Measurement Unit, Part B
(Level IV Graduate Math)
Draft
(NSSAL)
C. David Pilmer
©2010
(Last Updated: April 2015)
This resource is the intellectual property of the Adult Education Division of the Nova Scotia
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The following are permitted to use and reproduce this resource for classroom purposes.
 Nova Scotia instructors delivering the Nova Scotia Adult Learning Program
 Canadian public school teachers delivering public school curriculum
 Canadian non-profit tuition-free adult basic education programs
 Nova Scotia Community College instructors
The following are not permitted to use or reproduce this resource without the written
authorization of the Adult Education Division of the Nova Scotia Department of Labour and
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 Core programs at post-secondary institutions (exception: NSCC)
 Public or private schools outside of Canada
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Individuals, not including teachers or instructors, are permitted to use this resource for their own
learning. They are not permitted to make multiple copies of the resource for distribution. Nor
are they permitted to use this resource under the direction of a teacher or instructor at an
unauthorized learning institution.
Acknowledgments
The Adult Education Division would also like to thank the following NSCC instructors for
piloting this resource and offering suggestions during its development.
Eileen Burchill (IT Campus)
Elliott Churchill (Waterfront Campus)
Barbara Leck (Pictou Campus)
Suzette Lowe (Lunenburg Campus)
Floyd Porter (Strait Area Campus)
Brian Rhodenizer (Kingstec Campus)
Joan Ross (Annapolis Valley Campus)
Eric Tetford (Burridge Campus)
Tanya Tuttle-Comeau (Cumberland Campus)
Jeff Vroom (Truro Campus)
Table of Contents
Introduction ……………………………………………………………………………… ii
Negotiated Completion Date ……………………………………………………………….ii
The Big Picture ………………………………………………………………………….. iii
Course Timelines …………………………………………………………………………...iv
Introduction to Perimeter and Area ……………………………………………………….. 1
Perimeter and Area Formulas …………………………………………………………….. 4
More Perimeter and Area Formulas ……………………………………………………….. 8
Perimeter and Area of Composite 2D Shapes ……………………………………………...15
Introduction to Surface Area ……………………………………………………………… 24
Introduction to Volume …………………………………………………………………… 35
More Surface Area and Volume ………………………………………………………….. 43
Putting It Together ………………………………………………………………………… 50
Post-Unit Reflections………………………………………………………………………. 54
Soft Skills Rubric …………………………………………………………………………. 55
Answers……………………………………………………………………………………. 56
Reference Pages …………………………………………………………………………… 60
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Introduction
In this final unit, we will determine the area and perimeter of a variety of two dimensional
figures comprised of triangles, quadrilaterals, and/or circles. We will also measure and calculate
volumes and surface areas of three dimensional figures comprised of cylinders, prisms, cones
and/or spheres.
Negotiated Completion Date
After working for a few days on this unit, sit down with your instructor and negotiate a
completion date for this unit.
Start Date:
_________________
Completion Date:
_________________
Instructor Signature: __________________________
Student Signature:
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The Big Picture
The following flow chart shows the six required units and the four optional units (choose two of the four)
in Level IV Graduate Math. These have been presented in a suggested order. Instructors and students
may choose to alter this order to best serve the needs of the learner. Not all of the units take the same
amount of time to complete. For example, the Consumer Finance Unit is quite short. By contrast, the
Graphs and Function Unit and the Measurement Unit (A and B) tend to take the greatest amount of time.
Math in the Real World Unit (Required)
 Fractions, decimals, percentages, ratios, proportions, and signed
numbers in real world applications
 Math Games and Puzzles
Solving Equations Unit (Required)
 Solve and check equations of the form Ax  B  Cx  D ,
A  Bx 2  C , and A  Bx 3  C .
Consumer Finance Unit (Required)
 Simple Interest and Compound Interest
 TVM Solver (Loans and Investments)
Graphs and Functions Unit (Required)
 Understanding Graphs
 Linear Functions and Line of Best Fit
Measurement Unit (Required)
 Part A: Imperial and Metric Measures
 Part B: Perimeter, Area and Volume
Angles and Triangles Unit (Required)
 Angle and Line Relationships
 Similar Triangles
 Pythagorean Theorem
Choose two of the four.
Linear
Functions and
Systems of
Equations Unit
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Trigonometry
Unit
iii
Descriptive
Statistics Unit
Numeracy Unit
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Course Timelines
Graduate Level IV Math is a two credit course within the Adult Learning Program. As a two
credit course, learners are expected to complete 200 hours of course material. Since most ALP
math classes meet for 6 hours each week, the course should be completed within 35 weeks. The
curriculum developers have worked diligently to ensure that the course can be completed within
this time span. Below you will find a chart containing the unit names and suggested completion
times. The hours listed are classroom hours.
Unit Name
Minimum
Completion Time
in Hours
24
20
15
25
22
14
18
18
Total: 156 hours
Math in the Real World Unit
Solving Equations Unit
Consumer Finance Unit
Graphs and Functions Unit
Measurement Unit (A & B)
Angles and Triangles Unit
Selected Unit #1
Selected Unit #2
Maximum
Completion Time
in Hours
34
28
18
30
30
16
22
22
Total: 200 hours
As one can see, this course covers numerous topics and for this reason may seem daunting. You
can complete this course in a timely manner if you manage your time wisely, remain focused,
and seek assistance from your instructor when needed.
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Introduction to Perimeter and Area
Perimeter is the distance around the outside of a figure. To calculate the perimeter we merely
add the lengths of all sides.
Area is the number of square units needed to cover the inside of a two dimensional object.
Example 1
If each of the squares that comprise each of these figures is 1 cm by 1 cm (or 1 cm2), then
determine the perimeter and area of each figure.
(a)
(b)
(c)
(d)
(e)
(f)
Answers:
(a) The length of the rectangle is 4 cm. The width is 1 cm.
Perimeter  4  1  4  1
= 10 cm
The rectangle is comprised of four squares, each having an area of 1 cm2. The area of the
rectangle is therefore 4 cm2.
(b) Perimeter  3  2  3  2
= 10 cm
Area = 6 cm2
(c) Perimeter  3  1  1  1  1  1  1  3
= 12 cm
Area = 6 cm2
(d) Perimeter = 14 cm
Area = 6 cm2
(e) Perimeter = 14 cm
Area = 7 cm2
(f) Perimeter = 16 cm
Area = 9 cm2
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Questions
1. The square, shown on the right, measures 1 centimetre by 1 centimetre.
Therefore, it has an area of 1 square centimetre (cm2). Use this information to
figure out the area and perimeter of the figures below.
