Mathematical
Literacy
NQF 4
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Mathematical Literacy Level 4 Module 3
Level 4 Mathematical Literacy study guide, 46 pages. Read online or get the free PDF download.
Subject: Mathematical LiteracyMultiple grades46 pages
Mathematical Literacy. Study guide, 46 pages. Read online or download the PDF.
- Subject
- Mathematical Literacy
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- Multiple grades
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- Study guide
- Curriculum
- TVET
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- 46
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Solving space, shape and
orientation problems
Module 3
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Overview
3.1 Solve problems using the Theorem of
Pythagoras
3.2 Solving area and perimeter problems
3.3 Solve volume and surface area problems
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Solve problems using the
Theorem
Unit 3.1
of Pythagoras
• You are not always told which angles are right angles, but you can often
work this out from the context.
• For example, the angle between a horizontal structure and a vertical
structure is always a right angle.
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Solve problems using the
Theorem
Unit 3.1
of Pythagoras
• But don’t assume an angle is right angled unless you have facts about it.
• Wherever you see a right angle, you can use a right-angled triangle to
help you solve the problem.
• The triangle can be in different positions and facing different ways.
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The Theorem of Pythagoras
Unit 3.1
A B A
A C
B C C B
The Theorem of Pythagoras states that:
The square of the length of the hypotenuse is equal to
the squares of the other two sides added together.
In triangle ABC, we can write this as
AB2 + BC2 = AC2
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The Theorem of Pythagoras
Unit 3.1
A B A
A C
B C C B
• So if we are given the lengths of the other two sides, we can
calculate the length of the hypotenuse.
• We can also calculate any other side by manipulating the formula.
Important!
This theorem does not apply to triangles that do not have a right angle.
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Example 3.1 page 75 question 1
Unit 3.1
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Example 3.1 page 75 question 3
Unit 3.1
3. A metal bracket on a wall
supports a pot plant.
Calculate x.
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Exercise 3.1
Unit 3.1
Complete Exercise 3.1 on page 76 of your Student’s Book
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Solving area and perimeter
problems
Unit 3.2
Perimeter
• Perimeter is the total length all the way around a shape
• Add the lengths of all the sides to get the perimeter.
• The perimeter of a circle is called the circumference.
• Perimeter is measured in units of length, not square units.
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”
Solving area and perimeter
problems
Unit 3.2
Area
• Amount of space that a closed shape covers
• The units of area are always square units
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Maths Lit NQF Level 4
Solving area and perimeter
problems
Unit 3.2
Remember:
• A = lb means A = l × b
• A = ½ πr2 means A = ½ × π × r × r
• 3,14 is an approximate value of π.
• If you have a scientific calculator, use the π key to
be more accurate.
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Working with formulae
Unit 3.2
S
Cube S S
• Perimeter = 4s
S
• Area = s2
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Working with formulae
Unit 3.2
c a
Triangle h
b
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Working with formulae
Unit 3.2 b
Rectangle l l
• Perimeter = 2l + 2b
• Area = l × b b
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Working with formulae
Unit 3.2
r
Circle
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Working with formulae
Unit 3.2
Semi-Circle r
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Example 3.2 page 78
Unit 3.2
• Calculate the area and the perimeter of this shape.
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Example 3.2 page 78
Unit 3.2
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Example 3.3 page 79
Unit 3.2
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Changing the subject of a
formula
Unit 3.2
We can change a formula around to calculate a different quantity,
without changing the facts in the formula.
1 Write down the formula and any information you are given.
Change the formula so that the unknown value is on its own on the
2 left-hand side.
3 Put the known values into the formula and do the calculation.
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Example 3.4 page 80
Unit 3.2
• In this triangle, calculate the length of the base.
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Example 3.4 page 80
Unit 3.2
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Calculating costs
Unit 3.2
• Area and perimeter calculations are important in practical situations
such as creating objects, building, gardening, developing plans,
decorating.
• To calculate the costs, we need to calculate the quantities needed
first and then multiply by the cost rate of the materials.
