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4

197

4MeasurementThe ‘Gimli Glider’. A commercial aircraft runs completely out of fuel mid-flight. How could this happen?On 23 July 1983, Air Canada Flight 143 was about halfway through its journey from Montreal to Edmonton when alarms began to sound in the cockpit. No more fuel meant that both engines shut down and the pilots lost power to nearly all their instruments. Fortunately, the pilot, Captain Bob Pearson, was also an experienced glider pilot. He glided the plane to an old, disused airforce base at Gimli, where he managed to land it without causing any deaths or injuries to the 69 people aboard, or to the people on the ground racing go-karts on the old airstrip.

How did this near tragedy happen? At the time of the incident, Canada was in the middle of converting from the old imperial system of measurement to the new metric system. The amount of fuel required was calculated to be 22 300 kilograms. However,

an error made by the ground and flight crews meant that incorrect conversion factors were used and the plane ended up with 22 300 pounds of fuel, or just over 10 000 kilograms, less than half the required amount.

ForumWhen did Australia change from the imperial to the metric system of measurement?

Do any countries still use the imperial system?

Make a list of all the things (such as signs, labels or objects) that would have needed to be changed when the measurement system was changed.

Why learn this?The ability to accurately estimate and calculate length, surface area and volume is a valuable skill both at home and in the workplace. In the kitchen, cooked food cools quicker when it has a larger surface area. In the bedroom, the volume of your wardrobe is the space available to store all your clothes. In the garden, you may lay a path around the perimeter of a pond or pool. Tradespeople, such as painters, carpenters, hairdressers, chefs and mechanics, require good measurement skills to be successful. Other professionals who require good measurement skills include surveyors, interior designers and engineers.

After completing this chapter you will be able to:• calculate the perimeter of shapes with straight and curved edges• calculate the area of composite shapes made up of triangles, quadrilaterals, circles and

parts of circles• calculate the surface area of prisms and cylinders• calculate the volume and capacity of prisms and cylinders• convert units of length, area, volume and capacity.

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Curriculum LinksMeasurement and Geometry

Using units of measurement

Content descriptions and elaborations

ACMMG216

Calculate the areas of composite shapes

• understanding that partitioning composite shapes into rectangles and triangles is a strategy for solving problems involving area

ACMMG217

Calculate the surface area and volume of cylinders and solve related problems

• analysing nets of cylinders to establish formulas for surface area

• connecting the volume and capacity of a cylinder to solve authentic problems

ACMMG218

Solve problems involving the surface area and volume of right prisms.

• solving practical problems involving surface area and volume of right prisms

General capabilities

• Numeracy

• Literacy

• Critical and creative thinking

• Personal and social capability

• Information and communication technology (ICT) capability

Cross-curriculum priorities

• Sustainability

Big Ideas

• The area of composite shapes can be found by breaking them into simpler shapes and finding the combined area.

• The surface area of a solid is the combined area of its external faces.

• A net is a 2D representation of a solid that shows each of the individual faces.

• The volume of a prism or cylinder is found by multiplying the area of the cross-sectional base by the height of the solid.

• Capacity and volume are related measures.

• Volume is a measure of the space taken up by an object.

• Capacity is a measure of the fluid that a container can hold.

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198 PEARSON mathematics 9

Recall 4Prepare for this chapter by attempting the following questions. If you have difficulty with a question, go to Pearson Places and download the Recall Worksheet from Pearson Reader.

1 Calculate the following.

(a) 7.21 × 10 (b) 18.21 × 1000

(c) 1571.23 ÷ 100 (d) 0.74 ÷ 10 000

2 Copy and complete the following conversions.

(a) 4.2 cm = mm (b) 3.9 m = cm

(c) 7.1 km = m (d) 450 mm = m

(e) 18.6 cm = m (f) 5060 m = km

(g) 1100 cm = km (h) 3720 mm = km

3 What is the length of AB in each of the following diagrams? Where necessary, give your answers to one decimal place.

(a) (b) (c)

4 Match the shape to its name.

(a) (b) (c) (d)

