Your SlideShare is downloading.
×

×

Saving this for later?
Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.

Text the download link to your phone

Standard text messaging rates apply

Like this document? Why not share!

- Coaches: How Do You Stand Out Among... by AutoPilot Your Bu... 408 views
- Exel Transportation Sales Or Agent ... by guest24a85ae 1091 views
- We Are Social's Guide to Social, Di... by We Are Social Sin... 15074 views
- Maths A - Chapter 5 by westy67968 2380 views
- Grant e du bois by Serie Científica 779 views
- Mining Wikipedia For Awesome Data by Neil Crosby 28253 views
- Reports 6i by Senthamarai B 559 views
- Year 12 Maths A Textbook - Chapter 6 by westy67968 3090 views
- 家暴、性侵兒少辨識、通報與輔導 by moralityfire 7123 views
- Lithium: The Framework for People W... by Nate Abele 33237 views
- Maths A - Chapter 4 by westy67968 2546 views
- The Human Interface by Christopher Fahey 49037 views

3,886

Published on

No Downloads

Total Views

3,886

On Slideshare

0

From Embeds

0

Number of Embeds

0

Shares

0

Downloads

40

Comments

0

Likes

1

No embeds

No notes for slide

- 1. Pythagoras’theorem 2 George is a builder and needs to be sure that the walls in his building are square with the ﬂoor. This means that the wall and ﬂoor meet at right angles. The walls are 240 centimetres high. The builder runs a string from the top of the wall to a point on the ﬂoor 1 metre from the foot of the wall. The string is measured as being 260 centimetres long. Are the walls square with the ﬂoor? This chapter looks at working with right-angled triangles and, in particular, using a relationship between the sides of a right-angled triangle.
- 2. 44 Maths Quest 9 for Victoria Right-angled triangles This chapter investigates one of the most important ideas in geometry related to right- angled triangles. Triangles can be classiﬁed according to the length of their sides. They can be classiﬁed as either equilateral, isosceles or scalene. Triangles can be classiﬁed by their angles as well. A triangle with one right angle is said to be a right-angled triangle. Types of triangles Isosceles Equilateral Scalene Right-angled In any triangle, the longest side is opposite the largest angle; so Hy in a right-angled triangle, the longest side is opposite the right po ten us angle. This side has a special name. It is called the hypotenuse. e The theorem, or rule, that you will learn more about was named after an ancient Greek mathematician, called Pythagoras (580–501 BC). It will enable you to solve all kinds of practical problems related to right-angled triangles. The following exercise will help you to understand what Pythagoras has been credited with discovering 2500 years ago. remember remember 1. The longest side of a right-angled triangle is called the hypotenuse. 2. The hypotenuse is always situated opposite the right angle. 2A Right-angled triangles 2.1 1 For each of the following triangles, carefully measure the length of each side, in milli- HEET metres, and record your results in the table which follows. Note that the hypotenuse isSkillS always marked as c. a a b c c a Pythagoras’ c b b theorem b c a
- 3. Chapter 2 Pythagoras’ theorem 45 d a e f b a b a c c c b g a h a b c b c a b c d e f g h a b c a2 b2 a 2 + b2 c22 What do you notice about the results in the table in question 1?3 On a sheet of paper, carefully draw 6 right-angled triangles of different sizes. You will need to use a protractor, set square or template to make sure the triangles are right- angled. Carefully measure the sides of each triangle, and complete a table like the one in question 1. What do you notice about these results?4 Now draw some triangles that are not right-angled, measure their sides and complete the same kind of table. Remember to label the longest side as c. What do you notice this time?5 For this activity you will need graph paper, coloured pencils, glue and scissors. a On a sheet of graph paper, draw a right-angled triangle with a base of 4 cm and a height of 3 cm. b Carefully draw a square on each of the three sides of the triangle and mark a grid on each so that the square on the base is divided into 16 small squares, while the square on the height is divided into 9 small squares.
- 4. 46 Maths Quest 9 for Victoria c Colour the square on the base, and the square on the height in different colours as shown at right so that you can still see the grid lines. d Carefully cut out the two coloured squares from the triangle. e Now stick the larger of the coloured squares on the uncoloured square of the triangle (the 3 cm square on the hypotenuse). 4 cm f Using the grid lines as a guide, see if you can cut the smaller square up and ﬁt it on the remaining space. The two coloured squares should have exactly covered the third square. g What do you notice about the hypotenuse and the other two sides of a right-angled triangle?History of mathematics P Y T H AG O R A S ( c . 5 8 2 – 5 0 0 BC) During his life . . . about numbers, but it is impossible to sort out what Confucius is born. he discovered himself and what was discovered by The Persians invade others who carried on his work. His ideas inﬂuenced Egypt. the thinking of the Greek philosopher, Plato, who Greek city-states lived from about 430–350 BC and the followers of develop and their neoplatonism in the third century AD. citizens have a Pythagoras is credited with the discovery of what say in is now known as Pythagoras’ theorem, which states government. that in a right-angled triangle, the square on the hypotenuse (long side) is equal to the sum of the Pythagoras was a squares on the other two sides. This can be Greek expressed as c2 = a2 + b2. philosopher and Pythagorean triads (or mathematician. Little triples) are sets of is known about him numbers that obey c2 because he lived so long Pythagoras’ theorem.ago and he left no written work. We do know that Other people knew of a2 c ahe was born on the island of Samos and that as a this idea long before he byoung man he travelled to Egypt and Babylonia announced it and there is a(Mesopotamia) where he learned much of his Babylonian tablet known as ‘Plimpton 322’ (believed to b2mathematics and developed an interest ininvestigating it further. When he was 50 years of have been made about 1500age, he settled in Crotona, a Greek colony in years before Pythagoras wassouthern Italy, where he established a religious born) which has a set of the values we now callcommunity. The members believed in the doctrine Pythagorean triads.of the transmigration of souls and thought it was It is also thought that Pythagoras discovered thatimportant to lead a pure and moral life to raise the musical notes have a mathematical pattern.soul to a higher level for the next life. Questions Pythagoras and his followers believed that the way 1. Where was Pythagoras born?to understand the world was through numbers. The 2. What is the formula for his famous theorem?equilateral triangle seems to have been particularly 3. What is a Pythagorean triad?signiﬁcant and was treated as an object of religious 4. Which famous Greek philosopher wasveneration. He probably made several discoveries inﬂuenced by Pythagoras?
