Invoke Pythagoras and use:

L

^{2}+ W

^{2}+ H

^{2}= D

^{2}..........(with usual abbreviation of variables)

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Invoke Pythagoras and use:Can anyone please tell me the side length of a cube that has a longest diagnal of 11,400,576 feet??

L

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Do you know Pythagoras's Theorem?Thank you for your reply. I really don't know how to do that.

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You do not knowThankyou for your reply. I really don't know how to do that.

If you do not know how to do it, here is a clue: the height, width, and depth of a

\(\displaystyle H^2 + W^2 + D^2 = WHAT?\)

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Guess im not sure how to get that from the long diagonal inside a cubeYou do not knowhowto do this, or you do not understandwhyto do this?

If you do not know how to do it, here is a clue: the height, width, and depth of acubeare equal so

\(\displaystyle H^2 + W^2 + D^2 = WHAT?\)

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You are not supposed to get that from the length of the longest diagonal.Guess im not sure how to get that from the long diagonal inside a cube

The point here is that a cube is symmetric. The length of the longest side equals the length of the shortest side; the length of the longest diagonal equals the length of the shortest diagonal.

\(\displaystyle x = y = z \implies x^2 + y^2 + z^2 = x^2 + x^2 + x^2 = WHAT?\)

1 chicken plus 1 chicken plus 1 chicken = how many chickens.

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I think you're saying that you don't know that \(\displaystyle L^2 + W^2 + H^2 = D^2\), where L, W, H are the length, width, height, and D is the diagonal through the solid.Guess im not sure how to get that from the long diagonal inside a cube

To obtain that fact, first think about the diagonal of one face. If X is the diagonal of an L x W face, then \(\displaystyle L^2 + W^2 = X^2\). But then this diagonal is perpendicular to a W x H face, so \(\displaystyle X^2 + H^2 = D^2\). Put this together, and you have \(\displaystyle L^2 + W^2 + H^2 = D^2\).

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