Download chapter 2

A Mathematical Skills Fundamental for the Pulp and Paper Industry
PHYSICAL QUANTITIES AND
GEOMETRICAL RELATIONSHIPS 4
Facilitator Guide
NQF Level 4
Credits: 4
Unit Standard 9016
Compiled by:
Susann Richardson
Bredenkamp Bruwer
for
FIETA
Sparrow Research and Industrial Consultants © January 2006
Physical Quantities and Geometrical
Relationships 4
Learning Outcomes
Upon studying this module, the learner will be able to:

Identify a range of different plant measurements and the units in which these
measurements are made

Describe the difference between speed and acceleration and between mass and
weight

Recognise a wide variety of instruments, indicate the variable it’s used to measure
as well as the units in which it measures

Avoid the basic measurement errors commonly made

Estimate different quantities and calculate the error made

Understand geometric constructions

Draw parallel lines

Draw perpendicular lines

Draw a perpendicular bisector

Bisect an angle

Understand the Theorem of Pythagoras

Give directions on a map by using bearings

Determine distance using a scale

Identify a range of regular three dimensional forms

Calculate perimeter and areas of two dimensional shapes

Develop nets for regular three dimensional forms

Calculate surface area and volume of three dimensional forms

Understand time zones

Plot coordinates on a Cartesian Square
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
2
Sparrow Research and Industrial Consultants © January 2006
Unit Standard 9016
Represent analyse and calculate shape and motion in 2-and 3-dimensional space in
different contexts

Measure, estimate, and calculate physical quantities in practical situations relevant
to the adult

Explore, analyse & critique, describe & represent, interpret and justify geometrical
relationships
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
3
Sparrow Research and Industrial Consultants © January 2006
Table of Contents
CHAPTER 1:
WORKING WITH NUMBERS ..................................................................... 5
1
MEASUREMENT ....................................................................................................... 5
1.1
Units................................................................................................................... 5
1.2
SI base units ...................................................................................................... 6
1.3
Rules for writing units ......................................................................................... 6
2
DECIMAL MULTIPLIERS .......................................................................................... 7
2.1
Instruments ........................................................................................................ 9
2.2
Physical measurements ....................................................................................15
3
ESTIMATION ............................................................................................................16
CHAPTER 2:
BASIC GEOMETRY ..................................................................................19
1
BACKGROUND ........................................................................................................19
2
GEOMETRIC CONSTRUCTIONS ............................................................................22
3
DRAWING PARALLEL LINES ..................................................................................23
4
DRAWING A PERPENDICULAR LINE .....................................................................24
5
DRAWING A PERPENDICULAR BISECTOR ...........................................................24
6
BISECTING AN ANGLE ...........................................................................................25
7
THE THEOREM OF PYTHAGORAS ........................................................................31
8
CARTESIAN CO-ORDINATE SYSTEM ....................................................................33
CHAPTER 3:
BASIC GEOMETRY ..................................................................................35
1
BEARINGS ...............................................................................................................35
2
INTERNATIONAL TIME ZONES...............................................................................39
CHAPTER 4:
GEOMETRIC FORMS ...............................................................................41
1
SOLID FIGURES ......................................................................................................41
2
PERIMETER AND AREA ..........................................................................................48
2.1
Basic formulae for perimeters and areas ...........................................................48
3
SURFACE AREA AND VOLUME..............................................................................51
4
SCALE DRAWINGS .................................................................................................59
5
MAPS .......................................................................................................................64
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
4
Sparrow Research and Industrial Consultants © January 2006
CHAPTER 1:
WORKING WITH NUMBERS
On completion of this chapter, you will be able to:

Identify a range of different plant measurements and the units in which these
measurements are made

Describe the difference between speed and velocity as well as mass and weight

Recognise a wide variety of instruments, indicate the variable it’s used to measure
as well as the units in which it measures

