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In a cube of side 10, what is the length of the line joining the cente

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In a cube of side 10, what is the length of the line joining the cente  [#permalink]

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New post 25 May 2019, 19:33
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In a cube of side 10, what is the length of the line joining the center of any face and a vertex of the cube not contained in the plane of that face?

A. \(\sqrt{6}\)
B. \(3\sqrt{5}\)
C. \(5\sqrt{3}\)
D. \(5\sqrt{6}\)
E. \(10\sqrt{6}\)
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Re: In a cube of side 10, what is the length of the line joining the cente  [#permalink]

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New post 25 May 2019, 20:48
nick1816 wrote:
In a cube of side 10, what is the length of the line joining the center of any face and a vertex of the cube not contained in the plane of that face?

A. \(\sqrt{6}\)
B. \(3\sqrt{5}\)
C. \(5\sqrt{3}\)
D. \(5\sqrt{6}\)
E. \(10\sqrt{6}\)


Step 1: I drop a perpendicular from the center of the face (A) to the opposite face (B), which is also the center of the opposite face. This length is 10.
Step 2: I draw a st. line from A to the vertex of opposite face (C).

Distance between BC= 1/2 of diagonal of the face containing B
= (1/2) * 10*root(2)
Now Pythagoras theorem to find AC
AC^2 = [5(root 2)]^2 + 10^2

AC= 5 root(6), IMO D
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Re: In a cube of side 10, what is the length of the line joining the cente   [#permalink] 25 May 2019, 20:48
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