678
10.28
Chapter 10
Measurement
Technology
Connection
Area Relationships
Did you know you can find the
area of a kite by just knowing the
lengths of its diagonals? What is
this formula and can it be used to
find the areas of other quadrilaterals? Use Geometer’s Sketchpad® student modules available
at the companion website to
explore this and similar questions in this investigation.
Solution 1. Approximately 18 to 19 stamp units. 2. Approximately 6 to 7 business card units.
3. Approximately 23 to 24 eraser units.
Mathematics Investigation
Chapter 10, Section 2
www.mhhe.com/bbn
STANDARD UNITS OF AREA
To standardize area measurement, a square became the accepted area unit shape. However,
the size of the unit square differed in the two predominant systems of measurement: the
English system and the metric system.
1 in
Figure 10.25
English Units In the English system, area is measured by using squares whose sides have
lengths of 1 inch, 1 foot, 1 yard, or 1 mile. Each square unit is named according to the
length of its sides. The 1 inch 3 1 inch square in Figure 10.25 has an area of 1 square inch,
abbreviated 1 sq in or 1 in2 (think of the exponent 2 as indicating a square).
Similarly, we can measure larger areas by using a square foot (1 ft2), a square yard
(1 yd2), and a square mile (1 mi2).
E X AMPLE C
The different square units are related to one another.
1 in
1. How many square inches equal 1 square foot?
2. How many square feet equal 1 square yard?
1
Solution 1. 144 square inches. 2. 9 square feet. square
foot
(ft2)
1
square
inch
(in2)
1 yd
1 ft
1 ft (12 in)
(Note: The squares are not drawn to scale.)
1 yd (3 ft)
Section 10.2 Area and Perimeter
10.29
679
The common units for measuring area in the English system are shown in Figure 10.26.
English Units for Area
Figure 10.26
NCTM Standards
In grades 3–5, students should
learn about area more thoroughly,
as well as perimeter, volume,
temperature, and angle measure.
In these grades they learn that
measurements can be computed
using formulas and need not
always be taken directly with a
measuring tool. p. 44
Square inch
in2
Square foot
2
ft
square foot
144 square inches
2
Square yard
yd
Acre
ac
Square mile
1
144
mi
9 square feet
43,560 square feet
2
27,878,400 square feet
Metric Units In the metric system there is a square unit for area corresponding to each
unit for length. For example, a square meter (1 m2) is a square whose sides have a length
of 1 meter (shown in Figure 10.27, although not to scale). Square meters are used for measuring the areas of rugs, floors, swimming pool covers, and other such intermediate-size
regions. Smaller areas are measured in square centimeters. A square centimeter (1 cm2) is
a square whose sides have lengths of 1 centimeter. Even smaller areas are measured with
the square millimeter, a square whose sides have lengths of 1 millimeter. The actual sizes
of the square centimeter and the square millimeter are shown in Figure 10.27.
1 square meter (m2)
1 square centimeter (cm2)
1 square
millimeter (mm2)
1 cm
1m
1 cm
Figure 10.27
1m
(Note: The square meter is not drawn to scale.)
E X AMPLE D
Determine the following relationships between the metric units for area.
1. How many square millimeters are there in 1 square centimeter?
2. How many square centimeters are there in 1 square meter?
Solution 1. Since each side of a square centimeter [see figure (1) on page 680] has a length
of 1 centimeter, and 1 centimeter 5 10 millimeters, a square centimeter can be covered by 10 3 10 5
100 square millimeters. 2. Since 1 meter equals 100 centimeters, 100 square centimeters can be
680
10.30
Research Statement
Results involving questions on
perimeter from the 7th national
mathematics assessment reveal
that fourth-grade students have
difficulty with perimeter concepts, and eighth-grade students
show a lack of understanding
in more complex contexts that
require a conceptual understanding of perimeter.
Chapter 10
Measurement
placed along each side of a square with dimensions of 1 meter 3 1 meter [see figure (2) here], and
so 10,000 square centimeters will cover the square meter.
(1)
(2)
1 square meter (m2)
1 square centimeter (cm2)
1 cm
1 cm
(10 mm)
...
Martin and Strutchens
1m
...
1m
(100 cm)
The areas of countries, national forests, oceans, and other such large surfaces are measured with the square kilometer, a square whose sides each have a length of 1 kilometer.
Some metric units for area and their relationships are shown in Figure 10.28.
Metric Units for Area
Square millimeter
Square centimeter
Square meter
Figure 10.28
Square kilometer
mm2
cm
m
2
2
km
1
100
square centimeter
100 square millimeters
10,000 square centimeters
2
1,000,000 square meters
PERIMETER
Another measure associated with a region is its perimeter—the length of its boundary. The
perimeter of Figure 10.29 is 23 centimeters, which is greater than the width of this page.
