Lesson 3: Rectangles Inscribed in Circles
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Point of Concurrency the Three Perpendicular Bisectors of a Triangle Intersect at a Single Point
3.1 Special Segments and Centers of Classifications of Triangles: Triangles By Side: 1. Equilateral: A triangle with three I CAN... congruent sides. Define and recognize perpendicular bisectors, angle bisectors, medians, 2. Isosceles: A triangle with at least and altitudes. two congruent sides. 3. Scalene: A triangle with three sides Define and recognize points of having different lengths. (no sides are concurrency. congruent) Jul 249:36 AM Jul 249:36 AM Classifications of Triangles: Special Segments and Centers in Triangles By angle A Perpendicular Bisector is a segment or line 1. Acute: A triangle with three acute that passes through the midpoint of a side and angles. is perpendicular to that side. 2. Obtuse: A triangle with one obtuse angle. 3. Right: A triangle with one right angle 4. Equiangular: A triangle with three congruent angles Jul 249:36 AM Jul 249:36 AM Point of Concurrency The three perpendicular bisectors of a triangle intersect at a single point. Two lines intersect at a point. The point of When three or more lines intersect at the concurrency of the same point, it is called a "Point of perpendicular Concurrency." bisectors is called the circumcenter. Jul 249:36 AM Jul 249:36 AM 1 Circumcenter Properties An angle bisector is a segment that divides 1. The circumcenter is an angle into two congruent angles. the center of the circumscribed circle. BD is an angle bisector. 2. The circumcenter is equidistant to each of the triangles vertices. m∠ABD= m∠DBC Jul 249:36 AM Jul 249:36 AM The three angle bisectors of a triangle Incenter properties intersect at a single point. -
Regents Exam Questions G.GPE.A.2: Locus 1 Name: ______
Regents Exam Questions G.GPE.A.2: Locus 1 Name: ________________________ www.jmap.org G.GPE.A.2: Locus 1 1 If point P lies on line , which diagram represents 3 The locus of points equidistant from two sides of the locus of points 3 centimeters from point P? an acute scalene triangle is 1) an angle bisector 2) an altitude 3) a median 4) the third side 1) 4 Chantrice is pulling a wagon along a smooth, horizontal street. The path of the center of one of the wagon wheels is best described as 1) a circle 2) 2) a line perpendicular to the road 3) a line parallel to the road 4) two parallel lines 3) 5 Which equation represents the locus of points 4 units from the origin? 1) x = 4 2 2 4) 2) x + y = 4 3) x + y = 16 4) x 2 + y 2 = 16 2 In the accompanying diagram, line 1 is parallel to line 2 . 6 Which equation represents the locus of all points 5 units below the x-axis? 1) x = −5 2) x = 5 3) y = −5 4) y = 5 Which term describes the locus of all points that are equidistant from line 1 and line 2 ? 1) line 2) circle 3) point 4) rectangle 1 Regents Exam Questions G.GPE.A.2: Locus 1 Name: ________________________ www.jmap.org 7 The locus of points equidistant from the points 9 Dan is sketching a map of the location of his house (4,−5) and (4,7) is the line whose equation is and his friend Matthew's house on a set of 1) y = 1 coordinate axes. -
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. -
The Geometry of Diagonal Groups
The geometry of diagonal groups R. A. Baileya, Peter J. Camerona, Cheryl E. Praegerb, Csaba Schneiderc aUniversity of St Andrews, St Andrews, UK bUniversity of Western Australia, Perth, Australia cUniversidade Federal de Minas Gerais, Belo Horizonte, Brazil Abstract Diagonal groups are one of the classes of finite primitive permutation groups occurring in the conclusion of the O'Nan{Scott theorem. Several of the other classes have been described as the automorphism groups of geometric or combi- natorial structures such as affine spaces or Cartesian decompositions, but such structures for diagonal groups have not been studied in general. The main purpose of this paper is to describe and characterise such struct- ures, which we call diagonal semilattices. Unlike the diagonal groups in the O'Nan{Scott theorem, which are defined over finite characteristically simple groups, our construction works over arbitrary groups, finite or infinite. A