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CHAPTER VII--MEASUREMENT OF AREAS.

The rule for arriving at the minutely accurate value of the figure formed in the interspace caused by three (equal) circular figures touching each other:-

84${\tfrac {1}{2}}$.^[1] The minutely accurate measure of the area of an equilateral triangle, each side of which is equal in measure to the diameter (of the circles) is diminished by half the area of any of the (three equal) circles. The remainder happens to be the measure of the interspace area caused by three (mutually touching equal circles).

An example in illustration thereof.

85${\tfrac {1}{2}}$. What is the minutely accurate calculated value of a figure forming the interspace enclosed by three mutually touching (equal) circles the diameter (of each) of which is 4 in measure ?

The rule for arriving at the minutely accurate values of the diagonal, the perpendicular and the area in the case of a (regular) six-sided figure:-

86${\tfrac {1}{2}}$.^[2] In the case of a (regular) six-sided figure, the measure of the side, the square of the side, the square of the square of the side multiplied respectively by 2, 3 and 3 give rise, in that same order, to the values of the diagonal, of the square of the perpendicular, and of the square of the measure of the area.

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84${\tfrac {1}{2}}$. ^  Similarly the figure here elucidates at once the reason of the rule:-

86${\tfrac {1}{2}}$. ^  The rule seems to contemplate a regular hexagon. The formula given for the value of the area of the hexagon is ${\sqrt {3a^{4}}}$, where a is the length of a side. The correct formula, however is $a^{2}\times {\tfrac {3{\sqrt {3}}}{2}}$.