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# Ch 02

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### Ch 02

1. 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. 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. 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.
5. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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.