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Sphere Problem

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by samirpandeyit62 » Mon Nov 26, 2007 12:37 am
The answer choices are not very legible to be, but I would solve it as follows

Now min lenght from a vertice to a pt on the sphere will be when moving diagonally from a vertice & the diameter of sphere is 10

diagonal of base of square is rt(2) *10

so diagonal from side is rt(10^2 + 10*rt(2) ^2)

= 10* rt(3)

so min lenght is (10*rt(3) - 10)/2 = 5(rt(3) -1)
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Samir
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by camitava » Mon Nov 26, 2007 1:31 am
Sorry Samir! Probably I am missing something. I am getting some different answer at all.
Refer the figure below -

Image
We get OA = 5sqrt(2)
Now OD = 5
So AD = OA - OD = 5(sqrt(2) - 1)

I am not getting why you are talking about 5(sqrt(3) - 1)
Correct me If I am wrong


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Amitava
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by camitava » Mon Nov 26, 2007 1:32 am
Sorry Samir! Probably I am missing something. I am getting some different answer at all.
Refer the figure below -

We get OA = 5sqrt(2)
Now OD = 5
So AD = OA - OD = 5(sqrt(2) - 1)

I am not getting why you are talking about 5(sqrt(3) - 1)

Image
Correct me If I am wrong


Regards,

Amitava
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by samirpandeyit62 » Mon Nov 26, 2007 1:46 am
Hey Amitava the figures mentioned in the Q are 3D figures, ie cube, sphere, & diagonal of a cube is rt(3)*s (lenght of side)
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Samir
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by camitava » Mon Nov 26, 2007 1:51 am
Shi...! A miss again ... God pls help meeeeeeeeeeeeeee :cry: :oops:
Correct me If I am wrong


Regards,

Amitava
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by sankruth » Mon Nov 26, 2007 3:07 am
The distance between 2 opposite vertices of the cube =
SqRt(10^2 + 10^2 + 10^2) = 10*SqRt(3)

Since the Sphere is inscribed in the cude, radius = 5 and the centre will be the midpoint of the line joining the two opposite vertices i.e. [10SqRt(3)/2] = 5SqRt(3)

So, Shortest distance between the surface of Sphere and vertex of square = 5SqRt(3) - 5 = 5[SqRt(3)-1]

Hope that helps!
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