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Area of a Regular Octagon


Date: 10/27/1999 at 10:35:48
From: Nitin Tahiliani
Subject: Area of an octagon

How do you find the area of a regular octagon?


Date: 10/27/1999 at 17:19:39
From: Doctor Jeremiah
Subject: Re: Area of an octagon

Hi Nitin,

A good way is to split it up into triangles and then calculate the 
area of one of the triangles and multiply it by 8, since there are 
eight identical triangles.

Let's say you know the length of one of the sides:

        |<--N-->|
        +-------+
       / \     / \
      /   \   /   \
     +     \a/     +
     |      +      |
     |             |
     +             +
      \           /
       \         /
        +-------+

The angle a = 360/8 because there are 8 sides.

     a = 360/8
     a = 45

We can turn our triangle into a right angle triangle by cutting it in 
half:

     |<-N/2->|
     +-------+-------+
      \      |      /
       \     |     /
        \    h    /
         \   |   /
          \  |  /
         45/2| /
            \|/
             +

Now we can use trigonometry to find h:

     tan(45/2) = (N/2)/h
             h = (N/2)/tan(45/2)

Having found h, we can find the area of the right-angled triangle:

     RightAngledTriangleArea = h*b/2
                             = (N/2)/tan(45/2)*(N/2)/2
                             = (N/2)^2 / (2*tan(45/2))
                             = (N^2/4) / (2*tan(45/2))
                             = N^2/(8*tan(45/2))

Knowing the area of the right-angled triangle, we can find the area of 
the larger triangle (it's twice as big):

     LargeTriangleArea = 2*RightAngledTriangleArea
                       = 2*N^2/(8*tan(45/2))
                       = N^2/(4*tan(45/2))

Now we can multiply by eight to sum up all eight triangle areas:

     OctagonArea = 8*LargeTriangleArea
                 = 8*N^2/(4*tan(45/2))
                 = 2*N^2/tan(45/2)

And in general for any regular polygon the area can be found by

     PolygonArea = NumberSides*SideLength^2/(4*tan(180/NumberSides))

Which for an octagon with sides of length N is:

     OctagonArea = 8*N^2/(4*tan(180/8))
                 = 2*N^2/tan(180/8)
                 = 2*N^2/tan(45/2)

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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