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Scholium. Conceive a polyedron, all of whose faces touch the sphere; this polyedron may be considered as formed of pyramids, each having for its vertex the centre of the sphere, and for its base one of the polyedron's faces. Now it is evident that all these pyramids will have the radius of the sphere for their common altitude: so that each pyramid will be equal to one face of the polyedron multiplied by a third of the radius: hence the whole polyedron will be equal to its surface multiplied by a third of the radius of the inscribed sphere.
It is therefore manifest, that the solidities of polyedrons circumscribed about the sphere are to each other as the surfaces of those polyedrons. Thus the property, which we have shown to be true with regard to the circumscribed cylinder, is also true with regard to an infinite number of other bodies.
We might likewise have observed that the surfaces of polygons, circumscribed about the circle, are to each other as their perimeters.
PROPOSITION XVI. PROBLEM.
If a circular segment be supposed to make a revolution about a diameter exterior to it, required the value of the solid which it describes.
Let the segment BMD revolve about AC. On the axis, let fall the perpendiculars BE, DF; from the centre C, draw CI perpendicular to the chord BD; also draw the radii CB, CD.
The solid described by the sector BCA is equal to 37.CB2.AE (Prop. XIV. Sch. 1.); the solid described by the sector DCA .CB2.AF; hence the difference of these two solids, or the solid described
by the sector DCB=π.CB2. (AF-AE)=27.CB2.EF. But the solid described by the isosceles triangle DCB has for its measure 7.CI.EF (Prop. XII. Cor.); hence the solid described by the segment BMD=37.EF. (CB2-CI). Now, in the rightangled triangle CBI, we have CB2-CI-BI-BD2; hence the solid described by the segment BMD will have for its measure 7.EF.BD2, or 17.BD2.EF.
Scholium. The solid described by the segment BMD is to the sphere which has BD for its diameter, as 7.BD.EF is to 7.BD3, or as EF to BD.
PROPOSITION XVII. THEOREM.
Every segment of a sphere, included between two parallel planes, is measured by the half sum of its bases multiplied by its altitude, plus the solidity of a sphere whose diameter is this same altitude.
Let BE, DF, be the radii of the two bases of the segment, EF its altitude, the segment being described by the revolution of the circular space BMDFE about the axis FE. The solid described by the segment BMD is equal to 7.BD2.EF D (Prop. XVI.); and the truncated cone described by the trapezoid BDFE is equal to.EF.(BE2+DF+BE.DF) (Prop. VI.);
hence the segment of the sphere, which is the sum of those two solids, must be equal to 17.EF.(2BE2+2DF2+2BE.DF+BD) But, drawing BO parallel to EF, we shall have DO=DF-BE, hence DO2 DF2-2DF.BE+BE (Book IV. Prop. IX.); and consequently BD-BO+DO-EF+DF2-2DF.BE+BE. Put this value in place of BD2 in the expression for the value of the segment, omitting the parts which destroy each other we shall obtain for the solidity of the segment,
EF. (3BE+3DF2+ EF2),
an expression which may be decomposed into two parts; the л.ВЕ2+7.DF? 2
one ‡π.EF.(3BE2+3DF), or EF.("
half sum of the bases multiplied by the altitude; while the other .EF represents the sphere of which EF is the diameter (Prop. XIV. Sch.): hence every segment of a sphere, &c.
Cor. If either of the bases is nothing, the segment in question becomes a spherical segment with a single base; hence any spherical segment, with a single base, is equivalent to half the cylinder having the same base and the same altitude, plus the sphere of which this altitude is the diameter.
Let R be the radius of a cylinder's base, H its altitude: the solidity of the cylinder will be R2x H, or лR2H.
Let R be the radius of a cone's base, H its altitude: the solidity of the cone will be R2x H, or πR2H.
Let A and B be the radii of the bases of a truncated cone,
H its altitude: the solidity of the truncated cone will be 17.H. (A2+B2+AB).
Let R be the radius of a sphere; its solidity will be πR3. Let R be the radius of a spherical sector, H the altitude of the zone, which forms its base: the solidity of the sector will be R2H.
