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6 Geometry and space6 Geometry and spaceAustralian Curriculum linksYear 6Proficiency strandsProblem-solving includes calculating angles

Reasoning includes investigating new situations using known properties of angles

Content descriptions

Measurement and geometry

Shape Elaborations

Construct simple prisms and pyramids (ACMMG140)

• considering the history and significance of pyramids from a range of cultural perspectives including those structures found in China, Korea and Indonesia

• constructing prisms and pyramids from nets, and skeletal models

Year 7Content descriptions

Measurement and Geometry

Shape Elaborations

Draw different views of prisms and solids formed from combinations of prisms (ACMMG161)

• using aerial views of buildings and other 3-D structures to visualise the structure of the building or prism

Geometric reasoning Elaborations

Classify triangles according to their side and angle properties and describe quadrilaterals (ACMMG165)

• identifying side and angle properties of scalene, isosceles, right-angled and obtuse angled triangles

• describing squares, rectangles, rhombuses, parallelograms, kites and trapeziums

Demonstrate that the angle sum of a triangle is 1808 and use this to find the angle sum of a quadrilateral (ACMMG166)

• using concrete materials and digital technologies to investigate the angle sum of a triangle and quadrilateral

Identify corresponding, alternate and cointerior angles when two parallel straight lines are crossed by a transversal (ACMMG163)

• defining and classifying angles such as acute, right, obtuse, straight, reflex and revolution, and pairs of angles such as complementary, supplementary, adjacent and vertically opposite

• constructing parallel and perpendicular lines using their properties, a pair of compasses and a ruler, and dynamic geometry software

Investigate conditions for two lines to be parallel and solve simple numerical problems using reasoning (ACMMG164)

• defining and identifying alternate, corresponding and allied angles and the relationships between them for a pair of parallel lines cut by a transversal, including using dynamic geometry software

Year 8Proficiency strandsReasoning includes using congruence to deduce properties of triangles

Content descriptions

Measurement and geometry

Geometric reasoning Elaborations

Define congruence of plane shapes using transformations (ACMMG200)

• understanding the properties that determine congruence of triangles and recognising which transformations create congruent figures

• establishing that two figures are congruent if one shape lies exactly on top of the other after one or more transformations (translation, reflection, rotation), and recognising the equivalence of corresponding sides and angles

06 MW7NC_TRB Final.indd 24106 MW7NC_TRB Final.indd 241 19/12/11 5:12 PM19/12/11 5:12 PM

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Develop the conditions for congruence of triangles (ACMMG201)

• constructing triangles using the conditions for congruence

• solving problems using the properties of congruent figures, justifying reasoning and making generalisations

• investigating the minimal conditions needed for the unique construction of triangles, leading to the establishment of the conditions for congruence (SSS, SAS, ASA and RHS), and demonstrating which conditions do not prescribe congruence (ASS, AAA)

• plotting the vertices of two-dimensional shapes on the Cartesian plane, translating, rotating or reflecting the shape and using coordinates to describe the transformation

Establish properties of quadrilaterals using congruent triangles and angle properties, and solve related numerical problems using reasoning (ACMMG202)

• establishing the properties of squares, rectangles, parallelograms, rhombuses, trapeziums and kites

• identifying properties related to side lengths, parallelism, angles, diagonals and symmetry

Source: Australian Curriculum

Pre-testPre-testMost students should be able to recognise common geometric shapes by their appearance, even if they are not aware of geometric properties at this stage. Students need to be able to recognise shapes when they are represented in a non-standard position. For example, some students are unable to recognise a square as a square if it is tilted. Similarly, some students think shapes J and K in question 8 are triangles because they are ‘pointy’, rather than noticing that they each have more than three sides.

●1 Look at this set of letters.

a Which of the letters have perpendicular lines?b Which letters have parallel lines?c Which letters have both parallel and perpendicular lines?

●2 What word could be used to describe this set of lines?

●3 Which of these things are usually vertical and which are usually horizontal?a tabletopb floorc classroom wallsd ceilinge whiteboard

●4 Look again at the letters in question 1.a Which of the letters contain right angles?b In which of the letters can you see acute angles?c In which of the letters can you see obtuse angles?


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●5 For each of these angles

i name the angle using the given letters.

ii what is the size of each of these angles?

a b

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●6 Use your protractor to measure each of these angles and state whether the angle is an acute, obtuse or reflex angle.

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●7 How many degrees are in each of the following?a a straight angle b one revolution about a pointc a right angle.

●8 Which of the following shapes area triangles?b quadrilaterals?

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●9 Which of these shapes are squares?

A B C D

●10 Consider the shapes on the right.a This triangle has three equal sides.

What do we call this type of triangle?

b This quadrilateral has both pairs of opposite sides equal and all theangles are right angles. What do we call this type of quadrilateral?

●11 Name these three-dimensional shapes.

●12 Which of these shapes are prisms?

A B C D

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●1 a E, F, H, L, T b E, F, H, Z c E, F, H

●2 Parallel

●3 Horizontal: tabletop, floor, ceilingVertical: classroom walls, whiteboard

●4 a E, F, H, L, T b A, K, V, W, X, Y, Z c A, K, X, Y

●5 a i /PQR or /RQP ii 80°b i /JKL or /LKJ ii 145°

●6 a 808, acute b 3048, reflex c 1058, obtuse

●7 a 1808 b 3608 c 908

●8 a C, F, L b A, B, D, G, H, I, J

a b c

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●9 A, D

●10 a equilateral triangle b rectangle

●11 a cube b square pyramid c hexagonal prism

●12 B, C

Warm-up: parking problemWarm-up: parking problemChapter warm-ups are included in the teacher and student ebooks as separate worksheet for ease of printing. This chapter warm-up links with the analysis task Parking problem at the end of the chapter, where students investigate the advantages and disadvantages of each type of parking. The parking signs below indicate whether parallel parking or angle parking applies. Parallel parking is where the cars park end-to-end parallel to the kerb (the edge of the roadway), as shown in the photograph, and angle parking is where the cars park at an angle to the kerb, usually 458, 608 or 908.

a Find out the type of parking––angle parking or parallel parking––that applies in the road outside your school, in the shopping street of your town or suburb, or in the road where you live.

b Why do you think this type of parking has been used?



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Teaching note: the van Hiele theory of geometry learningWhen thinking about the teaching and learning of geometry, it is useful to gain an understanding of the levels of student understanding identified by Pierre van Hiele and Dina van Hiele-Geldof. In the 1950s this Dutch couple extensively explored children’s learning of geometry concepts. Russian mathematics educators found their work of interest, but it was not until the 1980s that the van Hiele theory came to the attention of American researchers.

The theory has now been the subject of research world-wide and various aspects of the theory have been criticised. There has been a renumbering of the original levels, so that the van Hieles’ Level 0 is now Level 1. John Pegg has also suggested the splitting of the current Level 2 (old Level 1) into two levels.

The van Hiele theory has also been criticised for its hierarchical nature of learning (that children do not progress to a higher level until they have mastered the previous level) and research has now shown that children can be at different levels for different geometry concepts. However, it does provide a useful framework for thinking about the development of geometric understanding. For further reading see Pegg, J (1995), ‘Learning and teaching geometry’, in: Grimison L & Pegg J, Teaching secondary school mathematics: theory into practice. Sydney, Harcourt Brace, 1995.

In the currently used numbering, the van Hiele levels may be summarised as follows:

Level Characteristic of the level Example

Level 1 Recognising shapes by their visual appearance alone

That shape is a square.

Level 2Level 2aLevel 2b

Recognising shapes by their propertiesRecognising one propertyRecognising more than one property

The shape is a square because it has four equal sidesThe shape is a square because it has four right angles, opposite sides are parallel, all sides are equal.

Level 3 Recognising relationships between properties

If all the angles of a square are right angles, then the opposite sides must be parallel.

Level 4 Deductive reasoning and understanding the minimum properties that will identify a shape

A rhombus with one right angle is sufficient to identify a shape as a square (why?); students are able to complete proofs such as the angle sum of a triangle.

Many students go through secondary school with a Level 1 understanding of geometry. The MathsWorld Australian Curriculum books place a strong emphasis on developing students’ thinking and understanding beyond Level 1.


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6.16.1 Lines, rays and segments Lines, rays and segments

Teaching note: parallel linesMany students will be familiar with parallel bars in the playground or in gymnastics. It is important for students to recognise that the definition ‘Parallel lines are lines that will never meet even if they are extended in either direction’ is incomplete. This is the case only if the lines are in the same plane. We can also say that the perpendicular distance between parallel lines is the same at all points along the lines.

Question 13 in exercise 6.1 is worth doing as a class activity, with each pair of students having a rectangular cardboard box, such as a shoe box or cornflakes packet. This provides an excellent means of conveying the notion that lines that never meet are not necessarily parallel. This idea can also be developed by looking at lines in the classroom, for example a vertical wall edge and its opposite horizontal ceiling-wall edge. However, the hands-on approach of looking at edges and drawing lines on a box helps students to see directly that there are line segments that will never meet, even if extended, that are not parallel. We must specify then that parallel lines are lines in the same plane that will never meet.

Introducing symbols Symbols for parallel and perpendicular are introduced in drawings as well as in written statements, for example AB || CD.

A four-sided shape is shown on the right.

a Which sides of this shape appear to be parallel?

b Mark the diagram with symbols to show this.

Working Reasoning

a Sides b and d appear to be parallel. These two sides

■ are the same distance apart at all positions.

■ are in the same plane.

