Archimedis Opera cum Apollonii Conicorum Libri IIII (1675) - Page 26

b 38. 1.
c const. & hyp.

to one of the perpendiculars V D, ^b is equal to the triangles A V B + B V C + C V A, ^c that is, the triangle M N O is equal to the surface of the pyramid excluding the base. Q. E. D.

Prop. XI.

Fig. 14.

If a pyramid (V A B C) is described about an Isosceles cone (V D E F), the surface of the Pyramid excluding the base is equal to a triangle having a base equal to the perimeter of the base A B C, and the height equal to the side of the cone (V D).

a 18. def. 11.
b 18. 11.
c 18. 3.
d 4. def. 11.
e 3. def. 11.

Let V Z be the axis of the cone, and from the center Z to the contact D let a straight line Z D be drawn; and because V Z ^a is a straight line to the plane A B C, the triangle V Z D is also ^b perpendicular to the plane A B C: but indeed the tangent A D ^c is perpendicular to Z D (the common section of the planes V Z D, A B C) ^d therefore A D is perpendicular to the plane V Z D, and ^e consequently to the line V D; therefore V D (the side of the cone) is the height of the triangle V A B. And by the same reasoning, the side of the cone is the height of the triangle V A C, and of all those which constitute the lateral surface of the cone. ^f Therefore, the triangle whose base is A B + B C + C A and whose height is the side of the cone, ^f is equal to the triangles constituting the surface of the cone. Q. E. D.
f 38. 1.

Prop. XII.

Fig. 15.

If a straight line (C D) falls within the circle (A B C D) which is the base of an Isosceles cone (V A B), and from its ends straight lines (C V, D V) are drawn to the vertex of the cone (V), the triangle (C V D) enclosed by the falling line and the lines drawn to the vertex is less than the conical surface (C A V B D) intercepted by the lines drawn to the vertex.

a 1. 6.
b 20. 1.
c 3. ax. huj.
d hyp.
e 8 hujus.

Let the arc C A B D be bisected at E, and let straight lines C E, D E, V E be drawn. And it is clear that triangle C V E + D V E ^a > C V D; because C E + D E ^b > C D, and the height is common: let the excess be X, first (suppose) not less than the segments C E, D E: and because the conical surface C V E + segment C E ^c is greater than the included triangle C V E (for the common limit is the chord C E); and similarly the conical surface D V E + segment D E > triangle D V E, it will be in sum the conical surface C V D + segments C E, D E > triangle C V E + D V E; and more so the conical surface C V D + X > triangle C V E + D V E ^d = triangle C V D + X. Whence by removing the common X, the conical surface C V D will be greater than the triangle C V D. If X should be less than the segments C E, D E, let the arcs C E, D E be bisected, and their halves, ^e until the remaining segments C A, A E, D B, B E are less...