Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
GEOMETRIC REASONING GEOMETRIC REASONING GEOMETRIC REASONING IN SOLVING PROBLEMS 4 CREDITS(91031) THE SKILLS YOU NEED TO KNOW: PYTHAGORAS THEORM p 94 1. Label the two short sides of the triangle a and b and the long side c 2. Substitute the known values into a 2 + b 2 = c 2 and solve TRIGONOMETRY: FINDING AN ANGLE p 102 Use to find an angle from two sides 1. Label each side (O, A or H) 2. Cross out one letter 3. Select SOH, CAH, or TOA 4. Draw triangle 5. Substitute lengths in with inverse sign and brackets 6. Write equation and solve to find the other length. Bearings: Problems involving bearings are very similar to trigonometry problems except angles are given as clockwise from north and written with three numbers e.g. 030 for 30° TRIGONOMETRY: FINDING A LENGTH p 98 Use to find a length from an angle and a length 1. Label each side (O, A or H) 2. Cross out one letter 3. Select SOH, CAH, or TOA 4. Draw triangle 5. Substitute numbers in 6. Write equation and solve to find the other length. Bearings: Problems involving bearings are very similar to trigonometry problems except angles are given as clockwise from north and written with three numbers e.g. 030 for 30° SIMILAR SHAPES p 106 1.Substitute in numbers and solve: Side 1 ( small ) Side 2 ( small ) = Side 1 (large) Side 2 (large) 2.If two shapes have the same angles then they are similar shapes and will be in proportion to each other PAGE 91 ANGLES OF POLYGONS p 110 • The interior angles of a polygon add to 180(n – 2)°, where n is the number of sides (int ∠, sum of polygon) • If a shape is regular (all the sides and angles are the same) to find each interior angle divide the sum of the interior angles by the number of sides. • The exterior angle is the angle between any side of a shape, and a line extended from the next side. • The exterior angles of a polygon add to 360° (ext ∠, sum of polygon) a + b + c + d + e + f = 360° ANGLES AROUND INTERSECTING LINES p 113 1.Adjacent angles on a straight line add to 180° ( ∠ s on str. line) a + b + c = 180° 2.Angles at a point add to 360° ( ∠ s at pt) a + b + c + d = 360° 3.Vertically opposite angles are equal (vert opp ∠ s) a = c and b = d ANGLES OF PARALLEL LINES • Corresponding angles on parallel lines are equal (corr ∠ s // lines) a=b • Alternate angles on parallel lines are equal (alt ∠ s, // lines) a=b • Co-interior angles on parallel lines are supplementary (add to 180°) (co-int ∠ s, // lines) a + b = 180° PAGE 92 p 116 GEOMETRIC REASONING ANGLES WITHIN CIRCLES p 119 • Angles on the same arc are equal a=b (∠s on same arc) • The angle at the centre is equal to twice the angle at the circumference on the same arc 2c = d (∠ at centre) • The angle in a semicircle is a right angle This is a special case of the above rule (∠ in semicircle) • The angle where the radius meets the tangent is 90° (rad ⊥ tgt) • Two tangents coming from the same point are equal (same length and angles) (tangs from a point) • The angle between a chord and a tangent equals the angle in the alternate segment. a = d, c = b. (∠ in alt seg) • Remember any chord forms an isosceles triangle with the centre. ABO and OBC are isosceles triangles. • Opposite angles in a cyclic quadrilateral add to 180° (opp ∠s cyclic quad) a + c = 180°, b + d = 180° • The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. a = b (ext ∠ cyclic quad) Note: Problems may rely on knowledge from earlier geometric reasoning sections PAGE 93 PYTHAGORAS THEORM SUMMARY a +b = c 2 2 2 1. Use this when 2 sides of a right angled triangle are known and the 3rd is required. a and b are the short sides and c is the long side. 2. Steps and example: 3. Label the two shorter sides a and b and the longest side c a 2 + b2 = c2 4. Substitute into a + b = c 32 + 42 = c 2 known values and solve to find the 9 + 16 = c 2 = 25 length of the other side c=5 For a complete tutorial on this topic visit www.learncoach.co.nz 2 2 2 NCEA QUESTIONS 1. The triangle FGH is part of a frame for a climbing net. OT is 90 cm long. OP is 70 cm long. Find the length of PT, x, the distance between the pole and a support along the ground. 3. Ali and Rob are designing a triathlon course. HF = 4.4m and the distance along the ground, HG = 6.2m Calculate the length of the side of the frame FG. a. The swim leg is around a triangular course ABC. 2. AB is 250 m. BC is 100 m. The angle at B is 90°. Calculate the length of AC. b. The cycle leg is around a triangular course DEF. A child’s practice goal post has one pole and two supports, as shown above, to the left. The two supports are each 90 cm long. The pole is always perpendicular to the ground. The diagram above right, shows the view from the side. PAGE 94 EF is 2.1 km. FD is 3.4 km. The angle at E is 90°. Calculate the length of DE. GEOMETRIC REASONING PRACTICE QUESTIONS 4. There is a bush walk in the Waipoua State Forest near the large kauri tree, Tane Mahuta (T). 