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# In the diagram above, triangle ABC is equilateral, figure SQRE is a sq

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Math Expert
Joined: 02 Sep 2009
Posts: 46167
In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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04 May 2015, 06:03
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Difficulty:

85% (hard)

Question Stats:

63% (02:35) correct 37% (02:58) wrong based on 207 sessions

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In the diagram above, triangle ABC is equilateral, figure SQRE is a square, and A is the midpoint of SQ. If the perimeter of triangle ABC is 6 inches, what is the length, in inches, of segment RY ?

A) 0.5

B) 1.5

C) $$\sqrt{3}-1.5$$

D) $$2-\sqrt{3}$$

E) $$\frac{\sqrt{3}}{2}$$

Attachment:

PS_3.gif [ 2.5 KiB | Viewed 4532 times ]

Kudos for a correct solution.

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In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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04 May 2015, 07:04
1
its C.

side of triangle is 2.draw a perpendicular from A to the bottom opf the square.Take it as point Z.

since AC =2,AE=sqrt3(30,60,90 trianghoe)

now AQ is sqrt3/2 (as it half side of sqare)

now AQY is 30,60,90 triangle too. with AQ=sqrt3/2

so QY=3/2(sqrt3 *sqrt3/2)

yr=qr-qy=sqrt3/2-3/2

option C
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Re: In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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04 May 2015, 14:40
1
find side of triangle: 6/3 = 2 inch per side
find area of equilateral triangle: 2²* sqrt(3) / 4 = sqrt (3)
find height: area of triangle = sqrt(3) = 1/2 * base * height -> solve for height: sqrt (3) = 1/2 * 2 * height -> height= sqrt (3)
find third side: its a 90° triangle -> ratio is 1:sqrt(3):2 -> third site = 1 inch.
find RB: (2 - sqrt(3)) / 2
find RY: use Thales' theorem -> 1 / sqrt(3) = RB / RY -> 1 / sqrt(3) = ((2 - sqrt(3)) / 2) / RY -> solve for RY -> RY = sqrt(3) - 1.5

C
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Re: In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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05 May 2015, 01:26
4

Perimeter of equilateral $$\triangle$$ = 6
Side of triangle = AB=BC=AC = 2

Side of Square = $$\sqrt{3}$$/2 * a = $$\sqrt{3}$$/2 * 2 = $$\sqrt{3}$$

Now $$\triangle$$YQA and $$\triangle$$YRB are similar triangles ; ----> As one angle $$90^{\circ}$$, $$\angle$$y vertically opposite; all angles equal (AAA).

As $$\triangle$$YQA ~ $$\triangle$$YRB ; implies ratio of sides equal $$\frac{YR}{RB} = \frac{YQ}{AQ}$$

let YR = x , YQ = $$\sqrt{3}$$ - x ; ----> ( as QR = $$\sqrt{3}$$ )

AQ = $$\sqrt{3}$$/2 , RB =(2-$$\sqrt{3}$$)/2 ;

$$\frac{YR}{RB} = \frac{YQ}{AQ}$$

$$\frac{x}{(2-\sqrt{3})/2}$$ = $$\frac{\sqrt3-x}{(\sqrt{3}/2)}$$

x = $$\frac{(2\sqrt3 - 3 )}{2}$$

x = $$\sqrt3 - 1.5$$

Ans : C
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Re: In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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01 Jun 2015, 10:11
hsbinfy :
Can you please explain how did you get QY=3/2.I am getting it as 1/2.
Please explain how did you apply 30-6-90 to that triangle and how did you got tht value.

Thanks
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Joined: 12 Nov 2014
Posts: 62
Re: In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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01 Jun 2015, 11:16
1
Shree9975 wrote:
hsbinfy :
Can you please explain how did you get QY=3/2.I am getting it as 1/2.
Please explain how did you apply 30-6-90 to that triangle and how did you got tht value.

Thanks

See the attachment.

For 30-60-90 triangle, sides are in the ratio 1 : √3 : 2.
You just have to apply this rule to the triangles in the figure.

