Prism And Cylinder

3. A lead pencil whose ends are regular hexagons was cut from a cylindrical piece of wood, with the least waste. If the original piece was 8 in. long and ½ in. in diameter, find the volume of the pencil. Solution: To illustrate the problem, it is better to draw the figure as follows

The first thing that we need to do is to consider the base of the pencil which is a regular hexagon and draw three diagonals that passes thru its center as follows

Since all the interior angles of a triangle are congruent, then it is an equilateral or equiangular triangle. All sides of an equilateral triangle are congruent. The base of a triangle is also equal to ¼ in.

Next, we need to consider a triangle in a regular hexagon so that we can solve for its area as follows

The area of a triangle is

5. The lateral surface of a concrete octagonal pier of height 10 ft. is to be resurfaced. If each of the base edges is 1 ft., how many square feet of surfacing will be required? Solution:

Hence, the area of the base of a regular hexagon is

To illustrate the problem, it is better to draw the figure as follows

Therefore, the volume of a pencil which is a regular hexagon prism is

Photo by Math Principles in Everyday life The lateral surface area of a concrete octagonal pier required to be resurfaced is

or

6. One part of a quartz crystal is a hexagonal prism with a right section of 1.29 sq. in., an edge of 2.31 in., and a base of 1.41 sq. in. Find the altitude of the prism.

7. A trench is 180 ft. long and 12 ft. deep, 7 ft. wide at the top and 4 ft. at the bottom. How many cubic yards of earth have been removed? Solution:

Solution: To illustrate the problem, it is better to draw the figure as follows

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life Photo by Math Principles in Everyday Life The volume of a quartz crystal is

The amount of earth removed is in the shape of a prism whose base is a trapezoid. The area of the base which is a trapezoid is

Therefore, the altitude of a quartz crystal is

The amount of earth removed which is the volume of a trapezoid prism is

Therefore, the amount of earth removed in cubic yards is

8. A trough is formed by nailing together, edge to edge, two boards 11 ft. in length, so that the right section is a right triangle. If 15 gals. of water are poured into the trough and if the trough is held level so that a right section of the water is an isosceles right triangle, how deep is the water? (231 cu. in. = 1 gal.)

Photo by Math Principles in Everyday Life Solution: Consider the cross section of a trough which is the base of a triangular prism

Photo by Math Principles in Everyday Life The cross section area of water in a trough is

9. A contractor agrees to build a dam, 180 ft. long and 20 ft. high, 12 ft. wide at the bottom and 8 ft. wide at the top, for $9.75 a cubic yard. Find his profit if his costs were $10,000. The length of a trough in inches is

Solution: To illustrate the problem, it is better to draw the figure as follows

The volume of water in a trough in cubic inches is Photo by Math Principles in Everyday Life The area of the base which is a trapezoid is

Therefore, the height or deep of water in a trough is

The volume of a dam which is a trapezoid prism is

The volume of a dam in cubic yards is

The labor cost of building a dam is

10. (a) Find the volume of water in a swimming pool with vertical ends and sides. The length measured at the water line is 50 ft. and the breadth is 20 ft. The bottom of the swimming pool is a plane slopping gradually downward so that the depth of the water at one end is 4 ft. and at the other end is 8 ft. (b) If the sides, ends, and bottom of the swimming pool are constructed of tile blocks whose glazed surface dimensions are 3 in. by 6 in., and if the ends and sides of the pool extend 2 ft. above the water level, find the number of blocks used if 1/20 of the surface area is covered by sealing material. Solution: (a) To illustrate the problem, it is better to draw the figure as follow

Therefore, the contractor's profit in building a dam is

Photo by Math Principles in Everyday Life Did you notice that the swimming pool is a trapezoid prism whose two opposite faces are right trapezoid? Well, the area of the base which is a right trapezoid is

Therefore, the volume of water in a swimming pool which is a trapezoid prism is Photo by Math Principles in Everyday Life By Pythagorean Theorem,

(b) If the ends and sides of the pool extend 2 ft. above the water level, the dimensions of the pool are as follows

Photo by Math Principles in Everyday Life

Hence, the total surface area of the pool is

We need to get the unknown dimensions of the sides of the pool first so that we can solve for the total surface area of the pool especially the area of the bottom of the pool. Let's analyze a surface of the pool as follows

The total area of the sealing material used is

The total area of the tile blocks used is

The area of a tile block is

Therefore, the number of tile blocks used for the pool is

The number of tile blocks must be in whole numbers. We have to round off to the nearest ones. The answer is

11. A football stand 150 yd. long has 20 tiers of seats; each tier has a rise of 2 ft. and tread of 3 ft. It is constructed of reinforced concrete with a cross section as shown. Find the amount of material used in its construction.

