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** Above: Some Pythagorean type Art; a "4:7:11 Eu du Cologne art for the eye" **equivalent of the one for the nose used by your granny. You have to know how to add/mix colours and the areas before you attempt such paintings! The one on the left is not really true, too much blue! The better mix one on the right also provides a clue to perhaps how, without further ado, you can see that the painting is mathematically true. Note: BCA is a right angle and D is halfway 'up' from BA to C. The ratios between the relevant areas appears above. However, can you prove it, i.e. that the ratios between the areas of the inside triangles is the same as that between the outside semicircles. Clue: Re-**Now further more have a good look **at the trirectangular tetrahedron 'roof' of the central prism of the model in the photograph below (or any other trirectangular tetrahedron). How would you subdivide it into four tetrahedrons so that the volumes of three ad up to that of the biggest one, their bases being the faces of the original tetrahedron? You should then have a little roof for each of the prisms. Would the volume ratio between these tetrahedrons be the same as that between the prisms? **By the way** -

**Museum:** The arrangement in the photograph on the left is what Swannie has in mind for his little science museum. It would have a 4,4m square base with the heights and areas in a 1:2:2:3 proportion and total height of 7,26m.

-** Perhaps ** you'l find an elegant way of turning a cube in to three cubes and so really line up with Pythagoras's all square classical set up? Theorem V for cuboids gets nearer there. Likewise, semi spheres into three smaller semi spheres, see Theorem II and III page, bottom right, that also gets near there. ** **

March 2012 addition of new theorems -

**ADD- ONN**

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