1.5 Measurement

1.5 MeasurementAS 90130 Internal (3 credits)

Calculate the area of the following shapes.

6 cm

3 cm

5 cm

5 cm

4 cm3 cm

4 cm

2 cm

A = 9 cm2

A = 12.6 cm2

A = 45 cm2

2 rugby fields side by side

5 mL8L

3760m750m2

55m2

1600cm3

60W

2m750mL

80kg

40L

-22°C

2000cm2

1m

9.85s

1.3kg

1500m160kPa

37.4°

50g

180mm

Try These – Metric Conversions

5.5 ha = _______m2

25 m = ________ cm345 m = _______km4321 mm = _______m0.12 m = ________ mm237 cm = _________ m998 km = ____________ cm199500 cm = ________ km

550002500.345

4.321 120 2.37

99 800 000 1.995

5.7 L = ______mL3400 kg = ______T0.42 T = _______kg4300 m2 = ______ha3 m2 = ________cm2

400000mm2 ____m2

4500cm3 _______ L

5700 3.4

420 0.430

30000 0.4

IWB FundamentalsEx. 8.02 pg 220

4.5

Starter12 ha of land is subdivided in 650 m2 lots. How many lots of this size can be created?

12 ha = _______ m2120000

120000 m2 = 184.6 650 m2

184 lots can be created.

A teaspoon holds 5 cm3. How many teaspoons are needed to measure out exactly 1 m3 of water?

5 cm3 = ____ mL = ____ L

1 m3 = ____ L

5.0051000

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 = The lower limit is 17 – 0.5 =

Therefore the limits of accuracy are 16.5 km ≤ Bluff ≤ 17.5 km

16.5 km17.5 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 kmThe 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:1.) 68 mm2.) 397 mm3.) 4 seconds4.) 50 g5.) 5890 kg6.) 820 cm7.) 92 kg8.) 89.1°

67.5 mm ≤ x ≤ 68.5 mm396.5 mm ≤ x ≤ 397.5 mm

3.5 seconds ≤ x ≤ 4.5 seconds45 g ≤ x ≤ 55 g5885 kg ≤ x ≤ 5895 kg815 cm ≤ x ≤ 825 cm91.5 kg ≤ x ≤ 92.5 kg89.05° ≤ x ≤ 89.15°

Exercise Questions

Note 2: AreasShape Area Formula Example

Rectangle Area = b × h 4 cm

2 cm

Area = 4 x 2 = 8 cm2

Square Area = l × l 3 cm 3 cm

Area = 3 x 3 = 9 cm2

Triangle Area = ½ b × h Area = ½ b×h

= ½ × 6 × 2 = 6 mm2

6 mm2 mm

Note 2: AreasShape Area Formula Example

Parallelogram Area = b × h 7 m

2 m

Area = 7 x 2 = 14 m2

Trapezium Area = 1/2 (a+b)× h 2 cm 6 cm 4 cmArea= ½ (4+2)×6= 18 cm2

Circle Area = πr2

Area = πr2

= π × (5)2

= 78.5 cm2

5 cm

Note 3: Perimeter & Circumference

• The perimeter of a figure is the total length of its sides.

11 cm

5 cmThe perimeter of this kite is:

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

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)

12.4 cm(π x 6.2) + (π x 6.2)

= 2 x π x 6.2

r = 6.2 (for large circle)

= 38.96 cm

Ex. 9.03 pg 250-254Ex. 9.04 pg 258Ex 9.05 pg 263-265

Find the perimeter if each square is exactly half the dimensions of the preceding square.

24 mm

24 m

m

(24 x 3) + (12 x 3) + (6 x 3) + (3 x 3) + (1.5 x 4)

= 141 cm

StarterA 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)

Hint – there are 21 uprights and 2 circles

2x

xLength of steel required = 21x + 2(2πx)

= 33.57 x

Amount of steel to purchase = $350 $4.35

= 80.46 m33.57 x = 80.46x = 2.40 m

= 4.8 m

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

4 cm

2 cm

5 cm

Area of compound shape:

= area of Rectangle + area of Triangle + area of semi-circle

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?How long are the straight sections?

Calculate the area enclosed by the track.

