Download Explanation and Activity - Specialist High Skills Major

Contextualized Learning Activities (CLAs)
For the “other required credits” in the bundle of credits, students in a Specialist High Skills Major
program must complete learning activities that are contextualized to the knowledge and skills
relevant to the economic sector of the SHSM. Contextualized learning activities (CLAs) address
curriculum expectations in these courses.
This CLA has been created by teachers for teachers.
It has not undergone an approval process by the Ministry of Education.
Contact Information
Board
Waterloo Catholic District School Board
Development date
August 2009
Contact person
Christine Stockie
Michael Sullivan
(519) 621-4050 ext. 623
(519) 621-4050 ext. 676
(519) 621-4057
Phone
Fax
E-mail
[email protected]
[email protected]
Specialist High Skills
Major
Manufacturing
Course code and title
MBF 3C – Foundations of College Mathematics
Name of
contextualized
learning activity
Measurement in Design
Brief description of
contextualized
learning activity.
Students will review and consolidate skills involving operations with
fractions and decimals, and converting between Metric and Imperial
measures. They will apply this knowledge to real-life examples from
the manufacturing sector, while studying and learning about 2- and 3dimensional design in a practical environment.
Duration
Approximately 6 hours:
Lesson 1 - Review - Operations with Fractions (60 min)
Lesson 2 - Review –Fractions and Decimals; Conversions with Metric
and Imperial Measure (60 min)
Lesson 3 – Proportional Reasoning, Ratios and Scales (60 minutes)
Lesson 4 – Trigonometry (60 minutes)
Lesson 5 – Designing and Building a Sawhorse (120 minutes)
Page 1 of 20
Overall expectations
Geometry and Trigonometry
The student will:
1. represent, in a variety of ways, two-dimensional shapes and threedimensional figures arising from real-world applications, and solve
design problems;
2. solve problems involving trigonometry in acute triangles using the
sine law and the cosine law, including problems arising from real-world
applications.
Specific expectations
Geometry and Trigonometry
The student will:
1.1 recognize and describe real-world applications of geometric
shapes and figures, through investigation, in a variety of contexts,
and explain these applications
1.3 create plans, and patterns from physical models arising from a
variety of real-world applications, by applying the metric and
imperial systems and using design or drawing software
1.4 solve design problems that satisfy given constraints using physical
models or drawings, and state any assumptions made
2.1 solve problems, including those that arise from real-world
applications, by determining the measures of the sides and angles
of right triangles using the primary trigonometric ratios
2.4 solve problems that arise from real-world applications involving
metric and imperial measurements and that require the use of the
sine law or the cosine law in acute triangles
Essential Skills:
Reading Text
Understanding text in the form of sentences or paragraphs
Document Use
Using information displays including blueprints, signs, drawings
Numeracy
Use of numbers and quantities
Writing
Completing solutions of multi-step problem-solving questions
Continuous Learning
Ongoing process of learning and acquiring skills
Work Habits:
Thinking Skills
Cognitive ability, problem solving
Teamwork
Work willingly and cooperatively with others
Initiative
Starts work with little or no prompting
Work Habits
Punctual, time effective, and able to follow directions
Organization
Written work is well laid out and neat
Working Independently
Accomplishes tasks independently
Page 2 of 20
Instructional/Assessment Strategies
Teacher’s notes
 The Math teacher should communicate with the Manufacturing teacher on a regular
basis. Both teachers should be kept up to date on developments that correspond to
each other’s course.
 The teacher should become familiar with the use of mathematics in the Manufacturing
sector.
 Providing applicable real life examples from the manufacturing sector can be beneficial
for student learning.
 Constant diagnostic feedback is important for consistent learning and student
development (ie. Through use of student worksheets #1-4).
 If the class is a split group (not all SHSM students) it may be advantageous to group the
SHSM students together, however, this CLA has benefits for all MBF students, not just
those enrolled in the SHSM program.
 Allow students to supplement their learning with applicable computer programs.
Context
This CLA is designed for students that plan on entering an apprenticeship or college program in
the Manufacturing sector.
Assessment and Evaluation of Student Achievement
Strategies/Tasks
1. Operations with Fractions
Purpose
Assessment for Learning (diagnostic, formative) or
Assessment of Learning (summative, evaluation)
Diagnostic Assessment (give consistent feedback on student
Progress)
2. Fractions & Decimals; Conversions
with Metric and Imperial Measures
Formative Assessment (give consistent feedback on student
Progress)
3. Proportional Reasoning and Scale
Diagrams
Formative Assessment (give consistent feedback on student
Progress)
4. Trigonometry
Formative Assessment (give consistent feedback on student
Progress)
5. Design and Build a Scale Model of a Summative Assessment (see rubric in Appendix B)
Sawhorse
Additional Notes/Comments/Explanations
Both Mathematics Text books and Manufacturing Texts and Manuals are a great source for additional
questions, practice, and examples.
See Attachments for further details
Page 3 of 20
Resources
Smith, Robert D. Mathematics for Machine Technology, 4th Edition, Delmar Publishers,1999
Cooke, Gordon et al, Pearson Math 11, Pearson Education Canada, 2008
Accommodations
 Individual Education Plans (IEP) should be followed at all times. Be sure to consult
the SERT for additional information and suggestions;
 additional time may be needed for quizzes, tests, assignments, and in-class work;
 the activities and lessons outlined in this CLA allow for flexibility in the delivery of the
material. Alternating teaching strategies can help students who are not progressing at
the appropriate level;
 font can be increased for those students that have a sight issue;
 class rules, behaviours, and due dates should be posted in the classroom and talked
about so that all students are aware of the expectations;
 if possible, more individual instruction time can be allotted to students in need;
 can account for student work habits when considering assignments;
 provide additional work for advanced students;
 provide time for peer-to-peer teaching;
 use audio aids if needed;
 provide additional assessment opportunities that are geared towards students
strengths;
 if available, many computer programs can be used to supplement student learning.
List of Attachments










