Grade 12 Mathematics Analytical Geometry Q&A

Total Page:16

File Type:pdf, Size:1020Kb

Grade 12 Mathematics Analytical Geometry Q&A Gr 12 Maths – Analytical Geometry GR 12 MATHS AANNAALLYYTTIICCAALL GGEEOOMMEETTRRYY QUESTIONS and ANSWERS Work through the Grade 11 Analytical downloads first to ensure your foundation is solid before attempting Grade 12 Analytical Geometry which includes circles. We wish you the best of luck for your exams. From The Answer Series team Copyright © The Answer Gr 12 Maths – Analytical Geometry: Questions y 15. PRQ is a tangent to the circle with 18. In the accompanying figure, y centre O at the point R(6; -2). the circle with the centre at the ANALYTICAL GEOMETRY origin passes through the point Q O x 15.1 Calculate the equation of Q ()-2 3; -2 T and cuts the P 15.1.1 the circle. (2) R(6; - 2) O x QUESTIONS x-axis and y-axis at P and S N 15.1.2 the tangent. (5) P respectively. T( -2 3 ; -2) OQR,ˆ 15.2 Calculate the size of 18.1 Show that the equation of S 2 2 GRADE 12 rounded off to one decimal digit. (3) the circle is x + y = 16 is. (3) 15.3 Calculate a if (2; a) is a point 18.2 Determine the coordinates of P and S. (2) ˆ 2 2 See why ABC = 90º in these 2 cases ... on the circle x + y = 40. (3) 18.3 Determine the equation of the line PS. (3) B • A diameter of a circle makes an angle of 90º 15.4 Determine the coordinates of Q. (2) 18.4 If the straight line QN which is perpendicular to PS at the circumference â The hypotenuse of A C passes through the origin, write down its equation. (2) Δ ? diameter a right-angled ABC is the diameter of ABC ! y 16. The straight line y = x - 1 cuts 18.5 Calculate the coordinates of N if QN meets PS at N. (4) 2 2 • A tangent to a circle is A the circle x + y = 25 at A and B. radius 18.6 Find the coordinates of a point E, which is the perpendicular to the radius B diameter B AB cuts the x-axis at S. The circle reflection of point T with respect to the line y = - x. (2) (or diameter) drawn to the point tangent O x intersects the x-axis at T. T S of contact, i.e. BC ⊥ AB. C 19. A circle with its centre O at the origin passes through the Calculate point P2() 3;2. 16.1 the coordinates of A A and B. (6) 19.1 Determine the equation of the circle. (2) CIRCLES – CENTRE AT THE ORIGIN 16.2 the equation of the circle with centre O and 19.2 Determine the gradient of line OP. (1) radius OS. (3) 19.3 Hence, without using a calculator, determine the size 16.3 the equation of the tangent to the circle at T. (2) Equation: x2 + y2 = r2 of the angle between OP and the positive x-axis. (2) y 19.4 Determine the equation of the tangent to the circle 17. In the figure the circle, with centre P2() 3;2 14. The origin O is the centre of y S(1; a) at the point in the form y = ax + q. (4) at the origin O, passes through the the circle in the figure. P(x; y) point K(- 3; 1). K(- 3; 1) R(k; 1) x O 20. In the figure, a circle is defined y P(x; y) and Q(3; -4) are T T O x ST is a tangent to the circle 2 2 two points on the circle and by the equation x + y = 9, A at S(1; a). Q E POQ is a straight line. two secants AB and BC meet at Q(3; - 4) 17.1 Determine the radius of the circle. (2) R is the point (k; 1) and RQ is a common point B(k; 0), a B(k; 0) a tangent to the circle. 17.2 Write down the equation of the circle. (1) second line DE passes through O x the origin and is perpendicular to Determine (leave answers in surd form if necessary): 17.3 Show clearly that the value of a = 3. (2) D AB with E a point on the circle. 