Let the lengths of the six sides be 1, root(3) x 2, 2, 2root(3), 3 respectively. The two longest sides, i.e., 2root(3) and 3, have no counterparts so that the reflection axe must pass through the both sides. Then it is not difficult to find the two solutions. One is obtained by placing the two right angles together and the two sides of root(3) colinear. The other is obtained by placing the two 30-deg angles together and the two sides 2 and 3 overlapped.

]]>Another one of my symmetry puzzles you might enjoy. This time there are only two pieces.

Make a 30-60-90 triangle (ie half an equilateral triangle)

Divide it into two 30-60-90 triangles of different sizes by dropping a perpendicular from the right angled corner to the opposite sidew.

Put these two pieces together to make a symmetrical shape.

There are two solutions.

Donald ]]>

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