(a)
Perimeter =
Area =
(b)
Perimeter =
Area =
(c)
Perimeter =
Area =
(d)
Perimeter =
Area =
(e)
Perimeter =
Area =
(f)
Perimeter =
Area =
(g)
Perimeter =
Area =
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(h)
Perimeter =
Area =
(i)
Perimeter =
Area =
(j)
Perimeter =
Area =
2. True or False:
(a) If two figures have the same perimeter, then they must have the same area.
True False
(b) If two figures have the same area, then they must have the same perimeter.
True False
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Perimeter and Area Formulas
In the previous section, we looked at figures that we made up of 1 cm by 1 cm squares. To find
the areas of these figures, we merely counted the number of squares. To find the perimeter we
merely counted the number of 1 cm units around the outside of the object. With all of the
questions we encountered in the previous section, the area and perimeter worked out to whole
number answers.
In reality, most things in the world cannot be divided easily into 1 cm by 1 cm squares. We
cannot simply count these squares to determine the area, or count the edges of squares on the
exterior of a figure to determine perimeter. Consider the examples below where we have placed
three different figures on a grid made up of 1 cm by 1 cm squares.
Rectangle
Triangle
Circle
The perimeter of this figure is
slightly less than 12 cm, but
how much less.
The perimeter of this triangle
would be easy to find if we
knew the length of the largest
side. Without that length, we
could estimate that the
perimeter is somewhere
between 10 cm and 14 cm.
The distance around the
outside of a circle is referred
to as the circumference,
rather than the perimeter. In
this case, we could say that
the circumference is slightly
less than 12 cm, but we don’t
know how much less.
The area of this rectangle is
obviously less than 9 cm2, but
how much less?
The area of this traingle is
obviously less than 9 cm2, but
how much less?
The area of this circle is
obviously less than 9 cm2, but
how much less?
Mathematicians studied these basics geometric figures and came up with formulas that allow us
to accurately determine the area of these figures.
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The First Three Sets of Formulas
Rectangle
Triangle
Circle
a
width, w
c
height, h
length, l
radius, r
base, b
P  2(l  w)
P  abc
Alw
A
C  d or C  2r
A  r 2
1
bh
2
The height is always at a right
angle to the base.
The diameter, represented by the
letter d, is the distance from one
side of the circle to the other side
which passes through the center
of the circle.
Pi, represented by the symbol  ,
is approximately equal to 3.14.
Example
Find the area and the distance around the outside of each of these figures. If an angle looks to be
a right angle (90o) in these diagrams, assume that it is a right angle.
(a)
(b)
(c)
11.7 ft
8.5 cm
4.5 cm
2.5 ft
12.8 mm
7.2 cm
Answers:
(a)
11.7 ft
2.5 ft
(b)
8.5 cm
P  2(l  w)
P  2(11.7  2.5)
A  lw
A  11.7  2.5
P  28.4 ft
A  29.25 ft 2
P  abc
1
A  bh
2
1
A   7.2  4.5
2
A  16.2 cm2
4.5 cm
P  8.5  4.5  7.2
7.2 cm
P  20.2 cm
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(c)
12.8 mm
Diameter = 12.8 mm, therefore radius = 6.4 mm
A  r 2
C  d
2
A  3.14  6.4 
C  3.1412.8
C  40.2 mm
A  128.6 mm2
Questions
If an angle looks to be a right angle (90o) in these diagrams, assume that it is a right angle.
1. Determine the area and perimeter (or circumference in the case of circles) of each of these
figures. Show all your work and remember to include the appropriate units of measure.
(a)
(b)
43 mm
26 cm
25 mm
(c)
(d)
2.5 ft
.
10.1 m
8.3 m
5.8 m
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(e)
(f)
8.3 in
5.5 in
14.3 cm
5.8 cm
9.1 in
6.2 in
2. Determine the perimeter and area of each of these figures in centimetres and square
centimetres respectively. You will have to use a ruler to determine the appropriate
dimensions so that the area can be calculated. Show your work.
(a)
(b)
(c)
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More Perimeter and Area Formulas
In the previous section, we looked at the perimeter and area formulas for rectangles, triangles
and circles. In this section, we will also look at the perimeter and area formulas for squares,
parallelograms and trapezoids.
Perimeter
Area
P  2(l  w)
Alw
P  4s
A  s2
Rectangle (two pairs of parallel sides,
and four right angles)
width, w
length, l
Square (two pairs of parallel sides,
four right angles, and all sides
of equal length)
side length, s
Triangle
c
h
a
P  abc
A
1
bh
2
b
Parallelogram (two pairs of parallel
sides)
P  2(a  b)
h
a
A  bh
b
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Perimeter
Area
Trapezoid (one pair of parallel sides)
a
c
h
A
P  abcd
d
a  b h
2
b
Circle
r
C  d or C  2r
d
A  r 2
Example
Determine the area and perimeter (circumference if it is a circle) of each of these figures.
Diagrams are not to scale.
(a)
(b)
(c)
4000 cm
15 in.
24 m
4.8 cm
3 ft.
32 m
(d)
(e)
(f)
150 cm
5 in.
7 in.
15.3 mm
8 in.
8 in.
1.9 m
15.3 mm
5 in.
3.1 m
1.1 m
1.7 m
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Answers:
(a) Triangle
Convert 4000 cm to 40 m.
P  abc
P  24  40  32
P  96 m
(b) Rectangle
Convert 3 ft. to 36 in.
P  2(l  w)
P  236  15
P  251
P  102 in
(c) Circle
1
bh
2
1
A  32 24 
2
A  384 m 2
A
Alw
A  36  15
A  540 in 2
C  d
C  3.144.8
C  15.1 cm
A  r 2
A  3.142.4 
2
A  3.145.76 
A  18.1 cm 2
(d) Parallelogram
P  2(a  b)
P  28  5
P  213
P  26 in
(e) Trapezoid
Convert 150 cm to 1.5 m
P  abcd
P  1.1  1.9  3.1  1.7
P  4s
P  415.3
P  61.2 mm
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A  5 7
A  35 in 2
P  7.8 m
(f) Square
A  bh
10
A
a  b h
2
1.1  3.11.5
A
2
4.21.5
A
2
A  3.2 m 2
A  s2
A  15.3
2
A  234 mm 2
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Questions
If an angle looks to be a right angle (90o) in these diagrams, assume that it is a right angle.
