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For the cube shown above, what is the degree measure of PQR

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Re: For the cube shown above, what is the degree measure of PQR [#permalink]

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New post 25 Jul 2016, 02:37
Keats wrote:
Bunuel wrote:
Keats wrote:
Bunuel - What will be the value of each diagonal in the equilateral triangle that is formed here ? (triangle PQR)


Do you mean angles instead of diagonals? Triangle does not have diagonals. If you mean angles, then each angle in equilateral triangle is 60 degrees.


Sorry that I didn't put it up correctly. I wanted to ask that what will be the value of each side of the equilateral triangle (which happens to be the diagonal of the cube), Bunuel


If we assume that the length of a side of the cube is x, then the length of each diagonal = \(\sqrt{x^2 + x^2} = \sqrt{2}x\)
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Re: For the cube shown above, what is the degree measure of PQR [#permalink]

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New post 25 Jul 2016, 02:39
Bunuel wrote:
If we assume that the length of a side of the cube is x, then the length of each diagonal = \(\sqrt{x^2 + x^2} = \sqrt{2}x\)


Thanks Bunuel. This helps :-D

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Re: For the cube shown above, what is the degree measure of PQR [#permalink]

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New post 01 Aug 2016, 10:36
nocilis wrote:
Attachment:
The attachment cube.gif is no longer available
For the cube shown above, what is the degree measure of PQR?

A. 30
B. 45
C. 60
D. 75
E. 90



Draw a construction (Pink Line) and

Answer: Option C
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Answer 3.jpg [ 58.96 KiB | Viewed 720 times ]


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Re: For the cube shown above, what is the degree measure of PQR [#permalink]

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New post 27 Aug 2016, 15:15
This question is now a part of GMAT Prep Exam Pack 2. Please update.

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Re: For the cube shown above, what is the degree measure of PQR [#permalink]

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New post 24 Dec 2016, 10:57
Bunuel wrote:
ferrarih wrote:
Hi, I understood the answer and I agree, but I can't find out why my logic isn't right.

Since it's a cube, all angles are 90 degrees, and since PO and RO are the bisectrix of the angle 90, than PQR would have to be 90, which is not the answer. What am thinking wrong?

Thanks.


Below solution might help.

For the cube shown above, what is the degree measure of PQR?
A. 30
B. 45
C. 60
D. 75
E. 90
Attachment:
cube.gif
Note that triangle PQR is equilateral: it's made by the diagonals of the adjacent faces of the given cube (and as faces of a cube are squares its diagonals are equal). Thus angle BEG=60 degrees.

Answer: C.

Bunuel and others - I still don't get it. PQ is in a different plane and QR is in a different plane and angle between two planes which are perpendicular to each other is 90 degrees irrespective of which lines we use to connect. What you and others have done here is created a new plane (plane in which the equilateral triangle resides). Please explain where am I going wrong in this thinking?

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Re: For the cube shown above, what is the degree measure of PQR [#permalink]

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New post 25 Dec 2016, 01:58
gmat2k17 wrote:
Bunuel wrote:
ferrarih wrote:
Hi, I understood the answer and I agree, but I can't find out why my logic isn't right.

Since it's a cube, all angles are 90 degrees, and since PO and RO are the bisectrix of the angle 90, than PQR would have to be 90, which is not the answer. What am thinking wrong?

Thanks.


Below solution might help.

For the cube shown above, what is the degree measure of PQR?
A. 30
B. 45
C. 60
D. 75
E. 90
Attachment:
cube.gif
Note that triangle PQR is equilateral: it's made by the diagonals of the adjacent faces of the given cube (and as faces of a cube are squares its diagonals are equal). Thus angle BEG=60 degrees.

Answer: C.

Bunuel and others - I still don't get it. PQ is in a different plane and QR is in a different plane and angle between two planes which are perpendicular to each other is 90 degrees irrespective of which lines we use to connect. What you and others have done here is created a new plane (plane in which the equilateral triangle resides). Please explain where am I going wrong in this thinking?


It's not clear what your question is. Yes, triangle PQR is in a plane, how else? We need the angle measure of PQR in that plane.

Please check similar questions to practice:
the-figure-above-is-a-cube-what-is-the-measure-of-angle-beg-126246.html
the-figure-above-is-a-cube-what-is-the-measure-of-angle-beg-129650.html
square-abcd-is-the-base-of-the-cube-while-square-efgh-is-the-56270.html
what-is-the-measure-of-the-angle-made-by-the-diagonals-of-th-88522.html

Hope it helps.
_________________

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Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: For the cube shown above, what is the degree measure of PQR [#permalink]

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New post 26 Dec 2016, 10:12
Bunuel wrote:
gmat2k17 wrote:
Bunuel and others - I still don't get it. PQ is in a different plane and QR is in a different plane and angle between two planes which are perpendicular to each other is 90 degrees irrespective of which lines we use to connect. What you and others have done here is created a new plane (plane in which the equilateral triangle resides). Please explain where am I going wrong in this thinking?


It's not clear what your question is. Yes, triangle PQR is in a plane, how else? We need the angle measure of PQR in that plane.

Please check similar questions to practice:

Hope it helps.


Apologies for the lack of clarity. Let me have a go at it again.

The front plane and the side plane of the cube are at 90 degrees to each other. For angle PQR, line PQ is in the front plane and line QR is in the adjacent plane and hence these two lines are at 90 degrees to each other since they reside on planes which are perpendicular to each other. However, in your answer and the answer posted by others we are taking a separate plane which is going inside the cube.

Is my question clear this time? I think I am lacking a fundamental understanding of 3D geometry here.

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Re: For the cube shown above, what is the degree measure of PQR [#permalink]

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New post 21 Aug 2017, 10:28
interesting! I have get an actual wooden box to understand this concept

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Re: For the cube shown above, what is the degree measure of PQR   [#permalink] 21 Aug 2017, 10:28

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