■ do not meet.

b We use arrowheads to show that two lines or line segments are parallel.

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The diagram on the right shows two line segments, AB and CD.

a What word can we use to describe AB and CD?

b Mark the diagram with a symbol to show this.

Working Reasoning

a AB and CD are perpendicular. Lines are perpendicular if there is an angle of 908 between them.

b The right angle symbol shows that the segments are perpendicular.

Teaching note: compass and ruler constructionsCompasses are not always appropriate in classrooms. The Math Open Reference Project website has excellent animations of compass and ruler constructions that can be paused and replayed. Scrolling down the page leads to a clearly set out proof of why each construction works.

www.mathopenref.com/tocs/constructionstoc.html

The following are direct links to constructing a line perpendicular to a given line passing through a point on the line (as in example 3 in the student book) and passing through an external point (as in example 4 in the student book).

www.mathopenref.com/constperplinepoint.html

www.mathopenref.com/constperpextpoint.html

exercise 6.1Below are the answers to the questions in exercise 6.1 in the student book.

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●2 There are parallel segments in the rows of bricks. The legs of the parallel bars are parallel to each other.

●3 a CD, EF b AB, CD, EF

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●6 In striped material, the stripes are parallel to each other. In check material the horizontal stripes are perpendicular to the vertical stripes.

●7

●8 Chris is building a house with rectangular rooms. He has checked that the floor is horizontal. The walls must be vertical and perpendicular to the floor. The walls on opposite sides of the room must be parallel to each other. Two walls meeting at a corner of the room must be perpendicular to each other.

●9 a perpendicular b vertical c parallel d horizontal

●10 a b

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●11 a meet b don’t meet, but not parallelc don’t meet, but not parallel d don’t meet and parallel

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●12 a Bay Street, Bank Street or Junction Streetb Main Street, Tower Street or River Street

●13

exercise 6.1 additional questions

●1 Explain why you think this shape is called a parallelogram.

●2 These two lines are parallel. a Measure how far apart the lines are.b Show on the diagram where you measured to

find how far apart the lines are.

●3 Jess is making a corduroy teddy bear for her little brother. She notices that each of the paper pattern pieces has a line labelled ‘Place on straight grain of fabric’. Using the word parallel, explain in a sentence what you think ‘Place on straight grain of fabric’ means.

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●1 Both pairs of opposite sides of the shape are parallel.

●2 a 2 cmb

●3 Place the arrow on the pattern parallel to the direction of the lines in the fabric.

2 cm


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6.26.2 Angles Angles

Teaching note: dual concept of angleIt is important that students understand the dual nature of the concept angle,

■ a dynamic concept of an amount of turning, for example as a door opens,

■ and the more familiar static concept of an angle at a vertex.

Plastic geostrips or strips of card and paper fasteners are useful in developing the concept of an angle as an amount of turning. Plastic geostrips are available from Modern Teaching Aids, www.teaching.com.au.

Another useful approach is to take students outside where there is room to spread out and ask them to make angles with their arms. If they start by putting both arms out horizontally together to the left, they can then move their right arm in a clockwise direction to make approximate representations of given angles.

Teaching note: developing angle senseStudents enjoy playing a game of ‘Simon Says’. Instruct them to close their eyes, turn clockwise through a right angle, turn anticlockwise through a straight angle, and so on.

Students can vary this by making the angles with outstretched arms rather than turning their body. By concentrating so hard on the direction and type of angle, students make enough errors for this to be fun.

Teaching note: symbol senseStudents should be encouraged to use correct mathematical language. They will see that we use the same symbol for perpendicular and a right angle because perpendicular means at right angles to.

Teaching note: estimating anglesAs well as being able to measure angles accurately, students should also be able to estimate angle sizes. A useful activity is for students to work in pairs, each carefully drawing a set of angles using pencil, ruler and protractor. Each student keeps a record of the sizes of the angles they have constructed. The angle sheets can then be swapped, with each student estimating the sizes of the angles drawn by the other student, then comparing their estimates with the constructed angle sizes. This can also be used as an exercise in accurate construction.

Teaching note: measuring anglesThree angle-measurers used by various trades people are shown on page 256 of the student book. There may be students in your class whose parents use one of these in their regular work. Students could think about situations where angles are important.

The Year 5 Australian Curriculum includes ‘measuring and constructing angles using both 1808 and 3608 protractors’. Measuring angles with each protractor is included here for those students who do not already have this skill.


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Students should think about the type of angle before they start, so they can check that they have used the correct scale on their protractors. The 3608 protractor is easier to use when working with reflex angles.

Use a 3608 protractor to measure the following angles.

a b c

Working Reasoning

a

The angle measures 738.

Place the protractor with its centre at the vertex of the angle, with the zero mark on one arm of the angle. Measure around from zero in a clockwise direction on the outer scale.

Check: the angle is an acute angle so its measure must be between 08 and 908.

b


Place the protractor with its centre at the vertex of the angle, with the zero mark on one arm of the angle. Measure around from zero in an anticlockwise direction on the inner scale.

Check: the angle is an obtuse angle so its measure must be between 908 and 1808.

continued

Extra example 3

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Working Reasoning

c


Place the protractor with its centre at the vertex of the angle, with the zero mark on one arm of the angle. Measure around from zero in a clockwise direction on the outer scale because it is the reflex angle that is required.

Check: the angle is a reflex angle so its measure must be between 1808 and 3608.

Use a 3608 protractor to draw the following angles.

a 1958 b 988 c 378

Working Reasoning

a Rule a line segment for one arm of the angle. Place the protractor with its centre at the right-hand end of the line segment as shown. Place a pencil mark at 1958. Join the pencil mark to the right hand of the line segment to make the other arm of the angle.

continued

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Working Reasoning

b Rule a line segment and place the centre of the protractor at the left-hand end of the segment, with the protractor line exactly along the segment. Measure around from 0 to 988 and place a small mark. Remove the protractor and join the mark to the left-hand end of the line segment to form the angle.

c Rule a line segment and place the centre of the protractor at the right-hand end of the segment, with the protractor line exactly along the segment. Measure around from 0 to 378 and place a small mark. Remove the protractor and join the mark to the right-hand end of the line segment to form the angle.


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exercise 6.2This exercise includes questions that reinforce the concept that we must be directly in front of an angle to measure it, and that angles can be important for safety (for example in wheel chair ramps) and in sport. A blackline master is included in the teacher and student ebooks providing a larger diagram for question 13.

Ans wersAns wers

Below are the answers to the questions in exercise 6.2 in the student book.

●1 a /TOP (or /POT), acute, 508 b /JCP (or /PCJ), obtuse, 1308

c /BUS (or /SUB), reflex, 2968 d /PAT (or /TAP), reflex, 2708

e /HDG (or /GDH), obtuse, 1638 f /XYZ (or /ZYX), acute, 258

g /MOP (or /POM), right, 908 h /KTP (or /PTK), obtuse, 1308

i /HFK (or /KFH), straight, 1808

●2 a obtuse angle b acute angle c reflex angle d right anglee obtuse angle f reflex angle g revolution h straight

●3 a line b perpendicular c obtuse angle d line segmente reflex angle f ray g acute angle h parallel

●4 a 668 b 3018 c 1258 d 2408

e 758 f 1538 g 2908 h 1058

●5 a 708 b 2608 c 1508 d 458 e 1328

f 958 g 2058 h 638 i 2858 j 3458

●6 a 338 b 408

●7 a The photograph at the left, because it has been taken from a position more directly in front of the beam.

b approximately 308

c 608

●8 a A wheelchair ramp cannot be too steep otherwise the person will go down the slope too quickly and will not be able to stop. The ski slope needs to be steep enough for the skier to build up enough speed to keep going. The escalator needs to be steep enough so that it does not take up too much horizontal space, but not so steep that people could fall over on it.

b i 38 ii 308 iii 308

c The photograph has been taken from a position side on to the escalator. If the angle is to be measured accurately, the photograph must be taken from a position directly in front of the escalator, at right angles to the tiled wall.

●9 a 408 b 538; yes, the angle is greater than 508

●10 a a = 49, b = 50, c = 51, d = 48, e = 41, f = 39, g = 36, h = 39b answers will vary

●11 a obtuse b acute

Question 13

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c reflex d reflex e acute

f acute g obtuse h acute

●12 Answers will match question 11.

●13 a b mid-on


●1 Match each of the terms listed on the left with their correct description from the list on the right.

Line segment an angle of 908

Ray lines that are at right angles to each other

Parallel lines an angle of 1808

Perpendicular lines an angle less than 908

Acute angle part of a line with a definite starting point but no finishing point

Obtuse angle an angle greater than 908 but less than 1808

Right angle part of a line with a definite starting point and finishing point

Straight angle an angle of 3608

Reflex angle two lines which are always the same distance apart

Revolution an angle greater than 1808 but less than 3608

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●2 In the following diagram name a a pair of parallel line segments.b a pair of perpendicular lines.c an acute angle.d an obtuse angle.e a right angle.

