U N SA C O M R PL R E EC PA T E G D ES Chapter 4 Measurement and introduction to Pythagoras’ theorem What you will learn Australian curriculum 4A Length and perimeter (Consolidating) 4B Circumference of a circle 4C Area 4D Area of special quadrilaterals 4E Area of a circle 4F Sectors and composite shapes (Extending) 4G Surface area of a prism (Extending) 4H Volume and capacity 4I Volume of prisms and cylinders 4J Time 4K Introduction to Pythagoras’ theorem (Extending) 4L Using Pythagoras’ theorem (Extending) 4M Finding the length of a shorter side (Extending) MEASUREMENT AND GEOMETRY Using units of measurement Choose appropriate units of measurement for area and volume and convert from one unit to another (ACMMG200) Find perimeters and areas of parallelograms, rhombuses and kites (ACMMG196) Investigate the relationship between features of circles such as circumference, area, radius and diameter. Use formulas to solve problems involving circumference and area (ACMMG197) Develop the formulas for volumes of rectangular and triangular prisms and prisms in general. Use formulas to solve problems involving volume (ACMMG198) Solve problems involving duration, including using 12- and 24-hour time within a single time zone (ACMMG199) NUMBER AND ALGEBRA Real numbers Investigate the16x16 concept of irrational numbers,32x32 including π (ACMNA186) Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Online resources U N SA C O M R PL R E EC PA T E G D ES • Chapter pre-test • Videos of all worked examples • Interactive widgets • Interactive walkthroughs • Downloadable HOTsheets • Access to HOTmaths Australian Curriculum courses The wheels are turning Civilisations in ancient and modern times have used measurement to better understand the world in which they live and work. The circle, for example, in the form of a wheel helped civilisations gain mobility, and modern society to develop machines. For thousands of years mathematicians have studied the properties of the circle including such measurements as its circumference and area. The ancient civilisations knew of the existence of a special number (which we know as pi) that links a circle’s radius with its circumference and area. It was the key to understanding the precise geometry of a circle, but they could only guess its value. We now know that pi is a special number that has an infinite number of decimal places with no repeated pattern. From a measurement perspective, pi is the distance a wheel with diameter 1 unit will travel in one full turn. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 176 Chapter 4 Measurement and introduction to Pythagoras’ theorem 4A Length and perimeter U N SA C O M R PL R E EC PA T E G D ES For thousands of years, civilisations have found ways to measure length. The Egyptians, for example, used the cubit (length of an arm from the elbow to the tip of the middle finger), the Romans used the pace (5 feet) and the English developed their imperial system using inches, feet, yards and miles. The modern-day system used in Australia (and most other countries) is the metric system, which was developed in France in the 1790s and is based on the unit called the metre. We use units of length to describe the distance between two points, or the distance around the outside of a shape, called the perimeter. CONSOLIDATING The Romans would have measured the perimeter of the Colosseum in paces. Let’s start: Provide the perimeter In this diagram some of the lengths are given. Three students were asked to find the perimeter. • Will says that you cannot work out some lengths and so the 6 cm perimeter cannot be found. • Sally says that there is enough information and the answer is 9 + 12 = 21 cm. 45 mm • Greta says that there is enough information but the answer is 90 + 12 = 102 cm. Who is correct? Discuss how each person arrived at their answer. Key ideas The common metric units of length include the kilometre (km), metre (m), centimetre (cm) and millimetre (mm) ×1000 km ×100 m ×10 cm ÷1000 ÷100 Perimeter is the distance around a closed shape. • All units must be of the same type when calculating the perimeter. • Sides with the same type of markings (dashes) are of equal length. x y mm ÷10 z P = 2x + y + z Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 177 Example 1 Converting length measurements U N SA C O M R PL R E EC PA T E G D ES Convert these lengths to the units shown in the brackets. a 5.2 cm (mm) b 85 000 cm (km) SO L U T I O N EX P L A N A T I O N a 1 cm = 10 mm so multiply by 10. 5.2 cm = 5.2 × 10 = 52 mm ×10 cm b 85 000 cm = 85 000 ÷ 100 ÷ 1000 = 0.85 km mm 1 m = 100 cm and 1 km = 1000 m so divide by 100 and 1000. Example 2 Finding perimeters Find the perimeter of this shape. 4 cm 3 cm SO L U T I O N EX P L A N A T I O N 6 cm P = 2 × (3 + 3) + 2 × 4 = 12 + 8 = 20 cm 4 cm 4 cm 3 cm 3 cm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 178 Chapter 4 Measurement and introduction to Pythagoras’ theorem Example 3 Finding an unknown length Find the unknown value x in this triangle if the perimeter is 19 cm. U N SA C O M R PL R E EC PA T E G D ES x cm P = 19 cm 5 cm SO L U T I O N EX P L A N A T I O N 2x + 5 = 19 2x + 5 makes up the perimeter. 2x = 14 2x is the difference between 19 and 5. x=7 If 2x = 14 then x = 7 since 2 × 7 = 14. 1 1–4 Write the missing number in these sentences. a There are mm in 1 cm. c There are m in 1 km. e There are mm in 1 m. 2 Evaluate the following. a 10 × 100 b 100 × 1000 3 State the value of x in these diagrams. a b xm b There are d There are f There are — cm in 1 m. cm in 1 km. mm in 1 km. c 10 × 100 × 1000 xm 10 cm 4 UNDERSTANDING Exercise 4A 7m c xm 12 m 14 m 3m 4 Find the perimeter of these quadrilaterals. a Square with side length 3 m. b Rectangle with side lengths 4 cm and 7 cm. c Rhombus with side length 2.5 m. d Parallelogram with side lengths 10 km and 12 km. e Kite with side lengths 0.4 cm and 0.3 cm. f Trapezium with side lengths 1.5 m, 1.1 m, 0.4 m and 0.6 m. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 5 Convert these measurements to the units shown in the brackets. a 3 cm (mm) b 6.1 m (cm) c 8.93 km (m) e 0.0021 km (m) f 320 mm (cm) g 9620 m (km) i 0.0043 m (mm) j 0.0204 km (cm) k 23098 mm (m) m 194 300 mm (m) n 10 000 mm (km) o 0.02403 m (mm) 5–7(½) d h l p 3 m (mm) 38 000 cm (km) 342 000 cm (km) 994 000 mm (km) 4A U N SA C O M R PL R E EC PA T E G D ES Example 1 5–7(½) FLUENCY 5–7(½) 179 Example 2 6 Find the perimeter of these shapes. a b 5m 7m 6m 3 cm 15 m 8m 10 cm d c 5 cm e f 1 cm 3 cm 4 cm 5 km 2 cm 1 km 8 km g h 4.3 cm i 7.2 mm 5.1 m 9.6 m 2.8 mm Example 3 7 Find the unknown value x in these shapes with the given perimeter (P). a b c xm 3m 4m 4m P = 12 m d 10 mm x cm xm P = 10 m e P = 22 cm xm f 7m x mm P = 46 mm 7 cm 13 m P = 39 m x km P = 26 km Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem 8, 9 4A 8–11 9–12 U N SA C O M R PL R E EC PA T E G D ES 8 Find the perimeter of these shapes. Give your answers in cm and assume that angles that look right-angled are 90◦ . a b PROBLEM-SOLVING 180 4 cm 30 mm 10 cm 15 mm c 7m d 1.1 cm 3m 20 mm 10 cm e 9 cm 12 m f 44 m 20 m 7 cm 9 Find the unknown value x in these diagrams. All angles are 90◦ . a b c 5 cm x cm 5 cm P = 24 cm x cm P = 34 cm 12 m xm P = 60 m 10 Jennifer needs to fence her country house block to keep her dog in. The block is a rectangle with length 50 m and width 42 m. Fencing costs $13 per metre. What will be the total cost of fencing? Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 PROBLEM-SOLVING Measurement and Geometry 11 Gillian can jog 100 metres in 24 seconds. How long will it take her to jog 2 km? Give your answer in minutes. 4A U N SA C O M R PL R E EC PA T E G D ES 12 A rectangular picture of length 65 cm and width 35 cm is surrounded by a frame of width 5 cm. What is the perimeter of the framed picture? 181 13 13, 14 REASONING 13 13 Write down rules using the given letters for the perimeter of these shapes, e.g. P = a + 2b. Assume that angles that look right-angled are 90◦ . b a b a c b b a a a d a e f b a b b 14 Write a rule for x in terms of its perimeter P, e.g. x = P – 10. a b x c 4 3 2 x 7 x 4 d e f x x x Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Disappearing squares — — 15 15 A square is drawn with a particular side length. A second square is drawn inside the square so that its side length is one-third that of the original square. Then a third square is drawn, with side length of one-third that of the second square and so on. U N SA C O M R PL R E EC PA T E G D ES 4A Chapter 4 Measurement and introduction to Pythagoras’ theorem ENRICHMENT 182 a What is the minimum number of squares that would need to be drawn in this pattern (including the starting square), if the innermost square has a perimeter of less than 1 hundredth the perimeter of the outermost square? b Imagine now if the situation is reversed and each square’s perimeter is 3 times larger than the next smallest square. What is the minimum number of squares that would be drawn in total if the perimeter of the outermost square is to be at least 1000 times the perimeter of the innermost square? Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 183 4B Circumference of a circle U N SA C O M R PL R E EC PA T E G D ES Since the ancient times, people have known about a special number that links a circle’s diameter to its circumference. We know this number as pi ( p ). Pi is a mathematical constant that appears in formulas relating to circles, but it is also important in many other areas of mathematics. The actual value of pi has been studied and approximated by ancient and more modern civilisations over thousands of years. The 256 Egyptians knew pi was slightly more than 3 and approximated it to be ≈ 3.16. The Babylonians 81 25 339 used = 3.125 and the ancient Indians used ≈ 3.139. 8 108 It is believed that Archimedes of Syracus (287–212 ) was the first person to use a mathematical technique to evaluate pi. He was able 22 223 and less than . In 480 , to prove that pi was greater than 71 7 the Chinese mathematician Zu Chongzhi showed that pi was close to 355 ≈ 3.1415929, which is accurate to seven decimal places. 113 22 was commonly used as 7 a good and simple approximation to pi. Interestingly, mathematicians have been able to prove that pi is an irrational number, which means that there is no fraction that can be found that is exactly equal to pi. If the exact value of pi was written down as a decimal, the decimal places would continue forever with no repeated pattern. Before the use of calculators, the fraction Archimedes of Syracus (287–212 BCE) Let’s start: Discovering pi Here are the diameters and circumferences for three Diameter d (mm) Circumference C (mm) C÷d circles correct to two decimal places. Use a calculator 4.46 14.01 11.88 37.32 to work out the value of Circumference ÷ Diameter 40.99 128.76 and put your results in the third column. Add your Add your own Add your own own circle measurements by measuring the diameter and circumference of circular objects such as a can. • What do you notice about the numbers C ÷ d in the third column? • Why might the numbers in the third column vary slightly from one set of measurements to another? • What rule can you write down which links C with d? Features of a circle fe rcum rence Ci • Diameter (d) is the distance across the centre of a circle. • Radius (r) is the distance from the centre to the circle. Note d = 2r. eter Diam Circumference (C) is the distance around a circle. Radius • C = 2 p r or C = p d Pi ( p ) ≈ 3.14159 (correct to five decimal places) 22 • Common approximations include 3.14 and . 7 • A more precise estimate for pi can be found on most calculators or on the internet. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Key ideas Chapter 4 Measurement and introduction to Pythagoras’ theorem Example 4 Finding the circumference with a calculator Find the circumference of these circles correct to two decimal places. Use a calculator for the value of pi. a b 4 cm U N SA C O M R PL R E EC PA T E G D ES 184 3.5 m SO L U T I O N a b EX P L A N A T I O N C = 2pr = 2 × p × 3.5 Since r is given, you can use C = 2 p r. = 7p Alternatively use C = p d with d = 7. = 21.99 m (to 2 d.p.) Round off as instructed. C = pd = p ×4 Substitute into the rule C = p d or use = 4p C = 2 p r with r = 2. = 12.57 cm (to 2 d.p.) Round off as instructed. Example 5 Finding circumference without a calculator Calculate the circumference of these circles using the given approximation of p . a b 10 m 14 cm p = 3.14 SO L U T I O N a b p = 22 7 EX P L A N A T I O N C = pd = 3.14 × 10 Use p = 3.14 and multiply mentally. Move the decimal point one place to the right. = 31.4 m Alternatively use C = 2 p r with r = 5. C = 2pr = 2× 22 × 14 7 = 88 cm 22 and cancel the 14 with the 7 7 before calculating the final answer. Use p = 2× 22 × 14 = 2 × 22 × 2 7 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 4 — Evaluate the following using a calculator and round to two decimal places. a p ×5 b p × 13 c 2× p ×3 d 2 × p × 37 U N SA C O M R PL R E EC PA T E G D ES 1 1–4 UNDERSTANDING Exercise 4B 185 2 Write down the value of p correct to: a one decimal place c three decimal places b two decimal places 3 Name the features of the circle as shown. a b c 4 A circle has circumference (C) 81.7 m and diameter (d) 26.0 m correct to one decimal place. Calculate C ÷ d. What do you notice? Example 4 5(½), 6, 7 5(½), 6, 7 5 Find the circumference of these circles correct to two decimal places. Use a calculator for the value of pi. a b c d 39 cm 18 m 2 mm e FLUENCY 5(½), 6, 7 f 5 cm 4m Example 5a 7 km 6 Calculate the circumference of these circles using p = 3.14. a b 100 cm 20 m c 3 km Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 186 Chapter 4 Measurement and introduction to Pythagoras’ theorem Example 5b 22 7 Calculate the circumference of these circles using p = . 7 a b FLUENCY 4B c U N SA C O M R PL R E EC PA T E G D ES 21 cm 70 m 7 mm 8–10 11–13 PROBLEM-SOLVING 9–12 8 A water tank has a diameter of 3.5 m. Find its circumference correct to one decimal place. 9 An athlete trains on a circular track of radius 40 m and jogs 10 laps each day, 5 days a week. How far does he jog each week? Round the answer to the nearest whole number of metres. 10 These shapes are semicircles. Find the perimeter of these shapes including the straight edge and round the answer to two decimal places. a 25 cm b c 4.8 m 12 mm 11 Calculate the perimeter of these diagrams correct to two decimal places. a b c 45° 14 m 2 cm 8 cm 12 Calculate the perimeter of these shapes correct to two decimal places. a b 4 cm c 5m 9m 10 m Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 13 Here are some student’s approximate circle measurements. Which students are likely to have incorrect measurements? C 25.1 cm 44 m 13.8 m 187 4B U N SA C O M R PL R E EC PA T E G D ES r 4 cm 3.5 m 1.1 m PROBLEM-SOLVING Measurement and Geometry 14 14, 15 15–17 14 Explain why the rule C = 2 p r is equivalent to (i.e. the same as) C = p d. 15 It is more precise in mathematics to give ‘exact’ values for circle calculations in terms of p , e.g. C = 2 × p × 3 gives C = 6 p . This gives the final exact answer and is not written as a rounded decimal. Find the exact answers for Question 5 in terms of p . REASONING Mick Svenya Andre 16 Find the exact answers for Question 12 above in terms of p . 17 We know that C = 2 p r or C = p d. a Rearrange these rules to write a rule for: i r in terms of C ii d in terms of C b Use the rules you found in part a to find the following correct to two decimal places. i The radius of a circle with circumference 14 m ii The diameter of a circle with circumference 20 cm — — 18 18 The box shows p correct to 100 decimal places. The Guinness World record for the most number of digits of p recited from memory is held by Lu Chao, a Chinese student. He recited 67 890 digits non-stop over a 24-hour period. ENRICHMENT Memorising pi 3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 Challenge your friends to see who can remember the most number of digits in the decimal representation of p . Number of digits memorised 10+ 20+ 35+ 50+ 100 000 Report A good show Great effort Superb Amazing memory World record Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 188 Chapter 4 Measurement and introduction to Pythagoras’ theorem 4C Area U N SA C O M R PL R E EC PA T E G D ES Area is a measure of surface and is often referred to as the amount of space contained inside a two-dimensional space. Area is measured in square units and the common metric units are square millimetres (mm2 ), square centimetres (cm2 ), square metres (m2 ), square kilometres (km2 ) and hectares (ha). The hectare is often used to describe area of land, since the square kilometre for such areas is considered to be too large a unit and the square metre too small. A school football oval might be about 1 hectare for example and a small forest might be about 100 hectares. Large wheat farms in Australia can range from 1000 to 15 000 hectares. Let’s start: Squares of squares Consider this enlarged drawing of one square centimetre divided into square millimetres. • How many square millimetres are there on one edge of the square centimetre? • How many square millimetres are there in total in 1 square centimetre? • What would you do to convert between mm2 and cm2 or cm2 and mm2 and why? • Can you describe how you could calculate the number of square centimetres in one square metre and how many square metres in one square kilometre? What diagrams would you use to explain your answer? Key ideas The common metric units for area include: • square millimetres (mm2 ) • square centimetres (cm2 ) • square metres (m2 ) • square kilometres (km2 ) • hectares (ha) ×10 000 1 cm = 10 mm 1 cm = 10 mm ×10002 ×1002 = 1 000 000 = 10 000 km2 m2 ×102 = 100 cm2 ÷10002 ÷1002 = 1 000 000 = 10 000 mm2 ÷102 = 100 m2 ha ÷10 000 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 189 Key ideas Area of squares, rectangles and triangles • Square A = l × l = l2 l U N SA C O M R PL R E EC PA T E G D ES • Rectangle A = l × w = lw w l 1 1 × b × h = bh 2 2 The dashed line which gives the height is perpendicular (at right angles) to the base. • Triangle A = h b Areas of composite shapes can be found by adding or subtracting the area of more basic shapes. 1 2 Example 6 Converting units of area Convert these area measurements to the units shown in the brackets. a 0.248 m2 (cm2 ) b 3100 mm2 (cm2 ) SO L U T I O N a EX P L A N A T I O N 0.248 m2 = 0.248 × 10 000 2 = 2480 cm 1 m2 = 1002 cm2 = 10 000 cm ×1002 2 Multiply since you are changing to a smaller unit. b 3100 mm2 = 3100 ÷ 100 = 31 cm 2 1 cm2 = 102 mm2 = 100 mm 2 Divide since you are changing to a larger unit. m2 cm2 cm2 mm2 ÷102 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem Example 7 Finding areas of rectangles and triangles Find the area of these shapes. a 2 cm b 7m U N SA C O M R PL R E EC PA T E G D ES 190 6 cm 13 m SO L U T I O N EX P L A N A T I O N a Write the formula for the area of a rectangle and substitute l = 6 and w = 2. A = lw = 6×2 = 12 cm2 b 1 A = bh 2 = 1 × 13 × 7 2 Remember that the height is measured using a line that is perpendicular to the base. = 45.5 m2 Example 8 Finding areas of composite shapes Find the area of these composite shapes using addition or subtraction. a b 4m 6m 1 mm 3 mm 1.2 mm 10 m SO L U T I O N a EX P L A N A T I O N 1 A = lw – bh 2 = 10 × 6 – 1 × 10 × 4 2 The calculation is done by subtracting the area of a triangle from the area of a rectangle. Rectangle – triangle = 60 – 20 6m = 40 m2 b 4m 10 m A = l2 + lw 2 = 3 + 1.2 × 1 The calculation is done by adding the area of a rectangle to the area of a square. Area = A1 + A2 = 9 + 1.2 = 10.2 mm 10 m 2 A1 A2 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 1 1–3 3 — UNDERSTANDING Exercise 4C By considering the given diagrams answer the questions. 1 cm = 10 mm U N SA C O M R PL R E EC PA T E G D ES a i How many mm2 in 1 cm2 ? 191 ii How many mm2 iii How many cm2 in 4 cm2 ? in 300 mm2 ? 1 cm = 10 mm b i How many cm2 in 1 m2 ? ii How many cm2 in 7 iii How many m2 in 40 000 cm2 ? c i How many m2 in 1 km2 ? ii How many m2 in 5 1 m = 100 cm m2 ? 1 m2 1 m = 100 cm 1 km = 1000 m km2 ? iii How many km2 in 2 500 000 m2 ? d i How many m2 in 1 ha? 1 km2 1 km = 1000 m 100 m ii How many m2 in 3 ha? iii How many ha in 75 000 m2 ? 1 ha 100 m 2 Which length measurements would be used for the base and the height (in that order) to find the area of these triangles? a b c 7m 10 cm 1.7 mm 6 cm 3m 5m 2.4 mm 2 mm 8 cm 3 One hectare is how many square metres? Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 192 Chapter 4 Measurement and introduction to Pythagoras’ theorem 4 Convert these area measurements to the units shown in the brackets. a 2 cm2 (mm2 ) b 7 m2 (cm2 ) c d 3 ha (m2 ) e 0.34 cm2 (mm2 ) f 2 2 2 2 g 3090 mm (cm ) h 0.004 km (m ) i j 450 000 m2 (km2 ) k 4000 m2 (ha) l 2 2 2 m 320 000 m (ha) n 0.0051 m (cm ) o p 4802 cm2 (m2 ) q 19 040 m2 (ha) r 2 2 s 0.0049 ha (m ) t 0.77 ha (m ) u 4–6(½) 0.5 km2 (m2 ) 700 cm2 (m2 ) 2000 cm2 (m2 ) 3210 mm2 (cm2 ) 0.043 cm2 (mm2 ) 2933 m2 (ha) 2.4 ha (m2 ) U N SA C O M R PL R E EC PA T E G D ES Example 6 4–6(½) FLUENCY 4–6(½) 4C Example 7 5 Find the areas of these squares, rectangles and triangles. a b 3m 7m 3 cm d c 13 cm 6 cm e 12 mm f 9 cm 11 m 4 cm 3m g h 3m 3 km i 7m 2m 4 km 10 km 18 m Example 8 6 Find the area of these composite shapes by using addition or subtraction. a b c 9m 14 cm 4m 5m 8 cm 5m 10 m d 3m 16 cm e 7 cm 3 cm 6 km 10 km f 7 km 2 km (Find the shaded area) 6 mm 4 mm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 7–10(½) 7–10(½), 11, 12 4C U N SA C O M R PL R E EC PA T E G D ES 7 Use your knowledge of area units to convert these measurements to the units shown in the brackets. a 0.2 m2 (mm2 ) b 0.000 043 km2 (cm2 ) c 374 000 cm2 (km2 ) d 10 920 mm2 (m2 ) e 0.000 000 2 ha (cm2 ) f 1 000 000 000 mm2 (ha) PROBLEM-SOLVING 7, 8 193 8 Find the area of these composite shapes. You may need to determine some side lengths first. Assume that angles that look right-angled are 90◦ . b a c 14 m 9 cm 3m 12 km 6 cm 3m 3m 20 m 15 km 3m (Find the shaded area) 9 Find the side length of a square if its area is: a 36 m2 b 2.25 cm2 10 a Find the area of a square if its perimeter is 20 m. b Find the area of a square if its perimeter is 18 cm. c Find the perimeter of a square if its area is 49 cm2 . d Find the perimeter of a square if its area is 169 m2 . 11 A triangle has area 20 cm2 and base 4 cm. Find its height. 12 Paint costs $12 per litre and can only be purchased in a full number of litres. One litre of paint covers an area of 10 m2 . A rectangular wall is 6.5 m long and 3 m high and needs two coats of paint. What will be the cost of paint for the wall? 13, 14 13–15 13 Write down expressions for the area of these shapes in simplest form using the letters a and b (e.g. A = 2ab + a2 ). a b c 2b a a REASONING 13 a b b b Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 194 Chapter 4 Measurement and introduction to Pythagoras’ theorem REASONING 4C 14 Using only whole numbers for length and width, answer the following questions. U N SA C O M R PL R E EC PA T E G D ES a How many distinct (different) rectangles have an area of 24 square units? b How many distinct squares have an area of 16 square units? 15 Write down rules for: a the width of a rectangle (w) with area A and length l b the side length of a square (l) with area A c the height of a triangle (h) with area A and base b — — 16 16 Two of the more important imperial units of length and area that are still used today are the mile and the acre. Many of our country and city roads, farms and house blocks were divided up using these units. Here are some conversions 1 square mile = 640 acres 1 mile ≈ 1.609344 km 1 hectare = 10 000 m2 a ENRICHMENT The acre Use the given conversions to find: i the number of square kilometres in 1 square mile (round to two decimal places) ii the number of square metres in 1 square mile (round to the nearest whole number) iii the number of hectares in 1 square mile (round to the nearest whole number) iv the number of square metres in 1 acre (round to the nearest whole number) v the number of hectares in 1 acre (round to one decimal place) vi the number of acres in 1 hectare (round to one decimal place) b A dairy farmer has 200 acres of land. How many hectares is this? (Round your answer to the nearest whole number.) c A house block is 2500 m2 . What fraction of an acre is this? (Give your answer as a percentage rounded to the nearest whole number.) Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 195 4D Area of special quadrilaterals U N SA C O M R PL R E EC PA T E G D ES The formulas for the area of a rectangle and a triangle can be used to develop the area of other special quadrilaterals. These quadrilaterals include the parallelogram, the rhombus, the kite and the trapezium. Knowing the formulas for the area of these shapes can save a lot of time dividing shapes into rectangles and triangles. Let’s start: Developing formulas These diagrams contain clues as to how you might find the area of the shape using only what you know about rectangles and triangles. Can you explain what each diagram is trying to tell you? • Parallelogram • Rhombus The area of each quadrilateral needs to be calculated to work out how many pavers are needed. • Kite • Trapezium 1 h 2 Area of a parallelogram Area = base × perpendicular height or A = bh Area of a rhombus or kite 1 Area = × diagonal x × diagonal y 2 1 or A = xy 2 b y x Area of a trapezium 1 Area = × sum of parallel sides × perpendicular height 2 1 or A = (a + b)h 2 Key ideas h x y b h a Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 196 Chapter 4 Measurement and introduction to Pythagoras’ theorem Example 9 Finding areas of special quadrilaterals b 3 mm c U N SA C O M R PL R E EC PA T E G D ES Find the area of these shapes. a 3m 8m 10 cm 5 mm 20 cm 11 mm SO L U T I O N EX P L A N A T I O N a The height is measured at right angles to the base. A = bh = 8×3 = 24 m2 b 1 A = xy 2 = 1 × 10 × 20 2 1 Use the formula A = xy since both diagonals 2 are given. This formula can also be used for a rhombus. = 100 cm2 c 1 A = (a + b)h 2 = 1 × (11 + 3) × 5 2 = 1 × 14 × 5 2 The two parallel sides are 11 mm and 3 mm in length. The perpendicular height is 5 mm. = 35 mm2 1 1, 2 2 Find the value of A using these formulas and given values. 1 a A = bh (b = 2, h = 3) b A = xy (x = 5, y = 12) 2 1 c A = (a + b)h (a = 2, b = 7, h = 3) 2 1 d A = (a + b)h (a = 7, b = 4, h = 6) 2 — UNDERSTANDING Exercise 4D Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 2 Complete these sentences. 197 4D U N SA C O M R PL R E EC PA T E G D ES a A perpendicular angle is degrees. . b In a parallelogram, you find the area using a base and the c The two diagonals in a kite or a rhombus are . 1 d To find the area of a trapezium you multiply by the sum of the two sides 2 and then by the height. e The two special quadrilaterals that have the same area formula using diagonal lengths x and y and the . are the UNDERSTANDING Measurement and Geometry 3(½), 4 3(½), 4 FLUENCY Example 9 3(½), 4 3 Find the area of these special quadrilaterals. First state the name of the shape. a b c 10 m 1.2 m 1.5 cm 5m 5m 3 cm d e 5c m 3 cm g f 11 km 22 km h 6.2 m 3.1 m i 2 cm 20 mm 4 cm 1 mm 1.8 mm 30 mm 7 cm j k 20 mm l 9m 8 cm 16 mm 5m 17 cm 50 mm 4m 4 These trapeziums have one side at right angles to the two parallel sides. Find the area of each. 2 cm a b c 3 cm 4m 2 cm 10 m 4 cm 13 cm 10 cm 5m Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem 5, 6 4D 5–7 6–8 PROBLEM-SOLVING 198 U N SA C O M R PL R E EC PA T E G D ES 5 A flying kite is made from four centre rods all connected near the middle of the kite as shown. What area of plastic, in square metres, is needed to cover the kite? 30 cm 60 cm 6 A parallelogram has an area of 26 m2 and its base length is 13 m. What is its perpendicular height? 7 A landscape gardener charges $20 per square metre of lawn. A lawn area is in the shape of a rhombus and its diagonals are 8 m and 14.5 m. What would be the cost of laying this lawn? 8 The parallel sides of a trapezium are 2 cm apart and one of the sides is 3 times the length of the other. If the area of the trapezium is 12 cm2 , what are the lengths of the parallel sides? 9 10, 11 REASONING 9 Consider this shape. 9, 10 a a What type of shape is it? b Find its area if a = 5, b = 8, and h = 3. All measurements are in cm. h b 10 Write an expression for the area of these shapes in simplest form (e.g. A = 2a + 3ab). a b c 3a b x a a 2a 2b 2x 5a 1 11 Would you use the formula A = xy to find the area of this rhombus? 2 Explain. 8 cm 10 cm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry — — 12, 13 12 Complete these proofs to give the formula for the area of a rhombus and a trapezium. Rhombus A = 4 triangle areas 1 2y 4D U N SA C O M R PL R E EC PA T E G D ES a ENRICHMENT Proof 199 = 4× 1 × base × height 2 = 4× 1 × 2 1 2 x × = a b Trapezium 1 A = Area (triangle 1) + Area (triangle 2) = = = c 1 × 2 × + 1 × 2 1 h 1 1 = × base1 × height1 + × base2 × height2 2 2 2 h b × + a Trapezium 2 A = Area (rectangle) + Area (triangle) = length × width + 1 × base × height 2 = × + 1 × 2 = = = + + – h b × = + 13 Design an A4 poster for one of the proofs in Question 12 to be displayed in your class. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 200 Chapter 4 Measurement and introduction to Pythagoras’ theorem 4E Area of a circle We know that the link between the perimeter of a circle and its radius has challenged civilisations for thousands of years. Similarly people have studied the link between a circle’s radius and its area. b U N SA C O M R PL R E EC PA T E G D ES h Archimedes (287–212 ) attempted to calculate the exact area of a circle using a particular technique involving limits. If a circle is approximated by a regular hexagon, then the approximate area would be the sum of the areas of 6 triangles with base b and height h. Hexagon (n = 6) So A ≈ 6 × 1 2 bh A=6× 1 2 bh If the number of sides (n) on the polygon increases, the approximation would improve. If n approaches infinity, the error in estimating the area of the circle would diminish to zero. h Proof b 1 A = n × bh 2 = 1 × nb × h 2 1 ×2pr×r 2 = p r2 = (As n approaches ∞, nb limits to 2 p r as nb is the perimeter of the polygon, and h limits to r.) Dodecagon (n = 12) A = 12 × 1 2 bh Let’s start: Area as a rectangle Imagine a circle cut into small sectors and arranged as shown. Now try to imagine how the arrangement on the right would change if the number of sector divisions was not 16 (as shown) but a much higher number. • • What would the shape on the right look like if the number of sector divisions was a very high number? What would the length and width relate to in the original circle? Try to complete this proof. A = length × width = 1 × 2 ×r Width Length = Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry The ratio of the area of a circle to the square of its radius is equal to p . A = p so A = p r2 r2 r 201 Key ideas U N SA C O M R PL R E EC PA T E G D ES A = π r2 A half circle is called a semicircle. 1 A = p r2 2 r A quarter circle is called a quadrant. 1 A = p r2 4 r Example 10 Finding circle areas without technology Find the area of these circles using the given approximate value of p . 22 a p = b p = 3.14 7 10 cm 7m SO L U T I O N a A = p r2 = 22 × 72 7 = 154 m2 b A = p r2 = 3.14 × 102 EX P L A N A T I O N Always write the rule. Use p = 22 and r = 7. 7 22 × 7 × 7 = 22 × 7 7 Use p = 3.14 and substitute r = 10. 3.14 × 102 is the same as 3.14 × 100 = 314 cm2 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem Example 11 Finding circle areas using a calculator Use a calculator to find the area of this circle correct to two decimal places. U N SA C O M R PL R E EC PA T E G D ES 202 2 cm SO L U T I O N EX P L A N A T I O N A = p r2 Use the p button on the calculator and enter p × 22 or p × 4. = p × 22 = 12.57 cm2 (to 2 d.p.) Example 12 Finding areas of semicircles and quadrants Find the area of this quadrant and semicircle correct to two decimal places. a b 5 km 3m SO L U T I O N a A= = 1 × p r2 4 1 × p × 32 4 EX P L A N A T I O N The area of a quadrant is 1 the area of a circle 4 with the same radius. = 7.07 m2 (to 2 d.p.) b r= 5 = 2.5 2 A= 1 × p r2 2 = The radius is half the diameter. 1 the area of a 2 circle with the same radius. The area of a semicircle is 1 × p × 2.52 2 = 9.82 km2 (to 2 d.p.) Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 1 1–3 3 — UNDERSTANDING Exercise 4E Evaluate without the use of a calculator. 22 ×7 7 c b 3.14 × 4 22 × 72 7 d U N SA C O M R PL R E EC PA T E G D ES a 3.14 × 10 203 2 Use a calculator to evaluate these to two decimal places. a p × 52 b p × 132 c p × 3.12 p × 9.82 d 3 What radius length (r) would be used to help find the area of these shapes? a b c 10 m 7 km 2.3 mm 4–6(½) 4–6(½) FLUENCY Example 10 4–6(½) 4 Find the area of these circles, using the given approximate value of p . a b c 7 cm m 28 m 14 km p = d 22 7 p = 22 7 p = f e 2m 200 m p = 3.14 p = 3.14 10 km 22 7 p = 3.14 Example 11 5 Use a calculator to find the area of these circles correct to two decimal places. a b c 1.5 mm 6m 3 cm d e 10 km f 3.4 cm 1.7 m Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 204 Chapter 4 Measurement and introduction to Pythagoras’ theorem Example 12 FLUENCY 4E 6 Find the area of these quadrants and semicircles correct to two decimal places. a b c U N SA C O M R PL R E EC PA T E G D ES 16 cm 17 mm d e 3.6 mm f 10 cm 7, 8 8m 9–11 10–12 7 A pizza tray has a diameter of 30 cm. Calculate its area to the nearest whole number of cm2 . 8 A tree trunk is cut to reveal a circular crosssection of radius 60 cm. Is the area of the cross-section more than 1 m2 and, if so, by how much? Round your answer to the nearest whole number of cm2 . PROBLEM-SOLVING 2 cm 9 A circular oil slick has a diameter of 1 km. The newspaper reported an area of more than 1 km2 . Is the newspaper correct? 10 Two circular plates have radii 12 cm and 13 cm. Find the difference in their area correct to two decimal places. 11 Which has the largest area, a circle of radius 5 m, a semicircle of radius 7 m or a quadrant of radius 9 m? 12 A square of side length 10 cm has a hole in the middle. The diameter of the hole is 5 cm. What is the area remaining? Round the answer to the nearest whole number. 10 cm 5 cm 10 cm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 13, 14 13 A circle has radius 2 cm. Find the area of the circle using p = 3.14. Find the area if the radius is doubled to 4 cm. What is the effect on the area if the radius is doubled? What is the effect on the area if the radius is tripled? What is the effect on the area if the radius is quadrupled? What is the effect on the area if the radius is multiplied by n? U N SA C O M R PL R E EC PA T E G D ES a b c d e f 4E 14, 15 REASONING 13 205 14 The area of a circle with radius 2 could be written exactly as A = p × 22 = 4 p . Write the exact area of these shapes. a b c 7 9 24 1 15 We know that the diameter d of a circle is twice the radius r, i.e. d = 2r or r = d. 2 1 Substitute r = d into the rule A = p r2 to find a rule for the area of a circle in terms of d. 2 b Use your rule from part a to check that the area of a circle with diameter 10 m is 25 p m2 . a — — 16 16 Reverse the rule A = p r2 to find the radius in these problems. a If A = 10, use your calculator to show that r ≈ 1.78. b Find the radius of circles with these areas. Round the answer to two decimal places. i 17 m2 ii 4.5 km2 iii 320 mm2 c ENRICHMENT Reverse problems Can you write a rule for r in terms of A? Check that it works for the circles defined in part b. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 206 Chapter 4 Measurement and introduction to Pythagoras’ theorem 4F Sectors and composite shapes EXTENDING U N SA C O M R PL R E EC PA T E G D ES A slice of pizza or a portion of a round cake cut from the centre forms a shape called a sector. The area cleaned by a windscreen wiper could also be thought of as a difference of two sectors with the same angle but different radii. Clearly the area of a sector depends on its radius, but it also depends on the angle between the two straight edges. θ Let’s start: The sector area formula Complete this table to develop the rule for finding the area of a sector. Angle Fraction of area Area rule Diagram 180◦ 180 = 1 360 2 A= 1 × pr2 2 180° 90◦ 90 = 360 A= × pr2 A= × pr2 90° 45◦ 30◦ q θ Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 207 Key ideas A sector is formed by dividing a circle with two radii. r θ U N SA C O M R PL R E EC PA T E G D ES θ r A sector’s area is determined by calculating a fraction of the area of a circle with the same radius. q • Fraction is 360 • Sector area = q × p r2 360 r θ r The area of a composite shape can be found by adding or subtracting the areas of more basic shapes. A = lw + 1 2 π r2 Example 13 Finding areas of sectors Find the area of these sectors correct to two decimal places. a b 120° 2 cm SO L U T I O N a 70° 5m EX P L A N A T I O N q × p r2 360 First write the rule for the area of a sector. = 120 × p × 22 360 = 1 × p ×4 3 Substitute q = 120 and r = 2. 120 1 Note that simplifies to . 360 3 A= = 4.19 cm2 (to 2 d.p.) b q = 360 – 70 = 290 A= = q × p r2 360 First calculate the angle inside the sector and remember that a revolution is 360◦ . Then substitute q = 290 and r = 5. 290 × p × 52 360 = 63.27 m2 (to 2 d.p.) Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 208 Chapter 4 Measurement and introduction to Pythagoras’ theorem Example 14 Finding areas of composite shapes 20 mm Find the area of this composite shape correct to the nearest whole number of mm2 . U N SA C O M R PL R E EC PA T E G D ES 10 mm SO L U T I O N a A = lw – EX P L A N A T I O N 1 2 pr 4 = 20 × 10 – The area can be found by subtracting the area of a quadrant from the area of a rectangle. 