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NCEA LEVEL 1 MATHEMATICS Part 5 - AS90153 Geometric Reasoning QUESTIONS & ANSWERS MAHOBE Published by Mahobe Resources (NZ) Ltd Distributed free at www.mathscentre.co.nz 2 NCEA Level 1 Mathematics, Questions & Answers Part 5 - AS90151 Geometric Reasoning Contributors: Robert Cole, Mohsen Davoudi, Kim Freeman, Dr Sophia Huang, Tara Kelly. This edition is Part 5 of a 6 Part eBook series designed to help you study towards NCEA. Published in 2009 by: Mahobe Resources (NZ) Ltd P.O. Box 109-760 Newmarket, Auckland New Zealand www.mahobe.co.nz www.mathscentre.co.nz © Mahobe Resources (NZ) Ltd ISBN(13) 9781877489075 This eBook has been provided by Mahobe Resources (NZ) Ltd to The New Zealand Centre of Mathematics. School teachers, University lecturers, and their students are able to freely download this book from The New Zealand Centre of Mathematics website www.mathscentre.co.nz. Electronic copies of the complete eBook may not be copied or distributed. Students have permission to print one copy for their personal use. Any photocopying by teachers must be for training or educational purposes and must be recorded and carried out in accordance with Copyright Licensing Ltd guidelines. The content presented within the book represents the views of the publisher and his contributors as at the date of publication. Because of the rate with which conditions change, the publisher and his contributors reserve the right to alter and update the contents of the book at any time based on the new conditions. This eBook is for informational purposes only and the publisher and his contributors do not accept any responsibilities for any liabilities resulting from the use of the information within. While every attempt has been made to verify the content provided, neither the publisher nor his contributors and partners assume any responsibility for errors, inaccuracies or omissions. All rights reserved. All the views expressed in this book are those of the author. The questions and suggested answers are the responsibility of the author and have not been moderated for use in NCEA examinations. 3 NCEA Level 1 Mathematics - Questions & Answers Contents About this Book 5 Geometric Reasoning 8 More Geometric Reasoning 9 Geometric Reasoning Example 10 Circle Geometry 11 Circle Geometry Example 12 Geometric Reasoning Achievement Example 13 Exercises 14 Geometric Reasoning - Merit Example 17 Geometric Reasoning - Excellence 21 Sample Exam 25 Answers 31 YEAR 11 MATHEMATICS MAHOBE 4 MAHOBE YEAR 11 MATHEMATICS 5 About This Book Q&A eResources are recognised as the leading study guides for NCEA. Each freely available title has been compiled by a team of experienced educators to meet the study and revision needs of NCEA students. They are proving to be valuable resources in the hands of students who want to work ahead of their regular class programme. They also serve as effective revision programmes in the run up to the final examinations. This book carefully explains the mathematical concepts that will be tested in NCEA then illustrates them with Achievement, Merit and Excellence examplars. It allows students to practise on NCEA-type questions and provides detailed solutions. After working through this programme, all students will be well prepared for their final assessments. 4cm The student who wrote the above answer on a recent assessment paper did not use a Q&A Level 1 Mathematics eResource. YEAR 11 MATHEMATICS MAHOBE 6 YOU BEAUTY Don’t FRY this summer get RAD! RAD is not just a suncreen but a vitamin enhanced lotion that will protect from both UVA and UVB radiation and will nourish and maintain a natural youthful and beautiful skin. Purchase Environ RAD online at: www.youbeauty.co.nz A SAFE TAN can also be a natural looking tan! For those who need a safe tan before summer we recommend Environ Self Tan. This is a moisturizing gel that leaves the skin moist and builds up a light, natural and healthy looking tan. Results will be noticeable after 2 hours. Purchase Environ Self Tan online at: www.youbeauty.co.nz 7 