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b Answers will vary. Two answers are: FD and BE; GC and BE.

c Answers will vary. Two answers are: /FGC; /DEF.

d one of /AGC or /AFD

e one of /EDF or /ECG

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Line segment part of a line with a definite starting point and finishing point

Ray part of a line with a definite starting point but no finishing point

Parallel lines two lines which are always the same distance apart

Perpendicular lines lines that are at right angles to each other

Acute Angle an angle less than 908

Obtuse Angle an angle greater than 908 but less than 1808

Right Angle an angle of 908

Straight Angle an angle of 1808

Reflex Angle an angle greater than 1808 but less than 3608

Revolution an angle of 3608


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6.36.3 Calculating angle sizes Calculating angle sizes

Teaching note: using geometric languageStudents should appreciate how appropriate geometric language (for example, ‘vertically opposite angles’, ‘supplementary angles’) facilitates communication. They should be given the opportunity to practise this language. This is one of the benefits of students using interactive geometry software such as GeoGebra, particularly when they are working in pairs. Using the names of the various software tools (for example, ‘perpendicular line’, ‘parallel line’, ‘midpoint’) reinforces correct language.

GeoGebra: complementary and supplementary angles

The GeoGebra (and HTML) file Complementary and supplementary angles in the student and teacher ebooks allows the size of angles to be changed to see pairs of angles that are complementary or supplementary.

GeoGebra: vertically opposite angles

The interactive GeoGebra file, Vertically opposite angles, is included in the teacher and student ebooks in both GeoGebra and HTML formats. Students will notice that any two adjacent angles make a straight angle. The reasoning about why vertically opposite angles are equal is based on the reasoning ‘if a 1 b 5 180 and a 1 c 5 180, then b and c must be equal’.

Complementaryand

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Teaching note: when do we put a degrees symbol?Pronumerals are introduced to refer to unknown angles as an alternative to using angle names such as /ABC. It should be noted that in this chapter the pronumerals used for angle sizes stand for a number—the number of degrees in the angle. The pronumerals already have a degrees sign after them on the diagram, so the answers will be numeric only, without the degrees sign; for example, a 5 50, not a 5 50°.

Teaching note: using algebraStudents could be introduced to equations informally as shown in extra example 5 part a and example 13 part b in the student book. Alternatively, they could use a visual or arithmetic approach and show their working on a copy of the diagram.

a Find the value of a. b ABC is a straight line. Find the size of /ABD.

a°36°

B CA

D

78°

Working Reasoning

a a 1 36 = 90 a = 90 2 36 a = 54

The two adjacent angles add to 908.

b /ABD 1 /DBC = 1808

/ABD 1 788 = 1808

/ABD = 1808 2 788

/ABD = 1028

ABC is a straight line. /ABD and /DBC are supplementary angles.

Worksheet: pool anglesThe worksheet Pool angles is included in the student and teacher ebooks.

This activity uses a pool table as its context. A diagram first shows how the ball bounces off the edge of the table at the same angle as it strikes the table. Students are then asked to use their ruler and protractor to draw the path of the ball after one, two, three and four bounces.

Pool angles

If you have ever played pool you will know the importance of angles. When a ball hits the edge, it bounces away at the same angle, as shown.

The drawing below shows a pool table. A ball is hit in the direction shown.

a Use your ruler and protractor to find where the ball is going to hit after its first bounce. Carefully draw the path of the ball.

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b Using your ruler and protractor, carefully draw the path of the ball for its second, third and fourth bounces. (Assume that the ball continues to bounce so that it forms equal angles with the edges of the table each time.)

55°

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a b

exercise 6.3Below are the answers to the questions in exercise 6.3 in the student book.

Ans wersAns wers

●1 a 1458 b 238 c 908 d 588 e 1718 f 428

●2 a 218 b 1098 c 728 d 1608 e 1158 f 1258

●3 a 768 b 658 c 948 d 1008 e 588 f 408

●4 a a = 38, b = 38, c = 114 b d = 26, e = 116, f = 64 c a = 61, b = 110, c = 110 d a = 48, b = 132, c = 42 e a = 82, b = 59, c = 39, d = 59 f a = 110, b = 70, c = 70 g a = 70, b = 39, c = 71, d = 70 h a = 36i k 5 72

●5

55° 55°

Angle Complement Angle Complement

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●6 a a = 50 b b = 10 c c = 32 d d = 45 e e = 64 f f = 24

●7

●8 a a = 32 b b = 86 c c = 138 d d = 45 e e = 65 f f = 90

●9 a 1078 b ADB is a straight line, so the angles must add to 1808

●10 a e = 46 b d = 70 c y = 107d m = 23 e h = 68 f w = 33g a = 45, b = 45, c = 85 h x = 66 i x = 40j x = 285 k x = 97 l x = 90

●11 a a = 60; b = 120 b c = 60; d = 120; e = 60; f = 120

●12 a 308 b 308; same size as angle at Ac 608 d 1208; supplementary angles e Queensberry Street f Elizabeth Street, Swanston Street

●13 a = 24

●14 a /AOB or /AOC or straight angle /AOD b /AOC

●15 808

●16 a 368

b Bec and Sarah’s total angle = 1088, Emma’s total angle = 1448

●17 a Blue 908; Green 1208; Yellow 728

b Red 788

c


Teaching noteIn question 2, students may need help in reasoning that at 10 minutes past 5, the hour hand will have moved towards 6. They have already calculated in part c that the hour hand moves 0.58 per minute, so in 10 minutes, the hour hand will have moved through 10 3 0.5°, that is, 58. Hence, the required angle is the angle between 2 and 5 (that is, 908) plus 58, making an angle of 958 between the two hands.

Angle Supplement Angle Supplement

1608 208 908 908

238 1578 1758 58

918 898 788 1028

1008 808 18 1798

78°

90°

72°

120°


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●1 a Through how many degrees does the minute hand of a clock turn each minute?

b What is the size of the obtuse angle between the hour hand and the minute hand of a clock at exactly 5 o’clock?

c Through how many degrees does the hour hand turn each minute?d What is the angle between the hour hand and the minute hand at

exactly 10 minutes past 5 o’clock?

●2 Justify your reasoning for each of the following. For example, you could copy the diagram and label the sizes of any angles you calculate along the way to finding the required angle.a Calculate the size of �MFX. b Find the value of x.

60°

x°

2x°55°

N

X

F

M

Y

G

●3 When rays of light meet a flat shiny surface, such as a mirror, they are reflected at the same angle as they meet the mirror. For example, the diagram below shows how the rays of light are reflected if they meet the mirror at an angle of 758.

75°75°

Mirror

Reflectedray of lightRay of light

meeting the mirror

A periscope is a device for seeing around corners. It consists of a bent tube with two mirrors arranged so that the rays of light are bent by 908 by each mirror and that the light rays end up parallel to their original direction.

Copy the drawing and show how you would place the two mirrors so that the rays of light follow the path shown through the periscope.

Lightrays

Eye

12

6

9 3

111

57

12 2

8 4



264

●4 Calculate the size of the angles between the blades of this wind generator.

Ans wersAns wers

●1 a 68 b 1508 c 0.58 d 958

●2 a 1458, /MFX = /NFY, as vertically opposite angles are equal.b x = 40, 2x 1 x 1 60 5 180° as supplementary angles.

●3

●4 120°

Lightrays

Mirror 1

Mirror 245°

Eye


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6.46.4 Angles and parallel lines Angles and parallel linesGeoGebra: parallel lines

A second interactive GeoGebra (and HTML) file Parallel lines is also included in the teacher and student ebooks. This time the lines AB and CD have been constructed so that they are parallel. Students can now confirm the angle relationships they observed in the Lines and transversals diagram when AB and CD were dragged until they appeared to be parallel. It is useful for students to see both interactive diagrams, as they then understand that any two lines may be cut by a transversal, but that the special angle relationships of equal alternate and corresponding angles and supplementary cointerior angles occur only when the lines are parallel. The converse is of course true, too: if alternate angles or corresponding angles are equal or if cointerior angles are supplementary, then the lines must be parallel.

GeoGebra: lines and transversals

The interactive GeoGebra file Lines and transversals is included in the student and teacher ebooks in both GeoGebra and HTML formats. Two lines and a transversal can be dragged on the screen to observe angle relationships. Students will see that vertically opposite angles are always equal. In this interactive diagram, the lines AB and CD have not been constructed to be parallel. As the two lines are dragged until they appear parallel, students will notice that other angle relationships become apparent.

Parallel lines

Lines and transversals



266

Class activity: angles and parallel linesThe class activity Angles and parallel lines includes four GeoGebra screen images of a pair of lines cut by a transversal. In the first two screens the lines are not parallel, but students can observe the relationship between vertically opposite angles. In the second pair of screens, the lines have been dragged until they are parallel. Students can now observe the additional angle relationships that occur for parallel lines cut by a transversal.

Find the values of the pronumerals.

Working Reasoning

a = 74 (Alternate angles) The angles marked a8 and 748 are alternate angles. As

the lines are parallel, the angles are equal.

b = 106 (Corresponding angles) The angles marked 748 and b8 are corresponding angles. As the lines are parallel, the angles are equal.Note also that the angles marked a8 and b8 are vertically opposite angles so they are equal.

Find the size of /PAB.

Working Reasoning

x 1 126 5 180 x 5 180 2 126 x 5 54

The angles marked 1268 and x8 are allied angles between parallel lines. Allied angles are supplementary, that is, they add to 1808.

Teaching note: the converse is also trueStudents have seen that when parallel lines are crossed by a transversal, alternate angles and corresponding angles are equal and allied angles are supplementary. The converse of this is also true: if alternate angles and corresponding angles are equal and allied angles are supplementary, then the lines must be parallel. Extra example 8 relates to this property.

Class activity

Angles and parallel lines

Extra example 6

a°

b°

74°

Extra example 7

x° 126°


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Are line segments AB and CD parallel?

Working Reasoning

The angles marked 508 and 518 are alternate angles.

