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A paper triangle with sides of lengths 3, 4, and 5 inches, as shown

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A paper triangle with sides of lengths 3, 4, and 5 inches, as shown  [#permalink]

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New post 10 Nov 2018, 10:28
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GMATbuster's Weekly Quant Quiz#8 Ques #9


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A paper triangle with sides of lengths 3, 4, and 5 inches, as shown, is folded so that point A falls on point B. What is the length in inches of the crease?


A) \(\frac{(1+ \sqrt{2})}{2}\)

B) \(\sqrt{3}\)

C) 7/4

D) 15/8

E) 2

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Re: A paper triangle with sides of lengths 3, 4, and 5 inches, as shown  [#permalink]

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New post 10 Nov 2018, 10:51
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Crease will be from midpoint of AB and perpendicular to AB. Let the perpendicular touch AC at D and start at E on AC.
Triangles AED and ABC are similar since one angle is common, other is 90 deg and one side is in a ratio of 1:2 (AE:AC)

So, ED/AE = BC/AB which gives, ED/(5/2) = 3/4 or ED = 15/8

Hence Option D
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Re: A paper triangle with sides of lengths 3, 4, and 5 inches, as shown  [#permalink]

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New post 10 Nov 2018, 11:14
Let us assume that the crease falls on vertex C.

So what is the max value that the crease can take:


If hypotenuse is 4:

16-6.25=9.75 (since A has been joined with B, we can say that the segment 5 has been divided into half 2.5)

So Max has to be close to 3


What is the minimum value:

If Hypotenuse is 3:

9-6.25= 2.75

So Square root of 2.75 lies between 1 and 2.
So we can say that the value has to be slighty greater than 1.5

So the closes value is root 3.

Hence answer should be Root 3
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Re: A paper triangle with sides of lengths 3, 4, and 5 inches, as shown  [#permalink]

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New post 10 Nov 2018, 11:26
Let midpoint of AB be D
intersection point with AC be E.
triangle ACB is similar to triangle ADE

using AA


BC/AC = DE/AD =3/4 = DE/(5/2)
DE = 15/8
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Re: A paper triangle with sides of lengths 3, 4, and 5 inches, as shown  [#permalink]

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New post 11 Nov 2018, 20:37
nkin wrote:
Crease will be from midpoint of AB and perpendicular to AB. Let the perpendicular touch AC at D and start at E on AC.
Triangles AED and ABC are similar since one angle is common, other is 90 deg and one side is in a ratio of 1:2 (AE:AC)

So, ED/AE = BC/AB which gives, ED/(5/2) = 3/4 or ED = 15/8

Hence Option D


Hi,

Can you please explain..."Crease will be from midpoint of AB and perpendicular to AB"...How is it perpendicular to AB?
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Re: A paper triangle with sides of lengths 3, 4, and 5 inches, as shown  [#permalink]

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New post 11 Nov 2018, 20:46
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Kezia9 wrote:
nkin wrote:
Crease will be from midpoint of AB and perpendicular to AB. Let the perpendicular touch AC at D and start at E on AC.
Triangles AED and ABC are similar since one angle is common, other is 90 deg and one side is in a ratio of 1:2 (AE:AC)

So, ED/AE = BC/AB which gives, ED/(5/2) = 3/4 or ED = 15/8

Hence Option D


Hi,

Can you please explain..."Crease will be from midpoint of AB and perpendicular to AB"...How is it perpendicular to AB?


Sure, consider a rectangular piece of paper. Now, if you fold the paper by joining one vertex to another adjacent vertex, lets say you do this on one of the shorter sides, then can you see that the crease will be perpendicular to that shorter side? Try imagining this with an equilateral triangle as well. Similarly, when you fold any figure along one side, then the resulting crease will be perpendicular. As a corollary, if a crease from a fold is not perpendicular, then the figure has not been folded along that side to which the crease is perpendicular.

Hope all this fluff makes sense!
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Re: A paper triangle with sides of lengths 3, 4, and 5 inches, as shown  [#permalink]

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Re: A paper triangle with sides of lengths 3, 4, and 5 inches, as shown  [#permalink]

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New post 12 Nov 2018, 23:36
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In my approach, I assumed that the crease line would bisect line AB as per the solution suggested by gmatbuster, however I also assumed that the crease line would intersect the triangle at angel C, however the solution here suggests that there is a new point of intersection at E. Can someone explain the reason why.

I applied Pythagoras theorem for a newly formed triangle as per the solution suggested by nitesh50

If hypotenuse is 4:

16-6.25=9.75 (since A has been joined with B, we can say that the segment 5 has been divided into half 2.5)

So Max has to be close to 3

Is this approach not right?
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Re: A paper triangle with sides of lengths 3, 4, and 5 inches, as shown  [#permalink]

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New post 12 Nov 2018, 23:41
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Hi
Query: Hi, I just have question regarding the solution you have suggested, can you explain why crease line does not bisect angle C in the diagram?
Any reason why the assumption is the perpendicular crease line intersects at a new point E.

Response: As it is clear from the legs of the right triangle ABC, the legs are unequal: 3 and4. So the crease will not be symmetric w.r.t angle C. The crease would be towards the longer side and hence intersect side AC at E.
If the triangle had been isosceles right angled triangle, the crease would bisect angle C too.

Before fold
Attachment:
2018_11_13 14_23 Office Lens.jpg
2018_11_13 14_23 Office Lens.jpg [ 88.52 KiB | Viewed 797 times ]


During fold
Attachment:
2018_11_13 14_27 Office Lens.jpg
2018_11_13 14_27 Office Lens.jpg [ 83.1 KiB | Viewed 793 times ]


Crease formed
Attachment:
2018_11_13 14_28 Office Lens.jpg
2018_11_13 14_28 Office Lens.jpg [ 66.59 KiB | Viewed 795 times ]

Hope, it is clear. Feel free to tag me again

Happy Learning.
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Re: A paper triangle with sides of lengths 3, 4, and 5 inches, as shown  [#permalink]

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New post 12 Nov 2018, 23:57
gmatbusters wrote:
Hi
Query: Hi, I just have question regarding the solution you have suggested, can you explain why crease line does not bisect angle C in the diagram?
Any reason why the assumption is the perpendicular crease line intersects at a new point E.

Response: As it is clear from the legs of the right triangle ABC, the legs are unequal: 3 and4. So the crease will not be symmetric w.r.t angle C. The crease would be towards the longer side and hence intersect side AC at E.
If the triangle had been isosceles right angled triangle, the crease would bisect angle C too.

Hope, it is clear. Feel free to tag me again

Happy Learning.


Thank you it is clear now.
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A paper triangle with sides of lengths 3, 4, and 5 inches, as shown  [#permalink]

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New post 14 Nov 2018, 08:02
[quote="gmatbusters"]

Official Solution



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A paper triangle with sides of lengths 3, 4, and 5 inches, as shown   [#permalink] 14 Nov 2018, 08:02
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