Area of a Regular Octagon: Proof of a FormulaDate: 08/31/98 at 00:35:37 From: Jin Park Subject: Area of Octagon Dear Dr. Math, My question pertains to the area of an octagon. My teacher gave me a formula for computing the area of an octagon. Suppose that you draw a line from the center of the octagon to any one of the vertices and call that length "r". Why then, is the area of the octagon: 2 * (r^2) * 2^(1/2) So basically, what I'm asking for is a proof for the formula I just gave. Thanks, Jin Park Date: 08/31/98 at 02:02:14 From: Doctor Pat Subject: Re: Area of Octagon Jin, Here is the reason the formula works. I'm not sure it can be called a proof at this level because we will use some ideas from geometry and trigonometry that, in a rigorus proof, we might have to prove as independent parts. That said, here is the way to get the formula. The formula you are using is for a regular octagon, one with all sides and angles the same. If you draw all 8 lines from the center to the vertices, you will get eight congruent triangles whose areas add up to the area of the octagon. Each triangle has two sides of length r and a 45 degree angle between them (since all eight center angles add up to 360). One formula for the area of a isosceles triangle is: Area = 1/2 * leg^2 * sin(vertex angle) In this case: Area = 1/2 * r^2 * sin(45) The sine of 45 degrees is exactly equal to sqrt(2)/2, which can also be given as 2^(1/2)/2. So the area of each triangle is sqrt(2)*r^2/4. Now there are eight such triangles in the octagon, so the area of the octagon is 8(sqrt(2)*r^2/4), which equals 2(sqrt(2))r^2, the formula suggested by your teacher. Thanks for the question. Hope that helps. - Doctor Pat, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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