A triangular-based pyramid (or tetrahedron) has three right angles meeting at one corner. How is the area of the face opposite the right angles related to the areas of the other three faces?

kevinatilusa

It seems to be a pythagorean theorem type deal: If D is the area of the face opposite the right angle and A,B,C are the areas of the other 3 faces: D^2=A^2+B^2+C^2. The quickest proof I can see for this is uses vectors and the formula that the area of the triangle with sides given by vectors a and b is half the length of a cross b. Apr 03 05, 2:35 AM

peasypod

Kevin is correct. D = A*3^(1/2) (root 3 times larger than the other faces, which are all the same area). A less quick proof of my own scribbling is that Pythagorus gives the base length "b" to be 2^(1/2) (root 2) times the vertex length "a". The side faces have area a*b*sin(45) and the base has area b*b*sin(60) = (2^1/2)*a*b*sin(60) The ratio of the areas (base/side) is then (2^1/2)*sin(60)/sin45 = 3^1/2. (since, sin(60) = root3 /2 and sin(45) = 1/root2) (These postings clearly require an equation editor....) Apr 03 05, 5:00 AM

ogicu8abruok

All four sides of a tetrahedron are exactly the same in shape and area. Am I missing something here? Apr 03 05, 1:54 PM

Baloo55th

In a regular tetrahedron, yes, the faces are all the same. If this one has three right angles in one corner, it's not a regular one. (Regular Tetrahedron would be a good name for one of these pop groups you get at the moment - they seem to have faces all the same...) Apr 03 05, 2:00 PM

gmackematix

Yay, Kevin. It is nice to see that Pythagoras' Theorem has variants in higher dimensions. My method was very long indeed as I took the sides meeting at right angles to be x,y and z and used Hero's formula for the area of the opposite triangle. The algebra did all simplify down as required. Peasy, dear, why have you assumed that the edges leading to the right angles are all the same length? Apr 03 05, 5:09 PM

peasypod

Gmack dearest, did I not say there was a requirement for an equation editor?? ;) My Pythagoran scribbles are not the greatest works of art let me tell you, but somehow I managed to end up with the same as Kevin. *Does Flynn's spontaneous stupid dance* and I guess half way through I conjected the possibility that I may or may not have been flipping between irregular and regular tetrahedron in my calculations being the doofas I am... Apr 03 05, 5:19 PM