Finding side length of cube from known diagnal

Nomadp

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

Subhotosh Khan

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

L2 + W2 + H2 = D2..........(with usual abbreviation of variables)
 

Nomadp

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Thankyou for your reply. I really don't know how to do that.
 

Subhotosh Khan

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JeffM

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Thankyou for your reply. I really don't know how to do that.
You do not know how to do this, or you do not understand why to do this?

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

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

Nomadp

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You do not know how to do this, or you do not understand why to do this?

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

\(\displaystyle H^2 + W^2 + D^2 = WHAT?\)
Guess im not sure how to get that from the long diagonal inside a cube
 

JeffM

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

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.
 

Dr.Peterson

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Guess im not sure how to get that from the long diagonal inside a cube
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.

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 \(\displaystyle L^2 + W^2 + H^2 = D^2\).
 

Mr. Bland

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All of the angles between the edges of a cube are 90 degrees. It's a three-dimensional object, so there are more of them, but it's ultimately still a set of perpendicular joints.

Pythagoras gave us a simple formula for finding the hypotenuse length of a triangle with a 90-degree angle in it, and this same formula is the basis of calculating the distance between two points. The cherry on top is that this formula works in n dimensions. The traditional distance formula, for some number of dimensions, is as follows:

\(\displaystyle d = \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2 + (z_{2} - z_{1})^2 + ...}\)​

This takes two points, \(\displaystyle (x, y, z, ...)_{1}\) and \(\displaystyle (x, y, z, ...)_{2}\), translates the first point to the origin, then calculates the magnitude of the resulting vector. If one corner of your cube is at the origin, what are the coordinates of the opposite corner?
 

hoosie

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Just in case you’re still not sure how to proceed have a look at an example I have provided for you:08830BDC-BE3D-4F2F-996B-F945930A98A1.jpeg
 
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