Operations with Fractions lesson
Student Worksheet #1
Imperial and Metric Measurement lesson
Metric/Imperial Conversion Table and Student Worksheet #2
Proportional Reasoning and Scale Diagrams lesson
Student Worksheet #3
Trigonometry lesson
Student Worksheet #4
Designing and Building a Sawhorse Assignment
Sawhorse Rubric
Page 4 of 20
LESSON 1: Operations with Fractions
In the manufacturing sector, workers are expected to have a solid working knowledge of both fractions and
decimals, and to be able to convert between the two types of numbers. It is important that students review
their basic skills of putting fractions in lowest terms, finding equivalent fractions and performing operations
(+, -, x, ÷) with fractions to accomplish this!
Lowest Terms:
Fractions are reduced to “lowest terms” by dividing both the numerator (top) and denominator (bottom) of the
fraction by the same number (make it the biggest value you can):
eg. a)
6 2 3


8 2 4
b)
12  4 3

=
16  4 4
Equivalent Fractions:
You can determine an “equivalent fraction” by multiplying the numerator and denominator by the same
number:
eg.
1 5 5


2  5 10
a)
b)
3  4 12


8  4 32
Improper and Mixed Fractions:
A fraction where the numerator is larger than the denominator is called an “improper fraction”. A “mixed
fraction” is a combination of a whole number and a fraction. To convert between the two, simply follow these
instructions:
eg. a) 1
3
(to change this mixed fraction into an improper fraction, multiply the whole number by the
4
denominator, then add the numerator, then place over the original denominator)

eg. b)
7
4
1X4+3
4
27
(to change this improper fraction into a mixed fraction, divide the numerator by the denomator.
16
The quotient is the whole number, and the remainder is the numerator of the fraction part.)
11
1
16
Page 5 of 20
27 ÷ 16 = 1 with a
remainder of 11
Adding and Subtracting Fractions:
When adding or subtracting two or more fractions, first find a common denominator, then add the numerators.
Once completed, fractions must always be put into lowest form.
eg. a)
1 3

3 4
4
9

=
12 12
13

12
1
1
12
3
1
2 1
8 16
19 17


8 16
38 17


16 16
21

16
5
1
16
b)
Change to improper fractions first
Final answer is a mixed fraction, in
lowest terms.
Multiplying Fractions:
When multiplying fractions, multiply the numerators to get the numerator of the answer, and then do the same
thing for the denominators. Remember that your final answer should be expressed as a mixed fraction in
lowest terms.
eg. a)
1
3
2
1 3

