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. 32. of this) But the Spheric is to the Inscrib'd as 8 is to

6, by this present Propofition. Therefore the CircumScrib'd is to the Inscrib'd as 12 is to 6, or 2 to 1.

PRO P. XXXV. Theorem.

HE Superficies of any Spherical Portion whatever Fig. 26, d

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perficies of the greatest inscribed Cone, which (BG) the Side of the Cone hath to (GO) the Radius of the Base.

Because (by' 25. of this) the Superficies of the Portion IL BG is equal to the Circle of the Radius B G; the Proportion thereof to QT, that is, to the Base of it self and of the Cone, will be duplicate to the Proportion (by 2. 1. 12.) of B G to GO; that is, (by 14. of this) of the Proportion of the conical Superficies Í BG to the same Base QT. Therefore it is manifeft (by Def. 10. l. 5.) that the Superficies I L BG is to the conical Superficies IBG, as the same conical Superficies IBG is to the Base Q T. Wherefore seeing the conical Superficies IBG, is to the Base QT, as B G (by 14. of this) is to GO, the Superficies of the Portion will also be to the conical Superficies I B G inscrib'din it, as BG is to GO, 2. E. D.

PRO P. XXXVI. Theorem.
HE Superficies of the Hemisphere (EOBD) Fig. 24.

hath that Proportion to (EBD) the Superficies of the greatest right infcribed Cone, which in a Square the Diameter hath to a Side ; and that Proportion to the Superficies of a like Cone circumscribed, as the Side in a Square hath to the Diameter.

Ì. The Demonstration of the firft Part is manifeft from the foregoing. For the Superficies of any Portion whatever, and consequently of the Hemisphere, EOBD, is to the conical Superficies inscrib'd, as BD is to D A. But BADK is a Square, whose Diameter is BD and the Side DA.

Part II. Let E BC be half of the Square circumscrib- Fig. 6.1.4. ed about the Circle (whose Centre is O); which EBC being turn'd about the Axis O B, let there from thence

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be produc'd

a Cone circumfcribed about the Hemisphere. Now because the Square E C is (by 47. I. 1.) double to the Square E B or ĜI, the Circle of the Diameter EC also is (by 2. l. 12.) double to the Circle whose Diameter is GI, that is, to the Circle HGDI.

But (by 24. of this) the Superficies of the Hemisphere included in the Cone EBC is double to the fame Circle. Therefore the Circle of the Diameter EC is equal to the hemispherical Surface. Wherefore seeing the conical Superficies EBC is (by 14. of this to the Circle of the Diameter EC, to wit, to its own Base, as the Side BE is to E O the Radius of the Base; it will be also to the chemispherical Superficies inscribed in it, as BE is to EO; that is, as the Diameter in a Square is to a Şide. 2. É. D.

PROP. XXXVII. Theorem.

gure with

The sameFi. Sphere hath the same Proportion to a Square coe Fig. 13-logo

nical Rhombus circumscribed about it, both in respect of the Solidity and Surface, which in a Square the Side bath to the Diameter.

Let the Square EBCF be circumscrib'd about HG DI, the greatest Circle of a Sphere, from which Square as turn'd round about the Axis BF, let a conical Rhombus encompassing the Sphere be produc'd.

As E B à Side of the Square (see Fig. 6, l. 4.) is to the Diameter EC, even lo let S be made to R; (see Fig. 13.1. 5.) and let this Proportion be continued thro' four Terms, S, R, Q, O; the Proportion then of $ to O

will be triplicate to the Proportion of S to R; that is, See def. 10. of EB to E C, and the Proportion of O to R will be 1. so

duplicate to the Proportion of O to Q, or of R to S; that is, of EC to EB; and consequently (by 20. 1. 6.) Ois to R as the Square of EC is to that of É B; from phence (by Schol. Pr. 6. and 7.1.4.) O is double to R.

These Things being thus settled, let the Sphere EBCF be understood to be circumscribed about the conical Rhombus. Thus the Sphere HGDI will be to the Sphere E BCF (by 18. 1. 12.) in the triplicate Proportion of the Diameter G I or E B to the Diameter EC; that is, (as I have already shew'd) it will be as S to 0.

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But the Sphere E BCF is to the conical' Rhombus in-
fcrib'd in it (by 30. of this) as 2 is to ry that is, (as I
have shew'd above) as O is to R. Therefore by Equa-
lity of Proportion, the Sphere HGDH is to the same
Rhombus which is describ'd about it, as S is to R, that
is, as in a Square the Side E B is to the Diameter E C.
Which was the first Part. Then from the second Part
of the foregoing, it appears that the Superficies of the
Hemisphere is to the Superficies of the Cone E BC, and
confequently the Superficies of the whole Sphere is to
the Superficies of the whole Rhombus E BCF, as in a
Square the Side is to the Diameter. Therefore the
Sphere as well in Solidity as' in Superficies is to the
square Rhombus EBCF, as in a Square the Side is to
the Diameter. 2. E. D.

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PROP. XXXVIII. Theorem.
HE Superficies of the Portion (BGKD) which Fig. 27.

contains an equilateral Cone (BKD) is double
to the Superficies of the same Cone.

This is manifest from 35. For the Superficies of the Portion BGKD is to the inscrib'd conic Superficies (by 35. of this) as BK is to B A. But because the Cone BKD is suppos'd to be equilateral, K B is equal to BD, and consequently double to BA. Therefore the Superficies BGKD is also double to the inscribed conical Superficies BK D. 2. E.D.