A rectangle B isosceles triangle

C parallelogram D trapezium

5 Calculate the perimeter and area of each of these shapes to the nearest whole number:

(a) (b) (c)

arc cross-section prism volume

area net sector

capacity perimeter subtended

circumference polyhedron surface area

Worksheet R4.1

Worksheet R4.2

Worksheet R4.3

A B

D C5 m

3 mC

BA

D

2.9 mm

1.6 mm

AD

BC

4.8 cm

16 cm

Worksheet R4.4

Worksheet R4.5 14 cm

8 cm12 cm

16 cm

20 cm

14 cm

Key Words

4 Measurement 199

PerimeterThe perimeter of a shape is the total length of its boundary, measured in units such as cm or km. In practical situations we can measure the individual lengths and add them together. If we have a diagram or a well-defined shape, we can use shortcuts (or formulas) where we have some lengths that are equal. Equal lengths are shown by markings such as and .

Perimeter formulas

Square

P = 4s

Rectangle

P = 2l + 2w or P = 2(l + w)

Rhombus

P = 4s

Parallelogram

P = 2a + 2b or P = 2(a + b)

Regular hexagon Equilateral triangle Isosceles triangle

P = 6s P = 3s P = 2a + b

Worked Example 1Find the perimeter of these shapes.

(a) (b)

Thinking(a) 1 Identify the shape and write down

the formula.(a) P = 2(a + b)

2 Substitute in the values and evaluate. P = 2(8.5 + 9.8)= 2 × 18.3

3 Write the answer with the given unit of length.

P = 36.6 m

s w

l

s a

b

ss a

b

1

8.5 m

9.8 m

6 cm

8 cm

4.1

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Resources

Worksheets

• R4.1 Multiplying and dividing decimals by powers of 10

• R4.2 Converting units of length• R4.3 Finding an unknown length

of a shape• R4.4 Classifying shapes• R4.5 Calculating area and perimeter

RECALL 4

1 (a)

72.1

(b)

18 210

(c)

15.7123

(d)

0.000 074

2 (a)

42 mm

(b)

390 cm

(c)

7100 m

(d)

0.45 mm

(e)

0.186 cm

(f)

5.06 km

(g)

0.011 km

(h)

0.003 72 km

3 (a)

5 m

(b)

1.6 mm

(c)

15.1 cm

4 (a)

D

(b)

A

(c)

C

(d)

B

5 (a)

Perimeter

=

44 cm, Area

=

112 cm

2

(b)

Perimeter

=

48 cm, Area

=

96 cm

2

(c)

Circumference

=

44 cm, Area

=

154 cm

2

WorksheetsWorksheets

Answers

Resources

Videos

• Finding pi from pie• Introduction to pi

Interactives

• Interactive Lesson: Circumference of a circle

Tutorials and quizzes

• Circumference of a circle

eWorked Examples

• eWorked Example 1: Perimeter• eWorked Example 2: Finding the

diameter or radius given the circumference

• eWorked Example 3: Arc length and perimeter of a sector

4.1

Question Answer

1

What is the perimeter of a square with side lengths of 34 cm?

136 cm

2

What is the perimeter of a rectangle with a length of 7 cm and a width of 4.5 cm?

23 cm

3

What is the name of a triangle with side lengths of different length?

scalene

4

Name four different types of quadrilaterals.

Any four of trapezium, rhombus, parallelogram, square, kite and rectangle.

5

Complete the following conversions.

(a)

5.6 cm

=

mm

(b)

0.85 km

=

m

(c)

27.06 m

=

cm

(d)

2350 mm

=

m

(a)

56 mm

(b)

850 m

(c)

2706 cm

(d)

2.35 m

Recap

199

Suggested Examples

1 Find the perimeters of the following shapes.

(a)

(b)

7 m

3 m

6.5 cm4.2 cm

Answers:

(a)

P

=

7

+

7

+

3

+

3

=

20 m

or

P

=

2

l

+

2

w

=

(2

×

7)

+

(2

×

3)

=

14

+

6

=

20 m

(b)

P

=

6.5

+

4.2

+

4.2

+

6.5

+

4.2

+

4.2

=

29.8 cm

or

P

=

2

a

+

4

b

=

2

×

6.5

+

4

×

4.2

=

13

+

16.8

=

29.8 cm

(cont.)


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4.1


Circumference of a circleYou should recall that the perimeter of a circle is called the circumference. The formula used to calculate the circumference of a circle is C = 2πr, where r is the radius, or C = πd, where d is the diameter. The plural of radius is radii.

If we know the circumference of a circle, we can use it to find the radius or diameter.

