Area of 15 inches=176.7146.

Then,

176.7146 x 32.5

3

=1914.4082 cubic inches.

Ex. 2. The diameter of the base of a cone is 20, and its height 24 inches; what are its contents in cubic inches? Ans. 2513.28 inches. Ex. 3. What are the contents of a cone when the diameter of its base is 1.5 feet, and its height 15 feet? Ans. 8.8358 feet. Ex. 4. The diameter of the base of a cone is 12.732 feet, and the height of it is 50 feet; what is its volume?

Ans. 2121.9386 feet.

Centre of Gravity. It is at a distance from the base of the line joining the vertex and centre of gravity of the base.

To ascertain the Contents of the Frustrum of a Cone, Fig. 89.

RULE.-Add together the squares of the diameters of the greater and less ends and the product of the two diameters; multiply their sum by .7854, and this product by the height; then divide this last product by three, and the quotient will give the contents required.

Or, add together the squares of the circumferences of the greater and less ends and the product of the two circumferences; multiply their sum by .07958, and this product by the height; then, divide this last product by three, and the quotient will give the contents required.

EXAMPLE.-What are the contents of the frustrum of a cone, the diameters of the greater and less ends, b d, a c, being respectively 5 and 3 feet, and the perpendicular height, e o, 9 feet?

52+32+5x3=49=the sum of the squares of the diameters and the product of the diameters.

49.7854-38.4846=the above sum by .7854.

38.4846 × 9

3

=115.4538=the last product the height and divided by

three, which is the result required.

Ex. 2. What are the contents of the frustrum of a cone, the diameters of the ends being respectively 2 and 4 feet, and the height 9 feet? Ans. 65.9736 feet.

Ex. 3. The frustrum of a cone is 12 inches in height, and has diameters of 7 and 9 inches; what are the contents of it? Ans. 646.3842 inches.

Centre of Gravity. It is at a distance from the base of (R+r)2+2R2 -; R and r radii of

the smaller end=height >

(R+r)2-Rr

the greater and less ends.

Pyramid.

Definition. A figure, the base of which has three or more sides, and the sides of which are plane triangles.

NOTE.--The volume of a pyramid is equal to one third of that of a prism having equal bases and altitude.

To ascertain the Contents of a Pyramid, Fig. 90.

RULE.-Multiply the area of the base by the perpendicular height, and one third of the product will be the contents required.

Fig. 90.

a

b

EXAMPLE.-What are the contents of a hexagonal pyramid, a b c, Fig. 90, a side, a b, being 40 feet, and its height, c e, 60 feet?

402×2.5981 (tabular multiplier, p. 60)=4156.96=area of base. 4156.96 × 60

3

=83139.2=one third of the area of the base the height=

the contents required.

Ex. 2. The height of a quadrangular pyramid is 67 feet, and the width of its base is 16.5 feet; what are its contents in cubic feet? Ans. 6080.25. Ex. 3. What are the contents of a pentagonal pyramid, its height being 12 feet, and each of its sides 2 feet? Ans. 27.528 feet.

Centre of Gravity. It is at a distance from the base of the line joining the vertex and centre of gravity of the base.

To ascertain the Contents of the Frustrum of a Pyramid, Fig. 91.

RULE. Add together the squares of the sides of the greater and less ends and the product of these two sides; multiply the sum by the tabular multiplier for areas in Table, p. 60, and this product by the height; then, divide the last product by three, and the quotient will give the contents required.

Or, (s2+2+8x8')x tab. mult. xS, 8 and s' representing

the lengths of the sides.

h

3

H

s

Or, a+a'+√a×a×

the ends.

h

=S, a and a' representing areas of

NOTE. When the areas of the ends are known, or can be obtained without reference to a tabular multiplier, use the following rule.

EXAMPLE.-What are the contents of the frustrum of a hexagonal pyramid, Fig. 91, the lengths of the sides of the greater and less ends, a b, c d, being respectively 3.75 and 2.5 feet, and its perpendicular height, e o, 7.5 feet?

3.752+2.52=20.3125=sum of the squares of sides of greater and less

ends.

20.3125+3.75×2.5=29.6875=above sum added to the product of the

two sides.

29.6875×2.5981×7.5=578.48=the last sum× tab. mult., and again by the height, which,÷3=192.82, the result required.

Ex. 2. The frustrum of a hexagonal pyramid has sides of 4 and 3 feet, and a height of 9 feet; what are its contents? Ans. 294.3891 feet.

When the Ends of a Pyramid are not those of a Regular Polygon, or when the Areas of the Ends are given.

RULE.

Add together the areas of the two ends and the square root of their product; multiply the sum by the height, and one third of the product will be the contents.

EXAMPLE. What are the contents of an irregular-sided frustrum of a pyramid, the areas of the two ends being 22 and 88 inches, and the length 20 inches?

22+88=110=sum of areas of ends.

22×88=1936, and ✓ 1936=44=square root of product of areas. 110+44 × 20

3

=

1026.66 one third of sum of above sum and product X

the height=the result required.

Ex. 2. The areas of the ends of an irregular-sided frustrum of a pyramid are 81 and 100 inches, and the length 25 inches; what are its contents? Ans. 2258.33 inches.

Centre of Gravity. It is at a distance from the centre of

A Spherical Pyramid is that part of a sphere included within three or more adjoining plane surfaces meeting at the centre of the sphere. The spherical polygon defined by these plane surfaces of the pyramid is called the base, and the lateral faces are sectors of circles.

To ascertain the Elements of Spherical Pyramids, see Docharty and Hackley's Geométry.

Cylindrical Ungulas.

Definition. Cylindrical ungulas are frustrums of cylinders. Conical ungulas are frustrums of cones."

*

To ascertain the Contents of a Cylindrical Ungula, Fig. 92. 1. When the section is parallel to the axis of the cylinder. RULE-Multiply the area of the base by the height of the cylinder, and the product will be the contents required. Or, axh=S.

*For Mensuration of Conical Ungulas, see Conic Sections, p. 253.