14.1 the equation of circle O. (2) 17.4 Write down the coordinates of a point Q which is C symmetrical to the point S with respect to the x-axis. (2) 20.1 Write down the value of k. (1) 14.2 the length of QT. (3) 20.2 Determine the length of AB. (2) 14.3 the length of PQ. (4) 17.5 Write down the gradient of OS. (1) ˆ 14.4 the equation of OQ. (2) 17.6 Hence, show that the equation of the tangent ST is 20.3 Calculate BAC, giving reasons. (2) given by + 3y = 10. (3) 14.5 the coordinates of P. (2) x 20.4 Calculate the area of ΔABC. (2) 14.6 the gradient of QR. (1) 17.7 Write down the coordinates of point T. (2) 20.5 Determine the equation of line DE. (2) 14.7 the equation of QR in the form y = mx + c. (3) 17.8 Hence, prove that TQ is a tangent to the circle. (3) 20.6 Hence, without using a calculator, determine the 14.8 the value of k. (3) 17.9 Write down the equation of another tangent to the coordinates of D. (4) circle (not shown in the figure above) which is also 14.9 whether the point S(3; 2) lies inside, outside or on the circle. Give a reason for your answer. (2) parallel to ST. (2) 1 Copyright © The Answer Gr 12 Maths – Analytical Geometry: Questions 28. The point M(2; 1) is the midpoint of chord PQ of the circle. 32. The vertices A and B of ΔABC lie y 2 2 CIRCLES – ANY CENTRE on the -axis. C x + y - x - 2y - 5 = 0 x DE 28.1 Determine the coordinates of the centre, A, of The centre of the inscribed circle Δ the circle. (2) of ABC is Q(4; 5). Q 2 2 2 Equation: (x - a) + (y - b) = r The circle touches AC at the 28.2 Determine the radius of the circle. (2) x point D(0; 8) and BC at the AB 28.3 If chord PQ ⊥ AM, determine the equation of point E(8; 8). 21. Determine the equation of the circle with chord PQ. (4) Determine: 21.1 centre (2; 3) and radius 7 units (3) 28.4 Calculate the coordinates of P and Q. (5) 32.1 the equation of the circle. (3) 21.2 centre (-1; 4) and radius 5 units. (3) 28.5 Determine the equation of the tangent to the 32.2 the equation of BC. (3) circle at the point (2; 3). (4) 32.3 the gradient of AC. (2) 22.1 Convert the following general form of the equation of a 2 2 32.4 Hence, or otherwise, prove that AC = BC. circle to the standard form: x - 6x + y + 8y - 11 = 0 (3) 29. A(3; -5) and B(1; 3) are two points in a Cartesian plane. Give reasons for your answer. (5) 22.2 Write down the centre and the radius of the circle. (3) NB: DRAW A PICTURE ! 33. A triangle with vertices A(-1; 7), B(8; 4) and C(7; 1) is given. ˆ 23. Determine the equation of the tangent that touches the 29.1 Calculate the length of AB and leave the 33.1 Show that ABC = 90º. (4) circle defined by: answer in simplified surd from if necessary. (2) 2 2 33.2 Determine the area of the triangle. (4) 23.1 (x - 1) + (y + 2) = 25 at the point (4; 2) (4) 29.2 Determine the equation of the circle with AB as 33.3 Determine the equation of the circle through 2 2 2 2 2 23.2 x - 2x + y + 4y = 5 at the point (-2; -1) (5) diameter in the form : (x - a) + (y - b) = r (4) A, B and C. (7) 29.3 Determine the equation of the tangent to the 24. A diameter AB of a circle with points A(-3; -2) and circle at A in the form: y = m + c. (5) Remember? x B(1; 4) is given. A tangent to a circle is perpendicular A radius to the radius (or diameter) drawn to 24.1 Determine the equation of the circle. (4) diameter B 30. The equation of a circle with radius 32 units is the point of contact, 24.2 Determine the equation of the tangent to the 2 2 tangent x + y - 6x + 2y - m = 0 i.e. ⊥ circle at A. (5) tangent radius. C 30.1 Determine the coordinates of the centre of 25.1 Prove that y = x + 7 is a tangent to the circle the circle. (4) y 2 2 34. In the diagram alongside, centre C x + y + 8x + 2y + 9 = 0. (6) 30.2 Determine the value of m. (3) of the circle lies on the straight line D 3x + 4y + 7 = 0. 25.2 Determine the point of contact of the tangent and the C P(0; 2) y The straight line cuts the circle circle in 25.1. (2) 31. A circle with centre M(-4; 2) has x N(- 2 ; y) at D and E(-1; -1). the points O(0; 0) and N(-2; y) on the circumference.
