<pre>
What is considered 'identical' for these questions?  If mirror-image shapes
are allowed, these are all pretty trivial.  If not, the problems are rather
more difficult...

1. Connect the center to every second vertex.

2. Connect the center to the midpoint of each side.

3. This is the hard one.  If you allow mirror images, it's trivial:
bisect the hexagon from vertex to vertex, then bisect with a
perpendicular to that, from midpoint of side to midpoint of side.

4. This one's neat.  Let the side length of the hexagon be 2 (WLOG).
We can easily partition the hexagon into equilateral triangles
with side 2 (6 of them), which can in turn be quartered into
equilateral triangles with side 1.  Thus, our original hexagon
is partitioned into 24 unit equilateral triangles.  Take the
trapezoid formed by 3 of these little triangles.  Place one such
trapezoid on the inside of each face of the original hexagon, so
that the long side of the trapezoid coincides with the side of the
hexagon.  This uses 6 trapezoids, and leaves a unit hexagon in the
center as yet uncovered.  Cover this little hexagon with two of
the trapezoids.  Voila.  An 8-identical-trapezoid partition.

5. Easy.  Do the rhombus partition in #1.  Quarter each rhombus by
connecting midpoints of opposite sides.  This produces 12 small
rhombi, each of which is equivalent to two adjacent small triangles
as in #4.

Except for #3, all of these partitions can be achieved by breaking up the
hexagon into unit equilateral triangles, and then building these into the
shapes desired.  For #3, though, this would require (since there are 24 small
triangles) trapezoids formed from 6 triangles each.  The only trapezoid that
can be built from 6 identical triangles is a parallelogram; I assume that the
poster wouldn't have asked for a trapezoid if you could do it with a special
case of trapezoid.  At any rate, that parallelogram doesn't work.
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