Swannies Theorems, Art in Maths and a follow-up to Pythagoras, De Tinseau and De Gua De Malves.

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IV and V

Swannies Fourth Theorem:
A theorem for the areas of Squares on  a trirectangular tetrahedron.
A Square and three smaller squares - `a la Pyhagoras
Method: (This was how the squares - yes they are all supposed to be squares - in the photograph below on the righ were found.) Draw a circle, prefferably the largest possible, on the base of a trirectangular tetrahedron. Draw three inscribed squares each with a side  parallel to a side of the base and call one of them the 'original'. Each one gives a right angled  projection on a rectangular face. Use the shorter side of each projection for completing a square on the face. Result: The areas of the three small  squares add up to that of the original one. In the photograph the square at the centre is shown in this position simply to shape a nice face.
Definition: The sum of the areas of squares on the rectangular faces of a trirectangular tetrahedron equals the area of a square on its base provided  the length of a shorter side of a rectangular projection of a square on the base onto a rectangular  face is used as the length of the side of the square on the face.

The photograph on the left: What you get if you don't use the circle in the method given above.  In the photograph below on the right  EF = DG/AD x BC.

Swannies Fifth Theorem:
A theorem for the volumes of Cuboids on  a trirectangular tetrahedron. A Cube and three cuboids - a la Swannie.  (cuboid, in this case: "like a high box with a square base") On the original square of Swannies Fourth Theorem  place a cube and on each of the other squares cuboids with height equal to the length of the side of the cube. Result: The volume of the cuboids add up to that of the cube.
Definition: Proceeding from Swannies Fourth Theorem for squares, the sum of the volumes of cuboids fitted onto the squares on the rectangularfaces of  the  trirectangular tetrahedron equals the volume of a cube fitted onto the square on the base,  provided the cuboids have a height equal in length to
the side of the cube. (Sorry no model yet. Also wish to turn them all into cubes without my calculation trick - but no dice!)

Below on the left: Extension of Theorem VI (next page) to a model. At the centre of the model is a trirectangular tetrahedron with an equilateral base. Stuck on the rectangular faces are three regular tetrahedrons/pyramids and below a tetrahedrom/pyramid with the same altitude. The volumes of the smaller pyramids add up to that of the one below.

Below on the righ: Just to fill the space. A 'which is more' - red or black?