The elements of geometrie of the most auncient philosopher Euclide of Megara. Faithfully (now first) translated into the Englishe toung, by H. Billingsley, citizen of London. Whereunto are annexed certaine scholies, annotations, and inuentions, of the best mathematiciens, both of time past, and in this our age. With a very fruitfull præface made by M. I. Dee, specifying the chiefe mathematicall scie[n]ces, what they are, and wherunto commodious: where, also, are disclosed certaine new secrets mathematicall and mechanicall, vntill these our daies, greatly missed

A Theoreme. 2.

Euery sphere, is quadrupl••, to that Cone, whose base is the greatest circle, & height, the semidia∣meter of the same sphere.

This is the 32. Proposition of Archimedes fi••st booke of the Sphere and Cylinder.

A Probleme. 1.

A Sphere being geuen, to make an vpright Cone, equall to the same: or in any other proportio••, geuen, betwene two right lines.

Suppose the Sphere geuen, to be A, his diameter being BC, and center D: with a line equall to the semidiameter BD (which let be NO) describe a circle NRP:* 1.1 whose diameter let be NP, and center O, it is euident, that NRP is equall to the greatest circle in A, contayned. At the center O, let a perpen∣dicular be reared equall to BD (the semidiameter of A) which suppose to be OQ: It is now plaine that to the Cone, whose base is the circle NRR, and height OQ, the Sphere A, is quadrupla: by th•• 2. Theoreme here, and by construction. Take a line equall to NP, which let be FE: and with the se∣midiameter FE (making th•• point F center) describe a circle: which suppose to be EKG, and dia∣meter EG. At the center F, reare a line perpendicular to EKG, by the 12. of the eleuenth: and make it equall to OQ. Let that line be ••L. I say that the Cone, whose base is the circle EKG, and height the line FL•• is equall to A. For seing FE, the semidiameter of EKG, is equall to N•• (the diameter of NRP) by construction:* 1.2 EG, the diameter of EKG, shall be double to NP. Wherfore the square of EG, is quadrupla to the square of NP: by the 4. of the second. But as the square of EG, is to the

square of NP, so is the circle EKG to the circle NRP, by the 2. of this twelueth. Wherefore the circl•• EKG, is quadruple to the circle NRP. And FL, the height is (by construction) equall to OQ the height. Wherefore the cone, whose base is the circle EKG, and height FL, is quadruple to the cone, whose base is NRP, and the height OQ: by the ••1. of this twelfth. But vnto the same cone whose base is NRP, and height OQ,

[illustration]

the Sphere A is likewise proued quadrupla••. Wherefore the cone whose base is EKG and height FL, is equall to the Sphere A: by the 7. of the fift. To a Sphere being geuen therefore, we haue made an vpright cone equall.

And as concerning the other part of this Probleme,* 1.3 it is now easie to execute, and that two wayes: I meane to A the sphere geuen, to make an vpright cone in any proportion geuen betwen two right lines. For, let the pro∣portion geuen, be that which is betwene X and Y. By the order [ 1] of my additions, vpon the 2. of this twelfth booke: to the circle EKG make an other circle in that proportion that X is to Y: which let be Z. Vpon the center of Z, reare a line perpendicular and equall to FL. I say that the cone, whose base is Z, and the height equall to FL, is to A, in the proportion of X to Y. For the cone vppon Z, by construction, hath height equall to the height of the cone LEKG: and Z, by construction, is to EKG•• as X is to Y: Wherefore, by the 11. of this twelfth, the cone vpon Z, is to the cone LEKG, as X is to Y. But the cone LEKG is proued equall to the sphere A. Wherfore the cone vpon Z, is to A, as X is to Y, by the 7. of the fift. To a Sphere geuen therefore, we haue made a cone in any proportion geuen, betwene two right lines. Secondly, as X is to Y, so to FL, let there be a fourth [ 2] line, by the 12. of the sixt: and suppose it to be W. I say that a cone, whose base is equall to EKG, and height the line W, is to A, as X is to Y. For, by the 14. of this twelfth, cones being set on equall bases, are one to the other, as their heightes are: But, by construction, the height W, is to the height FL, a•• X is to Y. Wherefore the cone which hath his base equall to EKG, and height the line W, is to the cone LEKG, as X is to Y. And it is proued, that to the cone LEKG, the Sphere A is equall: Wher∣fore, by the 7. of the fift, the cone, whose base is equall to EKG, and height the line W, is to A, as X is to Y. Therefore a Sphere being geuen, we haue made an vpright cone,* 1.4 in any proportion geuen be∣twene two right lines. And before, we made an vpright cone, equall to the Sphere geuen. Wherfore a Sphere being geuen, we haue made an vpright cone, equall to the same, or in any other proportion, geuen betwene two right lines. I call that an vpright cone, whose axe is perpendicular to his base.

Notes

• * 1.1

  Construction.
• * 1.2

  Demonstra∣tion.
• * 1.3

  The second part of the Probleme two wayes executed.
• * 1.4

  An vpright Cone.