After I had thus perceived that a sphere is four times as large as the cone whose base is the largest circle of the sphere and whose altitude is equal to its radius, it occurred to me that the surface of a sphere is four times as great as its largest circle, in which I proceeded from the idea that just as a circle is equal to a triangle whose base is the periphery of the circle and whose altitude is equal to its radius, so a sphere is equal to a cone whose base is the same as the surface of the sphere and whose altitude is equal to the radius of the sphere. \section*{Proposition III} By this method it may also be seen that a cylinder whose base is equal to the largest circle of a spheroid and whose altitude is equal to the axis of the spheroid, is one and one half times as large as the spheroid, and when this is recognized it becomes clear that if a spheroid is cut through its center by a plane perpendicular to its axis, one-half of the spheroid is twice as great as the cone whose base is that of the segment and its axis the same. For let a spheroid be cut by a plane through its axis and let there be in its surface an ellipse $\alpha\beta\gamma\delta$ [Fig.~3] whose diameters are $\alpha\gamma$ and $\beta\delta$ and whose center is $\kappa$ and let there be a circle in the spheroid on the diameter $\beta\delta$ perpendicular to $\alpha\gamma$; then imagine a cone whose base is the same circle but whose vertex is at $\alpha$, and producing its surface, let the cone be cut by a plane through $\gamma$ parallel to the base; the intersection will be a circle perpendicular to $\alpha\gamma$ with $\epsilon\zeta$ as its diameter. Now imagine a cylinder whose base is the same circle with the diameter $\epsilon\zeta$ and whose axis is $\alpha\gamma$; let $\gamma\alpha$ be produced so that $\alpha\theta = \gamma\alpha$; think of $\theta\gamma$ as a scale-beam with its center at $\alpha$ and in the parallelogram $\lambda\theta$ draw a straight line $\mu\nu \| \epsilon\zeta$, and on $\mu\nu$ construct a plane perpendicular to $\alpha\gamma$; this will intersect the cylinder in a circle whose diameter is $\mu\nu$, the spheroid in a circle whose diameter is $\xi o$ and the cone in a circle whose diameter is $\pi\rho$. Because $\gamma\alpha : \alpha\sigma = \epsilon\alpha : \alpha\pi = \mu\sigma : \sigma\pi$, and $\gamma\alpha = \alpha\theta$, therefore $\theta\alpha : \alpha\sigma = \mu\sigma : \sigma\pi$. But $\mu\sigma : \sigma\pi = \mu\sigma^2 : \mu\sigma \times \sigma\pi$ and $\mu\sigma \times \sigma\pi = \pi\sigma^2 + \sigma\xi^2$, for $ \alpha\sigma \times \sigma\gamma : \sigma\xi^2 = \alpha\kappa \times \kappa\gamma : \kappa\beta^2 = \alpha\kappa^2 : \kappa\beta^2$ (for both ratios are equal to the ratio between the diameter and the parameter [Apollonius, \emph{Con.} I, 21]) $ = \alpha\sigma^2 : \sigma\pi^2$ therefore $\alpha\sigma^2 : \alpha\sigma \times \sigma\gamma = \pi\sigma^2 :