Recommended publications
  • Analytic Geometry
    Guide Study Georgia End-Of-Course Tests Georgia ANALYTIC GEOMETRY TABLE OF CONTENTS INTRODUCTION ...........................................................................................................5 HOW TO USE THE STUDY GUIDE ................................................................................6 OVERVIEW OF THE EOCT .........................................................................................8 PREPARING FOR THE EOCT ......................................................................................9 Study Skills ........................................................................................................9 Time Management .....................................................................................10 Organization ...............................................................................................10 Active Participation ...................................................................................11 Test-Taking Strategies .....................................................................................11 Suggested Strategies to Prepare for the EOCT ..........................................12 Suggested Strategies the Day before the EOCT ........................................13 Suggested Strategies the Morning of the EOCT ........................................13 Top 10 Suggested Strategies during the EOCT .........................................14 TEST CONTENT ........................................................................................................15
    [Show full text]
  • Lesson 3: Rectangles Inscribed in Circles
    NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 3 M5 GEOMETRY Lesson 3: Rectangles Inscribed in Circles Student Outcomes . Inscribe a rectangle in a circle. Understand the symmetries of inscribed rectangles across a diameter. Lesson Notes Have students use a compass and straightedge to locate the center of the circle provided. If necessary, remind students of their work in Module 1 on constructing a perpendicular to a segment and of their work in Lesson 1 in this module on Thales’ theorem. Standards addressed with this lesson are G-C.A.2 and G-C.A.3. Students should be made aware that figures are not drawn to scale. Classwork Scaffolding: Opening Exercise (9 minutes) Display steps to construct a perpendicular line at a point. Students follow the steps provided and use a compass and straightedge to find the center of a circle. This exercise reminds students about constructions previously . Draw a segment through the studied that are needed in this lesson and later in this module. point, and, using a compass, mark a point equidistant on Opening Exercise each side of the point. Using only a compass and straightedge, find the location of the center of the circle below. Label the endpoints of the Follow the steps provided. segment 퐴 and 퐵. Draw chord 푨푩̅̅̅̅. Draw circle 퐴 with center 퐴 . Construct a chord perpendicular to 푨푩̅̅̅̅ at and radius ̅퐴퐵̅̅̅. endpoint 푩. Draw circle 퐵 with center 퐵 . Mark the point of intersection of the perpendicular chord and the circle as point and radius ̅퐵퐴̅̅̅. 푪. Label the points of intersection .
    [Show full text]
  • Squaring the Circle a Case Study in the History of Mathematics the Problem
    Squaring the Circle A Case Study in the History of Mathematics The Problem Using only a compass and straightedge, construct for any given circle, a square with the same area as the circle. The general problem of constructing a square with the same area as a given figure is known as the Quadrature of that figure. So, we seek a quadrature of the circle. The Answer It has been known since 1822 that the quadrature of a circle with straightedge and compass is impossible. Notes: First of all we are not saying that a square of equal area does not exist. If the circle has area A, then a square with side √A clearly has the same area. Secondly, we are not saying that a quadrature of a circle is impossible, since it is possible, but not under the restriction of using only a straightedge and compass. Precursors It has been written, in many places, that the quadrature problem appears in one of the earliest extant mathematical sources, the Rhind Papyrus (~ 1650 B.C.). This is not really an accurate statement. If one means by the “quadrature of the circle” simply a quadrature by any means, then one is just asking for the determination of the area of a circle. This problem does appear in the Rhind Papyrus, but I consider it as just a precursor to the construction problem we are examining. The Rhind Papyrus The papyrus was found in Thebes (Luxor) in the ruins of a small building near the Ramesseum.1 It was purchased in 1858 in Egypt by the Scottish Egyptologist A.
