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A sphere is inscribed in a cube with an edge of 10. What is [#permalink]
12 Feb 2012, 20:48

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Question Stats:

52% (01:38) correct
48% (00:38) wrong based on 303 sessions

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 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(\sqrt{3}- 1) (B) 5 (C) 10(\sqrt{2} - 1) (D) 5(\sqrt{3} - 1) (E) 5(\sqrt{2} - 1)

It would be easier if you visualize this problem.

As sphere is inscribed in cube then the edges of the cube equal to the diameter of a sphere --> Diameter=10.

Next, diagonal of a cube equals to Diagonal=\sqrt{10^2+10^2+10^2}=10\sqrt{3}.

Now half of (Diagonal minus Diameter) is a gap between the vertex of a cube and the surface of the sphere, which will be the shortest distance: x=\frac{Diagonal -Diameter}{2}=\frac{10*\sqrt{3}-10}{2}=5(\sqrt{3}-1)

Re: A sphere is inscribed in a cube with an edge of 10. What is [#permalink]
09 Jun 2013, 10:54

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Expert's post

WarriorGmat wrote:

i dint understood why diameter is 10? how smallest area is 1/2(diameter-diagonal) question was dreaded for me

bunuel help

Consider the cross-section as shown below:

Attachment:

square.png [ 3.86 KiB | Viewed 11459 times ]

The diameter = The edge.

As for your second question, check here: a-sphere-is-inscribed-in-a-cube-with-an-edge-of-10-what-is-127461.html#p1097531 The shortest distance from one of the vertices of the cube to the surface of the sphere is 1/2(diagonal of the cube - diameter of the circle). Diagonal of the cube - diameter of the circle, is the length of two little black arrows shown here:

Re: A sphere is inscribed in a cube with an edge of 10. What is [#permalink]
09 Jun 2013, 11:32

thanks bunuel for patiently providin g solution you rock!! too much and too little study is fatal thats what happening to me i have missed such a small stuff.

Re: A sphere is inscribed in a cube with an edge of 10. What is [#permalink]
15 Jun 2014, 11:29

Bunuel wrote:

sanjoo wrote:

Bunuel.. i cant understand the question ? can u elaborate it further..

And ya instead of sphere if it wud be a square inscribed in a cube then wat wud b the answer?

Look at the diagram below:

Attachment:

Sphere inscribed in a cube.png

The question asks about the lengths of the little black arrows shown.

As for the additional question, it doesn't make much sense: what does it mean a square is inscribed in a cube?

Hi Bunuel,

In your diagram, and in all solutions for questions of this kind, the distance is measured from the diagonal of the cube and not from any other place. Why is this so?

Re: A sphere is inscribed in a cube with an edge of 10. What is [#permalink]
17 Jun 2014, 21:10

question says "distance from one of the vertices of the cube to the surface of the sphere?".That's why it is calculated from diagnal Even i got similar doubt and even looked for answer Zero immediately. tricky one.

Re: A sphere is inscribed in a cube with an edge of 10. What is [#permalink]
06 Sep 2014, 10:01

enigma123 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?

formula-> diagonal of the cube- s*(\sqrt{3}) where s- side of the cube.

sphere inscribed in a cube will be having the diameter of the side of cube. so shortest distance will be (diagonal of the cube - diameter of the sphere) /2 .(when we subtract the diameter from diagonal of the cube we will be left with two side of the cube , hence divided by 2 )

give edge as 10 so sides(cube) as well diameter (sphere) =10.

Re: A sphere is inscribed in a cube with an edge of 10. What is [#permalink]
13 Jan 2015, 22:03

If you visualize the problem in your head, you realize that what you want is 1/2 the diagonal of the cube- the radius of the circle.

We know the radius of the circle is 5, because the circle touches the sided of the cube, which has a total length of 10.

If you memorized the diagonal of a cube, which I found helpful to do for my test, then you would know it is side*(sqrt(3)), but we want half of that so it is 5(sqrt(3)).

So, the distance of the vertice to the sphere is 5(sqrt(3))-5. That is not an answer, but we can see that D is the same thing, it just divided out the 5. _________________

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