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Diagonal of a cube

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Diagonal of a cube

by AleksandrM » Wed Jun 11, 2008 9:27 am
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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by AleksandrM » 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
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by egybs » 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)
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by Ian Stewart » 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.
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by egybs » 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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