MATHEMATICS 1.9 - AS90153 Geometric reasoning Always understand what the examiner wants! A past examination answer is shown below. The student who wrote this answer on a recent assessment paper did not use a Q&A Level 1 Mathematics eResource. Name the parts of a circle Santa lives here Intruding square Inside Circle Circle border YEAR 11 MATHEMATICS Running round in circles Outside Circle Bottom of circle MAHOBE 8 Geometric Reasoning The Facts to Learn Angles in a triangle add up to 180°. a + b + c = 180° Angles on a straight line add up to 180°. a + b + c = 180° Angles in a quadrilateral add up to 360°. a + b + c + d = 360° A cyclic quadrilateral has all its vertices (corners) touching a circle. Opposite angles of a cyclic quadrilateral add up to 180° a + c = 180°, d + b = 180° Angles round a point add up to 360°. a + b + c + d = 360° Exterior angles of a triangle = sum of the opposite interior angles a+b=d 60° 60° Isosceles triangle 2 sides the same length 2 angles the same size MAHOBE 60° Equilateral triangle 3 sides the same length 3 angles the same size (60°) YEAR 11 MATHEMATICS 9 More Geometric Reasoning Vertically opposite angles are equal. (Even when parallel lines are involved). b° b° « « b° b° b° b° « « « « Alternate angles (Z angles) are equal. « b° Corresponding angles (F angles) are equal. « Co-interior angles (C angles) add to 180°. Exterior angles of a polygon add up to 360°. Interior angles Exterior angle The sum of the interior angles of a polygon = (n - 2) × 180º (Where n = number of sides) This formula comes from splitting a polygon into triangles by using diagonals. There will always be 2 less triangles then there are sides. When given a geometry problem don’t just concentrate on the angle you have been asked to find. Try and find all the obvious angles according to the rules that you can remember. Sooner or later the answer you need will be found. YEAR 11 MATHEMATICS MAHOBE 10 Geometric Reasoning Example Find all the other angles in the diagram below. B C › 30° A 30º 15º › F E Triangle FAB is isosceles therefore FAB = D FBA = 75° AB and FC are parallel. ABFC is a Z-shape. Therefore if FBA = 75° then BFC = 75° (alternate or “Z” angles) EFC = 60° (straight line angles sum to 180°) BC and FE are parallel, therefore FCB = 60° (alternate or “Z” angles) FBC = 45° (triangle angles sum to 180°) FEC = 90° (triangle angles sum to 180°) Angles on a straight line mean that CED = 90° (straight line angles sum to 180°) If needed extend all the lines on the diagram. This can sometimes help you see all the related angles - especially when parallel lines are concerned. 45° 60° 75° 75° 60° 45° 90° › 60° 30° 30° 60° 90° 75° 30º 15º 60° MAHOBE 75° 60° 75° 30º 15º › 90° 90° 90° 90° YEAR 11 MATHEMATICS 11 tangent Circle Geometry radius Angle in a semicircle = 90º. Tangent and radius meet at 90º. bº A triangle formed using two radii is an isosceles triangle. bº The angle at the centre is twice the angle at the circumference. 2aº Angles in the same segment are equal. i.e angles formed at the circumference by chords are equal. A B C ACB = 61° When writing angles use three letters. The middle letter is where the angle is. YEAR 11 MATHEMATICS MAHOBE 12 Circle Geometry Examples 1. In the diagram below, O is the centre of the circle. Find the angles: U, V, W. The diagram is not drawn to scale. B Find all the obvious angles. A Angles in the same segment CAD = V O 70º CBD = 70º W Angles in a semicircle = 90º ABC = 90º, U C D DAB = 90º Isosceles Triangles. Sides are radii. AOD, BOC, BOA are all isosceles triangles. BDA = 70º, CAB = 20º and V = 20º (Angle in a semicircle, angles in an isosceles triangle.) Angle U = 70º (identical isosceles triangle to AOD) BOC and AOD = 40º (angles in a triangle = 180º) W = 140º (straight line = 180º or isosceles triangle AOB) 2. Find, with geometric reasons, the size of angle dº. B AB and CD are parallel lines. 4x + 5x add to 180º. (co-interior angles.) 