- 5. Chapter 2 Pythagoras’ theorem 47Using Pythagoras’ theorem Pythagoras was perhaps the ﬁrst mathematician to recognise and investigate a very important property of right-angled triangles. The theorem, or rule, named after him states that: In any right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. The rule is written as c2 = a2 + b2 where a and b are the two shorter sides and c is the hypotenuse. The hypotenuse is the longest side of a right-angled triangle c a and is always the side that is opposite the right angle. b Pythagoras’ theorem gives us a way of ﬁnding the length of the third side in a triangle, if we know the lengths of the two x other sides. 4 Finding the hypotenuse 7 We are able to ﬁnd the length of the hypotenuse when we are given the length of the two shorter sides by substituting into the formula c2 = a2 + b2. WORKED Example 1 For the triangle at right, ﬁnd the length of the hypotenuse, x, correct to 1 decimal place. x 4 THINK WRITE 7 1 Copy the diagram and label the sides a, b and c. Remember to label the c=x hypotenuse as c. a=4 b=7 2 Write Pythagoras’ theorem. c2 = a 2 + b 2 3 Substitute the values of a, b and c into x2 = 4 2 + 7 2 this rule and simplify. = 16 + 49 = 65 4 Calculate x by taking the square root x = 65 of 65. Round your answer correct to x = 8.1 1 decimal place. In many cases we are able to use Pythagoras’ theorem to solve practical problems. We can model the problem by drawing a diagram, and use Pythagoras’ theorem to solve the right-angled triangle. We then use the result to give a worded answer.
- 6. 48 Maths Quest 9 for Victoria WORKED Example 2A ﬁre is on the twelfth ﬂoor of a building. A child needs to be rescued from a window thatis 20 metres above ground level. If the rescue ladder can be placed no closer than 6 m fromthe foot of the building, what is the minimum length ladder needed to make the rescue?Give your answer correct to the nearest metre.THINK WRITE 1 Draw a diagram and label the sides a, b and c. a c Remember to 20 m x label the hypotenuse as c. 6m b 2 Write Pythagoras’ c2 = a2 + b2 theorem. 3 Substitute the x2 = 202 + 62 values of a, b and c = 400 + 36 into this rule and = 436 simplify. 4 Calculate x by taking x = 436 the square root of 436. = 20.8806 Round your answer ≈ 21 correct to the nearest metre. 5 Give a worded The ladder answer. needs to be 21 metres long. remember remember 1. The hypotenuse is the longest side of the triangle and is opposite the right angle. 2. The length of the hypotenuse can be found if we are given the length of the two shorter sides by using the formula: c2 = a2 + b2. 3. Worded problems can be solved by drawing a diagram and using Pythagoras’ theorem to solve the problem. 4. Worded problems should be given a worded answer.
- 7. Chapter 2 Pythagoras’ theorem 49 2B Using Pythagoras’ theoremWORKED 1 For the following triangles, ﬁnd the length of the hypotenuse, x, correct to 1 decimal 2.2Example place. HEET 1 SkillS a b c x x 7 x 3 5 L Spread XCE sheet E 24 4 12 Finding the d e 896 f 1.3 length of the x hypotenuse 6.2 Math 17.5 cad 742 x x 12.2 Pythagoras’ theorem 2 For each of the following triangles, ﬁnd the length of the hypotenuse, giving answers GC pro correct to 2 decimal places. gram a 4.7 b 19.3 c Pythagoras’ theorem 804 6.3 27.1 562 d e 0.9 f 152 7.4 87 10.3 2.7 3 A right-angled triangle has a base of 4 cm and a height of 12 cm. Find the length of the hypotenuse to 2 decimal places. 4 Find the lengths of the diagonals of squares that have side lengths of: a 10 cm b 17 cm c 3.2 cm 5 What is the length of the diagonal of a rectangle whose sides are: a 10 cm and 8 cm? b 620 cm and 400 cm? c 17 cm and 3 cm? 6 An isosceles triangle has a base of 30 cm and a height of 10 cm. Find the length of the two equal sides. 7 A right-angled triangle has a height of 17.2 cm, and a base that is half the height. FindWORKED the length of the hypotenuse, correct to 2 decimal places.Example 8 A ladder leans against a vertical wall. The foot of the ladder is 1.2 m from the wall, 2 and the top of the ladder reaches 4.5 m up the wall. How long is the ladder?
- 8. 50 Maths Quest 9 for Victoria 9 A ﬂagpole, 12 m high, is supported by three wires, attached from the top of the pole to the ground. Each wire is pegged into the ground 5 m from the pole. How much wire is needed to support the pole? 3.8 km me 10 Sarah goes canoeing in a large lake. She paddles 2.1 km to E ti the north, then 3.8 km to the west. Use the triangle at rightGAM 2.1 km Pythagoras’ to ﬁnd out how far she must then paddle to get back to her theorem — starting point in the shortest possible way. 001 Starting point 11 A baseball diamond is a square of side length 27 m. When a runner on ﬁrst base tries to steal second base, the catcher has to throw the ball from home base to second base. How far is that throw? Second base 27 m Catcher Finding a shorter side Sometimes a question will give you the length of the hypotenuse and ask you to ﬁnd one of the shorter sides. In such examples, we need to rearrange Pythagoras’ formula. Given that c2 = a2 + b2, we can rewrite this as: a2 = c2 − b2 or b2 = c2 − a2. It is easier to know what to do in this case if you remember that: ﬁnding a short side means subtract.
