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A sphere is inscribed in a cube with an edge of 10. What is the [#permalink]
Are sphere related questions included in the actual gre?
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Re: A sphere is inscribed in a cube with an edge of 10. What is the [#permalink]
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aishumurali wrote:
Are sphere related questions included in the actual gre?



\Yes

See our math book chapter https://gre.myprepclub.com/forum/gre-quant ... tml#p53168
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Re: A sphere is inscribed in a cube with an edge of 10. What is the [#permalink]
And what about hemisphere??

Posted from my mobile device
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Re: A sphere is inscribed in a cube with an edge of 10. What is the [#permalink]
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aishumurali wrote:
And what about hemisphere??

Posted from my mobile device


what you see it is perfectly bespoke for the GRE exam in terms of math concepts

See the scale on top on frequency
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Re: A sphere is inscribed in a cube with an edge of 10. What is the [#permalink]
Can someone please help me understand 2 questions:

1. Why is the diameter of the sphere equal to the edge of the cube?
2. Why is the shortest distance (diagonal of the cube - diameter of the sphere)/2?
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A sphere is inscribed in a cube with an edge of 10. What is the [#permalink]
2
Given that A sphere is inscribed in a cube with an edge of 10. And we need to find What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere

Shortest distance between any two points is achieved by dropping a perpendicular from one point to another.

To find the Shortest distance between the vertex of the cube (Refer point A in the below image) and the surface of the sphere (Refer point C), we need to drop a perpendicular from A to the surface of the sphere.

Attachment:
cube sphere.png
cube sphere.png [ 12.91 KiB | Viewed 3012 times ]


We know that the diagonals of a cube are Perpendicular bisectors of each other => The perpendicular from A to the surface of C will pass through O (the center of the cube and the sphere)

=> Distance AC = AO - CO = \(\frac{1}{2}\) * (Diagonal of cube) - \(\frac{1}{2}\) * (Diagonal of Sphere)

We know that Side of Cube = 10, and Diagonal of Cube = \(\sqrt{3}\) * Side = \(\sqrt{3}\) * 10

Since, sphere is touching the cube so if we look at the vertical diameter(DE) of the sphere then it will be same as the side of the cube = 10

=> Distance AC = \(\frac{1}{2}\) * (Diagonal of cube) - \(\frac{1}{2}\) * (Diagonal of Sphere) = \(\frac{1}{2}\) * (\(\sqrt{3}\) * 10 - 10) = \(5(\sqrt{3} – 1)\)

So, Answer will be B
Hope it helps!
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A sphere is inscribed in a cube with an edge of 10. What is the [#permalink]
kk117 : I tried explaining in detail in my post above.

Hope that helps!
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Re: A sphere is inscribed in a cube with an edge of 10. What is the [#permalink]
Got it! Thanks BrushMyQuant

My error was apparently in reading.

I thought the cube was inscribed inside the sphere and hence I got it wrong.

I just noticed it's the other way around
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Re: A sphere is inscribed in a cube with an edge of 10. What is the [#permalink]
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