The alternate angles are not equal so line segments AB and CD are not parallel.

Alternate angles formed by a transversal cutting across two lines are equal if the lines are parallel.

exercise 6.4

Ans wersAns wers


●1

●2 a 47 b 94 c m = 152, n= 152d d = 61, k = 119 e 48 f 88g x = 101, y = 101 h a = 36, b = 144 i a = 31, b = 149, c = 149

●3 a 56 b 124 c 143d 115 e e = 59, f = 59 f m = 42, n = 42g 134 h a = 57, b = 123, c = 57 i w = 112, x = 68, y = 68, z = 112

Extra example 8

50°

A

CB

D

51°

a b c d

e f g h



268

●4 a AB is parallel to CD because allied angles add to 1808.b AB is not parallel to CD because allied angles add to 1828.c AB is parallel to CD because allied angles add to 1848.d AB is parallel to CD because allied angles add to 1808.

●5 a /ACH b /BCGc /BCG d /BCHe /MBA f /CBMg /BCH h /ABN,/MBC,/BCH,/GCD

●6 358. Harry’s line of sight and the sea are parallel, so d8 and 358 are alternate angles, which are equal.


●1 Copy each of the following figures. On your drawing, label the sizes of all angles that you had to find in order to determine the value of the pronumeral, then write the value of the pronumeral.

a b

c d

36°

m° 118°

w°

37°

53°y° 50°

28°

k°

Ans wersAns wers

●1 a m = 36b w = 62c y = 90d k = 78


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6.56.5 Triangles Triangles

Teaching note: triangles as rigid shapesThe student text at the beginning of this section shows a triangle made from plastic geostrips and photographs of triangles used in constructions because of their rigidity. It is worth getting students to experience the rigidity of a triangle compared with a quadrilateral (which can be pushed into different shapes). Plastic geostrips or rolls of newspaper taped together can be used successfully for this. Plastic geostrips are slightly ‘bendy’, so even though the triangles are rigid in the plane, the slightly flexible plastic strips can be distorted by heavy-handed students who may then miss the point of the activity! In this case, the rigidity of stiff rolls of newspaper may be better. Alternatively, if your school has a dome kit, components of the kit could be used. Construction of the entire dome would then make a worthwhile additional activity.

The rigidity of the triangle may be compared with quadrilaterals, which can be made rigid by joining a diagonal (see beginning of section 6.6).

Diagonal

Teaching note: the angles in a triangleTearing the corners from a triangle and arranging them to form a straight line is an excellent visual reinforcement for the concept that the three angles of a triangle add to 180°. It is advisable that students tear, rather than cut, the corners from their triangle. If they cut the corners, they may become confused about which is the cut edge and which are the arms of the angle of the triangle. If they make a second copy of their triangle, students can paste the triangle and the arranged pieces into their mathematics books.

Students are sometimes not given the opportunity to progress from this teaching demonstration to a reasoned mathematical argument. It is important that students do not think the corner tearing demonstration actually proves the relationship. The following diagram (also included in the student book, shows that if we construct a line parallel to one side of the triangle, the same arrangement of the three angles to make a straight line can be used. This time, though, we are making use of alternate angles between parallel lines cut by a transversal to prove the relationship. Proof in this sense is an answer to the question ‘Why do the angles add to 180°?’

Class activity

Sum of the angles

of a triangle



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GeoGebra: angle sum of a triangle

The interactive GeoGebra file, Angle sum of a triangle, is included in the student and teacher ebooks. The vertices of the triangle can be dragged to see that the three angles always add to 180°. Again, this is not a proof, but a very convincing demonstration. Clicking on the Proof checkbox displays the parallel line and gives students the opportunity to reason why the relationship is true.

Angle sum ofa triangle


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GeoGebra: exterior angle of a triangle

The interactive GeoGebra file, Exterior angle of a triangle, is included in the student and teacher ebooks. The vertices of the triangle can be dragged to see that the three angles always add to 180°. After the students have had a chance to find the angle relationship for themselves, clicking on the check box displays the relationship and students are asked to explain it. In the triangle shown, we know that /ABC 1 /BCA 1 /CAB 5 180° (three angles of the triangle). But we also know that /ABC 1 /ABD 5 180° (straight line). So /BCA 1 /CAB must equal /ABD.

Using pronumerals for the angles may make the explanation clearer for students.

d°

b°c°

a°

a 1 b 1 c = 180 (angles of a triangle add to 1808)d 1 c = 180 (adjacent angles that make a straight line add to 1808)So a 1 b must equal d.

GeoGebra: types of triangles

The interactive GeoGebra file, Triangles, is included in the student and teacher ebooks in both GeoGebra and HTML formats. Three sliders allow the side lengths of the triangle to be changed. Students can see the side and angle properties of the different types of triangles and can also see that there are some sets of lengths for which it is not possible to make a triangle. The check box Show type can be clicked off to give students the opportunity to name the triangle type themselves.

Exterior angleof a triangle

Triangles



272

Teaching note: can any three side lengths make a triangle?The Triangles GeoGebra construction identifies side lengths that will not allow a triangle to be constructed. Making triangles with geostrips as shown on page 293 in the student book reinforces students understand why any three side lengths will not necessarily make a triangle. If pairs of students are given a random set of three geostrips (with three paper fasteners) some will find they can make triangles and others will not. This leads to a discussion of why some students were unable to make a triangle.

Class activity: constructing special trianglesThe class activity Constructing special angles is available in the teacher ebook as a student worksheet.

Required equipment: sharp pencil, ruler, pair of compasses

Construction 1

This construction encourages students to think about triangle properties. Because they have used the same compass opening for two sides of their triangle, the triangle must be isosceles.

Construction 2

In this construction, the same compass opening is used for all three sides of the triangle so the triangle must be equilateral. This is then linked with the angles which must, of course, be 608.

These constructions could also be completed using interactive geometry software such as Geogebra. The circle tool then takes the place of compasses.

Triangles

Class activity

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Constructing special triangles

Construction 1

a In your mathematics workbook, draw a line segment 10 cm long. ■ Label the ends A and B.

■ Using your ruler to measure, open your compass to 12 cm.

■ Place the compass point at A and draw an arc.

■ Place the compass point at B and draw another arc to intersect (cross) the first arc.

A10 cm

BA10 cm

12 cm

B A10 cm

B

■ Label the intersection point C.

■ Join each end of AB to C to form a triangle. Label the lengths of sides AC and BC.

10 cmA B

C

b What type of triangle have you constructed?

c Measure the three angles of your triangle and label them on the triangle.

Construction 2

Start with a 10 cm segment as in construction 1.

■ Label the ends of the segment D and E.

■ Place the compass point at the D and open the compass so that the pencil is exactly at E. Draw an arc.

10 cmD E 10 cmD E

10 cmD E

D E

F



274

■ Now do the opposite—place the point at E and place the pencil D. Draw another arc to intersect the first arc. Label the intersection point F.

■ Join each end of DE to F to make a triangle.

d What sort of triangle is it? Can you explain why?

e Carefully measure each of the angles of your triangle. If you have constructed your triangle accurately, each of the three angles should be 608. How close to 608 are your angles? Explain why the angles are 608.

Answersa No answer required.

b Isosceles triangle

c Students should find their angles close to 658, 658 and 508. (Note: by calculation, the angles are approximately 65.388, 65.388 and 49.248.)

d The triangle is an equilateral triangle because the length of two sides is determined by arcs that are equal in length to the 10 cm segment which forms the other side of the triangle.

e The three angles are equal and must add to 1808.

exercise 6.5

Ans wersAns wers


●1 a ÂBC (or ^CAB or ^BCA) b ^BOX (or ÔXB or ^XBO)c ^THA (or ^HAT or ÂTH) d ÊNP (or ^PEN or ^NPE)

●2 equilateral triangle: angles are 608, sides are equal

●3 a right-angled scalene triangle b acute-angled equilateral trianglec right-angled isosceles triangle d acute-angled isosceles trianglee obtuse-angled isosceles triangle f equilateral triangle (acute-angled)g obtuse-angled scalene triangle h acute-angled scalene trianglei acute-angled isosceles triangle

●4 a a = 63 b b = 49 c c = 76 d d = 36 e e = 86 f f = 53g g = 50 h h = 30 i i = 61 j j = 90 k k = 30 l l = 108

●5 a a = 32 b b = 64 c c = 70, d = 40d x = 60, y = 60, z = 60 e e = 74 f y = 45, z = 75 g g = 60 h h = 106 i t = 30, u = 120j d = 50, e = 80 k m = 45 l x = 60, y = 60

●6 a a = 65 b b = 77 c c = 90 d d = 108 e e = 125 f f = 52g g = 70 h h = 110 i m = 44 j j = 39 k a = 75 l x = 70

●7 a right-angled scalene triangle b


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●8 a i b i

ii acute-angled triangle ii right-angled triangleiii scalene triangle iii scalene triangle

c i d i

ii acute-angled triangle ii acute-angled triangleiii equilateral triangle iii isosceles triangle

e i f i

ii right-angled triangle ii obtuse-angled triangleiii scalene triangle iii isosceles triangle

g i h i

ii right-angled triangle ii acute-angled triangleiii scalene triangle iii isosceles triangle

●9 a no; 5 cm 1 3 cm , 9 cm b yes; 5 cm 1 6 cm . 7.5 cmc no; 2 cm 1 2 cm = 4 cm d yes; 3 cm 1 3 cm . 4 cme no; 5 cm 1 4 cm = 9 cm f no; 6 cm 1 8 cm , 16 cmg yes; 10 cm 1 3 cm . 11 cm h no; 7 cm 1 9 cm , 20 cm