Figure 10.29
Intuitively, it may seem that the area of a region should depend on its perimeter. For
example, if one person uses more fence to close in a piece of land than another person, it is
tempting to assume the first person has enclosed the greater amount of land. However, this
is not necessarily true.
Section 10.2 Area and Perimeter
E X AMPLE E
10.31
681
Each of the following figures has an area of 4 square centimeters. What is the perimeter of
each figure?
(1)
(2)
Solution 1. The perimeter is 8 centimeters. 2. The perimeter is 10 centimeters.
Example E shows that it is possible for two figures to have the same area but different
perimeters. It is also possible for two figures to have the same perimeter but different areas.
E X AMPLE F
For each figure determine the area in square centimeters and the perimeter in centimeters.
(1)
(2)
1 cm
Solution 1. The area is 3 square centimeters, and the perimeter is 10 centimeters. 2. The area
is 4 square centimeters, and the perimeter is 10 centimeters.
NCTM Standards
Rectangles Rectangles have right angles and pairs of opposite parallel sides, so unit squares
fit onto them quite easily. The rectangle in Figure 10.30 can be covered by 24 whole squares
and 6 half-squares. Its area is 27 square units. This area can be obtained from the product
6 3 4.5, because there are 4 12 squares in each of 6 columns. In general, if a rectangle has a
length l and a width w, the area of the rectangle is the product of its length and its width.
Area of rectangle 5 l 3 w 5 lw
Width (w)
Students should begin to develop
formulas for perimeter and area in
the elementary grades. Middlegrade students should formalize
these techniques, as well as
develop formulas for the volume
and surface area of objects like
prisms and cylinders. p. 46
AREAS OF POLYGONS
Figure 10.30
Length (I )
For a given perimeter, the dimensions of a rectangle affect its area. In the photographs
in Figure 10.31 on the next page, the same knotted piece of string has been formed into
three different rectangles. Using the lengths and widths of the rectangles, you can calculate
that each of their perimeters is 36 centimeters. Yet, the area decreases from 80 square centimeters (8 3 10) to 72 square centimeters (6 3 12) to 32 square centimeters (16 3 2), as the
682
10.32
Chapter 10
Measurement
shape of the rectangle changes. If we continue to decrease the width of the rectangle, we can
make its area as small as we please, although the perimeter will remain 36 centimeters.
Figure 10.31
8 cm 3 10 cm (left)
6 cm 3 12 cm (middle)
16 cm 3 2 cm (right)
Parallelograms Fitting unit squares onto a figure is a good way for schoolchildren to
acquire an understanding of the concept of area. However, actually placing squares on a
region is usually difficult because of the shape of the boundary (see part a of Figure 10.32).
One of the basic principles in finding area is that a region can be cut into parts and
reassembled without changing its area. This principle is useful in developing a formula for
the area of a parallelogram. The rectangle in part b of Figure 10.32 has been obtained from
the figure in part a by cutting triangle A from the left side of the parallelogram and moving
it to the right side to create a rectangle. The base of the rectangle is 5 centimeters, and its
height is 2 centimeters, so its area is 10 square centimeters. Since the rectangle was obtained
by rearranging the parts of the parallelogram, the area of the parallelogram is also 10 square
centimeters. Notice that the base of the parallelogram is 5 centimeters and its height, or
altitude (the perpendicular distance between opposite parallel sides), is 2 centimeters. This
suggests that the area of a parallelogram is the product of its base and its height.
2 cm
Area of parallelogram 5 b 3 h 5 bh
2 cm
A
Figure 10.32
NCTM Standards
The notion that shapes that look
different can have equal areas is
a powerful one that leads eventually to the development of general methods (formulas) for
finding the area of a particular
shape, such as a parallelogram.
p. 166
Figure 10.33
A
5 cm
5 cm
(a)
(b)
For a given perimeter, the area of a parallelogram depends on its shape. The two parallelograms formed by the inside edges of the linkages in the photograph in Figure 10.33
both have the same perimeter; but as the parallelogram is skewed more to the right, its
height decreases. Since the base of both parallelograms is the same and the area of a parallelogram is the base times the height, the parallelogram with the smaller height has the
smaller area. The area of the first parallelogram is approximately 72 square centimeters
(9 3 8), and the area of the second one is approximately 45 square centimeters (9 3 5). The
height of the parallelogram, and consequently its area, can be made arbitrarily small by
further skewing the linkages while the perimeter stays constant.
Section 10.2 Area and Perimeter
E X AMPLE G
683
10.33
Estimate the area of each parallelogram by visualizing the number of 1 centimeter 3 1 centimeter squares needed to cover the figure. Then compute the area. Was your estimate within
1 square centimeter of the correct area?