(

(b) 1 Write out the sum of the lengths. (b) P = 8 + 8 + 6 + 6 + 6

2 Simplify if possible. = 2 × 8 + 3 × 6

3 Write the answer with the given unit of length.

= 34 cm

Circumference of a circle

C = πd or C = 2πrwhere d = diameter and r = radius.

Worked Example 2(a) Find the diameter of a circle with a circumference of 8.5 cm, correct to one decimal place.

(b) Find the radius of a circle with a circumference of 40 mm, correct to one decimal place.

Thinking(a) 1 Write the circumference formula. (a) C = πd

2 Rearrange the formula to make d the subject, by dividing both sides by π.

= d

3 Substitute the value for C. = d

4 Calculate, rounding your answer to the specified number of decimal places. Remember to include the unit.

d = 2.705…d = 2.7 cm (1 d.p.)

(b) 1 Write the circumference formula. (b) C = 2πr

2 Rearrange the formula to make r the subject, by dividing by 2π.

= r

3 Substitute the value for C. = r

4 Calculate, rounding your answer to the specified number of decimal places.

r = 6.366…r = 6.4 mm (1 d.p.)

dr

2

Cπ---

8.5π

--------

C2π------

402π-------

4 Measurement

4.1

201

Arc length and perimeter of a sectorA sector of a circle is the area contained within two radii and the arc that joins them. An arc is part of the circumference of a circle. The distance along the arc of a sector (l) is in proportion to the angle subtended by (standing on) the arc at the centre of the circle.

To calculate the length of the arc, we multiply the entire circumference by the fraction represented by the arc. For example, an arc subtending an angle of 90° has a length that is one-quarter the circumference of the circle, because

= . The arc length is found by calculating × 2πr

(which can also be written as or ).

Similarly, an arc subtending an angle of 60° has a length

that is one-sixth the circumference, because =

An arc subtending an angle of 130° has a length that is or of the circumference.

The arc length l can be found by using the formula:

l = × 2πr

where θ (the Greek letter theta) is the angle subtended by the arc at the centre of the circle.

To find the perimeter of a sector of a circle, we need to add twice the radius to the length of the arc.

Arc length of a sector

l = × 2πr, where θ is the angle subtended at the centre.

Perimeter of a sector

P = l + 2r

Worked Example 3Calculate (i) the arc length and (ii) the perimeter of each of the following sectors, to the nearest whole number.

(a) (b)

Radii

↓

↓

r

r

l

θ

90360--------- 1

4--- 1

4---

2πr4

--------- πr2-----

60360--------- 1

6--- .

130360--------- 13

36------

θ360---------

θ360---------

3

10 cm

75°

235°

25.5 cm

200


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Suggested Examples (cont.)

2 (a) Find the diameter of a circle with a circumference of exactly 4 m. State your answer correct to one decimal place.

(b) Find the radius of a circle that has a circumference of 65 mm. State your answer correct to one decimal place.

Answers:

(a)

C

=

π

d

=

d

(divide both sides by

π

)

d

=

d

=

1.3 m (1 d.p.)

(b)

C

=

2

π

r

=

r

(divide both sides by 2

π

)

r

=

r

=

10.3 mm (1 d.p.)

3 Find the following, correct to two decimal places.

(a) The length of the arc.

(b) The perimeter of the sector.

Answers:

(a) Arc length

= ×

2

π

r

=

×

2

×

π ×

10

=

19.20 m (2 d.p.)

(b)

P

=

×

2

π

r

+

2

r

=

×

2

×

π ×

10

+

2

×

10

=

39.20 m (2 d.p.)

Cπ---

4π---

C2π------

652π-------

10 cm

110°

θ360-----------

110360-----------

θ360-----------

110360-----------

Class Activities

Making shapes revision

Equipment required

: 1 pipe cleaner per student

Pass out a pipe cleaner to each student and ask them to make the following shapes. You could join four pipe cleaners together to make a bigger model to use as a demonstration. Students could make a square, a rectangle, a circle, a kite, a rhombus, a parallelogram, or different triangles such as scalene, isosceles and equilateral.