6. a. The diagram shows a farm gate. Angle WXZ is a right angle. The width of the gate WX is 3.1 m. The height of the gate XZ is 1.25 m. Find the length of the brace WZ. 7. b. James stands at S, 8 m away from Tane Mahuta (T). FT, the height of the lowest branch is 12 m above ground level. Calculate the length of SF, the distance of James from the lowest branch. Lisa wanted to fly a remote control aeroplane from A to C but there was a tree in the way so instead she flew it from A to B to C. Calculate the distance she flew. 8. James walks 200 metres North from Tane Mahuta, point T, to point N. He then walks west until point W, ending up 600 metres from T. Calculate how far James walks to the West, WN. 5. Sam the dog likes pacing around the back yard. He goes along two fence lines then makes a diagonal back to where he started. What is the total distance he travels? 9. A flag pole FP is 22 metres high. A cable FC 36 metres long helps secure the pole. What is the distance of point C from the base of the tower P? A cuboid has a base measuring 10 cm by 12 cm. The cuboid is 14 cm high. The cuboid manages to fit exactly inside a sphere. What would the diameter of the sphere have to be? PAGE 95 ANSWERS NCEA 6. WZ 2 = WX 2 + XZ 2 WZ 2 = 3.12 + 1.252 1. HF 2 + FG 2 = HG 2 FG 2 = HG 2 − HF 2 FG 2 = 6.22 − 4.42 FG = 4.37 WZ = 3.12 + 1.252 = 3.34 m (2 dp) (Achieved) 2. PT 2 + OP 2 = OT 2 2 2 7. x + 70 = 90 x 2 = 902 − 702 = 3200 x = 3200 = 56.6 cm (1 dp) 2 CA = 250 + 100 Distance = 14.4 + 5 = 19.4 m (Merit) 2 2 CA = 250 + 100 8. Diag 2 = 7 2 + 92 2 = 269.3 m (1 dp) Diag = 7 2 + 92 = 11.4 m (1 dp) (Achieved) b. DE 2 + EF 2 = FD 2 Distance = 11.4 + 7 + 9 = 27.4 m DE 2 = FD 2 − EF 2 2 AB 2 = 82 + 122 AB = 42 + 32 =5 m (Achieved) 3. a. AB 2 + BC 2 = CA2 2 (Achieved) AB = 82 + 122 = 14.4 m (1 dp) 2 BC = 42 + 32 2 9. 2 DE = 3.4 − 2.1 = 2.67 km (2 dp) (Achieved) (Merit) The longest distance is between opposite corners of the box A and C. PRACTICE 4. a. SF 2 = ST 2 + FT 2 SF 2 = 82 + 122 SF = 82 + 122 (Achieved) = 14.4 m (1 dp) b. TW 2 = WN 2 + NT 2 WN 2 = TW 2 − NT 2 WN 2 = 6002 − 2002 2 2 WN = 600 − 200 (Achieved) = 565.7 m (1 dp) 5. FC 2 = FP 2 + CP 2 CP 2 = FC 2 − FP 2 CP 2 = 362 − 222 2 2 CP = 36 − 22 = 28.5 m (1 dp) PAGE 96 (Achieved) This can be found using two right angled triangles. First finding length AB then finding length AC. AB 2 = 102 + 122 AB = 102 + 122 = 15.62 cm (2 dp) 2 AC = 142 + 15.622 AC = 142 + 15.622 = 20.98 cm (2 dp) Therefore the diameter of the sphere is 20.98 cm. (Excellence) GEOMETRIC REASONING Study Tip: Last Minute Study If you are running out of time: • Focus on topics that came up often in class • Don’t bother studying topics that haven’t been covered in class. They probably won’t be in the exam • The LearnCoach summaries are a good place to start! PAGE 97 TRIGONOMETRY: FINDING A LENGTH SUMMARY 1. Use when 1 length and one angle of a right angled triangle is know and a second length is required. 2. Steps and example: 3. Label each side (O, A or H) 4. Cross out one side 5. Select SOH, CAH, or TOA SOH 6. Draw triangle 7. Substitute numbers in x = 10 × sin 30 8. Write equation and solve to find =5 m the other length. • Note: Problems involving bearings are very similar to trigonometry problems except angles are given as clockwise from north and written with three numbers e.g. 030 for 30° For a complete tutorial on this topic visit www.learncoach.co.nz OLD NCEA QUESTIONS 1. A balloon, A, is tied to the ground by the rope labelled TA. The wind is strong and causes the rope, TA, to make a straight line. The balloon is 40 m above the ground. The rope TA makes an angle of 26° with the ground. Calculate the length of the rope, TA. 2. 3. a. Calculate the height of the shed. b. Calculate the width of the shed. A ladder has two legs AB and AC. Each leg is 1.8m long. Angle ABD = 113° A shed in the playground has a roof that is 3.6 m long. 0.4 m of the roof overhangs the wall. The roof is at an angle of 40° to the horizontal. The walls of the shed are 3.8 m high. Calculate the length of c. PAGE 98 GEOMETRIC REASONING 4. A 35 m long bridge BE crosses a river. The width of the river is BC. Angle EBC is 18°. 