1.5 Measurement
AS 90130 Internal (3 credits)
Calculate the area of the following
shapes.
2 cm
3 cm
A = 9 cm2
4 cm
4 cm
6 cm
A = 12.6 cm2
5 cm
5 cm
A = 45 cm2
3 cm
2 rugby fields side by side
5 mL
8L
3760m
750m2
55m2
1600cm3
60W
2m
750mL
80kg
1.3kg
40L
-22°C
1500m
160kPa
37.4°
2000cm2
1m
9.85s
50g
180mm
Try These – Metric Conversions
55000 2
5.5 ha = _______m
2500 cm
25 m = ________
345 m = _______km
.345
4.321
4321 mm = _______m
120 mm
0.12 m = ________
237 cm = _________
2.37 m
998 km = ____________
99 800 000 cm
1.995 km
199500 cm = ________
5700
5.7 L = ______mL
3.4
3400 kg = ______T
420
0.42 T = _______kg
4300 m2 = ______ha
0.430
30000 2
3 m2 = ________cm
400000mm2 ____m
0.4 2
4.5 L
4500cm3 _______
IWB Fundamentals
Ex. 8.02 pg 220
Starter
12 ha of land is subdivided in 650 m2 lots.
How many lots of this size can be created?
120000 m2
12 ha = _______
184 lots can be created.
120000 m2 = 184.6
650 m2
A teaspoon holds 5 cm3. How many teaspoons are
needed to measure out exactly 1 m3 of water?
5 mL
5 cm3 = ____
= .005
____ L
1 m3 =1000
____ L
1000 L = 200 000 teaspoons
.005 L
Note 1: Limits of Accuracy
• Measurements are never exact. There is a limit to
the accuracy with which a measurement can be
made.
• The limits of accuracy of measurement refers to the
range of values within which the true value of the
measurement lies.
• The range of values is defined by an upper limit and
a lower limit.
Limits of Accuracy
• To find the upper limit, add 5 to the nearest
significant place.
• To find the lower limit, minus 5 to the nearest
significant place.
e.g. The distance to Bluff on a signpost reads 17 km.
The upper limit is 17 + 0.5 = 17.5 km
The lower limit is 17 – 0.5 = 16.5 km
Therefore the limits of accuracy are 16.5 km ≤ Bluff ≤ 17.5 km
e.g. From home to school it is 27.5 km.
What are the limits of accuracy for
my distance to school?
The upper limit is 27.5 + 0.05 = 27.55 km
The lower limit is 27.5 – 0.05 = 27.45 km
Therefore the limits of accuracy are:
27.45 km
≤ Distance ≤ 27.55 km
e.g. At the Otago vs Auckland game at Carisbrook
it was reported that 28500 people attended.
Give the limits of accuracy for the number of
people attending the game?
The upper limit is 28500 + 50 = 28550
The lower limit is 28500 – 50 = 28450
Therefore the limits of accuracy are:
28450 ≤ People ≤ 28550
Give the limits of accuracy for these measurements:
67.5 mm ≤ x ≤ 68.5 mm
1.) 68 mm
2.) 397 mm
396.5 mm ≤ x ≤ 397.5 mm
3.) 4 seconds
3.5 seconds ≤ x ≤ 4.5 seconds
4.) 50 g
45 g ≤ x ≤ 55 g
5.) 5890 kg
5885 kg ≤ x ≤ 5895 kg
6.) 820 cm
815 cm ≤ x ≤ 825 cm
7.) 92 kg
91.5 kg ≤ x ≤ 92.5 kg
89.05° ≤ x ≤ 89.15°
8.) 89.1°
Note 2: Areas
Shape
Area Formula
Example
4 cm
Rectangle
2 cm
Area = b × h
Area = 4 x 2 = 8 cm2
3 cm
Square
Area = l × l
3 cm
Area = 3 x 3 = 9 cm2
Triangle
Area = ½ b × h
Area = ½ b×h
=½×6×2
= 6 mm2
2 mm
6 mm
Note 2: Areas
Shape
Area Formula
Example
7m
Parallelogram
2m
Area = b × h
Area = 7 x 2 = 14 m2
2 cm
Trapezium
Circle
Area = 1/2 (a+b)× h
Area =
πr2
6 cm
4 cm
Area= ½ (4+2)×6= 18 cm2
Area = πr2
= π × (5)2
= 78.5 cm2
Note 3: Perimeter & Circumference
• The perimeter of a figure is the total length of
its sides.