●10 a b

i acute-angled triangle i acute-angled triangleii scalene triangle ii isosceles triangle

9 cm

5 cm

A

8 cm

B

C

10 cm

8 cm6 cm

A B

C

5 cm5 cm

A5 cm

B

C

6 cm7 cm

A7 cm

B

C

4 cm 5 cm

A3 cm

B

C

5 cm5 cm

A 8 cm B

C

5 cm12 cm

A13 cm

B

C6 cm6 cm

A8 cm

B

C

B CC

A

40°

7 cm

6 cmB C

A

35°

7 cm

7 cm



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c d

i acute-angled triangle i obtuse-angled triangleii isosceles triangle ii scalene triangle

e f

i obtuse-angled triangle i acute-angled triangleii scalene triangle ii scalene triangle

g h

i obtuse-angled triangle i right-angled triangleii isosceles triangle ii isosceles triangle

●11 a b

i obtuse-angled triangle i right-angled triangleii scalene triangle ii scalene triangle

c d

i obtuse-angled triangle i right-angled triangle\ii scalene triangle ii scalene triangle

e f

i obtuse-angled triangle i right-angled triangleii isosceles triangle ii scalene triangle

B C

A

30°

8 cm

8 cm

B C

A

45°

4 cm

6 cm

A B

C

60°

4 cm

9 cm

A B

C

55°

8 cm

10 cm

E F

G

30°30°8 cm

K L

M

5 cm

5 cm

B C

A

50°25°8 cm B C

A

40° 50°6 cm

B C

A

55° 20°6 cm

B C

A

50° 40°9 cm

A B

C

40° 40°5 cm

A

C

B70° 20°

7 cm


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g h

i right-angled triangle i obtuse-angled triangleii scalene triangle ii isosceles triangle

●12 base 1.73 m, slope 2 m scale : 10 cm = 1 metre

slope measures 20 cm = 2 metresbase measures 17.3 cm = 1.73 metres


●1 Find values of the pronumerals in each of the following diagrams.

a b c

d

64° 8x°

4x° 3x°y°

w°34°

22°

z°157°

e f62° 130°

r°

103°

126° x°t°

●2 This right-angled triangle has sides of length 3 cm and 4 cm as shown. a Measure the length of the third side of the triangle.b For many centuries builders have known that a triangle with side

lengths in these particular proportions is a right-angled triangle, and they have used this fact to construct right angles when laying out the foundations for buildings. Suggest how you could use a ruler and a long piece of rope (for example, about 15 metres long) to mark out a right angle on the ground. If possible, try out your suggestion.

●3 Recent technology provides rugby players with instantaneous information about their chances of kicking a goal when they are at different distances from the goal. The player’s distance from the goal, the angle of vision to the goal posts (shaded) and the apparent width between the goals is displayed.Calculate the angle of vision, g8, for the situation shown in the diagram.

P

R

Q35° 55°

8 cmP

R

Q40° 40°

10 cm

base

1 metre

30°

slope

3 cm

4 cm

45°

19°

g°



278

●4 Consider two triangles with the dimensions given.

i Side lengths 7 cm, 5 cm, 4 cm ii Side lengths 14 cm, 10 cm, 8 cma Use your ruler, pencil and compass to construct each of the triangles.b Notice that the triangles have the same shape but one is larger than the other. Measure the

three angles in each triangle and write them on your constructions. c Suggest another set of three side lengths that would give a triangle with the same shape as

these two triangles, but a different size.

●5 Consider two triangles with the dimensions given.

i Side lengths 5 cm, 4 cm, 3 cm ii Side lengths 10 cm, 8 cm, 6 cma Use your ruler, pencil and compass to construct each of the following triangles.b What sort of triangles are they?

●6 Harry and Kate are making a skateboard ramp. They want the ramp to make an angle of 30° with the ground and they want the top of the ramp to be exactly one metre above the ground. Harry and Kate have made a rough sketch of the side view of the ramp.They now need an accurate drawing of the side of the ramp so that they know how long to cut the wood for the base and the slope.Choose a suitable scale (for example, 10 cm on your drawing represents 1 m on the actual skateboard ramp) and make an accurate diagram using your ruler and protractor.Measure the length of the base and the slope and work out the lengths Harry and Kate will need to cut for each.

Ans wersAns wers

●1 a w = 120 b x = 12 c y = 67 d z = 46 e r = 112 f x = 23 t = 68

●2 a 5 cm b Knot or mark the rope into 3 m, 4 m and 5 m lengths and stretch into triangle shape.

●3 g 5 26

●4 a and bi ii

c Answers will vary. For example, 21 cm, 15 cm, 12 cm; 28 cm, 20 cm, 16 cm; 3.5 cm, 2.5 cm, 2 cm.

●5 a i ii

b right-angled triangles

●6 Slope = 2 mBase = 1.73 m

base

1 metre

30°

slope

7 cm

5 cm 4 cm

34° 44°

102°

14 cm

10 cm 8 cm

34° 44°

102°

5 cm

3 cm 4 cm90°

10 cm

6 cm 8 cm

90°


279

6.66.6 Quadrilaterals Quadrilaterals

Teaching note: angles in quadrilateralsAs for the angles of a triangle, the corner-tearing shown at the beginning of this section in the student book is a useful activity to reinforce the concept that the four angles of a quadrilateral add to 360°. Again, the visual demonstration can be extended to a consideration of why this is so. Dividing the quadrilateral into two triangles by drawing a diagonal readily shows that the 360° is made up of two lots of 180°.

GeoGebra: angle sum of a quadrilateral

The interactive GeoGebra (and HTML) file Angle sum of a quadrilateral provides another, perhaps even more convincing, demonstration for the angle sum.

Find the values of the pronumerals in this rhombus.

Working Reasoning

a 1 56 = 180a = 124

A rhombus is a special parallelogram.Adjacent angles in all parallelograms are supplementary (add to 1808).

b = 56 Opposite angles of parallelograms are equal.

c = 124 a = 124Opposite angles of parallelograms are equal.

Class activity

Sum of the angles

of a quadrilateral

Angle sum of aquadrilateral

Extra example 9

56°

a°

b°

c°



280

Identify each of these quadrilaterals. Draw a diagram for each to show the given information.

a The quadrilateral has four right angles. One pair of opposite sides has length 12 cm and other pair of opposite sides has length 10 cm.

b The quadrilateral has one pair of opposite sides parallel and no equal sides.

Working Reasoning

a The quadrilateral is a rectangle.

12 cm

12 cm

10 cm10 cm

The first sentence tells us that the quadrilateral is a rectangle (which includes squares).The second sentence tells us that it is a rectangle but not a square.

b The quadrilateral is a trapezium. Trapeziums are the only quadrilaterals with just one pair of parallel sides.

Teaching note: class inclusionAlthough we think of squares, rhombuses and rectangles as discrete shapes, they are all members of the family or class of parallelograms. They all satisfy the definition of a parallelogram as a plane shape with both pairs of opposite sides parallel.

In the same way we can consider squares as special cases of both rhombuses and rectangles as

■ a square is a special rectangle where all sides are equal.■ a square is a special rhombus where all angles are right-angles.

GeoGebra:

parallelogram family

The GeoGebra file Parallelogram family includes sliders for side lengths and angle. Students can make the special shapes of rhombus, rectangle and square simply changing the side lengths so that they are all equal, or by changing the angles to 908. Clicking on the Check box displays the special parallelogram name.

Extra example 10

Parallelogramfamily


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Class activity: how does a folding umbrella work?Required equipment: class set of umbrella ribs (three old umbrellas), GeoGebra file Umbrella

How does a folding umbrella work?

If you have an old folding umbrella, it is worth pulling it apart to separate the eight ribs. Three old umbrellas gives a class set! The construction of each rib is based on a parallelogram and students can move the hinged parallelogram to see how the shape remains a parallelogram, even though the angles change.

a What special quadrilateral is formed by the hinged shape in the umbrella rib?

b How is the hinged shape constructed so that this special quadrilateral is formed?

c What important function do you think this special quadrilateral plays in the opening and closing of the umbrella?

Ans wersAns wers

a parallelogram

b The ribs have been constructed so that both pairs of opposite sides are equal. This ensures that the shape is a parallelogram and, therefore, that both pairs of opposite sides stay parallel.Note that this is the converse of the statement that a plane shape with both pairs of opposite sides parallel also has both pairs of sides equal.

c This ensures that the umbrella folds neatly, with the folded parts moving parallel to each other as the umbrella is folded.

Class activity: hinged quadrilateralsRequired for this activity: GeoGebra or HTML files Toolbox and Car jack

Optional: several car jacks of the type shown below and an expanding box for tools, sewing or fishing.

Although rectangles are probably the most common quadrilateral (for example in books, tables, floors and windows), the special properties of the rhombus make it extremely useful in the design of tools and other items.

Class activity

Umbrella

Umbrella

Class activity

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282

Car jack

Car jacks of this type are readily available from ‘op shops’ or car wreckers’ yards. Based on a rhombus, the jack hinges neatly and compactly. But the particular rhombus property that is important for the operation of the jack is that the diagonals of a rhombus are perpendicular. The screw thread represents one of the diagonals of the rhombus. The design of the base of the jack ensures that this diagonal remains horizontal. The interactive GeoGebra file Car jack demonstrates how the other (invisible) diagonal remains perpendicular to the base, ensuring that the car will be lifted vertically, perpendicular to the ground.