5. An orienteering course is planned from point O. The first leg to a point marked A is 120 m on a bearing of 030°. The second leg begins at A and ends at point B. B is on a bearing of 120° and 110 m from A. Calculate the distance from O to B giving reasons for each step. Calculate the length of EC. PRACTICE QUESTIONS 6. Riggers were putting up a circus tent and the first job is to put up the main centre pole. The pole is 65 m high. 10 m from the top it has a support wire attached. The support wire is supposed to be at an angle of 48° to the main pole (ABC). The house is 6 m wide. The pitch of one side of the roof, angle EDF, is 27°. The owners want to put a solar panel on the side EF. What is the maximum length it could be? 9. The top of a cliff is 40 m above sea level. A person on a boat floating in the sea manages to spot a person standing on the cliff when they look at an angle of 17° Calculate the distance away from the pole that the support wire should be attached (AC). 7. When watching the circus Kim is sitting at K, 25 m away from the area under where the trapeze artists are performing, B. How far is the boat from the cliffs? 10. Ali, Rob, and Sarah are doing a triathlon course. Ali is doing the running leg of the course. From the start point, A, she runs 800 m at a bearing of 060°. She then runs a distance at a bearing of 150° until her bearing is 097° from her original position. She looks up at them with an angle of an angle of 63°. Calculate how high up the trapeze artists are, TB. 8. The triangle DEF shows a cross-section of the roof of a house. Calculate Ali’s distance from the start point A to her current position. PAGE 99 ANSWERS NCEA 1. Using SOH Substituting: 5. Angle OAB = 90° corresponding angles parallel lines Sum of angles on a straight line = 180° 40 sin 26 40 TA = = 91.25 m (2 dp) (Merit) sin 26° 2. a. Using SOH Substituting: Length OB = 1202 + 1102 = 162.8 m (1 dp) TA PRACTICE 6. O 3.6 sin 40 - 0.4 O = 3.2 × sin 40 = 2.06 (2 dp) Height = 2.06 + 3.8 = 5.86 m b. Using CAH Substituting: (Merit) Using TOA Substituting: AC 65 tan 48 -10 AC = 55 × tan 48 = 61.1 m (1 dp) 7. A Using TOA Substituting: (Achieved) TB tan 63 25 3.6 cos 40 - 0.4 A = 3.2 × cos 40 = 2.45 (2 dp) Width = 2 × 2.45 = 4.90 m = TB 25 = tan 63 49.1 m (1 dp) (Achieved) 8. (Achieved - a or b correct) (Merit - a and b correct, any method) (Excellence - a and b correct, correct working) 3. Using SOH Substituting: KM 6 sin 27 KM = 6 × sin 27 = 2.72 m (2 9. Using TOA Substituting: dp) (Achieved) A tan 17 40 Split into two right angled triangles and solve Angle ABC = 180 – 113 = 67° (angles on a line) Using CAH Substituting: A 40 = A = 130.8 m (1 dp) tan 17 (Achieved) 10. cos 67 1.8 A = 1.8 × cos 67 = 0.7 (1 dp) c = 2 × 0.70 = 1.4 m 4. (Merit - Answer without support) (Excellence - Answer with support) Using SOH Substituting: The change in direction is 90° (Co-interior angles, parallel lines and angles at a point) Using TOA Substituting: EC sin 18 35 EC = 35 × sin 18 = 10.8 m (1 x dp) (Achieved) cos 37 800 800 = x = 1001.7 m (1 dp) cos 37 PAGE 100 (Excellence) GEOMETRIC REASONING Study Tip: Evaluation After Each Exam ask yourself: • Where did most of the questions come from? • Which parts ate up most of my time? • Was I anxious during the exam? If so why? • What could I do differently next time? PAGE 101 TRIGONOMETRY: FINDING AN ANGLE SUMMARY 1. Use when 2 lengths of a right angled triangle are known and an angle is required 2. Steps and example: 3. Label each side (O, A or H) 4. Cross out one side 5. Select SOH, CAH, or TOA SOH 6. Draw triangle 7. Substitute lengths in with inverse sign and brackets (4 sin-1 10) 4 x = sin −1 8. Write equation and solve to find 10 the angle. = 23.6° (1 dp) • Note: Problems involving bearings are very similar to trigonometry problems except angles are given as clockwise from north and written with three numbers e.g. 030 for 30° For a complete tutorial on this topic visit www.learncoach.co.nz OLD NCEA QUESTIONS 1. The triangle FGH is part of a frame for a climbing net. 3. HF = 4.4 m and the distance along the ground, HG = 6.2 m. Calculate the angle the frame makes with the ground at FGH. 2. A cuboid has a base measuring 5 cm by 6 cm. The cuboid is 7 cm high. What is the angle between the line AB shown and the 5 cm edge of the cuboid? PAGE 102 A child’s practice goal post has one pole and two supports, as shown above, to the left. The two supports are each 90 cm long. The pole is always perpendicular to the ground. The diagram above right, shows the view from the side. OT is 90 cm long. OP is 70 cm long. Calculate the size of angle PTO. GEOMETRIC REASONING 4. The diagram shows a square pyramid, with base ABCD. 5. Each side of the base is 220 metres long. FG = 110 m The height of the pyramid EF is 140 m. Calculate the angle EGF. An orienteering course is planned from point O. The first leg to a point marked A is 120 m on a bearing of 030°. The second leg begins at A and ends at point B. B is on a bearing of 120° and 110 m from A. Calculate the bearing of the starting point O from the finish B. PRACTICE QUESTIONS 6. Ben bought a new fridge but