Research Statement
(1)
The 7th national mathematics assessment found that only
18 percent of the fourth-grade
students could draw a geometric shape with two side lengths
given.
(2)
3.5 centimeters
2 centimeters
4 centimeters
2 centimeters
Martin and Strutchens
Solution 1. 8 square centimeters (4 3 2 5 8). 2. 7 square centimeters (2 3 3.5 5 7).
Were your estimates close to the areas of these figures?
Triangles The triangle in Figure 10.34a is covered with 1 centimeter 3 1 centimeter
squares and parts of squares. Can you see why this shows that the area of the triangle is
more than 4 square centimeters? Since it is inconvenient to cover the triangle with squares,
we will use a different approach to find its area. Two copies of a triangle can be placed
together to form a parallelogram, as shown in part b. This can be accomplished by rotating
the triangle in part a about the midpoint of side AB. Since the parallelogram has a base of
5 centimeters and a height of 2 centimeters, its area is 10 square centimeters. Thus, the area
of the triangle is one-half as much, or 5 square centimeters.
A
B
C
2 cm
C
Figure 10.34
B
5 cm
5 cm
(a)
A
(b)
The preceding example suggests a general approach to finding the area of a triangle:
Place two copies of the triangle together to form a parallelogram, and then find the area of
the parallelogram. If the length of the base of the triangle is b and its height, or altitude
(the perpendicular distance to its base from the opposite vertex), is h, then the base and
altitude of the parallelogram are also b and h (Figure 10.35). So, the area of the parallelogram is b 3 h, and since the parallelogram is formed from two triangles,
Area of triangle 5 12 3 b 3 h 5 12 bh
h
Figure 10.35
b
Any side of a triangle may be considered the base, and each base has its corresponding
altitude. Regardless of the base and altitude chosen, as shown in Example H on the next
page, the triangle will have the same area.
684
10.34
E X AMPLE H
Chapter 10
Measurement
Triangle ABC is shown below in two positions, with two different bases and altitudes. One
of these altitudes falls outside the triangle. Determine the area of each triangle.
(1)
(2)
C
2 cm
2.9 cm
A
B
5.8 cm
A
C
B
4 cm
Solution 1. The area is 5.8 square centimeters: 12 3 5.8 3 2 5 5.8. 2. The area is 5.8 square
centimeters:
NCTM Standards
Students can develop formulas
for parallelograms, triangles, and
trapezoids using what they have
previously learned about how to
find the area of a rectangle, along
with an understanding that
decomposing a shape and
rearranging its component parts
without overlapping does not
affect the area of the shape. p. 244
1
2
3 4 3 2.9 5 5.8.
Trapezoids It is inconvenient to cover a trapezoid with square units because of its sloping
sides. However, as with a triangle, we can obtain a parallelogram by placing two trapezoids
together.
The trapezoid in Figure 10.36a has a lower base of length b and an upper base of
length u, and its height, or altitude (the perpendicular distance between its bases), is h. The
parallelogram in part b was obtained by placing two copies of the trapezoid side by side.
The parallelogram has a base of b 1 u and a height of h, so its area is (b 1 u) 3 h. Since
the parallelogram is formed from two trapezoids, this example suggests a general approach
for finding the area of a trapezoid:
Area of trapezoid 5 12 3 (b 1 u) 3 h 5 12 (b 1 u)h
u
u
b
h
h
b
Figure 10.36
E X AMPLE I
(a)
b+u
(b)
If the upper and lower bases of the trapezoid in Figure 10.36 are 2 centimeters and 4 centimeters, respectively, and the height is 2.5 centimeters, try to estimate the area to see if you can
come within 1 square centimeter of the correct area. Then use the formula for the area of a
trapezoid to exactly compute the area.
Solution The area of the trapezoid is 8.75 square centimeters: 12 3 (4 1 2) 3 2.5 5 7.5. Was your
estimate between 6 and 9 square centimeters?
Section 10.2 Area and Perimeter
10.35
685
CIRCUMFERENCES AND AREAS OF CIRCLES
Circles are part of our natural environment. The Sun, the Moon, flowers, whirlpools, and cross
sections of some trees have circular shapes. School children see many everyday examples of
circles, such as bottle caps, tops of cans, lamp shades, and toys like the hula hoop.
Figure 10.37
Hula hoops can be formed by
joining the ends of tubing
Circumference The perimeter or distance around a circle is called the circumference.
There is something deceptive about trying to estimate the circumference of a circle, as
illustrated in Example J.