1. Determine the area and perimeter (circumference if it is a circle) of each of these figures.
Diagrams are not to scale. Please note that in some cases we have used different units of
measure. Convert to the desired units of measure before doing any calculations. The desired
units are stated just above the diagram.
(a)
Desired Units: cm and cm2
(b)
Desired Units: m and m2
1100 cm
8m
119 mm
9m
9m
1100 cm
7.3 cm
(c)
Desired Units: in and in2
(d)
Desired Units: yd and yd2
7 yd
2 ft
13 yd
10 in
30 ft
11 yd
26 in
20 yd
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(e)
(f)
Desired Units: cm and cm2
9 ft
81 cm
23 cm
540 mm
(g)
Desired Units: mm and mm2
380 mm
(h)
4.2 cm
53 mm
47 mm
53 mm
7.2 yd
4.2 cm
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(i)
Desired Units: cm and cm2
(j)
Desired Units: ft and ft2
2.9 cm
6 ft
20 mm
5.1 cm
2 yd
3.5 cm
27 mm
2. Determine the perimeter and area of each of these figures in millimetres and square
millimetres respectively. You will have to use a ruler to determine the appropriate
dimensions so that the area can be calculated. Show your work.
(a)
(b)
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(c)
(d)
(e)
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Perimeter and Area of Composite 2D Shapes
Most shapes that you encounter in the real world are not a simple square, rectangle, trapezoid,
parallelogram, triangle, or circle. Most shapes are combinations of these basic shapes. Shapes
that are formed by the combinations of our basic shapes are referred to as composite shapes. In
this section we will learn how to determine the perimeter and area of such shapes.
Example 1
Determine the perimeter and area of the following figures.
(a)
(b)
9 cm
5m
11 cm
2m
6 cm
6.5 m
3.5 m
15 cm
6m
(c) This figure contains a quarter circle.
(d) This figure contains a half circle.
7 in
70 ft
3 in
28 ft
14 ft
23 ft
10 in
36 ft
24 ft
33 ft
Answers:
With many of these composite shape questions it may first appear that you are missing some
critical measurements, however, they can often be found by examining the other
measurements that have been provided.
(a) Composite Shape: Two Rectangles
9 cm
Two Missing Measurements:
15 – 9 = 6
11 – 6 = 5
5 cm
11 cm
6 cm
Perimeter = 11 + 9 + 5 + 6 + 6 + 15
= 52 cm
6 cm
Area (top rectangle)  l  w
 95
 45 cm2
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Area (bottom rectangle)  l  w
 15 6
 90 cm2
Area (total) = 45 + 90
= 135 cm2
(b) Composite Shape: Triangle and Rectangle
Two Missing Measurements:
6–2=4
6.5 – 3.5 = 3
2m
Perimeter = 6.5 + 6 + 3.5 + 2 + 5
= 23 m
Area (triangle)
5m
3m
6.5 m
4m
3.5 m
1
 bh
2
1
 43
2
 6 m2
6m
Area (rectangle)  l  w
 6  3.5
 21 m2
Area (total) = 6 + 21
= 27 m2
(c) Composite Shape: Rectangle and Quarter Circle
C  2r
 2  3.14  3
 18.84 in (full circle)
7 in
3 in
1
of 18.84
4
 0.25  18.84
 4.71 in (quarter circle)
10 in
Perimeter = 4.71 + 7 + 3 + 10
= 24.71 in
Area (circle)  r 2
2
 3.143
= 28.26 in2 (full circle)
Area (rectangle)  l  w
 73
= 21 in2
1
of 28.26  7.07 in2 (quarter circle)
4
Area (total) = 7.07 + 21 = 28.07 in2
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(d) Composite Shape: Half Circle and Trapezoid
Missing Measurement
70 – (28 + 14) = 28
70 ft
C  d
 3.14  28
 87.92 ft (full circle)
1
of 87.92
2
 0.5  87.92
 43.96 ft (half circle)
28 ft
28 ft
14 ft
23 ft
36 ft
24 ft
33 ft
Perimeter = 43.96 + 28 + 36 + 33 + 24 + 14
= 178.96 ft
Area (circle)  r 2
2
 3.1414
 3.14  196
 615.44 ft 2 (full circle)
1
of 615.44
2
 0.5  615.44
 307.72 ft 2 (half circle)
Area (trapezoid) 
a  b h
2
70  3323

2
 1184.5 ft 2
Area (total) = 307.72 + 1184.5
= 1492.22 ft2
Example 2
Determine the area of the shaded region in the following figure.
52 mm
22 mm
33 mm
28 mm
33 mm
52 mm
Answer:
In this case we need to find the area of the parallelogram and subtract the area of the circle.
Area (parallelogram)  bh
 52  28
 1456 mm 2
Area (circle)  r 2
2
 3.1411
 379.94 mm 2
Area (shaped portion) = 1456 -379.94
= 1076.06 mm2
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Example 3
Tylena is laying tile in her bathroom. The tile, mortar, and grout costs $4.25 per square foot.
The L-shaped floor is comprised of a rectangle measuring 6 feet by 3 feet connected to a square
with a side length of 2 feet. How much will it cost to tile the bathroom floor?
Answer:
This question is dealing with area, rather than perimeter.
We know this because the cost is dependent on the
amount of floor we must cover with tile. That is why we
are given the cost “per square foot.”
Based on the written description, a diagram of the floor
can be drawn.
Rectangle
Alw
A  63
Square
A  s2
A  18 ft 2
A  4 ft 2
3 ft
6 ft
2 ft
2 ft
A  22
Cost = 22 ft2  $4.25/ft2
= $93.50
Area (total) = 18 + 4 = 22 ft2
Questions
1. Determine the perimeter and area of each of these 2D composite figures.
(a)
(b) This figure is a half circle.
8m
11 m
6m
32.6 cm
20 m
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(c)
(d)
110 in
33 ft
28 in
25 ft
18 ft
48 in
31 ft
22 ft
14 ft
11 ft
66 in
22 ft
(e)
(f)
6.2 cm
12.0 m
4.0 m
6.0 m
5.0 cm
5.3 cm
5.0 m
4.0 m
1.3 cm
9.0 m
9.2 cm
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(g) This figure contains a half circle.
(h) This figure contains two quarter circles.
40 m
155 mm
20 m
16 m
20 m
15 m
40 mm
11 m
75 mm
(i) This figure contains two half circles.
(j) This figure contains one quarter circle.
