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3. Required the volume of a

27 feet, and the diameter of the

cone

whose altitude 18

base 10 feet.

Ans. 706.86 cu. ft.

4. Required the volume of a cone whose altitude is 10 feet, and the circumference of its base 9 feet.

Ans. 22.56 cu. ft.

5. Find the volume of the frustum of a cone, the altitude being 18, the diameter of the lower base 8, and that of the upper base 4.

6. What is the volume of the frustum altitude being 25, the circumference of the and that of the upper base 10 ?

Ans. 527.7888.

of a

cone, the lower base 20, Ans. 464.216.

7. If a cask, which is composed of two equal conic frustums joined together at their larger bases, have its bung diameter 28 inches, the head diameter 20 inches, and the length 40 inches, how many gallons of wine will it contain, there being 231 cubic inches in a gallon? Ans. 79.0613.

To find the volume of a sphere.

120. From the principle demonstrated in Book VIII., Prop. XIV., we may write the following

RULE.

Cube the diameter of the sphere, and multiply the result by, that is, by 0.5236; the product will be the volume required.

12?

EXAMPLES.

1. What is the volume of a sphere, whose diameter in Ans. 904.7808.

2. What is the volume of the earth, if the mean diameter be taken equal to 7918.7 miles.

Ans. 259992792083 cu. miles.

To find the volume of a wedge.

121. A WEDGE is a volume bounded by a rectangle ABCD, called the back, two trapezoids ABHG, DCHG, called faces, and two triangies ADG, CBH, called ends. The line GH, in which the faces meet, is called the edge. The two faces are equally inclined to the back, and so also are the two ends.

D

A

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There are three cases: 1st, When the length of the edge is equal to the length of the back; 2d, When it is less; and 3d, When it is greater.

In the first case, the wedge is a right prism, whose base is the triangle ADG, and altitude GH or AB: hence, its volume is equal to ADG multiplied by AB.

In the second case, through H, the middle point of the edge, pass a plane HCB perpendicular to the back and intersecting it in the line BC parallel to AD. This plane will divide the wedge into two parts, one of which is represented by the figure.

D

N

P

A

M

G

H

B

Through G, draw the plane GNM parallel to HCB, and it will divide the part of the wedge represented by the figure into the right triangular prism GNM - B, and the quadrangular pyr amid ADNM – G. Draw GP perpendicular to NM: it will also be perpendicular to the back of the wedge (B. VI., P. XVII.), and hence, will be equal to the altitude of the wedge.

Denote AB by L, the breadth AD by b, the edge GH by , the altitude by h, and the volume by V; then,

=

-

=

AM L-1, MB = GH=1, and area NGM bh: then
Prism = bhl; Pyramid = b(L — 1)}h =}bh(L—1), and
V = {bhl+}bh(L − 1) = 1⁄2bhl + {bhL — }bhl = 1bh(1+2L).

We can find a similar expression for the remaining part of the wedge, and by adding, the factor within the parenthesis becomes the entire length of the edge plus twice the length of the back.

In the third case, is greater than L, and denotes the altitude of the prism; the volume of each part is equal to the difference of the prism and pyramid, and is of the same form as before. Hence, the following

RULE. Add twice the length of the back to the length of he edge; multiply the sum by the breadth of the back, and that result by one-sixth of the altitude; the final product will be the volume required.

EXAMPLES.

1. If the back of a wedge is 40 by 20 feet, the edge 35 feet, and the altitude 10 feet, what is the volume?

Ans. 3833.33 cu.ft.

2. What is the volume of a wedge, whose back is 18 feet by 9, edge 20 feet, and altitude 6 feet?

To find the volume of a prismoid.

122. A PRISMOID is a frustum of a wedge.

Let L and B denote the length and breadth of the lower base, 1 and b the length and breadth of the upper base, M and m the length and breadth of the section equidistant from the bases, and h the altitude of the prismoid.

Through the edges L and ', let a plane be passed, and it will

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m

M

B

L

504 cu. ft.

divide the prismoid into two wedges, having for bases, the bases of the prismoid, and for edges the lines I and l'.

The volume of the prismoid, denoted by V, will be equal to the sum of the volumes of the two wedges; hence,

or,

V = †Bh(l + 2L) + дbh(L + 21);

V = jh(2BL + 2b1 + Bl + bL) ;

which may be written under the form,

V = h[(BL+b+ Bl+bL) + BL+ bl].

(A.)

Because the auxiliary section is midway between the bases,

we have,

2M = L + 1, and

2m = B + b ;

hence,

4Mm = (L+ 1) (B + b) = BL + bl + BL + bl.

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But BL is the area of the lower base, or lower section, bl is the area of the upper base, or upper section, and Mm is the area of the middle section; hence, the following

RULE.

To find the volume of a prismoid, find the sum of the areas of the extreme sections and four times the middle section; multiply the result by one-sixth of the distance between the extreme sections; the result will be the volume required.

This rule is used in computing volumes of earth-work in railroad cutting and embankment, and is of very extensive application. It may be shown that the same rule holds for every one of the volumes heretofore discussed in this work. Thus, in a pyramid, we may regard the base as one extreme section, and the vertex (whose area is 0), as the other extreme ; their sum is equal to the area of the base. The area of a section midway between between them is equal to one-fourth of the base: hence, four times the middle section is equal to the base. Multiplying the sum of these by onesixth of the altitude, gives the same result as that already found. The application of the rule to the case of cylinders, frustums of cones, spheres, &c., is left as an exercise for the student.

EXAMPLES.

One of the bases of a rectangular prismoid is 25 feet by 20, the other 15 feet by 10, and the altitude 12 feet; required the volume.

2. What is the volume of a stick whose ends are 30 inches by 27, and 24 length being 24 feet?

Ans. 3700 cu. ft.

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MENSURATION OF REGULAR

POLYEDRONS.

123. A REGULAR POLYEDRON is a polyedron bounded bv equal regular polygons.

The polyedral angles of any regular polyedron are all equal.

124. There are five regular polyedrons (Book VII., Page 208).

To find the diedral angle between the faces of a regular

polyedron.

125. Let the vertex of any polyedral angle be taken as the centre of a sphere whose radius is 1: then will this sphere, by its intersections with the faces of the polyedral angle, determine a regular spherical polygon whose sides will be equal to the plane angles that bound the polyedral angle, and whose angles are equal to the diedral angles between the faces.

It only remains to deduce formula for finding one angle of a regular spherical polygon, when the sides are given.

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