Apothem of a Hexagon

Date: 6/11/96 at 0:11:46
From: RS
Subject: Apothem of a hexagon

I need help in finding the apothem of a regular hexagon. Could you 
help me by telling me the formula.

Date: 6/13/96 at 23:48:28
From: Doctor Brian
Subject: Re: Apothem of a hexagon

Well, let's divide the regular hexagon into six isosceles triangles, 
and then we'll deal with one of those--they're all congruent anyways, 
so the apothem will be the same.

Now, since this is a hexagon, it's not too difficult to show that 
these are not just isosceles triangles, but equilateral.  (After all, 
the angle at the center of the circle, the vertex angle of the 
triangle, is 360/6 = 60)

Lets call the side length s.  Because these are equilateral triangles,
the lines from the center to the vertices are also of length s.

Now draw the apothem.  It will divide the side into two parts 
each with length s/2.

From here we use either the Pythagorean theorem (hypotenuse = s and 
one leg = s/2) or we use the 30-60-90 triangle relationship to get 
that the apothem is sqrt(3) * s / 2

If you've done anything with sine/cosine/tangent, then you may know 
that the sqrt(3) part comes from the tangent of 60.  The 60 in 
question is half the interior angle of 120.  It's *not* the 60 from 
the central angle.  What I'm getting at is that, if you know the 
interior angle of the n-gon, you can replace the sqrt(3) with the 
corresponding tangent value.

example:  with an octagon, each angle is 135 degrees, so half of that 
is 67.5, and the apothem is tan 67.5 * s / 2

anyways, that's a little extra that you might not need.

see ya

--Doctor Brian 

Date: 6/15/96 at 7:30:29
From: Anonymous
Subject: Re: Apothem of a hexagon

Thank you for your reply. Our situation is quite remote as we are a
missionary family working in the ROC with very little access to 
English material to assist in our two teenagers' education. They are 
currently home-schooled through an American correspondence course 
offered through A Beka. So we wish to thank you for you help.
Best Regards Bob

Date: 6/15/96 at 9:49:48
From: Doctor Ethan
Subject: Re: Geometry

Well please keep sending your questions to us we are thrilled to be 
able to help you.

-Doctor Ethan,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/