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The cube above has side length of 4.

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The cube above has side length of 4. [#permalink] New post   19 Sep 2017, 08:20
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The cube above has side length of 4.

Quantity A
Quantity B
After selecting one vertex of the cube, the
number of straight line segments longer than 4
that can be drawn from that vertex of the cube
to another vertex of the cube.
When the cube is placed on a flat surface, the
maximum number of edges of the cube that can
be touching the flat surface at once.


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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C
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IlCreatore
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Re: The cube above has side length of 4. [#permalink] New post   21 Sep 2017, 06:59
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From any of the vertices of a cube two kinds of segments depart: edges that we know being equal to 4 and diagonals that are either \(4sqrt(3)\) or \(4sqrt(2)\) that are both longer than 4. Thus we have to take in consideration this least kind of segment, which are 4 for each vertex. Then when a cube is put on a flat surface, there are at most 4 edges touching it. Thus the answer is C
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Re: The cube above has side length of 4. [#permalink] New post   16 Sep 2019, 10:48
IlCreatore wrote:
From any of the vertices of a cube two kinds of segments depart: edges that we know being equal to 4 and diagonals that are either \(4sqrt(3)\) or \(4sqrt(2)\) that are both longer than 4. Thus we have to take in consideration this least kind of segment, which are 4 for each vertex. Then when a cube is put on a flat surface, there are at most 4 edges touching it. Thus the answer is C


The endpoints of the perpendicular edges also "touches" the surface. Why did you discard that?
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Re: The cube above has side length of 4. [#permalink] New post   09 Oct 2022, 06:38
Can someone explain this please? From one vertex to another won't we use pythagoras theorem? so one diagonal would be 4*Square root iof 3 right?
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Re: The cube above has side length of 4. [#permalink]
09 Oct 2022, 06:38
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