three times, and all the rest of the surface of the hemisphere above the plane of reference once; also, that the area of this hemispherical surface is four times that of the trirectangular triangle. Then, by addition of the equations:

area 4 trirect. tri. + 2 area AEI=2(a+e+i) trir. tri. Transposing the first term, and dividing by 2,

area AEI=(a+e+i−2) trir. tri.

But (a+e+i-2) is the spherical excess of the given triangle, taking a right angle as a unit; that is, it is the ratio of the spherical excess of the given triangle to one right angle.

796. Corollary.—If the square of the radius be taken as the unit of area, then the area of any spherical triangle may be expressed (792),

(a+e+i-2)πR2.

AREAS OF SPHERICAL POLYGONS.

797. Theorem.—The area of any spherical polygon is equal to the area of the trirectangular triangle multiplied by the ratio of the spherical excess of the polygon to one right angle.

For the spherical excess of the polygon is evidently the sum of the spherical excess of the triangles which compose it; and its area is the sum of their areas.

EXERCISES.

798.-1. What is the area of the earth's surface, supposing it to be in the shape of a sphere, with a diameter of 7912 miles?

2. Upon the same hypothesis, what portion of the whole surface is between the equator and the parallel of 30° north latitude? 3. Upon the same hypothesis, what portion of the whole surface is between two meridians which are ten degrees apart?

4. What is the area of a triangle described on a globe of 13 inches diameter, the angles being 100°, 45°, and 53°?

VOLUME OF THE SPHERE.

799. Theorem.-The volume of any polyedron in which a sphere can be inscribed, is equal to one-third of the product of the entire surface of the polyedron by the radius of the inscribed sphere.

For, if a plane pass through each edge of the polyedron, and extend to the center of the sphere, these planes will divide the polyedron into as many pyramids as the figure has faces. The faces of the polyedron are the bases of the pyramids.

The altitude of each is the radius of the sphere, for the radius which extends to the point of tangency is perpendicular to the tangent plane (742). But the volume of each pyramid is one-third of its base by its altitude. Therefore, the volume of the whole polyedron is one-third the sum of the bases by the common altitude, or radius.

800. Theorem.-The volume of a sphere is equal to one-third of the product of the surface by the radius.

For, the surface of a sphere may be approached as nearly as we choose, by increasing the number of faces of the circumscribing polyedron, until it is evident that the sphere is the limit of the polyedrons in which it is inscribed. Then, this theorem becomes merely a corollary of the preceding.

801. Corollary. The volume of a spherical pyramid,

or of a spherical wedge, is equal to one-third of the product of its spherical surface by the radius.

802. A spherical SECTOR is that portion of a sphere which is described by the revolution of a circular sector about a diameter of the circle.

It

may have two or three curved surfaces.

Thus, if AB is the axis, and the generating sector is AEC, the sector has one spherical and one conical surface; but if, with the same axis, the gener

G

F

E

ating sector is FCG, then the sector has one spherical and two conical surfaces.

803. Corollary.—The volume of a spherical sector is equal to one-third of the product of its spherical surface by the radius.

804. The volume of a spherical segment of one base is found by subtracting the volume of a cone from that of a sector. For the sector

ABCD is composed of the segment ABC and the cone ACD.

E

B

D

The volume of a spherical segment of two bases is the difference of the volumes of two segments each of one base. Thus the segment AEFC is equal to the segment ABC less EBF.

805. All spheres are similar, since they are generated by circles which are similar figures. Hence, we might at once infer that their surfaces, as well as their volumes, have the same ratios as in other similar solids. These principles may be demonstrated as follows:

806. Theorem.

The areas of the surfaces of two spheres are to each other as the squares of their diameters ; and their volumes are as the cubes of their diameters, or other homologous lines.

For the superficial area of any sphere is equal to times the diameter multiplied by the diameter (786); that is TD2. But T is a certain or constant factor. Therefore, the areas vary as the squares of the diame

ters.

The volume is equal to the product of the surface by one-sixth of the diameter (800); that is, πD2 by D, or D3. But is a constant numeral. Therefore, the volumes vary as the cubes of the diameters.

USEFUL FORMULAS.

807. Represent the radius of a circle or a sphere, or that of the base of a cone or cylinder, by R; represent the diameter by D, the altitude by A, and the slant hight by H.

EXERCISES.

SOS.-1. What is the locus of those points in space which are at the same distance from a given point?

2. What is the locus of those points in space which are at the same distance from a given straight line?

3. What is the locus of those points in space, such that the distance of each from a given straight line, has a constant ratio to its distance from a given point of that line?

EXERCISES FOR GENERAL REVIEW.

809.-1. Take some principle of general application, and state all its consequences which are found in the chapter under review; arranging as the first class those which are immediately deduced from the given principle; then, those which are derived from these, and so on.

2. Reversing the above operation, take some theorem in the latter part of a chapter, state all the principles upon which its proof immediately depends; then, all upon which these depend; and so on, back to the elements of the science.

3. Given the proportion, a: b::c: d,

4. Form other proportions by combining the same terms. 5. What is the greatest number of points in which seven straight lines can cut each other, three of them being parallel; and what is the least number, all the lines being in one plane?

6. If two opposite sides of a parallelogram be bisected, straight lines from the points of bisection to the opposite vertices will trisect the diagonal.

7. In any triangle ABC, if BE and CF be perpendiculars to any line through A, and if D be the middle of BC, then DE is equal to DF.

8. If, from the vertex of the right angle of a triangle, there extend two lines, one bisecting the base, and the other perpen.