52.8 cm
7.0 ft
3.0 ft
11.0 ft
71.4 cm
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2. Determine the area of each of the shaded figures.
(a)
12.0 cm
10.0 cm
20.0 cm
6.0 cm
(b)
17.4 m
8.8 m
3.8 m
7.6 m
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(c)
13.4 cm
6.8 cm
12.8 cm
11.5 cm
(d)
8.0 in
20.0 in
12.0 in
12.0 in
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3. How many plants, spaced every 10 cm, are needed to surround a circular walkway with a
diameter of 4.0 metres?
4. Jacob is laying hardwood in his child's room. The room is rectangular and measures 13 feet
by 12 feet. He will also be laying hardwood in the closet that measures 5 feet by 2.5 feet. If
the hardwood costs $4 per square foot, how much will the hardwood cost to do the room and
closet?
5. A roll of sod (i.e. grass) covers 9 ft2. How many rolls
of sod would one need to surround a rectangular
building measuring 42 feet by 28 feet with a strip that
has a uniform width of 15 feet.
building
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Introduction to Surface Area
Surface area is the number of square units needed to cover the surface of a three dimensional
object.
Let’s Consider a Box
This box (i.e. a rectangular-based prism), shown on the right, has
been created by joining cubes measuring 1 cm by 1 cm by 1 cm.
There are six faces (front, back, right, left, top and bottom) to this
prism. The left and right faces are identical, the top and bottom
faces are identical, and the front and back faces are identical.
Surface Area of Our Prism
 Front and back faces each have an area of 12 cm2.
 Right and left faces each have an area of 8 cm2.
 Top and bottom faces each have an area of 6 cm2.
 Total Surface Area = 12 + 12 + 8 + 8 + 6 + 6
= 52 cm2
In reality, most things in the world cannot be easily divided into 1 cm by 1 cm by 1 cm cubes.
We cannot simply count the faces of these cubes to determine the surface area. For this reason,
mathematicians studied a variety of 3D shapes and developed formulas to determine surface
areas. Before we can look at these formulas, we have to learn about these basic three
dimensional shapes.
Prisms
 A solid three dimensional object that has two identical ends and all flat sides.
 The cross section is the same all along its length.
 The shape of the ends gives the prism its name. For example a triangular-based prism
has triangular ends.
 For this course we are only going to deal with right prisms; those whose side faces are
perpendicular to the base faces
Rectangular-based
Prism
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Cube: a special
type of squarebased prism
Triangular-based
Prism
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Cylinders
 A solid three dimensional object with two identical flat ends that are
circular and one curved side.
 It has the same cross-section from one end to the other.
Pyramids
 A solid three dimensional object where the base is a polygon (a straight-sided flat shape)
and the sides are triangles which meet at the top (the apex).
 The shape of the base gives the prism its name. For example a square-based prism has a
square base.
 For this course we are only going to deal with right pyramids those whose apex is directly
above the center of the base.
Square-based
Pyramid
Rectangular-based
Pyramid
Triangular-based
Pyramid
Cones
 A solid three dimensional object that has a circular base and one apex
 For this course we are only going to deal with right cones; those
whose vertex is directly above the center of the circular base.
Spheres
 A three dimensional object shaped like a ball where every point on its
surface is the same distance from the center.
Surface Area Formulas
Before we can look at the formulas for surface area, there is one other item we have examine;
nets of solids. Understanding nets helps one when working on surface area problems.
Nets of a Solid
A net of a solid is created by making cuts along some edges of a solid and opening it up to form
a plane figure. Based on this definition, a net can be folded up to create the original 3D solid.
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Original Solid
Corresponding Net
When working out the surface area of a solid we are just finding the area of each of the surfaces
on the corresponding net.
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If we consider a cube, its surface area is found by adding the areas of the six squares that make
up the six faces of the cube.
A6
A1
A2
Surface Area  A1  A2  A3  A4  A5  A6
A3
 s2  s2  s2  s2  s2  s2
A4
 6s 2
A5
Now we will apply the same line of thinking to determining the surface area of a triangular-based
prism.
A5
A1
A2
Surface Area  A1  A2  A3  A4  A5
A3
In this case you must find the area of three
different rectangles, the areas of two
identical triangles, and then add those areas
together.
A4
Most people find it easier to work out the surface area without using a defined formula, generally
because they understand it and remember it better if they take the common sense approach.
Example 1
Determine the surface area of the following prism.
Height = 3 m
Width = 2 m
Answer:
Method 1 – Common Sense Approach
Front and Back Faces
Right and Left Faces
(Both are 5 m by 3m)
(Both are 2 m by 3 m)
Alw
Alw
A  53
A  3 2
A = 15 m2
A = 6 m2
Length = 5 m
Top and Bottom Faces
(Both are 5 m by 2 m)
Alw
A  5 2
A = 10 m2
Surface Area = 15 + 15 + 6 + 6 + 10 + 10
= 62 m2
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Method 2 – Formula
Surface Area  2lw  2lh  2wh
 252  253  223
 20  30  12
= 62 m2
There are shapes, however, that cannot be solved using the common sense approach (e.g.
spheres, cones,…)
Surface Area Formula
Cube
A  6s 2
s
Rectangular-based Prism
h
A  2lw  2lh  2wh
w
l
Cylinder
(enclosed on the top and
bottom)
r
h
Cone
(enclosed on the bottom)
s
A  2rh  2r 2
A  rs  r 2
s – slant height
r
Sphere
r
A  4r 2


If the cylinder is not enclosed on the top and bottom (i.e. it is merely an open tube), then
the surface area formula is A  2rh .
If the cone is not enclosed on the top, then the surface area formula is A  rs .
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Example 2
Determine the surface area of the basketball if its diameter is 10 inches.
Answer:
Sphere with a diameter of 10 inches (or a radius of 5 inches)
A  4r 2
A  43.145
2
A  43.1425
A  314 in 2
40 mm
Example 3
Determine the surface area of this cylindrically-shaped cup.
Answer:
Since we are dealing with a cup, we know that the top is
open. That means the surface area formula will not be:
A  2rh  2r 2 (two circular surfaces)
Rather the formula will be:
(only one circular surface)
A  2rh  r 2
95 mm
Substitute 40 in for r and 95 in for h.
A  2rh  r 2
A  23.144095  3.1440
2
A  23864  5024
A  28888 mm 2
Example 4
Determine the surface area of following rectangular-based
pyramid. Please note that the heights of the triangular faces
are identified with a dotted line.
