Plane and Solid Geometry

781. The spherical excess of a spherical triangle is the sum of its angles less 180°. That is, E = A + B + C − 180°.

782. A spherical pyramid is a portion of a sphere bounded by a spherical polygon and the planes of its sides.

The vertex of a spherical pyramid is the center of the sphere.

The base of a spherical pyramid is the spherical polygon.

783. A spherical sector is the solid generated by the revolution of the sector of a circle about any diameter of the circle as an axis.

The base of the spherical sector is the zone generated by the arc of the circular sector.

A spherical cone is a spherical sector whose base is a zone of one base.

784. A spherical segment is a portion of a sphere included between two parallel planes that intersect the sphere.

The bases of a spherical segment are the circles made by the parallel planes.

The altitude of a spherical segment is the perpendicular distance between the bases.

A spherical segment of one base is a segment, one of whose bounding planes is tangent to the sphere.

A hemisphere is a spherical segment of one base, and that base is a great circle.

A spherical wedge is a portion of a sphere bounded by a lune and the planes of its sides.

Ex. 1. What is the spherical excess of a spherical triangle whose angles are 60°, 70°, and 100°?

Ex. 2. Distinguish between a zone and a spherical segment.

Ex. 3. Find the area of a spherical degree on a sphere whose surface is 3600 sq. in.

Ex. 4. Find the area of a spherical triangle containing 80 spherical degrees, on a sphere whose surface is 450 sq. ft.

PRELIMINARY THEOREMS

785. THEOREM. Either angle of a lune is measured by the arc of a great circle described with the vertex of the lune as a pole, and included between the sides of the lune. (See 726.)

786. COR. The angles of a lune are equal.

787. THEOREM. Every great circle of a sphere divides the sphere into two equal hemispheres, and the surface into two equal zones.

788. THEOREM. The spherical excess of a spherical n-gon is equal to the sum of its angles less (n − 2) 180°.

Proof By drawing diagonals from any vertex, the polygon is divided into n − 2 A.

For one A, E s - 180° (?) (781).

For another ▲, E' = s' — 180° (?).

.. by adding, the excess of all the ▲: 4-(n-2) 180° (Ax. 2).

Etc. for (n − 2) ▲.

= the sum of all their

That is, the excess of a spherical polygon = the sum of its -(n-2) 180°.

Q.E.D.

789. THEOREM. If a regular polygon having an even number of sides be inscribed in, or circumscribed about, a circle, and the figure be made to revolve about one of the longest diagonals of the polygon, the surface generated by the polygon will be composed of the surface of cones, cylinders, and frustums, and the surface generated by the circle will be a spherical surface.

790. THEOREM. If a regular polygon having an even number of sides be inscribed in, or circumscribed about, a circle, and the figure be made to revolve about one of the longest diagonals of the polygon, the surface generated by the perimeter of the polygon will approach the surface of the sphere generated by the circle, as a limit, if the number of sides of the polygon is indefinitely increased.

791. THEOREM. If a polyhedron be circumscribed about a sphere and the number of its faces be indefinitely increased, the surface of the polyhedron will approach the surface of the sphere as a limit, and the volume of the polyhedron will approach the volume of the sphere as a limit.

THEOREMS AND DEMONSTRATIONS

792. THEOREM. The area of the surface generated by a straight line revolving about an axis in its plane is equal to the product of the projection of the line upon the axis by the circumference of a circle whose radius is the line perpendicular to the revolving line at its midpoint, and terminating in the axis.

Given: Line AB revolving about axis XX'; CD= projection of AB on XX'; MP=a= erected at midpoint of AB and terminating in XX'; MO= radius of midsection.

To Prove:

Surface generated by ABCD 2 πа.

Proof: I. The surface generated by AB is

the surface of the frustum of a right circu

lar cone whose bases are generated by AC and BD, and the midsection, by MO.

Now, A ABH and MOP are similar (?) (321).

.. MO: AH = MP : AB (?).

Hence, MO · AB = AH · MP = CD· a (?).

... area of surface = 2 πCD a = CD⋅ 2 πа (Ax. 6).

II. If AB is to XX', the surface is cylindrical and equals CD · 2 πа (?) (665).

III. If AB meets XX' at C, the entire surface is conical and equals TBD AB (?) (693).

Now BD = 2 MO (?) (142); and MO. ABCD a (?).

· · π · BD · AB=π · 2 MO· AВ=π · 2 · CD · α = CD · 2 πа (Ax. 6). That is, the area of the surface = CD⋅ 2 πа (Ax. 6). Q.E.D.

Ex. 1. Find the spherical excess of a polygon whose angles are 80°, 110°, 140°, 130°, 160°.

Ex. 2. The spherical excess of a spherical polygon is the difference between the sum of its angles and the sum of the angles of a plane polygon having the same number of sides.

793.) THEOREM. The surface of a sphere is equivalent to four great circles, that is, to 4 πR2.

Given

Semicircle ACF; diameter AF;

s surface of sphere generated by revolving the semicircle about AF as an axis; R = radius of this sphere.

To Prove: S = 4 πR2.

Proof: Inscribe in this semicircle half of a regular polygon having an even number of sides. Draw the apothems and denote them by a. Draw the projections of the sides of the polygon on the diameter.

Now, if the figure revolve on AF as an axis,

the surface AB AP. 2πα

(AP+PS + ST + etc.) 2πa (Ax. 2). = AF. 2 πα (Ax. 6).

UFX

Now if the number of sides of the polygon is indefinitely increased, the entire surface generated by the polygon will approach s (?) (790), and a will approach R as a limit (437). Also AF 2 Tа will approach AF. 2 TR.

..S = AF· 2πR (?) (242). But AF = 2R (?).
..S= 4 πR2 (Ax. 6).

Q.E.D.

794. THEOREM. The area of a spherical degree equals

795. THEOREM. The areas of the surfaces of two spheres are to each other as the squares of their radii and as the squares of their diameters. (See 793.)

Ex. 1. What is the area of the surface of a sphere whose radius is 10 in.? What is the area of a spherical degree on this sphere?

796. THEOREM. The area of a zone is equal to the product of its altitude by the circumference of a great circle.

Proof: The area generated by chord BC (Fig. of 793) = PS · 2 πа (?) (792).

If the number of sides of the inscribed polygon is indefinitely increased, the length of chord BC will approach arc BC and the surface generated by chord BC will approach the area of a zone.

Also, PS 2 πα will approach PS.2πR (?).