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** SWANNIES** **FIRST** ** THEOREM - 2006**

The photograph

**How a model can be made**:

1. Mark out in straight lines joining at the edges any fair sized corner on an ordinary (cereal) box. 2. Neatly cut this corner from the box. This gives three rectangular triangles with a common vertex. 3. Cover up the triangular opening with cardboard and glue to form the triangular base of your trirectangular tetrahedron. 4. By measurement and calculation find the areas* of each of the four faces of the tetrahedron using a unit of choice, eg. cm. 5. From cardboard build pyramids to fit the faces of the tetrahedron with height in each case equal to the numerical value of the calculated area of the face and the same length unit type. 6. Stick the prisms onto the faces. *You may wish to use Herons formula for the finding the area of the non rectangular face:

Make the sides of the triangle a, b and c and let s = (a + b + c)/2. Then its Area wil be = square root of [s(s-** **

The following is the well-

3^2 + 4^2 = 5^2 (before Pythagoras)

with its triangle and squares.

The fo llowing is the prime example for

Formula:

3^3 + 4^3 + 5^3 = 6^3 (

A Projection type, after

and A Special case, after

as shown above in the photographs.

Note x, y, z and w do not represent any linear dimensions of a model. Specified models such as (3units)3 + (4units)3 + (5units)3 = (6units)3 needs the set up of three equations (see below) and calculation to find the lengths of the three rectangular sides of the tetrahedron required. The results of such calculation is used in an illustration that follows later. Note that a model with equal rectangular prisms e.g. (4.16u)3 + (4.16u)3 + (4.16u)3 = (6,00u)3 would obviously be an easier first attempt (see the star and others under More Models).

dimensions of the tetrahedron at its centre. Units not shown.

Formula: x^3 + y^3 + z^3 = w^3

Draw a trirectangular tetrahedron and mark the sides meeting at right angles a, b and c, in this order, from short to long. Therefore the areas of the three rectangular triangles are equal to ½ ab, ½ ac and ½ bc, from small to large. Now square these and set each equal to the volume of its prism.

¼ a^2 c^ 2 = 64 -

**Projection **and Reflection**: **For a model starting off as a shape marked out on the triangular 'hypotenuse' face of its tetrahedron, such as the cylindrical model in a photograph on the left, the clue comes from **De Tinseau** -* projection* on the white inside of an opposite face, is a fun way to do it. In the photograph on the left an ellips is formed from the circle on the tetrahedrons base. The sun is on your left at about 10:00. The base ( trihypotenuse?) with its circle of little holes appears as it is reflected in a mirror directly behind the tetrahedron and leaning a bit foreward.*Return to Horizontal Index at top of Page *

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