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

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II and III

Theorems II and III

Swannies Theorem as presented in the previous pages shall now be known as Swannies First Theorem.
Those that follow below are of a similar nature as the basic structure on which they are found remains the trirectangular tetrahedron.  Some diagrams are given to illustrate what follows, but only one model appears at the bottom of the next page.

General: Swannies Theorems now cover two and three dimentional shapes positioned on the large/hypotenuse bordered face/base of a trirectangular tetrahedron, relating to such a shape by projection shapes on the rectangular  faces such that areas and volumes, as the case may be, of these three shapes add up to that of the original one. The previous pages fully covered the First Theorem. A Summary and a definition is however included  here below.

Swannies First Theorem : Theorem for the volumes of prisms, cylinders and other shapes with vertical sides on a trirectangular tetrahedron.  The following is an example that deals with prisms:
A Prism and three smaller prisms.
Construct prisms to fit onto the faces of a trirectangular tetrahedron such that the height of a prism equals the numerical value of the area of the face, using the same length unit type. Result: The sum of the volumes of the prisms on the rectangular faces equals the volume of the prism on the base.
Definition (repeat): The sum of the volumes of prisms that fit onto the rectangular faces of a trirectangular tetrahedron equals the volume of a prism fitted onto  the base provided the height of a prism equals the numerical value of the area of its base using the same lenght unit type.

Swannies Second Theorem :
A theorem for the areas of  Circles on a trirectangular tetrahedron.
A Circle and three smaller circles - a la Pythagoras.
Method: Draw a circle on the base of a trirectangular tetrahedron, preferably the largest possible (as this, more accurately allows projection diameters to be found without actually drawing ellipses. Clue: Drop a perpendicular from the centre of the circle onto a side of the base and extend it both ways to the nearest sides of the adjacent triangles. Use the ratio between the two sections traversing the triangles and the diameter of the base circle to calculate the diameter of the circle in the rectangular triangle.). Use the short minor axis of its projected ellips on each of the rectangularfaces as the diameter for a circle drawn on the face. Result: The areas of the circles add up to that of the original circle.
Definition: The sum of the areas of circles on the rectangular faces of of a trirectangular tetrahedron equals the area of a circle on its base provided the minor axis of the projection of the circle on the base onto a rectangular face is used as the diameter for the circle on the face. (See Green to Red to Blue below on the right)          Formula:  pi(x^2 + y^2 + z^2) = pi(w^2) where x, y, z, and w are the radii of the circles.

See diagram and lay out thereof below:  The square shaped diagram, above on the right, shows an opened up trirectangular tetrahedron. This shape was chosen to show the nice ratios it gives, more so if for a specially shaped construction - see Museum under Personal -  with this square ground floor/base where the heights of the sections above ground level has a practical 1:2:3 ratio. Now, for a start, dealing  with the First theorem: The orange triangles represent one part, two parts, another two parts and three parts of eight parts of the orange square if so devided.(You agree? Also note whenever a colour is herewith mentioned it includes the colours/shapes within.) Therefore if u is a length
unit:       1u^2 + 2u^2 + 2u^2 + 3u^2 = 8u^2
and (1u^2)^2 + (2u^2)^2 + (2u^2)^2 = (3u^2)^2       De Gua
i.e.  1u^4     +    4u^4     +    4u^4    =    9u^4
i.e.  1u^3     +    4u^3     +    4u^3     =    9u^3           Swannie
A model conforming to these values is shown on the page Art.   The Second Theorem  is now used to find the circles. Going from green to red (by projection) to blue is shown. Likewise (or by the shorter method mentioned in the Method) the yellow circles are found.
So   yellow +  yellow  + a dash of blue =   green  and in this case, deviding the green circle - area wise - into twenty parts,
9 parts + 9 parts    +    2 parts      = 20 parts. General Formula:  pi(x^2 + y^2 + z^2) = pi(w^2) where x, y, z, and w are the radii of the circles.

Swannies Third Theorem :
A theorem for the volumes of a semi-Sphere and semi-Ellipsoids on a trirectangular tetrahedron.
A Semi-sphere and three semi-ellipsoids (semi-'rugby-footballoids') - `a la Swannie.
On the circle on the base of the tetrahedron in Swannies Second Theorem place a semi-sphere and on the smaller circles semi-elipsoids with their long axis equal to that of the radius of the circle on the base. Result: The volumes of the semi-ellipsoids  add up to that of the semi-sphere.
Definition: Proceeding from Swannies Second Theorem and its four related circles the sum of the volumes of semi-ellipsoids fitted onto the circles on the rectangular faces equals that of a semi-sphere fitted onto the circle on the base of the  tetrahedron provided the third axis length of the semi-ellipsoids equals the radius of the semi-sphere.
Formula:   2/3pi(ra^2 + rb^2 + rc^2) = 2/3pir^3 where r is the radius of the circle on the base and a, b and c the radii
of the circles on the other faces. As derived from 4/3pi r^3, the voume of a sphere.
Model: See below.  Above on the left: A  'thick' illustration of Pythagoras; an easy way to 'give Pythagoras volume', with formula a^2.d + b^2.d = c^2.d and in this case: a^3 + ab^2 = ac^2.  Here the thickness of the cuboids are the same; in theorems III and V the heights of the shapes are the same.
Above on the right: A trirectangular tetrahedron with an equilateral base fitted with a semisphere. On the rectangularfaces semi   polystyrene australorp shaped eggs were fitted (because you can buy them cheap and it takes much effort to make/shape the required items - and hens do it naturally and mathematically - close to what is required). They do have the same altitude/height above the faces as has the semisphere (its radius). The volumes of the three small semi 'stretched spheres'/ellipsoids add up to that of the semisphere, according to the Third Theorem.