Track tolerances

Running tracks that are to be used for high-level competitions, such as the Olympics or world championships, must be built to very strict guidelines. The ‘tolerance’ for an Olympic-standard 400 m running track is

+

0.04 m and -0.0 m, meaning that the length of the track can be up to 0.04 m longer than 400 m, but cannot be any shorter. Similarly, the tolerance for an Olympic-standard 50 m swimming pool is

+

0.03 m and -0.00 m.

Ask students to calculate how much further competitors in an 800 m or 1500 m race would run or swim if the track or pool was built to maximum tolerance. As an extension, they could research the speeds of runners or swimmers in the race and use these to calculate how much extra time would be taken to complete the distance.

Teaching Strategies

Conversion of units

Students should use the same unit for all measurements when calculating perimeter of shapes, so they need to be able to convert units. Discuss the history of units of length and the inconsistency of some of the older units, such as ‘hands’ (used to measure the height of a horse) and ‘cubits’.

Ask two different students to measure the length of the whiteboard in ‘hands’ and observe the difference in the measurements. Bring in a tape measure and measure the whiteboard, firstly in metres and then in centimetres.

The table shows the common length unit conversions.

Students should be fluent in converting units of length, so that they can develop an understanding of converting units of area and volume, where the conversion factors are the squares and cubes of the conversion factors for length.

Unit Conversionfactor

kilometre to metre × 1000

metre to centimetre × 100

centimetre to millimetre × 10

millimetre to centimetre ÷ 10

centimetre to metre ÷ 100

metre to kilometre ÷ 1000

Tired tyres

This is a challenging task that can be set for more able students.

The odometer on a car measures the number of kilometres travelled by counting the revolutions of the tyres. The pattern of marks and grooves on a tyre’s surface is known as the ‘tread’, and helps the tyre grip the road. As the car is driven, this tread is worn from the tyres, making them smoother and more dangerous. The depth of tread that can be worn from a tyre before it needs to be replaced is 8 mm. Suppose that the overall diameter of a wheel and a new tyre is 55 cm. If 8 mm of tread is worn evenly from all over the tyre, what effect will this have on the reading shown on the odometer?

(Students should find that the odometer would over-predict the distance travelled by the worn tyres by 5.02 cm per revolution, an error of 2.9%.)

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Thinking

(a) (i) 1 Write the appropriate formula. (a) (i) l = × 2πr

2 Write down the known values. r = 10 cm, θ = 75°

3 Substitute values into the formula. l = × 2 × π × 10

4 Simplify the expression by cancelling common factors and/or multiplying. (Here, we cancel common factors of 20 then 3).

5 Evaluate, writing your answer with the specified accuracy.

= 13 cm (nearest whole number)

(ii) 1 Write the appropriate formula.(ii) P = l + 2r

2 Write down the known values. (Use the exact value of l, rounding at the end of the calculation.)

r = 10 cm, l = cm

3 Substitute values and calculate. P = 2 × 10 +


(b) (i) 1 Write the appropriate formula. (b) (i) l = × 2πr

2 Write down the known values. r = 25.5 cm, θ = 235°

3 Substitute values into the formula. l = × 2 × π × 25.5

4 Simplify the expression by cancelling obvious common factors and/or multiplying. (Here, we cancel a common factor of 2.)

5 Evaluate, writing your answer with the specified accuracy.


(ii) 1 Write the appropriate formula. (ii) P = l + 2r

2 Write down the known values. (Use the exact value of l, rounding at the end of the calculation.)

r = 25.5 cm, l = cm

3 Substitute values and calculate. P = 2 × 25.5 +


θ360-----------

75360-----------

= 75360----------- 20 π××

1

18

= 75π18

---------25

6

= 25π6

----------

25π6

----------

25π6

----------

θ360-----------

235360-----------

= 235360----------- 2 π 25.5×××

1

180

= 5992.5π180

----------------------

5992.5π180

----------------------

5992.5π180

----------------------

4 Measurement

4.1

203

Perimeter

Equipment required: tape measure or ruler for Question 12

Fluency1 Find the perimeters of the following shapes using the measurements given.

