Diagonals and Axes of Symmetry
Date: 03/31/98 at 03:22:48 From: Paul Dobing Subject: Polygons - Diagonals and Symmetry My daughter is in 5th grade and working on diagonals and axes of symmetry of polygons. It was presented to her that a regular octagon has 8 diagonals and 8 axes of symmetry. Could you explain the concepts behind this and confirm that this answer is in fact correct? Regards, Paul Dobing
Date: 03/31/98 at 12:13:55 From: Doctor Rob Subject: Re: Polygons - Diagonals and Symmetry The number of diagonals is given by n*(n-3)/2 = 8*5/2 = 20, not eight. This is because if we count at each of the n vertices, each one can be connected to each of the other n-1 vertices. Two of these connections form edges of the polygon, not diagonals, so there are n-3 diagonals meeting at each of the n vertices. The total number of ends of diagonals is n*(n-3). Each diagonal has two ends. Thus the number of diagonals is just n*(n-3)/2. (Can you figure out why this is always a whole number, never a fraction?) Draw a regular octagon and see if you can find all 20 diagonals. There are eight axes of symmetry. Four of them are perpendicular bisectors of the sides of the octagon, and four of them are bisectors of the interior angles. (You figure out why there are only four of each, not eight!) If an axis of symmetry meets a line segment in the figure, the axis must be the perpendicular bisector of the segment, or else there must be a second line segment also meeting the axis at the same point and making an equal angle with the axis. In the first case, you have the perpendicular bisectors of the sides, and in the second case, you have the bisectors of the interior angles. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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