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Area of a Regular Octagon: Proof of a Formula

Date: 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 

Jin Park

Date: 08/31/98 at 02:02:14
From: Doctor Pat
Subject: Re: Area of Octagon


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
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Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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