wants to use a shelf from the old one as there are not many shelves in the new fridge. The problem is that the new fridge is narrower than the old one. The old shelf, AB, is 84 cm long. When he fits it into the new fridge it sits at an angle with one end 4 cm above the other. Calculate the angle the shelf AB makes with the horizontal AC. 7. Calculate Ali’s bearing from the start point A to the end point of the run. 9. A carver was given a wooden block and asked to replicate it. The base is a horizontal regular hexagon, with sides of 6 cm shown in the diagrams and a height of 25 cm. [A regular hexagon is a polygon made up from 6 equilateral triangles.] A pool player needs to hit a ball into a corner pocket. It sits 81 cm from the back wall and 62 cm from the side wall. What angle does the ball need to travel at to be sunk? 10. Calculate the angle between each triangular face of the block and the horizontal hexagon base. 8. Ali and Rob are designing a triathlon course. Ali is doing the running leg of the course. From the start point, A, she runs 800 m at a bearing of 060°. She then runs 600 m at a bearing of 150°. A plywood bike jump was created with a 1.5 m long angled ramp and that is 0.3 m high. At what angle does the rider ride up at? PAGE 103 ANSWERS 7. NCEA 1. Using SOH Substituting: (4.4 sin-1 6.2) 4.4 6.2 G = sin −1 = 45.2° 2. Using TOA Substituting: x = 62 − 32 = 5.20 m (2 dp) OR (Achieved) x tan 60 6 tan 5) 6 θ = tan −1 = 50.2° (1 dp) 5 Using SOH Substituting: x sin 60 (Achieved) Step 2: Using TOA Substituting: -1 −1 7 9 4. Using TOA Substituting: sin 90) dp) 25 θ = tan −1 = 78.3° (1 dp) (Excellence) 5.20 (Achieved) 8. tan-1110) −1 14 11 5. (1 dp) (Achieved) Angle is a right angle (Co-interior angles, parallel lines and angles at a point) Using TOA Substituting: (600 tan-1800) Angle OAB is a right angle (Co-interior angles, parallel lines and angles at a point) Using TOA Substituting: (120 3 θ = tan −1 = 36.9° (1 dp) 4 Bearing = 60 + 36.9 = 097° 9. tan-1110) 120 ABO = tan = 47.5° (1 dp) 110 Bearing = 360 − (60 + 47.5) = 252.5° Using TOA Substituting: −1 Using SOH Substituting: sin 84) 1 θ = sin = 2.73° (2 dp) 21 PAGE 104 −1 62 81 Using TOA Substituting: dp) (Achieved) (0.3 sin-1 1.5) -1 −1 (62 θ = tan = 37.4° (1 10. (4 (Merit) tan-1 81) (Excellence) PRACTICE 6. (25 tan-1 5.2) (140 EGF = tan = 51.8° 6 x = 6 sin 60 = 5.20 m (2 dp) (70 θ = sin = 51.1° (1 3 x = 3 tan 60 = 5.20 m (2 dp) OR -1 3. There are two steps to this. First calculate the distance from the edge of the base to the centre, then find the angle. Step 1: Either (Achieved) 0.3 θ = sin −1 = 11.5° (1 1.5 dp) (Achieved) GEOMETRIC REASONING Study Tip: Motivation When you study well for an exam: • Treat yourself for giving it your best shot • Watch a movie with friends or get some takeaways Small rewards motivate you to try your best in whatever you do. PAGE 105 SIMILAR SHAPES SUMMARY • If two polygons are similar, then: ▶▶ Corresponding angles are equal ▶▶ Corresponding sides are in proportion 1.Substitute in numbers and solve: Side 1 ( small ) Side 2 ( small ) = Side 1 (large) Side 2 (large) Side 1 ( small ) Side 2 ( small ) = Side 1 (large) Side 2 (large) • If two shapes have the same angles then 1.5 2 they are similar shapes and will be in = proportion to each other x 6 6 x = 1.5 × = 4.5 m For a complete tutorial on this topic2visit www.learncoach.co.nz OLD NCEA QUESTIONS 1. A ladder has two legs AB and AC. Each leg is 1.8m long. Angle ABD = 113° Express b in terms of c. PAGE 106 2. A child’s practice goal post has one pole and two supports, as shown to the right. The two supports are each 90 cm long. The pole is always perpendicular to the ground. The diagram below, shows the view from the side. OT is 90 cm long. OP is 70 cm long. A support bar, QR, is added at Q, where OQ = 30 cm. Calculate the distance of the Length OR. Show your working and explain your reasoning. GEOMETRIC REASONING PRACTICE QUESTIONS 3. 6. The Egyptian pyramids originally had a cap stone at the top which was made of a different material. The angled side of the pyramid is 78 m long and the base is 85 m wide. If the cap stone is 3.5 m wide, how far down the angled side of the pyramid does it extend? Two crosses are shown which are similar. Each arm of the large cross is 12 cm wide and 5 cm wide for the small cross. If each arm on the large cross is 33 cm long, how long are the arms on the small cross? 4. 7. At a BMX track there are two jumps next to each other. The smaller one has a ramp that is 2.5 m long and a 1.3 m drop off. If the larger ramp has a 6.5 m long ramp, how high is the drop off? Two arrows are drawn which are similar. The large arrow has an overall length of 30 cm and a tail length of 22 cm. If the small arrow has a tail length of 13.5 cm, what is its overall length? 