Photo by Math Principles in Everyday Life

Photo by Math Principles in Everyday Life Solution: Consider the given figure above

At the first layer, there is one block whose dimension is 2' x 3'. At the second layer, there are two blocks. At the third layer, there are three blocks, and so on. We need to get the total number of blocks so that we can get the area of the section. Since there's a sequence of number of blocks for each layer by arithmetic progression, then we can get the total number of blocks by using the sum of n terms as follows

Photo by Math Principles in Everyday Life The first thing that we need to do is to isolate the given figure which is the 20 tiers of seat section as follows

The total area of blocks is

The area of a triangle is

12. A railway cut 200 ft. and 30 ft. wide at the bottom is made with sloping sides, which are 80 ft. and 60 ft. in length. The 80 ft. side is inclined 45° and and the other side is inclined 30° to the horizontal. Find the cost of removing the earth at $2 per load, if the trucks have a capacity of 4 cu. yd.

Photo by Math Principles in Everyday Life Hence, the area of the 20 tiers of seat section which is the base of a prism is

Therefore, the amount of material used for the construction of a football stand which is a prism is

Solution: Consider the given figure above

Photo by Math Principles in Everyday Life To understand more the problem, it is better to analyze the front side which is a quadrilateral and the base of the prism, and then label further as follows

Photo by Math Principles in Everyday Life If you draw vertical lines from the two upper vertices and horizontal lines from the two lower vertices of a quadrilateral, then it becomes a trapezoid. Since the sides and angles of a quadrilateral are given, then we can solve the remaining sides of two right triangles.

The area of 30° right triangle is

Photo by Math Principles in Everyday Life The area of a trapezoid is The area of a quadrilateral which is the base of a prism is

The volume of the earth removed from a railway cut which is the volume of a prism is The area of 45° right triangle is

The volume of a prism in cubic yards is

13. The Pennsylvania Railroad found it necessary, owing to land slides upon the roadbed, to reduce the angle of inclination of one bank of a certain railway cut near Pittsburgh, PA, from an original angle of 40° to a new angle of 25°. The bank as it originally stood was 200 ft. long and had a slant length of 60 ft. Find the amount of the earth removed, if the top level of the bank remained unchanged.

Therefore, the cost of removing the earth from a railway cut is

Photo by Math Principles in Everyday Life Solution: Consider the given figure above

Photo by Math Principles in Everyday Life From the two upper vertices of a triangle, draw vertical lines perpendicular to the ground and then label further the figure as follows

Photo by Math Principles in Everyday Life

The length of the base of an obtuse triangle is

Consider a 40° right triangle,

The area of an obtuse triangle which is the base of the prism is

The value of h in a 25° right triangle is equal to the value of h in a 40° right triangle. Consider a 25° right triangle,

Therefore, the amount of earth removed which is the volume of a prism is

14. The trough shown in the figure has triangular ends which lie in parallel planes. The top of the trough is a horizontal rectangle 20 in. by 33 in., and the depth of the trough is 16 in. (a) How many gallons of water will it hold? (One gal. = 231 cu. in.) (b) How many gallons does it contain when the depth of the water is 10 in.? (c) What is the depth of the water when the trough contains 3 gals.? (d) Find the wetted surface when the depth of the water is 9 in.

