QUESTIONS

1. What is a pyramidal surface? A conical surface? Is the directrix necessarily closed?

2. What are the formulas or rules for finding the areas of pyramids, cones, and frustums of pyramids and cones?

3. What are the formulas or rules for finding the volumes of each of these solids?

4. Can you find the area of an oblique circular cone? Of an oblique pyramid? Can you find the volume of each of these?

5. If the altitude of a cone is divided into 10 equal parts by planes passed parallel to the base, how does the area of each section of the cone formed by these planes compare with the base of the cone?

6. Trace the steps in finding the volume of a triangular pyramid. Of any pyramid.

Of a cone.

7. What is the relation between two parallel sections of a pyramid or cone?

8. State theorems concerning the sections of a cone. Of a pyramid. 9. What effect does it have upon the volume of a pyramid or cone if the area of the base is doubled? If the altitude is doubled? If both base and altitude are doubled?

10. What effect does it have upon the lateral area of a right circular cone if the circumference of the base is doubled? If the area of the base is doubled?

11.

Point out forms of this chapter that occur in nature. That occur in the arts.

GENERAL EXERCISES

1. Given the following formulas for the right circular cone: (1) V= }πr2h, (2) S=πrs, (3) S=πr√h2+r2, (4) T=A+πr√h2+r2, (5) T= Tr2+πr Vh2+r2; solve each for each letter appearing.

2. In a right circular cone, A = 50π, and V=200π. Find r, h, and s. 3. If R and r are the radii of the two bases of a frustum of a cone,

4. Find the volume and the total area of a right circular cone whose radius of base is 6 in. and altitude is 5.3 in. Ans. 199.8 cu. in.; 263.9 sq. in.

5. Find the weight of a tapered brick smoke stack 175 ft. high in the form of a frustum of a right circular cone enclosing a cylinder. The inner diameter is 10 ft., the wall is 4 ft. thick at the base and 1 ft. 6 in. at the top. Ans. 1095.5 tons.

One cubic foot of brick weighs 112 pounds.

6. Find the lateral edge, lateral area, and volume of a regular pyramid each side of whose triangular base is 5 ft., and whose altitude is 9 ft. Ans. 9.45 ft.; 68.36 sq. ft.; 32.5 cu. ft.

7. A bin in a warehouse is 12 ft. square. A hopper is constructed on the base, which has a slope of 1:1. The distance from the vertex of the hopper to the top of the bin is 18 ft. Find the capacity of the bin if 1

bushel equals cu. ft.

8. The largest possible cylinder of diameter 6 in. is cut from a right circular cone having a diameter and altitude of 10 in. and 26 in. respectively. Find the volume of the cylinder.

9. Find the area and volume of the solid generated by revolving an isosceles triangle of base 10 in. and equal sides 16 in., about its base. Find the area and the volume of the solid when this triangle is revolved about one of its equal sides.

10. In the frustum of a pyramid whose base is 50 sq. ft. and altitude 6 ft., the basal edge is to the corresponding top edge as 5 to 3. Find the volume of the frustum. Ans. 196 cu. ft.

11. The base of a pyramid is a rectangle 10 in. by 8 in. Find the volume if each lateral edge makes with the base an angle of 45°.

12. The total surface of a regular quadrangular pyramid is T, and its eight edges are equal. Find the length of an edge.

13. Find the volume of a cone of revolution whose slant height is equal to the diameter of its base, and whose total area is T.

14. Show how to cut a pattern for the frustum of a right circular cone, if the upper and lower bases have diameters of 4 in. and 6 in., respectively, and the altitude is 8 in.

15. A right circular cone whose altitude is 8 in. and radius 6 in. rolls on a floor without slipping, making one complete revolution. What is the shape of the surface covered? Find its area.

16. A pyramid whose altitude is 12 in. weighs 30 pounds. At what distance from the vertex must it be cut by a plane parallel to its base so that the frustum cut off shall weigh 20 pounds?

17. A cube is cut by a plane passed through the other extremities of the three edges meeting at a vertex. What part of the volume of the cube is thus cut off?

CHAPTER IX

PRISMATOIDS AND POLYEDRONS

PRISMATOIDS

749. A polyedron all of whose vertices lie in two parallel planes is called a prismatoid. In either plane, the vertices may lie in any polygon, as Fig. (1); or, in one plane, the vertices may lie in a line, as in Fig. (2); or there may be a point only, as in a pyramid. The faces whose vertices are in the

parallel planes are evidently triangles, or quadrilaterals of types that have at least two parallel sides. Why?

750. The altitude of a prismatoid is the distance between the two parallel planes..

EXERCISES

1. Show that a prism is a prismatoid whose bases are congruent polygons.

2. Show that a frustum of a pyramid is a prismatoid whose bases are similar polygons.

3. Draw or describe a prismatoid that has a face that is a trapezoid. A parallelogram. A rectangle.

4. Compute the area of the midsection of a pyramid whose base is a regular hexagon having a side of 6 in.

5. Compute the areas of the midsection of the prismatoid in Fig. (2) if ABCD is a rectangle having AB=18 in. and BC=11 in., and EF =12 in.

751. Theorem. If B1 and B2 are the areas of the two bases, M the area of the midsection, and h the altitude, then the volume V of a prismatoid is given by the formula V=h(B1+B2+4M). Given a prismatoid of volume V, alti

tude h, bases B1 and B2, and midsection M.

1

To prove V=h(B1+B2+4M).

Proof. If any lateral face is not a triangle divide the face into triangles by a diagonal.

E

N

B

D

Take any point O in the midsection and connect it with the vertices of the prismatoid and the midsection. These lines form triangles with the edges of the prismatoid, which separate the prismatoid into pyramids that have the bases and faces of the prismatoid as bases.

(1) Pyramid O-ABCD=}/hB1.

(2) Pyramid O-NPQ=&hB2.

§ 737

Each of the pyramids like O-CDP, which may be called a lateral pyramid, can have its volume expressed in terms of the part of the midsection common to it. To prove this consider O-CDP apart from the prismatoid, and draw HD.

Altitude of each of pyramids P-OHI and D-OHI is th.
Pyramid P-OHI-h times area OHI.

N

Pyramid D-OHI=h times area OHI.

Pyramid O-CDH=2 times pyramid O-HDI.

Then pyramid O-CDH = 2h times area OHI.

Hence pyramid O-CDP=4h times area OHI.

$ 737

Why?

Why?

Why?

Similarly each lateral pyramid is equal to th times the area

of its part of the midsection.

(3) Therefore the sum of the lateral pyramids =hM = kh•4M.

Adding (1), (2), and (3),

=

Why?

V=†h(B1+B2+4M).

752. Other Applications of the Prismatoid Formula. The formula for the volume of a prismatoid can be applied also to various other solids such as cylinders, cones, spheres, combinations of these, and various other forms that cannot well be described here.

The prismatoid formula, because of its wide application, is used often by engineers and others in determining the volumes of embankments, abutments, and other irregular solids of various kinds.

EXERCISES

1. Derive the following formulas from the formula for the volume of a prismatoid:

(1) Prism and cylinder, V = Bh.

(2) Pyramid and cone, V=Bh.

(3) Frustum of pyramid or cone, V=}h(B1+B2+√B1B2). Derivation for frustum.

Given V=h(B1+B2+4M).

To derive V}h(B1+B2+√B1B2).

It is then necessary to express M in terms of B1 and B2.

Or 4M =B1+B2+2√ B1B2.

.. V=¿h(2B1+2B2+2√B1B2) =}{}h(B1+B2+√B,B2).