1 × p × 102 4 = 200 – 25 p = 121 mm2 (to nearest whole number) 1 1–3 3 — Simplify these fractions. a 180 360 b 90 360 c 60 360 45 360 d 2 Evaluate the following using a calculator. Give your answer correct to two decimal places. a 80 × p × 22 360 b 20 × p × 72 360 c UNDERSTANDING Exercise 4F 210 × p × 2.32 360 3 What fraction of a circle in simplest form is shown by these sectors? a b c 120° 4(½), 5, 6(½) Example 13a 4 Find the area of these sectors correct to two decimal places. a b 3 cm c 4(½), 5, 6(½) 2.5 cm 30° 20 mm 60° 13 mm d 4(½), 5, 6(½) FLUENCY 60° e f 270° 240° 5.1 m 11.2 cm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 g h i 315° 45° 21.5 m FLUENCY Measurement and Geometry 210° 4F U N SA C O M R PL R E EC PA T E G D ES 36.4 km 209 18.9 m Example 13b 5 Find the area of these sectors correct to two decimal places. a b c 115° 14.3 km 7.5 m Example 14 80° 6m 240° 6 Find the areas of these composite shapes using addition or subtraction. Round the answer to two decimal places. a b c 2m 10 cm 3m 20 cm 5m d 9 mm e f g h 3m 4m 24 km 20 mm i 2 mm 10 m 5 mm 7–9 1 cm 8–10 8–10 7 A simple bus wiper blade wipes an area over 100◦ as shown. Find the area wiped by the blade correct to two decimal places. 100° 1.2 m PROBLEM-SOLVING 3 cm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem 4F U N SA C O M R PL R E EC PA T E G D ES 8 At Buy-by-the-sector Pizza they offer a sector of a 15 cm radius pizza with an angle of 45◦ or a sector of a 13 cm radius pizza with an angle of 60◦ . Which piece gives the bigger area and by how much? Round the answer to two decimal places. PROBLEM-SOLVING 210 9 An archway is made up of an inside and outside semicircle as shown. Find the area of the arch correct to the nearest whole cm2 . 60 cm 60 cm 10 What percentage of the total area is occupied by the shaded region in these diagrams? Round the answer to one decimal place. a b c 249° 7.2 cm 3m 4 cm 11(½) 11(½), 12 11 An exact area measure in terms of p might look like p × 22 = 4 p . Find the exact area of these shapes in terms of p . Simplify your answer. a b c Find the shaded area REASONING 11(½) 1 mm 40° 5m 2 cm 3 mm d e 3 cm f 15 km 5m 10 m Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 REASONING Measurement and Geometry 12 Consider the percentage of the area occupied by a circle inside a square and touching all sides as shown. If the radius of the circle is 4 cm, find the percentage of area occupied by the circle. Round the answer to one decimal place. Repeat part a for a radius of 10 cm. What do you notice? Can you prove that the percentage area is always the same for any radius r? Hint: Find the percentage area using the pronumeral r for the radius. 4F U N SA C O M R PL R E EC PA T E G D ES a 211 Sprinkler waste — — 13 13 A rectangular lawn area has a 180◦ sprinkler positioned in the middle of one side as shown. B ENRICHMENT b c Lawn 2.5 m 30° O A 4.33 m 5m a Find the area of the sector OAB correct to two decimal places. b Find the area watered by the sprinkler outside the lawn area correct to two decimal places. c Find the percentage of water wasted, giving the answer correct to one decimal place. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 212 Chapter 4 Measurement and introduction to Pythagoras’ theorem 4G Surface area of a prism EXTENDING U N SA C O M R PL R E EC PA T E G D ES Many problems in three dimensions can be solved by looking at the problem or parts of the problem in two dimensions. Finding the surface area of a solid is a good example of this, as each face can usually be redrawn in two-dimensional space. The approximate surface area of the walls of an unpainted house, for example, could be calculated by looking at each wall separately and adding to get a total surface area. Let’s start: Possible prisms Here are three nets that fold to form three different prisms. • Can you draw and name the prisms? • Try drawing other nets of these prisms that are a different shape to the nets given here. Key ideas A prism is a polyhedron with a constant (uniform) cross-section. • The cross-section is parallel to the two identical (congruent) ends. • The other sides are parallelograms (or rectangles for right prisms). A net is a two-dimensional representation of all the surfaces of a solid. It can be folded to form the solid. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry The total surface area (TSA) of a prism is the sum of the areas of all its faces. Cube Rectangular prism Key ideas U N SA C O M R PL R E EC PA T E G D ES h 213 w l TSA = l TSA = 2lw + 2lh + 2wh 6l2 Example 15 Finding surface areas Find the surface area of this prism. 10 cm 8 cm 6 cm 15 cm SO L U T I O N EX P L A N A T I O N Area of 2 triangular ends One possible net is A = 2× = 2× 15 cm 1 × bh 2 1 ×6×8 2 6 cm 8 cm 8 cm 8 cm = 48 cm2 Area of 3 rectangles A = (6 × 15) + (8 × 15) + (10 × 15) = 360 cm2 Surface area 10 cm Work out the area of each shape or group of shapes and find the sum of their areas to obtain the total surface area. TSA = 48 + 360 = 408 cm2 The surface area of this chocolate can be estimated by a similar process. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 214 Chapter 4 Measurement and introduction to Pythagoras’ theorem 3 — How many faces are there on these prisms? Also name the types of shapes that make the different faces. a b c U N SA C O M R PL R E EC PA T E G D ES 1 1–3 UNDERSTANDING Exercise 4G A b c B C 3 How many rectangular faces are on these solids? a Triangular prism b Rectangular prism c Hexagonal prism d Pentagonal prism Pentagonal prism 4(½) Example 15 4(½), 5 4 Find the surface area of these right prisms. a b c 3 cm 2 cm 2 cm 8.2 m d 4 cm 8 cm 4(½), 5 FLUENCY 2 Match the net to its solid. a 12 cm e 1 cm 5m 3m 6m 4m f 9 cm 12 cm 15 cm 14 cm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 6 mm g h FLUENCY Measurement and Geometry 1.5 m i 4G U N SA C O M R PL R E EC PA T E G D ES 4 mm 215 6 mm 8m 4.2 m 3m 5 mm a Use A = bh to find the combined area of the two ends. b Find the surface area of the prism. 2 cm 1.5 cm 8 cm 6 cm 6, 7 6–8 7–9 6 An open box (with no lid) is in the shape of a cube and is painted on the outside including the base. What surface area is painted if the side length of the box is 20 cm? 7 A book 20 cm long, 15 cm wide and 3 cm thick is covered in plastic. What area of plastic is needed for to cover 1000 books? Convert your answer to m2 . 8 Find the surface area of these solids. a b 5m c 6m 3m PROBLEM-SOLVING 5 This prism has two end faces that are parallelograms. 5m 8m 7m 15 m 10 m 6m 3 cm 3m 9 The floor, sides and roof of this tent are made from canvas at a cost of $5 per square metre. The tent’s dimensions are shown in the diagram. What is the cost of the canvas for the tent? 2.8 m 2m 1.5 m Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem 10 4G 10, 11 10, 11 REASONING 216 U N SA C O M R PL R E EC PA T E G D ES 10 Write down the rule for the surface area for these right prisms in simplest form. a b c w h l l w l 11 A cube of side length 1 cm has a surface area of 6 cm2 . a What is the effect on the surface area of the cube if: i its side length is doubled? ii its side length is tripled? iii its side length is quadrupled? b Do you notice a pattern from your answers to part a? What effect would multiplying the side length by a factor of n have on the surface area? — — 12 12 An open box (with no lid) in the shape of a cube is made of wood that is 2 cm thick. Its outside side length is 40 cm. 2 cm ENRICHMENT The thick wooden box 40 cm a Find its total surface area both inside and out. b If the box was made with wood that is 1 cm thick, what would be the increase in surface area? Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 217 4H Volume and capacity U N SA C O M R PL R E EC PA T E G D ES Volume is a measure of the space occupied by a three-dimensional object. It is measured in cubic units. Common metric units for volume given in abbreviated form include mm3 , cm3 , m3 and km3 . We also use mL, L, kL and ML to describe volumes of fluids or gas. The volume of space occupied by a room in a house for example might be calculated in cubic metres (m3 ) or the capacity of fuel tanker might be measured in litres (L) or kilolitres (kL). A fuel tanker can carry about 32 000 litres. Let’s start: Packing a shipping container There are 250 crates of apples to be shipped from Australia to Japan. Each crate is 1 m long, 1 m wide and 1 m high. The shipping container used to hold the crates is 12 m long, 4 m wide and 5 m high. The fruit picker says that the 250 crates will ‘fit in, no problems’. The forklift driver says that the 250 crates will ‘just squeeze in’. The truck driver says that ‘you will need more than one shipping container’. • Explain how the crates might be packed into the container. How many will fit into one end? • Who (the fruit picker, forklift driver or truck driver) is the most accurate? Explain your choice. • What size shipping container and what dimensions would be required to take all 250 crates with no space left over? Is this possible or practical? Volume is measured in cubic units. Common metric units are: • cubic millimetres (mm3 ) ×10003 ×1003 ×103 • cubic centimetres (cm3 ) km3 m3 cm3 mm3 • cubic metres (m3 ) • cubic kilometres (km3 ) ÷10003 ÷1003 ÷103 Capacity is the volume of fluid or gas that a container can hold. Common metric units are: • millilitre (mL) • litre (L) • kilolitre (kL) • megalitre (ML) ×1000 ML ×1000 kL ÷1000 ×1000 L ÷1000 mL ÷1000 Some common conversions are: • 1 mL = 1 cm3 • 1 L = 1000 mL • 1 kL = 1000 L = 1 m3 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Key ideas 218 Chapter 4 Measurement and introduction to Pythagoras’ theorem Key ideas Volume of a rectangular prism • Volume = length × width × height V = lwh Volume of a cube V = l3 w U N SA C O M R PL R E EC PA T E G D ES • h l l Example 16 Finding the volume of a rectangular prism Find the volume of this rectangular prism. 2m 6m 4m SO L U T I O N EX P L A N A T I O N V = lwh First write the rule and then substitute for the length, width and height. Any order will do since 6 × 4 × 2 = 4 × 6 × 2 = 2 × 4 × 6 etc. = 6×4×2 = 48 m3 Example 17 Finding capacity Find the capacity, in litres, for a container that is a rectangular prism 20 cm long, 10 cm wide and 15 cm high. SO L U T I O N EX P L A N A T I O N V = lwh = 20 × 10 × 15 First calculate the volume of the container in cm3 . = 3000 cm3 Then convert to litres using 1 L = 1000 cm3 . = 3000 ÷ 1000 =3L 1 1–3 Count how many cubic units are shown in these cube stacks. a b 3 c — UNDERSTANDING Exercise 4H Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 2 Find the missing number. a 3×4×8 = c 8× × 12 = 192 UNDERSTANDING Measurement and Geometry 4H U N SA C O M R PL R E EC PA T E G D ES b × 5 × 20 = 600 d 20 × 2 × = 200 219 c 1000 kL = f 1 m3 = 4–6(½) Example 16 ML L 4–6(½) 4–6(½) FLUENCY 3 Write the missing number in the following unit conversions. a 1L= mL b kL = 1000 L 3 d 1 mL = cm e 1000 cm3 = L 4 Find the volume of these rectangular prisms. a 2 cm b c 5m 6 cm 3 cm 3 mm 4m 1m d e f 6m 4 mm 20 mm 4 km 5 Convert the measurements to the units shown in the brackets. a 2 L (mL) b 5 kL (L) c 0.5 ML (kL) e 4 mL (cm3 ) Example 17 f 50 cm3 (mL) 2m d 3000 mL (L) g 2500 cm3 (L) h 5.1 L (cm3 ) 6 Find the capacity of these containers, converting your answer to litres. a 20 cm b 30 cm 40 cm c 70 cm 60 cm 30 cm 10 cm d e 3m 3m 2m 1m 6m f 0.9 m 4m 0.8 m 0.5 m Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem 7–9 4H 8–10 9–11 PROBLEM-SOLVING 220 7 A oil tanker has a capacity of 60 000 m3 . a What is the ship’s capacity in: litres? ii kilolitres? U N SA C O M R PL R E EC PA T E G D ES i iii megalitres? b If the tanker leaks oil at a rate of 300 000 litres per day, how long will it take for all the oil to leak out? Assume the ship started with full capacity. 8 Water is being poured into a fish tank at a rate of 2 L every 10 seconds. The tank is 1.2 m long by 1 m wide by 80 cm high. How long will it take to fill the tank? Give the answer in minutes. 9 A city skyscraper is a rectangular prism 50 m long, 40 m wide and 250 m high. a What is the total volume in m3 ? b What is the total volume in ML? 10 If 1 kg is the mass of 1 L of water, what is the mass of water in a full container that is a cube with side length 2 m? 11 Using whole numbers only, give all the possible dimensions of rectangular prisms with the following volume. Assume the units are all the same. a 12 cubic units b 30 cubic units c 47 cubic units 12, 13 13, 14 12 Explain why a rectangular prism of volume 46 cm3 cannot have all its side lengths (length, width and height) as whole numbers greater than 1. Assume all lengths are in centimetres. REASONING 12 13 How many cubic containers, with side lengths that are a whole number of centimetres, have a capacity of less than 1 litre? Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 REASONING Measurement and Geometry 14 Consider this rectangular prism. How many cubes are in the base layer? What is the area of the base? What do you notice about the two answers from above? How can this be explained? If A represents the area of the base, explain why the rule V = Ah can be used to find the volume of a rectangular prism. Could any side of a rectangular prism be considered to be the base when using the rule V = Ah? Explain. 4H U N SA C O M R PL R E EC PA T E G D ES a b c 221 e Halving rectangular prisms 5 4 — — 15 15 This question looks at using half of a rectangular prism to find the volume of a triangular prism. a Consider this triangular prism. i Explain why this solid could be thought of as half a rectangular prism. ENRICHMENT d 3 ii Find its volume. b a Using a similar idea, find the volume of these prisms. i ii 8 cm 4 cm 5m 8m 7m 10 cm iii 2m iv 2 cm 1 cm 2 cm 4m 8m 5m 6 mm v vi 2 mm 7 mm 5 cm 3 mm 3 mm 3 cm 4 cm 9 cm 5 cm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 222 Chapter 4 Measurement and introduction to Pythagoras’ theorem 4I Volume of prisms and cylinders We know that for a rectangular prism its volume V is given by the rule V = lwh. Length × width (lw) gives the number of cubes on the base, but it also tells us the area of the base A. So V = lwh could also be written as V = Ah. U N SA C O M R PL R E EC PA T E G D ES h A w The rule V = Ah can also be applied to prisms that have different shapes as their bases. One condition, however, is that the area of the base must represent the area of the cross-section of the solid. The height h is measured perpendicular to the cross-section. Note that a cylinder is not a prism as it does not have sides that are parallelograms; however, it can be treated like a prism when finding its volume because it has a constant cross-section, a circle. l Here are some examples of two prisms and a cylinder with A and h marked. h h h A Cross-section is a triangle A Cross-section is a trapezium A Cross-section is a circle Let’s start: Drawing prisms Try to draw prisms (or cylinders) that have the following shapes as their cross-sections. • Circle • Triangle • Trapezium • Pentagon • Parallelogram The cross-section of a prism should be the same size and shape along the entire length of the prism. Check this property on your drawings. Key ideas A A prism is a polyhedron with a constant (uniform) cross-section. • The sides joining the two congruent ends are parallelograms. • A right prism has rectangular sides joining the congruent ends. h Volume of a prism = Area of cross-section × perpendicular height or V = Ah. Volume of a cylinder = Ah = p r2 × h = p r2 h r So V = p r2 h h A Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 223 Example 18 Finding the volumes of prisms b U N SA C O M R PL R E EC PA T E G D ES Find the volumes of these prisms. a A = 10 cm2 2m 3 cm 8m 4m SO L U T I O N EX P L A N A T I O N a Write the rule and substitute the given values of A and h, where A is the area of the cross-section. V = Ah = 10 × 3 = 30 cm3 b V = Ah ) ( 1 ×4×2 ×8 = 2 The cross-section is a triangle, so use A = 1 bh 2 with base 4 m and height 2 m. = 32 m3 Example 19 Finding the volume of a cylinder Find the volumes of these cylinders, rounding to two decimal places. a b 2 cm 14 m 10 cm SO L U T I O N a EX P L A N A T I O N V = p r2 h 2 = p × 2 × 10 = 125.66 cm3 (to 2 d.p.) b 20 m V = p r2 h 2 = p × 7 × 20 Write the rule and then substitute the given values for p , r and h. Round as required. The diameter is 14 m so the radius is 7 m. Round as required. = 3078.76 m3 (to 2 d.p.) Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 224 Chapter 4 Measurement and introduction to Pythagoras’ theorem 1 1–3 2 — UNDERSTANDING Exercise 4I For these solids: U N SA C O M R PL R E EC PA T E G D ES i state whether or not it looks like a prism ii if it is a prism, state the shape of its cross-section. a b c d e f 2 For these prisms and cylinder, state the value of A and the value of h that could be used in the rule V = Ah to find the volume of the solid. a b 10 mm c A = 6 m2 2 cm A=8 A = 12 mm2 cm2 1.5 m 3 Evaluate the following. b 1 ×3×4×6 2 c 1 (2 + 3) × 4 2 d 3.14 × 102 × 20 4–6(½) Example 18a 4 Find the volume of these solids using V = Ah. a b A = 4 m2 A = 20 cm2 4–6(½) 8 cm c 4–6(½) 11 mm FLUENCY a 2×5×4 11 m A = 32 mm2 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Example 18b FLUENCY Measurement and Geometry 5 Find the volume of these prisms. b 2m c 4I U N SA C O M R PL R E EC PA T E G D ES a 225 5 cm 10 cm 8 cm 20 cm 3m 7 cm 5m d e 7m f 3 mm 4 mm 2m 12 mm 3m 11 mm 6m 6m 2m 5m Example 19 6 Find the volume of these cylinders. Round the answer to two decimal places. a b c 10 m 20 cm 40 mm 10 mm 5m d 4 cm e 7 cm 10 m f 14 m 50 cm 7m 3m h 1.6 cm 10 m 15 m 17.6 km i 11.2 km 0.3 cm 7, 8 8–10 9–11 7 A cylindrical tank has a diameter of 3 m and height 2 m. a Find its volume in m3 correct to three decimal places. b What is the capacity of the tank in litres? 8 Jack looks at buying either a rectangular water tank with dimensions 3 m by 1 m by 2 m or a cylindrical tank with radius 1 m and height 2 m. a Which tank has the greater volume? b What is the difference in the volume correct to the nearest litre? PROBLEM-SOLVING g Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem 4I U N SA C O M R PL R E EC PA T E G D ES 9 Susan pours water from a full 4 L container into a number of water bottles for a camp hike. Each water bottle is a cylinder with radius 4 cm and height 20 cm. How many bottles can be filled completely? PROBLEM-SOLVING 226 10 There are 80 liquorice cubes stacked in a cylindrical glass jar. The liquorice cubes have a side length of 2 cm and the glass jar has a radius of 5 cm and a height of 12 cm. How much air space remains in the jar of liquorice cubes? Give the answer correct to two decimal places. 11 A swimming pool is a prism with a cross-section that is a trapezium as shown. The pool is being filled at a rate of 1000 litres per hour. a Find the capacity of the pool in litres. b How long will it take to fill the pool? 4m 8m 2m 3m 12, 13 13, 14 12 Using exact values (e.g. 10 p cm3 ) calculate the volume of cylinders with these dimensions. a Radius 2 m and height 5 m b Radius 10 cm and height 3 cm c Diameter 8 mm and height 9 mm d Diameter 7 m and height 20 m REASONING 12 13 A cylinder has a volume of 100 cm3 . Give three different combinations of radius and height measurements that give this volume. Give these lengths correct to two decimal places. 14 A cube has side length x metres and a cylinder has a radius also of x metres and height h. What is the rule linking x and h if the cube and the cylinder have the same volume? — — 15 15 Use your knowledge of volumes of prisms and cylinders to find the volume of these composite solids. Round the answer to two decimal places where necessary. a 5 mm b c 10 mm 16 cm 15 cm 2 cm 5 cm d 108 cm e f 8m 4m 2m ENRICHMENT Complex composites 10 m 12 m 8m 8 cm 24 cm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 227 Progress quiz 1 Convert: a 6.4 m into mm U N SA C O M R PL R E EC PA T E G D ES 38pt 4A b 180 cm into m 1 d 2 m into cm 2 c 97000 cm into km 38pt 4A 2 Find the perimeter of these shapes. a b 1m 6 mm 1.2 m 14 cm 3 d 12 cm 25 cm 38pt 4A 16 cm c 3m If the perimeter of a triangle is 24 cm, find the value of x. x cm 6 cm 38pt 4B/E 4 Find the circumference and area of these circles, correct to two decimal places. a b 7 cm 38pt 4C 5 38pt 4C/D 6 Find the area of these shapes. a 7.2 cm 26 mm Convert these area measurements to the units shown in brackets. a 4.7 m2 (cm2 ) b 4100 mm2 (cm2 ) 2 c 5000 m (ha) d 0.008 km2 (m2 ) b 1.4 m c 50 cm m 6c 8 cm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem 15 m d e U N SA C O M R PL R E EC PA T E G D ES 228 3.2 cm 8m 6.4 m 26 m f 2m g 3m 5m 1.5 m 38pt 4F Ext 38pt 4F 7 Find the area and perimeter of these sectors. Round to two decimal places. a b c 6 cm 50° 4 cm 2.5 cm 60° 8 Find the area of these composite shapes. Correct to two decimal places. a b 6 cm Ext 10 6 cm cm 8 cm 8 cm 4G/H/I 38pt 9 Find the surface area and volume of these prisms. a b 6 cm 8 cm Ext 8 cm 38pt 4I 8 cm c 12 cm 10 For this cylinder, find: a the volume in m3 correct to four decimal places b the capacity in litres to one decimal place. 4 cm 5.32 cm 4 cm 7 cm 6.5 cm 1.4 m 50 cm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 229 4J Time U N SA C O M R PL R E EC PA T E G D ES Time in minutes and seconds is based on the number 60. Other units of time, including the day and year, are defined by the rate at which the Earth spins on its axis and the time that the Earth takes to orbit the Sun. The origin of the units seconds and minutes dates back to the ancient Babylonians, who used a base 60 number system. The 24-hour day dates back to the ancient Egyptians, who described the day as 12 hours of day and 12 hours of night. Today, we use a.m. (ante The Earth takes 1 year to orbit the Sun. meridiem, which is Latin for ‘before noon’) and p.m. (post meridiem, which is Latin for ‘after noon’) to represent the hours before and after noon (midday). During the rule of Julius Caesar, the ancient 1 Romans introduced the solar calendar, which recognised that the Earth takes about 365 days to orbit 4 the Sun. This gave rise to the leap year, which includes one extra day (in February) every 4 years. Let’s start: Knowledge of time Do you know the answers to these questions about time and the calendar? • When is the next leap year? • Why do we have a leap year? • Which months have 31 days? • Why are there different times in different countries or parts of a country? • What do (or ) and (or ) mean on time scales? The standard unit of time is the second (s). Units of time include: • 1 minute (min) = 60 seconds (s) × 24 × 60 × 60 • 1 hour (h) = 60 minutes (min) day hour minute second • 1 day = 24 hours (h) ÷ 24 ÷ 60 ÷ 60 • 1 week = 7 days • 1 year = 12 months Units of time smaller than a second. • millisecond = 0.001 second (1000 milliseconds = 1 second) • microsecond = 0.000 001 second (1 000 000 microseconds = 1 second) • nanosecond = 0.000 000 001 second (1 000 000 000 nanoseconds = 1 second) a.m. or p.m. is used to describe the 12 hours before and after noon (midday). 24-hour time shows the number of hours and minutes after midnight. • 0330 is 3:30 a.m. • 1530 is 3:30 p.m. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Key ideas 230 Chapter 4 Measurement and introduction to Pythagoras’ theorem Key ideas U N SA C O M R PL R E EC PA T E G D ES The Earth is divided into 24 time zones (one for each hour). • Twenty-four 15◦ lines of longitude divide the Earth into its time zones. Time zones also depend on a country’s borders and its proximity to other countries. (See map on pages 231–232 for details.) • Time is based on the time in a place called Greenwich, United Kingdom, and this is called Coordinated Universal Time (UTC) or Greenwich Mean Time (GMT). • Places east of Greenwich are ahead in time. • Places west of Greenwich are behind in time. Australia has three time zones: • Eastern Standard Time (EST), which is UTC plus 10 hours. • Central Standard Time (CST), which is UTC plus 9.5 hours. • Western Standard Time (WST), which is UTC plus 8 hours. Example 20 Converting units of time Convert these times to the units shown in brackets. a 3 days (minutes) b 30 months (years) SO L U T I O N EX P L A N A T I O N a 1 day = 24 hours 3 days = 3 × 24 h = 3 × 24 × 60 min 1 hour = 60 minutes = 4320 min b 30 months = 30 ÷ 12 years =2 There are 12 months in 1 year. 1 years 2 Example 21 Using 24-hour time Write these times using the system given in brackets. a 4:30 p.m. (24-hour time) b 1945 (a.m./p.m.) SO L U T I O N EX P L A N A T I O N a Since the time is p.m., add 12 hours to 0430 hours. 4:30 p.m. = 1200 + 0430 = 1630 hours b 1945 hours = 7:45 p.m. Since the time is after 1200 hours, subtract 12 hours. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 11 10 9 8 7 5 6 4 6 3 4 2 1 231 0 0 GREENLAND 1 U N SA C O M R PL R E EC PA T E G D ES 3 ALASKA 9 6 ICELAND NORWAY CANADA K UNITED KINGDOM IRELAND N GERM L 4 Q 3½ FRANCE 8 P UNITED STATES 5 6 7 R 1 IT PORTUGALSPAIN MOROCCO ALGERIA MEXICO CUBA MAURITANIA MALI o NIGE NIGERI VENEZUELA COLUMBIA 5 PERU 4 BRAZIL BOLIVIA World cities key J K L M N P Q R N Auckland Edinburgh Greenwich Johannesburg London New York Vancouver Washington DC 3 ARGENTINA CHILE Sun 1:00 2:00 3:00 4:00 5:00 6:00 7:00 8:00 9:00 10:00 11:00 Sun 12:00 11 10 9 8 7 6 5 4 3 2 1 0 F Western Standard Time Western Australia Adelaide Alice Springs Brisbane Cairns Canberra, ACT Darwin Hobart Melbourne Perth I Eastern Standard D Time Northern Territory B Queensland AUSTRALIA 9½ Australian cities key A B C D E F G H I Central Standard Time C South Australia New South Wales A Victoria H 20:00 8 21:00 9 Tasmania E G 22:00 10 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem 2 1 4 3 5 6 8 7 9 10 12 11 11 12 U N SA C O M R PL R E EC PA T E G D ES 232 3 11 SWEDEN FINLAND 12 9 3 10 7 RUSSIA 5 9 4 8 4 MANY POLAND UKRAINE KAZAKHSTAN 6 10 MONGOLIA ROMANIA ITALY TURKEY GREECE SYRIA LIBYA EGYPT SAUDI ARABIA ER A CHAD 1 CHINA 8 AFGHANISTAN 4½ 5 PAKISTAN IRAN 3½ IRAQ 9 NEPAL 5¾ BURMA 6½ THAILAND INDIA 5½ 4 JAPAN SUDAN 2 PHILIPPINES 5½ ETHIOPIA MALAYSIA SRI LANKA DEM. REP. OF THE CONGO INDONESIA TANZANIA ANGOLA ZAMBIA MADAGASCAR NAMIBIA AUSTRALIA 3 M SOUTH AFRICA 11½ 9½ ZIMBABWE J NEW ZEALAND 12¾ 5 13:00 14:00 15:00 16:00 17:00 18:00 19:00 20:00 21:00 22:00 23:00 1 2 3 4 5 6 7 8 9 10 11 Sun Sun Sun 24:00 20:00 1:00 12 12 11 NT QLD WA SA NSW VIC Daylight Saving No Daylight Saving ACT TAS Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 233 Example 22 Using time zones U N SA C O M R PL R E EC PA T E G D ES Coordinated Universal Time (UTC) and is based on the time in Greenwich, United Kingdom. Use the world time zone map (on pages 231–232) to answer the following. a When it is 2 p.m. UTC, find the time in these places. i France ii China iii Queensland b iv Alaska When it is 9:35 a.m. in New South Wales, Australia, find the time in these places. i Alice Springs ii Perth iii London iv central Greenland SO L U T I O N EX P L A N A T I O N a i Use the time zone map to see that France is to the east of Greenwich and is in a zone that is 1 hour ahead. 2 p.m. + 1 hour = 3 p.m. ii 2 p.m. + 8 hours = 10 p.m. From the time zone map, China is 8 hours ahead of Greenwich. iii 2 p.m. + 10 hours = 12 a.m. Queensland uses Eastern Standard Time, which is 10 hours ahead of Greenwich. iv 2 p.m. – 9 hours = 5 a.m. Alaska is to the west of Greenwich, in a time zone that is 9 hours behind. b i 1 9:35 a.m. – hour = 9:05 a.m. 2 Alice Springs uses Central Standard Time, 1 which is hour behind Eastern Standard Time. 2 ii 9:35 a.m. – 2 hours = 7:35 a.m. Perth uses Western Standard Time, which is 2 hours behind Eastern Standard Time. iii 9:35 a.m. – 10 hours = 11:35 p.m. (the day before) UTC (time in Greenwich, United Kingdom) is 10 hours behind EST. iv 9:35 a.m. – 13 hours = 8:35 p.m. (the day before) Central Greenland is 3 hours behind UTC in Greenwich, so is 13 hours behind EST. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 234 Chapter 4 Measurement and introduction to Pythagoras’ theorem 3 — From options A to F, match up the time units with the most appropriate description. a single heartbeat A 1 hour b 40 hours of work B 1 minute c duration of a university lecture C 1 day d bank term deposit D 1 week e 200 m run E 1 year f flight from Australia to the UK F 1 second U N SA C O M R PL R E EC PA T E G D ES 1 1–3 UNDERSTANDING Exercise 4J 2 Find the number of: a seconds in 2 minutes d minutes in 4 hours g weeks in 35 days b minutes in 180 seconds e hours in 3 days h days in 40 weeks c hours in 120 minutes f days in 48 hours 3 What is the time difference between these times? a 12 p.m. and 6:30 p.m. b 12 a.m. and 10:45 a.m. c 12 a.m. and 4:20 p.m. d 11 a.m. and 3:30 p.m. Example 20 4 Convert these times to the units shown in brackets. a 3 h (min) b 10.5 min (s) d 90 min (h) e 6 days (h) g 1 week (h) h 1 day (min) j 20 160 min (weeks) k 2 weeks (min) m 5000 milliseconds (s) o 7 000 000 000 nanoseconds (s) q 0.000 002 7 s (microseconds) 5 Write the time for these descriptions. a 4 hours after 2:30 p.m. Example 21 4–11(½) c f i l 4–11(½) 240 s (min) 72 h (days) 14 400 s (h) 24 h (s) FLUENCY 4–10(½) n 2 500 000 microseconds (s) p 0.4 s (milliseconds) r 0.000 000 003 s (nanoseconds) b 10 hours before 7 p.m. 1 c 3 hours before 10 p.m. 2 1 d 7 hours after 9 a.m. 2 1 e 6 hours after 11:15 a.m. 4 f 3 1 hours before 1:25 p.m. 4 6 Write these times using the system shown in brackets. a 1:30 p.m. (24-hour) b 8:15 p.m. (24-hour) d 11:59 p.m. (24-hour) e 0630 hours (a.m./p.m.) g 1429 hours (a.m./p.m.) h 1938 hours (a.m./p.m.) c 10:23 a.m. (24-hour) f 1300 hours (a.m./p.m.) i 2351 hours (a.m./p.m.) Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 7 Round these times to the nearest hour. a 1:32 p.m. b 5:28 a.m. c 1219 hours 235 4J U N SA C O M R PL R E EC PA T E G D ES 8 What is the time difference between these time periods? a 10:30 a.m. and 1.20 p.m. b 9:10 a.m. and 3:30 p.m. d 10:42 p.m. and 7:32 a.m. e 1451 and 2310 hours d 1749 hours FLUENCY Measurement and Geometry c 2:37 p.m. and 5:21 p.m. f 1940 and 0629 hours Example 22a 9 Use the time zone map on pages 231–232 to find the time in the following places, when it is 10 a.m. UTC. a Spain b Turkey c Tasmania d Darwin e Argentina f Peru g Alaska h Portugal Example 22b 10 Use the time zone map on pages 231–232 to find the time in these places, when it is 3:30 p.m. in Victoria. a United Kingdom b Libya c Sweden d Perth e Japan f central Greenland g Alice Springs h New Zealand 11 What is the time difference between these pairs of places? a United Kingdom and Kazakhstan b South Australia and New Zealand c Queensland and Egypt d Peru and Angola (in Africa) e Mexico and Germany 14–17 17–21 12 A scientist argues that dinosaurs died out 52 million years ago, whereas another says they died out 108 million years ago. What is the difference in their time estimates? PROBLEM-SOLVING 12–14 13 Three essays are marked by a teacher. The first takes 4 minutes and 32 seconds to mark, the second takes 7 minutes and 19 seconds, and the third takes 5 minutes and 37 seconds. What is the total time taken to complete marking the essays? Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem 4J 14 Adrian arrives at school at 8:09 a.m. and leaves at 3:37 p.m. How many hours and minutes is Adrian at school? U N SA C O M R PL R E EC PA T E G D ES 15 On a flight to Europe, Janelle spends 8 hours and 36 minutes on a flight from Melbourne to Kuala Lumpur, Malaysia, 2 hours and 20 minutes at the airport at Kuala Lumpur, and then 12 hours and 19 minutes on a flight to Geneva, Switzerland. What is Janelle’s total travel time? PROBLEM-SOLVING 236 16 A phone plan charges 11 cents per 30 seconds. The 11 cents are added to the bill at the beginning of every 30-second block of time. a What is the cost of a 70-second call? b What is the cost of a call that lasts 6 minutes and 20 seconds? 17 A doctor earns $180 000 working 40 weeks per year, 5 days per week, 10 hours per day. What does the doctor earn in each of these time periods? a per day b per hour c per minute d per second (in cents) 18 A 2 hour football match starts at 2:30 p.m. Eastern Standard Time (EST) in Newcastle, NSW. What time will it be in United Kingdom when the match finishes? 19 If the date is 29 March and it is 3 p.m. in Perth, what is the time and date in these places? a Italy b Alaska c Chile 20 Monty departs on a 20 hour flight from Brisbane to London, United Kingdom, at 5 p.m. on 20 April. Give the time and date of his arrival in London. 21 Elsa departs on an 11 hour flight from Johannesburg, South Africa, to Perth at 6:30 a.m. on 25 October. Give the time and date of her arrival in Perth. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 22–23 4J 23–25 U N SA C O M R PL R E EC PA T E G D ES 22 When there are 365 days in a year, how many weeks are there in a year? Round your answer to two decimal places. REASONING 22 237 23 a b c d To convert from hours to seconds, what single number do you multiply by? To convert from days to minutes, what single number do you multiply by? To convert from seconds to hours, what single number do you divide by? To convert from minutes to days, what single number do you divide by? 24 Assuming there are 365 days in a year and my birthday falls on a Wednesday this year, on what day will my birthday fall in 2 years’ time? 25 a Explain why you gain time when you travel from Australia to Europe. b Explain why you lose time when you travel from Germany to Australia. c Explain what happens to the date when you fly from Australia to Canada across the International Date Line. — — 26 26 Use the internet to investigate how daylight saving affects the time in some places. Write a brief report discussing the following points. a i Name the States in Australia that use daylight saving. ENRICHMENT Daylight saving ii Name five other countries that use daylight saving. b Describe how daylight saving works, why it is used and what changes have to be made to our clocks. c Describe how daylight saving in Australia affects the time difference between time zones. Use New South Wales and Greece as an example. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 238 Chapter 4 Measurement and introduction to Pythagoras’ theorem 4K Introduction to Pythagoras’ theorem EXTENDING U N SA C O M R PL R E EC PA T E G D ES Pythagoras was a philosopher in ancient Greece who lived in the 6th century . He studied astronomy, mathematics, music and religion, but is most well known for the famous Pythagoras’ theorem. Pythagoras was known to provide a proof for the theorem that bears his name, and methods to find Pythagorean triples, which are sets of three whole numbers that make up the sides of right-angled triangles. The ancient Babylonians, 1000 years before Pythagoras’ time, and the Egyptians also knew that there was a relationship between the sides of a right-angled triangle. Pythagoras, however, was able to clearly explain and prove the theorem using mathematical symbols. The ancient theorem is still one of the most commonly used theorems today. c2 Pythagoras’ theorem states that the square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides. An illustration of the theorem includes squares drawn on the sides of the right-angled triangle. The area of the larger square (c2 ) is equal to the sum of the two smaller squares (a2 + b2 ). c a b2 b a2 Let’s start: Discovering Pythagoras’ theorem Use a ruler to measure the sides of these right-angled triangles to the nearest mm. Then complete the table. b b c c b a c a a a • • • • b c a2 b2 c2 Can you see any relationship between the numbers in the columns for a2 and b2 and the number in the column for c2 ? Can you write down this relationship as an equation? Explain how you might use this relationship to calculate the value of c if it was unknown. Research how you can cut the two smaller squares (with areas a2 and b2 ) to fit the pieces into the larger square (with area c2 ). Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry Key ideas U N SA C O M R PL R E EC PA T E G D ES The hypotenuse • It is the longest side of a right-angled triangle. Hyp c a oten • It is opposite the right angle. use Pythagoras’ theorem • The square of the length of the hypotenuse is the sum of b the squares of the lengths of the other two shorter sides. • a2 + b2 = c2 or c2 = a2 + b2 A Pythagorean triple (or triad) is a set of three numbers which satisfy Pythagoras’ theorem. 239 Example 23 Checking Pythagorean triples Decide if the following are Pythagorean triples. a 6, 8, 10 SO L U T I O N a b 4, 5, 9 EX P L A N A T I O N a2 + b2 = 62 + 82 = 36 + 64 Let a = 6, b = 8 and c = 10 and check that a2 + b2 = c2 = 100 (= 102 ) ∴ 6, 8, 10 is a Pythagorean triple. b a2 + b2 = 42 + 52 a2 + b2 = 41 and 92 = 81 so = 16 + 25 = 41 ( ≠ 92 ) a2 + b2 ≠ c2 ∴ 4, 5, 9 is not a Pythagorean triple. Example 24 Deciding if a triangle has a right angle Decide if this triangle has a right angle. 4m 7m 9m SO L U T I O N EX P L A N A T I O N a2 + b2 = 42 + 72 Check to see if a2 + b2 = c2 . In this case a2 + b2 = 65 and c2 = 81 so the triangle is not right angled. = 16 + 49 = 65 ( ≠ 92 = 81) Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 240 Chapter 4 Measurement and introduction to Pythagoras’ theorem 1–4 1 Calculate these squares and sums of squares. a 32 b 52 2 2 e 2 +4 f 32 + 72 4 d 1.52 h 122 + 152 U N SA C O M R PL R E EC PA T E G D ES c 122 g 62 + 112 — UNDERSTANDING Exercise 4K 2 Decide if these equations are true or false. a 22 + 32 = 42 b 62 + 82 = 102 2 2 2 d 5 –3 =4 e 62 – 32 = 22 c 72 + 242 = 252 f 102 – 52 = 52 3 Write the missing words in this sentence. is the longest side of a right-angled . 