Avoid the basic measurement errors commonly made

Estimate different quantities and calculate the error made
1
MEASUREMENT
In our daily lives and in the workplace we make many conscious and unconscious
measurements and estimated measurements – we may read the speed of the car’s
speedometer, estimate the length of a person, guess the outside temperature on a hot or
cold day and estimate the time it will take (or took) to complete a task.
1.1
Units
All these measurements are based on standardised units we use regularly – the units include
length (mm, cm, m, km), weight (kg, g), time (s, m, h), temperature (oC). The measurement
units themselves are quite useful in themselves, but when we start combining them with
other units (e.g. km/h, m3/h, kg/m3) the options and usefulness expands even further. In
many cases, new units based on a combination of the core units have been created to
simplify the use of units, for example N (Newton; kgxm.s2), J (Joule), (kgxm.s2), W (Work),
(kg/m2.s3) and Pa (Pascal), (kg/m3s2).
In a production environment, instruments are used to indicate important measurements of
different variables which need to be monitored and controlled. Typical examples of plant
measurements are temperature (oC), pressure (kPa) and flow rate (m3/h).
We therefore need to standardise measurement in order for it to be useful. All the “things”
that we can measure are called quantities which then give us quantitative information. To
measure a quantity we need to compare it to a standard unit. If we measure length, we
compare the length to the standard unit for length, which is the meter. The length is then
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
5
Sparrow Research and Industrial Consultants © January 2006
given as a number of times the standard unit: 5 m means that the object has a length five
times the standard unit.
All units are man-made. The kilogram is defined as the mass of a certain block of metal kept
in a laboratory in Paris. They could have made the block smaller, making the unit of mass
(kg) smaller or they could have made the block bigger, making the unit of mass larger. But,
as we all know, scientists have chosen that specific block of metal to indicate a mass of 1 kg.
1.2
SI base units
The system chosen by scientists for recording and reporting measurements are called the
Système International d’Unitès (International System of Units), abbreviated as SI. It is also
known as the metric system.
All SI Units are derived from seven base units. The seven base units are given in the
following table:
Physical Quantity
Unit
Symbol
Length
meter
m
Mass
kilogram
kg
Time
second
s
Area
square meters
m²
Temperature
Kelvin
K
Speed
meters per second
m/s
Acceleration
Meters per second square
m/s²
1.3
Rules for writing units
All units, when written out in full, start with a small letter.
When symbols are used, the symbol is written with a small letter, except when the unit is
named after a scientist. In that case, the first letter of the symbol is a capital letter. For
example, the symbol for second is “s”, but the symbol for Kelvin is K (after a scientist).
No other abbreviations are acceptable. You should not use sec for second – it must be s.
When two units are multiplied, a small dot ( · ) must be written between them. This is to
differentiate between, for example, ms (millisecond) and m·s (meters second).
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
6
Sparrow Research and Industrial Consultants © January 2006
2
DECIMAL MULTIPLIERS
It is difficult to measure very small objects and very large objects with the same unit. For
example, the meter is too large to measure the size of bacteria, while it is too small to
measure the distance between the stars. In the SI system, we form units that more closely
suit our needs by modifying the basic units with decimal multipliers. Most of us are familiar
with the kilometre (= 1 000 m) and the centimetre (= 0,01 m).
The following table give some of the most commonly used decimal multipliers:
Prefix
Symbol
Multiplication factor
mega
M
106
kilo
k
103
deci
d
10–1
centi
c
10–2
milli
m
10–3
micro
µ
10–6
The prefix “kilo” indicates 103 or 1 000.
So 1 km = 1 x 103 m (substitute k for 103) = 1 000 m
and 1 kg = 1 x 103 g = 1 000 g.
The prefix “centi” indicates 10–2.
So 1 cm = 1 x 10–2 m = 0,01 m.
When calculating physical quantities, it is often helpful to write all expressions in terms of
base SI units and scientific notation. This reduces the chance of getting confused with units.
For example, if you want to calculate the speed of an object that moved 9 Mm in 30 ms, you
cannot simply divide 9 by 30! Instead, it must first be converted to SI units (m and s):
9 x 106 m / 30 x 10-3s = 300 x 106 m/s = 300 Mm/s
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
7
Sparrow Research and Industrial Consultants © January 2006
Note that the answer is given as Mm/s, not m/µs (which amounts to the same thing) – this is
because a prefix is never written after the division sign. The only exception is the kilogram –
it is never written as “g”; this is because it is a base unit.
Study the following conversions for distance:
Kilometre
Metre
Centimetre
Millimetre
0,028 km
28 m
2 800 cm
28 000 mm
(km x 1000)
(m x 100)
(cm x 10)
In daily life we would use these terms weight and mass interchangeably to determine when it
is time to go on a diet or what the quantity of sugar is that we must buy. In science, mass is
defined as the quantity of a substance in kilograms while weight is defined as:
Weight = mass x gravitational acceleration (in N or kg/m.s2)
On the moon, a person will still have the same mass, but less weight (gravitational
acceleration is less), while in outer space a person becomes “weightless” (gravitational
acceleration = 0) despite the fact that the person has the same mass as anywhere else.
Mass therefore is the amount of matter of a body or object and weight is the force on the
same object due to the gravitational force of the earth.
When we compare the terms “speed” and “acceleration” they are defined differently in
scientific terms. Speed is the rate at which the distance changes during a certain time. The
formula used is distance  time. The unit is m/s or km/h.
If a distance of 240 km is completed in 2 hours the speed of the object will be calculated as
follows:
240  2= 120km/h
Acceleration on the other hand is the rate of change of speed. The unit is m/s² and the
formula used is speed divided by time. When the initial and end velocity differs, then the
following formula is used: (end velocity – initial velocity)  time.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
8
Sparrow Research and Industrial Consultants © January 2006
An object was accelerated from 46 m/s to 88 m/s in 7 seconds. Determine the acceleration of
the object.
Using the above formula of (end velocity – initial velocity)  time and substitute the values
given.
(88-46)  7
= (42)  7
= 6m/s²
2.1
Instruments
In the last two decades, digital instrument have largely replaced analogue readouts and dials
– apart from the flow meter on the left which is digital, the rest are all analogue meters.
Exercise 1 - Instruments
Match each of the following instruments with the appropriate quantity on the left and the units
on the right:
Measured quantity
Instrument
Units
Sound waves
Tachometer
A, Ω, v
Liquid volume
Scale
Pa
Current, resistance, potential difference
Pyrometer
Counts per minute (CPM)
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
9
Sparrow Research and Industrial Consultants © January 2006
Measured quantity
Instrument
Units
Chemical analysis
Geiger counter
Degrees (o)
Frequency
Speed
m
Weight
DP Cell
Degrees (o)
Flow
Angle
o
A person’s length
Direction
km/h
Pressure
Multi-meter
g / kg / tons
Thickness of a small object
Volumetric flask
%, ppm, ppb
Differential pressure
Oscilloscope
o
Temperature
Orifice plate
Hz
High temperature
Micrometer
Waveform
Compass
Chromatograph
μm
Radiation
Thermometer
ml
Parabola
Tape measure
Pa
Speedometer
Barometer
m3/h
2.1.1
C
C
Vernier Callipers
A Vernier Calliper is used to measure the thickness or the diameter of objects. It is calibrated
in centimetres and can measure accurate up to 0.01 cm or 0.1 mm.
A Vernier Calliper has an adjustable arm gliding over a shaft calibrated in centimetres and
millimetres.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
10
Sparrow Research and Industrial Consultants © January 2006
The reading on the Vernier below is 2,15. The zero on the adjustable arm lies between 2,1
and 2,2 on the calibrated shaft, therefore giving a reading of 21 millimetres or 2,1 cm. Now
find the line on the scale of the adjustable arm exactly in line with the line on the scale of the
shaft. In this example the five on the adjustable scale is exactly in line with one of the lines
on the scale of the shaft. Thus giving us a reading of 2,15.
Therefore, the zero on the adjustable arm always gives the first reading (i.e. 2,2; 4,7 etc.). To
find the second decimal, find the two lines on both scales exactly aligned.
Exercise 2 – Instruments
Find the reading indicated on the Vernier Calliper below:
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
11
Sparrow Research and Industrial Consultants © January 2006
Solution
5,09
2.1.2
Micrometer Screw
Obtaining a reading from a micrometer screw works on almost the same principle as a
vernier calliper. The ratchet is turned until both ends of the anvil are in contact with the
object.
Let’s take a reading on the micrometer below. Take the measurement on the calibrated shaft
first. The shaft is only calibrated every 0,5 cm, therefore the only accurate reading we can
take from the shaft is 3.5 cm. To have a more accurate reading, we have to read from the
thimble. This is where the micrometer screw and vernier calliper make use of the same
principle. The value on the thimble aligned with the line on the shaft is 0,46. If we now add
these two values the answer will be 3,96.
3,5 + 0,46 = 3,96
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
12
Sparrow Research and Industrial Consultants © January 2006
Exercise 3 – Instruments
Take the correct reading from the micrometer below.
Solution: 14,68
2.1.3
Spring Balance
In a spring balance the spring and its pointer are assembled in a casing which has a suitable
scale calibrated on it. The zero of the pointer can be adjusted. The balance has a hanger
from which to support it and a hook on which to hang the object to be weighted. The scale is
calibrated in Newton. The reading on the balance below is 5.5 Newton.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
13
Sparrow Research and Industrial Consultants © January 2006
2.1.4
Thermometer
The liquid-in-glass-thermometer is probably the most familiar type. The liquid, normally
mercury and ethanol, are contained in a thin, thick walled capillary tube. The principle on
which these thermometers operate is the expansion and subtraction of matter upon heating
and cooling respectively. The scale is calibrated in degrees Celsius (ºC) and normally
measures temperatures between 0 and 100 ºC.
2.1.5
Top-pan Balance
A top-pan balance is used in a laboratory to determine the mass of substances. The mass is
indicated on a digital readout.
2.1.6
Measuring Cylinder
A measuring flask is used to measure the volume of a liquid. These measurements can be
seen on the outside of the cylinder and are calibrated in millimetres (mm) or cubic
centimetres (cm³). When taking a measurement keep your eye level with the meniscus as
illustrated in the sketch below.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
14
Sparrow Research and Industrial Consultants © January 2006
2.2
Physical measurements
In a previous module, we investigated physical measurement in some depth – here we will
simply remind you of the principles. The most important concepts to bear in mind when
measuring are:

Ensure that the instrument is calibrated in order to obtain accurate measurements
with it.
A ruler or tape measure should generally be produced with sufficient
accuracy to serve its purpose. However, instruments used to perform chemical
analysis, physical testing or very accurate measurements, needs to be calibrated
before use. An error originating from not poor calibration is called a calibration error
– the sketch shows that the intervals on the top ruler are not the same as on the
other rulers.

The instrument should be zeroed properly. When there is no product to analyse, the
instrument should indicate zero – The sketch shows point “A” on the line not
properly lined up with the zero on the ruler.

The readings should be taken accurately. Where a sample (e.g. a line, temperature,
volume or pressure) is to be measured with an analogue instrument (ruler,
volumetric flask or measurement gauge), it is important to line the product, the
instrument and the eye up accurately. This error is called a parallax error – the last
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
15
Sparrow Research and Industrial Consultants © January 2006
reading at point “B” would be different if the eye is not brought in line with the ruler
and the line.
3
ESTIMATION
When a person works continually in a certain environment, he or she sometimes becomes
very accurate in estimating physical quantities such as lengths, distance, heights, weights,
temperature, speed, area and volume. The following exercise will assist you to determine
your own accuracy when estimate physical quantities by comparing your estimates with the
actual measured or calculated values.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
16
Sparrow Research and Industrial Consultants © January 2006
Exercise 4 – Estimation
Estimate, measure or calculate each of the following quantities and determine the accuracy
of your estimate with the actual value:
1. Estimate (without ever looking at the speedometer) the speed of a vehicle 20 seconds
after pulling away from a dead stop. As soon as your estimate is made, look at the
vehicle’s speedometer and determine the actual speed. Record four such values and
calculate the error with the given formula. Are your results improving in accuracy?
Test 1
Estimate of speed
km/h
Actual speed
km/h
Error = (Actual – Estimate) x 100
Actual
%
Test 2
Test 3
Test 4
2. Estimate the length of four colleagues and write down your estimate. Measure each of
these colleagues for their actual lengths or simply ask them. Record four such values
and calculate the error with the given formula.
Test 1
Estimate of height
m
Actual height
m
Error = (Actual – Estimate) x 100
Actual
%
Test 2
Test 3
Test 4
3. Estimate the surface area (in m2) of four rooms in your office or home. Measure each of
these for their actual lengths and width and calculate the actual size. Calculate the error
with the given formula.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
17
Sparrow Research and Industrial Consultants © January 2006
Room 1
Estimate of surface area
m2
Actual: length x width
m2
Error = (Actual – Estimate) x 100
Actual
%
Room 2
Room 3
Room 4
4. Estimate a process variable or quantity (e.g. filling rate, temperature, speed, weight) in
your workplace. Measure the actual value by using available instrumentation or other
measuring tools or methods. Calculate the error with the given formula.
Unit
………1
………2
………3
………4
Estimate of ……………………………….
Actual: …………………………………….
Error = (Actual – Estimate) x 100
Actual
Facilitator Guide
%
US 9016 – Physical Quantities and Geometrical Relationships 4
18
Sparrow Research and Industrial Consultants © January 2006
CHAPTER 2:
BASIC GEOMETRY
On completion of this chapter, you will be able to:

Understand geometric constructions

Draw parallel lines

Draw perpendicular lines

Draw a perpendicular bisector

Bisect an angle
1
BACKGROUND
In order to understand this unit, you need the background knowledge which has been dealt
with in a previous module:
Angles

Angle ABC is formed by arms AB and BC. The point B
is called the vertex of the angle. In this course, we will
measure angles in degrees, using a protractor.

A right angle is 90⁰

A straight angle (or straight line) is 180⁰

A full revolution is 360⁰

Vertically opposite angles are equal in size.
When two straight lines cross one each other, they
form four angles, a and c are vertically opposite,
and so are b and d.

a + b = 180º = c + d (angles on a straight line)

Parallel lines are straight lines that are always the same
distance apart, like railway-lines.
A straight line that
crosses these parallel lines is called a transversal. The
arrows on the lines indicate that they are parallel. In this
sketch, angles a and c are vertically opposite, so they are
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
19
Sparrow Research and Industrial Consultants © January 2006
equal in size. But angle a is also equal to angle d, they are corresponding angles.
Angles c and d are called alternate angles, and are equal for parallel lines. Another
pair of angles is b and d, and they are called co-interior angles. Co-interior angles
are not equal in size, but they sum to 180º, we say they are supplementary angles.

