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Can the area of the largest side be expressed in terms of the areas of the 3 smaller sides? Like the Pythagorean theorem in 3D.

How would I go about finding/proving the answer?

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Can the area of the largest side be expressed in terms of the areas of the 3 smaller sides? Like the Pythagorean theorem in 3D.

How would I go about finding/proving the answer?

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You mean a pyramidy thingy?Imagine the shape you would get if you chopped off the corner of a cube (the *smaller* corner piece with 4 sides and 6 edges).

Can the area of the largest side be expressed in terms of the areas of the 3 smaller sides? Like the Pythagorean theorem in 3D.

Well, the largest side is an equilateral triangle, and its edge is √2, so its area is … ?

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I am wanting to do it for areas, not distances. I already knew that.The distance in 3D is the same Pythagorus theorem = sqrt( x^{2}+ y^{2}+ z^{2})

You misunderstood. I never said that the "pyramidy thingy" would be perfectly cut off the corner (so it could, for example, be a very tall thin pyramidy thingy). So the edge would not have to be √2. And the triangle would not have to be equilateral. Sorry if I was unclear. Do you understand?Hi Georgepowell!

You mean a pyramidy thingy?

Well, the largest side is an equilateral triangle, and its edge is √2, so its area is … ?

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D^2 = A^2 + B^2 + C^2

Try to figure out how to prove it.

An analogous theorem holds in higher dimensions: for a right n-simplex, the square of the (n-1)-volume of the diagonal face is equal to the sum of the squares of the (n-1)-volumes of the other faces.

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so the area is the

which you

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Ahh cool! Just what I suspected! (although that is easy to say after I have the answer :tongue:) I will have a crack at proving it, although if I struggle I will ask for a clue.

D^2 = A^2 + B^2 + C^2

Try to figure out how to prove it.

An analogous theorem holds in higher dimensions: for a right n-simplex, the square of the (n-1)-volume of the diagonal face is equal to the sum of the squares of the (n-1)-volumes of the other faces.

I didn't understand any of that. Sorry. What are a b and c there? What do you mean "normal unit vector"?b-a) x (c-a) = 1/2 (bxc+cxa+axb) …

so the area is themagnitudeof that …

which youcanfind by using Pythagoras.

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Have you done vectors and cross-products?I didn't understand any of that. Sorry. What are a b and c there? What do you mean "normal unit vector"?

If not, ignore what I said.

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Vectors, yes a bit. Cross-products no, but I am have just skimmed over the wiki page on it and it is making sense.Have you done vectors and cross-products?

If not, ignore what I said.

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I don't think that's what George means. He never said that the chopped off pyramid has an equilateral base. Pythagoras doesn't only work for an isosceles right triangle, but for any right triangle. George is trying to prove a 3D analog for Pythagoras even when the three right triangles in his pyramid are not mutually congruent. Right?Hi Georgepowell!

You mean a pyramidy thingy?

Well, the largest side is an equilateral triangle, and its edge is √2, so its area is … ?

However, an illustrative first step would be to prove tiny-tim's special case, where all the right triangles are mutually congruent and the base triangle is therefore equilateral.

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[tex]D = \sqrt{ A^2 + B^2 + C^2} [/tex]

... to be perfectly analogous to the two dimensional Pythagorean. However, that's

Okay, here's a related one. Prove the

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Oh darn. [tex]2^4 = 16[/tex], not eight. So my root two factor was indeed spurious, and I'm now in agreement with Ben Niehoff's formula.However, that'snotwhat I got. I have an extra factor of [tex]\sqrt 2[/tex] in my calculation. It's worth someone double checking this, but I'm convinced that there is a 3D analog of the Pythagorean theorem, with a form similar but not identical to the 2D version. Nice question, George!

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Hi Georgepowell!

just got up … :zzz:

the area of a triangle is 1/2 bc sinθ, which happens to be the same as the magnitude of the cross product 1/2**b** x **c**, assuming the vertices are at **0** **b** and **c**.

**b** x **c**, of course, is a vector: its magnitude is twice the area of the triangle, and its direction is normal (perpendicular) to the triangle.

