A sphere is inscribed in a cube with an edge of 10. What is : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 18 Jan 2017, 05:22

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1

Kudos [?]: 43 [0], given: 0

A sphere is inscribed in a cube with an edge of 10. What is [#permalink]

### Show Tags

01 Nov 2006, 08:50
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 â€“ 1)

b. 5

c. 10(sqrt2 â€“ 1)

d. 5(sqrt3 â€“ 1)

e. 5(sqrt2 â€“ 1)
_________________

Impossible is nothing

Intern
Joined: 01 Nov 2006
Posts: 10
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

01 Nov 2006, 12:14
D

distance from the center to the vertex - x

x^2 = 5^2 + (5*sqrt2)^2

x = 5*sqrt3

distance that we're looking for is x-radius therefore x-5
Senior Manager
Joined: 01 Oct 2006
Posts: 497
Followers: 1

Kudos [?]: 32 [0], given: 0

### Show Tags

01 Nov 2006, 12:18
My answer is d.
The length of the diagonal of the face of the cube will be
10sqrt(3).
The radius for sphere will be 5
Smallest distance will be =(10sqrt(3)/2)-5 =5(sqrt(3)-1)
What is OA?
Manager
Joined: 12 Sep 2006
Posts: 91
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

02 Nov 2006, 02:05
I am getting E, where did i go wrong?

i see a right triangle with the radii as two sides and the third side =5sqrt2, therefore the answer is 5sqrt2-5

what's wrong in my thinking?
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

02 Nov 2006, 02:16
keeeeeekse wrote:
I am getting E, where did i go wrong?

i see a right triangle with the radii as two sides and the third side =5sqrt2, therefore the answer is 5sqrt2-5

what's wrong in my thinking?

The 2 sides, linked to the right angle, are :
> 5 (radius of the sphere)
> 10/2*sqrt(2) (the half part of line's length from 2 opposite vertex in a same side)

Thus, for the third, we have: x^2 = 25*2 + 25 = 25*3 that implies x = 5*sqrt(3)

Manager
Joined: 12 Sep 2006
Posts: 91
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

02 Nov 2006, 05:12
I see your point Fig, but i still don't see why the other side should be 10/2*sqrt(2), where does the sqrt(2) come from?
If the sphere is inscribed then, it is touching the sides of the cube in the middle so the 2 sides of the right angle should be identical.

Appreciate your further explanation, thanks!!!
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

02 Nov 2006, 05:23
keeeeeekse wrote:
I see your point Fig, but i still don't see why the other side should be 10/2*sqrt(2), where does the sqrt(2) come from?
If the sphere is inscribed then, it is touching the sides of the cube in the middle so the 2 sides of the right angle should be identical.

Appreciate your further explanation, thanks!!!

As u said, the contact point of the sphere is the center (gravity point) of a scare side. But, the distance is not the radius.

Actually, we can use the fact that the lenght of a line joining 2 opposite vertex on a same scare side is defined by hyp^2=10^2 + 10^2 => hyp = 10*sqrt(2).

Thus, to have the distance from 1 vertex to the contact point of the sphere at the center of a side, we divid by 2: hyp/2 = 10*sqrt(2)/2 = 5*sqrt(2)

The most simple is to draw 1 side of the cube... then, u can notice that each side measure 10.
Manager
Joined: 01 Nov 2006
Posts: 70
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

02 Nov 2006, 05:45
I don't understand that last answer, so maybe it's right.

Anyway, here's my solution.

Imagine the cube sitting with one corner at the origin and another corner at (10, 10, 10). The distance from these two points is the distance of the long diagonal. The distance formula says that this distance is Sqrt((10 - 0)^2 + (10- 0)^2 + (10-0)^2) = Sqrt(3*10^2) = 10*Sqrt(3).

The sphere sitting inside the cube has diameter 10. Since a sphere in a cube is completely symmetric, the distance from a vertex to the sphere is the same for all vertices. So along that long diagonal we run through the middle of the sphere. The distance outside the sphere is 10*Sqrt(3) - 10. The distance from one vertex to the sphere must be half that by symmetry. So the answer is 1/2*(10*Sqrt(3) - 10) = 5*(Sqrt(3) - 1)
Manager
Joined: 12 Sep 2006
Posts: 91
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

03 Nov 2006, 13:16
I just realised i wasn't looking at it 3 dimensionally, now all is clear. Merci beaucoup Fig
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

03 Nov 2006, 13:28
keeeeeekse wrote:
I just realised i wasn't looking at it 3 dimensionally, now all is clear. Merci beaucoup Fig

You are french as well? ... Nice surprise

Which school do u target?
Manager
Joined: 12 Sep 2006
Posts: 91
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

03 Nov 2006, 14:08
I haven't decided yet, i like Kellogg, LBS and INSEAD. My exam is in a few weeks and i want to see if i get a good score, otherwise everything changes!

U?
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

03 Nov 2006, 14:24
I would like to enter in HEC Some clues about it?

If I can not, INSEAD and LBS come after
Manager
Joined: 12 Sep 2006
Posts: 91
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

03 Nov 2006, 15:05
Excellent choice for the french market, even better than INSEAD i think.

when is your exam? and are you applying for 2007?
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

03 Nov 2006, 15:20
Already done it .... 680 The average of HEC is 660

I will visit the HEC campus on this 15 november ... I want to enter on January 2008

I have still the TOEFL to do : Have u already done it? ... Many people present it as easy, what do u think?
Manager
Joined: 12 Sep 2006
Posts: 91
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

03 Nov 2006, 15:44
I am lucky, don't have to do Toefl because i am part canadian...

but I heard if you did GMAT, Toefl is a piece of cake, so no need to worry.
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

03 Nov 2006, 16:01
It's nice... U have not the TOEFL in front of u, a step less

So, u think perhaps to do an MBA on fall 07... I hope u the best on it and for the GMAT as well

I have no ideas about the TOEFL level of difficulty... Anyway, I will work on it and do my best I need to cross the 100/120
03 Nov 2006, 16:01
Display posts from previous: Sort by

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.