E X AMPLE J
Obtain a circular object near you (a cup, glass, etc.). Estimate how many diameter lengths
it would take to wrap around the circumference of the object. Use string or the edge of a
piece of paper to measure the circumference and the diameter of the object, and then compare the diameter to the circumference to check your estimate.
Solution Were you surprised that it took a little more than 3 diameters to reach around the
circumference?
3.141592654
OFF
ON
DEG
Figure 10.38
RAD
There is a tendency to underestimate the circumferences of circles. The hula hoop in the
photograph above has a diameter of 3 feet and it took between 9 and 10 feet of tubing to
make it. Often people estimate the circumference of a circular object by doubling the diameter of the object. Actually, the circumference is a little greater than 3 times the diameter.
The exact ratio of the circumference of a circle to its diameter is the irrational number
p (pi), which is 3.1416 rounded to four decimal places. This ratio is expressed in the following equations, where C is the circumference of a circle, d is the diameter, and r is the radius.
C
5p
d
Technology
Connection
or
C 5 pd
or
C 5 2pr
Some calculators have a key for p. Pressing p on a calculator with 10 places for digits
will give the number shown in the display in Figure 10.38. However, since p is an irrational
number, the number in this display is only a rational number approximation of p. For most
purposes it is sufficient to approximate p by 3.1416. Note: For the exercises and problems
in this text, use a calculator value of p or approximate p by 3.1416.
686
10.36
E X AMPLE K
Chapter 10
Measurement
The photograph shows a piece of string being stretched around a tennis ball can.
Predict how the length of the string will compare to the height of the can. The can has
a diameter of approximately 7 centimeters and a height of approximately 20 centimeters.
Compute the length of the string, and compare it to the height of the can.
Solution It is common for people to predict that the length of the string is less than the height of
the can. However, since 3.1416 3 7 is approximately 22, the length of the string is approximately
22 centimeters, which is 2 centimeters greater than the height of the can.
Areas The area of a circle can be approximated by counting unit squares and parts of
squares that cover the circle.
E X AMPLE L
The following circle has been drawn on a centimeter grid. Approximate its area in square
centimeters.
Research Statement
Research shows that additional
attention needs to be given
to geometry and measurement across the curriculum,
as students consistently score
poorly in these areas.
Strutchens and Blume; Kenney and
Kouba; Lindquist and Kouba
Solution A first step might be to note that the circle is contained inside a 4 3 4 grid, which shows
that its area is less than 16 square centimeters. Next, we can count the squares inside the circle and
then combine the parts of the remaining squares, or we can estimate the parts of the squares outside
the circle and subtract their area from 16. A reasonable estimate for the area of the circle is 12 square
centimeters.
Section 10.2 Area and Perimeter
Figure 10.39
687
10.37
You might have noticed in Example L, on the previous page, that one-quarter of the circle
is contained in a square whose side has a length that is equal to the radius of the circle. This is
illustrated in Figure 10.39, which shows that the area of one-quarter of the circle with radius r
is less than the radius times itself, or r2. Thus, the area of the whole circle is less than 4 3 r2.
Let’s consider a method for determining the area of a circle. The circle in part a of
Figure 10.40 has been divided into 16 sectors. When these 16 sectors are rearranged and
placed together, as shown in part b, they form a figure whose shape is close to that of a parallelogram. The length of the base of the parallelogram-like figure is one-half the circumference of the circle, and the height of the figure is approximately the radius of the circle.
If the circle is cut into a greater number of sectors, the shape of the resulting parallelogramlike figure will be even closer to that of a parallelogram. Using the formula for the area of
a parallelogram, we can approximate the area of the figure in part b:
Area of parallelogram 5 b 3 h < 12 C 3 r 5 12 (2pr) 3 r 5 pr2
which is the formula for the area of a circle.
1
2
C
r
Figure 10.40
(a)
h
(b)
Area of circle 5 pr2
Notice that since p < 3.1416, the area of a circle is a little more than 3 times the square
of the radius of the circle (see Figure 10.39).
NCTM Standards
The Curriculum and Evaluation Standards for School Mathematics (p. 221) suggest that
students construct such a model to develop the formula for the area of a circle:
Students who can use the relationship between the shape of the “parallelogram”
and its area and the circumference of the circle to develop the formula for the area
of the circle are demonstrating plausible and deductive reasoning.
E X AMPLE M
Approximate the areas of 9-inch and 12-inch pizzas, using a value of 3 for p. If a 9-inch
pizza costs $10.80 and the unit cost per square inch of a 9-inch and a 12-inch pizza is the
same, what is the cost of the 12-inch pizza?
9 inches
12 inches
702
10.52
Research Statement
Students have difficulty in
determining the volume of a
three-dimensional array of cubes
that is presented as a diagram.
