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The Cone whose Basis is the hemispherical Superficies EOBD, and its Altitude the Radius AB, is to the Cone E B D (by II. l. 12.) as Base is to Base; that is, as the hemispherical Surface EOBD is to the greatest Circle PT. Therefore seeing the hemispherical Superficies EOBD is double to the greatest Circle (by_24. of this), the

Cone also which hath the Superficies EO BD for its Base, and the Radius A B for its Altitude, is double to the Cone EBD. But (by 28. of this the Hemisphere is equal to a Cone which hath the Rádius for its Altitude, and the hemispherical Superficies for its Base.

Therefore the Hemisphere is also double to the Cone E BD. 2. ED.

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Fig. 250

L

ET a Sphere be divided into two Segments IL BG, LISKG, by the Plane IQGT which doth not pass thro' the Centre A; and let the Diameter BOK be perpendicular to the cutting Plane.

As the Altitude o B of the Segment IL BG, is to the Radius of the Sphere A B: So let OK, the Altitude of the other Segment, be made to another Line KN

In like manner, As OK, the Altitude of the Segment IS KG, is to the Radius AK or A B, So let the Altitude o B of the other Segment be made to another Line B D. Which 'Things being suppos’d, I say,

1. The Cones ING and IDG, whose Altitudes are ON, OD, and I QG T their common Base, are equal to the spherical Segments,

2. There is the same Proportion of the Segments as there is of the right Lines DO, NO.

3. The Segment ISKG is to the greatest Cone IKG G inscrib'd in it, as NO is to KO; and the Segment IL BG is to the greatest Cone IBG inscrib'd in it, as DO is to BO.

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Part I. Let the Sphere and Cones be cut by a Plane thro' the Diameter B K. There will be produced in the Sphere the greatest Circle BLKG, and in the Cones 'the Triangles BIG, IKG. And because BOK the Diaineter is (by the Hypothesis) perpendicular to the Circle QT, IOB (by Def. 3.1. 11.) will be a right Angle. The Angle BI K 'in'the Semicircle is also a right ond (by 31. l. 3.) Because therefore in the Triangle BIK, there is drawn from the right Angle, 10 perpendicular to the Base BK; B I will be to TO, as (bý 8. l. 6.). BK to K I. Therefore the duplicate Proportion of BI to 10 is equal to the duplicate Proportion of BK to KI; that is, (because B K, KI, KO [by Corol. 2. Pr. 8. 1. 6.) are three Proportionals) equal to the Proportion of BK to KO.

Then because O B is (by the Hypothesis) to B D, as OK is to the Radius AB; by Inversion it will be always thus, DB is to BO, as AB to OK; and by Permutation thus, D B is to BA, as B O to OK; and by

. Compounding thus, D A is to BA, as BK is to OK. Becaufe therefore I have already shew'd the Proportion of B K to OK to be duplicate to the Proportion of BI to 10, and consequently (by 2. l. 12) equal to the Proportion betwixt the Circles describ'd by the Radius's BI, IO, DA will also be to BA, as the Circle of the Radius B I, to the Circle of the Radius 1 O. Therefore the Cone under the Altitude D'A, and for the Base, the Circle of the Radius 10, that is, the Circle QT, is equal to the Cone under the Altitude B A, (by 15.1. 12) which hath for its Base the Circle of the Radius BI; that is, (by Corol. Pr. 29. of this) the spherical Sector AIBG. Wherefore if the fame Cone I'A G be added as well to the Sector AIB G, as to the Cone under DA, and the Circle QT, the Wholes will be equal; to wit, the spherical Segment IL BG will be equal to two Cones, whereof one is that which is under the Base QT and the Altitude DA, and the other I A G is under the same Base QT, and the Altitude O A. But these two Cones (by. 14. 1. 12.) make up the Cone IDG.

Therefore the Segment IL BG will be equal to the Cone I DG. 2. ED. 3. By the same Reasoning, the Segment ISKG will be equal to the Cone ING, with this only Change, Q2

that

that the Cone I AG, which before was added, be now

taken away.

Part II. This is manifest from the first. For the Cones I D G and ING are betwixt themselves (by p. 14. 1. 12.) as are Do and NO. Therefore the Segments also I LBG, ISK G, equal to those Cones, are betwixt themselves, as the right Lines, DO, NO.

Part III. This likewise is manifeft from the first. For the Cone I DG is to the Cone I B G, (by the fame) as DO is to BO. Therefore the Segment also IL BG, which is equal to the Cone I DG, is to the Cone I BG, as D O is to BO.

Scholium.

FRom the first Part of this Propofition there arises

another Way of measuring spherical Segments, and that a very easy one; if, to wit, the Cones I DG, ING, be measured; which will be done if the third Parts of the right Lines DO, NO, be drawn into the Circle QT.

PROP. XXXII. Theorem. .