    [Show full text]
  • 10-1 Study Guide and Intervention Circles and Circumference
    NAME _____________________________________________ DATE ____________________________ PERIOD _____________ 10-1 Study Guide and Intervention Circles and Circumference Segments in Circles A circle consists of all points in a plane that are a given distance, called the radius, from a given point called the center. A segment or line can intersect a circle in several ways. • A segment with endpoints that are at the center and on the circle is a radius. • A segment with endpoints on the circle is a chord. • A chord that passes through the circle’s center and made up of collinear radii is a diameter. chord: 퐴퐸̅̅̅̅, 퐵퐷̅̅̅̅ radius: 퐹퐵̅̅̅̅, 퐹퐶̅̅̅̅, 퐹퐷̅̅̅̅ For a circle that has radius r and diameter d, the following are true diameter: 퐵퐷̅̅̅̅ 푑 1 r = r = d d = 2r 2 2 Example a. Name the circle. The name of the circle is ⨀O. b. Name radii of the circle. 퐴푂̅̅̅̅, 퐵푂̅̅̅̅, 퐶푂̅̅̅̅, and 퐷푂̅̅̅̅ are radii. c. Name chords of the circle. 퐴퐵̅̅̅̅ and 퐶퐷̅̅̅̅ are chords. Exercises For Exercises 1-7, refer to 1. Name the circle. ⨀R 2. Name radii of the circle. 푹푨̅̅̅̅, 푹푩̅̅̅̅, 푹풀̅̅̅̅, and 푹푿̅̅̅̅ 3. Name chords of the circle. ̅푩풀̅̅̅, 푨푿̅̅̅̅, 푨푩̅̅̅̅, and 푿풀̅̅̅̅ 4. Name diameters of the circle. 푨푩̅̅̅̅, and 푿풀̅̅̅̅ 5. If AB = 18 millimeters, find AR. 9 mm 6. If RY = 10 inches, find AR and AB . AR = 10 in.; AB = 20 in. 7. Is 퐴퐵̅̅̅̅ ≅ 푋푌̅̅̅̅? Explain. Yes; all diameters of the same circle are congruent. Chapter 10 5 Glencoe Geometry NAME _____________________________________________ DATE ____________________________ PERIOD _____________ 10-1 Study Guide and Intervention (continued) Circles and Circumference Circumference The circumference of a circle is the distance around the circle.
    [Show full text]
  • 20. Geometry of the Circle (SC)
    20. GEOMETRY OF THE CIRCLE PARTS OF THE CIRCLE Segments When we speak of a circle we may be referring to the plane figure itself or the boundary of the shape, called the circumference. In solving problems involving the circle, we must be familiar with several theorems. In order to understand these theorems, we review the names given to parts of a circle. Diameter and chord The region that is encompassed between an arc and a chord is called a segment. The region between the chord and the minor arc is called the minor segment. The region between the chord and the major arc is called the major segment. If the chord is a diameter, then both segments are equal and are called semi-circles. The straight line joining any two points on the circle is called a chord. Sectors A diameter is a chord that passes through the center of the circle. It is, therefore, the longest possible chord of a circle. In the diagram, O is the center of the circle, AB is a diameter and PQ is also a chord. Arcs The region that is enclosed by any two radii and an arc is called a sector. If the region is bounded by the two radii and a minor arc, then it is called the minor sector. www.faspassmaths.comIf the region is bounded by two radii and the major arc, it is called the major sector. An arc of a circle is the part of the circumference of the circle that is cut off by a chord.