In the more general case where the third vertex is at**a** (not **0**), the sides are **b** - **a** and **c** - **a**, and so the area times the normal unit vector is 1/2 (**b** - **a**) x (**c** - **a**) = 1/2 (**b** x **c** + **c** x **a** + **a** x **b**)

Square that, put the origin at the vertex of the cube, so that (**b** x **c**) x (**c** x **a**) = 0 and so on (can you see why?), and you get the result.

just got up … :zzz:

I didn't understand any of that. Sorry. What are a b and c there? What do you mean "normal unit vector"?

ok, now you've gt the basics:Vectors, yes a bit. Cross-products no, but I am have just skimmed over the wiki page on it and it is making sense.

the area of a triangle is 1/2 bc sinθ, which happens to be the same as the magnitude of the cross product 1/2

In the more general case where the third vertex is at

Square that, put the origin at the vertex of the cube, so that (

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It turned out that today in Maths we learnt about the dot product of two vectors, it looks very similar to the cross product. The way I understand it:

The cross product of two vectors**a**x**b** =|**a**|*|**b**|*sin(θ) [vector, perpendicular to **a** and **b**]

The dot product of two vectors**a**.**b** =|**a**|*|**b**|*cos(θ) [scalor]

The dot product**a**.**a** is equal to the area of a square with side lengths a. (because cos(0) = 1)

The size of the cross product**a**x**b** is equal to the area of a parallelogram made with the vectors **a** and **b**.

I was playing around with these two tools and I proved the original Pythagorean theorem (I think). Here goes:

A right angled triangle has two smaller sides represented by vectors**a** and **b**.

0.5*[**a**.**a**+**b**.**b**] is the sum of the squares of the two smaller sides.

0.5*[(**a-b**).(**a-b**)]=**a**.**a**+**b**.**b**+2***a**.**b** **a** and **b** are at right angles to each other so 2***a**.**b** = 0

And that is it, the rest is obvious!

The proof of the equation D²=a²+b²+c² for a pyramidy thingy with three sides given as vectors**a**, **b** and **c**:

The area of the larger side is given by 0.5*the area of a parallelogram formed by the two sides.

The area of the larger sides = 0.5*|(**b**-**c**)x(**a**-**c**)|

Expanding the cross product I got

=0.5*|**c**x**a**+**c**x**b**+**a**x**b**| The vectors are all perpendicular to each other, so |**c**x**a**|=|**c**|*|**a**|.

Then to find the size of this final vector I used Pythagoras' theorem (because each of the individual vectors are perpendicular to each other).

=0.5*√[(|**c**|*|**a**|)²+(|**b**|*|**a**|)²+(|**b**|*|**b**|)²]

The 0.5 on the outside can be turned into 0.25, and brought inside the square root. Then it can be brought into each individual square, to get:

0.5*√[(|**c**|*|**a**|)²+(|**b**|*|**a**|)²+(|**b**|*|**b**|)²]=√[(0.5*|**c**|*|**a**|)²+(0.5*|**b**|*|**a**|)²+(0.5*|**b**|*|**b**|)²]

And it is obvious that 0.5*|**c**|*|**a**| is the area of the triangle on the side of the pyramidy thing.

This post looks a bit messy, I think it makes sense though

The cross product of two vectors

The dot product of two vectors

The dot product

The size of the cross product

I was playing around with these two tools and I proved the original Pythagorean theorem (I think). Here goes:

A right angled triangle has two smaller sides represented by vectors

0.5*[

0.5*[(

And that is it, the rest is obvious!

The proof of the equation D²=a²+b²+c² for a pyramidy thingy with three sides given as vectors

The area of the larger side is given by 0.5*the area of a parallelogram formed by the two sides.

The area of the larger sides = 0.5*|(

Expanding the cross product I got

=0.5*|

Then to find the size of this final vector I used Pythagoras' theorem (because each of the individual vectors are perpendicular to each other).

=0.5*√[(|

The 0.5 on the outside can be turned into 0.25, and brought inside the square root. Then it can be brought into each individual square, to get:

0.5*√[(|

And it is obvious that 0.5*|

This post looks a bit messy, I think it makes sense though

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