4 Which letter represents the length of the hypotenuse in these triangles? a b c b x y a c s u w 5(½), 6, 7(½) Example 23 t 5(½), 6, 7(½) 5 Decide if the following are Pythagorean triples. a 3, 4, 6 b 4, 2, 5 d 9, 12, 15 e 5, 12, 13 g 9, 40, 41 h 10, 12, 20 5(½), 6, 7(½) FLUENCY The c 3, 4, 5 f 2, 5, 6 i 4, 9, 12 6 Complete this table and answer the questions. a 3 6 8 b 4 8 15 a2 c 5 10 17 b2 a2 + b2 c2 a Which two columns give equal results? b What would be the value of c2 if: i a2 = 4 and b2 = 9? ii a2 = 7 and b2 = 13? c What would be the value of a2 + b2 if: i c2 = 25? ii c2 = 110? 7 Check that a2 + b2 = c2 for all these right-angled triangles. a b 8 4 d 5 15 3 5 13 12 e 12 9 17 40 41 15 c f 6 9 6.5 2.5 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 8, 9 8 Write down an equation using the pronumerals given these diagrams. a b c x a b b 9, 10 4K U N SA C O M R PL R E EC PA T E G D ES x PROBLEM-SOLVING 8, 9 241 a d h d 9 A cable connects the top of a 30 m mast to a point on the ground. The cable is 40 m long and connects to a point 20 m from the base of the mast. a Using c = 40, decide if a2 + b2 = c2 . b Do you think the triangle formed by the mast and the cable is right angled? Give a reason. 10 (3, 4, 5) and (5, 12, 13) are Pythagorean triples since 32 + 42 = 52 and 52 + 122 = 132 . a Find 10 more Pythagorean triples using whole numbers less than 100. b Find the total number of Pythagorean triples with whole numbers of less than 100. Example 24 11, 12 11–13 11 If a2 + b2 = c2 , we know that the triangle must have a right angle. Which of these triangles must have a right angle? a b c 3 1 12 9 d 15 e 17 2 1 2 5 f 40 6 15 8 41 8 REASONING 11 9 5 12 If a2 + b2 = c2 is true, complete these statements. a c2 – b2 = b c2 – a2 = c c= Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 242 Chapter 4 Measurement and introduction to Pythagoras’ theorem 13 This triangle is isosceles. Write Pythagoras’ theorem using the given pronumerals. Simplify if possible. REASONING 4K U N SA C O M R PL R E EC PA T E G D ES c x — — 14 14 There are many ways to prove Pythagoras’ theorem, both algebraically and geometrically. a Here is an incomplete proof of the theorem that uses this illustrated geometric construction. Area of inside square = c2 1 Area of 4 outside triangles = 4 × × base × height 2 = Total area of outside square = ( + ENRICHMENT Pythagoras proof )2 = a2 + 2ab + b2 Area of inside square = Area (outside square) – Area of 4 triangles = – = c2 = b a Comparing results from the first and last steps gives b a c c c b c a b a b Use the internet to search for other proofs of Pythagoras’ theorem. See if you can explain and illustrate them. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 243 Measurement and Geometry 4L Using Pythagoras’ theorem EXTENDING From our understanding of equations, it may be possible to solve the equation to find an unknown. This is also the case for equations derived from Pythagoras’ theorem, where, if two of the side lengths of a right-angled triangle are known, then the third can be found. ? U N SA C O M R PL R E EC PA T E G D ES 3 4 So if c2 = 32 + 42 then c2 = 25 and c = 5. We also notice that if c2 = 25 then √ c = 25 = 5 (if c > 0). This application of Pythagoras’ theorem has wide range of applications wherever right-angled triangles can be drawn. √ sign may Note that a number using a not always result in a whole number. √ √ For example, 3 and 24 are not whole numbers and neither can be written as a fraction. These types of numbers are called surds and are a special group of numbers (irrational numbers) that are often approximated using rounded decimals. The exterior of the Australian centre for the Moving Image is created from many right-angled triangles. Let’s start: Correct layout Three students who are trying to find the value of c in this triangle using Pythagoras’ theorem write their solutions on a board. There are only very minor differences between each solution and the answer is written rounded to two decimal places. Which student has all the steps written correctly? Give reasons why the other two solutions are not laid out correctly. Student 1 Student 2 Student 3 c 2= a 2 + b 2 c 2 = a 2 +b 2 c=a2 +b2 =42 +92 =42 +92 =42 +92 = 97 √ = 97 = 97 √ ∴ c = 97 = 97 √ = 97 = 9.85 = 9.85 = 9.85 9 4 c Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 244 Chapter 4 Measurement and introduction to Pythagoras’ theorem √ Surds are numbers that have a sign when written in simplest form. • They are not a whole number and cannot be written as a fraction. • Written as a decimal, the decimal places would continue forever with no repeated pattern (just like the number pi). Surds are therefore classified as irrational numbers. √ √ √ √ 2, 5, 2 3 and 90 are all examples of surds. • Using Pythagoras’ theorem. √ c If c2 = a2 + b2 then c = a2 + b2 . a Note √ √ • a2 + b2 ≠ a + b, for example, 32 + 42 ≠ 3 + 4 b √ • If c2 = k then c = k if c ≥ 0. U N SA C O M R PL R E EC PA T E G D ES Key ideas Example 25 Finding the length of the hypotenuse Find the length of the hypotenuse for these right-angled triangles. Round the answer for part b to two decimal places. a b 9 c 6 c 8 7 SO L U T I O N a EX P L A N A T I O N c2 = a2 + b2 2 2 =6 +8 = 100 √ ∴ c = 100 Write the equation for Pythagoras’ theorem and substitute the values for the shorter sides. Find c by taking the square root. = 10 b c2 = a2 + b2 First calculate the value of 72 + 92 . = 72 + 92 = 130 √ ∴ c = 130 = 11.40 (to 2 d.p.) √ 130 is a surd, so round the answer as required. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 245 Example 26 Applying Pythagoras’ theorem A rectangular wall is to be strengthened by a diagonal brace. The wall is 6 m wide and 3 m high. Find the length of brace required correct to the nearest cm. ce Bra U N SA C O M R PL R E EC PA T E G D ES 3m 6m SO L U T I O N EX P L A N A T I O N c2 = a2 + b2 c = 32 + 62 = 45 √ ∴ c = 45 a=3 b=6 = 6.71 m or 671 cm (to nearest cm) Decide if these numbers written with a √ √ a 9 b 11 √ 3 sign simplify to a whole number. Answer Yes or No. √ √ c 20 d 121 2 Round these surds correct to two decimal places using a calculator. √ √ √ a 10 b 26 c 65 3 Copy and complete this working out. a b c2 = c2 = a2 + b2 c = = 92 + 122 = √ ∴c= = 4–5(½) Example 25a √ 230 d = 92 + 402 = √ ∴c= = = 52 + 122 = √ ∴c= = — 4–5(½) 4–5(½) 4 Find the length of the hypotenuse (c) of these right-angled triangles. a b c 24 c 9 3 7 c 4 d e c 12 16 c 40 60 f 36 c c FLUENCY 1 1–3 UNDERSTANDING Exercise 4L 11 27 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 246 4L 5 Find the length of the hypotenuse (c) of these right-angled triangles correct to two decimal places. a b c 7 5 c 4 U N SA C O M R PL R E EC PA T E G D ES Example 25b FLUENCY Chapter 4 Measurement and introduction to Pythagoras’ theorem 6 d 2 c c 1 3 e 4 19 f 2.5 c c c 32 3.5 Example 26 6 7 7–9 A rectangular board is to be cut along one of its diagonals. The board is 1 m wide and 3 m high. What will be the length of the cut, correct to the nearest cm? 7–9 1m 3m PROBLEM-SOLVING 6, 7 The size of a television screen is determined by its diagonal length. Find the size of a television screen that is 1.2 m wide and 70 cm high. Round the answer to the nearest cm. 8 Here is a diagram showing the path of a bushwalker from Camp 1 to Camp 2. Find the total distance calculated to one decimal place. 3 km Camp 1 2 km 1.5 km Camp 2 9 A 20 cm straw sits in a cylindrical glass as shown. What length of straw sticks above the top of the glass? Round the answer to two decimal places. 14 cm 4 cm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 10 10, 11 REASONING 10 10 Explain the error in each set of working. c2 = 22 + 32 b c2 = 32 + 42 c c2 = 22 + 52 ∴c=2+3 = 72 = 4 + 25 =5 = 49 = 29 √ = 29 4L U N SA C O M R PL R E EC PA T E G D ES a 247 ∴c=7 11 Prove that these are not right-angled triangles. a b 2 10 1 3 12 c 8 21 24 5 — — 12 ENRICHMENT Perimeter and Pythagoras 12 Find the perimeter of these shapes correct to two decimal places. a b 10 cm c 18 cm 2m 7 cm 3m d 4 cm 6 mm e 4 mm 8 mm f 5m 2m Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 248 Chapter 4 Measurement and introduction to Pythagoras’ theorem 4M Finding the length of a shorter side EXTENDING U N SA C O M R PL R E EC PA T E G D ES We know that if we are given the two shorter sides of a right-angled triangle we can use Pythagoras’ theorem to find the length of the hypotenuse. Generalising further, we can say that if given any two sides of a right-angled triangle we can use Pythagoras’ theorem to find the length of the third side. Let’s start: What’s the setting out? The triangle shown has a hypotenuse length of 15 and one of the shorter sides is of length 12. Here is the setting out to find the length of the unknown side a. Fill in the missing gaps and explain what is happening at each step. a a2 + b2 = c2 a2 + 2 a2 + 2 = 12 = a2 = ∴a= √ (Subtract from both sides) 15 (Hypotenuse) = Key ideas Pythagoras’ theorem can be used to find the length of the shorter sides of a right-angled triangle if the length of the hypotenuse and another side are known. Use subtraction to make the unknown the subject of the equation. For example: 24 a2 + b2 = c2 a2 + 242 = 252 a 25 a2 + 576 = 625 a2 = 49 (Subtract 576 from both sides.) √ ∴ a = 49 =7 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 249 Example 27 Finding the length of a shorter Find the value of a in this right-angled triangle. a U N SA C O M R PL R E EC PA T E G D ES 5 4 SO L U T I O N EX P L A N A T I O N a2 + b2 = c2 2 2 2 a +4 =5 Write the equation using Pythagoras’ theorem and substitute the known values. a2 + 16 = 25 a2 = 9 √ ∴a= 9 Subtract 16 from both sides. =3 Example 28 Applying Pythagoras to find a shorter side A 10 m steel brace holds up a concrete wall. The bottom of the brace is 5 m from the base of the wall. Find the height of the concrete wall correct to two decimal places. 10 m Wall 5m SO L U T I O N EX P L A N A T I O N Let a metres be the height of the wall. 2 2 a +b =c 2 a2 + 52 = 102 Choose a letter (pronumeral) for the unknown height. Substitute into Pythagoras’ theorem. 2 a + 25 = 100 a2 = 75 √ ∴ a = 75 = 8.66 (to 2 d.p.) The height of the wall is 8.66 metres. Subtract 25 from both sides. √ 75 is the exact answer. Round as required. Answer a worded problem using a full sentence. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 250 Chapter 4 Measurement and introduction to Pythagoras’ theorem 2 — Find the value of a in these equations. (Assume a is a positive number.) a a2 = 16 b a2 = 49 c a2 + 16 = 25 2 2 d a + 9 = 25 e a + 36 = 100 f a2 + 441 = 841 g 10 + a2 = 19 h 6 + a2 = 31 i 25 + a2 = 650 U N SA C O M R PL R E EC PA T E G D ES 1 1(½), 2 UNDERSTANDING Exercise 4M 2 Copy and complete the missing steps. a 15 9 b b 7 25 a a2 + b2 = c2 a2 + b2 = c2 a2 + 92 = 72 + b2 = + b2 = = 225 a2 = ∴a= b2 = 576 √ ∴b= √ = = 3–4(½) Example 27 3–4(½) 3–4(½) 3 Find the length of the unknown side in these right-angled triangles. a b c 12 5 3 9 a 15 a d e a b 30 34 17 FLUENCY a2 + b 41 f b 11 61 8 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 251 4M U N SA C O M R PL R E EC PA T E G D ES 4 Find the length of the unknown side in these right-angled triangles, giving the answer correct to two decimal places. a b c 2 14 8 2 5 FLUENCY Measurement and Geometry 3 d e 22 18 f 50 14 100 9 Example 28 5 A yacht’s mast is supported by a 12 m cable attached to its top. On the deck of the yacht, the cable is 8 m from the base of the mast. How tall is the mast? Round the answer to two decimal places. 5–7 6–8 PROBLEM-SOLVING 5, 6 12 m 8m Deck 6 A circle’s diameter AC is 15 cm and the chord AB is 9 cm. Angle ABC is 90◦ . Find the length of the chord BC. C 15 cm A 9 cm B 14 cm 7 A 14 cm drinking straw just fits into a can as shown. The diameter of the can is 7 cm. Find the height of the can correct to two decimal places. 7 cm 8 Find the length AB is this diagram. Round to two decimal places. 