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Homework Help: Simple Math question. (Geometry)

  1. Jan 28, 2008 #1
    This is for body-centered cubic crystal structure.

    1. The problem statement, all variables and given/known data

    Prove a = 4R/sqrt(3). R is the radius of sphere and a is the edge of the cell (cubic).
    The image attached will be cut into cube so that 4 outter spheres' centers will be the edges of the cube, a.

    2. Relevant equations


    3. The attempt at a solution

    Well, the disagnal from the farthest two points are 4R.
    I tried to use one face to calculate the diagnal edge of the face and form a triangle with 4R and sqrt(2) but it won't work.
    Something is wrong...please help!

    Attached Files:

  2. jcsd
  3. Jan 28, 2008 #2


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    Homework Helper

    Look at it this way, you've got the find the length of the cube diagonal which you know to be 4R. To find the cube diagonal you just have to apply pythagoras theorem twice. Firstly to find the base square diagonal, which you should get as sqrt(2). Then you find the cube diagonal using the base square diagonal and the height of the cube. Then just equate the 2 expressions.

    Your image hasn't been approved yet so this is all I can say.
  4. Jan 28, 2008 #3


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    Science Advisor

    [itex]\sqrt{2}[/itex]? No, of course that won't work. Were did you get [itex]\sqrt{2}[/itex]? Perhaps you were thinking that the diagonals of a square, of side length a, have diagonal length [itex]a\sqrt{2}[/itex]. But here you are dealing with a cube. To find the length of a diagonal, you have to apply the Pythagorean theorem again. The length of the diagonal from bottom vertex to the opposite bottom vertex is the length of the diagonal of one face, a square: [itex]a\sqrt{2}[/itex]. But now you have a right triangle with base leg of lenth [itex]a\sqrt{2}[/itex] and vertical leg (up to the opposite point of the cube) of length s. By the Pythagorean theorem, the diagonal length is [itex]\sqrt{2a^2+ a^2}= a\sqrt{3}[/itex].

    Use that instead.
  5. Jan 28, 2008 #4
    Wow...I've been thinking that a*sqrt(2) was hypotenuse...
    I apologize for such a thread.
    Thank you very much.
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