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A sphere is inscribed in a cube with an edge of x centimeter [#permalink]

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20 Jul 2013, 05:03

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A sphere is inscribed in a cube with an edge of x centimeters. In terms of x what is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

A. \(x(\sqrt{3}- 1)\) B. \(\frac{x}{2}\) C. \(x(\sqrt{2} - 1)\) D. \(\frac{x}{2}(\sqrt{3} - 1)\) E. \(\frac{x}{2}(\sqrt{2} - 1)\)

A sphere is inscribed in a cube with an edge of x centimeters. In terms of x what is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

A. \(x(\sqrt{3}- 1)\) B. \(\frac{x}{2}\) C. \(x(\sqrt{2} - 1)\) D. \(\frac{x}{2}(\sqrt{3} - 1)\) E. \(\frac{x}{2}(\sqrt{2} - 1)\)

M28-01

Say \(x=10\) centimeters.

Then, since a 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 minus Diameter is gap between the vertex of a cube and the surface of the sphere --> \(gap=\frac{Diagonal -Diameter}{2}=\frac{10*\sqrt{3}-10}{2}=5(\sqrt{3}-1)\).

Since \(x=10\) then \(5(\sqrt{3}-1)=\frac{x}{2}(\sqrt{3} - 1)\).

Re: A sphere is inscribed in a cube with an edge of x centimeter [#permalink]

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20 Jul 2013, 06:48

Wow awesome explanation !! +1 to you..By any chance are there any questions which are similar to this other apart from the link provided by you? _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

Re: A sphere is inscribed in a cube with an edge of x centimeter [#permalink]

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12 Nov 2013, 14:36

Why do we have to go along the diagonal of the cube?

I am getting E by going straight from the corner towards the sphere. The way I'm doing it would be the same if it were a circle/square rather than sphere/cube.

Why do we have to go along the diagonal of the cube?

I am getting E by going straight from the corner towards the sphere. The way I'm doing it would be the same if it were a circle/square rather than sphere/cube.

Why is this wrong? I can't figure it out

That's not correct. Draw a cube and see how you will inscribe a sphere in it. Note that the sphere will not touch any edges/corners of the cube. It will touch only the 6 faces of the cube at one point each. This point will lie in the center of the face of the cube. If you go the usual two dimensional way, you are assuming that the sphere is lying flat on the face of the cube which is not correct. The sphere only touches the face of the cube on one point i.e. the point where the diagonals of the square face intersect. Hence, actually the distance of this diagonal to the sphere will be half the length of the diagonal. On the other hand, the diagonal of the cube (from one vertex to the opposite vertex across the cube will go right through the center of the sphere. It will stick a little bit out on both sides close to the vertex but will predominantly lie within the sphere on its diameter. So we find the length of the cube diagonal, subtract the sphere diameter out of it and divide the rest of the diagonal by 2 to get length of each little piece.

Think of a globe and its inclined axis. Imagine making a cube around it such that the globe touches each face of the cube. The shortest distance between a vertex of the cube and the globe will be the part of the inclined axis sticking out of the globe touching a vertex of the cube. _________________

Re: A sphere is inscribed in a cube with an edge of x centimeter [#permalink]

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13 Nov 2013, 12:37

Wow! You made the visualization perfectly clear. I was thinking cylinder not sphere, however your explanation made the second half of solving the problem make the algebra come together easily.

Thanks so much for that reply, I really appreciate it.

Re: A sphere is inscribed in a cube with an edge of x centimeter [#permalink]

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18 Feb 2015, 21:18

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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A sphere is inscribed in a cube with an edge of x centimeter [#permalink]

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25 Feb 2015, 14:46

I did it using the relationship between the diagonal of the cube and the side of the cube.

So, if the side is x, the diagonal is xSQRT3.

So, to find the gap, we need to subtract the diameter of the circle from the diagonal of the square (which would leave us with the two small gaps between the circle and the square, across the diagonal of he square) and divide this by 2, to get only the length of one of the gaps.

