The Complete Works of Archimedes

Cones of equal height have to each other the same ratio as their bases. And those having equal bases have the same ratio as their heights. If a cylinder is cut by a plane parallel to the base, the cylinder is to the cylinder as the axis is to the axis. And cones having the same bases as the cylinders are in the same ratio as the cylinders. In equal cones, the bases are reciprocally proportional to the heights; and those whose bases are reciprocally proportional to the heights are equal. And cones whose diameters of the bases have the same ratio as the axes—that is, as the heights—are to one another in the triplicate ratio of the diameters in the bases. All these things indeed were demonstrated by those who came before.

17

Geometric diagram for Proposition 17 showing two triangular cross-sections of cones with labeled points α, β, γ, κ and δ, ε, ζ, η, θ.
If there are two isosceles cones, of which the surface of one cone is equal to the base of the other, and the perpendicular drawn from the center of the base to the side of the cone is equal to the height [of the other], the cones will be equal. Let there be two isosceles cones, one $α β γ$, the other $δ ε ζ$. And let the base of $α β γ$ be equal to the surface of $δ ε ζ$. And let the height $α κ$ be equal to the perpendicular $η θ$ drawn from the center of the base $θ$ to one side of the cone, such as to $δ ε$. I say that the cones are equal. For since the base of $α β γ$ is equal to the surface of $δ ε ζ$, and equals have the same ratio to the same thing, therefore as the base of $α β γ$ is to the base of $δ ε ζ$, so is the surface of $δ ε ζ$ to the base of $δ ε ζ$. But as the surface is to its own base, so is $δ ε$ to $η θ$. For it was shown above that the surface of every isosceles cone has the same ratio to the base as the side of the cone has to the radius of the base, that is, $δ ε$ to $η θ$. As $δ ε$ is to $η θ$, so is $δ η$ to $θ κ$; for the triangles are equiangular. And $θ κ$ is equal to $α κ$. Therefore, as the base of $α β γ$ is to the base of $δ ε ζ$, so is the height of $δ ε ζ$ to the height of $α β γ$. Therefore, the bases of $α β γ$ and $δ ε ζ$ are reciprocally proportional to their heights. Therefore, cone $α β γ$ is equal to cone $δ ε ζ$.

18

Geometric diagram for Proposition 18 depicting a rhombus αβγδ (representing two cones sharing a base) and a separate triangular section of a third cone labeled θ, π, κ.
To any rhombus composed of isosceles cones, there is an equal cone having its base equal to the surface of one of the cones containing the rhombus, and its height equal to the perpendicular drawn from the vertex of the other cone to one side of the [first] cone. Let there be a rhombus composed of isosceles cones $α β γ δ$, whose base is the circle about the diameter $β γ$, and its height $α δ$. And let there be set out another cone $κ θ κ$, which has its base equal to the surface of cone $α β γ$, and its height equal to the perpendicular drawn from the point [vertex $δ$] to $α β$, or to its extension; and let it be $δ ζ$. And let the height of the cone $θ π κ$ be $θ λ$, and let $θ λ$ be equal to $δ ζ$. I say that the cone is equal to the rhombus. For let there be set out another cone $μ ν ξ$, having its base equal to the base of cone $α β γ$, and its height