8. 5. ABCEG has similar shape DEFHI inside it. DE = 8 cm, EF = 6 cm, and FG = 4 cm. Find the length CD. These two shapes are similar. Use the given lengths to find the length BC. PAGE 107 ANSWERS NCEA 1. a. Side 1 ( small ) = Side 2 ( small ) 5. Side 1 (large) Side 2 (large) b 0.475 = 1.8 c 0.86 (Merit) b= c 2. OQR and OPT are similar triangles: Side 1 ( small ) Side 2 ( small ) = Side 1 (large) Side 2 (large) OR OQ = OT OP OR 30 = 90 70 30 y = 90 × = 38.6 cm (1 dp) (Merit) 70 PRACTICE 3. 4. Side 1 ( small ) Side 2 ( small ) = Side 1 (large) Side 2 (large) 5 x = 12 33 5 (Merit) 33 × = 13.75 cm x = 12 Side 1 ( small ) Side 2 ( small ) = Side 1 (large) Side 2 (large) 13.5 l = 22 30 13.5 l = 30 × = 18.41 cm (2 dp) (Merit) 22 PAGE 108 6. 7. 8. Side 1 ( small ) Side 2 ( small ) = Side 1 (large) Side 2 (large) EF DE = EG CE 6 8 = 10 CE 10 CE = 8 × = 13.33 cm (2 dp) 6 CD = 13.3 - 8 = 5.3 cm (Merit) Side 1 ( small ) Side 2 ( small ) = Side 1 (large) Side 2 (large) 3.5 x = 85 78 3.5 x = 78 × = 3.21 m (2 dp) (Merit) 85 Side 1 ( small ) Side 2 ( small ) = Side 1 (large) Side 2 (large) 2.5 1.3 = 6.5 h 6.5 h = 1.3 × = 3.38 m (Merit) 2.5 Side 1 ( small ) Side 2 ( small ) = Side 1 (large) Side 2 (large) 13.5 6 = 22 AC 22 AC = 6 × = 9.78 cm (2 dp) 13.5 BC = 9.8 - 6 = 3.8 cm (Merit) GEOMETRIC REASONING Study Tip: Study Enhancement If you are spending long hours studying or working remember to: • Drink fluids • Eat well • Sleep well • Do regular exercise and move around occasionally while studying It’s the basics that can make some of the biggest differences. PAGE 109 ANGLES OF POLYGONS SUMMARY • The angles inside a polygon are called interior angles. • A Polygon with n sides has n interior angles • The Exterior Angle is the angle between any side of a shape, and a line extended from the next side. • The interior angles of a polygon add to 180(n – 2)°, where n is the number of sides (int ∠, sum of polygon) e.g. to the right n = 5 Sum of interior angles = 180(5 - 2) = 540° • If a shape is regular (all the sides and angles are the same) to find each interior angle divide the sum of the interior angles by the number of sides. • The exterior angles of a polygon add to 360° (ext ∠, sum of polygon) a + b + c + d + e + f = 360° For a complete tutorial on this topic visit www.learncoach.co.nz OLD NCEA QUESTIONS 1. ABCDE is a regular pentagon. a. Calculate the size of angle ABC giving reasons for each step. b. If many objects of the same shape fit together to form a pattern, without leaving any spaces, the shape is said to tessellate. Explain whether or not a regular pentagon will tessellate, giving reasons for your answer. PAGE 110 GEOMETRIC REASONING PRACTICE QUESTIONS 2. 6. A fire tower is placed at a high place overlooking a huge pine plantation. In the regular pentagon shown, what is the value of a and all exterior angles and b and all interior angles. 3. 4. Calculate both the interiors and exterior angles of a a. regular hexagon b. regular decagon (10 sided figure) c. regular nine sided figure In the STOP sign below, what is the value of any interior angle (assume it is a regular figure). Give reasons. a. Given angle b = 140°, calculate angle a, giving reasons. b. Given angle a = 160°, calculate angle b, giving reasons. c. Find angle c. d. In a differently sized fire tower, x = 152° and y = 5. In the soccer ball calculate angles a and b, assuming all panels are regular. 74°, find angle c. 7. Fencing mesh is commonly made up of either 4 or 6 sided figures. Although in reality this is not the case, assume the figures are regular and calculate angles a, b, c, d, and e giving reasons. f. Give geometric reasons why both these geometric shapes tessellate. PAGE 111 ANSWERS b. Sum of interior angle of a pentagon is: NCEA 1. a. Sum of interior angles of a pentagon = 3 ×180° = 540° 540 = 108° Each interior angle is = 5 ABC = 108° (Achieved - Attempted to find interior angle) (Merit - correct angle) b. For a shape to tessellate the interior angles must be factors of 360°. 