technology (Computer Software) Re: Figures X and Y above show how eight identical triangular pieces of ca [#permalink] ### Show Tags 21 Sep 2017, 22:44 This is a good question Take √2 is a length of one side of the square Then we have perimeter of the square = 4√2 perimeter of the rectange is 6 ten ration = 4√2 : 6 = 2√2: 3 _________________ Regards, Naveen email: nkmungila@gmail.com Please press kudos if you like this post Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7988 Location: Pune, India Re: Figures X and Y above show how eight identical triangular pieces of ca [#permalink] ### Show Tags 18 Dec 2017, 05:44 Bunuel wrote: Figures X and Y above show how eight identical triangular pieces of cardboard were used to form a square and a rectangle, respectively. What is the ratio of the perimeter of X to the perimeter of Y? (A) 2:3 (B) $$\sqrt{2}:2$$ (C) $$2\sqrt{2} :3$$ (D) 1:1 (E) $$\sqrt{2}:1$$ Kudos for a correct solution. [Reveal] Spoiler: Attachment: 2015-10-20_1410.png Check out our detailed video solution to this problem by Chris here: https://www.veritasprep.com/gmat-soluti ... olving_138 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Figures X and Y above show how eight identical triangular pieces of ca   [#permalink] 18 Dec 2017, 05:44

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# Figures X and Y above show how eight identical triangular pieces of ca

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