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approximation to that of the circle, but we can improve it by replacing the square by a regular
octagon, with all its points on the circle. Now, this octagon can by divided into eight triangles,
following the same procedure as for the square. The height of each of these triangles equals the
distance from the center of the circle to the middle of one side of the octagon. Just as for the
square case, the total area of these eight triangles is equal to that of a long thin triangle of the
same height, and with base length equal to the perimeter of the octagon.
It is evident that the height of the octagon’s triangles is closer to the radius of the circle than the
height of the square’s triangles, and the perimeter of the octagon is closer to the circumference
of the circle than the perimeter of the square was.
The process is repeated: the octagon is replaced by a regular 16sided polygon, with all its points
on the circle. This polygon is equal in area to the sum of the 16 triangles formed by drawing lines
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This note was uploaded on 01/08/2012 for the course PHY 322 taught by Professor Daser during the Spring '09 term at SUNY Stony Brook.
 Spring '09
 DASER

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