In the beginning was the circle, not the square. The square came later. (Background: “In The Beginning Was The Circle”, Ersjdamoo’s Blog entry of August 16, 2013.)

But before there was the circle, there was the point.

The great planetary Lha is subservient to the Solar Lha. The Solar Lha, in turn, is subservient to the Universal Solar Lha, the Solar Emperor, the Raja Sun.

Hidden from our sight, behind some planets, Raja Suns are concealed. One such King Star is said to be hidden behind Jupiter.

Higher than the Raja Suns, the Universe evolves from the Central Sun, the POINT, the ever-concealed germ. (Source: *Tales Of The Holy Lance*, by Brian Redman. Published by Lulu.com)

The Universe evolves from the POINT. In the beginning was the point. Then came the circle, and eventually squares and other shapes.

John A. Parker, in the 1874 edition of his book, *The Quadrature of the Circle*, may have introduced a new equation for finding the areas of circles, squares, and other shapes. I at least had never before seen the equation: Area equals half the circumference multiplied by the least radius which the shape can contain. (Background: “Curved Lines Are Different”, Ersjdamoo’s Blog entry of August 14, 2013.)

If you are trying to “square the circle”, it might be helpful to have such an equation, one which is common for all shapes including squares and circles.

Seen above (hopefully) is Parker’s 4th proposition from chapter 2 of his book, labeled by me as Proposition 2-4. You should be able to see a larger view by clicking on the image, which includes Parker’s Plate 10 illustration.

Looking at Plate 10 (you might try right-clicking on the image and opening the enlargement in a new tab), Parker begins by assigning a value of 3 to the circumference of circle E. Using his half-circumference times least radius formula, Parker computes the area of circle E to be 0.7183315… (However my own calculation using the Parker formula returns an area of 0.7161972…)

Looking at Plate 10, you will see depicted the equilateral triangle F. Parker assigns a value of 1 to each side of triangle F, making its perimeter (or circumference) total to 1+1+1=3. Does the Parker formula of Area equals half-circumference times least radius work on this triangle? Parker uses a different formula, multiplying the diameter of triangle F (square root of 0.75) by half of one side (0.5) and gets an area of 0.4330127… for triangle F. (Using an area formula for equilateral triangles which I found on Internet (Area equals the square root of 3 multiplied by one of the sides squared, and that divided by 4), my answer for the area of triangle F is the same as Parker’s.)

Next comes a circle G having the same area as triangle F. The circle G is inscribed in a triangle H. Parker calculates that the area of triangle H will be the same as the area of the original circle E.

Hence we have John A. Parker’s Proposition 2-4: “The circumference of any circle being given, if that circumference be brought into any other shape formed of straight lines and of equal sides and angles, the area of that shape is equal to the area of another circle, which circle being circumscribed by another and similar shape, the area of such shape circumscribing the last named circle is equal to the area of the circle whose circumference is given.”

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