108° is not a factor of 360°, so a pentagon will not tessellate. This is based on the principal of angles at a point adding to 360°. (Excellence) c. d. PRACTICE 2. Sum of the angles = 360° (ext ∠, sum of polygon) 360 = 72° a(and all other exterior angles) = 5 Sum of the interior angles of a pentagon = 180(5 − 2) = 540° Regular pentagon has 5 equal interior angles 540 b (and all other interior angles) = = 108° 5 (Merit) 3. a. Hexagon Interior angle = 180(6 − 2) = 720 = 120° 6 6 Hexagon exterior angle = b. Decagon interior angle = 180(10 − 2) 1440 = = 144° 10 10 360 Decagon exterior angle = c. 360 = 60° 6 10 = 36° Nonagon interior angle = 180(9 − 2) 1260 = = 140° 9 9 360 = 40° 9 (Merit - 2 out of 3 correct) 4. Octagon interior angle = 180(8 − 2) = 1080 = 135° 8 8 int ∠, sum of polygon (Merit) Nonagon exterior angle = 5. a. a = 180(5 − 2) = 540 = 108° 5 5 b. b = 360 -108 = 252° int ∠, sum of polygon ∠s at pt (Merit) 6. a. Sum of interior angle of a pentagon is: = 180(5 − 2) = 540° 90 + 90 + 110 + 140 + a = 540 a = 540 − 140 − 110 − 90 − 90 a = 110° PAGE 112 (Merit) = 180(5 − 2) = 540° 90 + 90 + 110 + 160 + b = 540 b = 540 − 160 − 110 − 90 − 90 b = 90° (Merit) Sum of interior angle of a pentagon is: = 180(5 − 2) = 540° 108 + 108 + 120 + 120 + c = 540 c = 540 − 120 − 120 − 108 − 108 c = 84° (Merit) c above = c below as they are vertically opposite. Sum of interior angle of a pentagon is: = 180(5 − 2) = 540° x + x + y + y + c = 540° c = 540 − (2 × 152) − (2 × 74) = 88° (Merit) 7. a. 180(6 − 2) 720 = = 120° 6 6 int ∠, sum of polygon (Merit) b. b = 360 - a = 360 - 120 = 240° ∠s at pt (Merit) c. c = 360 - 2a = 360 - 240 = 120° ∠s at pt (Merit) 180(4 − 2) 360 = = 90° d. d = 4 4 (int ∠, sum of polygon) (Merit) e. e = 360 - d = 360 - 90 = 270° ∠s at pt (Merit) f. For a shape to tessellate the interior angles must be factors of 360°. For the hexagon the interior angles are 120° which is a factor of 360°. For the diamond the interior angles are 90° which are also a factor of 360°. This is based on the principal of angles at a point adding to 360°. (Excellence) a= GEOMETRIC REASONING ANGLES AROUND INTERSECTING LINES SUMMARY Example: Rules: • Adjacent angles on a straight line add to 180° ( ∠ s on str. line) a + b + c = 180° x = 180 – 112 = 68° ( ∠ s on str. line) • Angles at a point add to 360° ( ∠ s at pt) a + b + c + d = 360° x = 360 – 98 – 155 = 107° ( ∠ s at pt) d • Vertically opposite angles are equal (vert opp ∠ s) a = c and b = d x = 139° (vert opp ∠ s) For a complete tutorial on this topic visit www.learncoach.co.nz PRACTICE QUESTIONS Find the size of the angles shown and give a reason (the diagrams are not to scale): 1. 2. 3. 4. 5. 6. 7. 8. PAGE 113 ANSWERS PRACTICE 1. x = 180 – 95 – 45 = 40° ∠ s on str. line 2. x = 360 – 139 – 132 = 89° ∠ s at pt 3. x = 180 – 90 – 68 = 22° ∠ s on str. line 4. a = 180 – 30 = 150° b = 30° c= 150° ∠ s on str. line vert opp ∠ s ∠ s on str. line OR vert opp ∠ s (Achieved) 5. a = 180 – 162 = 18° b = 180 – 108 = 72° ∠ s on str. line ∠ s on str. line 6. x = 360 – 90 – 59 = 211° ∠ s at pt 7. x = 180 – 80 – 55 = 45° ∠ s on str. line and vert opp ∠ s (Achieved) 8. x = (180 – 90) / 2 = 45° ∠ s on str. line (Achieved) (Achieved) (Achieved) (Achieved) (Achieved) (Achieved) Study Tip: Pre-Exam Before an Important Exam: Don’t: spend a lot of time talking with classmates who haven’t studied. Do: Avoid negative vibes and focus on your own preparation and goals. PAGE 114 GEOMETRIC REASONING Study Tip: Cramming Cramming helps some people. But don’t lose sleep. Getting your normal sleep will ensure you are your best physically and mentally for the exam. PAGE 115 ANGLES OF PARALLEL LINES SUMMARY Example: Rules: • Corresponding angles on parallel lines are equal (corr ∠ s // lines) a=b x = 63° (corr ∠ s // lines) • Alternate angles on parallel lines are equal (alt ∠ s, // lines) a=b x = 56° (alt ∠ s, // lines) • Co-interior angles on parallel lines are supplementary (add to 180°) (co-int ∠ s, // lines) a + b = 180° x = 180 – 40 = 140° (co-int ∠ s, // lines) For a complete tutorial on this topic visit www.learncoach.co.nz OLD NCEA QUESTIONS 1. ABCD is an isosceles trapezium. Angle CBA = 78°. AD = BC. 3. Metal railings are fitted to the edge of a deck. XZ is parallel to AD. BY is parallel to CZ. One section of railing is shown in the diagram below: Calculate the size of angle EDA giving reasons for each step of your answer. 2. The diagram shows part of a climbing frame. a. Find the size of angle YBC. LM = LN. KL is parallel to NM. LM is parallel to KN. Angle LNK = 54°. Calculate the size of angle LMN, explaining the reason for each step of your answer PAGE 116 Give geometric reasons for each step in your solution. b. XYB is an isosceles triangle. Use geometric reasoning for each step to show that XB and YC cannot be parallel. GEOMETRIC REASONING 4. a. If x is 110°, find the size of angle PRQ. PR and QR are the same length. Angle RQS is x. RT is parallel to PS. Give geometric reasons. b. Prove that angle PRT and angle RQS are equal for all values of x. Explain your geometric reasoning clearly and logically. PRACTICE QUESTIONS Find the size of the angles shown and give a reason (the diagrams are not to scale): 5. 6. 7. 8. 9. 10. 11. 