Fig. 24•

A

Right Cylinder (GK) is both in Solidity and the

whole Superficies to the Sphere about which it is circumscrib'd as 3 to 2.

Let BQ be the common Axis of the Sphere and Cy. linder, and EBD the greateft Cone infcrib'd in the Hemisphere EOBD. Because the Cylinder E K (half of GK) is (by 10. I. 12.) triple to the Cone EBD, while the Hemisphere is double to the same Cone (by 30 of this), it is manifest that the Cylinder E K is to the Hemisphere as 3 to 2. Therefore also the whole Cylinder GK is to the whole Sphere QE BD, as 3 to 2. Which was the first Part.

Then because the side of the Cylinder K N is equal to GN the Diameter of the Base, its Superficies without the Bases will be fourfold (by Corol. Pr. 12. of this) of the Base MI, and consequently taken together with the Bases, that is, the whole Superficies of the Cylinder, will be fixfold of the Base MI," which is equal to the greatest Circle of the Sphere. But the Superficies of the Sphere is fourfold of that greatest Circle.

Therefore

the

the whole Superficies of the Cylinder GK is to the Su-
perficies of the Sphere, as 6 to 4, or as 3 to 2. Which
was the other Part.

Therefore a Cylinder is both in Solidity and the whole
Superficies to the Sphere, about which it is circumscribid,
as 3 to 2. 2.E.D.

Schulium.

IT is an Argument what a great Value

Archimedes puts upon this Theorem, that he would have a Sphere infcribd in a Cylinder fet upon his Tomb. And perhaps amongst so many other famous Discoveries, this chiefly and above all others pleas'd him, for this Reason, to wit, because there was one and the same rational Proportion both of Bodies, and of the Surfaces which contain them. We have demonstrated a like Identity of Affections betwixt Rings, and the Surfaces of Rings, in the 4th Book of our Cylindricks and Annularies, Prop. 13, 14, 15: And another famous Example of the same hath also offer'd it self to me in the Sphere it self. For I have found, that like as a Sphere is to a right Cylinder which encompasseth it (which will necessarily be equilateral) as 2 is to 3, and this both in respect of Solidity and Surface; fo likewise the Sphere hath to an equilateral Cone encompassing it, that Proportion which 4 hath to 9;

4

and this both in regard of Solidity and Superficies. From which this also follows, That the fefquialteral Proportion found by Archimedes in the Sphere and Cylinder, is continued in three Solids, a Sphere, Cylinder, and equilateral Cone. The Demonstration of both which Things, withsome other Theorems of my own, in which thewonderful Nature of the Sphere will more appear, I fall subjoin in the thirteen following Propofitions.

PROP. XXXIII. Theorem.

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HE Superficies of a Sphere is double to the Su- Fig. 26.

Sphere.

Let AKLD be the Square inscrib'd in the greateft
Çircle of a Sphere, from which turn'd round, there is
R. 3

describ'd

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describ'd a square Cylinder, and let AL be drawn as 4 Diameter common to the Square and Sphere. Because the Square of A L is (by 47. l. 1.) equal to the equal Squares of A K, KL, it will be double to one A K. Therefore also the Circle of the Diameter A L, is (by 2. I. 12.) double to the Circle, whose Diameter is AR; to wit, to the Circle CN.

But the Superficies of the Sphere is (by 24. of this) fourfold to the Circle whose Diameter is AL; for that is the greatest Circle of the Sphere, seeing AL is the Diameter of the Sphere. Therefore the Superficies of the Sphere is eightfold of the Circle CN. But because LKKA (by the Hypothefis) are equal, the cylindrical Superficies A CL is (by Corol

. . 12. of this) quadruple of the Circle CN. T'here fore fince the Superficies of the Sphere is eightfold of the same Circle, it will be double to the cylindrical Superficies. 'Q.E.D.

PRO P. XXXIV. Theorem.

Fig. 26.

THA

a

'HE Superficies of a Sphere bath that Propor

tion to the whole Superficies of a square Cylinder infcrib'd in it, which 4 bath to 3.

Let the fame Things be suppos'd which were in the foregoing Demonstration.

Because by the Hypothefis LK the Side of the Cylinder, and AK the Diameter of the Base thereof are equal, the cylindrical Superficies CL will be quadruple (by Corol. Pr. 12. of this) to the Base CN, and consequently the whole Superficies of the Cylinder is to both Bases CN and $ L, as o is to 2. But the Superficies of the Sphere is to both Bases together CN, SL, as 8 is to 2, seeing in the foregoing it was shew'd that it is to one Base as 8 to 1. Therefore the Superficies of the Sphere is to the cylindrical Superficies CL as 8 is to 6, or 4 to 3. 2.- É. D.

Corollary.
THE
HE whole Superficies of a right Cylinder describ'd

about a Sphere, is to the whole Superficies of an equilateral Cylinder inscrib’d, as 2 is to i. For the Çircumfcribd is to the spheric Superficies as 12 is to 8 (by

23.

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