Archimedis Opera cum Apollonii Conicorum Libri IIII

(LR : Rq b cor. 2. 12.

) S : circle R. Q.E.D.
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### Prop. XIX.

Note: In this and following [propositions], triangles representing cones are understood to be traversed by axes.
Fig. 23. 24.

If an isosceles cone (ABC) is cut by a plane (DQE) parallel to the base (BPC), the surface of the cone (DBCE) intercepted by the parallel planes is equal to a circle whose radius (Z) is the mean proportional between the side of the cone (DB) intercepted between the parallel planes and the sum of the radii (FB + GD) of the circles (BPC, DQE) which are in the parallel planes.

a hyp.
b 17. 6.
c sch. 1. 2.
d 16. 6.
e 4. 6.
f cor. 2. 12.
g 17. of this.
Fig. 25.

For because DB : Z a hyp.

AD : DG). f cor. 2. 12. Therefore circle radius Z = circle radius $\sqrt{AB \times BF}$ - circle radius $\sqrt{AD \times DG}$. But circle radius $\sqrt{AB \times BF}$ g 17. of this = surface ABC; and circle radius $\sqrt{AD \times DG}$ g 17. of this = surface ADE. Therefore circle radius Z is equal to the surface DBCE. Q. E. D.
Coroll. Hence, if the straight line YX bisects the sides DB, EC; the circle with radius $\sqrt{DB \times YX}$ will be equal to the conical surface DBCE.
For, having drawn the straight lines YG, BE, FX; because DG = DE/2 and DY = DB/2, YG and BE will be parallel; therefore in parallelogram GYUE, YV = GE. By similar reasoning, VX = BF. Whence YX = GE + BF.
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### Prop. XX.

Fig. 26.
a hyp.
b 7. 5.
c 4. 6.
d 18 of this.
e 7. 5. & hyp.
f 11. 5.

If there be two isosceles cones (BAC, XZ), and the surface of one (BAC) be equal to the base of the other (Z); and the perpendicular (DE) drawn from the center of the base (D) to the side of the cone (AC) be equal to the height (X); then the cones (BAC, XZ) will be equal.
For, having drawn AD, because X a hyp. = DE, it will be AD : X b 7. 5.

AC : CD (for triangles ADE, ACD are similar right-angled triangles) d 18 of this

Z : base BFC f 11. 5.