(a) (b)

(c) (d)

(e) (f)

2 (a) Find, correct to one decimal place, the diameter of a circle with a circumference of:

(i) 15 m (ii) 32 m (iii) 141.39 mm (iv) 188.52 mm

(b) Find, correct to one decimal place, the radius of a circle with a circumference of:

(i) 93 km (ii) 71 km (iii) 804.248 cm (iv) 1.2566 cm

3 Find (i) the arc length and (ii) the perimeter of each of the following sectors, to the nearest whole number.

(a) (b) (c)

NavigatorQ1, Q2, Q3, Q4, Q5 (a)–(e), Q6, Q7, Q10, Q12

Q1, Q2, Q3 (a)–(d), Q5, Q6, Q7, Q8, Q10, Q12, Q13

Q1 Column 2, Q2 (a)(i)–(ii), Q2 (b)(i)–(ii), Q3 (d)–(f), Q5, Q6, Q8, Q9, Q10, Q11, Q13, Q14

4.1Answerspage 614

1

9 cm

8 cm

10 cm

8 cm

6 cm

15 cm

10 cm

15 cm

8 mm

20 mm

25 mm

12 mm

30 cm

40 cm

2

3

60°

14 cm

3 m

135°

371 mm

227°

202


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Class Activities

Finding Pi

Equipment required

: piece of string, ruler, scissors, circular objects such as hula hoop, filter paper, drink bottles, Frisbee, cans, plates etc.

This activity could help to develop a more concrete understanding of pi. Students may have done this activity in Year 8 (Exercise 5.2

Pearson Mathematics 8

), but if they have not, it is a valuable exericse. Give students the following instructions:

• Use the string to measure the circumference around each object.

• Mark out the length or cut the string.

• Stretch out the string and place the string across the ruler/measuring tape and record the measurement. This is the circumference.

• Measure the diameter of the object measuring from edge to edge through the centre.

• Make sure that you use the same unit: mm, cm or m.

• Record your results and complete the following table.

• Ask students if they notice any pattern to their results in the final column. The pattern may be clearer if they round their results to one decimal place. Results that are closer to the value of

π

are often obtained with larger objects.

Alternatively, this activity could be done using a spreadsheet application to find

π

.

Object Circumference (C) diameter (D) C ÷ D

Frisbee

Filter paper

Hoop

Bangle

Drink bottle

Ring

Top of a can

Egg ring

203

Classification of shapes using a Venn diagram

Pass out cards with the names of shapes from the list below. Ask students to draw the shape on the card and to list two properties of that shape. Draw a large Venn diagram with three sets on the whiteboard. Ask three students to name a set from one of the properties on their card. For example, has a 90

°

angle, has 4 sides or has 2 pairs of parallel sides. Each of the three students place their card into the Venn diagram, but have an opportunity to move it once the other sets are named. The rest of the class then come up to stick their cards on the Venn diagram.

This activity provides a good opportunity for revision of Venn diagrams. Some students may find that their shape doesn’t fit inside any of the circles, and so should be placed on the outside. Others may find their shape could fit in either of the two circles, and so be placed in the intersections of the two circles.

Students may choose to make polygons regular or irregular.

Venn diagram example:

Square Circle Sector Rectangle Trapezium Kite

Parallelogram Rhombus Equilateral triangle

Scalene triangle

Isosceles triangle

Irregular quadrilateral

Hexagon Octagon Pentagon Heptagon Nonagon Decagon

Curvedsides

Rightangle

Pair ofparallelsides

(b) (i)

14.8 km

(ii)

11.3 km

(iii)

128.0 cm

(iv)

0.2 cm

3 (a) (i)

15 cm

(ii)

43 cm

(b) (i)

7 m

(ii)

13 m

(c) (i)

1470 mm (ii) 2212 mm

EXERCISE 4.1

1 (a) 34 cm (b) 24 cm

(c) 60 cm (d) 45 cm

(e) 130 mm (f) 140 cm

2 (a) (i) 4.8 m (ii) 10.2 m

(iii) 45.0 mm (iv) 60.0 mm

Answers


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4.1


(d) (e) (f)

Understanding4 Choose the correct answer.

The perimeter of the triangle at right is:

A 12 cm B 17 cm

C 19 cm D 35 cm

5 Find the perimeters of these shapes, giving your answers to one decimal place where necessary.

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

6 Find the diameter of a CD if its circumference is 37.7 cm. Write your answer correct to one decimal place.

7 Find the circumference of Earth if the diameter, to the nearest 100 km, is 12 800 km. Give your answer correct to the nearest 100 km.

26 m

307°

26.3 cm

149° 53° 164 cm

5 cm

7 cm

Dotted lines are there only to indicate measurements. Make sure you don’t include them in your perimeter calculation.

