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

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

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20 Oct 2015, 03:13
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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 [ 11.87 KiB | Viewed 23292 times ]

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

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28 Apr 2016, 07:15
13
7
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.

Attachment:
2015-10-20_1410.png

We are given that square X and rectangle Y were created with 8 identical triangles. We use the fact that the two diagonals of a square always form right angles where they intersect. Thus, we see that square X is composed of 4 identical 45-45-90 right triangles and rectangle Y is also composed of 4 of these 45-45-90 right triangles. Since 45-45-90 right triangles have a side ratio of x: x: x√2, we can label the sides of our triangles in the two figures. Notice that the sides of the square are the hypotenuses of the triangles and the sides of the rectangle are the legs of the triangles.

Letting n be the length of the side of the 45-45-90 triangle, we label the diagram as shown:

We have what we need to determine the perimeter of square X and rectangle Y.

Perimeter of X = n√2 + n√2 + n√2 + n√2 = 4n√2

Perimeter of Y = n + n + n + n + n + n = 6n

Finally we must determine the ratio of the perimeter of square X to rectangle Y.

(Perimeter of X)/(Perimeter of Y) = (4n√2)/(6n) = 2√2/3, or, equivalently, 2√2 : 3

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

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29 Nov 2015, 06:08
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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 [ 6.35 KiB | Viewed 19385 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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Re: Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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20 Oct 2015, 04:10
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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

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

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20 Oct 2015, 04:23
13
6
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.

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 = $$4\sqrt{2}/6$$ = $$2\sqrt{2}/3$$

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

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20 Oct 2015, 07:26
2
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.

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

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

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21 Oct 2015, 04:48
2
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.

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

kudos, if you like explanation
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Re: Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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11 Dec 2015, 00:28
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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Re: Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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11 Dec 2015, 04:01
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ruchisankrit wrote:
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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Re: Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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11 Dec 2015, 04:22
Why should this be considered only as isoscles triangles and not as equilateral triangles ?
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Re: Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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11 Dec 2015, 04:28
2
Swaroopdev wrote:
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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Re: Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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21 Dec 2015, 20:30
Hi i used diagonal as a√2 so in square because of diagonal bisector it becomes a√2/2. In rectangle we have six a√2/2. So perimeter will be sum of these six sides? Is it wrong as i am not getting correct ans by this method
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Re: Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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16 Mar 2016, 15:54
GMATinsight wrote:
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.

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 = $$4\sqrt{2}/6$$ = $$2\sqrt{2}/3$$

can you show how the 6 for rectangle is calculated exactly?
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Re: Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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17 Mar 2016, 06:08
sagnik242 wrote:
GMATinsight wrote:
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.

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 = $$4\sqrt{2}/6$$ = $$2\sqrt{2}/3$$

can you show how the 6 for rectangle is calculated exactly?

I n my solution I mentioned

Every piece is an isosceles right angle triangle with side = 1 (assumed)

i.e. the Rectanle have two sides of Dimension 2 unit each (horizontal lines) and two sides of Dimension 1 unit each (Vertical lines)

therefore perimeter = Sum of all 4 sides = 2+2+1+1 = 6

I hope that helps!!!
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Re: Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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04 Apr 2016, 16:43
This is one of the best question that is present in the official guide.
Here the triangle involved is an isosceles right triangle
let the sides be => x,x,x√2
so the perimeter of the square => 4 * x√2 => 4√2 * x
now the perimeter of the rectangle => 6x
hence the ratio => 2√2:3
Hence C
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Re: Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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10 Apr 2016, 07:09
vaidhaichaturvedi wrote:
Hi i used diagonal as a√2 so in square because of diagonal bisector it becomes a√2/2. In rectangle we have six a√2/2. So perimeter will be sum of these six sides? Is it wrong as i am not getting correct ans by this method

Hello everyone! I understand the explanation to get to answer C, but I got it wrong on the first time because of the same thing above! Can anyone please help us with this method?

Thanks!
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Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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10 Apr 2016, 11:58
let sides opposite 90°∠=√2
sides opposite 45°∠=1
perimeter of x=4√2
perimeter of y=6
4√2:6=2√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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24 Apr 2016, 15:29
1
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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Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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25 Apr 2016, 04:12
1
karim2982 wrote:
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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Re: Figures X and Y above show how eight identical triangular pieces of ca  [#permalink]

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28 Jun 2016, 19:07
Hi,

I think I am going to ask a very dumb question.
The side of a square is say = a (root 2),
diagonal = a ---- Understood

However,
Why don't we take the side of the rectangle = side of square (length) * 2*(side of square) (width)
why do we consider the side of rectangle (length) = 1/2 diagonal of square
Re: Figures X and Y above show how eight identical triangular pieces of ca &nbs [#permalink] 28 Jun 2016, 19:07

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