The perimeter of this kite is:
5 cm
11 cm
11 cm + 5 cm + 11 cm +5 cm
= 32 cm
• The circumference of a circle is the total
distance around it.
C = 2πr
or
C = πd
e.g. Calculate the circumference of
a circle which has a radius of 32 cm
C = 2πr
= 2 × π × 32
= 201.1 cm (4 sf)
• To calculate the radius, when given the
circumference, we need to rearrange the
formula to make r the subject.
C = 2πr
r= C
2π
e.g. Calculate the radius of a circle that has a
circumference of 11.5 m
r = 11.5
2π
r = 1.83 m
Find the perimeter of this shape that is formed
using 3 semicircles (2dp)
r = 6.2 (for large circle)
(π x 6.2) + (π x 6.2)
= 2 x π x 6.2
= 38.96 cm
12.4 cm
Find the perimeter if each square is exactly half the
dimensions of the preceding square.
24 mm
24 mm
Ex. 9.03 pg 250-254
Ex. 9.04 pg 258
Ex 9.05 pg 263-265
(24 x 3) + (12 x 3) + (6 x 3) + (3 x 3) + (1.5 x 4)
= 141 cm
Starter
A cage is to be constructed entirely of steel bars as
shown. Steel bar costs $4.35/m and you have $350 to
spend on steel. The cage consists of steel bar uprights
and a circular hoop top and bottom.
The structure is to be twice as wide as it is high.
Calculate the height (x) and diameter (2x)
Length of steel required = 21x + 2(2πx)
2x = 4.8 m
x
= 33.57 x
Amount of steel to purchase = $350
$4.35
33.57 x = 80.46
x = 2.40 m
= 80.46 m
Hint – there are 21 uprights
and 2 circles
Note 5: Compound Areas
• Compound Areas are made up of more than
one mathematical shape
• To find the area of a compound shape, find
the areas of each individual shape and either
add or subtract as required.
e.g. Find the area of this shape
2 cm
Area of compound shape:
= area of Rectangle + area of Triangle
+ area of semi-circle
5 cm
4 cm
Area = b × h + ½ b × h + ½ π × r2
= 4 × 5 + ½ 4 × 2 + 0.5 π × (2)2
= 30.3 cm2 (1 dp)
e.g.
Think of a typical running track.
What is the perimeter? 400 m
How long are the straight sections? 100 m
Calculate the area
enclosed by the track.
Acircle + Arectangle = Atotal
3183.1 + 6366.2 = 9549.3 m2 (1dp)
100 m
d=?
Starter
A glass porthole on a ship has a diameter
of 28 cm. It is completely surrounded by
a wooden ring that is 3 cm wide.
a.) Calculate the area of glass in the
porthole A = πr2
r = 14 cm
A = π (14)2
A = 616 cm2
b.) Calculate the area of the wooden ring
Area of porthole = πr2 , r = 17 cm
(including frame) = 908 cm2
Area of frame = 908-616
= 292 cm2
Note 6: Area of a Sector/ Arc Length
Recall: Area of a circle = πr2
• The sector is part of a circle
• Area of a sector = x° × πr2
360°
e.g.
35°
6 cm
x°
r
Area of a sector = 35° × π × (6 cm)2
360°
=
11.0 cm2
Note 6: Area of a Sector/ Arc Length
• Arc length = x° × 2πr
360°
e.g.
75°
6 cm
x°
r
Arc length = 75° ×
360°
=
2π × 6
7.85 cm (3 sf)
e.g. The length of the minor arc of a circle is 3π
cm and the length of the major arc is 15π cm
a.) Find the radius of the circle
3π cm
Total circumference = 3π + 15π
= 18π
60°
So 18π = 2 × π × r
18 = 2 r
r = 9 cm
b.) Find the angle of the minor sector
= length of minor arc × 360°
total circumference
3π
= 18 π × 360°
15π cm
= 1 × 360°
6
= 60°
Note 7: Calculating the Radius/ Diameter
from the Area or Circumference
• When we know the circumference or area of a
circle, we can rearrange the equation to
calculate the diameter or radius of the circle.
e.g.
