Home Study Modules KS4 Foundation Level Pythagoras Theorem MathSphere material is used in over 15 000 schools in the UK and abroad There are 14 Foundation Level GSE Revision Modules altogether. You may purchase the other thirteen by going to the www.gcsemathsphere.co.uk website and clicking the UY NOW button. You will then be given a username and password for instant download access to both the foundation and the higher modules. egin working now towards the grade you want and deserve. The 14 Foundation Level Modules are: lgebra 1 lgebra 2 lgebra 3 rithmetic non calculator Graphs Positive and Negative Numbers Powers and Standard Form Probability Pythagoras Theorem Ratios and Proportion Statistics Transformations Trigonometry Types of Number MathSphere www.gcsemathsphere.co.uk is the GSE website for MathSphere Ltd
How to Use the MathSphere GSE Modules The modules are written in two parts: Questions and nswers. ll the questions come at the beginning of the module, followed by all the answers. The best way to work is to determine which questions you are going to tackle in a particular revision session and then have a good go at them without looking at the answers. When you have done as many as you can, or if you need to check an answer or your working, have a look at the answers. If you have had problems, try to see how to solve the problem and then have another go at it. There are so many questions here, that if you work through them systematically and ensure that you can have a very good shot at them all, it will be very difficult for you not to get the grade you deserve! Remember to begin your revision early to give yourself plenty of time to absorb all the ideas in the syllabus. There is just one small problem with this approach and that is that there may be a few topics within a module that you have not yet covered at school. Don t panic. Just mark these for attention later and have a go at them when you have covered the work at school. There are plenty of other questions in the other modules you can be practising! Remember too that by revising well in the months before the examinations, you will be a long way ahead of most people who always leave their revision to the last minute (and nearly always regret doing that as the examination looms closer and closer!). Good luck with your revision and examinations. lan Young MathSphere Page 2
Normally we do not give you any clues in the question section, but there is one point that is so important we want to include it here at the beginning of this section. It is this: When solving right angled triangles (that is, when you are trying to find angles or sides given some information about the triangle) you only have two sets of tools available Pythagoras Theorem and Trigonometry there is nothing else. If you have this type of question in the examination, which you are almost certain to have, don t sit there scratching your head. Decide whether to use Pythagoras or Trigonometry and get on with the question. How do you decide? Easy! If the question does not involve angles you are not given angles and you are not asked to find angles given the three sides of a triangle it must be Pythagoras Theorem. If you are given angles to use or are asked to find an angle, you should use Trigonometry. Simple as that! In some more complicated examples, you may be given angles that you don t need to use straight away (they are ready for a later part of the question) as well as the lengths of sides, but by the time you get to those you will, hopefully, have had plenty of practice and will be able to pick your way through the questions. 1. a) State Pythagoras Theorem. b) Which is the longest side in a right angled triangle? 2. alculate the distance in each of the following triangles. Round the answers to three significant figures where appropriate. a) b) c) 11.8 m 4 m 6 cm 12.9 m 3 m 5 cm d) e) f) 13 m 8.5 mm 12 m 6.33 mm 8.7 cm 3.2 cm Page 3