2 1
3

2
1
1
2
3 1
2 
8 2
19 1


8 2
19

16
3
1
16
b)
Change the first mixed fraction to an
improper fraction first
numerator X numerator
denominator X denominator
Dividing Fractions:
When dividing fractions, you need to multiply the first fraction by the reciprocal of the second one. (Note: a
“reciprocal” is a “flipped” fraction – one where the numerator becomes the denominator, and vice versa)
eg. a)
15
2
16
15 1


16 2
15

32
Page 6 of 20
1 3

2 4
7 4
 
2 3
28

6
4
4
6
2
4
3
b) 3
Change this improper fraction to a
mixed fraction, in lowest terms
Student Worksheet #1
Complete the following exercises to demonstrate your mastery of operations with fractions:
1.
2.
?
Page 7 of 20
3.
4.
5.
1
1
A metal rod with a diameter of 1 inches is 19 feet long. It needs to be divided into 8 equal pieces.
4
2
What is the length of each piece?
You need 12 pieces of 3
3
4
Page 8 of 20
inch square bar. What is the total length of steel that needs to be ordered?
LESSON 2: Decimals and Fractions; Imperial and Metric Measurements
In the manufacturing sector, as well as in most other trades, workers use their knowledge of fractions AND
decimals everyday. Fractions are used extensively with Imperial measurement, and decimals are used mainly
when measuring in Metric, however, it is important for students to be able to convert from one type of number
into another, thus using fractions and decimals interchangeably.
Converting Fractions to Decimals:
To change a fraction to a decimal, divide the numerator by the denominator.
27
30
eg. a)
b)
= 27 ÷ 30
= 0.9
21
6
= 21 ÷ 6
= 3.5
Converting Fractions to Percent:
To convert a fraction to a percent, first change the fraction into a decimal, then multiply by 100.
12
20
eg. a)
b)
= (12 ÷ 20) x 100
= (0.6) x 100
= 60%
29
24
= (29 ÷ 24) x 100
= (1.21) x 100
= 121%
Converting Between Metric and Imperial Units of Measure:
Refer to the tables below to help you complete the following conversions:
eg. a) 15 yds = _______ ft
b) 15 yds = _________ in
15 × 3
15 × 3 × 12
= 45 ft
d) 28 cm = _________ m
e)
28 cm = _________ mm
= 0.28 m
g) 4 m = ________ in
4 × 100 ÷ 2.54
= 157.48 in
h) 1
15 yds = __________ m
15 × 0.9144
= 540 in
28 ÷ 100
Page 9 of 20
c)
= 13.716 m
f)
28 cm = __________ in
28 × 100
28 ÷ 2.54
= 280 mm
= 11.0236 in
3
in = ________ cm
8
1.375 × 2.54
= 3.4925 cm
i)
12.5 ft = __________ mm
12.5 × 12 × 2.54 ×10
= 3810 mm
Student Worksheet #2
Metric and Imperial Conversion Charts:
Length:
÷
Metric
Imperial
Metric  Imperial
10mm = 1cm
100cm = 1m
1000m = 1km
12 inches = 1 foot
3 feet = 1 yard
1760 yards = 1 mile
1 inch  2.54 cm
1 foot  30.48 cm
1 yard  0.9144m
1 mile  1.609 km
×
Mass:
Metric
Imperial
Metric  Imperial
1000 mg = 1 g
1000 g = 1 kg
1000 kg = 1 t
16 ounces = 1 lbs.
2000 lbs. = 1 ton
1 ounce  28.35 g
1 pound  0.454 kg
1 ton  0.907 t
Metric
Imperial
Metric  Imperial
1000mL = 1L
16 fl. ounces = 1 pint
2 pints = 1 quart
8 pints = 1 gallon
1 fl. ounce  29.574 mL
1 pint  0.473 L (473 mL)
1 gallon  3.785 L
Volume:
6. Complete the following conversions, using the tables provided above:
b) 3⅛ ft = ________ in
c) 186 in = ___________ ft
e) 96 in = ________ yd
f) 78 ft = ___________ yd
g) 6.377 cm = __________mm
h) 3.47 m = ____________cm
i) 300.3 mm = __________ m
j) 0.93 mm = _________ cm
k) 5.032 cm = ___________ m
l) 0.45 m = ____________ mm
m) 0.75 ft = __________ m
n) 2.400 m = ___________ in
o) 17.300 cm = __________ in
p) 8.000 m = __________ ft
q) 0.360 in = ____________ mm
r) 84.00 cm = ____________ ft
s) 780.00 mm = __________ ft
t) 3.50 yd = ____________ m
u) 3
a) 0.4 yd = _________ in
d)
1
4
yd = _________ ft
Page 10 of 20
1
4
in = ______________ cm
7.
A screw measures
5
in long. How long is this in millimeters (use decimals)?
8
8. You have a piece of lumber that is 3 yards long, but you only need a length that measures 2.3 m.
How much is left over (in m)? What percent of the entire piece of lumber is left over?
9. You need 36 steel rods that each measure 18 inches long. How much steel do you need to order in feet?
In yards? In metres?
Page 11 of 20
Lesson 3: Proportional Reasoning and Scale Diagrams
In architecture, construction or any kind of design and manufacturing, a SCALE is used to make accurate
drawings that show sizes of rooms, placement of furniture, or the dimensions of an item. The scale is
presented as a RATIO – where the first number represents the length in the drawing and the second number
represents the actual length of the object. For example, a scale of 1:10 means that 1 cm on the drawing
represents a distance of 10 cm on the actual object (or that the drawing is 10 times smaller than the object).
eg. a)
Length on drawing =
3.3
cm
Now, using the scale of 1:7, we multiply the drawing length by 7 to find the
actual length of the hammer…..
3.3 × 7 = 23.1 cm
Scale 1 : 7
eg. b)
If the real bee measures 0.8 cm in length, find the scale for this scale diagram.
Length on drawing =
3.7
cm
Actual length =
0.8
cm
Now, set up a ratio using the drawing length first, then the actual length:
3.7
3.7 : 0.8
=
n

