67 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 16-sided 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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