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

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Joined: 07 Jun 2004
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A sphere is inscribed in a cube with an edge of 10. What is  [#permalink]

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10 Sep 2010, 07:39
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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) $$10(\sqrt{3}- 1)$$
(B) $$5$$
(C) $$10(\sqrt{2} - 1)$$
(D) $$5(\sqrt{3} - 1)$$
(E) $$5(\sqrt{2} - 1)$$
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Joined: 02 Sep 2009
Posts: 59720
A sphere is inscribed in a cube with an edge of 10. What is  [#permalink]

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10 Sep 2010, 11:41
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TehJay wrote:
How did you come up with that formula for the shortest distance from the vertex to the sphere?

It would be easier if you visualize this problem.

As sphere is inscribed in cube then the edge of the cube equals 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-Diameter is gap between the vertex of a cube and the surface of the sphere:

$$x=\frac{Diagonal -Diameter}{2}=\frac{10*\sqrt{3}-10}{2}=5(\sqrt{3}-1)$$
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10 Sep 2010, 07:51
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rxs0005 wrote:
A sphere is inscribed in a cube with edge 10. The sphere touches the edges of the cube

What is the shortest distance from any vertex of the cube to the surface of the sphere

5

10

5*root(2) -1

10*root(2) - 1

2*root(5) - 1

Original question, with correct answer choices and correct OA is as follows:

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) $$5(\sqrt{2} - 1)$$

Shortest distance=(diagonal of cube-diameter of sphere)/2= $$\frac{10*\sqrt{3}-10}{2}=5(\sqrt{3}-1)$$

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10 Sep 2010, 11:20
How did you come up with that formula for the shortest distance from the vertex to the sphere?
Manager
Joined: 06 Aug 2010
Posts: 148
Location: Boston

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10 Sep 2010, 11:48
Bunuel wrote:
TehJay wrote:
How did you come up with that formula for the shortest distance from the vertex to the sphere?

It would be easier if you visualize this problem.

As sphere is inscribed in cube then the edge of the cube equals 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-Diameter is gap between the vertex of a cube and the surface of the sphere --> $$x=\frac{Diagonal -Diameter}{2}=\frac{10*\sqrt{3}-10}{2}=5(\sqrt{3}-1)$$

Thanks, I get it now!
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Joined: 07 Feb 2010
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A sphere is inscribed in a cube with an edge of 10  [#permalink]

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14 Dec 2010, 07:32
Attachment:
Book1.pdf [7.98 KiB]

Attachment:
Book1.pdf [7.98 KiB]

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?
Math Expert
Joined: 02 Sep 2009
Posts: 59720
Re: A sphere is inscribed in a cube with an edge of 10. What is  [#permalink]

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16 Oct 2013, 02:48
rxs0005 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) $$10(\sqrt{3}- 1)$$
(B) $$5$$
(C) $$10(\sqrt{2} - 1)$$
(D) $$5(\sqrt{3} - 1)$$
(E) $$5(\sqrt{2} - 1)$$

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Hope it helps.
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08 Feb 2014, 03:53
Bunuel wrote:
TehJay wrote:
How did you come up with that formula for the shortest distance from the vertex to the sphere?

It would be easier if you visualize this problem.

As sphere is inscribed in cube then the edge of the cube equals 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-Diameter is gap between the vertex of a cube and the surface of the sphere --> $$x=\frac{Diagonal -Diameter}{2}=\frac{10*\sqrt{3}-10}{2}=5(\sqrt{3}-1)$$

What i didn't comprehend was why 1/2. How can we evaluate that it is 1/2 ? is it a formula that we have to cram?
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Joined: 02 Sep 2009
Posts: 59720

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08 Feb 2014, 06:19
Abheek wrote:
Bunuel wrote:
TehJay wrote:
How did you come up with that formula for the shortest distance from the vertex to the sphere?

It would be easier if you visualize this problem.

As sphere is inscribed in cube then the edge of the cube equals 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-Diameter is gap between the vertex of a cube and the surface of the sphere --> $$x=\frac{Diagonal -Diameter}{2}=\frac{10*\sqrt{3}-10}{2}=5(\sqrt{3}-1)$$

What i didn't comprehend was why 1/2. How can we evaluate that it is 1/2 ? is it a formula that we have to cram?

It would be easier if you visualize. Check the diagrams here: a-sphere-is-inscribed-in-a-cube-with-an-edge-of-10-what-is-127461.html and here: a-sphere-is-inscribed-in-a-cube-with-an-edge-of-10-what-is-127461.html#p1234095

Hope it helps.
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Joined: 13 Jan 2015
Posts: 1
Re: A sphere is inscribed in a cube with an edge of 10. What is  [#permalink]

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20 Jan 2015, 16:30
Can someone let me know what is wrong with my approach please.

1. I take the corner of the square to the centre of the square
2. Then subtract out the radius of the circle (all in 2dimension)
3. Use 3d Pythagorus to get distance to the sphere

I do not arrive at the correct answer and instead get 5(rt(3)*rt(2)-1)

Thanks
Intern
Joined: 01 May 2015
Posts: 1
Re: A sphere is inscribed in a cube with an edge of 10. What is  [#permalink]

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09 Aug 2015, 10:42
Why can you not think of the problem as a pyramid formed by a vertex of the cube, the closest point where the cube's diagonal intersects the sphere, and an adjacent tangent point between the cube and sphere? Then apply 45/45/90 logic? It doesn't seem to yield the same answer, but IDK where I went wrong. Thx!
Intern
Joined: 20 Dec 2014
Posts: 36
Re: A sphere is inscribed in a cube with an edge of 10. What is  [#permalink]

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09 Apr 2018, 09:06
bigfoot wrote:
Can someone let me know what is wrong with my approach please.

1. I take the corner of the square to the centre of the square
2. Then subtract out the radius of the circle (all in 2dimension)
3. Use 3d Pythagorus to get distance to the sphere

I do not arrive at the correct answer and instead get 5(rt(3)*rt(2)-1)

Thanks

Hey

Can u upload a photo of your calculations.. I went by a similar method and got the right answer.
Though you can refer the problems highlighted by Bunuel
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Re: A sphere is inscribed in a cube with an edge of 10. What is  [#permalink]

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21 Aug 2019, 21:41
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Re: A sphere is inscribed in a cube with an edge of 10. What is   [#permalink] 21 Aug 2019, 21:41
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