Let P and Q be the two bases of a spherical segment, H its
altitude: the solidity of the segment will be P+Q.H+.H3.
If the spherical segment has but one base, the other being nothing, its solidity will be PH+д¬H3.
1. A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles.
These arcs are named the sides of the triangle, and are always supposed to be each less than a semi-circumference. The angles, which their planes form with each other, are the angles of the triangle.
2. A spherical triangle takes the name of right-angled, isosceles, equilateral, in the same cases as a rectilineal triangle. 3. A spherical polygon is a portion of the surface of a sphere terminated by several arcs of great circles.
4. A lune is that portion of the surface of a sphere, which is included between two great semi-circles meeting in a common diameter.
5. A spherical wedge or ungula is that portion of the solid sphere, which is included between the same great semi-circles, and has the lune for its base.
6. A spherical pyramid is a portion of the solid sphere, included between the planes of a solid angle whose vertex is the centre. The base of the pyramid is the spherical polygon intercepted by the same planes.
7. The pole of a circle of a sphere is a point in the surface equally distant from all the points in the circumference of this circle. It will be shown (Prop. V.) that every circle, great or small, has always two poles.
PROPOSITION I. THEOREM.
In every spherical triangle, any side is less than the sum of the other two.
Let O be the centre of the sphere, and ACB the triangle ; draw the radii OA, OB, OC. Imagine the planes AOB, AOC, COB, to be drawn; these planes will form a solid angle at the centre O; and the angles AOB, AOC, COB, will be measured by AB, AC, BC, the sides of the spherical triangle. But each of the three plane angles forming a solid angle is less than the sum of the other two (Book VI. Prop. XIX.); hence any side of the triangle ABC is less than the sum of the other two.
PROPOSITION II. THEOREM.
The shortest path from one point to another, on the surface of a sphere, is the arc of the great circle which joins the two given points.
Let ANB be the arc of the great circle which joins the points A and B ; and without this line, if possible, let M be a point of a shorter path between A and B. Through the point M, draw MA, MB, arcs of great circles; and take BN=MB.
By the last theorem, the arc ANB is shorter than AM+MB; take BN=BM, from both; there will remain AN AM. Now, the distance of B from M, whether it be the same with the arc BM or with any other line, is equal to the distance of B from N; for by making the great circle BM to revolve about the diameter which passes through B, the point M may be brought into the position of the point N; and the shortest line between M and B, whatever it may be, will then be identical with that between N and B: hence the two paths from A to B, one passing through M, the other through N, have an equal part in each, the part from M to B equal to the part from N to B. The first path is the shorter by hypothesis; hence the distance from A to M must be shorter than the distance from A to N; which is absurd, the arc AM being proved greater than AN; hence no point of the shortest line from A to B can lie out of the arc ANB; hence this arc is itself the shortest distance between its two extremities.*
*Note. This demonstration of Legendre appears to be inconclusive.-ED.
PROPOSITION III. THEOREM.
The sum of the three sides of a spherical triangle is less than the circumference of a great circle.
Let ABC be any spherical trianangle; produce the sides AB, AC, till they meet again in D. The arcs ABD, ACD, will be semicircumferences, since two great circles always bisect each other (Book A VIII. Prop. VII. Cor. 2.). But in the triangle BCD, we have the side BC<BD+CD (Prop. I.); add AB +AC to both; we shall have AB+ AC+BC<ABD+ACD, that is to say, less than a circumference.
PROPOSITION IV. THEOREM.
The sum of all the sides of any spherical polygon is less than the circumference of a great circle.
Take the pentagon ABCDE, for example. Produce the sides AB, DC, till they meet in F; then since BC is less than BF+CF, the perimeter of the pentagon ABCDE will be less than that of the quadrilateral AEDF. Again, produce the sides AE, FD, till
they meet in G; we shall have ED<EG+DG; hence the perimeter of the quadrilateral AEDF is less than that of the triangle AFG; which last is itself less than the circumference of a great circle; hence, for a still stronger reason, the perimeter of the polygon ABCDE is less than this same circumference.