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

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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
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 $$= 4*\sqrt{2} : 6 = 2\sqrt{2} : 3$$

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

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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) $$\sqrt{2}:1$$

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[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

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

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11 Sep 2017, 13:25
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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

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

[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)

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

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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

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

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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
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Re: Figures X and Y above show how eight identical triangular pieces of ca   [#permalink] 21 Sep 2017, 22:44

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