! Warning

3.6 km

1.2 km

30 cm

20 cm

12 cm

20 cm

25 cm

50 cm

40 cm

15 cm

15 cm15 cm

40 cm55°

55°

15 cm

4 Measurement

4.1

205

8 Andrea is flying her model aeroplane on a wire 18 m long, as shown.

(a) How far does it fly in one circuit? (Answer to the nearest metre.)

(b) How long will it take to complete one circuit if it is flying at a speed of 5 metres per second? (Answer to the nearest tenth of a second.)

(c) How long would the wire need to be if the distance for one circuit was 200 m? (Answer to the nearest centimetre.)

9 Every night Nicole walks her dog twice around the park. The plan of the park is shown below.

How far does she walk in km? (Answer to the nearest metre.)

Reasoning10 If a plane travels halfway round the world at an altitude of 11 300 m, what extra distance,

in km, will it fly compared with the distance measured at the surface of the Earth? (The radius of the Earth is approximately 6400 km.) Round your answer to one decimal place.

11 A sheet of corrugated iron has corrugations that are based on the arc of a circle of radius 3 cm and a central angle of 120°. Four corrugations are shown below.

How many corrugations are formed when a flat sheet of steel 1 m wide is bent to form a sheet of corrugated iron?

Open-ended12 Estimate first, then use a tape measure or ruler to measure the perimeter of:

(a) the front cover of this book (b) the top of your desk

(c) the whiteboard in your room (d) the floor of the classroom.

13 Find whole number diameters of three different circles whose circumferences are between 50 cm and 90 cm.

14 Sketch and label a shape with a boundary containing straight as well as curved sections whose perimeter is between 30 and 40 cm.

18 m

Drawing a diagram can help you ‘picture’ a problem.

Hint

50 m

150 m

150 m

50 m

100 m

100 m

120°

3 cm

204


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3 (d) (i) 139 m (ii) 191 m

(e) (i) 68 cm (ii) 121 cm

(f) (i) 152 cm (ii) 480 cm

4 B (5 + 5 + 7) cm

5 (a) P = 3.6 + 1.2 + (3.6 – 1.2) + × 2× π × 1.2

= 9.084…

Perimeter is 9.1 km

(b) P = 20 + 2 × 30 + × π × 20

= 111.415…

Perimeter is 111.4 cm

(c) P = 2 × 12 + × π × 12 = 42.8495…


(d) P = 2 × 25 + π × 20

= 112.831…


(e) P = 2 × π × 50

= 314.1592…


(f) P = 2 × 40 + × 2 × π × 15 = 127.123…


(g) P = π × 15 + 0.5 × π × 30 = 94.247…


Worked Solutions

14---

12---

12---

12---

(h) P = 0.75 × π × 80 + 2 × 40= 268.4955…


(i) P = 4 × 15 + × π × 30 = 88.797…


6 C = π × d37.7 = π × d

d = 37.7 ÷ π= 12.000…

Diameter is 12.0 cm

7 C = π × d = π × 12 800= 40 212.385…

Circumference is 40 200 km

8 (a) C = 2π × r= 2π × 18= 113.097…

Circumference is 113 m

(b) Time = C ÷ 5= 22.619…

Time taken is 22.6 s

(c) 200 = 2πr

so r = 200 ÷ 2π = 31.8309…

Radius is 31.83 m

9 P = [0.75 × 2π × 100 + 50 + 50 +50 + 100 + 3 × 150] × 2

= 2342.477…

Distance walked is 2342 km.

110360---------

10 Cearth = π × d= π × 2 × 6400= 40 212.3859…

Cflight = π × d= π × 2(6400 + 11.3)= 40 283.3859

Difference = 40 283.3859 – 40 212.3859 = 71 m

Halfway round = 71 ÷ 2 = 30.5 m

11 arc length = × 2π × 3

= 6.283…

No. corrugations = 100 ÷ arc length= 15.915…= 15 (almost 16)

Open-ended – Sample answers12 The perimeter of the front cover of

the book is 100 cm; the remaining perimeters are students’ own answers.

13 Any whole number value from 16 to 28 cm (e.g. a diameter of 20 cm gives a circumference of ≈ 62.83 cm; a diameter of 25 cm gives a circumference of ≈ 78.54 cm; a diameter of 28 cm gives a circumference of ≈ 87.96 cm).