25 11 24 A B Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem 9 4M 9, 10 10, 11 9 Describe what is wrong with the second line of working in each step. a a2 + 10 = 24 b a2 = 25 c a2 + 25 = 36 U N SA C O M R PL R E EC PA T E G D ES a2 = 34 =5 a+5=6 √ 10 The number 11 is an example of a surd that is written as an exact value. Find the surd that describes the exact lengths of the unknown sides of these triangles. a b c 100 5 REASONING 252 2 7 120 1 11 Show how Pythagoras’ theorem can be used to find the unknown length in these isosceles triangles. Complete the solution for part a and then try the others. Round to two decimal places. a2 + b2 = c2 a x2 + x2 = 52 5 2x2 = 25 x2 = x ∴x= x b 10 Pythagorean families √ x c x d 34 x 61 — — 12 12 (3, 4, 5) is called a Pythagorean triple because the numbers 3, 4 and 5 satisfy Pythagoras’ theorem (32 + 42 = 52 ). a b c d e Explain why (6, 8, 10) is also a Pythagorean triple. Explain why (6, 8, 10) is considered to be in the same family as (3, 4, 5). List 3 other Pythagorean triples in the same family as (3, 4, 5) and (6, 8, 10). Find another triple not in the same family as (3, 4, 5), but has all 3 numbers less than 20. List 5 triples that are each the smallest triple of 5 different families. ENRICHMENT x Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 253 Investigation U N SA C O M R PL R E EC PA T E G D ES GMT and travel As discussed in Section 4J, the world is divided into 24 time zones, which are determined loosely by each 15◦ meridian of longitude. World time is based on the time at a place called Greenwich near London, United Kingdom. This time is called Coordinated Universal Time (UTC) or Greenwich Mean Time (GMT). Places east of Greenwich are ahead in time and places west of Greenwich are behind. In Australia, the Western Standard Time is 2 hours behind Eastern Standard Time and Central Standard 1 Time is hour behind Eastern Standard Time. Use the world time zone map on pages 232–233 to answer 2 these questions and to investigate how the time zones affect the time when we travel. East and west 1 Name five countries that are: a ahead of GMT b behind GMT Noon in Greenwich 2 When it is noon in Greenwich, what is the time in these places? a Sydney b Perth c Darwin d Washington, DC e Auckland h Japan f France g Johannesburg 2 p.m. EST 3 When it is 2 p.m. Eastern Standard Time (EST) on Wednesday, find the time and day in these places. a Perth b Adelaide c London d western Canada e China f United Kingdom g Alaska h South America Adjusting your watch 4 Do you adjust your watch forwards or backwards when you are travelling to these places? a India b New Zealand 5 In what direction should you adjust your watch if you are flying over the Pacific Ocean? Flight travel 6 You fly from Perth to Brisbane on a 4 hour flight that departed at noon. What is the time in Brisbane when you arrive? 7 You fly from Melbourne to Edinburgh on a 22 hour flight that departed at 6 a.m. What is the time in Edinburgh when you arrive? 8 You fly from Sydney to Los Angeles on a 13 hour flight that departed at 7:30 p.m. What is the time in Los Angeles when you arrive? Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Chapter 4 Measurement and introduction to Pythagoras’ theorem 9 Copy and complete the following table. Departing Brisbane Melbourne Hobart London New York Beijing Arriving Broome London Adelaide Tokyo Sydney Vancouver Departure time 7 a.m. 1 p.m. Flight time (hours) 3.5 23 1.5 12 15 Arrival time 4 p.m. 11 p.m. 3 a.m. 7:15 p.m. U N SA C O M R PL R E EC PA T E G D ES 254 3:45 p.m. 10 Investigate how daylight saving alters the time in some time zones and why. How does this affect flight travel? Give examples. Pythagorean triples and spreadsheets Pythagorean triples (or triads) can be grouped into families. The triad (3, 4, 5) is the base triad for the family of triads (3k, 4k, 5k). Here are some triads in this same family. k Triad 1 (3, 4, 5) (base triad) 2 (6, 8, 10) 3 (9, 12, 15) 1 Write down three more triads in the family (3k, 4k, 5k). 2 Write down three triads in the family (7k, 24k, 25k). 3 If (3k, 4k, 5k) and (7k, 24k, 25k) are two triad families, can you find three more families that have whole numbers less than 100. 4 Pythagoras discovered that if the smaller number in a base triad is a then the other two numbers in the triad are given by the rules: 1 2 (a + 1) 2 and 1 2 (a – 1) 2 Set up a spreadsheet to search for all the families of triads of whole numbers less than 200. Here is how a spreadsheet might be set up. Fill down far enough so that c is a maximum of 200. 5 List all the base triads using whole numbers less than 200. How many are there? Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 Measurement and Geometry 255 Up for a challenge? If you get stuck on a question, check out the 'Working with Unfamiliar Questions' poster at the end of the book to help you. U N SA C O M R PL R E EC PA T E G D ES Problems and challenges 1 A cube has capacity 1 L. What are its dimensions in cm correct to one decimal place? 2 A fish tank is 60 cm long, 30 cm wide, 40 cm high and contains 70 L of water. Rocks with a volume of 3000 cm3 are placed into the tank. Will the tank overflow? 3 What proportion (fraction or percentage) of the semicircle does the full circle occupy? 4 What is the distance AB in this cube? (Pythagoras’ theorem is required.) B 1m A 5 By what number do you multiply the radius of a circle to double its area? 6 Find the exact value (as a surd) of a in this diagram. (Pythagoras’ theorem is required.) a 1 1 1 1 7 1 1 1 A cube of side length 3 cm has its core removed in all directions as shown. Find its total surface area both inside and out. 3 cm 1 cm 1cm 3 cm 8 A square just fits inside a circle. What percentage of the circle is occupied by the square? Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 256 Perimeter Triangle 4 cm Quadrilaterals Square A = l 2 Rectangle A = lw Parallelogram A = bh 1 Rhombus A = 2 xy 1 Kite A = 2 xy 1 Trapezium A = 2 (a + b)h 3 cm 10 cm 6 cm P = 2 × 10 + 2 × 4 = 28 cm A = 12 bh U N SA C O M R PL R E EC PA T E G D ES Chapter summary Chapter 4 Measurement and introduction to Pythagoras’ theorem 1 × 2 = 6×3 = 9 cm2 Units 1 km = 1000 m 1 m = 100 cm 1 cm = 10 mm 2 ×100 2 ×10 2 cm 2 2 ÷10 ×1000 2 km m 2 ÷1000 ÷100 A = πr 2 = π × 72 = 49π cm2 2 mm2 Sectors (Ext) A= 3m 7 cm 2 1 ha = 10 000 m2 Circumference C = 2πr or πd =2×π ×3 = 6π m2 Circle Units = θ 360 280 360 × πr 2 × π × 22 = 9.77 m2 Area 2m 280° Surface area (Ext) Length Net Measurement 3m 1m 2m Volume Units km3, m3, cm3, SA = 2×(1×2)+2×(1×3)+2×(2×3) = 4 + 6 + 12 = 22 m2 Time mm3 1 min = 60 s 1 h = 60 min 0311 is 03:11 a.m. 2049 is 08:49 p.m. Pythagoras’ theorem (Ext) ×1000 ×1000 ×1000 ML kL L mL Theorem ÷1000 ÷1000 ÷1000 a 1 mL = 1 cm3 1 m3 = 1000 L b a2 + b2 = c2 Rectangular prism V = lwh = 10 × 20 × 30 = 6000 cm3 =6L c Prism and cylinders V = Ah = 12 × 3 × 1 × 2 30 cm = 3 m3 2m 10 cm 1m 3m 6 cm 7 + 72 = 74 — ∴ c = √ 74 5 c Finding a shorter side V = πr 2h = π × 22 × 6 = 75.40 cm3 2 cm 20 cm Finding c c 2 = 52 a 12 = 22 a2 + a2 + 1 = 4 a2 = 3 – a = √3 1 2 Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 257 Multiple-choice questions 1 The perimeter of this rectangle is 20 cm. The unknown value x is: A 4 B 16 C 5 D 10 E 6 4 cm U N SA C O M R PL R E EC PA T E G D ES 38pt 4A Chapter review Measurement and Geometry x cm 38pt 4B 2 A wheel has a diameter of 2 m. Its circumference and area (in that order) are given by: A p, p2 B 2p, p C 4p, 4p D 2, 1 E 4, 4 38pt 4C 3 The area of this triangle is: A 27.5 m2 B 55 m D 110 m E 16 m2 38pt 4E 38pt 4F Ext Ext 38pt 4D Ext 11 m 4 Using p = 3.14, the area of a circular oil slick with radius 100 m is: A 7850 m2 B 314 m2 C 31 400 m2 D 78.5 m2 E 628 m2 5 A sector of a circle is removed as shown. The fraction of the circle remaining is: 29 7 A 290 B C 36 36 D 38pt 4G 5m C 55 m2 7 180 E 70° 3 4 6 A cube has a surface area of 24 cm2 . The length of its sides is: A 12 cm B 8 cm D 4 cm E 2 cm C 6 cm x 7 The rule for the area of the trapezium shown is: 1 1 1 A xh B (x + y) C xy 2 2 2 1 D p xy2 E (x + y)h 2 h y 38pt 4H 8 The volume of a rectangular prism is 48 cm3 . If its width is 4 cm and height 3 cm, its length would be: A 3 cm B 4 cm C 2 cm D 12 cm E 96 cm 38pt 4I 9 A cylinder has radius 7 cm and height 10 cm. Using p = A 1540 cm2 38pt 4K B 440 cm3 C 440 L 22 , its volume would be: 7 D 1540 cm3 E 220 cm3 10 The rule for Pythagoras’ theorem for this triangle would be: A a2 – b2 = c2 B x2 + y2 = z2 C z2 + y2 = x2 √ D x2 + z2 = y2 E y = x2 – z2 x y z Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 258 Short-answer questions 38pt 4A 1 Convert these measurements to the units given in the brackets. a 2 m (mm) b 50 000 cm (km) c 3 cm2 (mm2 ) e 0.01 km2 (m2 ) f 350 mm2 (cm2 ) g 400 cm3 (L) d 4000 cm2 (m2 ) h 0.2 m3 (L) U N SA C O M R PL R E EC PA T E G D ES Chapter review Chapter 4 Measurement and introduction to Pythagoras’ theorem 38pt 4A/B 2 Find the perimeter/circumference of these shapes. Round the answer to two decimal places where necessary. a b c 5m 6 cm 8m 3m 8 cm d 12 m e f 20 mm 3 cm 8m 2 cm 4C/D/E 38pt 3 Find the area of these shapes. Round the answer to two decimal places where necessary. a b c 5 cm 18 m 2 cm 7m 6 cm 11 cm d e 20 km f 8 km 10 cm 16 m 4 cm 14 km 8m g h 110° 3 cm 38pt 4F i 4m 4 Find the area of these composite shapes. a b 10 cm c 5 cm 5 cm Ext 2 cm 4 cm 9 cm 8 cm 14 cm 6 cm Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 259 38pt 4G Ext 5 Find the surface area of each prism. a b 3m 8 mm 10 mm c 5 cm 7 cm U N SA C O M R PL R E EC PA T E G D ES 10 m Chapter review Measurement and Geometry 15 mm 6 cm 6 mm 38pt 4H/I 3 cm 6 Find the volume of each prism, giving your answer in litres. Remember 1 L = 1000 cm3 and 1 m3 = 1000 L. a b c 3 cm 8 cm 40 cm 12 cm 20 cm 1m 10 cm 38pt 4I 7 Find the volume of these cylinders, rounding the answer to two decimal places. a b c 3 mm 10 m 14 cm 8m 38pt 4J 7.5 mm 20 cm 8 An oven is heated from 23◦ C to 310◦ C in 18 minutes and 37 seconds. It then cools by 239◦ C in 1 hour, 20 minutes and 41 seconds. a Give the temperature: i increase ii decrease b What is the total time taken to heat and cool the oven? c How much longer does it take for the oven to cool down than heat up? 38pt 4J 9 a b c 38pt 4J 10 When it is 4:30 p.m. in Western Australia, state the time in each of these places. a New South Wales b Adelaide c United Kingdom d China e Finland f South Korea g Russia (eastern mainland) h New Zealand What is the time difference between 4:20 a.m. and 2:37 p.m.? Write 2145 hours in a.m./p.m. time. Write 11:31 p.m. in 24-hour time. Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 260 38pt 4L 11 Use Pythagoras’ theorem to find the length of the hypotenuse in these right-angled triangles. Round the answer to two decimal places in part c. a b c 8 7 Ext c 6 c U N SA C O M R PL R E EC PA T E G D ES Chapter review Chapter 4 Measurement and introduction to Pythagoras’ theorem c 24 3 38pt 4M 12 Use Pythagoras’ theorem to find the unknown length in these right-angled triangles. Round the answer to two decimal places in parts b and c. a b c 20 8 Ext 8 5 17 23 Extended-response questions 1 A company makes square nuts for bolts to use in building construction and steel structures. Each nut starts out as a solid steel square prism. A cylinder of diameter 2 cm is bored through its centre to make a hole. The nut and its top view are shown here. 2 cm 4 cm 2 cm 4 cm 4 cm 4 cm The company is interested in how much paint is required to paint the nuts. The inside surface of the hole is not to be painted. Round all answers to two decimal places where necessary. a b c Find the area of the top face of the nut. Find the total outside surface area of the nut. If the company makes 10 000 nuts, how many square metres of surface needs to be painted? The company is also interested in the volume of steel used to make the nuts. d Find the volume of steel removed from each nut to make the hole. e Find the volume of steel in each nut. f Assuming that the steel removed to make the hole can be melted and reused, how many nuts can be made from 1 L of steel? Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400 261 2 A simple triangular shelter has a base width of 2 m, a height of 2 m and a length of 3 m. a 2m 3m U N SA C O M R PL R E EC PA T E G D ES b Use Pythagoras’ theorem to find the hypotenuse length of one of the ends of the tent. Round the answer to one decimal place. All the faces of the shelter including the floor are covered with canvas material. What area of canvas is needed to make the shelter? Round the answer to the nearest whole number of square metres. Chapter review Measurement and Geometry c d 2m Every edge of the shelter is to be sealed with a special tape. What length of tape is required? Round to the nearest whole number of metres. The shelter tag says that is occupies 10 000 L of space. Show working to decide if this is true or false. What is the difference? Uncorrected 3rd sample pages • Cambridge University Press © Greenwood et al., 2015 • 978-1-107-56885-3 • Ph 03 8671 1400

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