Answer:
Area of Rectangular Base
Alw
A  36  18
41 m
44 cm
36 cm
A  648 cm 2
18 cm
Area of Triangular Face with a Base of 36 cm and a Height of 41 cm
1
A  bh
2
1
A  36 41
2
A  738 cm 2
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Area of Triangular Face with a Base of 18 cm and a Height of 44 cm
1
A  bh
2
1
A  1844 
2
A  396 cm 2
There are five faces on this figure (one rectangular face, two identical triangular faces, and
another two identical triangular faces).
Surface Area  A1  A2  A3  A4  A5
 648  738  738  396  396
 2916 cm 2
Questions
1. Both of these rectangular-based prisms is created using cubes measuring 1 cm by 1 cm by 1
cm. Determine the surface area of each of these prisms without doing any formal
calculations.
(a)
(b)
2. Draw the nets for the following figures.
(a)
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(b)
(c)
(d)
3. Calculate the surface area of each of these figures.
(a)
5.8 m
6.1 m
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(b) Note: The figure is enclosed on both ends.
10 ft
16 ft
(c) Note: The figure is enclosed at the top.
9.8 cm
20.3 cm
(d)
84 in
(e) Cube
4.6 m
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(f)
29 yd
20 yd
25 yd
21 yd
(g) Square-based Pyramid: Please note that the height of the triangular face is identified with
a dotted line.
23.6 cm
14.9 cm
(h) Triangular-based Pyramid: The triangular base is an equilateral triangle (i.e. all sides are
of equal length). Please note that the heights of the triangular faces are identified with a
dotted line.
3.2 m
2.8 m
5.9 m
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(i) Note: This figure is open on the top.
12.4 m
13.1 m
(j) Square-based Prism
18 ft
12 ft
(k) Triangular-based Prism: The triangular base is an isosceles triangle (i.e. two sides are of
equal length). Please note that the heights of the triangular faces are identified with a
dotted line.
2.6 m
2.5 m
1.4 m
4.4 m
2.6 m
4.3 m
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Introduction to Volume
In the previous section we learned that surface area is the number of square units needed to
cover the surface of a three dimensional object.
In this section we will learn that volume is the number of cubic units needed to fill the inside of a
three dimensional object.
Let’s Consider a Box
This box (i.e. a rectangular-based prism), shown on the right,
has been created by joining cubes measuring 1 cm by 1 cm by 1
cm (or each cube has a volume of 1 cm3).
Volume of Our Prism
Since this prism is comprised 24 cubes each having a
volume of 1 cm3, then the volume of the prism is 24 cm3.
In reality, most things in the world cannot be easily divided into 1 cm by 1 cm by 1 cm cubes.
We cannot simply count these cubes to determine the volume. For this reason, mathematicians
studied a variety of 3D shapes and developed formulas to determine their volumes.
Volume Formulas
For prisms and cylinders, the volume is found by taking the area of the base and multiplying by
the height.
Volume prism or cylinder = Area of Base  Height
Based on this the volume of a square-based prism will be:
Volume prism or cylinder = Area of Base  Height
Volume square-based prism = Area of Square Base  Height
Volume square-based prism  s 2  h
Volume square-based prism  s 2 h
Based on this the volume of a triangular-based prism will be:
Volume prism or cylinder = Area of Base  Height
Volume triangular-based prism = Area of Triangular Base  Height
1
Volume triangular-based prism  bh  H
2
1
Volume triangular-based prism  bhH
We changed height of prism to H so it would not
2
be confused with the height, h, of the triangle
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Based on this the volume of a cylinder will be:
Volume prism or cylinder = Area of Base  Height
Volume cylinder = Area of Circular Base  Height
Volume cylinder  r 2  h
Volume cylinder  r 2 h
In general the volume formula for prisms is written in the form:
V  Bh where B is the base area
Before we can look at the formulas for volume for pyramids and cones, we have to examine the
relationships between volumes of prisms and the volumes of pyramids, and similarly between the
volumes of cylinders and the volumes of cones. Look at the figures below. We have a squarebased pyramid within a square-based prism, a triangular-based pyramid within a triangular-based
prism, and finally a cone within a cylinder.
When mathematicians studied the relationships between these figures, they discovered
something very interesting.
They learned that the volume of the pyramid or cone is one-third that of the corresponding prism
or cylinder (i.e. a figure with the same base area). Based on this, we can conclude the following.
1
 Volume square-based prism
3
1
 s2h
3
Volume square-based pyramid =
1
 Volume triangular-based prism
3
11

  bhH 
3 2

1
 bhH
6
Volume triangular-based pyramid =
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1
 Volume cylinder
3
1
 r 2 h
3
Volume cone =
In general the volume formula for pyramids is written in the form:
V 
1
Bh where B is the base area
3
Volume Formula
V  Bh
- where B is the base area
Prism
h
h
Rectangular-based Prism:
V  lwh
Cube:
V  s3
Pyramid
h
Cylinder
1
Bh
3
- where B is the base area
V
h
r
V  r 2 h
h
Cone
1
V  r 2 h
3
h
r
Sphere
r
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V  r 3
3
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Example 1
A particular fish tank is the shape of a rectangular-based prism. How much water will this tank
hold if it has a height of 40 cm, a width of 35 cm, and a length of 70 cm?
Answer:
Rectangular-based Prism:
V  lwh
V  703540
V  98000 cm 3
Example 2
A company must make solid steel ball bearings that have a diameter of 23.8 mm. How much
steel is in each ball bearing?
Answer:
Sphere:
Diameter = 23.8 mm, therefore radius = 11.9 mm
4
V  r 3
3
4
3
V  3.1411.9
3
V  7055.2 mm3 of steel
Example 3
Determine the volume of each figure.
(a)
(b)
6 ft
9.84 m
4.26 m
8 ft
(c)
(d)
6.5 in
11.2 cm
9.4 cm
14.8 cm
14.25 in
9.75 in
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Answers:
(a) Cylinder: Diameter = 6 ft, therefore Radius = 3 ft
V  r 2 h
V  3.143 8
2
V  226.1 ft 3
(b) Cone
1
V  r 2 h
3
1
2
V  3.144.26 9.84
3
V  186.9 m 3
(c) Triangular-based Prism
V  Bh where B is the base area
1
We changed height of prism to H so it would not
V  bh  H
2
be confused with the height, h, of the triangle
1
V  9.7514.256.5
2
V  451.5 in 3
(d) Rectangular-based Pyramid
1
V  Bh where B is the base area
3
1
V   lw  h
3
1
V  11.29.414.8
3
V  519.4 cm 3
Example 4
A concrete footing with a trapezoidal
cross-section is being poured. Its
measurements are shown on the
accompanying diagram. How much
concrete is needed for the footing?
