Diagonal of a cube

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AleksandrM Legendary Member
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Diagonal of a cube

Post Wed Jun 11, 2008 9:27 am
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    The diagonal of a cube of side x is xsqroot3.

    This can be found by applying the Pythagorean Theorem twice (first to find the diagonal of a face of the cube, xsqroot2, and then to find the diagonal through the center, xsqroot3).

    Can someone please demonstrate for me the latter part (xsqroot3).

    Assume we are dealing with a cube with side 4.

    Thanks.

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    AleksandrM Legendary Member
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    Post Wed Jun 11, 2008 9:31 am
    Never mind I got it. Just imagine a cube that has been cut through the side

    You have base 4 and hight 4sqroot2:

    (4sqroot2)^2 + 4^2 = x^2 diagonal from one end of cube to ther other

    x = 4sqroot3

    egybs Master | Next Rank: 500 Posts Default Avatar
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    Post Wed Jun 11, 2008 9:46 am
    Diag1 = Diagonal of one side of the cube
    Diag2 = Diagonal of the entire cube


    Diag1^2 = x^2+x^2
    Diag1 = (2x^2)^.5 = x(2^.5)
    Diag2^2 = (x(2^.5))^2 + x^2
    Diag2^2 = 2x^2 + x^2 = 3x^2
    Diag2 = (3x^2)^.5 = x(3^.5)

    Post Wed Jun 11, 2008 12:09 pm
    You could also use the three-dimensional version of the Pythagorean Theorem:

    d^2 = x^2 + y^2 + z^2

    where d is the diagonal inside a box ('rectangular prism', in math terms) of dimensions x by y by z.

    Thus in the special case of a cube measuring x by x by x,

    d^2 = x^2 + x^2 + x^2
    d^2 = 3x^2
    d = root(3)*x

    This won't be useful very often on the GMAT, but occasionally it does help.

    Thanked by: joiedevivre
    egybs Master | Next Rank: 500 Posts Default Avatar
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    Post Wed Jun 11, 2008 12:24 pm
    Heh good point Ian... that's a lot nicer!


    Ian Stewart wrote:
    You could also use the three-dimensional version of the Pythagorean Theorem:

    d^2 = x^2 + y^2 + z^2

    where d is the diagonal inside a box ('rectangular prism', in math terms) of dimensions x by y by z.

    Thus in the special case of a cube measuring x by x by x,

    d^2 = x^2 + x^2 + x^2
    d^2 = 3x^2
    d = root(3)*x

    This won't be useful very often on the GMAT, but occasionally it does help.

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