- 9. Chapter 2 Pythagoras’ theorem 51 WORKED Example 3Find the length, correct to 1 decimal place, of the unmarked side ofthe triangle at right. 14 cmTHINK WRITE 8 cm1 Copy the diagram and label the sides a, b and c. Remember to label the hypotenuse a as c. c = 14 b=82 Write Pythagoras’ theorem for a shorter side. a2 = c2 − b23 Substitute the values of a, b and c into this a2 = 142 − 82 rule and simplify. = 196 − 64 = 1324 Find a by taking the square root of 132. a = 132 Round to 1 decimal place. = 11.5 cm Practical problems may also involve you being required to ﬁnd the shorter side of a right-angled triangle. WORKED Example 4A ladder that is 4.5 m long leans up against a vertical wall. The foot of the ladder is 1.2 mfrom the wall. How far up the wall does the ladder reach? Give your answer correct to1 decimal place.THINK WRITE1 Draw a diagram and label the sides a, b and c. Remember to label the hypotenuse as c. c = 4.5 m a b = 1.2 m2 Write Pythagoras’ theorem for a shorter side. a2 = c2 − b23 Substitute the values of a, b and c into this a2 = 4.52 − 1.22 rule and simplify. = 20.25 − 1.44 = 18.814 Find a by taking the square root of 18.81. a = 18.81 Round to 1 decimal place. = 4.3 m5 Give a written answer. The ladder will reach a height of 4.3 m up the wall. Some questions will require you to decide which method is needed to solve the problem. A diagram will help you decide whether you are ﬁnding the hypotenuse or one of the shorter sides.
- 10. 52 Maths Quest 9 for Victoria remember remember 1. A shorter side of the triangle can be found if we are given the length of the hypotenuse and the other shorter side. 2. When ﬁnding a shorter side, the formula used becomes a2 = c2 − b2 or b2 = c2 − a2. 3. On your diagram check whether you are ﬁnding the length of the hypotenuse or one of the shorter sides. 2C Finding a shorter side reads L Sp he WORKED 1 Find the length, correct to 1 decimal place, of the unmarked side in each of the Example etEXCE 3 following triangles. Finding a b c the length of 14 3.2 the shorter side 10 d hca 17 8.4Mat 8 Pythagoras’ theorem d e 1.2 f 51.8 ogram 382GC pr Pythagoras’ 457 theorem 75.2 3.8 2 Find the value of the pronumeral, correct to 2 decimal places. a a b c c 1.98 8.4 30.1 47.2 2.56 17.52 b d 0.28 e f 2870 d 468 1920 f e 0.67 114 For questions 3 to 13, give your answers correct to 2 decimal places.
- 11. Chapter 2 Pythagoras’ theorem 53 3 The diagonal of the rectangular sign at right is 34 cm. If the height of this sign is 25 cm, ﬁnd the width. 4 The diagonal of a rectangle is 120 cm. One side has a length of 70 cm. Find: a the length of the other side b the perimeter of the rectangle c the area of the rectangle. 5 An equilateral triangle has sides of length 20 cm. Find the height of the triangle. 6 The road sign shown at left is in the form of an equilateral triangle. Find the height of the sign and, hence, ﬁnd its area. 76 cmWORKED 7 A ladder that is 7 metres long leans up against a verticalExample 4 wall. The top of the ladder reaches 6.5 m up the wall. How far from the wall is the foot of the ladder? 8 A tent pole that is 1.5 m high is to be supported by ropes attached to the top. Each rope is 2 m long. How far from the base of the pole can each rope be pegged? 9 A kite is attached to a string 150 m long. Sam holds the 150 m end of the string 1 m above the ground, and the horizontal distance of the kite from Sam is 80 m as 80 m shown at left. How far above the ground is the kite? 1m 10 Ben’s dog ‘Macca’ has wandered onto a frozen pond, and is too frightened to walk back. Ben estimates that the dog is 3.5 m from the edge of the pond. He ﬁnds a plank, 4 m long, and thinks he can use it to rescue Macca. The pond is surrounded by a bank that is 1 m high. Ben uses the plank to make a ramp for Macca to walk up. Will he be able to rescue his dog? 11 Penny, the carpenter, is building a roof for a new house. The roof has a gable end in the form of an isosceles triangle, with a base 7.5 m 7.5 m of 6 m and sloping sides of 7.5 m. She decides to put 5 evenly spaced vertical strips of wood as decoration on the gable as shown at right. How many metres of this decorative wood does she need? 6m 12 Wally is installing a watering system in his garden. The pipe is to go all around the edge of the rectangular 56 cm garden, and have a branch diagonally across the garden. The garden measures 5 m by 7.2 m. If the pipe costs 40 cm $2.40 per metre (or part thereof), what will be the total cost of the pipe? 13 The size of a rectangular television screen is given by the length of its diagonal. What is the size of the screen at left?
- 12. 54 Maths Quest 9 for Victoria Shortest path A surf lifesaving contest involves swimming A from marker A to the shoreline BC, then on to D 100 m 80 m marker D as shown in the diagram at right. Suppose a lifesaver swims the course A–E–D B E C where E is 60 m from B. 60 m 150 m 1 What total distance will the lifesaver swim? 2 Suppose point E was 100 m from B. Determine how far the lifesaver would swim in total. 3 Suppose point E is halfway between B and C. How far would the lifesaver swim in total? 4 Complete the table below. Length BE Length AE Length ED Total length (m) (m) (m) (AE + ED) (m) 50 60 70 80 90 100 110 5 Use the data in the table to plot a graph of (AE + ED) (m) length BE (horizontal axis) versus total Total length 240 length (AE + ED) (vertical axis). 238 6 What is the shortest distance the lifesaver 236 could swim from marker A to the shoreline 234 then to marker D? 7 Draw a scale diagram showing the path taken 50 70 90 110 Length BE (m) by the lifesaver to swim this shortest distance. Measure the angles AEB and DEC. What do you notice?
- 13. Chapter 2 Pythagoras’ theorem 55 The shortest distance covered by the lifesaver obeys the Law of Reﬂection. The angle of A D incidence (AEB) equals the angle of reﬂection 100 m 80 m (DEC). We can use this law and a scale diagram to ﬁnd the point, E, for situations B E C where the markers are at different distances D from the shoreline. Reﬂect the line segment, DC, about the line, BC (shown dotted), then connect A to D′. The intersection point with the shoreline gives the position of E. 8 Use the Law of Reﬂection to solve A the following problem. Remember G to use a scale diagram. D This time the race involves 100 m 80 m touching the shoreline in two places E and F. What is the shortest 60 m distance the lifesaver should swim? B E F C 100 m 200 m 1 1 True or false? a The hypotenuse is the longest side of a right-angled triangle. x b The hypotenuse forms part of the right angle. 2 cm c Pythagoras’ theorem involves the formula a2 − b2 = c2. 