One study showed that fewer
than 25 percent of fifth-grade
students could solve such a
problem.
Ben-Haim, Lappan, and Houang
E X AMPLE B
Chapter 10
Measurement
2. If 6 layers of the die on page 701 fill the box, what is the volume of the box?
3. If 4.5 layers of the cube on page 701 fill the box, what is the volume of the box?
Solution 1. Approximately 33 dice will cover the base, and approximately 18 of the larger cubes
will cover the base. 2. The volume of the box is approximately 198 dice units. 3. The volume
of the box is approximately 81 cube units.
The difficulty that children at all grade levels have in understanding the concept of
volume is indicated in the next example. Every 4 years the NAEP (National Assessment of
Educational Progress) administers mathematics tests in schools throughout the United
States. Example B contains a question on volume from one of these tests.
In a national assessment on mathematics, students were shown the figure at the left and
asked how many cubes the box contained. What is the correct answer, and what do you
think was the incorrect answer most commonly given by students?
Solution The box contains 12 cubes. Only 6 percent of the 9-year-olds, 21 percent of the 13-yearolds, and 43 percent of the 17-year-olds answered the question correctly.* The most common incorrect answer was 16. Can you see how students might have obtained this answer?
STANDARD UNITS OF VOLUME
For each English unit of length (inch, foot, etc.) and each metric unit of length (centimeter,
meter, etc.) there is a corresponding unit of volume, a cube whose three dimensions are the
given length.
English Units Cubic units are named according to the length of their edges. The most
commonly used units for measuring nonliquid volume in the English system are the cubic
inch, the cubic foot, and the cubic yard. The cubes for these units are illustrated in Figure 10.43 on the next page. These cubes are not drawn to scale.
* T. P. Carpenter, T. G. Coburn, R. E. Reys, and J. W. Wilson, “Results and Implications of the NAEP Mathematics
Assessment: Elementary School,” Arithmetic Teacher: 438–450.
Section 10.3 Volume and Surface Area
703
10.53
1 yd
1 ft
1 in
1 in 1 in
Figure 10.43
1 yd
1 ft
1 ft
Cubic inch
1 yd
Cubic foot
Cubic yard
The volume of a microwave oven might be measured in cubic inches. A larger volume,
such as that of a room or freezer, might be measured in cubic feet. Larger volumes, such as
the volume of a truckload of gravel or crushed rock, are measured by the cubic yard.
E X AMPLE C
Determine the following English unit relationships.
1. How many cubic inches equal 1 cubic foot?
2. How many cubic feet equal 1 cubic yard?
3. How many cubic inches equal 1.4 cubic feet?
Solution 1. Imagine a box in the shape of a cube whose dimensions are each 1 foot. The base
of the box is 12 inches 3 12 inches, so the floor of the box can be covered by 144 cubes, each of
which is 1 inch 3 1 inch 3 1 inch. Since 12 such layers will fill the box, its volume is 1728 cubic
inches (12 3 144 5 1728).
12 in
12 in
12 in
2. Similarly, the floor of a cube-shaped box whose dimensions are each 1 yard can be covered with
9 cubes, each of which is 1 foot 3 1 foot 3 1 foot. Three such layers will fill the box, so its volume
is 27 cubic feet. 3. Since 1 cubic foot 5 1728 cubic inches, 1.4 cubic feet 5 2419.2 cubic inches
(1.4 3 1728 5 2419.2).
The English units for volume and their relationships are summarized in Figure 10.44.
English Units for Volume
Figure 10.44
Cubic inch
in3
Cubic foot
Cubic yard
ft3
yd3
1
1728
cubic foot
1728 cubic inches
27 cubic feet
704
10.54
Chapter 10
Measurement
Metric Units The common metric units for measuring nonliquid volume are the cubic
millimeter, the cubic centimeter, and the cubic meter. Each unit is named according to
the length of the edges of its cube. For example, the edges of the cube for the cubic centimeter each have a length of 1 centimeter. The cubes for these three metric units are shown
in Figure 10.45. These cubes are not drawn to scale.
1m
100
100
1 cm
1 mm
1 mm
Figure 10.45
E X AMPLE D
1 cm
1 mm
(a)
1m
100
1 cm
1m
(b)
(c)
Determine the following relationships.
