855. Corollary. 1. If there are lunes of angles A, B, etc., we have Area (lune 4+ lune B+ etc.) area lune (A+B+ etc.). A+ =

=

856. Cor. 2. The area of a hemisphere is that of a lune of 180°.

THEOREM X.

85%. The area of a spherical triangle is proportional to its spherical excess.

Proof. Let ABC be the triangle. Continue any one side, as BC, so as to form the great circle

BCEF. Continue the other two sides B

till they meet this circle in E and F. Then

1. Area ABC + area ACE = lune

BAEC lune of angle B.

Area ABC area ABF = lune CAFB F = lune of angle C.

Area ABC area AFE = lune ACA'B = lune of angle A.

(§ 853)

2. Because BCEF is a great circle, the sum of the four areas ABC, ACE, ABF, and AFE is a hemisphere. Therefore, if we add the three equations (1), we have

2 area ABC+ hemisphere lune angle (A+B+C). (§855) 3. Because the hemisphere is the same as a lune of 180° (§856), we may write the last equation.

2 area ABC+ lune 180° = lune angle (A + B + C); whence, by transposition,

and

2 area ABC = lune (A + B + C − 180°)

--

Area ABClune (A+B+C' — 180°). 4. A + B + C 180° is, by definition, the spherical excess of the triangle ABC. Because the area of the lune is proportional to its angle, the area ABC is proportional to the same angle or to A+B+C-180°. Q.E.D.

Corollary. If the three vertices of a triangle are on one great circle, its three sides will coincide with that circle, and each of its angles will be 180°. Its area will then be a hemisphere, and its spherical excess 3.180° - 180° = 360°.

Doubling these quantities, we have the area of the whole sphere corresponding to a spherical excess of 720°. Hence

858. The area of a spherical triangle is to that of the whole sphere as its spherical excess is to 720°.

Scholium. Every spherical triangle divides the surface of the sphere into two parts, of which one may be considered within the triangle, and the other without it. We may consider either of these parts as the area of the triangle, if, in .applying the preceding proposition, we measure the angles through that part of the spherical surface whose areas we are considering. If the inner angles are A, B, and C, the angles measured on the outer surface will be 360° — A, 360° — B, and 360° — C (cf. §§ 25-27). Subtracting 180° from the sum of these three angles gives 900° − (A + B + C) as the outer spherical excess. If the inner area becomes indefinitely small, A+B+C will approach to 180°, and the outer spherical excess will approach to 900° - 180° 720°, which is therefore the spherical excess for the outer angles of the triangle whose outer area is that of the whole sphere:

=

859. Def. A zone is that portion of the surface of a sphere contained between two parallel circles of the sphere.

860. Def. The altitude of a zone is the perpendicular distance between the planes of its bounding circles.

THEOREM XI.

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

Proof. Let HKRSLI be a zone of a sphere whose centre is at O, and let the plane of the paper be a section of the sphere through the centre, perpendicular to the base HI of the zone.

The zone is then generated by the motion of the arc HKR around the axis OT, perpendicular to its base.

In the arc HKR inscribe the chords HK, KR. Let M be the middle point of HK. Join OM, and from M drop the perpendicular MD upon OT. Then

1. By the revolution around the axis OT, the chord HK will describe the lateral surface of the frustrum of a cone of

also,

Area of frustrum

2. Because OMK is a right angle,

Angle OMD comp. KMD = comp. KHI;

Angle Gangle D = right angle.

Therefore the triangles OMD and HKG are similar, and HK: KG:: MO : MD;

whence

MD.KHMO.KG.

3. Hence, from (1),

Area of frustrum 27. OM. KG.

4. In the same way,

Area of frustrum KRSL = 2πOP × alt. of frustrum. Inscribe in the arc HKR an indefinite number of equal chords and consider the frustra they describe. Then

5. The perpendiculars OM, OP, etc., will approach the radius of the sphere as their limit.

6. The sum of the lateral surfaces of all the frustra will approach the surface of the zone as their limit.

7. Because the area of each frustrum approaches the limit 2π radius of sphere X alt. of frustrum,

the sum of all the areas will approach the limit

2π radius of sphere X sum of altitudes of frustra
2 radius of sphere X alt. of zone,

which limit is the area of the zone.

8. But 27 radius of sphere = circumf. of great circle. Hence:

-B

Area of zone = alt. of zone X circ. of great circle. Q.E.D. Corollary 1. Let AB and CD be two parallel tangent planes to a sphere. Since the preceding Atheorem is true of zones of all altitudes, it will remain true how near soever we suppose the bases of the zones to the tangent planes. If, then, we suppose the bases of the zones to approach the tangent planes as their limit, the alti- C

D

tude of the zone will approach to the diameter of the sphere, and its surface to the surface of the sphere. Hence:

862. The entire surface of a sphere is equal to the pro

duct of its diameter into its circumference.

Cor. 2. If we put r for the radius of the sphere, we have
Diameter = 2r,

Now we have found the area of a circle of radius r to be πr (§480). Hence:

863. The area of the surface of a sphere is equal to the area of four great circles.

The area of a hemisphere is equal to twice that of a great circle.

CHAPTER II.

VOLUMES OF SOLIDS.

864. Def. The volume of a solid is the measure of its magnitude.

865. Def. The base of a solid is that one of its faces which we select for distinction.

866. Def. The altitude of a solid is the perpen

dicular dropped from its highest point upon the plane of the base.

867. A right parallelopiped is a parallelopiped of which one pair of opposite faces are perpendicular to the other four faces.

REMARK. A right parallelopiped differs from a rectangular one (§ 685) in that two of the faces of the former may make any angle with each other, whereas the rectangular parallelopiped has all its faces perpendicular to each other.

868. Def. The unit of volume is the volume of a cube of which each edge is the unit of length.

Volumes of Polyhedrons.

THEOREM XII.

869. Right prisms having equal altitudes and identically equal bases are identically equal.

Hypothesis. ABCDE-FGHIJ and A'B'C'D'E'-F'G'H' I'J', two right prisms in which

1. Because AF and A'F' are each perpendicular to the

base, they will coincide (§ 585).

2. Because AF = A'F',

Point F point F".

3. In the same way, every vertex of the one will coincide

with a corresponding vertex of the other.