400 m100 m

d = ?

100 m

Acircle + Arectangle = Atotal

3183.1 + 6366.2 = 9549.3 m2 (1dp)

StarterA 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

b.) Calculate the area of the wooden ring

A = πr2 r = 14 cmA = π (14)2

A = 616 cm2

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

• The sector is part of a circle• Area of a sector = x° × πr2

360°

x°r

e.g.6 cm

35° Area of a sector = 35° × π × (6 cm)2

360°

= 11.0 cm2

Recall: Area of a circle = πr2

Note 6: Area of a Sector/ Arc Length

• Arc length = x° × 2πr 360° x°

r

e.g.6 cm

75° Arc length = 75° × 2π × 6 360°

= 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

= length of minor arc × 360° total circumference

a.) Find the radius of the circle3π cm

15π cm

Total circumference = 3π + 15π = 18π

So 18π = 2 × π × r 18 = 2 r

r = 9 cm

b.) Find the angle of the minor sector

= × 360°183π

π

= × 360°61

= 60°

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 = 2πr25.4 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

π

r = 35.6 π√

r = 3.37 cm

Diameter = 6.7 cm (1 dp)

IWB FundamentalsEx 9.04 pg 258-259

StarterA 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.

18 m

24 m

2 m

2.5 m

Merit - StarterThe 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.

18 m

24 m

2 m

2.5 m

Area courtyard = (24 ×18) – 4(π×1.252) – 22

No. of moulds = 408.37 m2/ (.25 × .15)

Total cost = 10890 × $ 0.83

≈ $9100

= 408.37 m2

= 10 890

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

SA (cube) = 6a2

aa

Surface Area

4 cm

5 cm7 cm

Calculate the surface area of this triangular prism

How many faces are there? 5

2 Triangles: 2 × (1/2 b × h) = 2 × 0.5 × 5cm × 4 cm = 20 cm2

Left Rectangle: b × h = 7 cm × 4 cm = 28 cm2

Base Rectangle: b × h = 7 cm × 5 cm = 35 cm2

Right Rectangle: b × h = 7 cm × 6.4 cm = 44.8 cm2

6.4 cm

Total surface Area = 127.8 cm2

Surface Area SA (cone) = the curved surface + the circular base

= π r l + π r2

SA (sphere) = 4π r2

l

r r

Surface Area (cylinder)

SA (cylinder) = area of rectangle + 2 ×(area of circle) = 2πr × h + 2 ×(πr2) = 2πrh + 2πr2

SA (cylinder) = 2πr (h + r)

C = 2πr h

r

Surface area depends on whether the ends are open or not. Open cylinders are called pipes.

TextbookEx 12.04 pg 156-157Ex 12.05 pg 157-160

IWB FundamentalsEx 9.06 pg 271-273Ex 9.07 pg 275-279

StarterGrass 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 the gardener buy to ensure he has enough to cover the required land, and at the best possible price?

22.0 m

15 m

4.1 m

9.2 m 7.0 m

7.6 m

Total Area = 22×15 – (4.1×9.2) – (7.0×7.6) = 239.08 m2

239.08 m2 / 26 = 9.2 1kg bags9 x $8.95 + 1 x $5.95 = $86.50

A gardener wishes to sow grass seed on his land.

= 239 m2 (3 sf)

Note 9: Volume of Prisms

• Volume = Area of cross section × Length

Area

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

4 m 10 m

4 cm

5 cm7 cm

Volume = (1/2 b×h) × L = ½ × 4cm ×5cm

× 10 cm = 70 cm3

Cylinder – A circular Prism

Volume (cylinder) = πr2× hVolume = Area of cross section × Length

8 cm1.2 cm

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.

8.0 m

13 m

12 m

Volume = ½ × 13 × 12 × 8 = 624 m3

624 m3 × 5 = 3120 m3/hr

3 hr

m3120×

min 60hr 1

= 52 m3/min

Starter

125 cm

B1 cm

8 cm

10cm

10cm

10cm

A

1.) Calculate the volume of these two cubes /cuboids of ice

2.) Which would melt faster if left outside on a hot day?

3.) Calculate the total area of the six faces for both pieces.