    [Show full text]
  • Geometry Course Outline
    GEOMETRY COURSE OUTLINE Content Area Formative Assessment # of Lessons Days G0 INTRO AND CONSTRUCTION 12 G-CO Congruence 12, 13 G1 BASIC DEFINITIONS AND RIGID MOTION Representing and 20 G-CO Congruence 1, 2, 3, 4, 5, 6, 7, 8 Combining Transformations Analyzing Congruency Proofs G2 GEOMETRIC RELATIONSHIPS AND PROPERTIES Evaluating Statements 15 G-CO Congruence 9, 10, 11 About Length and Area G-C Circles 3 Inscribing and Circumscribing Right Triangles G3 SIMILARITY Geometry Problems: 20 G-SRT Similarity, Right Triangles, and Trigonometry 1, 2, 3, Circles and Triangles 4, 5 Proofs of the Pythagorean Theorem M1 GEOMETRIC MODELING 1 Solving Geometry 7 G-MG Modeling with Geometry 1, 2, 3 Problems: Floodlights G4 COORDINATE GEOMETRY Finding Equations of 15 G-GPE Expressing Geometric Properties with Equations 4, 5, Parallel and 6, 7 Perpendicular Lines G5 CIRCLES AND CONICS Equations of Circles 1 15 G-C Circles 1, 2, 5 Equations of Circles 2 G-GPE Expressing Geometric Properties with Equations 1, 2 Sectors of Circles G6 GEOMETRIC MEASUREMENTS AND DIMENSIONS Evaluating Statements 15 G-GMD 1, 3, 4 About Enlargements (2D & 3D) 2D Representations of 3D Objects G7 TRIONOMETRIC RATIOS Calculating Volumes of 15 G-SRT Similarity, Right Triangles, and Trigonometry 6, 7, 8 Compound Objects M2 GEOMETRIC MODELING 2 Modeling: Rolling Cups 10 G-MG Modeling with Geometry 1, 2, 3 TOTAL: 144 HIGH SCHOOL OVERVIEW Algebra 1 Geometry Algebra 2 A0 Introduction G0 Introduction and A0 Introduction Construction A1 Modeling With Functions G1 Basic Definitions and Rigid
    [Show full text]
  • Find the Radius Or Diameter of the Circle with the Given Dimensions. 2
    8-1 Circumference Find the radius or diameter of the circle with the given dimensions. 2. d = 24 ft SOLUTION: The radius is 12 feet. ANSWER: 12 ft Find the circumference of the circle. Use 3.14 or for π. Round to the nearest tenth if necessary. 4. SOLUTION: The circumference of the circle is about 25.1 feet. ANSWER: 3.14 × 8 = 25.1 ft eSolutions Manual - Powered by Cognero Page 1 8-1 Circumference 6. SOLUTION: The circumference of the circle is about 22 miles. ANSWER: 8. The Belknap shield volcano is located in the Cascade Range in Oregon. The volcano is circular and has a diameter of 5 miles. What is the circumference of this volcano to the nearest tenth? SOLUTION: The circumference of the volcano is about 15.7 miles. ANSWER: 15.7 mi eSolutions Manual - Powered by Cognero Page 2 8-1 Circumference Copy and Solve Show your work on a separate piece of paper. Find the diameter given the circumference of the object. Use 3.14 for π. 10. a satellite dish with a circumference of 957.7 meters SOLUTION: The diameter of the satellite dish is 305 meters. ANSWER: 305 m 12. a nickel with a circumference of about 65.94 millimeters SOLUTION: The diameter of the nickel is 21 millimeters. ANSWER: 21 mm eSolutions Manual - Powered by Cognero Page 3 8-1 Circumference Find the distance around the figure. Use 3.14 for π. 14. SOLUTION: The distance around the figure is 5 + 5 or 10 feet plus of the circumference of a circle with a radius of 5 feet.