The circumference of a
circle is 25.4 cm. Calculate the
radius.
C = 2πr
25.4 cm = 2π × r
r = 25.4 cm
2π
r = 4.0 cm (1 dp)
• e.g.
The area of a circle is 35.6 cm2
Calculate the diameter.
A = πr2
35.6 = π × r2
35.6 = r2
r = 3.37 cm
Diameter = 6.7 cm (1 dp)
π
√
r = 35.6
π
IWB Fundamentals
Ex 9.04 pg 258-259
Starter
A concrete courtyard is designed as in the diagram below. There are
four circular gardens (all the same size) and one square garden.
The rest of the courtyard is to have a pattern moulded into the
concrete to look like bricks. The cost of each brick mould is 83¢ and a
moulded paver measures 250 mm x 150 mm.
Estimate the minimum number of
moulds for the entire courtyard
(assume no wastage), giving your
answer to the appropriate level of
precision and then calculate the
total price.
2.5 m
2m
24 m
18 m
Merit - Starter
The rest of the courtyard is to have a pattern moulded into the
concrete to look like bricks. The cost of each brick mould is 83¢ and a
moulded paver measures 250 mm x 150 mm.
Estimate the minimum number of moulds for the entire courtyard
(assume no wastage), giving your answer to the appropriate level of
precision and then calculate the total price.
Area courtyard = (24 ×18) – 4(π×1.252) – 22
= 408.37 m2
No. of moulds = 408.37 m2/ (.25 × .15)
= 10 890
2.5 m
2m
Total cost = 10890 × $ 0.83
≈ $9100
24 m
18 m
Note 8: Surface Area
• The surface area of a solid is the sum of the
areas of each face or curved surface.
It is helpful to picture
surface area as the net
of the shape
a
a
SA (cube) = 6a2
Surface Area
6.4 cm
Calculate the surface area of
this triangular prism
How many faces are there?
4 cm
5 cm
Left Rectangle: b × h
= 7 cm × 4 cm
= 28 cm2
Base Rectangle: b × h
= 7 cm × 5 cm
= 35 cm2
5
2 Triangles: 2 × (1/2 b × h)
= 2 × 0.5 × 5cm × 4 cm
= 20 cm2
Right Rectangle: b × h
= 7 cm × 6.4 cm
= 44.8 cm2
Total surface Area = 127.8 cm2
Surface Area
SA (cone) = the curved surface + the circular base
= π r l + π r2
l
r
SA (sphere) = 4π r2
r
Surface Area (cylinder)
SA (cylinder) = area of rectangle + 2 ×(area of circle)
= 2πr × h + 2 ×(πr2)
= 2πrh + 2πr2
r
SA (cylinder) = 2πr (h + r)
Surface area depends on whether the
ends are open or not. Open cylinders
are called pipes.
Textbook
IWB Fundamentals
Ex 12.04 pg 156-157 Ex 9.06 pg 271-273
Ex 12.05 pg 157-160 Ex 9.07 pg 275-279
C = 2πr
h
Starter
A gardener wishes to sow grass seed on his land.
Grass seed is sold in two sizes. 500 g bags which cost $5.95/bag and
cover an area of 13 square metres and 1 kg bags which cost $8.95 and
cover double the area of the 500g bag.
What quantity and
combination of bags should
15 m
7.0 m
the gardener buy to ensure
9.2 m
he has enough to cover the
7.6 m
required land, and at the
4.1 m
best possible price?
22.0 m
Total Area = 22×15 – (4.1×9.2) – (7.0×7.6)
= 239.08 m2
239.08 m2 / 26 = 9.2 1kg bags
= 239 m2 (3 sf)
9 x $8.95 + 1 x $5.95 = $86.50
Note 9: Volume of Prisms
• Volume = Area of cross section × Length
L
The volume of a solid
figure is the amount of
space it occupies. It is
measured in cubic
centimetres, cm3 or cubic
metres, m3
Examples:
• Volume = (b × h) × L
= 4 m × 4 m × 10 m
= 160 m3
 Volume = (1/2 b×h) × L
= ½ × 4cm ×5cm
× 10 cm
= 70 cm3
4 cm
5 cm
Cylinder – A circular Prism
Volume = Area of cross section × Length
Volume (cylinder) = πr2× h
V =πr2× h
= π(1.2cm)2× 8cm
= 36.19 cm 3 (4 sf)
Merit
This tent-shaped plastic hothouse is to change its air five times
every hour. What volume of air per minute is required from the fan
to achieve this? Round appropriately.