The area of the base is

Therefore, the volume of a trough when filled with water which is the volume of a prism is

The volume of a trough in gallons is Photo by Math Principles in Everyday Life Solution: (a) Consider the given figure above

(b) Consider the front side of a trough

Photo by Math Principles in Everyday Life

The volume of a trough with water in gallons is Photo by Math Principles in Everyday Life Bu using similar triangles, the area of the base with water is

(c) The volume of a trough with water in cubic inches is

The area of the base of a trough with water is Therefore, the volume of a trough with water is

By using similar triangles (d) Consider the front side of a trough

Photo by Math Principles in Everyday Life Therefore, the depth of water in a trough is

Photo by Math Principles in Everyday Life By using similar triangles, the length of the base is

Let's assume that the base of a trough is an isosceles triangle.

The lateral area of the wetted surface of a trough is

Photo by Math Principles in Everyday Life By using Pythagorean Theorem, the length of the wetted edge is

The area of the wetted base of a trough is

Therefore, the total area of the wetted surface of a trough is

16. A dam 100 ft. long has a cross section which is a trapezoid whose altitude is 16 ft., and whose upper base is 5 ft. If the lower base angles of the cross section are 50° and 65°, find the volume of material the dam contains.

Consider the 65° right triangle,

Solution: To illustrate the problem, it is better to draw the figure as follows

The length of the lower base of a trapezoid which is the cross section of a dam is

Since the length of the lower base is not given, then we have to isolate the base first and then label further so that we can solve for its length. Hence, the area of the base is

Consider the 50° right triangle, Therefore, the volume of a trapezoid prism which is a dam is

1. A vertical stone column 12.5 ft. high has an elliptical base with the longer axis twice the shorter. If the weight of the column is 12,400 lbs. and if the stone weighs 160 lbs. per cu. ft., find the area of the largest and smallest axial section of the column.

The volume of the column which is a cylinder is

Solution: To illustrate the problem, it is better to draw the figure as follows

The length of b which is the semi-minor axis of the ellipse is

The area of the ellipse which is the base of the column is

but

Hence, the above equation becomes

and the area of the smallest axial section of the column which is a rectangle is The length of a which is the semi-major axis of the ellipse is

Therefore, the area of the largest axial section of the column which is a rectangle is

3. During a rain, ¼ in. of water fell. Find how many gallons of water fell on a level 10-acre park. (Take 1 cu. ft. = 7.48 gals. and 1 acre = 43,560 sq. ft.)

The volume of the rainwater which is a cylinder is

Photo by Math Principles in Everyday Life Solution: Consider the given figure above

Photo by Math Principles in Everyday Life Let's assume that the thickness of the rainwater within the level 10-acre park is ¼ in. Since the thickness of the rainwater is uniform, then the given figure is considered as a cylinder. In this case, the area of the base which is the level 10-acre park in square feet is

Therefore, the volume of the rainwater in gallons is

4. An ice-storage plant removed from the center of a pond a mass of ice covering an area of 2 acres. If the ice had a uniform thickness of 2 ft., find the weight in tons of the ice removed. (Ice weighs 56 lbs. per cu. ft.; 1 ton = 2,240 lbs.; 1 acre = 43,560 sq. ft.)

The weight of an ice covering is

Solution: To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life The area of the base which is the top of ice covering from the center of a pond is

The volume of an ice covering which is a cylinder is

Therefore, the weight of an ice covering in metric tons is

5. The average depth of a lake is estimated to be 40 ft. If the surface area is 15 acres, find the volume of water in the lake. (1 acre = 43,560 sq. ft.)

6. An indoor roller skating rink with an area of 1,500 sq. yd. has a concrete flooring 3 in. thick. Find the amount of concrete used in laying the floor. Solution:

Solution: Let's assume that the lake is in a shape of a cylinder. Since the surface area of a lake as well as the estimated depth are given in the problem, then we can get the volume of water in the lake as follows

Therefore, the volume of water in the lake is

The floor for the roller skating rink must be smooth and uniform in thickness. If the thickness of the floor is the same within the roller skating rink, then it is considered a cylinder. Since the surface area of a roller skating rink as well as the thickness are given in the problem, then we can get the amount of concrete which is the volume of a cylinder used in laying the floor as follows

