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**Theorem VI: **A theorem for the areas of **similar triangles** on** **the faces of a trirectangular tetrahedron. The base and three smaller triangles similar to it as in all three above illustrations. The one in the centre shows the triangles in their proper positions.

From an edge between the base and a rectangular face of a trirectangular tetrahedron draw lines to represent the heights of the two triangles (they join to form one straight line). Around the line on the rectangular face mark out a triangle similar to the base, using angles taken from the base. Likewise draw similar triangles on the other faces. See the diagram of an ''opened up" tetrahedron above on the left. (Please note in my photographs vertical dimensions sometimes show up shortened, rendering angles and lenghts inaccurate. The drawing on the right is an example where the total figure is a square and also serves as a proof for this case ) *Definition:***Regarding a ****trirectangular tetrahedron; let its base in turn be mirrored, in shape and orientation, across each base line as an inscribed triangle of a rectangular face. The sum of the areas of the three similar triangles thus found ads up to that of the similarly shaped base.)****Theorem VII** A theorem for projections of the rectangular faces of a trirectangular tetrahedron onto the base.

The base and three smaller triangles.

It is obvious that the projections of the rectangular faces of a trirectangular tetrahedron onto the base equal the base in area. ** Definition: **(Trivial)

It covers some of the above theorems.

Mark out any shape on the base of a trirectangular triangle. Determine the projections of this shape onto the rectangular faces. From a selected point on an edge of the base draw a vertical line through the corresponding sections of the shapes on the adjoining faces. Use the section through the projection as a base line around which to draw a similar shape to that on the base. Likewise draw similar shapes on the other faces.

**On the left** is a trirectangular tetrahedron with an equilateral base. On the rectangular faces are shown, in darker shades, triangles also of equilateral shape, in accordance with the definintion of Theorem VI. **On the right**: An ad infinitum/infinitesimal proof for a special case of Theorem VI?

The tetrahedron is of the same shape as the one on the left but with its rectangular faces flattened out. Positioning the red tringles inside the black triangle as shown gives overlaps shown as green triangles and leaves an open triangle at the centre. Flipping the green overlaps onto the open triangle in turn gives overlaps shown as blue traingles and again leaves an open triangle at the centre. If accurately done the blue triangles can in turn be flipped over onto the open triangle to see what turns up. Better perhaps to to enlarge the blue triangles and the open triangle back to the size of the green triangles and the triangle between them and repeat this doing, again and again.

**On the left **a proof for Theorem VI. As above ABCD is a flattened out trirectangular tetrahedron with an equilateral base . The red, blue and green triangles are found as per definition of the Theorem. To be proven that they fit, area wise, into the black triangle. Flipped over they overlap in triangle EFG with open areas left between A and E, B and F, and G and C. The similar little triangles appropiatly marked with coloured dots shows how the fit can be achieved.

See the Add-

**Theorem IX **(a quasi type!):** **(May 2013) A theorem on the **volumes of inscribed Semispheres** on the faces of a trirectangular tetrahedron.

Draw the inscribed circles of the faces of the tetrahedron. Use the radii of these circles to build four semispheres to fit onto onto the tetrahedron. ** Definition:** The sum of the volumes of the inscribed semispheres on the rectangular faces of a trirectangular tetrahedron near equals the volume of the inscribed semisphere on its base.

(2/3)pi{(ra)3 + (rb)3 + (rc)3} ≈ (2/3)pi{(rd)3}

The volumes my well give a few % volume error. However using the error and the volume obtained with rd and now reculculating new 'true' ra, rb and rc's, call them rat, rbt and rct, to turn ≈ above into = the latter radii are found to differ from the originals by the likes of 0,3mm that should be hardly noticeable in models with dimensions of say ≤40 cm.

For more irregular models/examples the following may be required to caculate a radius: (i) Heron's (or Hero's?) formula: Area of triangle = √{s(s -

Perhaps there is a best and/or a worst nearness example for the theorem? "Why so near" begs an answer? Could this theorem be adaptable to all/some similar shapes that fit nicely into the semispheres (or their full spheres if you make them sit proud on on the faces)?**Theorem X:** (i) A theorem (also of a quasi kind) on the **volumes of Cylinders** on a trirectangular tetrahedron.

In essence this is a variation of Theorem IX with 'half cylinders' of hight equal to the radius of a base, replacing the semispheres.The volume of a semisphere being ½ (4/3 x pi x r3) and that of half of a cylinder with base pi x r2 and height 2r being ½ (pi x r2 x 2r) or ½ (2 x pi x r3) the formula for such a model is simply: pi{(ra)3 + (rb)3 + (rc)3} = pi{(rd)3}** Definition:** Given the height of a cylinder equals the radius of its base, the sum of the volumes of inscribed cylinders on the rectangular faces of a trirectangular tetrahedron near equals the volume of the inscribed cylinder on the base.

(ii) Another variarion of Theorem IX with

(iii) Others? -

(iv) As (iii) but with a

The Lament below was generated before I realized IX and X were only of a 'quasi' nature, but is left here to show mood swings!

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