9x = 180º therefore x = 20º. A 5x D If x = 20º, 5x = 100º. This means that d = 80º. Opposite angles in a cyclic 4x C triangle are supplementary. MAHOBE YEAR 11 MATHEMATICS 13 Geometric Reasoning - Achievement Examples 1. The diagram shows the side view of a bridge over a stream. B 125º A AB = BC, BCD = 125º. Calculate the size of angle BCA = 180° - 125° C D ABC (supplementary angles sum to 180°) = 55° BAC = 55° (base angles of an isosceles triangle are equal) ABC = 70° (triangle angles sum to 180°) The diagram below shows the side view of a water slide. Find the angle CAB the angle between the ladder and the slide. A Slid e Support 2. 126° C E 160° Level Ground B ACE = 54° as angles on a straight line sum to 180° ABE = 20° as angles on a straight line sum to 180° CAB = 180° - 20° - 54° = 106° as angles in a triangle sum to 180° YEAR 11 MATHEMATICS MAHOBE 14 Exercises 1. a. A Calculate the size of angle CDE . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. C 67º 42º E B . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. D b. H I Calculate the size of angle HIJ. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. 52º K . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. J c. Calculate the size of angle c. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. 84º . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. cº d. Calculate the size of VUW . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. U V . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. X 55º Z MAHOBE W Y . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. Note: Diagrams are not drawn to scale. YEAR 11 MATHEMATICS 15 2. Shape A is a regular pentagon. Shape H is a regular heptagon. Both shapes share the side AB. a. Calculate the size of the angle CBD. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. A Shape P C . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. Shape H B . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . .. D The diagram below gives the angles of a playground slide. b. Calculate the size of the shaded angle XYZ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35º U Y 15º Z W Two cyclic quadrilaterals are shown below. The size of angle c. Find the size of the angles OMN, and OPN. d. How do we know that JK is parallel to OP? JKN is 95º. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . M J O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95º K N YEAR 11 MATHEMATICS P MAHOBE 16 3. a. PQRS is a cyclic quadrilateral. O is the centre of the circle. Find the size of angles PRQ and PSQ. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . R .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . 120º P Q .. . . . . . . . . . . . . . . . . . . . . . . . . . . . S b. In the diagram below (which is not drawn to scale), C is the centre of the circle, and XY is a tangent to the circle. Angle ABY = 70º. Complete the sentences below to find, in 4 logical steps the angle which equals 50º. A C 70º X B Y Angle Size Reason XBC 90º .......................................... CBA ...... Adjacent angles on a line add up to 180º. CAB 20º .......................................... XCB 40º .......................................... ...... 50º .......................................... MAHOBE YEAR 11 MATHEMATICS 17 Geometric Reasoning - Merit Examples 1. The diagram below shows a pentagram - a five sided star. The shaded shape formed by ABCDE is a regular pentagon. Z Calculate the angle VCW. Give geometric reasons with each of your calculations. The total of all the interior angles B A Y V of a regular pentagon can be calculated by the formula: C E (n-2) × 180 where n is the D number of sides (5). Therefore 3 × 180° = 540° X W Each angle = 108º (as 540° ÷ 5 = 108°). VCW = 108º as it is vertically opposite. N Axis of symmetry This means that P O 42º 2. M R Q The figure above shows a semicircle with centre R. RO is an axis of symmetry. The angle NMP is 42º. Find the angle NQM. Support your answer with geometric reasons. QNM = MPQ = 90º (angle in a semicircle) This means that NOP and NOM = 48º (angles in a triangle sum to 180º) MOQ = 132º (angles around a point sum to 360º) (Vertically opposite angles are equal) OQM = OMQ = 24º (base angles of an isosceles triangle) YEAR 11 MATHEMATICS MAHOBE 18 4. In the diagram below the points A, B, C and D lie on a circle with centre O. Calculate the size of angle BCD. Give geometric reasons for your answer. B ....... . . . . . . . . . . . . . . . .. . . . . . . 