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Sphere Plm - pls Help

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Alchemist-mba
 Post subject: Sphere Plm - pls Help
 Post Posted: Fri Sep 12, 2008 3:47 pm 
Hi,

I have a question on the answer of one of the question that appeared in GMAT practive exam

The problem is - A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

1. vertex and where the sphere touches the cube - 5
2. vertex and toward inside of cube .. 5(Sqrt(3) – 1)
3. vertex and towards the cube's face or base - 5(Sqrt(2) – 1)

of all these .. '3' yields the minimal value... hence i picked 3.. but the original answer picked '2' - Can you please explain?
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JK
 Post subject:
 Post Posted: Fri Sep 12, 2008 7:43 pm 
it said vertex and stuff?

Are 1 2 and 3 the answer choices? Pls provide
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Alchemist-mba
 Post subject:
 Post Posted: Fri Sep 12, 2008 11:49 pm 
The given solutions are:

A. 10(sqrt(3) – 1)
B. 5
C. 10(sqrt(2) – 1)
D. 5(sqrt(3) – 1)
E. 5(sqrt(2) – 1)

I picked E - the O.A is D.... cant figure why it is D
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JK
 Post subject:
 Post Posted: Sat Sep 13, 2008 12:51 pm 
All you're trying to do is find the difference between the 1/2 the diagonal of the cube and the radius of the cube. Or the diagonal - diameter.

So if a cube has an edge of 10. Radius = 5. Diagonal = 10sqrt3. Half the diagonal is 5sqrt3

The difference is 5sqrt3-5
simplified, 5(sqrt3-1)

I would have said A but realized then you're double counting the gap between the cube and the circle.

PEACE
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RonPurewal
 Post subject:
 Post Posted: Thu Oct 09, 2008 7:19 am 
Offline
ManhattanGMAT Staff


Posts: 7146
JK wrote:
All you're trying to do is find the difference between the 1/2 the diagonal of the cube and the radius of the cube. Or the diagonal - diameter.

So if a cube has an edge of 10. Radius = 5. Diagonal = 10sqrt3. Half the diagonal is 5sqrt3

The difference is 5sqrt3-5
simplified, 5(sqrt3-1)

I would have said A but realized then you're double counting the gap between the cube and the circle.

PEACE


perfection.

if anyone needs to visualize this, here's a visual explanation:
imagine balancing the cube perfectly on one of its corner points (i.e., spinning it like a top).
now, draw a line from the vertex that's on top, straight down (into the interior of the cube), until it hits the sphere. this line, which runs along one of the main diagonals of the cube (i.e., it would drill straight down to the bottom vertex if continued far enough), will hit the surface of the sphere at perfectly right angles.
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tsabed
 Post subject: Re: Sphere Plm - pls Help
 Post Posted: Sat Nov 13, 2010 11:02 pm 
Offline
Students


Posts: 1
Hi for those of you who were just as confused as I was about the same thing: as to why it was root 3 and not root 2
go here for a visual representation and a better understanding of how to calculate the diagonal of a cube:

http://mathcentral.uregina.ca/QQ/databa ... rett1.html

I realised, that the shortest distance is the (INSIDE, MIDDLE diagonal) not the surface diagonal.
Therefore, the (diagonal of the surface of the cube) is root 2
you need to do the pythag theorem again to get
the (inside, middle diagonal) which is root 3 (the diagonal along which the shortest distance lies)

hope that helps.

-t
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jnelson0612
 Post subject: Re: Sphere Plm - pls Help
 Post Posted: Sun Nov 14, 2010 12:05 am 
Offline
ManhattanGMAT Staff


Posts: 1857
Thanks everyone! Sounds like this one has been successfully explained.

_________________
Jamie Nelson
ManhattanGMAT Instructor
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