n
0.8 1
n = 3.7 ÷ 0.8
n = 4.625
:1
The scale is 4.625 : 1.
A PROPORTION is a comparison of two ratios. The order of the numbers in a proportion is important, since it
is often used to calculate for missing values. For example, if a piece of metal has a length : width ratio of 3 : 2,
and the length is actually 15ft, the matching width value can be calculated using equivalent fractions:
eg. c)
3

2
15
This proportion can be solved by “cross-multiplying”… 3m = 30
m
m = 10
eg. d)
Given the dimensions in the diagram below, find the width and height of a cinder block with a length
of 24 in.
l = 4.2 cm
h = 1.0cm
4.2
24

1.0
h

1.2
w
Note that the first fraction in the proportion
represents length values, the second fraction
represents height, and the third is width. In all
cases, the numerators are the dimensions from
the diagram, and the denominators are the
actual dimensions of the block (Imperial).
w = 1.2 cm
Working with 2 fractions at a time, and cross multiplying, you get:
4.2 h = 24
h = 5.714 in
Page 12 of 20
4.2 w = 28.8
w = 6.857 in
Student Worksheet #3
10. Use a ruler to determine the actual length of this bolt, if the scale is 1:3.
11. Determine the actual height of this height gauge, if the scale is 7.5:1.
12. Given the floor plan below, use a ruler and the dimensions shown to determine the scale.
Page 13 of 20
13.
B
C
D
A
Page 14 of 20
Lesson 4: Trigonometry
Trigonometry is the branch of mathematics that deals with the relationships between the sides and angles in
triangles. Since pretty much any polygon can be divided up into a series of triangles, trigonometry has widereaching uses in professions such as construction, architecture, aerospace, engineering… to name just a few.
Accuracy is key in the manufacturing field when you are designing and machining objects, and using
trigonometry is the way to get there!
Right-Angled Triangles:
The sides of a right-angled triangle are related using the Pythagorean Theorem, whereby the square of the
hypotenuse is equal to the sum of the squares of the other two sides.
Referring to the diagram, then, the Pythagorean
Theorem states that:
c2 = a2 + b2.
The Primary Trigonometry Ratios are used to relate the angles of a right-angled triangle. To review briefly, the
ratios are:
sin ө =
cos ө =
tan ө =
length of the side opposite from 
length of the hypotenuse
length of the side adjacent to 
length of the hypotenuse
length of the side opposite from 
length of the adjacent to 
eg. a) Solve the following triangle for V, W and r :
V
18.200 in
7.620 in
W
r
c2 = a2 + b2
18.2 2 = 7.62 2 + r 2
331.24 = 58.0644 + r 2
331.24 – 58.0644 = r 2
273.1756 = r 2
sin W =
r  273.1756
r = 16.528 in
tan V =
7.620
18.200
W = 24.751 o
tan V =
r
7.620
16.528
7.620
V = 65.249 o
Page 15 of 20
eg. b) A ramp needs to be built at the loading dock of the school.
Find the length of the ramp (k) and it’s angle of inclination ( ө ).
r 2 = 0.765 2 + 1.6065 2
r 2 = 3.1661
r = 3.1661
r = 1.7793 m
h = 0.765 m
ө
tan ө =
j = 1.6065 m
0.765
1.6065
ө = 25.463 o
Oblique Triangles:
An oblique triangle is one without any right angles. Since a triangle such as this has no right angle, it has no
hypotenuse. Without a hypotenuse, the Pythagorean Theorem and Primary Trig Ratios are useless. The
tools available to solve these types of triangles are the Sine Law or the Cosine Law….
Sine Law: The Sine Law can be used to solve for any missing information in an oblique triangle where
“opposite information” is given (meaning you know the measure of at least one angle and the matching
sinA sinB sinC


