I have the right to do this problem by illustration a picture and present of symmetry. My question about this difficulty is what if it is no an octagon, but any regular polygon. What is a simple means to deal with the problem?

Problem: How numerous lines the symmetry walk a constant octagon have?

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I"d say there is 8 .Drawing one octagon is best but if you desire to carry out without the , imagine drawing symmetric present inbetween the present of the Octagon or you deserve to imagine drawing lines in ~ the suggest where the the 2 currently meet.

Lines drawn inbetween present = 4Lines drawn where 2 points meet = 4Total = 8


For n-gons over there are constantly $2n$ symmetries in total; $n$ reflections and also $n$ rotations. So in this case there space 16 symmetries in total, 8 reflections and also 8 rotations.

A nice means to think around this is to consider where you have the right to put each vertex. A symmetry is any kind of permutation that preserves adjacency that vertices. Label the vertices 1 with to 8, climate you have 8 selections for wherein to put the first vertex, 2 for the next and also only 1 after ~ that. For this reason we have 16 symmetries.

If you want more information on this look increase Dihedral groups.


My answer I got was 10 due to the fact that if you draw lines threw the octagon because it will explain much more to you. - forth grader advice


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