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A sphere is inscribed in a cube with an edge of 10. What is

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A sphere is inscribed in a cube with an edge of 10. What is [#permalink] New post 18 Dec 2005, 01:46
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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?

A. 10(sqrt3 – 1)

B. 5

C. 10(sqrt2 – 1)

D. 5(sqrt3 – 1)

E. 5(sqrt2 – 1)
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Re: PS MGMAT 1-28 [#permalink] New post 18 Dec 2005, 02:28
GMATT73 wrote:
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?

A. 10(sqrt3 – 1)

B. 5

C. 10(sqrt2 – 1)

D. 5(sqrt3 – 1)

E. 5(sqrt2 – 1)



The needed distance lies on the line connecting two vertices which are symmetric through the centre of the sphere( this centre is also that of the cube)
Let D be the distance between the two symmetric vertices and d be the needed distance:
We have : 2*d= D - the diameter of the sphere

D can be calculated = sqrt( 10^2 + [10sqrt2]^2 ) = 10 sqrt3
the diameter of the sphere= 10 since the sphere is tangently inscribed in the cube
---> 2d= 10sqrt3 - 10 ---> d= 5( sqrt3-1)

D it is.
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 [#permalink] New post 18 Dec 2005, 06:39
diameter of the sphere is 10. Diagonal of the cube is 10sqrt2. So the required distance is (10sqrt2-10)/2 which gives 5(sqrt2-1), or E)
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 [#permalink] New post 18 Dec 2005, 09:08
The answer is D.

The diagonal to be considered is the BODY diagonal which is 10*sqrt(3)
Half of it is 5*sqrt(3) and radius of the sphere is 5.
Hence the answer D.
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 [#permalink] New post 18 Dec 2005, 13:48
yes D indeed,

what would be the answer if the question ask for longest distacne?
is it 5sqrt2????
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 [#permalink] New post 18 Dec 2005, 19:41
SunShine wrote:
yes D indeed,

what would be the answer if the question ask for longest distacne?
is it 5sqrt2????


The longest distance will be from a vertix to the tangent point of the sphere with the cube's side which contributes to the formation of the vertix ---> the distance = 1/2 side of the cube= 5
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 [#permalink] New post 19 Dec 2005, 23:12
Good work gang. OA is D.
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 [#permalink] New post 22 Dec 2005, 03:08
I understand your point however the longest distance =1/2 side of the cube =5 is REALLY longest?


What about the distance from the corner of the cube to the tangent of the sphere to one of the surfaces of the square?

in that case it will be 5 sqrt2.


Do you agree???



laxieqv wrote:
SunShine wrote:
yes D indeed,

what would be the answer if the question ask for longest distacne?
is it 5sqrt2????


The longest distance will be from a vertix to the tangent point of the sphere with the cube's side which contributes to the formation of the vertix ---> the distance = 1/2 side of the cube= 5
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 [#permalink] New post 22 Dec 2005, 03:13
SunShine wrote:
I understand your point however the longest distance =1/2 side of the cube =5 is REALLY longest?


[b]What about the distance from the corner of the cube to the tangent of the sphere to one of the surfaces of the square?[/b]
in that case it will be 5 sqrt2.


Do you agree???


Agree! :)
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 [#permalink] New post 22 Dec 2005, 19:12
-->Formula for the longest diagonal of the cube=sqrtl^2+l^2+l^2=l*sqrt3
==>diagonal of the circle inscribe in square=side of the square

The shortest distance=Length (longest diagonal of the cube)-length (diameter of the sphere)/2=10sqrt3-10/2=10(sqrt3-1)

A it is.
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 [#permalink] New post 22 Dec 2005, 19:13
Sorry, forgot to divide by 2.
-->Formula for the longest diagonal of the cube=sqrtl^2+l^2+l^2=l*sqrt3
==>diameter of the circle inscribe in square=side of the square

The shortest distance=Length (longest diagonal of the cube)-length (diameter of the sphere)/2=10sqrt3-10/2=10(sqrt3-1)/2=5(sqrt3-1)

D it is.
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  [#permalink] 22 Dec 2005, 19:13
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