3. Draw these triangles very accurately. In each case measure the hypotenuse as accurately as you can. Find the hypotenuse using Pythagoras Theorem (rounding to the nearest millimetre) and compare your answer with the measured value. Is your measured value within one millimetre of the calculated value? a) b) c) 5.8 cm 7 cm 9 cm 8.3 cm 6 cm 5 cm 4. ladder is 15 m long and leans against a wall. The foot of the ladder is 4 m from the bottom of the wall. How far up the wall does the ladder reach? If the ladder is moved so that its foot is only 3 m from the base, how much further up the wall will the ladder reach? 5. What is the length of a diagonal of a rectangle 16.2 Km by 5.7 Km? 6. ship left a port and travelled 152 Km due east and then 193 Km due north. How far was it from its starting point? 7. flagpole is 22 m high and is constrained from bending in high winds by cables each 31 m long fixed into the ground. Each cable is stretched tight. How far is each cable fixed into the ground from the foot of the flagpole? 8. Two points and, are drawn on a co-ordinate grid. is ( 2, 5) and is (7, 9). What is the distance between and? 9. The top of a chimney on one house is 9.3 metres above the ground. The top of a chimney on another house is 6.9 metres above the ground. The chimneys are 15.6 metres apart, measured horiontally. bird flies directly from the top of the higher chimney pot to the lower. How far does it fly? 10. Two points and, are drawn on a co-ordinate grid. is (a, b) and is (c, d). What is the distance between and? 11. Show that a triangle with sides 3367, 3456 and 4825 units respectively is right angled. 12. What is the height of the vertical line of symmetry of this isosceles triangle? 11 cm 11 cm 9 cm Page 4
13. What is the area of each of these isosceles triangles? a) b) 8.4 m c) 12 cm 12 cm 8.4 m 10.2 m 15 cm 6.7 mm 4.3 mm 6.7 mm 14. ship sails from a port P to an island Q, 100 Km south, then on to another island R, 142 Km east of Q. From R it travels south again 110 Km to another island S and finally east a distance of 205 Km to its destination T. a) How far is it from P to R as the crow flies? b) How far is it from R to T as the crow flies? c) How far is it from P to T as the crow flies? 15. X Y line is drawn diagonally on a can from X to Z via Y, as shown in the diagram. (Z is directly below X.) Z The can has a diameter of 13 cm and a height of 15 cm. What is the length of the line XYZ? 16. The diagram shows the end of a shed. alculate the length XY. W 2.4 m Z 3.1 m 2.2 m X Y 17. n escalator rises 8 metres over a distance 18 metres measured horiontally. If a person standing still on the escalator takes 22 seconds to go from the bottom to the top, what is the speed of the escalator to the nearest tenth of a metre per second? Page 5
18. man standing on the top of a cliff sees a boat on the sea. The distance of the man from the boat, measured in a straight line is 125 metres. The distance from the bottom of the cliff to the boat is 93 metres. How high is the cliff? 19. Simon, Ma and eryl were discussing a triangle whose sides were 42 m, 33 m and 27 m long. Each made a statement as follows: Simon said the triangle was right angled. Ma said the triangle was not right angled. eryl said that if the two shorter sides had been measured accurately, the hypotenuse was within 0.5 metres of making the triangle right angled. Which of the people made correct statements? 20. Here is a partly finished proof of Pythagoras Theorem. We draw a square with a smaller, rotated square drawn inside it as in the diagram. The distances a, b and c are as shown. a b b c c a a c c b b a y filling in the boxes, complete the proof: rea of one of the small triangles = ½ ab rea of all four small triangles = rea of large square = (a + b) 2 = rea of smaller square = = ut area of small square is also, therefore = Page 6
nswers 1. a) Pythagoras Theorem states that: In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. There are several ways of phrasing this; as long as the examiners agree that you understand what the theorem is all about and can understand your particular phraseology, you will get the marks. 2. a) b) The hypotenuse. {s the square on the hypotenuse is equal to the sum of the squares on the other two sides, the hypotenuse must always be the longest side.} 4 m 3 m 2 = 4 2 + 3 2 = 16 + 9 = 25 i.e. = 25 = 5 m b) 6 cm On those questions where you need to round to three significant figures when you take a square root, you need only write down the first four significant figures because you are only going to look at the fourth one to see if it is 5 or more. ny other figures you write down are going to be ignored, so what s the point? 5 cm 2 = 6 2 + 5 2 = 36 + 25 = 61 i.e. = 61 = 7.810 = 7.81 cm (corr to 3 S.F.) c) 11.8 m 12.9 m 2 = 12.9 2 + 11.8 2 = 166.41 + 139.24 = 305.65 i.e. = 305.65 = 17.48 = 17.5 m (corr to 3 S.F.) Page 7
In parts d), e) and f), we are given the hypotenuse and one other side. s the hypotenuse is the longest side, we need to subtract the square of one of the shorter sides from the hypotenuse to find the square of the other shorter side. d) 13 m 12 m 2 = 13 2 12 2 = 169 144 = 25 i.e. = 25 = 5 m e) 8.5 mm 6.33 mm 2 = 8.5 2 6.33 2 = 72.25 40.0689 = 32.1811 i.e. = 32.1811 = 5.672 = 5.67 mm (corr to 3 S.F.) f) 3.2 cm 8.7 cm 2 = 8.7 2 3.2 2 = 75.69 10.24 = 65.45 i.e. = 65.45 = 8.090 = 8.09 cm (corr to 3 S.F.) Page 8