So, the diagonal is xSQRT3 The diameter of the circle is x, as the side of the square ( this is obvious if you draw the diagonal in the middle of the square, where the circle is touching the sides of the square).

(xSQRT3 - x) / 2 = x/2 (SQRT3 - 1).

Sorry for the ugly formatting, but I couldn't do the square roots in preview...

Now half of Diagonal minus Diameter is gap between the vertex of a cube and the surface of the sphere

Does GMAC require this knowledge about spheres and these distances?

This question requires no particular knowledge about spheres. It needs you to just visualise - nothing wrong with that. It is certainly suitable for GMAT. You could get a volume of sphere kind of question too but you will be given the formula used to find the volume of sphere. For a regular 3-D figure such as a cylinder or prism (where Volume = Area of base * Height), you could be required to find the volume without the formula. _________________

Why do we have to go along the diagonal of the cube?

I am getting E by going straight from the corner towards the sphere. The way I'm doing it would be the same if it were a circle/square rather than sphere/cube.

Why is this wrong? I can't figure it out

That's not correct. Draw a cube and see how you will inscribe a sphere in it. Note that the sphere will not touch any edges/corners of the cube. It will touch only the 6 faces of the cube at one point each. This point will lie in the center of the face of the cube. If you go the usual two dimensional way, you are assuming that the sphere is lying flat on the face of the cube which is not correct. The sphere only touches the face of the cube on one point i.e. the point where the diagonals of the square face intersect. Hence, actually the distance of this diagonal to the sphere will be half the length of the diagonal. On the other hand, the diagonal of the cube (from one vertex to the opposite vertex across the cube will go right through the center of the sphere. It will stick a little bit out on both sides close to the vertex but will predominantly lie within the sphere on its diameter. So we find the length of the cube diagonal, subtract the sphere diameter out of it and divide the rest of the diagonal by 2 to get length of each little piece.

Think of a globe and its inclined axis. Imagine making a cube around it such that the globe touches each face of the cube. The shortest distance between a vertex of the cube and the globe will be the part of the inclined axis sticking out of the globe touching a vertex of the cube.

Responding to a pm:

Quote:

CAN YOU please explain why is the diagonal root-square 10^2*10^2*10^2 and not just 10^2*10^2 (applying P.theor.)?

Diagonal of a square will be \(\sqrt{(10^2 + 10^2)}\) (shown by 'd' in the diagram) Diagonal of a cube will be 3 dimensional (shown by 'D' in the figure - the green line). We will need to use pythagorean theorem again on it. It will be the hypotenuse when the legs are height of the cube (a) and the diagonal of the square face (d).

A sphere is inscribed in a cube with an edge of x centimeters. In terms of x what is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

A. \(x(\sqrt{3}- 1)\) B. \(\frac{x}{2}\) C. \(x(\sqrt{2} - 1)\) D. \(\frac{x}{2}(\sqrt{3} - 1)\) E. \(\frac{x}{2}(\sqrt{2} - 1)\)

A sphere is inscribed in a cube with an edge of x centimeter [#permalink]

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17 Jul 2016, 11:58

skamal7 wrote:

A sphere is inscribed in a cube with an edge of x centimeters. In terms of x what is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

A. \(x(\sqrt{3}- 1)\) B. \(\frac{x}{2}\) C. \(x(\sqrt{2} - 1)\) D. \(\frac{x}{2}(\sqrt{3} - 1)\) E. \(\frac{x}{2}(\sqrt{2} - 1)\)

M28-01

Attachment:

download.png [ 7.18 KiB | Viewed 288 times ]

look at the fig. diagonal of cube=[square_root]3*x but sphere of radius x is in between cube so left out distance from two opposite vertices =[square_root]3*x-x but we need only any one side distance --->1/2([square_root]3*x-x) x/2([square_root]3*-1) Ans D

gmatclubot

A sphere is inscribed in a cube with an edge of x centimeter
[#permalink]
17 Jul 2016, 11:58

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