12. a. In the diagram find angles x and y b. Find any sets of parallel lines and give reasons 13. Toby has bought some expanding trellis. All pieces of wood pointing in the same direction are parallel. In the diagram below the trellis is fully expanded. What is angle a? Give a reason 14. Explain why c = a + b PAGE 117 ANSWERS NCEA 1. ∠ BAD = 78° PRACTICE Isosceles trapezium Alt ∠ s, // lines (Achieved - Both angles correct) (Merit - Angles and explanations correct) ∠ EDA = 78° 2. ∠ NLM = 54° Alt ∠ s, // lines (or rotation about mid point of LN) ∠ LMN = 63° Base ∠ , isos Δ (Achieved - Both angles correct) (Merit - Angles and explanations correct) 3. a. ∠ XYB = 62° ∠ YBC= 62° ∠ s, // lines alt ∠ s, // lines corr (Achieved - Both angles correct) (Merit - Angles and explanations correct) b. XYB is an isosceles triangle, so two of its interior angles must be equal. If XB and YC are parallel then: ∠ XBA= 64° corr ∠ s, // lines ∠ YBX = 180 - ∠ YBC - ∠ XBA =180 - 62 - 64 =54° ∠ sum Δ ∠ BXY = 180 - ∠ YBX - ∠ XYB =180 - 54 - 62 ∠ sum Δ =64° If this is the case then no two interior angles of XYB are equal, which is a contradiction. XYB is actually an isosceles triangle, therefore, XB and YC cannot be parallel. (Achieved - two angles correct) (Merit - Angles and explanations correct) 5. x = 65° corr 6. x = 75° alt b. ∠QRT = 180 − x co-int ∠ s, // lines ∠PRQ = 180 − 2 × ∠RQP Base ∠ , isos Δ = 180 − 2 × (180 − x) ∠ s on str. line = 180 − 360 + 2 x = −180 + 2 x Therefore: ∠PRT = ∠PQR + ∠QRT = −180 + 2 x + 180 − x =x = ∠RQS (Achieved - 2 angles correct) (Merit - Angles and explanations correct) PAGE 118 ∠ s, // lines (Achieved) 60° vert opp ∠ s (Achieved) b. b = 180 – 60 = 120° alt ∠s, // lines OR adj ∠s on str. Line (Achieved) 8. x = 180 – 102 = 78° corr ∠ s, // lines and adj ∠ s on str. line (Achieved) 9. x = 180 – 37 = 143° alt. ∠ s, // lines (Achieved) 10. a. a = 180 – 39 = 141° co-int ∠ s, // lines (Achieved) b. b = 60° alt ∠ s, // lines (Achieved) 11. a. a = 180 – 42 – 59 = 79° adj ∠ s on str line (Achieved) b. b = 59° alt. ∠ s, // lines (Achieved) c. c = 59 + a = 59 + 79 = 138° alt. ∠ s, // lines OR co-int ∠ s, // lines (Achieved) 12. a. x = 180 – 98 = 82° adj, ∠ s on str. line ∠ s on str. line ∠ s on str. line int. ∠ s of quadrilateral ∠ s on str. line y = 99° (Merit) b. Lines AB and CD are parallel alt. ∠ s, // lines Lines FG and HI are parallel corr ∠ s // lines (Merit) ∠ LJK = 87° ∠ JKM = 93° ∠ LMK =88° ∠ RQP = 180 - x = 70° ∠ s on str. line ∠ PRQ = 180 - 2 x 70 Base ∠ , isos Δ = 40° (Achieved - Both angles correct) (Merit - Angles and explanations correct) (Achieved) 7. a. a = 4. a. ∠ s, // lines 13. a = 180 – 68 = 112° alt. 14. ∠ XYU ∠ s, // lines (Merit) = a and ∠ ZYU = b alt. ∠ s, // lines ∠ XYZ = ∠ XYU + ∠ ZYU = a + b vert opp ∠ s ∠ XYZ = c = a + b (Merit) GEOMETRIC REASONING ANGLES WITHIN CIRCLES SUMMARY The parts of a circle are shown below: Rules: Example: • Angles on the same arc are equal a=b (∠s on same arc) x = 35° (∠s on same arc) • The angle at the centre is equal to twice the angle at the circumference on the same arc 2c = d (∠ at centre) x = 43 x 2 = 86° (∠ at centre) • The angle in a semicircle is a right angle This is a special case of the above rule (∠ in semicircle) x = 90 – 72 = 18° (∠ in semicircle) • The angle where the radius meets the tangent is 90° (rad ⊥ tgt) x = 180 – 56 – 90 = 34° (rad ⊥ tgt) • Two tangents coming from the same point are equal (same length and angles) (tangs from a point) • The angle between a chord and a tangent equals the angle in the alternate segment. a = d, c = b. (∠ in alt seg) x = (180 – 52) / 2 = 64° (tangs from a point and base ∠ isos Δ) • Remember any chord forms an isosceles triangle with the centre. ABO and OBC are isosceles triangles. x = 180 – 2 x 34 = 112° ( ∠ sum isos Δ) • Opposite angles in a cyclic quadrilateral add to 180° (opp ∠s cyclic quad) a + c = 180°, b + d = 180° • The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. a = b (ext ∠ cyclic quad) x = 32° y = 73° (∠ in alt seg) x = 72° (opp ∠s cyclic quad) x = 96° (ext ∠ cyclic quad) For a complete tutorial on this topic visit www.learncoach.co.nz PAGE 119 OLD NCEA QUESTIONS 1. DAC is a tangent. O is the centre of the circle. Calculate the size of angle DAB, explaining the reason for each step of your answer. 2. A, B, and C are points on the circumference of the circle. O is the centre of the circle. AB is parallel to OC. Angle CAO = x°. 5. EFGD is a cyclic quadrilateral with angle EDG = 82°. O is the centre of the circle. Find the size of angle HFJ. Explain your geometric reasoning clearly and logically. 6. a. Calculate the size of angle ACB in terms of x. 3. For the diagram above: a. Find the size of angle reflex COA, x. Explain your reasoning. b. Find the size of angle DCO, y. Give geometric reasons for each step in your solution. 