2 Find the length of the hypotenuse of the triangle above right, 5 cm correct to 2 decimal places. 11 m 3 Find the value of the pronumeral in the triangle at right, correct to 1 decimal place. x 16 m 4 multiple choice The length of a diagonal of a square which has sides of 8 mm is: A 10.5 mm B 11.31 mm C 17.3 mm D 19.6 mm E 22.23 mm 5 A roof is 3 m above the ground. A ladder is placed 1 m from the foot of the wall. How long does the ladder need to be in order to reach the roof? (Give your answer correct to the nearest cm.) 6 A ﬂagpole is supported by a wire that is 5 m long. The wire is pegged 2.5 m from the foot of the pole. Find the height of the pole, correct to 2 decimal places. 7 The diagonal of a rectangle is 9.2 cm. The length is 6.3 cm. Find the width, correct to 1 decimal place. 8 Find the area of the rectangle in question 7. 9 Keith drives 5.3 km east, then 9.2 km north. How far in, metres, is he from his starting point? (Give your answer correct to 1 decimal place.)10 An isosceles triangle has sloping edges equal to 10 cm and a base equal to 5 cm. Calculate the height of the triangle, correct to 1 decimal place.
- 14. 56 Maths Quest 9 for VictoriaWorking with different units When we use Pythagoras’ theorem, we are usually working with a practical situation where measurements have been given. In any calculation, it is essential that all of the measurements are in the same units (for example, cm). Do you remember the relationship between the units of length? 10 mm = 1 cm 100 cm = 1 m To change to a larger unit, divide by the conversion factor. To change to a smaller unit, multiply by the conversion factor. 1000 m = 1 km For example, to change kilometres to metres, multiply by 1000. To change centimetres to metres, divide by 100. The following chart shows all the conversions of length that you are likely to need. ÷ 10 ÷ 100 ÷ 1000 millimetres centimetres metres kilometres (mm) (cm) (m) (km) × 10 × 100 × 1000 When using Pythagoras’ theorem, always check the units given for each measurement. If necessary, convert all measurements to the same units before using the rule. WORKED Example 5 Find the length, in mm, of the hypotenuse of a right-angled triangle if the 2 shorter sides are 7 cm and 12 cm. THINK WRITE 1 Draw a diagram and label the sides a, b and c. Remember to label the c hypotenuse as c. a = 12 cm b = 7 cm 2 Check that all measurements are in the same units. They are. 3 Write Pythagoras’ theorem for the c2 = a2 + b2 hypotenuse. 4 Substitute the values of a and b into this c2 = 122 + 72 rule and simplify. = 144 + 49 = 193 5 Find c by taking the square root. Give c = 193 the units in the answer. = 13.89 cm 6 Check the units required in the answer = 13.89 × 10 mm and convert if necessary. = 138.9 mm
- 15. Chapter 2 Pythagoras’ theorem 57 WORKED Example 6 The hypotenuse and one other side of a right-angled triangle are 450 cm and 3.4 m respectively. Find the length of the third side, in cm, correct to the nearest whole number. THINK WRITE 1 Draw a diagram and label the sides a, b and c. Remember to label the c = 450 cm hypotenuse as c. a b = 3.4 m 2 Check that all measurements are in 3.4 m = 3.4 × 100 cm the same units. They are different, = 340 cm so convert 3.4 m into cm. 3 Write Pythagoras’ theorem for a a2 = c2 − b2 shorter side. 4 Substitute the values of b and c a2 = 4502 − 3402 into this rule and simplify. = 202 500 − 115 600 = 86 900 5 Find a by taking the square root of a = 86 900 86 900. Round to the nearest whole = 295 number. 6 Give a worded answer. The third side will be approximately 295 cm long. remember remember 1. When using Pythagoras’ theorem, always check the units given for each measurement. 2. If necessary, convert all measurements to the same units before using the rule. Working with different 2D units Where appropriate, give answers correct to 2 decimal places.WORKED 1 Find the length, in mm, of the hypotenuse of a right-angled triangle, if the two shorter 2.3Example HEET sides are 5 cm and 12 cm. SkillS 5 2 Find the length of the hypotenuse of the following right-angled triangles, giving the answer in the units speciﬁed. a Sides 456 mm and 320 mm, hypotenuse in cm. XCE L Spread b Sides 12.4 mm and 2.7 cm, hypotenuse in mm. sheet E c Sides 32 m and 4750 cm, hypotenuse in m. Pythagoras’ d Sides 2590 mm and 1.7 m, hypotenuse in mm. theorem e Sides 604 cm and 249 cm, hypotenuse in m. f Sides 4.06 km and 4060 m, hypotenuse in km. g Sides 8364 mm and 577 cm, hypotenuse in m. h Sides 1.5 km and 2780 m, hypotenuse in km.
- 16. 58 Maths Quest 9 for VictoriaWORKED 3 The hypotenuse and one other side of a right-angled triangle are given for each caseExample below. Find the length of the third side in the units speciﬁed. 6 a Sides 46 cm and 25 cm, third side in mm. b Sides 843 mm and 1047 mm, third side in cm. c Sides 4500 m and 3850 m, third side in km. d Sides 20.3 cm and 123 mm, third side in cm. e Sides 6420 mm and 8.4 m, third side in cm. f Sides 0.358 km and 2640 m, third side in m. g Sides 491 mm and 10.8 cm, third side in mm. h Sides 379 000 m and 82 700 m, third side in km. 4 Two sides of a right-angled triangle are given. Find the third side in the units speciﬁed. The diagram shows how each triangle c is to be labelled. Remember: c is always the hypotenuse. a a a = 37 cm, c = 180 cm, ﬁnd b in cm. b a = 856 mm, b = 1200 mm, ﬁnd c in cm. b c b = 4950 m, c = 5.6 km, ﬁnd a in km. d a = 125 600 mm, c = 450 m, ﬁnd b in m. e a = 0.0641 km, b = 0.153 km, ﬁnd c in m. f a = 639 700 cm, b = 2.34 km, ﬁnd c in m. 4 cm 5 multiple choice a What is the length of the hypotenuse in this triangle? 3 cm A 25 cm B 50 cm C 50 mm D 500 mm E 2500 mm 82 cm b What is the length of the third side in this triangle? A 48.75 cm B 0.698 m C 0.926 m 43 cm D 92.6 cm E 69.8 mm c The most accurate measure for the length of the third side in the triangle at right is: A 4.83 m B 23.3 cm C 3.94 m 5.6 m D 2330 mm E 4826 mm 2840 mm d What is the length of the third side in this triangle? A 34.71 m B 2.97 m C 5.89 m D 1722 cm E 4.4 m 394 cm 4380 mm 6 A rectangle measures 35 mm by 4.2 cm. Find the length of the diagonal in mm. 7 A sheet of A4 paper measures 210 mm by 297 mm. Find the length of the diagonal in centimetres. 8 A rectangular envelope has a length of 21 cm and a diagonal measuring 35 cm. Find: a the width of the envelope b the area of the envelope. 9 A right-angled triangle has a hypotenuse of 47.3 cm and one other side of 30.8 cm. Find the area of the triangle.
