The Complete Works of Archimedes

Therefore, the remaining cylindrical surface cut off by the straight lines $α γ, β δ$ is greater than the parallelogram $α γ β δ$.

If on the surface of a right cylinder there be two straight lines, and from the extremities of these lines there be drawn certain tangents to the circle which forms the base of the cylinder, being in the same planes as the bases, and these tangents meet, then the parallelograms contained by the tangents and the sides of the cylinder shall be greater than the surface of the cylinder between the straight lines on the surface of the cylinder. Let the circle $α β γ$ be the base of a right cylinder, and let there be two straight lines on its surface, whose extremities are $α, γ$. From $α, γ$ let tangents to the circle be drawn in the same plane, and let them meet at $κ$. Let there also be conceived tangents to the circle drawn from the extremities of the lines on the surface at the other base. It must be shown that the parallelograms contained by the tangents and the sides of the cylinder are greater than the surface of the cylinder along the arc $α β γ$. For let the tangent $ε ζ$ be drawn, and from points $ε, ζ$ let certain straight lines be drawn parallel to the axis of the cylinder as far as the surface of the other base. Now the parallelograms contained by $α κ, κ γ$ and the sides of the cylinder are greater than the parallelograms contained by $α ε, ε ζ, ζ γ$ and the sides of the cylinder. For since $α κ, κ γ$ are greater than $α ε, ε ζ, ζ γ$, let $α ε, ζ γ$ be added as common; therefore the whole $α κ, κ γ$ are greater than $α ε, ε ζ, ζ γ$. Let the area $κ$ be the amount by which they are greater. By which the convex [surface] bisecting [it] is greater than the figures contained by the straight lines $α, ε, ζ, β, ζ, γ$ and the arcs $α δ β γ$, the whole being greater than before. Of the surface composed of the parallelograms along $α ε, ε ζ, ζ γ$, and the trapezium $α ε ζ γ$, and its opposite on the other base of the cylinder, the boundary is the perimeter of the parallelogram along $α γ$. So also the surface composed of the cylindrical surface along arc $α β γ$ and the segments $α β γ$ and their opposites has the same boundary. The aforesaid surfaces happen to have the same boundary, which is in a plane; and both are concave in the same direction. And one of them encloses the other, while they have some parts in common. Therefore the enclosed surface is smaller. Thus, when the common segments $α β γ$ and their opposites are subtracted, the surface of the cylinder along the arc $α β γ$ is less than the composite surface consisting of the parallelograms along $α ε, ε ζ, ζ γ$ and the segments $α ε β, β ζ γ$ and their opposites. But the areas of the said figures are less than the surface composed of the parallelograms along $α κ, κ γ$; for with the area $κ$ being greater than the figures, they were equal to them. It is clear, therefore, that the parallelograms contained by $α κ, κ γ$ and the sides of the cylinder are greater than the surface of the cylinder along the arc $α β γ$. If it were not the case that half of the area $κ$ is greater than the said figures, tangents to the circle would be drawn in some shape so that the remaining shapes become less than half of $κ$. And the rest will be demonstrated in the same manner as before.

These things being granted, it is manifest from what has been said that if a pyramid be inscribed in an isosceles cone, the surface of the pyramid excluding the base is less than the conical surface; for each of the triangles containing the pyramid is less than the conical surface between the sides of the triangle. So also the whole surface of the pyramid excluding the base is less than the surface of the cone excluding the base. And that if a pyramid be circumscribed about an isosceles cone, the surface of the pyramid excluding the base is greater than the surface of the cone excluding the base, on account of what follows from that. It is also manifest from what has just been granted that if a prism be inscribed in a right cylinder, the surface of the prism composed of the parallelograms is less than the surface of the cylinder excluding the base; for each parallelogram of the prism is less than the surface of the cylinder corresponding to it. And that if a prism be circumscribed about a right cylinder, the surface of the prism composed of the

A geometrical diagram positioned on the right side of the text block. It shows a circle labeled with points alpha (α), beta (β), and gamma (γ) representing the base of a cylinder. A point kappa (κ) lies outside the circle, with tangent lines drawn to alpha and gamma. An inscribed polygon with vertices including epsilon (ε) and zeta (ζ) is shown along the arc. Straight vertical lines rise from these points to indicate the cylindrical surface and its relationship to the surrounding parallelograms and inscribed/circumscribed shapes.