1. How many cubic centimeters equal 1 cubic meter?
2. How many cubic millimeters equal 1 cubic centimeter?
3. How many cubic millimeters equal 3.4 cubic centimeters?
Solution 1. Visualize a cube-shaped box that measures 1 meter on each edge (see part c of Figure 10.45), and imagine filling it with cubes whose edges measure 1 centimeter. The floor of the large
cube is 100 centimeters 3 100 centimeters, so it can be covered by 10,000 of the smaller cubes. Since
there are 100 such layers, the volume of the box is 1,000,000 cubic centimeters (100 3 10,000). 2. A
cube whose edges each have a length of 1 centimeter has dimensions of 10 millimeters 3 10 millimeters
3 10 millimeters. So, its volume is 1000 cubic millimeters (10 3 10 3 10). 3. Since 1 cubic centimeter 5
1000 cubic millimeters, 3.4 cubic centimeters 5 3400 cubic millimeters.
Some of the metric units for volume and their relationships are shown in the table in
Figure 10.46.
Metric Units for Volume
Cubic millimeter
Figure 10.46
Cubic centimeter
Cubic meter
mm3
3
cm
m3
1
1000
cubic centimeter
1,000 cubic millimeters
1,000,000 cubic centimeters
SURFACE AREA
Another important measure associated with objects in space is their amount of surface.
Surface area is expressed as the number of unit squares needed to cover the surface of a
three-dimensional figure.
Section 10.3 Volume and Surface Area
E X AMPLE E
10.55
705
The area of the top of the box in the figure is 9 square centimeters. What is the total surface
area, including the base of the box?
Solution The top and bottom faces of the box each have an area of 9 square centimeters. The right and
left faces each have an area of 6 square centimeters, and the front and back faces each have an area of
6 square centimeters. So, the total surface area is 42 square centimeters (2 3 9 1 2 3 6 1 2 3 6).
The surface area of an object cannot be predicted on the basis of its volume, any more
than the perimeter of a figure is determined by the area of the figure.
E X AMPLE F
The box in Example E and the box below each have a volume of 18 cubic centimeters. How
do the surface areas of the two boxes compare?
Solution The top and bottom faces of the box in this figure each have an area of 18 square centimeters, the right and left sides each have an area of 3 square centimeters, and the front and back
faces each have an area of 6 square centimeters. So the total surface area is 54 square centimeters.
This is 12 square centimeters greater than the surface area of the box in Example E.
Examples E and F show that figures in space can have the same volume but different
surface areas. The amount of material you would need to build the box in Example F is
about 130 percent of the amount of material you would need to build the box in Example E
(130 percent of 42 5 1.3 3 42 < 54), although both have a volume of 18 cubic centimeters.
VOLUMES AND SURFACE AREAS OF SPACE FIGURES
Prisms In Figure 10.47, on the next page, the length (20) times the width (10) gives the number
of cubes (200) on the floor of the box (or base of the rectangular prism). Since the box can be
filled with 6 levels of cubes, it will hold 1200 cubes. This volume of 1200 cubic centimeters
706
10.56
NCTM Standards
Students in grades 3–5 should
develop strategies for determining
surface area and volume on the
basis of concrete experiences.
They should measure various rectangular solids using objects such
as tiles and cubes, organize the
information, look for patterns, and
then make generalizations. p. 175
Chapter 10
Measurement
can be obtained by multiplying the three dimensions of the box: length 3 width 3 height. In
general, a rectangular prism with length l, width w, and height h has the following volume:
Volume of rectangular prism 5 length 3 width 3 height
V 5 lwh
6 cm
20 cm
10 cm
Figure 10.47
The volume of any prism can be found in a similar way. The base of the prism in
1
Figure 10.48 is a right triangle, which is covered by 4 2 cubes. Since 6 levels each with
1
4 2 cubes fill the prism, its volume is 27 cubic centimeters (6 3 4.5).
6 cm
Figure 10.48
3 cm
3 cm
The number of cubes that cover the base of this prism is the same as the area of the
base. Therefore, the volume of the prism can be computed by multiplying the area of the
base by the height, or altitude, of the prism. In general, the volume of any right prism having a base of area B and a height of h can be computed by the formula:
Volume of prism 5 area of base 3 height
V 5 Bh
The formula for the volume of an oblique prism is suggested by beginning with a stack
of cards, as in part a of Figure 10.49, and then pushing them sideways to form an oblique
prism, as in part b on page 708. If each card in part a is 12.5 centimeters 3 7.5 centimeters
and the stack is 5 centimeters high, the volume of the right prism in Figure 10.49a is 12.5
centimeters 3 7.5 centimeters 3 5 centimeters, or 468.75 cubic centimeters. The base of
the oblique prism in part b is also 12.5 centimeters 3 7.5 centimeters, and its height,
or altitude (the perpendicular distance between its upper and lower bases), is also
5 centimeters. Since both stacks contain the same number of cards, their volumes are both
708
10.58
Chapter 10
Measurement
468.75 cubic centimeters. This means that the volume of the oblique prism can be computed
by multiplying the area of its base by its height. In general, the volume of any prism, right
or oblique, is the area of its base times its height.