Expanding toolbox

The design of the expanding sewing box shown here is also found in tool boxes and fishing boxes. The interactive GeoGebra file Expanding box shows how the trays stay parallel to the base as they are pulled out. The brackets attached to the drawers form parallelograms that stay parallelograms because their opposite sides are of fixed equal lengths. There may be a student in the class who could bring a similar box to demonstrate how it works.

Car jack

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Hinged quadrilaterals

1 Car jack

a What is a car jack designed to do?Open the GeoGebra file Car jack. Drag point P to simulate operating the car jack.

b What special quadrilateral is formed by the four equal sides of the quadrilateral?

c What property of this special quadrilateral is useful for storing the jack?

d The horizontal screw thread of the jack represents one of the diagonals of the quadrilateral. What do you know about the two diagonals of this special quadrilateral?

e Why is this property useful for the way the jack is designed to work?

2 Toolbox

a What special quadrilateral shape is formed by the hinged brackets connecting the trays?

b How is the box constructed so that this special quadrilateral shape is formed?

c Can you explain why the trays stay parallel to each other (and to the base of the toolbox) as they lift out?

Ans wersAns wers

1 Car jack

a A car jack is designed to lift a car to change a wheel.

b rhombus

c The sides are equal, so it can be folded flat.

d The diagonals are at right angles (perpendicular).

e The care moves at right angles to the ground, that is, vertically upwards.

2 Toolbox

a A parallelogram (or in this case the parallelogram may be a rhombus)

b The important aspect of the design is that the bracket strips are connected to the trays so that a parallelogram is formed.

c The design is using the property that a shape with both pairs of opposite sides equal will also have both pairs of opposite sides parallel.

Teaching note: exercise 6.6 question 9Question 9 relates to the scissor lift which incorporates the rhombus properties of parallel and equal sides for compactness of closing. Again, as in the case of the car jack, the perpendicular diagonals of the rhombus ensure that the hinged vertices of the parallel bars move vertically, perpendicular to the ground. This ensures that the work platform remains horizontal. Students could consider whether any parallelogram would work, or whether it is important that the special case rhombus is used.

If possible, it is worth getting students to construct the scissor lift model from geostrips and paper fasteners. They enjoy the tactile feel of opening and closing the model while observing how the sides remain parallel and the pivot points move up and down perpendicular to the horizontal. One previously disengaged student became totally involved with this activity. He had been sitting operating the rhombus geostrip linkage for a short time then suddenly announced: ‘My dad drives one of those scissor lifts’. This one experience seemed to generate an interest in mathematics for this student. Students also enjoy operating the virtual scissor lift in the interactive GeoGebra file Scissor lift.

Class activity

Hingedquadrilaterals

Scissor lift



284

What property of the rhombus makes it useful in the scissor lift?

Move point P horizontally.

Scissor lift

exercise 6.6

Ans wersAns wers


●1 a a = 55 b b = 49 c c = 124 d d = 113 e e = 103 f f = 56g g = 108 h h = 114 i i = 110 j j = 132 k k = 58 l l = 82

●2 a square b rhombus c trapeziumd rectangle e parallelogram f kiteg parallelogram h square i rhombus

●3 rectangle, trapezium, square

●4 a k = 93 b p = 90, q = 90, r = 90 c t = 45d d = 69 e e = 101, f = 50 f x = 62, y = 118g m = 110 h u = 40, v = 140 i m = 72, n = 108 j x = 118, y = 62, z = 118 k p = 40, q = 140, r = 40 l x = 95, y = 97m a = 121, b = 121 n d = 70 o a = 125, b = 60

●5 a

b Both pairs of opposite sides of a parallelogram are parallel and equal.c The opposite angles of a parallelogram are equal. The adjacent angles of a parallelogram add

to 1808.

66°

114°

114°

66°

5.4 cm

5.4 cm

5.4 cm4 cm


6.6

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285

●6 a Yes, they all have both pairs of opposite sides parallel and equal.b A rhombus is a parallelogram with four equal sides. A rectangle is a parallelogram with four

right angles. A square is a parallelogram, with four equal sides and four right angles.

●7 a, c

b The kite has two pairs of equal sides and one pair of equal angles.

●8 a b two

●9 a rhombusb The bars stay parallel to each other. Both pairs of opposite sides of the rhombus are parallel

and equal, so the bars close up neatly and compactly. The diagonals of each rhombus are at right angles to each other so the rhombuses move up and down (or in and out) at right angles to the base. This is particularly important in the case of the scissor lift to ensure that the work platform remains parallel to the ground.

●10 a kite b /DCB 5 90° /CBA 5 110° /BAD 5 50°

●11 a trapeziumb It is the best shape to fit a bicycle with the least amount of wasted space. The two trapeziums

fit beside each other to make a rectangle.

●12 608 and 1208

●13 h = 72, s = 74

●14 a Examples are shown below.i ii iii

66°

90°

102° 102°

4.2 cm

3.3 cm

4.2 cm

3.3 cm

axis of symmetry

66°

114°

75°

105° 74°

74°

106°90° 90°

102°78°

106°



286

b No. The angles in a quadrilateral always add up to 3608, so if you had three right angles then the fourth angle would also have to be a right angle.

●15 a a = 29 b b = 120 c c = 123, d = 57 d x = 106

●16 James is correct. Both a square and a rhombus have both pairs of opposite sides parallel and all sides equal. They also both have opposite angles equal. The difference is that for a square all the angles must be 908, but for the rhombus there is no such limitation.


●1 a Using your ruler and protractor, carefully draw a quadrilateral which has exactly

i four right angles.

ii two right angles.

iii one right angle.b Is it possible to draw a quadrilateral that has exactly three right angles? Explain.

●2 Find the value of the pronumerals in each of the following figures.

a°

b° x°d°

c°

94°

42° 65°

57°

77°

92°

23°

72° 67°

59°

a b c d

●3 When a square is folded in half from corner to corner as shown, two triangles are formed.What are the sizes of the three angles in each triangle?

Ans wersAns wers

●1 a i ii iii

b No, because three right angles add to 2708, leaving 908 for the fourth angle, so there would be four right angles.

●2 a a = 29 b b = 120 c c = 123, d = 57 d x = 106

●3 458, 458, 908


287

6.76.7 Representing three-dimensional Representing three-dimensional objects in two dimensionsobjects in two dimensions

Teaching note: oblique and isometric drawingsThe advantages and disadvantages of oblique and isometric representations can be discussed with students. The oblique drawing has the advantage that front and back faces are accurately depicted, but the disadvantage is that the depth cannot be accurately shown. The isometric drawing has the advantage that all three dimensions—length, width and depth—can be shown accurately, but the drawing appears distorted to our eye because we are accustomed to a perspective view.

Use 1 cm graph paper to construct an oblique drawing of a box with a front face of 9 cm by 7 cm.

Working Reasoning

Draw a rectangle 9 cm by 7 cm, then draw line segments at 458 to the horizontal from three vertices of the rectangle. The length of these segments depends on how deep you wish the box to appear.

Draw the remaining horizontal and vertical line segments to complete the box.

Use isometric grid paper to construct an isometric drawing of a box 20 mm high with a base 50 mm by 20 mm.

continued

Extra example 11

Extra example 12



288

Working Reasoning

Draw a vertical line segment 2 units long to represent the height of the box. Draw segments 5 and 2 units long to represent two edges of the base of the box.

Draw the remaining line segments to complete the box, making sure the vertical edges are all 2 units long. The box could have been drawn the other way around, with the 5 unit segment to the left and the 2 unit segment to the right.

Teaching note: plans and elevationsThere is an excellent isometric drawing tool provided at the National Council of Teachers of Mathematics website at www.nctm.org.

Use this isometric drawing to draw the

a plan.

b front elevation.

c side elevation.

continued


Extra example 13

Front


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289

Working Reasoning

a The plan shows the ‘floor’ area that the shape covers.

b The front elevation shows what we would see if we looked directly at the front of the shape without seeing the depth. It appears as if the blocks have been moved forward so that they are all in the same plane.

c The side elevation shows what we would see if we looked directly at the side of the shape without seeing the depth.

exercise 6.7

Ans wersAns wers


●1 ●2

●3 ●4




290

●5 a i ii iii

b i ii iii

c i ii iii

d i ii iii

●6 a i ii iii

b i ii iii

c i ii iii

d i ii iii

4 cm

5 cm

4 cm

4 cm

5 cm

4 cm

4 cm

3 cm

4 cm

2 cm

3 cm

2 cm

5 cm

4 cm5 cm

2 cm

4 cm

2 cm

1.5 cm

4 cm

1.5 cm

3.5 cm

4 cm

3.5 cm


6.7

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ap


291

e i ii iii

f i ii iii

●7 a i ii iii

b i ii iii

●8

●9

●10 Answers will vary according to prisms used. Check with your teacher.

●11 a plan b side elevation c isometric drawing d oblique drawing

4 cm

6 cm4 cm

2.5 cm 3 cm

6 cm

3 cm

50 cm

40 cm

50 cm

25 cm

40 cm

25 cm

3.5 m

4 m

5 m

a b c d



292


●1 Make an isometric drawing of a box 2 units high, 4 units wide and 6 units long.

●2 Make an oblique drawing of a box 2 units high, 4 units wide and 6 units long.

●3 Use this isometric drawing to draw thea plan. b front elevation. c side elevation.

