The areas of circles are to each other as the
squares of their diameters.
If you have a circle 2 in. in diameter and
another 4 in. in diameter, then one circle will be four times as great
in area as the other, because the square of 4 is four times as great as
the square of 2.
Now, if we refer to Diagram 1, we see how two equal
squares may be cut into four pieces that will form one larger square;
from which it is self-evident that any square has just half the area of
the square of its diagonal.
In Diagram 2, I have introduced a square as
it often occurs in ancient drawings of the Monad; which was my reason
for believing that the symbol had mathematical meanings, since it will
be found to demonstrate the fact that the area of the outer ring or
annulus is exactly equal to the area of the inner circle.
Diagram 2 with Diagram 1, and you will see that as the square of the
diameter CD is double the square of the diameter of the inner circle,
or CE, therefore the area of the larger circle is double the area of
the smaller one, and consequently the area of the annulus is exactly
equal to that of the inner circle.
This answers our first question.
In Diagram 3, I show the simple solution to the
It is obviously correct, and may be proved by the
cutting and superposition of parts.
The dotted lines will also serve to
make it evident.
The third question is solved by the cut CD in Diagram
2, but it remains to be proved that the piece F is really one-half of
the Yin or the Yan.
This we will do
in Diagram 4.
The circle K has one-quarter the area of the circle
containing Yin and Yan, because its diameter is just one-half the
Also L in Diagram 3 is, we know, one-quarter the area.
therefore evident that G is exactly equal to H, and therefore half G is
equal to half H.
So that what F loses from L it gains from K, and F
must be half of Yin or Yan.