A closed shape whose boundary consists of straight lines is called a polygon. The
most important polygons for our course are triangles and quadrilaterals. A triangle
has three sides and three angles. The angles in a triangle add up to 180º. A
quadrilateral is any flat shape consisting of four sides and four
interior angles, which add to 360º.

A triangle with two sides equal is called an isosceles triangle.
In the diagram, the lines drawn through them indicate the equal
sides (AC = AB). The angles B and C are called the base angles of the isosceles
triangle. The base angles of an isosceles triangle are equal.

An equilateral triangle is a triangle with all three sides equal. In an equilateral
triangle, all the angles are equal and, since they sum to 180º, each angle is 60º.

The theorem of Pythagoras: In a right angled triangle,
the square on the hypotenuse is equal to the sum of the
squares on the two other sides.
∴If  = 90º, then BC2 = AB2 + AC2

The following table shows some special types of quadrilaterals and their properties:
Name
Facilitator Guide
Defining property
Other properties
One pair of sides
parallel
Other two sides not parallel and
need not be same length
Two pairs of adjacent
sides equal
Diagonals intersect at right angles
One pair opposite angles equal
The longer diagonal bisects the
other and it bisects the angles
Both pairs opposite
sides parallel
Both pairs opposite sides equal
Both pairs opposite angles equal
Diagonals bisect each other
US 9016 – Physical Quantities and Geometrical Relationships 4
20
Sparrow Research and Industrial Consultants © January 2006
Name

Defining property
Other properties
A parallelogram with all
sides equal
All properties of parallelograms
Diagonals bisect each other at right
angles
Diagonals bisect the angles
A parallelogram with all
angles 90º
All properties of parallelograms
Diagonals are equal in length
A rectangle with all
sides equal
All properties of parallelograms,
rectangles and rhombus
Polygon: a closed flat shape bounded by straight lines. It always has the same
number of angles as sides.

If all sides are the same length, it is called a regular polygon. Some polygons have
special names, referring to the number of sides it has:
Number of sides / angles
Name
3
triangle
4
quadrilateral
5
pentagon
6
hexagon
7
heptagon
8
octagon
9
nonagon
10
decagon
The angles of a polygon sum to (n-2) x 180º, where n is the number of sides in the
polygon. In a regular polygon, all angles are the same size, i.e. divide (n-2) x 180º
by n to find one angle.

Circles: A line from the centre of a circle to the
circumference is called the radius. A line joining
two points on the circumference is a chord.
Facilitator Guide
A
US 9016 – Physical Quantities and Geometrical Relationships 4
21
Sparrow Research and Industrial Consultants © January 2006
diameter cuts the circle in half, creating two semi-circles

An angle subtended by a diameter is a right angle, i.e.
AĈB = 90º

A tangent to a circle is a line which touches the circle in one
point only. If a radius is drawn to the point of contact with the
tangent, then the radius will be perpendicular to the tangent.
Exercise 5 – Background knowledge
1. Complete the table by making ticks  for TRUE, and crosses x for FALSE.
Property
Trapezium Kite Parallelogram Rectangle Rhombus Square
One axis of
symmetry only
✗
✓
✗
✗
✗
✗
2 pairs of sides
parallel
✗
✗
✓
✓
✓
✓
All sides equal
✗
✗
✗
✗
✓
✓
Diagonals bisect
each other
✗
✗
✓
✓
✓
✓
Diagonals are equal
in length
✗
✗
✗
✓
✗
✓
No axes of
symmetry
✓
✗
✓
✗
✗
✗
2
GEOMETRIC CONSTRUCTIONS
In a previous module you were introduced to scale
drawings, and mainly learned how to interpret them.
Now you must learn to make a few scale drawings.
These need to be very accurate, so your instruments
must be in good working order and pencils very sharp.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
22
Sparrow Research and Industrial Consultants © January 2006
Necessary instruments

A ruler, marked in cm and mm (a transparent ruler works best)

A protractor (to measure angles)

A compass (to draw circles, and other uses)

A set square (transparent triangle) for drawing parallel lines

Sharp pencils

A rubber
Make sure that the compass holds the pencil tightly, if you can find one using lead, like a
“pacer pencil”, that works best. When using a compass, hold it by the top part only, not the
two “arms”, and make sure the sharp point is properly anchored on the paper.
3
DRAWING PARALLEL LINES
If you have a line AB and you have to draw a line parallel to it, you need your ruler and the
triangle.
Place the triangle next to the line as shown, and the ruler against the triangle. Now move the
triangle up along the ruler to where you want the next line to be. Draw the line.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
23
Sparrow Research and Industrial Consultants © January 2006
4
DRAWING A PERPENDICULAR LINE
If you want to draw a line perpendicular to line AB, simply put the triangle on the line, and
draw the perpendicular line:
5
DRAWING A PERPENDICULAR BISECTOR
A perpendicular bisector is a line which is perpendicular to, and bisects (cuts in half) another
line. For this you need your ruler and compass.
Start with your line AB. Open your compass so that it is slightly wider than half the length of
AB. You have to guess this, but it need not be an exact length! As long as it’s shorter than
AB itself. Put the sharp point at A, and draw an arc through AB:
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
24
Sparrow Research and Industrial Consultants © January 2006
Now do the same from B, making sure that you don’t change the width of the compass. A
line through the two points where your arcs cross will be the required line:
6
BISECTING AN ANGLE
Remember that “bisect” means to cut in half. Always make sure that you understand what is
bisected, an angle, or a line? Students often confuse the two because they don’t read the
question properly.
In this case you are given an angle DEF:
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
25
Sparrow Research and Industrial Consultants © January 2006
Put the point of your compass on E and draw an arc through DE and EF:
Now keep your compass at the same width, i.e. your radius remains the same. Put the
compass on F and draw a small arc, then do the same from D:
Draw a straight from E through the cross:
Now you must practise these constructions. Remember to leave your construction lines
showing, you get marks for the correct construction lines and markings.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
26
Sparrow Research and Industrial Consultants © January 2006
Exercise 6 - First practise these basics
1. Use your ruler to draw a line AB, 5 cm long. Now draw a line above it, parallel to AB.
Make this line 5 cm long and call it DE.
2. Draw a line EF, 5 cm long. Now draw DE perpendicular to EF, i.e. the lines must meet
at E.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
27
Sparrow Research and Industrial Consultants © January 2006
3. Draw a line AB, 8 cm long. Now draw the perpendicular bisector of AB.
4. Draw a line ST, 4 cm long. Use your protractor to measure an angle of 70º from S and
draw the line PS. Now you have PST = 70º. Use your compass to bisect this angle. Use
your protractor to check that the angles are 35º each.
1. Use a ruler and compass to construct a triangle ABC with AB = 5 cm, BC = 6 cm and AC
= 2 cm.
Solution
Start by drawing one line, say BC, using your ruler. Open the compass to a length of 5 cm
for AB.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
28
Sparrow Research and Industrial Consultants © January 2006
Put the point on B and draw an arc with radius 5 cm.
Using your compass to measure 2 cm, draw an arc from C. Where the two arcs intersect, is
A.
2. From the top of a cliff, 45 m above sea level, the angle of depression of a boat is 32〬.
Using a scale of 1 cm = 5 m, make a scale drawing to determine how far the boat is from
the foot of the cliff.
Solution
The angle of depression is the angle measured from the horizontal down, whereas the angle
of elevation would be measured from the horizontal up.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
29
Sparrow Research and Industrial Consultants © January 2006
First draw a rough sketch of what you have in mind:
Angle FTB = 58º, because the horizontal line is perpendicular to the vertical cliff.
In the scale drawing, the cliff of 45 m must be represented by 9 cm. To make the scale
drawing you must:

Draw a horizontal line representing FB, of unknown length, just make it fairly long for
starters

Draw FT perpendicular to TB, using the triangle.

Use your ruler to mark point T 9 cm above F.

Use the protractor to measure an angle of 58〬 = FTB.

Draw this line, where it cuts FB, will be the correct point B.

Use your ruler to measure FB and multiply by 5, this will give the distance FB in
meters.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
30
Sparrow Research and Industrial Consultants © January 2006
If your sketch was accurate, you should get FB = 14,4 cm, which gives the distance of the
boat from the foot of the cliff as 72 m.
Exercise 7 Basic geometry
1. Draw an equilateral triangle SDF, with sides = 10 cm each. Draw the perpendicular
bisectors of SD and DF. Name the point of intersection T. Draw the perpendicular
bisector of SF. What do you notice?
Perpendicular bisector of SF goes through T.
7
THE THEOREM OF PYTHAGORAS
In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on
the two other sides.
If  = 90°, then BC2 = AB2 + AC2
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
31
Sparrow Research and Industrial Consultants © January 2006
A light house, 10m high, is on top of a cliff which is
70m above sea level. A small boat, 192m from the
foot of the cliff, can see the light from the light house.
How long is the beam of light?
Solution
AC2 = AB2 + BC2
= 802 + 1922
= 43 264m2
 AC = 208m
What is the radius of the circle shown?
Solution
Inside the quarter circle is a rectangle, thus opposite sides are equal. Radius is OB:
OB2 = 32 + 42 = 25
 OB = 5 units
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
32
Sparrow Research and Industrial Consultants © January 2006
8
CARTESIAN CO-ORDINATE SYSTEM
The Cartesian square is the perpendicular intersection of two number lines. The vertical line
is called the y-axis and the horizontal line the x-axis. The point of intersection is called the
origin and has coordinates of (0;0). The position of any point is given as an ordered pair of
numbers, written in brackets. For example, the vertexes of triangle PQR on the Cartesian
square below will have the following coordinates: P (-2;2) Q(1;1) R(-1;-2)
Note that the order is important. The x-coordinate is always first and the y-coordinate
second.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
33
Sparrow Research and Industrial Consultants © January 2006
On the Cartesian square below, draw triangle ABC with vertexes A (-2;3), B (3;1) and C (-2;2).
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
34
Sparrow Research and Industrial Consultants © January 2006
CHAPTER 3:
BASIC GEOMETRY
On completion of this chapter, you will be able to:

Give directions on a map by using bearings
1
BEARINGS
To give directions on a map, we use bearings. Start with the 8 compass points:
Directions can be given in two ways:

Kimberley is 48º northeast of Cape Town, or

The bearing of Kimberley from Cape Town is 048º
In the first case, you must draw a vertical line representing North from Cape Town and
measure an angle of 48º to the east of the north line:
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
35
Sparrow Research and Industrial Consultants © January 2006
In the second case you would do the same, only this time the direction is always clockwise
from north. So the bearing of Cape Town from Kimberley would read as:

42º south of west , or

180º + 48º = 228º (see sketch below)
1. Three towns, P, Q and R have the following bearings: Q is 130 km east of P; R is 45º
south of east of P and 70º south of west of Q. Make a scale drawing of this information
and determine the distances between the towns. Use a scale of 1 cm : 10 km.
Solution
Start with line PQ = 13 cm. Then use your protractor to measure an angle of 45º south of
this line, starting from P, draw a line. Starting from Q, measure 70º to the south, draw a line.
Where the two lines meet, is R.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
36
Sparrow Research and Industrial Consultants © January 2006
Measuring PR and QR, you should get PR = 13,5 cm and RQ = 10,2 cm. So the distances
are PR = 135 km and QR = 102 km.
Exercise 8 - Bearings
1. Bloemfontein is 45 km south and 140 km east of Kimberley. Use a scale drawing with
scale 1 cm : 20 km to answer the questions:
a)
What is the bearing of Kimberley from Bloemfontein?
290° or
Facilitator Guide
70° west of north
US 9016 – Physical Quantities and Geometrical Relationships 4
37
Sparrow Research and Industrial Consultants © January 2006
b)
What is the distance from Kimberley to Bloemfontein?
On the scale drawing, 7,4 cm
 distance = 7,4 x 20 = 148 km
2. Themba drives 50 km in the direction 30° south of east, then he drives 94 km in the
direction 28° south of west.
a)
Make a scale drawing, using scale 1 cm : 10 km.
b)
Use the drawing to calculate how far he is from his starting point.
8cm x 10 km = 80 km
c)
In which direction must he drive to go straight home?
30° east of north
d)
What is the bearing of his endpoint from his
starting point?
210°
3. Two lighthouses are 50 km apart. The bearing of A
from B is 0°. A lies on bearing 300° from the boat,
while the bearing of B from the boat is 230°. Use a
scale of 1cm : 4km to answer the questions:
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
38
Sparrow Research and Industrial Consultants © January 2006
a)
How far is the boat from A?
10,8 cm x 4 km = 43,2 km
b)
How far is the boat from B?
11,7 cm x 4 km = 46,8 km
c)
What is the bearing of the boat from B?
50°
2
INTERNATIONAL TIME ZONES
The earth is divided into lines of latitude and longitude. The lines of latitude starts from the
equator, which is the 0º line of latitude, moving north and south to the North and South poles
respectively. The lines of longitude originate from the 0º longitude, the Greenwich line. They
are displaced eastward and westward to the 180º line of longitude. This is known as the
International Date Line. The interval between two lines of longitude and latitude respectively
are one degree. Every 15º longitude is known as one time zone. A 15º longitude interval on a
map is equivalent to sixty minutes or one hour. If you cross the International Date Line from
east to west you need to subtract one day. When crossing the line from west to east you will
add one day. Every 15º you move west, subtract one hour and add one when moving east
over a line of longitude.
In the sketch below every brown line indicates a time interval of one hour and is therefore 15º
apart. It is also important to note that the International Date Line (the green line) never
crosses the land.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
39
Sparrow Research and Industrial Consultants © January 2006
Exercise 9: Time Zones
If the current time at location D is 15:45, find the times at the following locations:
a)
A; B; C; E; F.
Solution: A: 19:45 B: 15:45 C: 18:45 E: 12:45 F: 14:45
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
40
Sparrow Research and Industrial Consultants © January 2006
CHAPTER 4:
GEOMETRIC FORMS
On completion of this chapter, you will be able to:

Identify a range of regular three dimensional forms

Calculate perimeter and areas of two dimensional shapes

Develop nets for regular three dimensional forms

Calculate surface area and volume of three dimensional forms
1
SOLID FIGURES
In a previous module (US 9013) you have worked with solid figures, 3-dimensional shapes.
Also, you have worked with 2-dimensional shapes and polygons.
The 3-dimensional
equivalent of a polygon is a polyhedron, i.e. a solid whose surface consists of polygons.
We need to work in 3 dimensions. Here is a list of solids that will be discussed in this course:
Figure
Sketch
Name
Description/properties
A ball: Every point on the surface is
the same distance from the centre
Sphere
Its surface consists of 6 squares.
All sides are the same length
Cube
A rectangular block
Cuboid / rectangular
prism
Base is a polygon. Sides are
triangles
Pyramid
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
41
Sparrow Research and Industrial Consultants © January 2006
Right triangular prism
Base is a triangle. Sides are
rectangles, which are perpendicular
to the base
Circular base and top, like a can of
cool drink.
Cylinder
The following sketch shows a rectangular prism:
Its surface consists of 6 rectangles; each rectangle is called a face of the solid. The line
between two faces is called an edge, and edges meet at a vertex of the solid. A prism is a
polyhedron where the top and base faces are identical polygons. If the other faces are
vertical to the base, the prism is called a right prism.
In a pyramid all faces (except for the base) must be triangles. The base could also be a
triangle, but it may be any polygon.
A pyramid is usually named for its base, e.g.
a
triangular pyramid, or a square-based pyramid.
You have also been taught how to draw the net of a solid. A solid can have several different
nets, as long as they could be cut out and folded into the solid. It is not possible to draw a
net of a sphere, because one cannot flatten the curved shape without changing (stretching or
compressing) the actual shape of the sphere.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
42
Sparrow Research and Industrial Consultants © January 2006
Look at the following nets. If you should cut them out and fold them into 3-dimensional
shapes, what would they be? Choose your answers from the following: cone; cube; cuboid;
cylinder; pyramid; sphere; triangular prism.
1.
2.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
43
Sparrow Research and Industrial Consultants © January 2006
3.
4.
5.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
44
Sparrow Research and Industrial Consultants © January 2006
Solutions
1. Triangular prism
2. Cone
3. Rectangular prism/cuboid
4. Pyramid
5. Cylinder
It is possible to draw a net in different ways, for example, a cube could have the following
nets, and more!
Exercise 10 – Solid figures
1. The diagram shows the net of a solid.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
45
Sparrow Research and Industrial Consultants © January 2006
a)
What type of solid is this?
Rectangular pyramid
b)
How many faces has the solid?
5 faces
c)
How many edges has the solid?
8 edges
d)
In an accurately drawn net, which of the edges must be equal?
AB=BC=CD=DE=EF=FG=GH=HA and BD=FH and BH=DF
2. This diagram represents an octahedron (a solid with 8 faces). Each face is an equilateral
triangle (all 3 sides the same length).
a)
How many edges does the octahedron have?
12 edges
b)
How many vertices does the octahedron have?
6 vertices
c)
Draw a net of the solid.
The diagram shows one possibility.
The octahedron consists of 8 equilateral
triangles, but remember that it must be possible to cut out the shape and fold it
into the solid.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
46
Sparrow Research and Industrial Consultants © January 2006
3. The diagram shows a match box:
a)
What is this shape called?
Rectangular prism
b)
Draw an accurate net of the box
c)
How many boxes would fit into a carton of 7 cm by 5 cm by 5,5 cm? Find the
answer by calculating volumes first, and then draw a sketch to determine if this is
actually possible.
Volume of match box = 1 x 3,5 x 5,5 = 19,25 cm3
Volume of carton = 5,5 x 7 x 5 = 192,5 cm3
10 boxes will fit
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
47
Sparrow Research and Industrial Consultants © January 2006
2
PERIMETER AND AREA
In a previous course you were taught to find the perimeter and area of 2-dimensional shapes.
2.1
Basic formulae for perimeters and areas
Shape
Perimeter
2 (L + B)
add all 4 sides
Area
length × breadth
=LB
½ × sum of parallel sides ×
perpendicular height
= ½ (a + b) h
add all 3 sides
2πr
2 (b + c)
Facilitator Guide
½ × base × perpendicular height
=½bh
π × radius × radius
= π r2
base × perpendicular height
US 9016 – Physical Quantities and Geometrical Relationships 4
=bh
48
Sparrow Research and Industrial Consultants © January 2006
Exercise 11 – Perimeters and areas
1. In the following diagrams, find the area and perimeter:
a)
Diagonal side of triangle:
x2 = (3,5)2 + 82
∴ x = 8,7 cm
Perimeter of triangle = 2(8,7) + 7 = 24,4 cm
Perimeter of rectangle = 2(4) + 8 = 16 cm
Circumference of semi-circle = ½(2πx4) = 12,57 cm
∴ Total perimeter = 52,97 cm
Area of triangle = ½(7)(8) = 28 cm²
Area of rectangle = 8 x 4 = 32 cm²
Area of semi-circle = ½π(4)² = 25,13 cm²
∴ Total area = 85,13 cm²
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
49
Sparrow Research and Industrial Consultants © January 2006
b)
Perimeter of trapezium = 3 + 3 + 5 + 4,5 = 15,5 m
Area of trapezium = ½(5 + 3)(3) = 12 m²
2. A rectangular metal plate measuring 10 cm by 1 m, has circular holes cut in it. There are
4 holes of diameter 3 cm each, and 7 holes of diameter 1 cm each. Calculate the area of
metal.
Area of rectangle = 10 x 100 = 1000 cm²
Area of large circle = π(1,5)² = 7,07 cm²
Area of small circle = π(0,5)² = 0,79 cm²
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
50
Sparrow Research and Industrial Consultants © January 2006
∴ area of metal = 1000 – 4(7,07) – 7(0,79) = 966,19 cm²
3. If a trundle wheel does 2000 revolutions in 1 km, what is its radius?
1 km = 1000 m
∴ Circumference = 1000 ÷ 2000 = 0,5 m = 50cm
∴ 50 = 2πr
∴ r = 50 ÷2π = 8 cm
3
SURFACE AREA AND VOLUME
As you know by now, the surface area of a solid is the area of its net. We give the
formulae for surface areas and volumes, but remind you that you can draw the net (except
for a sphere!), and calculate the surface area this way too.
The following solids are important in this course, and their surface areas and volumes are
given:
SOLID
VOLUME
base area × height
= πr2 H
lxbxH
TOTAL SURFACE AREA
(closed surface)
(2 × base area) + (base
perimeter × height)
= 2πr2+ 2πrH
2(l + b)H + 2(l x b)
s×s×s
= s3
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
6s2
51
Sparrow Research and Industrial Consultants © January 2006
SOLID
VOLUME
TOTAL SURFACE AREA
(closed surface)
(perimeter of base = add 3
sides)
½bh × H
(a + b + c)H + 2 × ½bh
= (a + b + c)H + bh
⁄3 πr3
4πr2
4
πrl + πr2
⅓πr2 h
⅓ area of base ×
perpendicular height
= πr(r + l)
area of base + areas of
triangular sides
The formulae for the pyramid were general formulae, because the base may be any polygon.
1. The pyramid has square base ABCD with sides
= 8 cm. Altitude PT = h = 3 cm and TE =
½AB.
a)
Explain why T is the midpoint of AC.
b)
What kind of triangle is PTE?
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
52
Sparrow Research and Industrial Consultants © January 2006
c)
What is the volume of the pyramid?
d)
What is the length of PE?
e)
Find the surface area.
Solution
a)
The diagonals of a square bisect each other.
b)
Right angled triangle, PT  TE.
c)
V = ⅓ x area of base x perpendicular height
= ⅓ x 8 2 x 3 = 64 cm3
d)
PE2 = PT2 + TE2
8 + 16 = 25
PE = 5 cm
e)
SA = area of base + areas of triangular sides
= 82 + 4(½ x 8 x 5)
= 144 cm2
2. In the sketch is an ice cream cone with a round ball of ice
cream, 6 cm in diameter. It is topped with a cherry which has a
diameter of 10 mm. The cone has a perpendicular height of 9,5
cm and slant height of 10,5 cm. The radius of the cone is 28
mm.
a)