4. b. The points A, Q, Z, N lie on the circumference of a circle centre O. AQ is parallel to NZ. Find the size of angle NZQ, in terms of x. Explain your geometric reasoning clearly and logically. For the diagram below, prove that angle C + angle A = 180°. What angle properties does a cyclic parallelogram have? Explain your answer with geometric reasoning. Use the blank diagram above (where O is the centre of the circle) if you wish. Give geometric reasons for each step in your solution. PAGE 120 GEOMETRIC REASONING PRACTICE QUESTIONS You must give a geometric reason for each step leading to your answer in every question. 11. 7. In the diagram, WX and YZ are parallel. WZ and XZ are equal length. Angle WXZ = 37°. Calculate the size of angle YXZ. In the diagram above, AB is a tangent to the circle and TD is a diameter of the circle. ∠ABC = 42°. Calculate the size of ∠CED. 12. 8. In the diagram above, points A and T lie on a circle with centre C. TCF is a straight line. BF = BT. DE is a tangent that touches the circle at T. ∠TFB = 54°. Find the size of ∠ADT. 9. O is the centre of the circle. Angle BAC = 39°. Lines CB and BA are the same length. Line OB is perpendicular to line AC. Find the size of angle OCA. In the diagram P, Q, R and S lie on a circle, centre, O. TR is a tangent to the circle at R. Angle PQR = 29°. Angle SRT = 22°. Find the size of angle SPO. 13. ABE and ACF are two triangles. Angle BCD = 25°. Angle DEF = 38°. Find the size of angle FDE 14. 10. O is the centre of the circle. TQ and RQ are tangents to the circle. Angle RST is 42°. Lines SR and ST are the same length. Find the size of angle QRS. The corners of ABCD lie on a circle. The angle AOC is 132° and the angle FAB is 101°. Find the angle CED. PAGE 121 ANSWERS 4. NCEA Let c = angle C and a = angle A, then: ∠ DOB (reflex) = 2c 1. ∠ CAO = 90° ∠ AOC = 58° ∠ AOB = 122° ∠ OAB = 29° ∠ DAB = 61° base ∠ , isos Δ and 2. base ∠ , isos Δ ∠ sum Δ (Merit) 5. ∠ EFG = 180 - 82 = 98° ∠ HFJ= ∠ EFG = 98° opp ∠s cyclic quad vert opp ∠s (Merit) 6. a. ∠ sum Δ ∠ s at pt ∠ at centre alt ∠s, // lines ∠ s at pt 2a + 2c = 360° 2(a + c) = 360° 360 a+c = = 180° 2 ∠ sum Δ rad ⊥ tangent (Achieved - Two angles correct) (Merit - Two angles and explanations correct) (Excellence - Full answer with coherent steps and explanations) ∠ ACO = x Obtuse ∠ AOC = 180 - 2x Reflex ∠ AOC = 180 + 2x ∠ ABC = 90 + x ∠ CAB = x ∠ ACB = 90 - 2x ∠ at centre ∠ BOD (reflex) = 2a rad ⊥ tangent ∠ sum Δ ∠s on a line ∠ AQZ = 180 - ∠ ZNA opp ∠s cyclic quad ∠ NZQ = 180 - ∠ AQZ Co-int ∠s, // lines = 180 - x b. = 180 - (180 - x) = x° (Merit) (Achieved - Two angles correct) (Merit - Two angles and explanations correct) (Excellence - Full answer with coherent steps and explanations) 3. a. ∠ AOC = 2 x 72 = 144° ∠ at centre ∠ s at pt x = Reflex ∠ COA = 360 - 144 Obtuse = 216° b. (Merit) Draw line segment OD. ODA is an isosceles triangle: ∠ AOC = 38° base ∠ , isos Δ ∠ ODC = 72 - 38 = 34° y = ∠ DCO = 34° base ∠ , isos Δ PAGE 122 (Excellence) Let ANZQ be a cyclic parallelogram. From (a) above we know that: ∠ ZNA = ∠ NZQ and ∠ NAQ = ∠ AQZ AN and QZ are also parallel so we also know: ∠ ZNA = ∠ NAQ and ∠ AQZ= ∠ NZQ Therefore all interior angles are equal. Since the interior angles of a quadrilateral add to (4 - 2) x 180 = 360°, each interior angle is 90°. Therefore a cyclic parallelogram must be a rectangle. (Excellence) GEOMETRIC REASONING PRACTICE 7. ∠ BTD = 90° ∠ BDT = 90 – 42 = 48° ∠ CET = 48° ∠ DET = 90° ∠ CED = ∠ DET - ∠ CET rad ⊥ tangent 12. ∠ sum Δ ∠ on same arc ∠ in a semi-circle (Excellence) h = 90 – 48 = 42° 8. = 83° x= ∠ RSO = 42 / 2 = 21° ∠ SRO = 21° ∠ QRS = 90 + 21 = 111° ∠ DBC = 180 – 25 – x ∠ sum Δ ∠ DBA = 180 –(155 – x) ∠s on a line opp Solve: 38° + x + 25° + x = 180° 2x = 117° x = 58.5° 14. symmetry base ∠ s, isos Δ rad ⊥ tangent (Excellence)h 11. ∠ XZY = 37° ∠ XZW = (180 – 37) / 2 = 71.5° ∠ WZY = 71.5 + 37 = 108.5° ∠ WXY = 180 - 108.5 = 71.5° ∠ YXZ = 71.5 - 37 = 34.5° ∠s on a line ∠ DFA + ∠ DBA = 180° (Excellence) h 10. ∠ DFA = 180 –(142 – x) = 25° + x base ∠ , isos Δ ∠ s, // lines base ∠ s, isos Δ sum adj ∠ opp ∠s cyclic alt ∠s ∠ sum Δ = 155 – x base ∠ , isos Δ vert opp ∠ DFE = 180 – 38 – x = 38° + x ∠ at centre = 51 - 39 = 12° ∠ FDE = ∠ BDC = 142 – x (Excellence) h ∠ BOC = 39 x 2 = 78° ∠ OCB = (180 – 78) / 2 = 51° ∠ BCA = 39° ∠ OCA = ∠ OCB- ∠ BCA (Excellence) h 13. ∠ ABT = 180 - 2 x 54 = 72° ∠ sum isos Δ ∠ at centre ∠ ACT = 72 x 2 = 144° ∠ ATF = (180 – 144) / 2 = 18° base ∠ , isos Δ ∠ ADT = 90 - 18 = 72° Rad ⊥ tan 9. ∠ at centre ∠ POR = 2 x 29 = 58° ∠ PSR = 180 – 29 = 151° opp ∠s cyclic quad ∠ SRO = 90 – 22 = 68° rad ⊥ tangent ∠ SPO = 360 – 151 - 68 - 58 ∠ sum quad ∠ ABC = 132 / 2 = 66° ∠ CDE = 66° ∠ BAD = 180 –101 = 79° ∠ DCE = 79° ∠ CED = 180 – 66 – 79 =35° ∠ s cyclic quad (Excellence) ∠ at centre ext ∠ s cyclic quad ∠s on a line ext ∠ s cyclic quad ∠ sum Δ (Excellence) h quad (Excellence) h PAGE 123