opposite side). The Sine Law states that
.
a
b
c
eg. c)
sin64 
P
3.125 cm
3.125

sinP
1.901

sinQ
y
solve for P first using the first 2 fractions in
the proportion
y
64o
1.901 cm
Q
1.901 × sin64 o = sin P × 3.125
sin P = 1.901 × sin 64 o
3.125
Sin P = 0.54675
P = 33.145 o
To solve for angle Q, 180 – 64 – 33.145 = 82.855 o
Now solve for the final fraction in the proportion:
3.125 × sin Q = sin 64 o × y
y = 3.125 × sin 64 o
sin 82.855 o
y = 2.831 cm
Page 16 of 20
Cosine Law: The Cosine Law is used to solve for missing information in an oblique triangle where no opposite
information is given. The Cosine Law states that c 2 = a 2 + b 2 – 2ab cosC, where C is the one angle that
you know, or are trying to find.
eg. d) A hiker enters a park (at P) and walks 3.28km North. She then turns (at N) and heads Southeast for
2.13km. How far is she from her original position?
First, draw a triangle to organize the information from the question.
Note: “Southeast” indicates a heading of exactly 45 o from south to
east, as pictured on the diagram.
N
45 o
N 2 = S 2 + P 2 – 2SP cos N
N 2 = 3.28 2 + 2.13 2 – 2(3.28)(2.13) cos 45 o
N 2 = 15.2953 – 9.8803
2.13 km
3.28 km
S
P
Page 17 of 20
N = 5.4150
N = 2.33 km
Student Worksheet #4
14.
x
A bridge MN is to be built across a river. Point L is 121 ft from M, and at an
angle of 83 o. Angle L measures 38 o.
Calculate the length of the bridge.
15.
M
N
L
U
16.
A steel support beam must be fabricated, according to the diagram,
along length ST. Calculate the length of the beam.
12 ft
29 o
S
11 ft
T
Page 18 of 20
17.
You are working in a manufacturing shop, and the belt breaks
on the lathe. To order the new belt, you need to fill in the missing
information from the schematic diagram below.
Belt
p
a
Hints:
The belt consists of a continuous
loop around both circular drives.
ø This symbol represents diameter.
All measurements are in inches.
Angle a = ___________
Slanted length p = ___________ in
Page 19 of 20
Summative Design Assignment:
Design and Build a Scale Model of a Sawhorse
You are required to design a sawhorse to be used in manufacturing classes. The design specifics have been
provided for you below (all measurements are in inches).
a) Calculate all missing lengths and angles.
b) Create a scale diagram of your sawhorse.
c) Create a 3-dimensional drawing OR a 3-dimensional model of your piece of equipment.
?
?
?
?
?
Supports are added at the midpoint of the horizontal supports (as indicated below), to offer additional stability.
Determine the lengths of these pieces.
30
30
?
Page 20 of 20