Therefore, the amount of concrete required for laying the floor is

7. The outer protective smokestack of a steamship is streamlined so that it has a uniform oval section parallel to the deck. The area of this oval section is 48 sq. ft. If the length of the stack is 15 ft. and the stack is raked aft so that its axis at its upper end is horizontally 4.2 ft. from the lower end, find the volume enclosed by the stack. (Raked means inclined.) Solution: To illustrate the problem, it is better to draw the figure as follows Therefore, the volume of a smokestack which is a cylinder is

Photo by Math Principles in Everyday Life The area of a base which is an oval section is already given in the problem. By Pythagorean Theorem, the altitude of the smokestack is

8. The crown of a straw hat has a base of 38 sq. in. The depth of the crown is 3 in. (Inside dimensions are given.) If the head occupies two-thirds of the space enclosed by the crown, find the volume remaining for ventilation. Solution:

9. Find the volume of the largest cylinder with circular base that can be inscribed in a cube whose volume is 27 cu. in. Solution: To illustrate the problem, it is better to draw the figure as follows

In this problem, the volume of a cube is given. The length of the edge of a cube is Since the area of the base or crown of a straw hat as well as the depth of the crown are already given, then we can get the volume of the crown of a straw hat as follows

If the head occupies two-thirds of the space or volume of the crown, then the volume remaining for the ventilation is one-third of the space of the crown. Therefore,

If the circular arc of the base of a cylinder is tangent to the bottom edges of a cube, then it follows that

where x is the length of the edge of a cube and r is the radius of a circle. The radius of a circle is

Therefore, the volume of the largest cylinder that can be inscribed in a cube is

10. A cylinder whose base is a circle is circumscribed about a right prism of altitude 12.6 ft. Find the volume of the cylinder if the base of the prism is a square of edge 3 ft.

But

Solution: To illustrate the problem, it is better to draw the figure as follows

Hence, the above equation becomes,

or

Consider the base of a cylinder which is a circle that circumscribed about a square as follows

Photo by Math Principles in Everyday Life The length of the diagonal of a square is equal to the diameter of a circle. By Pythagorean Theorem, the diameter of a circle is

Therefore, the volume of a circular cylinder is

But

Hence, the above equation becomes

or

CIRCULAR CYLINDER 1. A pine log 40 ft. long has a uniform circular cross section of radius 11 in. The ends lie in parallel planes which are not perpendicular to the axis of the log. How much paint will be required to give the log (ends not included) two coats, if 1 gal. of paint covers 800 sq. ft. of surface? If pine weighs 31 lbs. per cu. ft., find the weight of the log.

Therefore, the amount of paint required to paint the surface of a pine logs in two coats is

Solution: To illustrate the problem, it is better to draw the figure as follows The volume of a circular cylinder which is a pine log is

where K is the area of a cross section and e is the length of an element or the slant height. Substitute the values, we have

The surface or lateral area of a circular cylinder which is a pine log is

where pk is the circumference or perimeter of the cross section and e is the length of an element or the slant height. Substitute the values, we have

Therefore, the weight of a pine log is

2. A straight stairway has a balustrade, with dimensions as shown, containing 2 circular balustrade to a step. The foot and hand rails are parallel. Find the total amount of wood used and the total area to be painted.

two balustrades and there are 14 steps in a stairway. Hence, there are 28 balustrades in a stairway. The surface area of a right circular cylinder which is a balustrade is

Therefore, the total area of balustrades to be painted is

Photo by Math Principles in Everyday Life

The volume of a right circular cylinder which is a balustrade is

Solution: Consider the given figure above

Therefore, the amount of wood used for balustrades is

Photo by Math Principles in Everyday Life A balustrade in a stairway is in a shape of right circular cylinder. Each step has

3. A circular concrete conduit, whose inside diameter is 10 ft., is 1 ft. thick. It rises 16 ft. per 1000 horizontal feet. The vertical plane which contains the axis is perpendicular to the two vertical planes which contain the ends of the conduit. If the ends are 3000 ft. apart, find the amount of concrete used in the construction of the conduit.