- 17. Chapter 2 Pythagoras’ theorem 5910 A horse race is 1200 m. The track is straight, and 35 m wide. How much further than 1200 m will a horse run if it starts on the outside and ﬁnishes on the inside as shown? Finishing Starting gate post 1200 m 35 m11 A ramp is 9 metres long, and rises to a height of 250 cm. What is the horizontal distance, in metres, between the bottom and the top of the ramp?12 Sarah is making a gate, which has to be 1200 mm wide. It must be braced with a diagonal strut made of a different type of timber. She has only 2 m of this kind of timber available. What is the maximum height of the gate that she can make?13 A rectangular park is 260 m by 480 m. Danny usually trains by running around the edge of the park. After heavy rain, two adjacent sides are too muddy to run along, so he runs a triangular path along the other two sides and the diagonal. Danny does 5 circuits of this path for training. How far does he run? Give your answer in km.14 A swimming pool is 50 m by 25 m. Peter is bored by his usual training routine, and decides to swim the diagonal of the pool. How many diagonals must he swim to com- plete his normal distance of 1200 m?15 A hiker walks 4.5 km west, then 3.8 km south. How far in metres is she from her SHE ET 2.1 Work starting point?16 A square has a diagonal of 10 cm. What is the length of each side?
- 18. 60 Maths Quest 9 for VictoriaWhat is Bagheera in ‘The Jungle Book’? Bagheera Jungle Book’ Calculate the lengths of the lettered sides in the triangles. Place the lengths with their letters above them, in ascending order in the boxes below. The letters will spell out the puzzle answer. Give lengths accurate to 2 decimal places. 130 cm 6m 1.2 m R 1m B 9m L 1.5 m 6m 7m A T 6.31 m E 520 cm 155 cm 180 cm 7.2 m N 170 cm A 2m 6.2 m 4m 13 cm C 12.98 cm 5.8 m A 4m K 3m H 3m 578 cm P 621 cm 567 cm
- 19. Chapter 2 Pythagoras’ theorem 61Composite shapes In all of the exercises so far, we have been working with only one right-angled triangle. Many situations involve more complex diagrams, where the right-angled triangle is not as obvious. A neat diagram is essential for these questions. WORKED Example 7 8 cmFind the length of the side, x. Give your x 4 cmanswer correct to 2 decimal places.THINK 10 cm WRITE 1 Copy the diagram. On the diagram, create a right-angled triangle 8 and use the given measurements to work out the lengths of 2 sides. 2 Label the sides of your right-angled triangle as a, b and c. c a 4 Remember to label the hypotenuse as c. b 2 8 3 Check that all measurements are in the same units. They are the same. 4 Write Pythagoras’ theorem for the hypotenuse. c 2 = a2 + b2 5 Substitute the values of a, b and c into this rule and simplify. x2 = 4 2 + 2 2 = 16 + 4 = 20 6 Find x by taking the square root of 20. Round your answer correct x = 20 to 2 decimal places. = 4.47 cm Some situations will involve shapes that contain more than one triangle. In this case it is a good idea to split the diagram into separate right-angled triangles ﬁrst. WORKED Example 8For the diagram at right, ﬁnd 15the length of the sides marked 6 xx and y to 2 decimal places.THINK 2 WRITE y 1 Copy the diagram. 15 a 2 Find and draw any right-angled triangles c a x 6 x contained in the diagram. Label their sides. c 3 To ﬁnd an unknown side in a right-angled b b y 2 triangle, we need to know 2 sides, so ﬁnd x ﬁrst. 4 For the triangle containing x, write down a2 = c2 − b2 Pythagoras’ theorem for a shorter side. x2 = 62 − 22 Use it to ﬁnd x. = 36 − 4 = 32 x = 32 = 5.66 5 We now know 2 sides for the other triangle because we can substitute x = 5.66. 6 For the triangle containing y, write down b2 = c2 − a2 Pythagoras’ theorem for a shorter side and y2 = 152 − 5.662 use it to ﬁnd y. = 225 − 32 = 193 y = 193 = 13.89
- 20. 62 Maths Quest 9 for Victoria remember remember 1. To examine more complex situations involving Pythagoras’ theorem, a diagram is essential. 2. When a diagram is given, try to create or separate any right-angled triangles using further diagrams. 2E Composite shapes 8 cm Where appropriate, give answers correct to 2 decimal places. WORKED Example 1 Find the length of the side x in the ﬁgure at right. x reads 7 cm L Sp he 7 etEXCE 2 For the following diagrams, ﬁnd the length of the sides Pythagoras’ 12 cm theorem marked x. a 12 b x c x d x hca 7 37 20 120Mat 200 Pythagoras’ 15 52 theorem 100 WORKED 3 For each of the following diagrams, ﬁnd the length of the sides marked x and y. d Example hca a b 8Mat Pythagoras’ 10 x 5 theorem DIY 12 8 x 3 y y 4 c 5 d y 18 x x 5 12 20 10 2 7 4 multiple choice a The length of the diagonal of the rectangle at right is: 12 A 9.7 B 19 C 13.9 D 12.2 E 5 b The area of the rectangle at right is: A 12.1 B 84.9 C 15.7 D 109.6 E 38 14 7
- 21. Chapter 2 Pythagoras’ theorem 63 20 c The value of x in this shape is: x A 24 B 24.7 C 26 D 38.4 E 10 24 30 d What is the value of x in this ﬁgure? A 5.4 B 7.5 C 10.1 D 10.3 E 4 x 5 2 75 A hobby knife has a blade in the shape of a right-angled trapezium with the sloping edge 20 mm, and parallel sides of 32 mm and 48 mm. Find the width of the blade and, hence, the area.6 Two buildings, 10 and 18 m high, are directly opposite each other on either side of a 6 m wide European street. What is the distance between the top of the two buildings? 18 m 10 m 6m7 Jess paddles a canoe 1700 m to the west, then 450 m south, and then 900 m to the east. She then stops for a rest. How far is she from her starting point?8 A yacht race starts and ﬁnishes at A and consists of 6 legs; AB, BC, CA, AE, EC, CA, in that order as shown in the B 4 km A ﬁgure at right. If AB = 4 km, BC = 3 km and CE = 3 km, ﬁnd: 3 km a AE b AC C c the total length of the race. 3 km E
