CHAPTER 13:

Three-Dimensional Figures

“Your ultimate success will depend on your ability to see in different dimensions. ”

This chapter is relatively short. There are only a few figures we need to know. Since these are three-dimensional figures, we discuss their volumes and surface areas (areas of all of the sides). The diagonal is the distance from one corner internally to an opposite corner.

BOX

This figure is also known as a rectangular solid, and if that isn’t a mouthful enough, its correct name is a rectangular parallelepiped. But essentially, it’s a box.

• Volume = V = lwh
• Surface area = SA = 2lw + 2lh + 2wh
• , + known as the 3-D Pythagorean theorem

Example 1:

For the given figure, find V, SA, and d.

Solution:

V = lwh = (3)(4)(2) = 24 cubic feet; SA = 2lw + 2lh + 2wh = 2(3)(4) + 2(3)(2) + 2(4)(2) = 52 square feet; feet

CUBE

A cube is a box for which all of the faces, or sides, are equal squares.

• V = e³ (say “e cubed”). Cubing comes from a cube!
• SA = 6e²
• 
• A cube has 6 faces, 8 vertices, and 12 edges.

Example 2:

For a cube with an edge of 10 meters, find V, SA, and d.

Solution:

V = 10³ = 1,000 cubic meters; SA = 6e² = 6(10)² = 600 square meters; meters

CYLINDER

A cylinder is shaped like a can. The curved surface is considered as a side, and the top and bottom are equal circles.

• V = πr²h
• SA = top + bottom + curved surface = 2πr² + 2πrh

Once a neighbor of mine wanted to find the area of the curved part of a cylinder. He wasn’t interested in why, just the answer. Of course, being a teacher I had to explain it to him. I told him that if you cut a label off a soup can and unwrap it, the figure is a rectangle; neglecting the rim, the height is the height of the can and the width is the circumference of the circle. Multiply this height and width, and the answer is 2πr × h. He waited patiently and then soon moved out of the neighborhood. Just kidding!

 

In general, the volume of any figure for which the top is the same as the bottom is V = Bh, where B is the area of the base. If the figure comes to a point, the volume is . The surface area is found by adding up all the sides.

Example 3:

Find V and SA for a cylinder of height 10 yards and diameter of 8 yards.

Solution:

We see that since d = 8, r = 4. Then V = πr²h = π(4² × 10) = 160π cubic yards; SA = 2πr² + 2πrh = 2π4² + 2π(4)(10) = 112π square yards.

Let’s do some exercises.

Use this figure for Exercises 1 through 3. It is a pyramid with a square base. WX = 8, BV = 3, and B is in the middle of the base.

Exercise 1:

The volume of the pyramid is

1. . 192
2. . 96
3. . 64
4. . 32
5. . 16

Exercise 2:

The surface area of the pyramid is

1. . 72
2. . 112
3. . 144
4. . 224
5. . 448

Exercise 3:

VY =

1. . 6
2. .
3. . 7
4. . 9
5. .

Exercise 4:

In the given rectangular solid, the perimeter of ΔABC =

1. .
2. . 30
3. .
4. . 37
5. . 41
   

Exercise 5:

The volume of the cylinder shown is:

1. 640π
2. 320π
3. 288π
4. 144π
5. 72π

Exercise 6:

ABKL is the face of a cube with AB = 10, and box BCFG has a square front with BC = 6.

 

The surface area that can be viewed in this configuration is

1. 300
2. 356
3. 396
4. 400
5. 1360
   

Exercise 7:

A cylinder has volume V. If we triple its radius, by what factor should we multiply the height in order that the volume stays the same?

1. 
2. .
3. 1
4. 3
5. 9

Exercise 8:

The numerical value of which of the following is sufficient to determine the volume of a cube? Indicate all correct choices.

1. Edge
2. Total surface area
3. Surface area of one face
4. Diagonal of one face
5. Diagonal of the cube
6. Distance between the middle of one face and the middle of the opposite face

Exercise 9:

The numerical value of which of the following is sufficient to determine the volume of a cone? Indicate all correct choices.

1. Radius of the base
2. The vertical height
3. The diameter of the base and the vertical height
4. The diameter of the base and the slant height
5. The value of r²h
6. The volume of a cylinder whose radius and height match those of the cone

Exercise 10:

The area of the base of a pyramid is 20 and its vertical height is 6. What is the volume?

Exercise 11:

The volume of a cube with edge 6 is the same as the volume of a rectangular box with a height of 4 and a width of 3. What is the length of the box?

Let’s look at the answers.

Answer 1:

The answer is (C). .

Answer 2:

The answer is (C). ; ∆ABV is a 3-4-5 right triangle with AB = 4 and BV = 3, so AV = 5. AV is the height of each triangular side, h, and b = XY = 8. So SA = 8² + 2(8)(5) = 144.

Answer 3:

The answer is (B). ΔAVY is a right triangle with right angle at A. AY = 4 and AV = 5, so .

Answer 4:

The answer is (C). In the given figure, we have to use the 2-D Pythagorean theorem three times to find the sides of ∆ABC. ∆BCY is a 6-8-10 triple, so BC = 10, and ∆ACX is a 8-15-17 triple, so AC = 17.

For ∆ABZ, we actually have to calculate the missing side AB = . So the perimeter is .

Answer 5:

The answer is (E). The diameter of the base is 6, and again we have a Pythagorean triple; so h = 8. The volume is π(3²)(8) = 72π.

Answer 6:

The answer is (C). The areas are: ABKL = 100; IJLK = 100; BCFG = 36; CDEF = 60; EFGH = 60; GHIK = 40. The total is 396.

Answer 7:

The answer is (A). The volume V = πr²h. For simplicity, let r = 1 and h = 1. So V = π. If we triple the radius, V = π(3)²h. For the original volume to still be π, 9h = 1 or .

Answer 8:

The correct answers are (A), (B), (C), (D), (E), and (F). Since the volume of a cube is e³, where e is the length of one edge, any information that yields the length of one edge will be sufficient for finding the volume. Choice (A) becomes obviously correct. Choices (B) and (C) are correct because the total surface area is given by 6e², and the surface area of one face is e². Choices (D) and (E) are correct because the diagonal of one face equals an edge multiplied by and the diagonal of the cube equals an edge multiplied by . Choice (F) describes a length equivalent to an edge, so it must be correct.

Answer 9:

The correct answers are (C), (D), (E), and (F). The volume of a cone is , where r is the radius of the base and h is the vertical height, so we need to be able to find both r and h. Choice (C) is correct because the radius is just one-half the diameter. Choice (D) is correct because the vertical height can be determined from the slant height and the radius by using the Pythagorean theorem ((radius)² + (vertical height)² = (slant height)²). Choice (E) is correct because of the known variables. Choice (F) is correct because the volume of a cylinder is πr²h, so the volume of a cone is simply one-third this value. Choices (A) and (B) give one value (r or h) but not the other.

Answer 10:

The correct answer is 40. The volume of a pyramid is , where B is the area of the base and h is the vertical height. Thus, the volume is

Answer 11:

The correct answer is 18. The volume of the cube is 6³ = 216. The volume of a rectangular solid is given by the formula V = lwh, where l = length, w = width, and h = height. Then 216 = (/)(3)(4) = 12/. Thus,

Let’s go to the last review chapter of the book.