PROP. XXXIX. Theorem.

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HE Superficies of a Sphere is to the whole Super- Fig. 27. ficies of an equilateral Cone inscril'd in it, as

16 to 9.

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were

10

Let Z be the Center of the Sphere, and BKD the
equilateral Cone inscribed, and K ZA O the A
mon to the Sphere and Cone. If the Sphere a
be cut thro' this, there will be produce in the
the greatest Circle OBKD, and in the Cone e'.. cqui-
lateral Triangle BKD, one side whereof BAD will be
the Diameter of the Basis of the Cone QT. And be-

cause

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causethe Axis of the Cone Ķ A is perpendicular to the Base QT, BAK (Def. 3. l. 11.) will be a right Angle.

. Therefore the Square of B A is equal to the Rectangle KAO. (Corol. 1. Pr. 17.1.6.) Now because the side of the equilateral Triangle cuts off (Corol. 5. Pr. 15.b. 4.) a 4th Part of the Axis AO, the Rectangle K A O, that is, the Square of B A, will be triple to the Square of AO (by 1. 1. 6.) Wherefore seeing the Square of the Radius Zo is (Corol. 3. Pr. 4.1. 2.) quadruple of the Square of AO, the Square of the Radius Zo will be to the Square of the Radius B A, as 4 is to 3. Therefore the Circle OBKD is also (by 2.1. 12.) to the Circle QT, as 4 is to 3. Therefore four Circles O BKD, that is (by 24. of this) the whole spherical Superficies DG is to the Circle QT, as 16 is to 3. But (Corol. 1. Pr. 14. of this) the Superficies of the equilateral Cone BKD is to the Circle QT, to wit, its own Bafe, as 2 is to 1 ; and

, confequently the whole Superficies of the Cone BKD, including its Base, is to the Base, to wit the Circle QT, as 3 is to r, or 9 to 3: Therefore seeing I have shew'à that the Superficies of a Sphere is to the same Circle, as 16 is to 3, the Superficies of the Sphere DG will be to the whole Superficies of the equilateral Cone, as '16 is to 9. 2.E.D.

or otherwise thus :

1

а

B Ecause. (by Corol. s. Pr. 15.1. 4.) the Side BD of the

equilateral Triangle cuts off a 4th Part of the Axis AO, the spherical Superficies BOD will be a 4th Part by 27. of this, and consequently the Superficies BGKD, three 4th Parts of the Superficies of the whole Sphere. Wherefore if the whole Superficies be suppos’d to be 16, the Superficies BGKD will be 12. But (by the fore

( going) the Superficies BGKD is double to the conical Superficies BKD, and consequently is to it, as 12 to 6. Therefore the whole Superficies of the Sphere is to the conical BKD, as 16 is to 6. Then because the Superficies of the Cone BKD (as being equilateral) is (by Corol. 1. Pr. 14. of this) double to the Bafe QT, it is manifeft that the conical Superficies BKD (to wit, without the. Base) is to the whole Superficies of the Cone, as 2 is to 3 ; that is, as 6 to 9. Therefore by equality of Proportion the whole Superficies of the

Sphere

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Sphere is to the whole Superficies of the equilateral
Cone infcrib’d, as 16 to 2. 2. E. D.

PROP. XL. Theorem.

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THE Superficies of a Sphere bears that Propor-Fig. 28.

tion to the whole Superficies of an equilateral Cone circumscrib'd about its which 4 doth to 9,

Let there be circumscrib'd about the greatest Circle of a Sphere BPM, the equilateral Triangle DOF; by which, as turn'd round about the Axis OAB, let there be produc'd an equilateral Cone, circumscrib'd about the Sphere. And let there also be circumscrib'd about the equilateral Triangle DOF the Circle NDL OF, which, as is manifeft, is concentrical to the former; and let the Axis O AB be produc'd to N. Because BN is a 4th Part of the Axis ON, (as is manifest from Corol. 5. Pr. 15.1. 4.) ON is double to BK. Wherefore the Proportion betwixt Circles being duplicate (by 2. l. 12.) of the Proportion of the Diameters, the Circle BPM will be to the Circle NDLOF, as I to 4. But it hath already been shew'd in the first foregoing Demonstration, that the Circle NDLO Fis to the Circle QT, the Base of the equilateral Cone inscrib'd in the Sphere FL, as 4 is to 3. Therefore by equality of Proportion the Circle BPM is to the Circle OT, as I is to 3. But the whole Surface of the Cone DOF is (by Cor. 1. Pr. 14. of this) triple to QT. Therefore the whole Superficies of the Cone is ninefold of the Circle BPM. Wherefore feeing the Superficies of the Sphere TP is quadruple (by 24. of this) of the fame Circle BPM, the who e Superficies of the equilateral Cone DOF is to the Superficies of the Sphere to which it is circumscrib'd, as 9 is to 4. 2. E. D.

Coroll. 1. From this Demonstration it is manifest that the Axis

. BO of an equilateral Cone circumscrib'd about a Sphere, is one and a half of the Diameter of the Sphere BK, or as 3 to 2.

2. That QT the Base of the Cone DOF is also one and an half of both Bases of the Cylinder circumscrib'd about the fame Sphere. For 2 T is to BPM, as 3 to 1. Tberefore QT is to B P M iwice, as 3 is to 2.

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