47º ....... . . . . . . . . . . . . . . . .. . . . . . . A ....... . . . . . . . . . . . . . . . .. . . . . . . 58º O ....... . . . . . . . . . . . . . . . .. . . . . . . ....... . . . . . . . . . . . . . . . .. . . . . . . D ....... . . . . . . . . . . . . . . . .. . . . . . . ....... . . . . . . . . . . . . . . . .. . . . . . . 5. The diagram below shows a circle with two tangents at W and X. O is the centre of the circle. Calculate the angle WYX. Give geometric reasons for your answer. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . O W 150º X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . MAHOBE YEAR 11 MATHEMATICS C 19 6. In the figure below ABDC is an isosceles trapezium. Find the length of BD. Give geometric reasons for your answer. 12 cm E A B 10 cm C D 15 cm ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .... 7. In the diagram below ABC is an isosceles triangle. O is the centre of the circle. The angle OBC = 31º a. Calculate the size of BAC. b. Calculate the size of ODC. Give reasons for your answers. A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... O B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... 31º C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... YEAR 11 MATHEMATICS MAHOBE 20 8. The diagram below is NOT drawn to scale. O is the centre of the circle. P, K, Q and N are points on the circumference. QT is the tangent to the circle at Q. Angle POQ = 108° Angle NPQ = 35° Calculate the angles x and y. P 35º N 108º O K T Q ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .... MAHOBE YEAR 11 MATHEMATICS 21 Geometric Reasoning - Excellence Examples 1. The diagram below shows a circle with points A, B and C on the circumference. Centre of the circle is O. B Calculate the size of CAB. Give geometric reasons for your answer. C 32º A Triangle AOC is isosceles (OA = OC) OAC = 32º O (base angles of an isosceles triangle) AOC = 116º (angle sum of a triangle = 180º) The reflex angle about the centre is 360º - 116º = 244º ABC = 122º (angle at the circumference = ½ angle at centre) BCA = 32º. (alternate angles are equal i.e. BCA and OAC) BAC = 26º (angle sum of a triangle = 180º) 2. In the figure below, O is the centre of the circle. XZ is the diameter. Prove that c = 2b. Give geometric reasons for your answer. XYZ = 90º (angle in a semicircle) Y OX, OZ and OY are radii therefore there Z are two isosceles triangles formed, OYZ and OYX. This means that O and OYZ = aº OXY = bº. b = 90 - a (angle in a semicircle) X c = 180 - 2a (angles in a triangle sum) Therefore c = 2b YEAR 11 MATHEMATICS MAHOBE 22 9. In the figure below, line KM forms an axis of symmetry. Length QN = Length QK = Length QL. Angle NQM = 120°. Angle NMQ = 30°. Prove that the quadrilateral KLMN is cyclic. K L N Q 120º 30º M ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... MAHOBE YEAR 11 MATHEMATICS 23 10. The circle below has a centre O and a tangent XY at point B. The points A and D lie on the circle. Prove that YBC = BAC X B A Y O C ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... ....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ... YEAR 11 MATHEMATICS MAHOBE 24 MAHOBE YEAR 11 MATHEMATICS 25 MAHOBE Level 1 Mathematics - Sample Exam AS90153 Geometric Reasoning Published by Mahobe Resources (NZ) Ltd Distributed free at www.mathscentre.co.nz YEAR 11 MATHEMATICS MAHOBE 26 MAHOBE YEAR 11 MATHEMATICS 27 You are advised to spend 25 minutes answering the questions in this section. QUESTION ONE Pauline’s garden table is illustrated in the diagram. A ABEF is a straight and horizontal table top. XY is an axis of symmetry. Angle BED = 110° Diagram is not to scale. X B E F 110° Y C D Calculate the size of angle ABC giving reasons for your answers. ................................................................ ................................................................ ................................................................ ................................................................ ..................................... QUESTION TWO Stan’s step-ladder is shown in the diagram. L Angle LMN = 122° KL = LN Calculate the size of the inside angle of the top of the step-ladder KLM giving reasons. ..................................... ..................................... ..................................... 122° . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .K YEAR 11 MATHEMATICS M MAHOBE N 28 QUESTION THREE The diagram shows a regular pentagon inside a circle. A B E 36° C D Calculate the size of BEA and DCB giving reasons for your answers. ................................................................ ................................................................ ................................................................ ................................................................ QUESTION FOUR In the diagram, PQRO is a rhombus QR is parallel to ST. Angle TSR = 63°. Calculate the angle of ROT. Give a geometric reason for each step leading to your answer. Q S P R W O 63° T ................................................................ ................................................................ MAHOBE YEAR 11 MATHEMATICS 29 QUESTION FIVE In the diagram V, W, X and Y lie on a circle, centre O. ZW is a tangent to the circle at W. Angle YVW = 65° Angle XWZ = x° V 65° Find the size of angle OYX. Y O X x° W Z ................................................................ ................................................................ ................................................................ ................................................................ ................................................................ ................................................................ ................................................................ ................................................................ ................................................................ YEAR 11 MATHEMATICS MAHOBE 30 QUESTION SIX KLM and MNO are two triangles. Angle KLM = 32° Angle MNO = 24° L K J 32° M O Find the size of angle OMN. 45° N Give geometric reasons for each step leading to your answer. ................................................................ ................................................................ ................................................................ ................................................................ ................................................................ ................................................................ ................................................................ ................................................................ MAHOBE YEAR 11 MATHEMATICS 31 MAHOBE The Answers YEAR 11 MATHEMATICS MAHOBE 32 The Answers Page 14 1. A a. d. 67º E B 55º 70º C 42º 67º U 71º V 55º D X 55º CED = 67° - vertically opposite W Y Z Angles in a triangle sum to 180° 180° - 67° - 42° = 71° See diagram above for the corresponding angles. The 70° is b. then calculated from the isosceles triangle WUV. 52º H I 52º 64º Page 15 2. a. Sum interior angles = (n-2) × 180º Sum interior angles of pentagon 52º K (5 - 2) × 180º = 540º 64º Each angle = 108º J Sum interior angles heptagon (7-2) × 180º = 900º Each angle = 128.57º Angle IHJ = 52° alternate angles, Angle CBD = 360 - (108 + 128.57) parallel lines. = 123.43º HIJ = 64° - Angles in an isosceles b. triangle. X c. 84º 75º 35º 96º 84º 84º U 90º Y 15º Z W Angle UYW = 90º right angled rectangle. See the diagram above for all the corresponding angles. c = 96° angles on a straight line MAHOBE The angle ZYW = 75º angle sum of a triangle. XYZ = 360º - (75º + 90º + 35º) = 160º YEAR 11 MATHEMATICS 33 2. C. M J XBC = Angle between a tangent b. Page 15 (cont) 85º and a radius O CBA = 20° 95º CAB = Base angles of an isosceles triangles are equal K 85º 95º N XCB = 40° angles on straight line P CXB = Angles in a triangle sum to 180° NMJ = 85º opposite angles of a cyclic