Volume = ½ × 13 × 12 × 8
= 624 m3
12 m
624 m3 × 5 = 3120 m3/hr
13 m
1 hr
3120m 3
×
60 min
hr
= 52 m3/min
Starter
B
A
10cm
1 cm
8 cm
10cm
10cm
125 cm
1.) Calculate the volume of these two cubes /cuboids of ice
A = 1000 cm3
B = 1000 cm3
2.) Which would melt faster if left outside on a hot day?
3.) Calculate the total area of
the six faces for both pieces.
Merit
The walls of a lounge are to be wallpapered.
The room’s dimensions are depicted below.
Each roll of wallpaper is 50 cm wide and 8 m long. Each roll of
wallpaper costs $34.95. Calculate how many rolls of wallpaper
are needed to wallpaper the room, and the cost.
You must hang complete strips
of wallpaper to cover the 2.5 m
height, part pieces cannot be
pasted together. Ignore door,
windows and pattern matching.
Strips per roll – 8 m ÷ 2.5 m
= 3 / roll
2.5 m
9m
4.2 m
Long Wall – 9 m ÷ 0.5 m
= 18 strips across
18 ÷3 = 6 rolls per side = 12 rolls
Merit
The walls of a lounge are to be wallpapered.
The room’s dimensions are depicted below.
Each roll of wallpaper is 50 cm wide and 8 m long. Each roll of
wallpaper costs $34.95. Calculate how many rolls of wallpaper
are needed to wallpaper the room, and the cost.
You must hang complete strips of
wallpaper to cover the 2.5 m
height, part pieces cannot be
pasted together. Ignore door,
windows and pattern matching.
2.5 m
9m
Short Wall – 4.2 m ÷ 0.5 m
= 8.4 (9) strips across
4.2 m
9 ÷ 3 (strips per roll) = 3 rolls per side
= 6 rolls + 12 rolls (long) 18 x $34.95= $629.10
= 18
Note 10: Volume of
Pyramids, cones & Spheres
V = 1/3 A × h
Apex
A = area of base
h = perpendicular height
V = 1/3 A × h
= 1/3 (5m×4m)×8m
= 53.3 m3
Volume of Cones
A cone is a pyramid on a circular base
V = 1/3 × πr2× h
A = πr2 (area of base)
h = perpendicular height
Vertex
9 cm
V = 1/3 × πr2× h
= 1/3 × π×(1.5cm)2 × 9cm
= 21.21 cm3 (4 sf)
Spheres
• A sphere is a perfectly round
ball.
• It has only one measurement:
the radius, r.
• The volume of a sphere is:
V=
3
4/3πr
IWB Fundamentals
Ex 10.02 pg 296-299
Ex 13.02 pg 165-167
Starter
The sonar of a whale
can be heard within a
sphere of diameter
0.150 km. How many
litres of water are
contained in this
sphere?
V =4/3
3
πr
V = 4/3 π(75 m)3
V = 1767146 m3
= 1.77 x 106 kL
= 1.77 x 109 L
Merit
An excavator is digging a drainage trench. The shape is
twice as wide as it is deep. This particular trench has a
width across the top of 3.2 m and a length 245 m.
What is the best model to calculate the volume of
material removed?
What volume of material must be moved to make this trench?
Possible models
Half cylinder
Trapezoidal Prism
Volume = ½ x π
(1.6)2
= 985.2 m3
x 245
IWB Fundamentals
Ex 10.03 pg 301-304
3.2 m
1.6 m
* diagram not to scale
Starter
mm.
Vwith peel = 4/3 π (45)3
= 381 704 mm3
Vno peel = 4/3 π (40)3
= 268 083 mm3
Vpeel = Vwith peel – Vno peel
= 113 621 mm3
= 114 000 mm3
Note 11: Liquid Volume (Capacity)
 There are 2 ways in which we measure volume:
 Solid shapes have volume measured in cubic units
(cm3, m3 …)
 Liquids have volume measured in litres or millilitres
(mL)
Metric system – Weight/volume conversions for water.