3. Don t forget to label your diagrams with appropriate letters. In question 2, we were given the letter referring to the length we had to calculate, so there was no need to use, and. In this question we are given no letters, so we must add our own. a) 7 cm 6 cm 2 = 2 + 2 = 7 2 + 6 2 = 49 + 36 = 85 = 85 = 9.21 = 9.2 cm (to nearest millimetre). ompare your measured value with 9.2 cm. How close were you? b) 9 cm 5 cm 2 = 2 + 2 = 9 2 + 5 2 = 81 + 25 = 106 = 106 = 10.29 = 10.3 cm (to nearest millimetre). ompare your measured value with 10.3 cm. How close were you? c) 5.8 cm 8.3 cm 2 = 2 + 2 = 5.8 2 + 8.3 2 = 33.64 + 68.89 = 102.53 = 102.53 = 10.12 = 10.1 cm (to nearest millimetre). ompare your measured value with 10.1 cm. How close were you? Page 9
4) Now, here s a rule you should never break: LWYS DRW DIGRM IF YOU RE NOT GIVEN ONE IN THE QUESTION. I am not sure if I can make this any clearer, but you should always draw a diagram unless the examiners provide one in the question and, even then, you may benefit from re-drawing a small part of their diagram if it helps you understand the question. Why? 1. because it is very easy without a diagram to make a simple mistake and lose marks. For example, with Pythagoras Theorem it is quite easy in the heat of an examination to add the two squares instead of subtracting them as sometimes needs to be done. What s the point in throwing away marks in an examination? You may as well stay at home and have a cup of coffee! 2. Even if you make a bit of a mess with a difficult question, you can sometimes get a mark for drawing a diagram because this shows you have been able to translate the information given in words into diagrammatic form. 2 nd position h 2 h 1 1 st position 3m 4 m In first position: h 1 2 = 15 2 4 2 = 225 16 = 209 h 1 = 209 = 14.45 = 14.5 m (corr to 3 S.F.) Make sure you answer the question here. 14.7 m is not the answer to the second part it is just part of the working. When the ladder is moved closer: h 2 2 = 15 2 3 2 = 225 9 = 216 h 2 = 216 = 14.69 = 14.7 m (corr to 3 S.F.) Ladder will now reach 14.7 14.5 = 0.2 m further up the wall. s a safety note, if you are thinking of using a tall ladder, this question shows that a ladder that is pretty vertical to start with won t reach much further up a wall if you move the base closer to the wall. In this case, you gain only 20cm of extra height by moving the base a full metre in. If the ladder is too close, it may well tip backwards with you on it and all for not real gain! Page 10
5. 16.2 Km 5.7 Km D D 2 = 16.2 2 + 5.7 2 = 262.44 + 32.49 = 294.93 i.e. D = 294.93 = 17.17 = 17.2 Km (corr to 3 S.F.) 6. R 193 Km P 152 Km Q PR 2 = 152 2 + 193 2 = 23 104 + 37 249 = 60 353 i.e. PR = 60 353 = 245.6 Distance of ship from starting point was 246 Km (orr to 3. S.F.) 7. 31 m 22 m x x 2 = 31 2 22 2 = 961 484 = 477 i.e. x = 477 = 21.84 Distance of cable anchor from pole = 21.8 m (orr to 3. S.F.) e careful about whether the question involves a subtraction or an addition and don t forget to correct your answer. Page 11
8. ( 2, 5) y x 2 = 14 2 + 9 2 = 196 + 81 = 277 i.e. = 277 = 16.64 = 16.6 units (orr to 3. S.F.) 14 When there are no units as in this case you can just leave them out or put units as we have done here. 9 (7, 9) 9. X 9.3 6.9 = 2.4 m Y 15.6 m XY 2 = 2.4 2 + 15.6 2 = 5.76 + 243.36 = 249.12 i.e. XY = 249.12 = 15.78 These questions are good examples of it being essential to draw diagrams! ird flies 15.8 m (orr to 3. S.F.) 10. (c, d) (a, b) 2 = (a c) 2 + (b d) 2 i.e. = (a c) 2 + (b d) 2 If you had and in different positions, you may have: = (c a) 2 + (d b) 2 which is equally valid since (c a) 2 = (a c) 2 and (d b) 2 = (b d) 2. Page 12
11. We need to show, using Pythagoras Theorem, that 3 367 2 + 3 456 2 = 4 825 2 3 367 2 + 3 456 2 = 11 336 689 + 11 943 936 = 23 280 625 4 825 2 = 23 280 625 Therefore sides of 3367, 3456 and 4825 units form a right angled triangle. (Q.E.D.) Q.E.D.? Quite easily done! 12. s soon as you realise that an isosceles triangle is just two right angled triangles joined together, life becomes a piece of cake. If the isosceles triangle does not have a line of symmetry drawn to reveal the two right angles triangles, it is probably a good idea to draw one yourself. 11 cm 11 cm In any case it is always a good idea to label a more complicated diagram like this so that you can refer to the points and put the examiner in a good mood as he/she reads through your work! M M 9 cm 4.5 cm M 2 = 11 2 4.5 2 = 121 20.25 = 100.75 i.e. M = 100.75 = 10.03 = 10.0 cm (orr to 3 S.F.) Page 13