triangle sum to 180º Page 18 This means angle NMO = 95º 4. B 47º angles on a straight line and NPO = 85º as it is the opposite 47º 61º A angle in the cyclic triangle. d. JK is parallel to OP as all the O 72º C 61º angles indicate that they are D corresponding. R Page 16 86º 58º OB, OA and OD are all radii. This means that there are 2 isosceles 60º triangles AOB (47°, 47°, 86°) and AOD (61°, 61°, 58°) 120º P BAO + DAO: 61° + 47° = 108° Q 120º BCD S 3. a. = 72° opposite angles cyclic quad. or BOA + AOD: 86° + 58° = 144° BCD = 72° angle at circumference PRQ = 60º angle at the centre PSQ = 120º opposite angles in a 5. cyclic quadrilateral. B. W O 150º X 90º 90º Isosceles triangle (2 radii) A C 20º 140º 30º 20º 70º X Y Y B 90º angle between tangent and radius OWY AND OXY are 90° as the angle between the tangent and radius are perpendicular. WYX = 30° interior angles of a quadrilateral sum to 360° YEAR 11 MATHEMATICS MAHOBE 34 Page 19 Page 20 E 6. P 12 cm 35º N A 126º B 10 cm C T 108º O 54º K 19º 35º D 15 cm 36º 36º Q EAB and ECD are similar triangles. EA 12 This means = 10 15 8. Lines OP and OQ are radii. This means that EA = 8 cm This means that OPQ is an isosceles BD = CE - EA triangle and the base angles OPQ and (CE = DE, AE = BE) = 12 cm - 8 cm OQP = 36° = 4 cm A 7. Angle PKQ = 54° as it is angle at the circumference (½ POQ) Angle x (PNQ) = 126° as opposite angles 59° 118º O B 31º 31º C of a cyclic quadrilateral sum to 180°. OQT is the angle between a radius and a tangent = 90°. NQP = 19° angle sum of a triangle D NQT (y° ) = 90° - 36° - 19° = 35° BCO = 31° base angles isosceles triangle. COB = 118° angle sum of a triangle BAC = 59° angle at circumference is half that at the centre. BAC is isosceles triangle BDC is an isosceles triangle Angle BDC = 121° opposite angles cyclic quadrilateral. ODC is half this 60.5°. MAHOBE YEAR 11 MATHEMATICS 35 K Page 22 L Page 27 60º 60º 60º 60º 60º 60º Q 30º N Question One FED = 70° angles on a straight line 30º ABC = 70° symmetry through XY 120º 120º Question Two LMK = 58° angles on a straight line LKM = 58° as triangle is isosceles 30º30º KLM = 64° angle sum of a triangle Page 28 M Question Three 9. KQN = 60° angles on a straight line BEA = 36° angles in the same segment QKN = 60° isosceles triangle QN = QK DCB = 108° interior angles of a regular LKN = 120° line KM is a line of pentagon. symmetry Question Four LKN = 2 × QKN OTS is an isosceles triangle LMN = 60° due to line of symmetry KM STO = 63° base angles isosceles triangle LKN + LMN = 180° ROT = 180° - 2 × 63° = 54° KLMN = a cyclic triangle (opposite angles of a cyclic quadrilateral sum to 180°) Page 29 Page 23 Question Five X OW is at 90° to WZ 10. It is a radius and tangent B This means that 0WX = 90° - x ß A O Y YXW = 115° opposite angles of a cyclic quadrilateral WOY = 130° angle at centre is twice the C angle at the circumference OB is a radius and the tangent XY is 90° Angle sum of a quadrilateral is 360° to the tangent. OYX = 360 - 130 - 115 - (90 - x) Let YBC = ß = (25 + x)° OBC = 90 - ß tangent 90º to radius OCB = 90 - ß base angles of isos triangle OB = OC BOC Page 30 Question Six = 180 - (90 - ß) - (90 - ß) Let KML = x°, then OMN = x° as they are = 2ß vertically opposite. BAC = ß angle at circumference is half JOM = x + 45°, JKM = x + 32° as the that at the centre exterior angles of a triangle = the sum of BAC = ß, YBC = ß the two opposite interior angles Therefore YBC = BAC JOM + JKM = 180° (cyclic quadrilateral) x° + 45° + x° + 32° = 180° x = 51.5° YEAR 11 MATHEMATICS MAHOBE 36 MAHOBE YEAR 11 MATHEMATICS 37 YEAR 11 MATHEMATICS MAHOBE 38 The Mahobe RedBack Calculator The NZ Centre of Mathematics recommends the Mahobe RedBack scientific calculator for NCEA. This calculator has advanced features, such as equation solving, to help take the sweat out of such calculations. Purchase it direct from the Mahobe website www.mahobe.co.nz. 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Q&A Science is $19.95 and can be purchased from Whitcoulls, Paper Plus or from the Mahobe website www.mahobe.co.nz MAHOBE YEAR 11 MATHEMATICS