Diagonals of shapes

[etspets]

I, after being bored by sitting, decided to revise for my upcoming exams to end the 9th grade. I found an equation. This equation told me how to calculate the diagonal of a square. But there wasn't anything about how to find the diagonal (line connecting two furthest points) of a cube. So I decided to figure it out myself, and this is what I got:

diagonal = side * sqrt 3

After comparing the result with the equation for a square (d = side * sqrt 2). I noticed a similiarity. If you think about it, the equation could be written as followed:

diagonal = side * sqrt dimension of shape

This also is true for dimensions 0 and 1 (in 0 dimension the answer is always 0 and in dimension 1 the diagonal = length of the side).

So, my question is, does this work in the higher dimensions? I'm too lazy to think that much (at least for now) and I'll just hope someone else figures it out.

Also, has this been discovered already?

[KvickFlygarn87]

Hm. I think Pythagora's Theorem has this covered.

If you don't know, it's "A^2*B^2=C^2". A is one of the catheti (is it catheti or cathetuses? I'm Swedish so my English isn't all too perfect ) in a right triangle. B is the other cathetus and C is the hypothenuse.

EDIT: By "has this covered" means that you can use it to calculate that. Your way is faster though

[Nuke]

its sqrt(x^2+y^2+z^2)

i think pythagorean theorem works no matter how many dimentions you have, just keep adding the squares of each dimention.

Edited April 19, 2014 by Nuke

[Fr8monkey]

I believe that is true only with right angles.

[Superfluous J]

...find the diagonal (line connecting two furthest points) of a cube.

I believe that is true only with right angles.

Cubes are all right angles by definition.

[Fr8monkey]

Double post

[Nuke]

I believe that is true only with right angles.

it gets the length of any vector. in the case of a cube assumes one point is centered at 0'0'0 and the opposite point's coordinates go into the formula. you could use vector subtraction to find the distance between the two points and use that instead. its a pretty nifty little formula, ive used it several times as part of the vector library for my game engine.

[etspets]

To be honest I did just kind of simplyfy the Pythagoras theorem. But I was very happy when I got to this result all by myself.

[TheDarkStar]

To be honest I did just kind of simplyfy the Pythagoras theorem. But I was very happy when I got to this result all by myself.

Yeah, math is fun when you start figuring things out on your own .

If you don't know, it's "A^2*B^2=C^2". A is one of the catheti (is it catheti or cathetuses? I'm Swedish so my English isn't all too perfect ) in a right triangle. B is the other cathetus and C is the hypothenuse.

In English (at least American English, not sure about British), what you called catheti are called legs of the triangle. The hypotenuse is still named the same way.

[BigFatStupidHead]

So, my question is, does this work in the higher dimensions? I'm too lazy to think that much (at least for now) and I'll just hope someone else figures it out.

Well, supposing a unit square of sides length =1, we have sqrt(1^2+1^2) for the length of the diagonal. A unit cube will be sqrt(1^2+(the diagonal of the square)^2 , or sqrt(1+2). A tesseract will be sqrt(1^2+(the diagonal of the cube)^2) , or sqrt(1+3) which = 2, and so on. It's a really neat relation. Good work noticing it on your own!

[KvickFlygarn87]

In English (at least American English, not sure about British), what you called catheti are called legs of the triangle. The hypotenuse is still named the same way.

I know what it is, I was just unsure of its plural form