    [Show full text]
  • 3.1 the Robertson-Walker Metric
    M. Pettini: Introduction to Cosmology | Lecture 3 RELATIVISTIC COSMOLOGY 3.1 The Robertson-Walker Metric The appearance of objects at cosmological distances is affected by the curvature of spacetime through which light travels on its way to Earth. The most complete description of the geometrical properties of the Universe is provided by Einstein's general theory of relativity. In GR, the fundamental quantity is the metric which describes the geometry of spacetime. Let's look at the definition of a metric: in 3-D space we measure the distance along a curved path P between two points using the differential distance formula, or metric: (d`)2 = (dx)2 + (dy)2 + (dz)2 (3.1) and integrating along the path P (a line integral) to calculate the total distance: Z 2 q Z 2 q ∆` = (d`)2 = (dx)2 + (dy)2 + (dz)2 (3.2) 1 1 Similarly, to measure the interval along a curved wordline, W, connecting two events in spacetime with no mass present, we use the metric for flat spacetime: (ds)2 = (cdt)2 − (dx)2 − (dy)2 − (dz)2 (3.3) Integrating ds gives the total interval along the worldline W: Z B q Z B q ∆s = (ds)2 = (cdt)2 − (dx)2 − (dy)2 − (dz)2 (3.4) A A By definition, the distance measured between two events, A and B, in a reference frame for which they occur simultaneously (tA = tB) is the proper distance: q ∆L = −(∆s)2 (3.5) Our search for a metric that describes the spacetime of a matter-filled universe, is made easier by the cosmological principle.
    [Show full text]
  • Hyperbolic Geometry
    Flavors of Geometry MSRI Publications Volume 31,1997 Hyperbolic Geometry JAMES W. CANNON, WILLIAM J. FLOYD, RICHARD KENYON, AND WALTER R. PARRY Contents 1. Introduction 59 2. The Origins of Hyperbolic Geometry 60 3. Why Call it Hyperbolic Geometry? 63 4. Understanding the One-Dimensional Case 65 5. Generalizing to Higher Dimensions 67 6. Rudiments of Riemannian Geometry 68 7. Five Models of Hyperbolic Space 69 8. Stereographic Projection 72 9. Geodesics 77 10. Isometries and Distances in the Hyperboloid Model 80 11. The Space at Infinity 84 12. The Geometric Classification of Isometries 84 13. Curious Facts about Hyperbolic Space 86 14. The Sixth Model 95 15. Why Study Hyperbolic Geometry? 98 16. When Does a Manifold Have a Hyperbolic Structure? 103 17. How to Get Analytic Coordinates at Infinity? 106 References 108 Index 110 1. Introduction Hyperbolic geometry was created in the first half of the nineteenth century in the midst of attempts to understand Euclid’s axiomatic basis for geometry. It is one type of non-Euclidean geometry, that is, a geometry that discards one of Euclid’s axioms. Einstein and Minkowski found in non-Euclidean geometry a This work was supported in part by The Geometry Center, University of Minnesota, an STC funded by NSF, DOE, and Minnesota Technology, Inc., by the Mathematical Sciences Research Institute, and by NSF research grants. 59 60 J. W. CANNON, W. J. FLOYD, R. KENYON, AND W. R. PARRY geometric basis for the understanding of physical time and space. In the early part of the twentieth century every serious student of mathematics and physics studied non-Euclidean geometry.