0.60 m
0.40 m
Answer:
Trapezoidal-based Prism:
V  Bh where B is the base area
 a  b h 
V 
H
 2 
 0.60  0.950.40 
V 
3.85
2


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0.95 m
3.85 m
We changed height of prism to H so it would not
be confused with the height, h, of the trapezoid.
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V  0.313.85
V  1.19 m 3 of concrete
Questions
1. Both of these rectangular-based prisms is created using cubes measuring 1 cm by 1 cm by 1
cm (or having a volume of 1 cm3). Determine the volume of each of these prisms
(a)
(b)
2. Calculate the volume of each of these figures.
(a)
5.8 m
6.1 m
3.2 m
(b)
10 ft
16 ft
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(c)
9.8 cm
17.8 cm
(d)
84 in
(e) Cube
4.6 m
(f)
29 yd
Warning:
There is a measurement in this diagram which is
not needed to work out the volume of the figure.
20 yd
25 yd
21 yd
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(g) Square-based Pyramid: Please note that the height of the pyramid is 22.4 cm.
22.4 cm
14.9 cm
(h) Triangular-based Pyramid: The triangular base is an equilateral triangle (i.e. all sides are
of equal length). The height of the pyramid is 5.8 m.
3.2 m
2.8 m
5.8 m
(i) Parallelogram-based Prism
12.4 cm
32.8 cm
14.7 cm
3. Brittany has a cylindrically shaped above ground pool. It has a diameter of 15 feet. The
maximum depth of the water in the pool is 4 feet. How much water is needed to fill the
pool?
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More Surface Area and Volume
Example 1
You are given the following figure which is enclosed at the
base.
(a) Determine its surface area.
(b) Convert the surface area from square yards to square feet.
(c) Determine its volume.
(d) Convert the volume from cubic yards to cubic metres.
Answers:
(a) Cone
A  rs  r 2
5 yd
4 yd
6 yd
A  3.1435  3.143
A  47.1  28.26
A = 75.4 yd2
2
(b) When making conversions, use dimensional analysis (a technique we learned earlier in
this unit.)
yd2  ft2
ft 2
yd2  2  ft 2
yd
75.4 yd2 
9 ft 2
678.6 2

ft  678.6 ft 2
2
1
1 yd
(c) Cone
1
V  r 2 h
3
1
2
V  3.143 4
3
V  37.7 yd3
(d) Use dimensional analysis.
yd3  m3
m3
yd3  3  m 3
yd
37.7 yd3 
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0.7646 m 3 28.8 3

m  28.8 m 3
3
1
1 yd
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Example 2
You are given the following composite
3D figure. It’s comprised of a squarebased prism and a square-based pyramid.
It is 157 mm high. The slant height for
one of the triangular faces is 136 mm.
(a) Determine its surface area.
(b) Convert the surface area from square
millimetres to square inches.
(c) Determine its volume.
(d) Convert the volume from cubic
millimetres to cubic feet.
157 mm
136 mm
39 mm
134 mm
Answers:
(a) This figure has four identical triangular faces, four identical rectangular faces, and one
square face.
Area of One
Trianglular Face
1
A  bh
2
1
A  134 136 
2
A  9112
Area of One
Rectangular Face
A  lw
A  13439
A  5226
Area of Square Face
(Base of Figure)
A  s2
A  134 2
A  17956
Surface Area = 9112 + 9112 + 9112 + 9112 + 5226 + 5226 + 5226 + 5226 + 17956
= 75 308 mm2
(b) Use dimensional analysis.
mm2  cm2  in2
cm 2 in 2
mm 2 

 in 2
mm 2 cm 2
1 cm 2
1 in 2
75308 2
75308 mm 2 


in  116.8 in 2
2
2
645
100 mm
6.45 cm
(c) This figure is comprised of a square-based prism with a height of 39 mm and a squarebased pyramid with a height of 118 mm (157 – 39).
V  Bh
V  s h
2
V  134 2  39
V  700284
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Bh
3
1
V   s2  h
3
1
2
V  134  118
3
V  706269
V 
44
Volume = 700284 + 706269
= 1 406 553 mm3
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(d) Use dimensional analysis.
mm3  cm3  in3  ft3
cm 3 in 3 ft 3
mm 3 

 3  ft 3
3
3
mm cm in
1 cm 3
1 in 3
1 ft 3
1406553 3
1406553 mm 3 



ft  0.04966 ft 3
3
3
3
28321920
1000 mm 16.39 cm 1728 in
Example 3
A large water tank is created by joining two
hemispheres (two halves of a sphere) to a
cylinder.
(a) One gallon of paint covers 30 m2 with
one coat. How much paint is required to
give this tank two coats of paint?
(b) What is the maximum amount of water
this tank can hold?
Answer:
(a) Cylinder (open at both ends)
A  2rh
A  23.141.96.7 
A  79.9
6.7 m
3.8 m
Sphere (created by combining two hemispheres)
A  4r 2
A  43.141.9
A  45.3
2
Surface Area = 79.9 + 45.3
= 125.2 m2
Two coats will be covering 250.4 m2, therefore we will need 8.34 gallons of paint.
(b) Cylinder
V  r 2 h
V  3.141.9 6.7 
V  75.9
2
Sphere
4
V  r 3
3
4
3
V  3.14 1.9 
3
V  28.7
Volume = 75.9 + 28.7
= 104.6 m3 of water
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Example 4
A cube of steel having a side length of 32 mm has a hole with a
diameter of 18 mm drilled through it (See diagram.). What is the
volume of the resulting steel figure?
Answer:
We are dealing with a cube that a cylindrical piece removed
from it.
V  s3
V  r 2 h
V  32 3
V  32768
V  3.149 32
V  8139
2
Volume = 32768 – 8139
= 24629 mm3
Questions
1. You are given the following cube.
(a) Determine its surface area.
(b) Convert the surface area from in2 to cm2.
(c) Determine the figure’s volume.
(d) Convert the volume from in3 to mm3.
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2. You are given the following triangularbased prism.
(a) Determine its surface area.
(b) Convert the surface area from cm2 to
mm2.
(c) Determine the figure’s volume.
(d) Convert the volume from cm3 to in3.
