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Alchemist-mba
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Post subject: Sphere Plm - pls Help  Posted: Fri Sep 12, 2008 3:47 pm |
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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
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Post subject: Posted: Fri Sep 12, 2008 7:43 pm |
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it said vertex and stuff?
Are 1 2 and 3 the answer choices? Pls provide
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Alchemist-mba
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Post subject: Posted: Fri Sep 12, 2008 11:49 pm |
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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
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Post subject: Posted: Sat Sep 13, 2008 12:51 pm |
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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
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Post subject: Posted: Thu Oct 09, 2008 7:19 am |
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| ManhattanGMAT Staff |
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Posts: 8087
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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
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Post subject: Re: Sphere Plm - pls Help Posted: Sat Nov 13, 2010 11:02 pm |
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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.htmlI 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
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Post subject: Re: Sphere Plm - pls Help Posted: Sun Nov 14, 2010 12:05 am |
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| ManhattanGMAT Staff |
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Posts: 2395
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Thanks everyone! Sounds like this one has been successfully explained.
_________________
Jamie Nelson
ManhattanGMAT Instructor
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miteshsholay
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Post subject: Re: Sphere Plm - pls Help Posted: Mon Aug 20, 2012 5:17 am |
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 why are we taking the longest diagonal? and subtracting the radius from it sounds awkward.they two go separate ways if we refer to the attached figure.shouldnt BX be the shortest distance? if we take surface diagonal and subtract diameter of cylinder from that, we can half the obtained distance to get even shorter distance between cylinder and vertex. why isnt 5(root2 -1) the ans? plz somebody explain.
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miteshsholay
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Post subject: Re: Sphere Plm - pls Help Posted: Mon Aug 20, 2012 5:23 am |
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oh.its a sphere.
i thought it was a cylinder.
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tim
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Post subject: Re: Sphere Plm - pls Help Posted: Tue Aug 21, 2012 1:24 pm |
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Location: Southwest Airlines, seat 21C
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okay. let us know if there are any other questions on this one..
_________________
Tim Sanders
Manhattan GMAT Instructor
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ag1816
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Post subject: Re: Sphere Plm - pls Help Posted: Sun Mar 17, 2013 5:50 pm |
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To find the shortest distance from the vertex of the cube to the sphere, why would we subtract only 1/2 of the diagonal from the radius of the sphere. I don't understand why we wouldn't subtract the entire diagonal length from the diameter of the sphere?
Is it because by doing the above you are actually finding the distance distance on both sides of the sphere - i.e. the gap between the sphere and the cube, instead of finding the distance between just 1 vertice and sphere?
May have just answered my own question, but want to make sure I understood correctly.
thanks
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RonPurewal
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Post subject: Re: Sphere Plm - pls Help Posted: Mon Mar 18, 2013 10:26 am |
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ag1816 wrote: Is it because by doing the above you are actually finding the distance distance on both sides of the sphere - i.e. the gap between the sphere and the cube, instead of finding the distance between just 1 vertice and sphere? precisely.
_________________
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C.F. Forbes
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