Front


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293

Ans wersAns wers

●1

●2

●3 a b c

plan front elevation side elevation



294

Analysis tasks

Answers to the student book task, A parking problem, are included in this section together with two additional analysis tasks: Polyominoes and Boom angles.

Student book: a parking problemA parking problem links to the chapter warm-up worksheet Parallel and angle parking and looks at how angles are important in safe street (kerbside) parking. Students will discover that the greater the angle the cars make with the kerbside, the greater the number of cars that can fit. However, there are many other issues to take into account, such as the safety of cars backing out in front of traffic. Discussion of these issues could be at whole class level or students could work in groups.

The teacher ebook includes a blackline master template (A parking problem) of 1:100 scale parking space models of the 2.7 3 5.5 m spaces for angle parking and the slightly longer 2.7 3 6 m spaces for parallel parking. It is easier if the two different sizes of parking space are photocopied onto different coloured paper. These pieces could be saved and used again by other classes to save class time—in this case, thin card may be preferable. The pieces could also be laminated for future use. For the slightly longer parking spaces for parallel parking, each group should be given at least 10 pieces. (Actually, only six are needed, but we don’t want to give them the answer before they start!). For the angle parking spaces, each group should have at least 30 pieces so that they can set out the 908 and 608 parking at the same time.

AnswersAnswersa

Parallel parking: 6 cars

b Spaces can be shorter for 608 or 908 angle parking because there is space in the road behind the parked cars, so cars do not need extra space for getting in or out of the parking space.

c

908 angle parking: 14 cars

d

608 angle parking: 12 carsNote: Students may also like to see how many cars would fit for 458 angle parking.

458 angle parking: 9 cars

A parking problem

BLM


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295

e

f Factors affecting choice of parking type include■ width of the road■ number of parking spaces needed■ length of the available parking strip■ amount of traffic on the road, which could make angle parking more dangerous■ depth of gutters in many towns, where parallel parking would not allow car doors to be

opened on the passenger side ■ how busy the area is in terms of parking requirements.

Additional task: polyominoesThis analysis task develops spatial visualisation skills. The task starts with triominoes, shapes made from three squares joined by sides, then leads on to tetrominoes and pentominoes. Students are asked to work out how many different tetrominoes and pentominoes are possible, drawing each of the possible arrangements of squares. They are then asked to use the shapes to fill certain regions.

The game of dominoes has rectangle shaped pieces made up of two squares joined together.

We could join together different numbers of squares to make triominoes (three squares), tetrominoes (four squares) and pentominoes (five squares).

If we have two squares there is only one way of putting them together, as shown above.

If we imagine that we can pick up the triominoes and turn them over (reflect them) and turn them around (rotate them), there are two different ways of putting three squares together.

These two triominoes are the same The two different triominoes

Advantage Disadvantage

Parallel parking Parallel parking is suited to narrower roads and is safer for cars leaving the parking space.

Fewer cars will fit.

Angle parking More cars can fit in a certain length of roadway than for parallel parking. 458 and 608 parking fit more cars than parallel parking, but require less road width than 908 parking.

Angle parking spaces extend out further into the roadway, and it is harder for drivers to see traffic coming when they are backing out of an angle parking space.



296

a Fill this 6 3 3 rectangle using a combination of the two different triominoes.

b There are five different tetrominoes. Draw the five tetrominoes, being careful to check whether some of your tetrominoes are actually the same.

c Fill this 8 3 3 rectangle using i two different tetrominoes. ii four different tetrominoes.

d There are 12 different pentominoes. Draw the 12 tetrominoes, again being careful to check whether some are actually the same.

e Fill this 10 3 3 rectangle using i two different pentominoes. ii four different pentominoes.

f Fill this square with five different pentominoes.

AnswersAnswersa

b


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297

c i ii

d

e i

ii

f



298

Additional task: boom anglesThis task uses the angles displayed on the boom of a crane together with the safe loads and the horizontal reach of the crane. The task gives practice in drawing angles and in interpreting information in a table.

Boom angles

The part of a crane that can be tilted at different angles is called the boom. The angle, /ABC, that the boom makes with the horizontal is called the boom angle.

a Use your protractor to find the boom angle for this crane.

Cranes have an angle measurer resembling a 908 protractor on the boom which tells the crane driver the angle the boom makes with the horizontal.

30°

90°80° 70° 60° 50° 40°

20° 10°0°

The crane shown in the photographs below was working on a building construction site and was lifting concrete walls into position. The boom is shown in two different positions.

Position 1 Position 2

b Measure the boom angle (marked) for the crane in the two different positions, and compare your answers with the crane’s angle measurer in each position.

C

A

B

Boom


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299

c The boom is telescopic and its length can be changed, depending on how far it must reach to set down its load. If the boom is fully extended and the angle is too small, the crane may tip over with very heavy loads. The capacity chart in the cabin tells the driver the boom angle that is safe for a particular load, and how far the crane can reach horizontally for that angle and that boom length.

The capacity chart on the right shows the loads that can be safely lifted for different boom angles when the boom is fully extended to 31.5 m. It also shows how far the crane can reach for each angle.

For each of the two positions of the crane in part b, use the capacity chart to find the maximum load the crane could safely lift.

d What load could the crane safely lift if the boom angle is 408?

e For each of the two positions of the crane in part b, use the capacity chart to find how far the boom could reach horizontally.

f Using your protractor and ruler, make a careful drawing of the boom to show the angle it would need to make with the horizontal for a load of 8100 kg.

AnswersAnswersa 44°b positio n 1 = 48°; position 2 = 80° (so the crane’s measurer is wrong in both positions: it says

53° instead of 48° in position 1 and 51° instead of 80° in position 2) c position 1 = 2200 kg; position 2 = 10 000kg d 1300 kge position 1 = 21 m; position 2 = 6 m f

Further investigations in chapter 15The following investigations in chapter 15 are suitable as additional or alternative tasks for chapter 6.

■ Catching the sun’s heat■ Star polygons

The investigation Catching the sun’s heat involves drawing angles accurately with a protractor and interpreting given data to show the ideal tilt angles for solar panels in various places in Australia.

The investigation Star polygons builds on students’ familiarity with a 5-pointed star to look at other star polygons. The task incorporates fractions as well as interesting geometry. A blackline master file in the student and teacher ebooks provides the templates for the vertices.

Horizontal reach

Boom length 31.5 m

Boom angle (8)

Load (kg)

Horizontal reach (m)

79.0 10 000 6

77.0 10 000 7

75.0 10 000 8

73.5 9000 9

71.5 8100 10

67.5 6550 12

63.5 5000 14

59.0 4000 16

55.0 3200 18

50.0 2500 20

45.5 1800 22

40.0 1300 24

33.5 900 26

25.5 650 28

71.5°


300


Review Geometry and space

Visual map suggestion

Lines,segmentsand rays

Angles

ParallelSame distance apart; in the same plane and never meet

Acute< 90˚

Right90˚

Straight180˚

Revolution360˚

Vertically opposite angles are equal

Parallel lines Alternate angles are equal

Corresponding angles are equal Allied angles are supplementary

Complementary angles add to 90˚

Supplementary angles add to 180˚

Obtuse� 90˚

� 180˚

Reflex� 180˚� 360˚

PerpendicularAt right angles

Triangles3 sides

Sum of angles = 180°

Sides

IsoscelesScalene EquilateralObtuse-angled

Right-angled

Acute-angled

Angles


301

ch

ap


Quadrilaterals4 sides

Sum of angles = 360°

SpecialQuadrilaterals

RectanglesAll angles 90°.

Diagonals equal in length.

RhombusesAll sides equal.

Diagonals intersect at

right angles.

Trapezia1 pair of opposite

sides parallel.2 pairs of

supplementaryallied angles.

ParallelogramsBoth pairs of opposite sides

parallel and equal.Adjacent angles supplementary.

Diagonals bisect each other.

KitesTwo pairs of

adjacent sides equal.One pair of opposite

angles equal.Diagonals

intersect atright

angles.

SquaresAll angles 90° and

all sides equal.Diagonals equal and

intersect atright angles.

Note that the boxes for rectangles, squares and rhombuses have been included in the box for parallelograms to indicate that they all belong to the family of parallelograms. Similarly squares have been shown overlapping the boxes for rectangles and rhombuses.

Revision answers●1 C ●2 E ●3 C ●4 B ●5 E

●6 a The line PQ is parallel to the line RS.b i AB is perpendicular to CD. ii EF is parallel to GH.

●7 a 144° b 215°

●8 a i /MLG or /GLM ii acute angle iii 65°b i /PHC or /CHP ii reflex angle iii 280°

●9

BAC

D


302


●10

●11 a complementary: 53° and 37° b supplementary: 139° and 41°

●12 a b 45°

●13 a k 5 28 b /EXY 5 73° c a 5 80; b 5 100; c 5 43

●14 a a 5 50 (cointerior angles in parallel lines add to 1808)b b 5 146 (corresponding angles in parallel lines are equal)

●15 a a 5 124 b b 5 22

●16 a rectangle; a 5 30 b parallelogram; b 5 98 c kite; c 5 115d square; d 5 90 e rhombus; e 5 105 f trapezium; f 5 90, g 5 130

●17 a Yes, the sum of the two shortest sides is more than the third side, so the sides will meet.b Yes, the sum of the two shortest sides is more than the third side, so the sides will meet.c No, the sum of the two shortest sides is equal to the third side, so the sides will not meet.

