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

Question

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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Official Answer

D

nkin · Answer

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

nitesh50 · Answer

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

mishugarg · Answer

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

Kezia9 · Answer

Hi,

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

nkin · Answer

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!

GMATBusters · Answer

Official Solution https://gmatclub.com/forum/download/file.php?id=45054 https://gmatclub.com/forum/download/file.php?id=45055 Solution.jpg WhatsApp Image 2018-11-13 at 00.03.53.jpeg

Preethi88 · Answer

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?

GMATBusters · Answer

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 2018_11_13 14_23 Office Lens.jpg During fold 2018_11_13 14_27 Office Lens.jpg Crease formed 2018_11_13 14_28 Office Lens.jpg Hope, it is clear. Feel free to tag me again Happy Learning.

Preethi88 · Answer

Thank you it is clear now.

Kezia9 · Answer

Official Solution https://gmatclub.com/forum/download/file.php?id=45054 https://gmatclub.com/forum/download/file.php?id=45055 Solution.jpg WhatsApp Image 2018-11-13 at 00.03.53.jpeg [/qu

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