Apothem of a HexagonDate: 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/ |
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