6. In a sphere whose diameter is 20 feet, what is the solidity of a segment, the bases of which are on the same side of the center, the first at the distance of 3 feet from it, and the second of 5 feet; and what is the solidity of a second segment of the same sphere, whose bases are also on the same side of the center, and at distances from it, the first of 5 and the second of 7 feet?

Ans.

{Solidity of first segment, 525.7 cu. ft.

"second

7. If the diameter of the single base of a spherical segment be 16 inches, and the altitude of the segment 4 inches, what is its solidity? *

Ans. 435.6352 cubic inches. 8. The diameter of one base of a spherical segment is 18 inches, and that of the other base 14 inches, these bases being on opposite sides of the center of the sphere, and the distance between them 9 inches; what is the volume of the segment, and the radius of the sphere? Vol. seg., 2219.5 cubic inches. Rad. of sphere, 9.4027 inches. 9. The radius of a sphere is 20, the distance from the center to the greater base of a segment is 10, and the distance from the same point to the lesser base is 16; what is the volume of the segment, the bases being on the same side of the center? Ans. 4297.7088.

Ans.

{

10. If the diameter of one base of a spherical segment be 20 miles, and the diameter of the other base 12 miles, and the altitude of the segment 2 miles, what is its solidity, and what is the diameter of the sphere ?

*First find the radius of the sphere.

NOTE.-The KEY to this work contains full solutions to all the problems in the Geometry and Trigonometry, and the necessary diagrams for illustration.

BOOK VIII.

PRACTICAL GEOMETRY.

APPLICATION OF ALGEBRA TO GEOMETRY, AND ALSO PROPOSITIONS FOR ORIGINAL INVESTIGATION.

No definite rules can be given for the algebraic solution of geometrical problems. The student must, in a a great measure, depend on his own natural tact, and his power of making a skillful application of the geometrical and analytical knowledge he has thus far obtained.

The known quantities of the problem should be represented by the first letters of the alphabet, and the unknown by the final letters; and the relations between these quantities must be expressed by as many independent equations as there are unknown quantities. To obtain the equations of the problem, we draw a figure, the parts of which represent the known and unknown magnitudes, and very frequently it will be found necessary to draw auxiliary lines, by means of which we can deduce, from the conditions enunciated, others that can be more conveniently expressed by equations. In many cases the principal difficulty consists in finding, from the relations directly given in the statement, those which are ultimately expressed by the equations of the problem. Having found these equations, they are treated by the known rules of algebra, and the values of the required magnitudes determined in terms of those given.

PROBLEM I.

Given, the hypotenuse, and the sum of the other two sides of a right-angled triangle, to determine the triangle.

A.

Let ABC be the ▲. Put CB = y, AB

C

Reducing these two equations, and we have

x = 38±†√2h3 — s2 ;

If h = 5 and 8 = 7, x =

REMARK. In place of putting x to represent one side, and y the other, we might put (x+y) to represent the greater side, and (x-y) the less side; then,

Given, the base and perpendicular of a triangle, to find the

side of its inscribed square.

Let ABC be the A. Put

AB b, the base, CD = p,

=

the perpendicular.

Draw EF parallel to AB,

and suppose it equal to EG,

A

G

D H

=X.

a side of the required square; and put EF: Then, by similar A's, we have

CI: EF:: CD: AB.

That is, the side of the inscribed square is equal to the product of the base and altitude, divided by their sum.

PROBLEM III.

In a triangle, having given the sides about the vertical angle, and the line bisecting that angle and terminating in the base, to find the base.

Let ABC be the ▲, and let a circle be circumscribed about it. Divide the arc AEB into two equal parts at the point E, and draw EC. This line bisects the vertical angle, (Cor., Th. 9, B. III).

Put AD CB = b, CD

=

=

x, DB

Draw BE.

=y, AC:

= α,

E

D

B

c, and DE=w. The two A's, ADC and EBC, are equiangular; from which we have

w + c : b :: a : c; or, cw + c2 = ab; (1)

But, as EC and AB are two chords that intersect each other in a circle, we have

cw= xy,

(Th. 17, B. III).

Therefore, xy + c2 = ab.

(2)

But, as CD bisects the vertical angle, we have

ab xy, (Th. 24, B. II).

Now, as x and y are determined, the base is deter mined.

REMARK.-Observe that equation (2) is Theorem 20, Book III

PROBLEM IV.

To determine a triangle, from the base, the line bisecting the vertical angle, and the diameter of the circumscribing circle.

Describe the circle on the given diameter, AB, and divide it into two

H

parts, in the point D, so that AD x

DB shall be equal to the square of

one half the given base, (Th. 17, B. III). Through D draw EDG, at right

B

angles to AB, and EG will be the given base of the

triangle.

Put
Then,

AD=n, DB = m, AB = d, DG = b.

and these two equations will determine n and m; therefore, we shall consider n and m as known.

Now, suppose EHG to be the required A; and draw HIB and HA. The two A's, ABH, DBI, are equiangular; and, therefore, we have

AB : HB :: IB : DB.

But HI is a given line, that we will represent by c; and if we put IB · =w, we shall have HB = c + w; then the above proportion becomes,

d: c + w :: w: m.

Now, w can be determined by a quadratic equation; and, therefore, IB is a known line.

In the right-angled ▲ DBI, the hypotenuse IB, and the base DB, are known; therefore, DI is known, (Th. 39, B. I); and if DI is known, EI and IG are known. Lastly, let EH = x, HGy, and put EI=p, and IG = q.

Then, by Theorem 20, Book III, pq + c2 = xy (1) x: y :: p q (Th. 24, B. II).

But,