You will see that RY = √3 - 1.5

Attachments

tria.jpg [ 35.17 KiB | Viewed 3559 times ]

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Re: In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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02 Jun 2015, 07:15
1
Here's the step-by-step guide to thinking through this question.

Let the length of RY be x inches.

We are given that Triangle ABC is equilateral. This means, Angle ABC = 60 degrees

In right triangle BRY,

$$\frac{RY}{BR} = tan60 =\sqrt{3}$$

That is, $$\frac{x}{BR} = \sqrt{3}$$ . . . (1)

If we can find the value of BR, we will be able to find the value of x.

So, let's try to find more about BR now.

We are given that the perimeter of the equilateral triangle ABC = 6 inches

This means, each side of triangle ABC = 6/3 = 2 inches

Now, let each side of square QRES be 2a units.

Since we are given that A is the mid-point of side QS, this means that equilateral triangle ABC is placed symmetrically about the square QRES.

Therefore, $$CE = BR = \frac{(2 - 2a)}{2} = 1 - a$$ . . . (2)

Substituting (2) in (1), we get:

$$\frac{x}{(1-a)} = \sqrt{3}$$ . . . (3)

Equation 3 contains two unknowns: x and a. So, to find a unique value of x, we now know that we should try to find another relation involving x and/or a.

We get it by dropping a perpendicular from A on side BC.

In right triangle APB,

$$\frac{AP}{BP} = tan60 = \sqrt{3}$$

That is, $$\frac{2a}{1} = \sqrt{3}$$

That is, $$a = \frac{(sqrt3)}{2}$$ . . . (4)

By solving (3) and (4), we get $$x = \sqrt{3} - 1.5$$

Hope this helped!

Best Regards

Japinder
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Re: In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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02 Jun 2015, 07:53
Bunuel wrote:

In the diagram above, triangle ABC is equilateral, figure SQRE is a square, and A is the midpoint of SQ. If the perimeter of triangle ABC is 6 inches, what is the length, in inches, of segment RY ?

A) 0.5

B) 1.5

C) $$\sqrt{3}-1.5$$

D) $$2-\sqrt{3}$$

E) $$\frac{\sqrt{3}}{2}$$

Attachment:
The attachment PS_3.gif is no longer available

Kudos for a correct solution.

Attachments

File comment: www.GMATinsight.com

sol1.jpg [ 240.74 KiB | Viewed 3518 times ]

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Re: In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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02 Jun 2015, 08:37
Bunuel wrote:

In the diagram above, triangle ABC is equilateral, figure SQRE is a square, and A is the midpoint of SQ. If the perimeter of triangle ABC is 6 inches, what is the length, in inches, of segment RY ?

A) 0.5

B) 1.5

C) $$\sqrt{3}-1.5$$

D) $$2-\sqrt{3}$$

E) $$\frac{\sqrt{3}}{2}$$

Attachment:
PS_3.gif

Kudos for a correct solution.

Similar questions to practice:
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in-the-figure-above-sqre-is-a-square-ab-ac-and-as-aq-161814.html
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Re: In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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19 Nov 2015, 15:28
I can not understand how we get 1.5 i understand how we got Square root 3 please expert help. thank you in advance.
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Re: In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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20 Jun 2016, 12:43
1
All of you are doing this problem incorrectly if you even begin to do the algebra or geometry.

If you know that the side of the triangle is 2 inches, then you know the length of RY has to be significantly less than that, so take a look at the answer choices.

A) 1/2 - way too big.
B) 1.5 - way too big.
C) .2 - close, keep this in mind
D) .3 - close, but still too big.
E) .85 - way too big.

The answer is between C and D, and because that segment looks a lot closer to 1/5 than to 1/3, go with answer C.

Time ~20 seconds
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Re: In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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21 Sep 2017, 01:04
1
1
Here is my solution. Since, this one had a diagram and lot of rules, I posted an image. Hope this helps.
Attachments

21903586_10207991228218545_79220063_n.jpg [ 74.7 KiB | Viewed 1179 times ]

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Re: In the diagram above, triangle ABC is equilateral, figure SQRE is a sq   [#permalink] 21 Sep 2017, 01:04
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