h
h
Figure 10.49
E X AMPLE G
(a)
(b)
Sometimes a prism will have more than one pair of bases, as shown in the following figures. Figure (1) is a right prism whose base is a parallelogram. By turning it so that one of
its lateral faces becomes the base, as in figure (2), we can classify the prism as an oblique
prism. Determine the volume of each prism, given the following dimensions:
1. Figure (1): area of base, 280 square centimeters; altitude, 6 centimeters
2. Figure (2): area of base, 120 square centimeters; altitude, 14 centimeters
(1)
(2)
Right prism
Oblique prism
Solution 1. 1680 cubic centimeters. 2. 1680 cubic centimeters.
The faces of right prisms are rectangles; and the faces of oblique prisms are rectangles and
parallelograms.
The surface area of a prism is the sum of the areas of its bases and faces.
Cylinders The cylindrical buildings in Figure 10.50 are skyscrapers in Los Angeles. Just
as with conventional rectangular buildings, architects need to know the volumes and surface areas of these glass-walled cylinders.
Figure 10.50
Los Angeles skyscrapers and
street lights
Section 10.3 Volume and Surface Area
NCTM Standards
Although [middle-grade] students may have developed an
initial understanding of area and
volume, . . . some measurement
of area and volume by actually
covering shapes and filling
objects can be worthwhile for
many students. p. 242
10.59
709
To compute the volumes of cylinders, we continue to use unit cubes even though they
do not conveniently fit into a cylinder. More than 33 cubes are needed to cover the base of
the cylinder in Figure 10.51. Furthermore, since the cylinder has a height of 12 centimeters,
it will take at least 12 3 33, or 396, cubes to fill the cylinder.
Figure 10.51
If we used smaller cubes in Figure 10.51, they could be packed closer to the boundary
of the base and a better approximation would be obtained for the volume of the cylinder.
This suggests that the formula for the volume of a cylinder is the same as that for the
volume of a prism. For a cylinder with a base of area B and a height h,
Volume of cylinder 5 area of base 3 height
V 5 Bh
A right cylinder without bases can be formed by joining the opposite edges of a rectangular sheet of paper (Figure 10.52). The circumference of the base of the cylinder is the
length of the rectangle, and the height of the cylinder is the height of the rectangle. Therefore, the surface area of the sides of a cylinder is the circumference of the base of the cylinder times its height.
h
h
r
Figure 10.52
2πr
For any right cylinder whose base has a radius r and whose height is h, the base has a circumference of 2pr, and the surface area of the side of the cylinder is 2pr 3 h. Adding the
area of both bases, 2pr2, to the area of the side of the cylinder produces the total surface
area:
Surface area of cylinder 5 2prh 1 2pr2
710
10.60
E X AMPLE H
Research Statement
The 7th national mathematics
assessment found that questions
assessing familiarity with
volume and surface area were
difficult for fourth and eighthgrade students.
Martin and Strutchens
Chapter 10
Measurement
Compute the volume and surface area of a tennis ball can if the diameter of its base is
7 centimeters and the height of the can is 20 centimeters.
Solution Surface area: The radius of the base is 3.5 centimeters, so the top and the base of the
can each have an area of p (3.5)2 square centimeters, which to two decimal places is 38.48 square
centimeters. The circumference of the can is 7p centimeters, which to two decimal places is 21.99
centimeters; so the lateral surface of the can has an area of approximately 21.99 3 20 square centimeters, which to two decimal places is 439.80 square centimeters. Thus, the total surface area of
the can is approximately 516.76 square centimeters, or 517 square centimeters when rounded to the
nearest whole number. Volume: 20 3 p(3.5)2 cubic centimeters, which to the nearest whole number
is 770 cubic centimeters.
Pyramids The Pyramid of Cheops, also known as the Great Pyramid of Egypt, was built
about 2600 b.c.e. and is one of the seven wonders of the ancient world. It has a height of
148 meters. The Transamerica Pyramid in San Francisco (Figure 10.53) has a height of
260 meters and was built in 1972. Even though the Egyptian pyramid is the shorter of these
giant pyramids, its volume is several times the volume of the taller pyramid. (See exercise
23 in Exercises and Problems 10.3.)
Figure 10.53
Transamerica Pyramid in
San Francisco
The red pyramid in Figure 10.54d on the next page is inside the cube in part b, which
has a square base EFGH and height GC. The red pyramid, together with the green pyramid
in part a and the blue pyramid in part c divide the cube into three congruent pyramids. (Can
you see how these fit together?) Therefore, the volume of each pyramid is one-third of the
volume of the cube. Since the volume of the cube is the area of its base EFGH times
the height of the cube, the volume of each pyramid is one-third of the area of the base of
the cube times the height of the cube.
