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605-655 (Medium)|   Geometry|                           
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Bunuel

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.

Attachment:
2015-10-20_1410.png

Every piece is an isosceles right angle triangle with side = 1 (assumed)
Hypotenuse of that triangular piece = \(\sqrt{2}\)

Perimeter of Square = 4*Hypotenuse = \(4\sqrt{2}\)

Perimeter of Rectangle = 6*Side = \(6\)

Ratio of Perimeter of X to perimeter of Y = \(4\sqrt{2}/6\) = \(2\sqrt{2}/3\)

Answer: option C
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Square -
The diagonals of a square bisect at right angles and are of equal length
The 4 triangles are 45-45-90 triangles i.e isosceles right triangle.
If total length of diagonal of square is 2a
=> a will be length of the equal sides in the triangle
Therefore , side of square = a * (2)^(1/2)
Perimeter of square = 4 a * (2)^(1/2)

Perimeter of rectangle = a*2 + 2a * 2 = 6a
Ratio of perimeter of square X to perimeter of rectangle Y = (2)^(1/2) /3

Answer C
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Bunuel

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.

Attachment:
2015-10-20_1410.png

By spliting a square into 4 equal isosceles triangles you have those triangles with x:x:x(\(\sqrt{2}\)) ratio.

Regarding the square you have therefore a perimeter of \(4*\sqrt{2}\).

Since the question stem states that all 8 triangles are the same, the rectangular region therefore has perimeter = 6*1.

Hence the ratio X:Y is \(2*\sqrt{2}\):3

Answer C
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Bunuel

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.

Attachment:
2015-10-20_1410.png

let side of square be a
then the perimeter of square = 4 a

and the diagonal of square =\(\sqrt{2} a\)

therefore each of other two sides of cardboard \(= (\sqrt{2} * a)/2 = a/\sqrt{2}\)

so, perimeter of rectangle = \(6 a/\sqrt{2} = 3\sqrt{2} a\)

required ratio = \((4 a) / 3\sqrt{2} a = 2\sqrt{2} / 3\)

Answer choice C

kudos, if you like explanation
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Given Info: We are given two figures, one rectangle and another square which are formed by identical triangular pieces. We need to find the ratio of perimeters of these 2 figures.

Interpreting the Problem: In order to find the ratio of the perimeter of these 2 figures, we have to first workout the sides of the identical triangles which form the square and triangle.

Solution: Finding the sides of the identical triangle in terms of sides of the square. Let us assume the side of the square to be a. Now the diagonal of the square will be \(a\sqrt{2}\). Now since the diagonals of the square bisect each other, side of the identical triangle will be \(a\sqrt{2}/2\).
The other side of the triangle will be the side of the square i.e. a.

The sides of identical triangles is shown in the figure.

Attachment:
5.png
5.png [ 6.35 KiB | Viewed 91168 times ]

Now Calculating the perimeter of Figure X.
Perimeter of the square will be 4a

Calculating the perimeter of Figure Y.
Perimeter of rectangle will be \(2a\sqrt{2}/2\) + \(4a\sqrt{2}/2\) = \(3a\sqrt{2}\)

Ratio of perimeter of both figures

\(Perimeter X/Perimeter Y\) = 4a:\(3a\sqrt{2}\) =\(2\sqrt{2}:3\)
Hence, option C is correct.
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I have a slight confusion. I got the answer wrong , as I considered the diagonal to be square root 2a, the sides of square as a, and therefore perimeter to be 4a. Isn't the diagonal making two Isosceles right triangle as well? Why are we considering smaller 4 triangles and not the the other 2 triangle?
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ruchisankrit
I have a slight confusion. I got the answer wrong , as I considered the diagonal to be square root 2a, the sides of square as a, and therefore perimeter to be 4a. Isn't the diagonal making two Isosceles right triangle as well? Why are we considering smaller 4 triangles and not the the other 2 triangle?

If you assume the diagonal to be \(\sqrt{2}\)a
So the sides of the square will be a
Hence the perimeter of square = 4a

Now coming to the rectangle,
In the rectangle, the placements of the triangle is in different way.
Hence the breadth of the rectangle = a/ \(\sqrt{2}\)
Length = 2a / \(\sqrt{2}\)
Perimeter = 6a / \(\sqrt{2}\)

Ratio = 4a* \(\sqrt{2}\)/6 = 2 \(\sqrt{2}\)/3
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Why should this be considered only as isoscles triangles and not as equilateral triangles ?
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Swaroopdev
Why should this be considered only as isosceles triangles and not as equilateral triangles ?
Because the diagonals of a square bisect each other at 90 degrees.
Hence we need to have 4 right angled isosceles triangle to make a square

The equilateral triangle will not do the trick. You cannot make a square using 4 equilateral triangles.
Does this help?
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Bunuel
I am not sure what I am doing wrong but I just simply CANNOT get to the result.
My logic is: perimeter of X = 4a and perimeter of Y = 3a√2. So then the ratio is: perimeter of X / perimeter of Y which is: 4a/3a√2 = 4/3√2..... and that simply is not equivalent to what the answer is: 3/2√2. Can someone please explain????? THANK YOU ALL!
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Bunuel
I am not sure what I am doing wrong but I just simply CANNOT get to the result.
My logic is: perimeter of X = 4a and perimeter of Y = 3a√2. So then the ratio is: perimeter of X / perimeter of Y which is: 4a/3a√2 = 4/3√2..... and that simply is not equivalent to what the answer is: 3/2√2. Can someone please explain????? THANK YOU ALL!

Hi,
you are taking opposite sides..
the square sides are composed of Hypotenuse...so each side is \(\sqrt{2}a\), so P= \(4a*\sqrt{2}\)..
whereas sides of rectangle is a.. so P=6a..
ratio = \(4a*\sqrt{2}/6a\) = \(2*\sqrt{2}/3\)
C
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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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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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Bunuel

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.

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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Bunuel

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.

Attachment:
2015-10-20_1410.png

Check out our detailed video solution to this problem by Chris here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#-soluti ... olving_138
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First, let’s focus on Square X. By definition, diagonals of a square are perpendicular to each other and divide each other equally.

So, if we let \(a\) the length of half the diagonal, we have:



Since the triangles are isosceles right triangles, the ratio of the three sides is \(1:1:a\sqrt{2}\).

Hence, we have the triangles’ ratio of sides as \(a:a:a\sqrt{2}\). This means that Square X has a length of \(a\sqrt{2}\).


Square X has a perimeter of \(= 4 (a√2) = 4a\sqrt{2}\) units.

Since Rectangle Y is made out of the same triangles, we have the following dimensions:


The perimeter of Rectangle Y is \(6a\) units

Hence, the ratio of the perimeters is
\(4a\sqrt{2}\):\(6a\)
\(4\sqrt{2}\):\(6\)
\(2\sqrt{2}\):\(3\)

The final answer is .
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Hi All,

This prompt shows us two shapes (a square and a rectangle) that were made up using 8 IDENTICAL triangles. We’re asked for the ratio of the PERIMETER of the square to the PERIMETER of the rectangle. It’s worth noting that the 4 sides of the square are the HYPOTENEUSE of the triangles and the 6 segments that make up the perimeter of the rectangle are the LEGS of the triangles. This question can be solved in a couple of different ways, including by TESTing VALUES.

IF… the triangles are Isosceles triangles with sides of 3, 3 and 3√2, then…

The perimeter of the square is 4(3√2) = 12√2

The perimeter of the rectangle is 6(3) = 18

Thus, the ratio is 12√2 : 18, which can be reduced by dividing both pieces by 6… This gives us a ratio of 2√2 : 3

Final Answer:
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