its center of gravity, and the center of gravity of the triangle $\zeta\alpha\gamma$ is $\chi$; hence triangle $\alpha\zeta\gamma : $ segm. $\alpha\beta\gamma$ when transferred to $\theta$ as its center of gravity $= \theta\kappa : \kappa\chi$. But $\theta\kappa = 3\kappa\chi$; hence also triangle $\alpha\zeta\gamma = 3$ segm. $\alpha\beta\gamma$. But it is also true that triangle $\zeta\alpha\gamma = 4\Delta\alpha\beta\gamma$ because $\zeta\kappa = \kappa\alpha$ and $\alpha\delta = \delta\gamma$; hence segm. $\alpha\beta\gamma = \frac{4}{3}$ the triangle $\alpha\beta\gamma$. This is of course clear. It is true that this is not proved by what we have said here; but it indicates that the result is correct. And so, as we have just seen that it has not been proved but rather conjectured that the result is correct we have devised a geometrical demonstration which we made known some time ago and will again bring forward farther on. \section*{Proposition II} That a sphere is four times as large as a cone whose base is equal to the largest circle of the sphere and whose altitude is equal to the radius of the sphere, and that a cylinder whose base is equal to the largest circle of the sphere and whose altitude is equal to the diameter of the circle is one and a half times as large as the sphere, may be seen by the present method in the following way: \begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig02.png} \begin{center} {\small Fig.~2.} \end{center} \end{wrapfigure}