eral equations together, we find that the entire surface described by the revolution of the regular polygon about its diameter, is equal to the product of the circumference whose radius is CI, by the diameter AB.

This being true as to the surface described by the perimeter of any regular polygon, it is therefore true of the surface described by the circumference of a circle. But this surface is that of a sphere, and the radius CI then becomes the radius of the sphere. Therefore, the area of the surface of a sphere is equal to the product of the diameter by the circumference of a great circle.

787. Corollary.-The area of the surface of a sphere is four times the area of a great circle. For the area of a circle is equal to the product of its circumference by one-fourth of the diameter.

788. Corollary.-The area of the surface of a sphere is equal to the area of the curved surface of a circumscribing cylinder; that is, a cylinder whose bases are tangent to the surface of the sphere.

AREAS OF ZONES.

789. A ZONE is a part of the surface of a sphere included between two parallel planes. That portion of the sphere itself, so inclosed, is called a segment. The circular sections are the bases of the segment, and the distance between the parallel planes is the altitude of the zone or segment.

One of the parallel planes may be a tangent, in which case the segment has one base.

790. Theorem.-The area of a zone is equal to the product of its altitude by the circumference of a great

circle.

This is a corollary of the last demonstration (786). The area of the zone described by the arc AD, is equal to the product of AH by the circumference whose radius is the radius of the sphere.

AREAS OF LUNES.

791. Theorem.-The area of a lune is to the area of the whole spherical surface as the angle of the lune is to four right angles.

It has already been shown that the angle of the lune is measured by the arc of a great circle whose pole is at the vertex. Thus, if AB is the axis of the arc DE, then DE measures the angle DAE, which is equal to the angle DCE. But evidently the lune varies exactly with the angle DCE or DAE. This may be rigorously demonstrated in the same manner as the principle

D

C

E

that angles at the center have the same ratio as their intercepted arcs.

Therefore, the area of the lune has the same ratio to the whole surface as its angle has to the whole of four right angles.

TRIRECTANGULAR TRIANGLE.

792. If the planes of two great circles are perpendicular to each other, they divide the surface into four equal lunes. If a third circle be perpendicular to these

Geom.-23

two, each of the four lunes is divided into two equal triangles, which have their angles all right angles and their sides all quadrants. Hence, this is sometimes called the quadrantal triangle.

This triangle is the eighth part of the whole surface, as just shown. Its area, therefore, is one-half that of a

great circle (787).

the circle is

Since the area of

times the square of

the radius, the area of a trirectangu

lar triangle may be expressed by R2.

The area of the trirectangular triangle is frequently assumed as the unit of spherical areas.

AREAS OF SPHERICAL TRIANGLES.

793. Theorem.-Two symmetrical spherical triangles are equivalent.

Let the angle A be equal to B, E to C, and I to D. Then it is known that

the other parts of the triangle are respectively equal, but not superposable; and it is to be proved that the triangles are equivalent.

B

Let a plane pass through the three points A, E, and I; also, one through B, C, and D. The sections thus made are small circles, which are equal; since the distances between the given points are equal chords, and circles described about equal triangles must be equal. Let O be that pole of the first circle which is on the same side of the sphere as the triangle, and F the corre

sponding pole of the second small circle. Let O be joined by arcs of great circles OA, OE, and OI, to the several vertices of the first triangle; and, in the same way, join FB, FC, and FD.

Now, the triangles AOI and BFD are isosceles, and mutually equilateral; for AO, IO, BF, and DF are equal arcs (753). Hence, these triangles are equal (772). For a similar reason, the triangles IOE and CFD are equal; also, the triangles AOE and BFC. Therefore, the triangles AEI and BCD, being composed of equal parts, are equivalent.

The pole of the small circle may be outside of the given triangle, in which case the demonstration would be by subtracting one of the isosceles triangles from the sum of the other two.

794. It has been shown that the sum of the angles of a spherical triangle is greater than the sum of the angles of a plane triangle (771). Since any spherical polygon can be divided into triangles in the same manner as a plane polygon, it follows that the sum of the angles of any spherical polygon is greater than the sum of the angles of a plane polygon of the same number of sides.

The difference between the sum of the angles of a spherical triangle, or other polygon, and the sum of the angles of a plane polygon of the same number of sides, is called the spherical excess.

795. Theorem.-The area of a spherical triangle is equal to the area of a trirectangular triangle, multiplied by the ratio of the spherical excess of the given triangle to one right angle.

That is, the area of the given triangle is to that of the trirectangular triangle, as the spherical excess of the given triangle is to one right angle.

Let AEI be any spherical triangle, and let DHBCGF be any great circle, on one side of which is the given triangle. Then, considering this circle as the plane of reference of the figure, produce the sides of the triangle AEI around the sphere.

Now, let the several angles of the given triangle be represented by a, e, and i; that is, taking a right angle for the unit, the angle EAI is equal to a right angles, etc. Then, the area

of the lune AEBOCI is to the whole surface as a is to 4 (791). But if the trirectangular triangle, which is one-eighth of the spherical surface, be taken as the unit of area, then the area of this lune is 2a. But the triangle BOC, which

H

E

is a part of this lune, is equivalent to its opposite and symmetrical triangle DAF. Substituting this latter, the area of the two triangles ABC and DAF is 2a times the unit of area.

In the same way, show that the area of the two triangles IDH and IGC is 2i, and that the area of the two triangles EFG and EHB is 2e times the unit of These equations may be written thus:

area.

area (ABC + ADF)=2a times the trirectangular triangle;

area (IDH+IGC)=2i times the trirectangular tri

angle;

area (EFG+EHB)=2e times the trirectangular tri

angle.

In adding these equations together, take notice that the triangles mentioned include the given triangle AEI