Section 10.3 Volume and Surface Area
711
10.61
A
A
A
B
D
B
C
C
E
D
C
E
E
F
(c)
F
(a)
(b)
H
H
G
C
E
F
(d)
Figure 10.54
H
G
In general, if B is the area of the base of a pyramid and h is the height, or altitude
(perpendicular distance from the apex to the base), of the pyramid, then
Volume of pyramid 5 1 3 area of base 3 height
3
1
V 5 Bh
3
The faces of a pyramid are its base and triangular lateral sides. The surface area of the
pyramid is the sum of the areas of the faces. Exercises and Problems 10.3 has several questions involving the surface area of pyramids.
The surface area of a pyramid is the sum of the
area of its base and its triangular lateral sides.
E X AMPLE I
Determine the volume of the Pyramid of Cheops to the nearest cubic meter. Its square base
has sides of length 232.5 meters, and its height is 148 meters.
Solution The area of the base is 54,056.25 square meters (232.5 3 232.5), and
1
3 54,056.25 3 148 5 2,666,775
3
So, the volume of the pyramid is 2,666,775 cubic meters.
Cones The pile of crude salt in Figure 10.55, on the next page, has the shape of a cone
with a circular base. The conical shape forms as salt is poured from the conveyor belt at the
top left. The salt has been evaporated from ocean water and awaits further purification.
712
10.62
Chapter 10
Measurement
Figure 10.55
The volume of a cone can be approximated by the volume of a pyramid inscribed in
the cone. The hexagonal pyramid in Figure 10.56 has a volume of approximately 430 cubic
centimeters, which is slightly less than the volume of the cone. As the number of sides in
the base of the pyramid increases, the volume of the pyramid becomes closer to the volume
of the cone. This intuitively indicates that the pyramid and cone formulas are the same.
Since the volume of the pyramid is one-third the area of its base times its height, we can
use the same formula to calculate the volume of a cone. In general, for any cone whose base
has area B and whose height is h,
Volume of cone 5 1 3 area of base 3 height
3
1
V 5 Bh
3
20 cm
5 cm
Figure 10.56
The surface area of the cone in Figure 10.56 is the area of its base, p(5)2, plus the
area of its lateral surface. The area of the lateral surface can be approximated by the
area of the triangular sides of the hexagonal pyramid. The altitude of one of these triangular sides (see Figure 10.57 on the next page) is approximately equal to the slant
height of the cone, which by the Pythagorean theorem is 152 1 202 . If s is the length
Section 10.3 Volume and Surface Area
10.63
713
of one side of the hexagonal base, the area of one triangle is 12 s 152 1 202 , and the area of
the lateral surface is 6112 s 152 1 202 2. Notice in Figure 10.57 that 6s is the perimeter of the base
of the pyramid. Closer approximations can be found by increasing the number of sides of
the polygonal base of the pyramid. As the number of sides of the polygon increases, its
perimeter 6s gets closer to the circumference of the circle, 2p(5), and the area of the
lateral sides of the pyramid gets closer to
兹5 2 +
20 2
1 (2p)(5) 152 1 202 5 p(5) 152 1 202
2
20
5
Figure 10.57
s
In general, for a cone of radius r and altitude h, the area of the lateral surface is pr 1r2 1 h2
and the surface area of the cone is the sum of the area of its base and its lateral face.
Surface area of cone 5 pr 1r2 1 h2 1 pr2
E X AMPLE J
The height of the cone of salt, in Figure 10.55 on the previous page, is 12 meters, and the
diameter of its base is 32 meters.
1. Determine the number of railroad boxcars this salt will fill if each boxcar has a volume
of 80 cubic meters.
2. Determine the number of square meters of plastic sheet needed to cover the lateral
surface of this cone of salt.
Solution 1. The area of the base of the cone of salt is approximately 804 square meters. Since
1
3 804 3 12 5 3216
3
the volume of the cone to the nearest cubic meter is 3216 cubic meters. This amount of salt will fill
about 40 boxcars. 2. The area of the lateral surface of the cone is approximately
p (16) 1162 1 122 ¯ 1005.312
So, to the nearest square meter, 1005 square meters of plastic sheet is needed to cover the lateral
surface of the cone.
Spheres Figure 10.58, on the next page, shows a view of Earth as seen from the Galileo
spacecraft. Earth, the planets, and their moons are all spherical shapes, spinning and orbiting about a spherical Sun. There is considerable variation in the volumes of these objects.
Earth has about 18 times the volume of the smallest planet, Mercury. The volume of the
largest planet, Jupiter, is 10,900 times the volume of Earth.