ARCHIMETES: The Squaring of the Circle and the Calculation of

The Babylonians (2000 BC) knew that there was a relationship between the Circumference of a Circle and its Diameter, i.e., X = Circumference / Diameter. This relationship was defined by the symbol in the 18 th century AD so that:

= Circumference (C) / Diameter (D) or
C = 2r, where r is the radius of the Circle (1/2 the Diameter)
Its value was estimated (by direct measurement) by the Babylonians to be: = 3.125
The value of used today is: 3.142

We now know that the Area of a Circle is: Image Source

Squaring the Circle:

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The Square within the Circle has 4 angles equal to 90o each. The diagonal of the square (which is the same as the diameter of this circle, equal to 2r) bisects one of these angles resulting in two right triagles each with three angles equal to 45o, 45o, and 90o. The side of each triangle (in a ratio of 1:1; x:x) is found by:

c2 = x2 + x2
(2r)2 = x2 + x2
4r2 = 2x2
x2 = 4r2/2
x2 = 2r2
each side x = Ö2r
The area the the square is:
Area (A) = length x width
A = Ö2r x Ö2r = 2r2.

Archimetes (287-212 BC) provided a mathematical method for the calculation of by using a circle bounded by two regular polygons (N = number of sides of the polygon). He based his formulation on the fact that the perimeter of the outer regular polygon was larger than the circumference of the circle, while the perimeter of the inner regular polygon was less that the circumference of the circle. By incrasing the number of sides (N) of the regular polygon, the two perimeters would approximate the value of the circumference of the circle.

The perimeter, P, of a polygon with "N" sides of size "b" is
P = b1 + b2 + b3 + b4 +.......bN = Nb
When N is sufficiently large, Archimetes found that the perimeter of the regular polygon is equal to the circumference of the inscribed circle, i.e.,
Nb = 2r

Archimetes method:

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  1. Archimetes started with an inscribed hexagon within a circle and progressively increased the sides of the regular polygon. Initially, he found the area of the hexagon by dividing it into six equal triangles each with three angles equal to 60o, or 180o for each equalateral triangle.

    The area of each triangle is: A = 1/2(b.h), where "h" is the altitude (a) of the triangle and "b" is the value of one side = r/2.
    From r2 = a2 + b2, Archimetes calculated the value of "h":
    r2 = a2 + b2
    r2 = a2 + (r/2)2
    a2 = r2 - r2/4
    a2 = 4r2/4 - r2/4
    h2 = 3r2/4
    h = Ö3r/2.
    Since there are six equalateral triangles within the hexagon, Archimetes calculated the area of the hexagon to be: A = 6((r)(Ö3r/2)/2) = 2.59r2.

  2. By progressively increasing the number of sides, N, (Archimetes used a 96 sided regular polygon) and triangles within the regular polygon, Archimetes calculated the area of the regular polygon to be: A = 3.14r2; a value equal to the area of the circle .....and without the aid of a calculator or computer!!!!