?
A quantity surveyor is someone who works in
the construction industry. Their main job is to
calculate and manage the costs of a whole
project.
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Example 3.4 page 81
Unit 3.2
1. A landscaper wants to fence parts of a garden as
shown in the diagram here.
a) Work out the total quantity of fencing needed.
b) If fencing costs R75 per metre, what will the total cost
be?
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Example 3.4 page 81
Unit 3.2
a) Circumference of herb garden = 2πr
= 2π (2,5)
= 15,7 m
Circumference of rose garden = 2πr
= 2π (3)
= 18,85 m
Perimeter of ornamental trees and lawn
= 2(4 + 3) = 14 m
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Example 3.4 page 81
Unit 3.2
Total quantity of fencing
= 15,7 m + 18,85 m + 14 m
= 48,55 m
b) 48,55 m × R75/m
= R3 641,25
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Example 3.4 page 81
Unit 3.2
2. The landscaper wants to apply lawn dressing to
the trees and lawn area and compost to the herb
and rose gardens.
He needs 5 kg of compost or lawn dressing to each
square metre of garden.
He has the following prices:
Product 20 kg 50 kg 100 kg
Compost R200 R420 R800
Lawn dressing R250 R650 R1 695
Which bags should he buy for this season, and
how much will it cost in total?
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Example 3.4 page 81
Unit 3.2
2. Area of herb garden = πr2 = 19,63 m2
Area of rose garden = πr2 = 28,27 m2
Area of lawn and trees garden = 3 × 4 = 12 m2
5 kg of compost or lawn dressing per m2
Amount of compost needed
= (19,63 + 28,27) × 5 = 239,5 kg
Amount of lawn dressing needed
= 12 × 5 = 60 kg
He needs to buy the 50 kg bag of compost and the
20 kg bag of lawn dressing:
Cost = R250 + R420 = R670
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Exercise 3.2
Unit 3.2
Complete Exercise 3.2 on page 82 of your Student’s Book
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Solve volume and surface area
problems
Unit 3.3
height
length
width
• All objects have 3 dimensions: length, width and height.
• We use volume to measure and compare how much space things take
up.
• Liquids (water, milk, cool drink, etc) are measured in units such as litres
and millilitres.
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Solve volume and surface area
problems
Unit 3.3
height
length
width
• We can measure the volume of solid objects in cubic units.
• We can convert between liquid volume and solid units. (see Topic 1).
• Volume is calculated by multiplying 3 dimensions together, so it is
measured in cubic units
• Example: Volume of a box = length × breadth × height (in cubic units)
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Formulae for surface area and
volume
Unit 3.3
s
s
Cube s
Surface area = 6 × side2 = 6s2
Volume = s3
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Formulae for surface area and
volume
Unit 3.3
h
b
l
Rectangle
Surface area = 2(lb + lb + bh)
Volume = l × b × h
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Formulae for surface area and
volume
Unit 3.3 r
h
Cylinder
Surface area = area of 2 circles + area of rectangle
= 2πr2 + 2πrh
Volume = πr2h
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Formulae for surface area and
volume
Unit 3.3
r
r
r
Sphere
Surface area = 4πr2
Volume = πr3
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Formulae for surface area and
volume
Unit 3.3
Cone
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Volume and surface area
Unit 5.3
Triangular prism
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Nets of 3-D objects
Unit 5.3
• To find the surface area of 3-D objects, it helps to imagine or sketch their
nets (when they are ‘opened up’ and laid flat).
• The total surface area is the sum of the areas of the flat shapes.
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Exercise 3.3
Unit 3.3
Complete Exercise 3.3 on page 86 of your Student’s Book
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Assignment 1
Unit 3.3
Complete Assignment 1 on page 87 of your Student’s Book
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Assignment 2
Unit 3.3
Complete Assignment 2 on page 89 of your Student’s Book
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VIDEO: Application in the
Workplace
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Module assessment
Module 3
Complete Module assessment on page 91 of your Student’s Book
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