- 22. 64 Maths Quest 9 for Victoria 9 A painter uses a trestle to stand on in order to paint a ceiling. It consists of 2 stepladders connected by a 4 m long plank. The inner feet of the 2 stepladders are 3 m apart, and each ladder has sloping sides of 2.5 m. How high off the ground is the plank? 10 A feature wall in a garden is in the shape of a 4.7 m trapezium, with parallel sides of 6.5 m and 4.7 m. The wall is 3.2 m high. It is to have fairy lights around the 3.2 m perimeter (except for the base). How many metres of 6.5 m lighting are required? 10 m 11 A garden bed is in the shape of a rectangle, with a triangular pond at one end as shown in the ﬁgure at right. 5m Garden Tony needs to cover the garden to a depth of 20 cm with Pond topsoil. How much soil (in cubic metres) does he need? 3m 12 Katie goes on a hike, and walks 2.5 km north, then 3.1 km east. She then walks 1 km north and 2 km west. How far is she, in a straight line, from her starting point? 13 A rectangular gate is 3.2 m long and 1.6 m high, and 3.2 m consists of three horizontal beams, and ﬁve vertical beams as shown in the diagram at right. Each section is 1.6 m braced with diagonals. How much timber is needed for the gate? 14 The diagram at right shows the cross-section through a roof. 5200 mm a Find the height of the roof, h, to the nearest millimetre. h b The longer supports are 5200 mm long. Find the length of the shorter supports, to the nearest millimetre. 9000 mm A 10 15 Find the distance, AB, in the plan of the paddock at right. 10 B 17 1 16 Calculate the value of a, b, c, d. Leave your answers in surd 1 (square root) form. Can you see a pattern? 1 a b c 1 d 1 QUEST SM AT H GE 1 If your watch is running 2 minutes late and EN loses 5 seconds each hour, how many CH L hours will it take for your watch to AL be running 5 minutes late? 2 Sixteen matches are used to create the pattern shown at right. Remove 4 matches to leave exactly 4 triangles. Draw your answer.
- 23. Chapter 2 Pythagoras’ theorem 65Career profile ROB BENSON — Sheet metal w orker and found the work appealed to me. I make food-processing machinery for food manufacturers (like Cadbury). Projects can take from a week to a month to complete. A typical day is spent completing part of a larger project. Occasionally I go on-site to install machinery. I need to use mathematics in my job every day for many tasks; such as interpreting draftsmen’s plans and calculating costs associated with each job. In particular, I use measurements to check that angles are ‘square’. A key formula that helps me with Name: Rob Benson this is Pythagoras’ theorem. Profession: Sheet metal worker Questions Qualiﬁcations: First Class Sheet Metal 1. What type of machinery does Rob make at apprenticeship Tripax Engineering? Employer: Tripax Engineering 2. Explain how Rob can check that angles are ‘square’. I was offered an apprenticeship in the sheet 3. Find out how you can obtain information metal working trade soon after I left school about apprenticeships that are available.Pythagorean triads Pythagorean triads, or triads for short, are 3 whole numbers that satisfy Pythagoras’ theorem. This means that the 3 numbers could be the sides of a right-angled triangle. We test Pythagorean triads by ﬁrst calculating c2, then calculating a2 + b2. If both calculations yield the same result, the original 3 numbers form a Pythagorean triad. WORKED Example 9 Do the numbers 5, 7 and 10 form a Pythagorean triad? THINK WRITE 1 Arrange the values so that the largest Let a = 5, b = 7, and c = 10. will be c, the hypotenuse. 2 Write down Pythagoras’ theorem. c2 = a2 + b2 3 Substitute the value of c in the left-hand c2 = 102 side of the equation and evaluate. = 100 4 Substitute the values of a and b in the a 2 + b2 = 5 2 + 72 right-hand side of the equation and = 25 + 49 evaluate. = 74 5 If both sides of the equation are equal, c2 ≠ a2 + b2 the numbers form a Pythagorean triad. The numbers do not form a Pythagorean triad.
- 24. 66 Maths Quest 9 for Victoria The numbers 3, 4 and 5 form a Pythagorean triad because 32 + 42 = 25 and 52 = 25. If we multiply each of these 3 numbers by another number, the resulting numbers will also form a Pythagorean triad. For example, if we multiply 3, 4 and 5 by 5, we get 15, 20 and 25. These 3 numbers satisfy Pythagoras’ theorem. Check: c2 = 252 = 625 a2 + b2 = 152 + 202 = 225 + 400 = 625 so c = a2 + b2. 2 Algebra can be used to ﬁnd sets of numbers that are Pythagorean triads. There are 2 ways in which this can be done. Method 1 Start with any odd number, and make this the shortest length of the triangle. S2 – 1 Use the formula M = ------------- where S = shortest side and M = middle side, to - 2 calculate the middle side of the triangle. You now have two sides of the triangle and can use c2 = a2 + b2 to calculate the third side (hypotenuse, c). WORKED Example 10 If the smallest number of a Pythagorean triad is 7, ﬁnd the middle number and, hence, ﬁnd the third number. THINK WRITE 1 Write down the given information, and S=7 the formula to ﬁnd the middle side, M. S2 – 1 M = ------------- - 2 Substitute S = 7 into the formula and 72 – 1 2 = ------------- - evaluate. 2 49 – 1 = -------------- - 2 48 = ----- - 2 = 24 The middle number is 24. 3 Use Pythagoras’ theorem to ﬁnd the c2 = a2 + b2 third number. = 72 + 242 = 49 + 576 = 625 c = 625 = 25 The third number is 25. 4 State the solution. The Pythagorean triad required is 7, 24, 25. Why do you think this rule works only when the smallest side is an odd number?
- 25. Chapter 2 Pythagoras’ theorem 67 Method 2 This method takes any 2 numbers, and applies a set of rules to them, in order to generate Pythagorean triads. Select 2 numbers, x and y, with the following rules: Rule 1 The number chosen for x must be larger than y; that is x > y. Rule 2 One number must be odd and the other even. Rule 3 The numbers chosen for x and y must have no common factors. For example, we could not choose 6 and 9 since they have a common factor of 3. The Pythagorean triad is given by: 2xy, x2 − y2, x2 + y2. WORKED Example 11Find a Pythagorean triad using the values x = 7 and y = 2.THINK WRITE1 Write down the values given in the x = 7, y = 2 question.2 Substitute these values into the 2xy = 2 × 7 × 2 expression for each term. = 28 x2 − y2 = 72 − 22 = 49 − 4 = 45 x2 + y2 = 72 + 22 = 49 + 4 = 533 Check that the numbers obtained satisfy c = 532 2 Pythagoras’ theorem. = 2809 Does c2 = a2 + b2? a + b = 282 + 452 2 2 = 784 + 2025 = 2809 Therefore, c2 = a2 + b2.4 State the solution. The Pythagorean triad required is 28, 45, 53. remember remember 1. Pythagorean triads consist of three whole numbers that satisfy Pythagoras’ theorem. S2 – 1 2. Starting with an odd number, S, we can use the formula M = ------------- to ﬁnd - 2 the middle number, M, of a Pythagorean triad. The third number can be found using Pythagoras’ theorem (c2 = a2 + b2). 3. We can produce a Pythagorean triad using chosen values of x and y by substituting these into the expressions 2xy, x2 – y2, x2 + y2.
