Archimedis Opera cum Apollonii Conicorum Libri IIII (1675) - Page 25

construction

is less than the ratio of X to Y squared, that is, the ratio of X to Z, which is less than the ratio of A to B. Therefore, it is done.

Cor. Prop. VII.

That we may similarly demonstrate that, given two unequal magnitudes and a sector, a polygon can be described about the sector, and another similar one inscribed in it, so that the circumscribed has a smaller ratio to the inscribed than the greater magnitude has to the smaller.

Lemma. Prop. VIII.

It is also manifest that, if a circle or sector is given, and some space, by inscribing equilateral polygons in the circle or sector, and still continuously [bisecting] the remaining segments, there remain certain segments of the circle or sector which are smaller than the proposed space.
* For these things are handed down in the Elements.

* see 2. 12.

Prop. IX.

* that is, the excess of the polygon over the circle.

It must be demonstrated that to a given circle (or sector) (A) and a space (B), a polygon can be circumscribed about the circle (or sector), * so that the segments left from the circumscription are smaller than the given space.

a 6 of this.

Let a figure be circumscribed about the circle, which let be called C; and let another be inscribed,
b

b 9. ax. 1.

which let be called I, such that C : I ⊏ A + B : A. I say it is done. For because
c

c 8. 5.

A ⊐ I; it will be C : A ⊏ (C : I ⊏) A + B : A. Whence by dividing
d

d const.

C — A : A ⊏ B : A.
e

e 10. 5.

and thus C — A ⊏ B. Therefore it is done.

Prop. X.

Fig. 12. 13.

If a pyramid (V A B C) having an equilateral base (A B C) is inscribed in an Isosceles cone (V A X B Y C Z), its surface, excluding the base, is equal to a triangle (M N O) having a base (N O) equal to the perimeter of the base (A B C), and a height (M N) equal to the perpendicular (V D) let down from the vertex (V) to one side of the base (A B).

For let V E, V F also be drawn perpendicular to the sides BC, CA, and because the triangles AVB, BVC, CVA are mutually ᵃ equilateral to each other, * the perpendiculars V D, V E, V F will be equal to one another. Therefore the triangle whose base is equal to AB, BC, CA taken together, and the height...