What is the volume of the ball of ice cream? Give your
answer in cm3.
b)
Find the surface area of the cone.
c)
Calculate the total volume of the cone, ice cream and cherry in cm3.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
53
Sparrow Research and Industrial Consultants © January 2006
Solution
3
π(3)3 = 113,10 cm3
4
a)
Volume of sphere =
b)
Surface area of cone = π(2,8)(10,5) = 92,36 cm2
c)
Volume of ice cream = 113,10 cm3
Volume of cherry = =
3
π(1)3 = 4,19 cm3
4
Volume of cone = ⅓π(2,8)2 (9,5) = 78,00 cm3
 total volume = 195,29 cm3
3. A solid sphere of diameter 10,5 cm and a cylinder
15,75 cm in height have equal volumes. Find the
diameter of the cylinder.
Solution
Volume of sphere =
=
3 3
πr
4
3
π (5,25)3
4
= 606,13 cm3
Volume of cylinder = π r2 H
606,13 = π r2 x 15,75
606,13 = 49,48 r2
r2 = 12,25
r = 3,5 cm
 diameter = 7 cm
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
54
Sparrow Research and Industrial Consultants © January 2006
Exercise 12 – Surface area and volume
12 cm
1. A tissue box has dimensions 15 cm by 12 cm by
12 cm
12 cm. It has a circular hole of diameter 6 cm
where the tissues can be removed.
Find the
surface area of the box.
15 cm
Surface area = 2(12 x 12) + (4 x 12 x 15)
– π(3)²
= 1008 – 28,27
= 979,73 cm²
2. A circular kiddies’ swim pool has radius 1,5 m and is filled with water to a depth of 30 cm.
Calculate the volume of water in the pool.
Give your
answer in litres. (1000 cm³ = 1ℓ)
Volume = πr² x H
= π(150)² x 30
= 2120575 cm³
1000 cm³ = 1 ℓ,
 pool has 2120 ℓ of water.
3. A rectangular block of metal has dimensions 25 cm by 16 cm by 10cm. It is melted down
and cylindrical bullets of diameter 1 cm and height 2 cm are made. How many bullets
can be made? How much metal will be left as scrap?
Volume of block = 25 x 16 x 10 = 4000 cm³
Volume of 1 bullet = πr² x H
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
55
Sparrow Research and Industrial Consultants © January 2006
= π(0,5)² x 2
= 1,57 cm³
 number of bullets = 4000 ÷ 1,57 = 2547,8
 2547 bullets can be made. 1,21 cm³ of metal is left.
4. A pendant is made out of wood, in the form of a
cross with longer sides 1 cm each and shorter sides
0,5 cm each.
The pendant is 3 mm thick.
A
diamond shape (rhombus) with diagonals 1 cm and
0,7 cm each, is cut from a semi-precious stone and
placed in the middle of the cross. It is 2 mm thick.
What is the total volume of the pendant?
Volume of wood = [(1 x 2) + 2(1 x 0,5)] x 0,3
= 0,9 cm³
To find area of rhombus, divide it into two triangles with base = 7 mm and height = 5
mm. (diagonals bisect at right angles)
area = 2 x (½ x 0,7 x 0,5) = 0,35 cm²
volume = 0,35 x 0,2 = 0,07 cm³
 Total volume = 0,97 cm³
5. A toilet roll core has a diameter of 5 cm and a width
of 10 cm. Find its external surface area (ignore the
internal surface area) and the volume inside the
core.
A length of toilet paper is wound around it 500
times. How long is the toilet paper if one can assume the average diameter of the
roll is 8 cm? Round your answer to the nearest metre.
External surface area = 2πrH
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
56
Sparrow Research and Industrial Consultants © January 2006
= 2π x 2,5 x 10
= 157,05 cm²
Volume = πr² x H
= π x (2,5)² x 10
= 196,31 cm³
Average circumference = 2πraverage
= 2π x 4
= 25,128 cm
Length = 500 x 25,128 = 12 564 cm ~ 125 m long
6. A lightweight tent is in the form of half a sphere
(hemisphere); with radius 2 m. Find the area of
outer canvass (surface area) and volume of
space in the tent.
Surface area = ½(4πr²)
= 2π x 2² = 25,13 m²
Volume = ½(⁴⁄₃πr³)
= ⅔π x 2³ = 16,75 m³
7. Find the volume and total surface area of each solid
a)
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
57
Sparrow Research and Industrial Consultants © January 2006
V = base area x H
= ½ x 3 x 4 x 7 =42 cm3
Surface area = (2 x base area) + (base perimeter x height)
For perimeter of base, we need the length of AC. Use the theorem of Pythagoras:
AC2 = AB2 + BC2
=9 + 16 = 25
 AC = 5 cm
 Surface area = (2 x ½ x 4 x 3) + [(3 + 4 + 5) x 7]
= 96 cm2
b)
The T is the base, because then all sides are perpendicular to it.
V = base area x height
Base area = (4 x 1) + (5 x 1) = 9 m2
 V = 9 x 6 = 54 m3
Surface area = ( 2 x 54) + [(5 + 1 + 2 + 4 + 1 + 4 + 2 + 1) x 6]
= 108 + 120 = 228 m2
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
58
Sparrow Research and Industrial Consultants © January 2006
4
SCALE DRAWINGS
Scale drawings of three dimensional figures can also be constructed giving the front, side
and top views of the object as illustrated in the figures below.
Three dimensional figures like these are drawn using isometric paper. An example of two
such three dimensional figures are given below. Try to redraw these two figures on the
isometric paper provided. Remember that all measurements on scale drawing are made in
millimetres to ensure accurate drawings.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
59
Sparrow Research and Industrial Consultants © January 2006
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
60
Sparrow Research and Industrial Consultants © January 2006
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
61
Sparrow Research and Industrial Consultants © January 2006
Different views (front, top and side) of three dimensional objects can also be used to illustrate
the operation mechanisms of some objects.
The following figure is a front view of a car jack. It has a very simple, but effective operation
mechanism.
The upper and lower lever arms rotate around four axils, two each fixed in the base block
and the upper block respectively. When the cork screw is turned clockwise the axils at point
9 and 10 will move closer together thus increasing the height between the upper and lower
axils. This action will cause the jack to lift the object it supports. To lower the object again the
cork screw must be turned in the opposite direction.
The plan of a house or a factory is another example of scale drawings. A drawn plan of a
house is an exact replica of the house. The scale on a plan is of utmost importance. Using
the scale, all the measurements of the house can be calculated correctly. Take a close look
at the following example of a scale:
1:120
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
62
Sparrow Research and Industrial Consultants © January 2006
This scale tells us that one unit on the plan represents 120 units in real life. Thus, if the width
of a room measured on the plan equals 3 cm, then the width of that same room in the house
can be measured as follows:
Width measured on plan x scale = Real width
30 mm x 120 = 360 cm or 3.6 metres
Study the following scale drawing (plan) of a house and complete the questions to follow:
a) Measure the width of the door opening at the bottom of the plan marked ENT.
b) Calculate the length and width of the kitchen.
c) Determine the area of the kitchen.
NB: All your answers must be given in metres.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
63
Sparrow Research and Industrial Consultants © January 2006
5
MAPS
A map is a special kind of scale drawing. It is a representation of reality (distance, relief,
roads, rivers, etc.) Just like the plan for a house, a map also has a scale. This scale is used
to determine the distance between two places on the map and if calculated correctly will give
you the real distance between the two places.
Study the following example of a scale map: Map Scale = 1:50 000
Determine the distance between town A and B if the distance measured on the map was 63
mm. (Remember always to measure in millimetres on a map.)
Use this equation: Real distance = (Map distance x Map scale)  (1000)
This equation will give the answer in meters. If you divide by one million in stead of a
thousand, the answer will be in kilometres.
Our example: (63 x 50 000)  (1000) = 3150 m
Calculate the real distance (in kilometres) between town A and town B if the distance
measured on the map was 105mm. The scale of the map is 1:100 000.
Solution:
105 100000 1000000  10,5km
Remember to divide by 1000 000 to convert mm to km.
Facilitator Guide
US 9016 – Physical Quantities and Geometrical Relationships 4
64