5.7 cm
8.1 cm
3.8 cm
4.1 cm
11.2 cm
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3. You are given the following composite figure.
(a) Determine its surface area.
(b) Convert the surface area from m2 to in2.
(c) Determine the figure’s volume.
(d) Convert the volume from m3 to yd3.
10 m
23 m
15 m
12 m
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4. One gallon of paint covers 30 m2 with one coat. How
much paint is required to give this grain silo one coat of
paint?
4m
20 m
5. A rectangular-based prism of steel has two
holes each with a diameter of 2.4 cm drilled
through it (See diagram.). What is the volume
of the resulting steel figure?
3.8 cm
3.6 cm
8.3 cm
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Putting It Together
Questions:
1. Determine the perimeter and area of the following figures.
(a) This figure contains a half circle.
14 ft
13 ft
12 ft
19 ft
(b)
51 m
22 m
11 m
22 m
28 m
28 m
20 m
18 m
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2. A concrete deck of uniform width (8 feet) is placed
around a rectangular pool measuring 18 ft by 32 ft.
(a) If a fence is placed around the outside of the
concrete deck, how much fencing is required?
(b) What is the area of the concrete deck?
(c) Express the area of the concrete deck in square
yards.
8 ft
18 ft
Pool
32 ft
3. Determine the surface area and volume of the following figures.
(a)
29 in
54 in
49 in
61 in
64 in
88 in
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(b) This figure is comprised of eight identical isosceles triangle faces.
Height of
triangular
face: 26 cm
48 cm
20 cm
4. Determine the surface area and volume of the
following figure.
5.0 m
7.2 m
8.3 m
6.5 m
14.8 m
13.2 m
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5. (a) Determine the volume of the following
figure. The figure is created by cutting
the top portion off of a cone. The
portion that has been removed is also a
cone.
(b) Convert the volume to cubic feet.
4.0 m
4.0 m
6.0 m
12.0 m
6. Determine the amount of concrete needed to create the following pipe. Express your answer
in both cubic centimetres and cubic inches.
30 cm
40 cm
150 cm
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Post-Unit Reflections
What is the most valuable or important
thing you learned in this unit?
What part did you find most interesting or
enjoyable?
What was the most challenging part, and
how did you respond to this challenge?
How did you feel about this math topic
when you started this unit?
How do you feel about this math topic
now?
Of the skills you used in this unit, which
is your strongest skill?
What skill(s) do you feel you need to
improve, and how will you improve them?
How does what you learned in this unit fit
with your personal goals?
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Soft Skills Rubric
Look back over the module you have just completed and assess yourself using the following
rubric. Use pencil or pen and put a checkmark in the column that you think best describes your
competency for each description. I will look at how accurately you have done this and will
discuss with you any areas for improvement.
You will be better prepared for your next step, whether it is work or further education, if you are
competent in these areas by the end of the course. Keep all of these rubrics in one place and
check for improvement as you progress through the course.
Date:
Competent
demonstrates the concept
fully and consistently
Throughout this module, I…
Approaching
Competency
Developing
Competency
demonstrates the concept
most of the time
demonstrates the concept
some of the time
 Attended every class
 Let my instructor know if not
able to attend class
 Arrived on time for class
 Took necessary materials to class
 Used appropriate language for
class
 Used class time effectively
 Sustained commitment
throughout the module
 Persevered with tasks despite
difficulties
 Asked for help when needed
 Offered support/help to others
 Helped to maintain a positive
classroom environment
 Completed the module according
to negotiated timeline
 Worked effectively without close
supervision
Comments:
(Created by Alice Veenema, Kingstec Campus)
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Answers
Introduction to Perimeter and Area (pages 1 to 3)
1. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
Perimeter = 6 cm, Area = 2 cm2
Perimeter = 10 cm, Area = 5 cm2
Perimeter = 14 cm, Area = 6 cm2
Perimeter = 16 cm, Area = 8 cm2
Perimeter = 12 cm, Area = 8 cm2
Perimeter = 14 cm, Area = 8 cm2
Perimeter = 16 cm, Area = 7 cm2
Perimeter = 18 cm, Area = 9 cm2
Perimeter = 18 cm, Area = 8 cm2
Perimeter = 18 cm, Area = 11 cm2
2. (a) False
(b) False
Perimeter and Area Formulas (pages 4 to 7)
1. (a) Perimeter = 136 mm
(b) Circumference = 81.64 cm
(c) Perimeter = 24.2 m
(d) Circumference = 15.7 ft
(e) Perimeter = 40.2 cm
(f) Perimeter = 23.6 in
Area = 1075 m2
Area = 530.66 cm2
Area = 24.07 m2
Area = 19.6 ft2
Area = 82.94 cm2
Area = 25.025 in2
2. Answers will vary slightly.
(a) Perimeter = 14.4 cm
(b) Circumference = 10.99 cm
(c) Perimeter = 16.6 cm
Area = 11.27 cm2
Area = 9.62 cm2
Area = 9.92 cm2
More Perimeter and Area Formulas (pages 8 to 14)
1. (a) Perimeter = 38.4 cm
(b) Perimeter = 40 m
(c) Perimeter = 60 in
(d) Perimeter = 51 yd
(e) Circumference = 56.52 ft
(f) Perimeter = 173 cm
(g) Perimeter = 190 mm
(h) Circumference = 22.6 yd
(i) Perimeter = 13.5 cm
(j) Perimeter = 24 ft
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Area = 86.87 cm2
Area = 88 m2
Area = 120 in2
Area = 135 yd2
Area = 254.34 ft2
Area = 931.5 cm2
Area = 1974 mm2
Area = 40.7 yd2
Area = 9.585 cm2
Area = 36 ft2
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2. Answers will vary slightly.