14 Possible shape with P ≈ 37.4 cm:

120360---------

3 cm

10 cm

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PEARSON mathematics 9 teacher companionStudent Book

Veterinary maths

The job of a vet involves treating and performing surgery on sick animals. De-sexing and vaccinations are the two most common procedures. Dogs are the most common animals dealt with, along with wombats, koalas, lizards, horses, cows, cats, sheep, birds, pigs and rats.

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When an animal has become dehydrated the amount of water to be replaced (the deficit) is estimated as a percentage of the body weight.

For example, a 35-kg calf with 10% dehydration has a water deficit of 3.5 kg (10% of 35 kg).

1 Find the deficit for an 80-kg Great Dane that is 5% dehydrated.

Each litre of water has a mass of 1 kg. For the calf, 3.5 kg of water is the same as 3.5 L, or 3500 mL.

2 Convert the deficit found in 1 to the number of mL needed for the Great Dane.

Animals that are severely dehydrated require additional fluid of 60 mL per day for each kilogram of body mass. For the calf, this would be: 60 mL × 35 = 2100 mL of additional fluid per day.

3 Find the additional fluid needed per day for the Great Dane.

The total amount of fluid in the first day of treatment is the total of the deficit and the additional fluid. For the calf, this would be 3500 mL + 2100 mL = 5600 mL.

4 Find the total amount of fluid needed in the first day for the Great Dane.

The fluid is given via an intravenous drip, for which 10 drops = 1 mL. For the calf, the number of drops needed in a day is 10 drops × 5600 = 56 000 drops.

5 Find the number of drops needed in a day for the Great Dane.

The drip needs to be set at a rate involving seconds.

The number of seconds in a day is: 24 × 60 × 60 = 86 400 s.

The flow rate for the calf would be: 56 000 drops

86 400 s = 0.65 drops per second.

6 Find the number of drops per second for the Great Dane.

7 Repeat 1 to 6 for a 12-kg Kelpie that is 10% dehydrated.

Research• Find out some of the common causes of dehydration

in animals.

• When a dehydrated cat is brought into a surgery, water is sometimes injected directly under the skin, where it is quickly absorbed. Find out why a drip is used in some cases and not others.

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PEARSON mathematics 9 teacher companionTeacher Support

MATHS 4 REAL: VETERINARY MATHS

1 Great Dane

5% of 80 kg = × 80 = 4 kg

Water deficit is 4 L

Kelpie

10% of 12 kg = 1.2 kg

Water deficit is 1.2 L

2 Great Dane

4 L = 4000 mL

Kelpie

1.2 L = 1200 mL

3 Great Dane

Additional fluid needed = 60 mL × 80= 4800 mL

Kelpie

60 mL × 12 = 720 mL

4 Great Dane

Total fluid required = 4000 + 4800 mL = 8.8 L

Kelpie

1200 + 720 mL = 1920 mL

5 Great Dane

Number of drops required = 10 × 8800 = 88 000

Kelpie

10 × 1920 = 19 200 drops

Worked Solutions

5100---------

Teaching Strategies

6 Great Dane

Flow rate = 88 000 ÷ 86 400 = 1.0185…

Kelpie

1.02 drops per sec

13 200 ÷ 86 400 = 0.152 777…

0.15 drops per second

For students experiencing difficulty with the Maths 4 RealStudents may benefit from having the instructions for each question written into a consolidated list or flow chart. A summary of the steps involved is as follows:

1–2 Find the water deficit: Calculate the percentage of the animals body mass. Mass in kg = volume of water in L

3 Find the additional fluid required per day: Calculate body mass × 60 mL

4 Total amount of fluid required in the first day = water deficit (1) + additional fluid (3)

5 Number of drops in one day = Total amount of fluid in first day (4, in mL) × 10 (10 drops = 1 mL)

6 Number of drops per second = Number of drops in one day (5) ÷ 86 400

7 Convert number of drops per second (6) to a whole number, by multiplying by 2, 3, 4 etc.

For students who complete the Maths 4 Real quicklyStudents could use the research questions in the task as a starting point for further research about the mathematics involved in veterinary science (e.g. as the calculation of medicine dosages). They could contact the local vet to ask questions, or invite them to address the class.

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