- 26. 68 Maths Quest 9 for Victoria 2F Pythagorean triads reads L Sp he WORKED 1 Do the numbers given below form Pythagorean triads? Example etEXCE 9 a 6, 8, 10 b 5, 12, 13 c 4, 5, 6 Right d 24, 7, 25 e 16, 20, 12 f 14, 16, 30 angle tester 2 Use the triads below to create three other triads, and check that they satisfy Pythagoras’ d theorem. hcaMat a 3, 4, 5 b 5, 12, 13 Pythagorean triads WORKED 3 The smallest numbers of four Pythagorean triads are given below. Find the middle Example 10 number and, hence, ﬁnd the third number. a 9 b 11 c 13 d 29 4 What do you notice about the triads in question 3 above? WORKED 5 For each of the following, ﬁnd a Pythagorean triad using the given values of x and y. (In Example 11 each case check that the 3 numbers found satisfy Pythagoras’ theorem.) a x = 6 and y = 1 b x = 7 and y = 4 c x = 8 and y = 3 d x = 11 and y = 6 e x = 14 and y = 9 f x = 15 and y = 2 ET 2.2 6 Challenge! SHE a Does the method used in question 5 work if both x and y are even? Work b Does it work if both x and y are odd? Try some examples of your own. Will the house stand up? At the beginning of the chapter, we saw how George checked if the walls of the house he was building were ‘square’ with the ﬂoor. The wall was 240 cm high and a string was pegged to the ﬂoor 1 m from the foot of the wall. The string was then measured as being 260 cm. 1 Draw a diagram of this problem. 2 Convert all measurements in the problem to centimetres. 3 Do the numbers in part 2 form a Pythagorean triad? 4 Are the walls square with the ﬂoor of the house? 5 George checks another wall in a room with a height of 3320 mm. Again a string is pegged to the ﬂoor 1 m from the foot of the wall. If the wall is square with the ﬂoor, what measurement would you expect for the length of the string? 6 Give 3 more examples of situations where Pythagoras’ theorem would be useful.
- 27. Chapter 2 Pythagoras’ theorem 69 2 3 cm 1 Find the value of the hypotenuse, correct to 1 decimal place. 7 cm 17 m 2 Find the value of the pronumeral, correct to 1 decimal place. x 3 multiple choice 11 m A road sign is in the form of an equilateral triangle. Each side measures 54 cm. The height of the triangle is: A 31.4 cm B 46.8 cm C 55.3 cm D 61.2 cm E 72.3 cm 4 A rectangular garden bed is 190 cm long and the diagonal is 3 m. Find the width of the garden, in metres, correct to 2 decimal places. 5 Find the area of the garden bed in question 4. 6 True or false? In a right-angled triangle if the hypotenuse is 890 cm and one other side is 0.5 m, then the third side is 8.91 m. 7 Do the numbers 5, 12, 13 form a Pythagorean triad? 8 Are the numbers 9, 17, 19 a Pythagorean triad? 9 If the smallest number of a Pythagorean triad is 7, ﬁnd the middle number and, hence, ﬁnd the third number. 10 Use x = 8 and y = 5 to form a Pythagorean triad.Pythagoras in 3-D Many real-life situations involve 3-dimensional (3-D) shapes: shapes with length, width and height. Some common 3-D shapes used in this section include boxes, pyramids and right-angled wedges. Box Pyramid Right-angled wedge The important thing about 3-D shapes is that in a diagram, right angles may not look like right angles, so it is important to redraw sections of the diagram in 2 dimensions, where the right angles can be seen accurately.
- 28. 70 Maths Quest 9 for Victoria WORKED Example 12Find the length AG in this box. A B 6 C D E F 5 H 10 GTHINK WRITE 1 Draw the diagram in 3-D. A B 6 C D E F 5 H 10 G 2 Draw, in 2-D, a right-angled triangle A that contains AG and label the sides. Only 1 side is known, so we need to ﬁnd another right-angled triangle to 6 use. E G 3 Draw EFGH in 2-D and show the E F diagonal EG. Label the side EG as x. 5 x 5 H 10 G 4 Use Pythagoras’ theorem to ﬁnd EG. c =a +b 2 2 2 x2 = 52 + 102 = 25 + 100 = 125 x = 125 = 11.18 5 Place this information on triangle AEG. A Label the side AG as y. y 6 E 11.18 G 6 Solve this triangle for AG. c =a +b 2 2 2 y2 = 62 + 11.182 = 36 + 125 = 161 y = 161 = 12.69 7 Write your answer. The length of AG is 12.69.
- 29. Chapter 2 Pythagoras’ theorem 71 WORKED Example 13A piece of cheese in the shape of a right-angled wedge sits on a table. It has a rectangularbase measuring 14 cm by 8 cm, and is 4 cm high at the thickest point. An ant crawlsdiagonally across the sloping face. How far, to the nearest millimetre, does theant walk?THINK WRITE1 Draw a diagram in 3-D and label the B C vertices. E 4 cm Mark BD, the path taken by the ant, with F A 8 cm a dotted line. 14 cm D x2 Draw in 2-D a right-angled triangle that B contains BD, and label the sides. Only one side is known, so we need to ﬁnd 4 another right-angled triangle to use. D E3 Draw EFDA in 2-D, and show the E F diagonal ED. Label the side ED as x. 8 x 8 A 14 D4 Use Pythagoras’ theorem to ﬁnd ED. c =a +b 2 2 2 x2 = 82 + 142 = 64 + 196 = 260 x = 260 = 16.125 Place this information on triangle BED. B Label the side BD as y. y 4 D E 16.126 Solve this triangle for BD. c2 = a2 + b2 y2 = 42 + 16.122 = 16 + 260 = 276 y = 276 = 16.61 cm7 Check the answer’s units: we need to = 166.1 mm convert cm to mm; so multiply by 10.8 Give the answer in worded form, The ant walks 166 mm. rounded to the nearest mm.
- 30. 72 Maths Quest 9 for Victoria remember remember 1. Pythagoras’ theorem can be used to solve problems in 3 dimensions (3-D). 2. Some common 3-D shapes include boxes, pyramids and right-angled wedges. 3. To solve problems in 3-D it is helpful to draw sections of the original shape in 2 dimensions (2-D). 2G Pythagoras in 3-D Where appropriate, give answers correct to 2 decimal places. 2.4 WORKED 1 Find the length, AG. HEET Example a b cSkillS 12 A B A B A B C 10 D C C reads D D L Sp he 10.4 etEXCE Pythagoras’ E 10 E F F theorem 10 7.3 hca d H 10 G H 8.2 G E FMat Pythagoras’ 5 theorem H G 5 A B E 4 2 Find CB in the wedge at right and, hence, ﬁnd AC. F D 10 C 7 A B 3 If DC = 3.2 m, AC = 5.8 m, and CF = 4.5 m in the ﬁgure E at right, ﬁnd AD and BF. F D C V 8 A 4 Find BD and, hence, the height of the pyramid at right. B 8 D 8 C E EM = 20 cm 5 The pyramid ABCDE has a square base. The pyramid is 20 cm high. Each sloping edge measures 30 cm. Find the A B length of the sides of the base. M D C