(a) Perimeter = 148 mm
(b) Perimeter = 177 mm
(c) Perimeter = 192 mm
(d) Perimeter = 234 mm
(e) Perimeter = 164
Area = 1288 mm2
Area = 1428 mm2
Area = 2304 mm2
Area = 2886 mm2
Area = 1578.5 mm2
Perimeter and Area of Composite 2D Shapes (pages 15 to 23)
1. (a) Perimeter = 62 m
(b) Perimeter = 83.8 cm
(c) Perimeter = 316 in
(d) Perimeter = 159 ft
(e) Perimeter = 27 cm
(f) Perimeter = 44 m
(g) Perimeter = 128.0 m
(h) Perimeter = 355.6 mm
(i) Perimeter = 308.6 cm
(j) Perimeter = 34.3 ft
2
(a)
(b)
(c)
(d)
Area = 180 m2
Area = 417.1 cm2
Area = 4400 in2
Area = 1056 ft2
Area = 38.9 cm2
Area = 80 m2
Area = 716.9 m2
Area = 5512 mm2
Area = 5958.4 cm2
Area = 73.6 ft2
160 – 28.3 = 131.7 cm2
66.12 – 16.72 = 49.4 m2
77.05 + 147.2 – 36.30 = 187.95 cm2
(240 – 96) + (157 – 56.5) = 244.5 in2
3. 1256 cm therefore 125 or 126 plants
4. 168.5 ft2 therefore $674
5. 3000 ft2 therefore 334 rolls of sod (Have to round up to ensure that lawn is completely
covered.)
Introduction to Surface Area (pages 24 to 34)
1. (a) Surface Area = 32 cm2
(b) Surface Area = 76 cm2
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2. (a)
(b)
(c)
(d)
3. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
146.9 m2
659.4 ft2
926.2 cm2
22 155.8 in2
127.0 m2
210 + 210 + 725 + 525 + 500 = 2170 yd2
222.01 + 175.82 + 175.82 + 175.82 + 175.82 = 925.3 cm2
4.48 + 9.44 + 9.44 + 9.44 = 32.8 m2
255.0 m2
1152 ft2
1.75 + 3.08 + 5.59 + 5.59 = 16.01 m2
Introduction to Surface Area and Volume (pages 35 to 42)
1. (a) Volume = 12 cm3
(b) Volume = 40 cm3
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2. (a)
(c)
(e)
(g)
(i)
113.2 m3
1789.3 cm3
97.3 m3
1657.7 cm3
5978.8 cm3
(b)
(d)
(f)
(h)
1256 ft3
310 181.8 in3
5250 yd3
8.7 m3
3. 706.5 ft3
More Surface Area and Volume (pages 43 to 49)
1. (a) 18.375 in2
(c) 5.359 in3
(b) 118.5 cm2
(d) 87 834 mm3
2. (a) 145.06 cm2
(c) 87.25 cm3
(b) 14506 mm2
(d) 5.32 in3
3. (a) 866.6 m2
(c) 1997.0 m3
(b) 1 342 745 in2
(d) 2611.9 yd3
4. 20.1 gallons of paint
5. 81 cm3
Putting It Together (pages 50 to 53)
1. (a) Perimeter = 64.84 ft, Area = 30 + 168 + 56.52 = 254.52 ft2
(b) Perimeter = 139 m, Area = 242 + 586.5 = 828.5 m2
2. (a) 164 ft
(b) 1056 ft2
(c) 117.3 yd2
3. (a) Surface Area = 3456 + 1856 + 3904 + 5632 + 2866.5 + 2866.5 = 20 581 in2
Volume = 183 456 in3
(b) Surface Area = 2080 cm2
Volume = 6400 cm3
4. Surface Area = 957.2 m2, Volume = 1568.6 m3
5. (a) 263.8 m3
(b) 9316 ft3
6. 82 425 cm3 or 2030 in3
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Reference Pages
Conversions:
Length
1 inch (in.) = 2.54 centimetres (cm)
1 foot (ft) = 12 inches (in)
1 yard (yd) = 3 feet (ft)
1 mile (mi) = 5280 feet (ft)
1 metre = 100 centimetres (cm)
1 kilometre (km) = 1000 metres (m)
Weight
1 pound (lb) = 454 grams (g)
1 pound (lb) = 16 ounces (oz)
1 ton = 2000 pounds (lb)
1 kilograms (kg) = 1000 grams (g)
1 metric ton = 1000 kilograms (kg)
Liquid Capacity
1 quart (qt) = 1.137 litres (L)
1 pint (pt) = 20 fluid ounces (fl oz)
1 quart (qt) = 2 pints (pt)
1 gallon (gal) = 4 quarts (qt)
1 litre (L) = 1000 millilitres (mL)
Time
1 minute (min) = 60 seconds (sec)
1 hour (hr) = 60 minutes (min)
1 day = 24 hours (hr)
1 week = 7 days
Area
1 ft2 = 144 in2
1 yd2 = 9 ft2
1 acre = 4840 yd2
1 mi2 = 640 acres
1 cm2 = 100 mm2
1 m2 = 10 000 cm2
1 km2 = 1 000 000 m2
1 hectare (ha) = 10 000 m2
1 in2 = 6.45 cm2
1 m2 = 10.76 ft2
1 mi2 = 2.59 km2
Volume
1 ft3 = 1728 in3
1 yd3 = 27 ft3
1 m3 = 1 000 000 cm3
1 cm3 = 1000 mm3
1 in3 = 16.39 cm3
1 yd3 = 0.7646 m3
Metric System Prefixes
Prefix
Symbol
tera
giga
mega
kilo
hecto
deka
deci
centi
milli
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T
G
M
k
h
da
d
c
m
Multiple or
Submultiple
1 000 000 000 000
1 000 000 000
1 000 000
1000
100
10
0.1
0.01
0.001
Power of
10
1012
10 9
10 6
10 3
10 2
101
10 1
10 2
10 3
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Perimeter
Area
P  2(l  w)
Alw
P  4s
A  s2
Rectangle
width, w
length, l
Square
side length, s
Triangle
c
a
h
P  abc
A
1
bh
2
b
Parallelogram
h
a
P  2(a  b)
A  bh
b
a
Trapezoid
c
h
d
P  abcd
A
a  b h
2
b
Circle
C  d or C  2r
A  r 2
d
r
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Surface Area Formula
Cube
A  6s 2
s
Rectangular-based Prism
h
A  2lw  2lh  2wh
w
l
Cylinder
(enclosed on the top and
bottom)
r
h
Cone
(enclosed on the bottom)
s
A  2rh  2r 2
A  rs  r 2
s – slant height
r
Sphere
r
A  4r 2


If the cylinder is not enclosed on the top and bottom (i.e. it is merely an open tube), then
the surface area formula is A  2rh .
If the cone is not enclosed on the top, then the surface area formula is A  rs .
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Volume Formula
V  Bh
- where B is the base area
Prism
h
h
Rectangular-based Prism:
V  lwh
Cube:
V  s3
Pyramid
h
Cylinder
1
Bh
3
- where B is the base area
V
h
r
V  r 2 h
h
Cone
1
V  r 2 h
3
h
r
Sphere
r
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V  r 3
3
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