By Pythagorean Theorem, the length of a concrete conduit is

Solution: To illustrate the problem, it is better to draw the figure as follows The cross section of a concrete conduit is a two concentric circles. The area of the region between the two concentric circles is

The first thing that we need to do is to solve for the altitude or vertical distance of the ends of a conduit which is a right circular cylinder by similar triangles as follows Therefore, the amount of a concrete used for the construction of a conduit which is the volume of a circular cylinder is

4. A smokestack of a ship is 25 ft. long with a rake aft (angle of the stack's inclination from the vertical) so that its top rises 24 ft. above the deck. The cross section of the flue is a circle with a diameter of 3½ ft. Completely encircling the flue is a protective stack. The distance between the two stacks is 6 in. Find the space enclosed between the two stacks and also the outside painting surface of the protective stack. (Neglect the thickness of the metal).

Therefore, the amount of space enclosed between the two stacks of a smokestack which is the volume of a circular cylinder is

and the amount of outside painting the surface of a smokestack is

Solution: The cross section of a smokestack of a ship is a two concentric circles. The area of the region between the two concentric circles is

RIGHT CIRCULAR CYLINDER 1. The diameter of a well is 6 ft., and the water is 7 ft. deep. How many gallons of water are there in the well, reckoning 7.48 gal. to the cubic foot?

2. A paint manufacturer desires a cylindrical steel drum to hold 50 gal. of roof paint. For convenience in handling, it is found necessary to limit the inside diameter to 2½ ft. Find the height of the drum desired. (1 gal. = 231cu. in.)

Solution: Solution: To illustrate the problem, it is better to draw the figure as follows

Since the diameter of a well as well as the depth of water inside the well are given, then we can calculate the volume of water as follows

The first thing that we need to do is to convert the given volume which is the amount of paint inside the cylindrical drum in cubic feet as follows

Therefore, the height of the drum desired to hold 50 gal. of roof paint is

Therefore, the amount of water inside the well in gallons is

3. Find the amount of waste in making the largest possible cylindrical rod from a bar of iron 3 ft. long which has a square cross section whose diagonal is 6 in. Solution: To analyze the given word problem, let's label further the given figure as follows

The first that we need to do is to solve for the area of the cross section of the waste material. The cross section of the waste material is the difference of a square and a circle. By using Pythagorean Theorem, the length of one side of a square which is also the diameter of a circle is

Therefore, the amount of waste material removed from a bar of iron which is the difference of the volume of a square prism and a right circular cylinder is

or

Hence, the area of the cross section of the waste material is

4. A cubic foot of water weighs about 62.4 lb. What must be the diameter of a cylindrical pail 1 ft. high in order for it to hold the water from 25 lb. of ice? How many square inches of sheet tin are required to make the pan? (Neglect waste in cutting and lapping.)

as the volume of a cylindrical pail. Therefore, the diameter of a cylindrical pail is

Solution: To illustrate the problem, it is better to draw the figure as follows

or

The total amount of tin sheet required to make a cylindrical pan is

Since the weight of water or ice as well as its density are given, then we can get the volume of water or ice as follows

The volume of water or ice is the same

5. If a lead pipe ¼ in. thick has an inner diameter of 1½ in., find the number of cubic inches of lead in a piece of pipe 10 ft. long. Solution:

6. A certain factory manufactures tin cans. It received an order for 10,000 cylindrical tin cans of height 5 in. and diameter 3 in. How many square feet of tin did this order require if 7 sq. in. are allowed for waste and seams in each can? Solution:

The first thing that we need to do is to get the area of the cross section which is the section between the two concentric circles as follows

The area of the base of a cylindrical tin can is

Therefore, the amount of lead in a piece of lead pipe which is the volume of a circular cylinder is The surface area of a cylindrical tin can is

paint used to paint the top and sides of a gasholder as follows

The area of the excess tin used for waste and seams of a tin can is The total area of the gasholder to be painted is

Therefore, the total amount of tin used to manufacture 10,000 tin cans is

7. A gas company erected a new gasholder. The gasholder (top and sides) was given three coats of paint. It took 50 bbl. of paint of 50 gal. each to complete the work. The height of the tank was 160 ft. and the diameter of the base was 218 ft. If the paint cost $1.00 per gallon, what was the cost per square yard of surface?

Therefore, the cost per square yard of a paint required to paint a gasholder is

Solution: The first thing that we need to do is to calculate the total cost of

or