Page:The Algebra of Mohammed Ben Musa (1831).djvu/88

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thousand; the quotient is the periphery. Both methods come very nearly to the same effect.^[1]

If you divide the periphery by three and one-seventh, the quotient is the diameter.

The area of any circle will be found by multiplying the moiety of the circumference by the moiety of the diameter; since, in every polygon of equal sides and (52) angles, such as triangles, quadrangles, pentagons, and so on, the area is found by multiplying the moiety of the circumference by the moiety of the diameter of the middle circle that may be drawn through it.

If you multiply the diameter of any circle by itself, and subtract from the product one-seventh and half one-seventh of the same, then the remainder equal to the area of the circle. This comes very nearly to the same result with the method given above.^[2]

Every part of a circle may be compared to a bow. It must be either exactly equal to half the circumference, or less or greater than it. This may be ascertained by the arrow of the bow. When this becomes equal to the moiety of the chord, then the arc is

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1. The three formulas are,
   ${\begin{aligned}&\quad 1st,3{\tfrac {1}{7}}d=p\;{\text{i.e.}}\;3.1428d\\&\quad 2d,{\sqrt {10d^{2}}}=p\;{\text{i.e.}}\;3.16227d\\&\quad 3d,{\tfrac {d\times 62832}{20000}}=p\;{\text{i.e.}}\;3.1416d\end{aligned}}$
2. The area of a circle whose diameter is $d$ is $\pi {\tfrac {d^{2}}{4}}={\tfrac {22}{7\times 4}}d^{2}=(1-{\tfrac {1}{7}}-{\tfrac {1}{2\times 7}})d^{2}$.