- 31. Chapter 2 Pythagoras’ theorem 73 6 The sloping side of a cone is 10 cm and the height is 8 cm. Find the radius of the base. 8 cm 10 cm 7 An ice-cream cone has r a diameter across the top of 6 cm, and sloping side of 13 cm. How deep is the cone?WORKED 8 A piece of cheese in the shape of a right-angled wedge B CExample 13 sits on a table. It has a base measuring 20 mm by 10 mm, 4 mm E and is 4 mm high at the thickest point, as shown in the F ﬁgure. A ﬂy crawls diagonally across the sloping face. A 20 mm D 10 mm How far, to the nearest millimetre, does the ﬂy walk? 30 cm 9 Jodie travels to Bolivia, taking with her a suitcase as shown in the photo. She buys a carved walking stick 1.2 m long. Will she be able to ﬁt it in her suitcase for the ﬂight 65 cm home? 90 cm 10 A desk tidy is shaped like a cylinder, height 18 cm and diameter 10 cm as shown. A pencil that is 24 cm long rests inside. What length of the pencil is above the top of the cylinder? 18 cm 10 cm 11 A 10 m high ﬂagpole is in the corner of a rectangular park that measures 240 m by 150 m. 10 m 240 m A a Find: iii the length of the diagonal of the park 150 m iii the distance from A to the top of the pole B iii the distance from B to the top of the pole. b A bird ﬂies from the top of the pole to the centre of the park. How far does it ﬂy?
- 32. 74 Maths Quest 9 for Victoria 12 A candlestick is in the shape of two cones, joined at the vertices as shown. The smaller cone has a diameter and sloping side of 7 cm, and the larger one has a diameter and sloping side of 10 cm. How tall is the candlestick? 13 The total height of the shape at right is 15 cm. Find the length 15 cm of the sloping side of the pyramid. 6 cm 14 cm 14 cm 14 A sandcastle is in the shape of a truncated cone as shown. Find the diameter of the base. 20 cm 30 cm 32 cm me E ti 15 A tent is in the shape of a triangular prism, with a height ofGAM 120 cm Pythagoras’ 120 cm as shown at right. The width across the base of the theorem — door is 1 m, and the tent is 2.3 m long. 002 Find the length of each sloping side, in metres. Then 2.3 m ﬁnd the area of fabric used in the construction of the 1m SHE ET 2.3 sloping rectangles which form the sides.Work 16 A cube has sides of x cm. Find, in terms of x, the length of the diagonal from a top corner to the opposite bottom corner.
- 33. Chapter 2 Pythagoras’ theorem 75 Electrical cableWhen electricians need to place electrical cable in a room built on a concrete slab,they need to connect it along the walls or the ceiling. To keep the costs down, theywould try to use the least amount of cabling. For each of the two rooms shown below, ﬁnd the shortest length of electricalcable that could be used. A1 The electrical cable needs to run from one top corner 3m (point A) of the room to another corner (point B). (Hint: First draw the nets of the two rectangles that contain points A and B. There are two possibilities.) 4m 6m B2 In this second room the electrical cable needs to 10 m run from point A (which is situated in the middle A of the wall where it meets the ceiling) to point B. 8m 5m B
- 34. 76 Maths Quest 9 for Victoriasummary Copy the sentences below. Fill in the gaps by choosing the correct word or expression from the word list that follows. 1 The longest side of a right-angled triangle is called the . 2 The hypotenuse is always situated opposite the . 3 theorem states that in any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: c2 = a2 + b2. 4 Pythagoras’ theorem gives us a way of ﬁnding the of the third side in a triangle, if we know the lengths of the 2 other sides. 5 When using Pythagoras’ theorem, always check the given for each measurement. 6 If necessary, all measurements to the same units before using the rule. 7 To examine more complex situations involving Pythagoras’ theorem, a is essential. 8 When a diagram is given, try to create or separate any using further diagrams. 9 consist of three whole numbers that satisfy Pythagoras’ theorem. 10 Starting with an odd number, S, we can use the formula to ﬁnd the middle number, M, of a Pythagorean triad. The third number can be found using Pythagoras’ theorem (c2 = a2 + b2). 11 We can produce a Pythagorean triad using chosen values of x and y by substituting these into the 2xy, x2 − y2, x2 + y2. 12 Pythagoras’ theorem can be used to problems in three dimensions (3-D). 13 Some common 3-D shapes include boxes, and right-angled wedges. 14 To solve problems in 3-D, it is helpful to draw sections of the shape in two dimensions. WORD LIST Pythagoras’ S2 – 1 original solve hypotenuse M = -------------- - units right-angled 2 length Pythagorean triangles convert right angle triads pyramids expressions diagram
- 35. Chapter 2 Pythagoras’ theorem 77CHAPTER reviewWhere appropriate, give your answers correct to 2 decimal places. 1 What is the relationship of the hypotenuse, c, and the two shorter sides, a and b, of a right- angled triangle. 2A 2 Calculate x to 2 decimal places. a 7.2 m b 2B x 8.2 cm 8.4 m x 9.3 cm 3 A broomstick leans against a wall. The stick is 1.5 m long and reaches 1.2 m up the wall. How far from the base of the wall is the bottom of the broom? 2B 4 The top of a kitchen table measures 160 cm by 90 cm. A beetle walks diagonally across the table. How far does it walk? 2B 160 cm 90 cm
- 36. 78 Maths Quest 9 for Victoria 5 Calculate x, correct to 2 decimal places. 2C a b 117 mm x 82 mm x 123.1 cm 48.7 cm 6 How high up a wall can a 20 m ladder reach when it is placed 2 m from the foot of a wall? 2C Give your answer correct to 1 decimal place. 7 A rectangular garden bed measures 3.7 m by 50 cm. Find the length of the diagonal, in cm. 2D 8 A road sign is in the shape of an equilateral triangle with sides of 600 mm. Find the area of 2D the sign, in cm2. 9 Calculate x in the ﬁgure at right. 17 2E x 8 4 10 Find the perimeter of the shape below. 2E 20 8 14 75 m 15 Beach 11 An ironwoman race involves 3 swim legs and a beach run as shown 2E in the ﬁgure. Sea 100 m What is the total distance covered in the race? 250 m 12 Which of the following are Pythagorean triads? 2F a 15, 36, 39 b 50, 51, 10 c 50, 48, 14 13 Calculate the height of this pyramid. 2G 10 mm 8 mm 8 mm 2G 14 A drink container, in the shape of a box, measures 20 cm by 10 cm by 4 cm. How long must a drinking straw be so as to reach diagonally across the box, and have 5 cm outside the box?testyourself CHAPTER 2

Be the first to comment