13. a) 12 cm 12 cm M 15 cm M 2 = 12 2 7.5 2 = 144 56.25 = 87.75 i.e. M = 87.75 = 9.3674 rea of triangle = ½ hb = 0.5 x 9.3674 x 15 = 70.2555 = 70.3 cm 2 (orr to 3 S.F.) Earlier we said that when you take the square root, only write down four significant figures and correct to three. You will have noticed that we have not done that here because we then carry on and use that number in another calculation to find the area. We save the correcting to three significant figures to the very end. b) 8.4 m 8.4 m M 10.2 m M 2 = 8.4 2 5.1 2 = 70.56 26.01 = 44.55 i.e. M = 44.55 = 6.6745 rea of triangle = ½ hb = 0.5 x 6.6745 x 10.2 = 34.03 = 34.0 m 2 (orr to 3 S.F.) c) 6.7 mm M 4.3 mm 6.7 mm M 2 = 6.7 2 2.15 2 = 44.89 4.6225 = 40.2675 i.e. M = 40.2675 = 6.3456 rea of triangle = ½ hb = 0.5 x 6.3456 x 4.3 = 13.643 = 13.6 mm 2 (orr to 3 S.F.) Page 14
14. 210 Km P 100 Km It is possible that you are not working on this module in the UK and have not heard of the expression as the crow flies. This simply means in a straight line as that is what crows are supposed to do, I guess. Q 142 Km R 110 Km X S 205 Km T 347 Km a) PR 2 = 100 2 + 142 2 = 10 000 + 20 164 = 30 164 i.e. PR = 30 164 = 173.6 = 174 Km (orr to 3 S.F.) b) RT 2 = 110 2 +205 2 = 12 100 + 42 025 = 54 125 i.e. RT = 54 125 = 232.6 = 233 Km (orr to 3 S.F.) There is another possible origin (as there always seems to be, of course) and that is that when sailors were lost or not sure of their position, they would release a crow from a cage. The crow would climb high in the sky and fly straight towards any land it could see. c) PT 2 = 210 2 + 347 2 = 44 100 + 120 409 = 164 509 i.e. PT = 164 509 = 405.5 = 406 Km (orr to 3 S.F.) s you probably realised, the shortest distance from P to T is not the sum of the distances PR and RT which is 407 Km. In this case it is just one kilometre out, but could be a lot more in a different situation. Page 15
15. X Y Z We aren t normally expected to find the length of a line spiralling around a cylinder directly, so immediately we have to devise a way to make this into a straight line. We do this simply by cutting the cylinder open with a direct cut from X to Z and then laying the cylinder out flat. With a very large hammer if necessary! X 15 cm Y Z 1 π x 13 cm Z 2 Notice that the point Z has been split into two points, Z 1 and Z 2. Notice also that Z 1 Z 2 is the circumference of the cylinder which is π x 13 cm long. Now it is just a matter of applying Pythagoras Theorem. XZ 2 2 = XZ 1 2 + Z 1 Z 2 2 = 15 2 + (π x 13) 2 = 225 + 1667.96 = 1892.96 i.e. XZ 2 = 1892.96 = 43.50 = 43.5 cm (orr to 3 S.F.) Page 16
16. The diagram shows the end of a shed. alculate the length XY. W P 2.4 m Z We need a right angled triangle, so draw a horiontal line from Z to a point P. Then we can find the distance XY by finding PZ. 3.1 m 2.2 m X Y PW = XW XP = XW YZ = 3.1 2.2 = 0.9 metres. PZ 2 = WZ 2 PW 2 = 2.4 2 0.9 2 = 5.76 0.81 = 4.95 PZ = 4.95 = 2.224 m = 2.22 m (orr to 3. S.F.) Therefore XY = 2.22 m (orr to 3. S.F.) 17. 8 m 18 m 2 = 2 + 2 = 18 2 + 8 2 = 324 + 64 = 388 = 388 = 19.697 metres. Speed of the escalator = 19.697 22 = 0.8945 = 0.9 metres per second (orr to nearest tenth of a metres per second) 18. 125 m 93 m 2 = 2 2 = 125 2 93 2 = 15625 8649= 6976 Height of cliff = = 6976 = 83.52 = 83.5 metres (orr to 3 S. F.) Page 17
19. 33 m 42 m? 27 m Find length of hypotenuse assuming measurement of two shorter sides to be accurate: 2 = 33 2 + 27 2 = 1089 + 729 = 1818 Therefore = 1818 = 42.63 = 42.6 m (orr to nearest tenth of a metre) onclusions: Simon is incorrect, the triangle is not perfectly right angled. Ma is therefore correct. eryl is also incorrect as the hypotenuse is just over 0.5 m of what it would need to be for the triangle to be right angled. Notice how, in this question, some explanation of your answer is needed for each person. If you do not give this, the examiner would be justified in saying that you had guessed and not give you the full marks, even if you had worked out the hypotenuse accurately. Page 18
20. a b b c c a a c c b b a rea of one of the small triangles = ½ ab rea of all four small triangles = 2ab rea of large square = (a + b) 2 = a 2 + 2ab + b 2 rea of smaller square = a 2 + 2ab + b 2 2ab = a 2 + b 2 ut area of small square is also c 2, therefore a 2 + b 2 = c 2 Pythagoras Theorem Page 19