Weight
Liquid Volume
Equivalent Solid Volume
1 gram
1 kg
1 mL
1 litre
1 cm3
1000 cm3
e.g.
600 ml = $ 0.83/0.6 L
= $ 1.383 / L
1 L = $ 1.39/ L
2 L = $ 2.76/ 2 L
= $ 1.38 / L
e.g.
a.) Calculate the area of glass required for the fish tank
S.A. = 2(55 × 42) + 2(18 × 42) + (55 x 18) = 7122 cm2
b.) Calculate the volume of water in the tank. Give your answer
to the nearest litre
Vtank = 55 × 42 × 18
= 41580 cm3
42 cm
Vwater = 4/5 (41580)
= 33264 cm3
= 33 L
55 cm
Example:
• Estimate the weight of a
333 mL can of soda.
333 grams + can
• How many cans would it
take to fill a container
which has a volume of
4300 cm3 ?
Number of cans = 4300
333
= 12.91
≈ 13 cans
IWB Fundamentals
Ex 10.05 pg 315-318
Ex 14.01 pg 173-175
Starter
A time capsule is buried in the foundations of a new
classroom block at JMC. It consists of a 20 cm cylinder,
fitted at each end with a hemisphere. The total length is
28 cm. What is the capacity (in L) of the time capsule?
28 cm
Volume (sphere) = 4/3πr3
= 268.08 cm3
20 cm
Volume (cylinder) = πr2 x h
= 1005.31 cm3
Radius = 4 cm
Total volume = 268.08 + 1005.31
= 1273.39 cm3
Capacity = 1273.39 mL = 1.273 L
Which of these buckets has the largest mass?
(assume volumes are the same)
Water
Popcorn
Gold Coins
Note 12: Density
 The density of an object describes the ratio of its
mass to a standard volume.
 The density of water is exactly 1. (1 cm3 = 1 g)
M=V×D
M
V
D
V=M
D
D=M
V
M = mass, V = Volume, D = Density
Density Examples:
 Concrete has a density of 2.3 g/cm3. What is the mass of a
block of concrete that measures 15 cm x 12 cm x 10 cm?
Ans: Volume = 15 x 12 x 10
M=V×D
M
= 1800 cm3
Mass = V x D
= 1800 x 2.3
= 4140 g (4.14 kg)
V
D
Density Examples:
 A 10 kg block of gold has a volume of 518 cm3. Calculate
the density of gold in g/cm3.
10 kg = 10000 g
D=M
V
M
Density = 10000 g
518 g/cm3
= 19.3 g/cm3
IWB Fundamentals
Ex 11.03 pg 331 - 334
Ex 11.01 pg 323 - 325
V
D
Gamma Textbook
Ex 14.02 pg 176-177
Starter
A gold wedding band with diameter 16 mm & a cross section
as shown below shows the band is semi circular with a
radius of 4 mm. Estimate the volume by imagining the ring
cut and opened up.
Density of gold is 19.3 g/cm3
Estimate the value of the ring if it
was melted down and recovered.
Assume gold is currently traded
for $59.70/g
16 mm
4 mm
Mass = 1.263 cm3 × 19.3g/cm3
= 24.38 g
Length of ‘opened up’ ring = π × 1.6 cm
= 5.0265 cm
Value = 24.38 g x $59.70
= $1456
Volume of ‘opened up’ ring = ½ × π × 0.42 × 5.0265 cm
= 1.263 cm3
Note 13: Time
• Equivalent times (in seconds)
1 minute = 60 seconds
1 hour = 60 minutes = 60 x 60 seconds = 3600 seconds
1 day = 24 hours = 24 x 60 x 60 seconds
= 86 400 seconds
24 hour time is represented using 4 digits e.g 0630 hours
6:30 am
12 hour clock times are followed by am or pm
There are 365.242 days in a year. We account for this by
adding an extra day every 4 years so that this error does
not build up.
International Time Zones
International Time Zones
Measuring and Modelling Practical
Situations.
• Apply the Skills we have learned
• Represent the situation using a Diagram
– Assign appropriate measurements
– Simplify and Clarify
• Use appropriate Units
– Be consistent
• Identify regular shape(s) from your diagram
• Do Calculations
• Relate your answer back to the original problem