A = 1000 cm3 B = 1000 cm3

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 9 m

4.2 mLong Wall – 9 m ÷ 0.5 m

= 18 strips across

Strips per roll – 8 m ÷ 2.5 m = 3 / roll

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 9 m

4.2 m

Short Wall – 4.2 m ÷ 0.5 m = 8.4 (9) strips across

9 ÷ 3 (strips per roll) = 3 rolls per side= 6 rolls + 12 rolls (long)= 18

18 x $34.95= $629.10

Note 10: Volume of Pyramids, cones & Spheres

V = 1/3 A × h

A = area of base h = perpendicular height

V = 1/3 A × h = 1/3 (5m×4m)×8m

5 m 4 m

8 m(altitude)

Apex

= 53.3 m3

Volume of Cones

V = 1/3 × πr2× h A = πr2 (area of base) h = perpendicular height

= 1/3 × π×(1.5cm)2 × 9cm = 21.21 cm3 (4 sf)

Vertex

A cone is a pyramid on a circular base

1.5 cm

9 cm V = 1/3 × πr2× h

Spheres

• A sphere is a perfectly round ball.• It has only one measurement: the radius, r.

• The volume of a sphere is: V = 4/3πr3

Ex 13.02 pg 165-167


StarterThe 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?

= 1.77 x 109 L

V = 4/3 π(75 m)3

V = 1767146 m3

= 1.77 x 106 kL

V =4/3 πr3

MeritAn 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?

1.6 m

3.2 m

* diagram not to scale

Possible models Half cylinderTrapezoidal Prism

Volume = ½ x π (1.6)2 x 245

= 985.2 m3


Starter

Vwith peel = 4/3 π (45)3

= 381 704 mm3

mm.

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)

Weight Liquid Volume Equivalent Solid Volume

1 gram 1 mL 1 cm3

1 kg 1 litre 1000 cm3

Metric system – Weight/volume conversions for water.

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 1

23

e.g.

S.A. = 2(55 × 42) + 2(18 × 42) + (55 x 18)

55 cm

42 cm

a.) Calculate the area of glass required for the fish tank

b.) Calculate the volume of water in the tank. Give your answer to the nearest litre

= 7122 cm2

Vtank = 55 × 42 × 18= 41580 cm3

Vwater = 4/5 (41580)= 33264 cm3 = 33 L

Example:• Estimate the weight of a

333 mL can of soda.

• How many cans would it take to fill a container which has a volume of 4300 cm3 ?

333 grams + can

Number of cans = 4300 333 = 12.91 ≈ 13 cans

Ex 14.01 pg 173-175


StarterA 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

20 cm

Radius = 4 cm

Volume (sphere) = 4/3πr3

Volume (cylinder) = πr2 x h

= 268.08 cm3

= 1005.31 cm3

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)

WaterPopcorn 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)

DV

MM = 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

= 1800 cm3 Mass = V x D

= 1800 x 2.3

DV

M

= 4140 g (4.14 kg)

M = V × D

Density Examples:A 10 kg block of gold has a volume of 518 cm3. Calculate

the density of gold in g/cm3.

Density = 10000 g 518 g/cm3

10 kg = 10000 g

= 19.3 g/cm3

DV

MD = M V

Gamma TextbookEx 14.02 pg 176-177

IWB FundamentalsEx 11.03 pg 331 - 334Ex 11.01 pg 323 - 325

StarterA 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.

16 mm

4 mm

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/gLength of ‘opened up’ ring = π × 1.6 cm

= 5.0265 cm

Volume of ‘opened up’ ring = ½ × π × 0.42 × 5.0265 cm = 1.263 cm3

Mass = 1.263 cm3 × 19.3g/cm3

= 24.38 g

Value = 24.38 g x $59.70 = $1456

Note 13: Time

• Equivalent times (in seconds)1 minute =

1 hour = 1 day =

60 seconds60 minutes = 60 x 60 seconds = 3600 seconds

24 hours = 24 x 60 x 60 seconds

24 hour time is represented using 4 digits12 hour clock times are followed by am or pm

e.g 0630 hours 6:30 am

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.

= 86 400 seconds

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

Documents

1.5 Measurement