    [Show full text]
  • 101.1 CP Lines and Segments That Intersect a Circle.Notebook April 17, 2017 Chapter 10 Circles Part 1 - Lines and Segments That Intersect a Circle
    101.1 CP Lines and Segments that Intersect a Circle.notebook April 17, 2017 Chapter 10 Circles Part 1 - Lines and Segments that Intersect a Circle In this lesson we will: Name and describe the lines and segments that intersect a circle. A chord is a segment whose endpoints are on the circle. > BA is a chord in circle O. A diameter is a chord that intersects the center of the circle. > BA is a diameter in circle P. A radius is a segment whose endpoints are the center of the circle and a point on the circle. > PA is a radius in circle P. A radius of a circle is half the length of the circle's diameter. 1 101.1 CP Lines and Segments that Intersect a Circle.notebook April 17, 2017 A tangent is a line or segment that intersects a circle at one point. The point of tangency is the point where the line or segment intersects the circle > line t is a tangent line to circle O and P is the point of tangency A common tangent is a line that is tangent to two circles. A secant is a line that intersects a circle at two points. The segment part of a secant contained inside the circle is a chord. > line AB is a secant in circle O > segment AB is a chord in circle O 2 101.1 CP Lines and Segments that Intersect a Circle.notebook April 17, 2017 Examples: Naming Lines and Segments that Intersect Circles Name each line or segment that intersects 8L.
    [Show full text]
  • Chapter 14 Hyperbolic Geometry Math 4520, Fall 2017
    Chapter 14 Hyperbolic geometry Math 4520, Fall 2017 So far we have talked mostly about the incidence structure of points, lines and circles. But geometry is concerned about the metric, the way things are measured. We also mentioned in the beginning of the course about Euclid's Fifth Postulate. Can it be proven from the the other Euclidean axioms? This brings up the subject of hyperbolic geometry. In the hyperbolic plane the parallel postulate is false. If a proof in Euclidean geometry could be found that proved the parallel postulate from the others, then the same proof could be applied to the hyperbolic plane to show that the parallel postulate is true, a contradiction. The existence of the hyperbolic plane shows that the Fifth Postulate cannot be proven from the others. Assuming that Mathematics itself (or at least Euclidean geometry) is consistent, then there is no proof of the parallel postulate in Euclidean geometry. Our purpose in this chapter is to show that THE HYPERBOLIC PLANE EXISTS. 14.1 A quick history with commentary In the first half of the nineteenth century people began to realize that that a geometry with the Fifth postulate denied might exist. N. I. Lobachevski and J. Bolyai essentially devoted their lives to the study of hyperbolic geometry. They wrote books about hyperbolic geometry, and showed that there there were many strange properties that held. If you assumed that one of these strange properties did not hold in the geometry, then the Fifth postulate could be proved from the others. But this just amounted to replacing one axiom with another equivalent one.
    [Show full text]
  • Lesson 2: Circles, Chords, Diameters, and Their Relationships
    NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 2 M5 GEOMETRY Lesson 2: Circles, Chords, Diameters, and Their Relationships Student Outcomes . Identify the relationships between the diameters of a circle and other chords of the circle. Lesson Notes Students are asked to construct the perpendicular bisector of a line segment and draw conclusions about points on that bisector and the endpoints of the segment. They relate the construction to the theorem stating that any perpendicular bisector of a chord must pass through the center of the circle. Classwork Opening Exercise (4 minutes) Scaffolding: Post a diagram and display the steps to create a Opening Exercise perpendicular bisector used in Construct the perpendicular bisector of line segment below (as you did in Module 1). Lesson 4 of Module 1. ���� Label the endpoints of the segment and . Draw circle with center and radius . Draw another line that bisects but is not perpendicular to it. Draw circle with center ���� and radius . List one similarity and one difference���� between the two bisectors. . Label the points of Answers will vary. All points on the perpendicular bisector are equidistant from points and 46T . ���� Points on the other bisector are not equidistant from points A and B. The perpendicular bisector intersection as and . meets at right angles. The other bisector meets at angles that are not congruent. Draw . ���� ⃖����⃗ You may wish to recall for students the definition of equidistant: EQUIDISTANT:. A point is said to be equidistant from two different points and if = . Points and can be replaced in the definition above with other figures (lines, etc.) as long as the distance to those figures is given meaning first.
    [Show full text]