●18 a p 5 30 b 1508 c 4 pieces d 908

●19 a b isosceles c acute-angled

●20 a b isosceles c right-angled

CE

A BD

C D

E

135°

CB

A

6 cm

6 cm

5.5 cm

ED

F

5 cm

5 cm


303

ch

ap


●21 a b scalene c obtuse-angled

●22 i ii

●23 i

●24

40°30°J K

L

8 cm


304


Practice quiz answers●1 C ●2 E ●3 E ●4 C ●5 D

●6 A ●7 A ●8 D ●9 A ●10 B

●11 D ●12 A ●13 C ●14 D ●15 B

Chapter tests

Test A

Multiple-choice questions

●1 Two lines which are at right angles to each other areA adjacent. B perpendicular.C complementary.D parallel.E vertically opposite.

●2 The size of /AFD is A 378

B 538

C 1278

D 1438

E 2338

●3 Which one of the following is a pair of complementary angles?A 818 and 98 B 908 and 1808 C 648 and 1168 D 778 and 238 E 608 and 208

●4 Which one of the following statements is not correct?A Two line segments that are parallel could not be perpendicular.B Two lines that are in the same plane and do not meet must be parallel.C Complementary angles add to 908.D Vertically opposite angles are always equal.E Adjacent angles are always supplementary.

●5 Which of the following sets of three lengths could be the sides of a triangle?A 5 cm, 8 cm, 15 cmB 6.5 cm, 6.5 cm, 13 cmC 8.1 cm, 9.8 cm, 17.9 cmD 6.2 cm, 12.4 cm, 18.6 cmE 11.5 cm, 8.1 cm, 16.7 cm [5 3 2 = 10 marks]

Practice quiz

Chapter 6

Chapter test A

Chapter 6

37°

E

D

B

FC

A


305

ch

ap


Short-a nswer questions

●6 Carefully draw diagrams using a ruler and a pencil, then add the correct symbols to showa two intersecting line segments that are perpendicular to each other.b two lines that are parallel to each other. [1 1 1 = 2 marks]

●7 Find the size of the following angle and state its type.

270 260 250 240 230 220 210200

190

280290

300

310

320

330

340

350

9080706050

4030

2010

100110

120

130

140

150160

170

90 100 110 120 130 140 150160

170180

8070

6050

4030

2010

0

27028029030031032033

034

035

0

260250

240

230

220

210200

190

●8

a Measure the size of /WAY. b Name the angle that is the supplement of /WAY.

[1 1 1 1 1 1 1 = 4 marks]

●9 Philip Street, Hamilton Road, Bay Road and Market Street meet at a junction as shown. a What is the size of the acute angle that Philip Street

makes with Hamilton Road?b Which two roads or streets form a pair of vertically

opposite angles?c What is the size of the acute angle that Market Street

makes with Hamilton Road?

[1 1 1 1 1 = 3 marks]

P

A

S

Y

W

PhilipStreet

Market Stre

et

Bay

Roa

d

Ham

ilton

Roa

d

76°

24°

43°


306


●10 A pair of parallel lines is cut by a transversal as shown. a Name a pair of alternate angles.b Name a pair of corresponding angles.c Name a pair of cointerior angles.d What is the size of /DGH?

[1 1 1 1 1 1 1 = 4 marks]

●11 A quadrilateral has two sides of length 7 cm and two sides of length 10 cm. Jason says that the quadrilateral must be a rectangle. Draw diagrams to show that there are two other possibilities. In each case name the special quadrilateral you have drawn.

[2 1 2 = 4 marks]

●12 Find the value of the pronumerals in each of the following.a b

c d

e f

g h

[8 3 2 = 16 marks]

C

A

E

H

F

143°

B

GD

3b°68°

b°

43°

55°

a°

48°

b°

125°

82°

d°

40°

145°

a°

80°

65°

112°

x°

104°

a°98°

a°


307

ch

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Extended-response questions

●13 Follow these steps.a Use your compass, ruler and pencil to construct ÂBC with the following sides: BC = 7 cm,

AB = 3.5 cm and CB = 5 cm.b Classify ÂBC according to its side lengths.c Classify ÂBC according to its angle sizes. [2 1 1 1 1 = 4 marks]

●14 Follow these steps.a Starting with the edge shown, construct an isometric drawing of 5 cm cube. (Assume the

isometric grid is a 1 cm grid.)

b Draw a front elevation of this three-dimensional shape. [2 1 1 = 3 marks]

[Total = 50 marks]

Tes t A ans wersTes t A ans wers

●1 B ●2 D ●3 A ●4 D ●5 E

●6 a b

●7 2578, reflex

●8 a 298 b /PAW

●9 a 808 b Hamilton Road and Philip Street c 578

●10 a /AFG and /DGF or /BFG and /CGF

b /AFG and /CGH or /BFG and /DGH or /EFB and /FGD or /EFA and /FGC

c /AFG and /CGF or /BFG and /DGF

d 378


308


●11 b parallelogram kite

10

10

77

10 10

7 7

●12 a b 5 28 b a 5 82 c b 5 42 d d 5 27e a 5 105 f x 5 103 g a 5 76 h a 5 82

●13 a b Scalene c Obtuse-angled

●14 a b

Test B

Multiple-choice questions

●1 Two angles that add to 908 areA complementary.B supplementary.C acute.D obtuse.E reflex.

B C

A

7 cm

3.5 cm 5 cm

Chapter test B

Chapter 6


309

ch

ap


●2 The value of x is A 108

B 508

C 1008

D 1408

E 2808

●3 Which of the following is a pair of supplementary angles?A 1638 and 278 B 598 and 1218 C 418 and 498 D 2408 and 608 E 908 and 1808

●4 Which one of the following statements is correct?A Adjacent angles are always supplementary.B Two line segments that intersect at 908 are parallel.C Vertically opposite angles are always complementary.D Two lines are parallel if they are in the same plane and they never meet.E Two line segments are parallel if they are in the same plane and they never meet.

●5 Which of the following could not be the sides of a triangle?A 3 cm, 4 cm, 5 cmB 8.2 cm, 9.8 cm, 17.9 cmC 8.1 cm, 9.9 cm, 17.9 cmD 5 m, 12 m, 16 mE 5 m, 8 m, 15 m [5 3 2 = 10 marks]

Short-answer questions

●6 Name a pair of a supplementary angles.b complementary angles.c vertically opposite angles.

[1 1 1 1 1 = 3 marks]

43°

37°x°

x°

EF

A

B

DC

O


310


●7 a Find the size of the following angle.

27026025024023022021

020

019

0

280290

300

310

320

330340

350

90 80 70 6050

4030

2010

100110

120

130

140

150

160

170

9010011012013014015

016

017

018

0

8070

6050

4030

2010

0

270 280 290 300 310 320 330340

350

260250

240

230

220

210

200

190

b State its type. [1 1 1 = 2 marks]

●8 Measure the size of /POM.

R

Q P

M

O

[1 mark]

●9 Find the value of the pronumerals.a b

35°y°x°

3a°2a°

a°


311

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ap


c d

e f

g h

[8 3 2 = 16 marks]

Extended-response questions

●10 Follow these steps.a Use a compass and a ruler to construct a triangle with sides of length 5 cm, 5 cm and 7 cm.b Classify the triangle according to its side lengths. [2 1 1 = 3 marks]

●11 Follow these steps.a Use a pencil and a ruler to draw a diagram showing a line AB that is parallel to a line CD. Use

the correct symbols to show that the lines are parallel.b On the same diagram, draw a line segment EF that cuts both AB and CD. c Label the point M where EF cuts AB and the point N where EF cuts CD.d Name a pair of alternate angles.e Name a pair of corresponding angles.f Name a pair of cointerior angles. [2 1 1 1 1 1 1 1 1 1 1 = 7 marks]

●12 Complete the following.a State two properties that rectangles, rhombuses, parallelograms and kites have in common.b Draw and name three different types of quadrilaterals with two sides of length 3 cm and two

sides of length 5 cm. [2 1 3 = 5 marks]

130°

c°

115° 30°

x°

100°

145°y°40°

d°

100°

60°

P Q

R

S

x°d°

48°

79°

95°


312


●13 Follow the steps.a Use the isometric grid paper below to make an isometric drawing of a box 3 units high, 4 units

wide and 6 units long. Start with the edge shown.

b Draw a front elevation of this 3-dimensional shape.

[2 1 1 = 3 marks]

[Total = 50 marks]

Tes t B ans wersTes t B ans wers

●1 A ●2 C ●3 B ●4 D ●5 E

●6 a /FOA and /AOD or /FOB and /BOD or /FOC and /COD or /EOF and /FOC or /EOA and /AOC or /EOB and /BOC or /FOE and /EOD or /EOD and /DOC

b /AOB and /BOC or /EOF and /FOA or /COD and /FOA

c /EOF and /COD

●7 a 75° b acute

●8 1358

●9 a x 5 145,y 5 35 b a 5 60 c c 5 130 d x 5 35e y 5 45 f d 5 70 g x 5 110 h d 5 42


313

ch

ap


●10 a b isosceles

●11 a, b, c

●12 a four sides; angles add to 3608

b

rectangle parallelogram kite

●13 a b

5 cm 5 cm

7 cm

E

M

N

F

C D

A B

d /AMN and /DNM or /BMN and /CNM

e /AME and /CNM or /BME and /DNM or /AMN and /CNF or /BMN and /DNF

f /AMN and /CNM or /BMN and /DNM

5 cm

3 cm

5 cm

3 cm

5 cm 5 cm

3 cm 3 cm


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