The tie up with Pythagoras:

The illustration above on the **left** shows *box dimensions* as vectors while the model on the **right** shows vectors mounted on the vertex of its tetrahedron that sits atop its base prism. The photograph at the **centre** shows Plato's values vectors representing both area and height for prisms: Yellow and red results in orange; orange is replicated at the vertex where with the blue vector the resultant is the black one; and pointing downward is its negative the really relevant one as that is where the base prism of a model will go. The square of the diagonal d of a rectangular box with sides/dimensions a, b and c is given by Pythagoras as

d2 = a2 + b2 + c2 . However one of each of a,b,c and d can be construed as the area vectors A,B,C and D of the faces of a trirectangular tetrahedron, one a representing for example say 2,279^2 cm2 (or 5,2 in the box above) for the area of face A and the other a, 2,279^2 cm for the height of a prism on face A, thus giving the volume of the prism on A as 2,279^4 cm3 = 27cm3 = (3cm)3 as .fits in with x3 of the equation of Swannies Theorem. Likewise done with b, c and d gives the set of Plato's values: 3^3 + 4^3 + 5^3 = 6^3 . Note: A length dimension in the Pythagoras equation above squared is equivalent to an area vector times a height vector in Swannies theorem. Thus the line-up between the theorems - Pythagoras serving as a proof for Swannies Theorem.

The tie up with De Tinseau/De Gua De Malves :

*According to De Gua the sum of the squares of the areas of the three rectangular triangles of a trirectangular tetrahedron add up to the square of the area of its 'hypotenuse'/large/remaining face.* That is like a 4 + b 4 + c 4 = d 4 and with length unit, say cm, also to the power of four, gives for example a 4 = 2,279^4 cm4 = 27cm4 , whatever cm4 means. However if a2 of the a4 is taken for the area of a face, area = a2 = 2,279^2 cm2 = 5,194cm2 , which it is, and the remaining a2 of the a4 is used for the height of a prism on that face, ie. height = 2,279^2 cm2 + assistance of the maths fairy that makes the one cm dissappear = 5,194 cm, the volume of the prism becomes 27 cm3 . De Tinseau/De Malves theorem therefore also serves as a proof for Swannies theorem.

**The model above on the right:**

In this model the prisms are stacked next to each other with its tetrahedron capping the large/base prism. On top 'vectors' representing both the areas of its faces and the heights of the pyramids are arranged. More about this set up as a possible model for a museum under Arty Models.

**The Photographs below (non theorem- just for fun)**:

The photograph on the **left**: Just another way of showing "three vectors" at right angles and their resultant. The areas of the smaller semicircles also do add up to that of the one on the resultant.

The "Arrow head" at the **centre** is a re-arrangement of four triangular sections of a square (and also of the faces of a trirectangular tetrahedron) which says that angle ACB must be a right angle. Furthermore CD = 2 AC. Now if AB = 2 length units see if you can find (i)the area of ABC (in units squared, u^2 if you wish) , (ii) The area of BCD and (iii) making use of De Gua's Theorem (See above under "**The tie up** --- ") the area of FAB. When done have another look at the model in the above right-hand corner.

To fill this space: On the **right **in 2006: Maybe? but No - a nice try, but this 'Heron' would not fly. Here, afterwards it was then just presented, complete with stand, as a curiosity. Then,** **checked** **again** **11** **May 2013: Yes! - what an oversight some years ago!But it does fly/work!. This one, perhaps the simplest of them all(?) evaded me to now only become Theorem IX. Then 1 June 2013, developing some doubts, and going for calculation to higher decimals only : No but very close, good enough for models! - rather then Close/Quasi and interesting enough to make them Theorems IX and X! (?)

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