diagonal semilattice depends on a dimension m and a group T . For m = 2, it is a Latin square, the Cayley table of T , though in fact any Latin square satisfies our combinatorial axioms. However, for m > 3, the group T emerges naturally and uniquely from the axioms. (The situation somewhat resembles projective geometry, where projective planes exist in great profusion but higher- dimensional structures are coordinatised by an algebraic object, a division ring.) A diagonal semilattice is contained in the partition lattice on a set Ω, and we provide an introduction to the calculus of partitions. Many of the concepts and constructions come from experimental design in statistics. We also determine when a diagonal group can be primitive, or quasiprimitive (these conditions turn out to be equivalent for diagonal groups). -
Simple Polygons Scribe: Michael Goldwasser
CS268: Geometric Algorithms Handout #5 Design and Analysis Original Handout #15 Stanford University Tuesday, 25 February 1992 Original Lecture #6: 28 January 1991 Topics: Triangulating Simple Polygons Scribe: Michael Goldwasser Algorithms for triangulating polygons are important tools throughout computa- tional geometry. Many problems involving polygons are simplified by partitioning the complex polygon into triangles, and then working with the individual triangles. The applications of such algorithms are well documented in papers involving visibility, motion planning, and computer graphics. The following notes give an introduction to triangulations and many related definitions and basic lemmas. Most of the definitions are based on a simple polygon, P, containing n edges, and hence n vertices. However, many of the definitions and results can be extended to a general arrangement of n line segments. 1 Diagonals Definition 1. Given a simple polygon, P, a diagonal is a line segment between two non-adjacent vertices that lies entirely within the interior of the polygon. Lemma 2. Every simple polygon with jPj > 3 contains a diagonal. Proof: Consider some vertex v. If v has a diagonal, it’s party time. If not then the only vertices visible from v are its neighbors. Therefore v must see some single edge beyond its neighbors that entirely spans the sector of visibility, and therefore v must be a convex vertex. Now consider the two neighbors of v. Since jPj > 3, these cannot be neighbors of each other, however they must be visible from each other because of the above situation, and thus the segment connecting them is indeed a diagonal. -
And Are Congruent Chords, So the Corresponding Arcs RS and ST Are Congruent
9-3 Arcs and Chords ALGEBRA Find the value of x. 3. SOLUTION: 1. In the same circle or in congruent circles, two minor SOLUTION: arcs are congruent if and only if their corresponding Arc ST is a minor arc, so m(arc ST) is equal to the chords are congruent. Since m(arc AB) = m(arc CD) measure of its related central angle or 93. = 127, arc AB arc CD and . and are congruent chords, so the corresponding arcs RS and ST are congruent. m(arc RS) = m(arc ST) and by substitution, x = 93. ANSWER: 93 ANSWER: 3 In , JK = 10 and . Find each measure. Round to the nearest hundredth. 2. SOLUTION: Since HG = 4 and FG = 4, and are 4. congruent chords and the corresponding arcs HG and FG are congruent. SOLUTION: m(arc HG) = m(arc FG) = x Radius is perpendicular to chord . So, by Arc HG, arc GF, and arc FH are adjacent arcs that Theorem 10.3, bisects arc JKL. Therefore, m(arc form the circle, so the sum of their measures is 360. JL) = m(arc LK). By substitution, m(arc JL) = or 67. ANSWER: 67 ANSWER: 70 eSolutions Manual - Powered by Cognero Page 1 9-3 Arcs and Chords 5. PQ ALGEBRA Find the value of x. SOLUTION: Draw radius and create right triangle PJQ. PM = 6 and since all radii of a circle are congruent, PJ = 6. Since the radius is perpendicular to , bisects by Theorem 10.3. So, JQ = (10) or 5. 7. Use the Pythagorean Theorem to find PQ. -
Squaring the Circle in Elliptic Geometry
Rose-Hulman Undergraduate Mathematics Journal Volume 18 Issue 2 Article 1 Squaring the Circle in Elliptic Geometry Noah Davis Aquinas College Kyle Jansens Aquinas College, [email protected] Follow this and additional works at: https://scholar.rose-hulman.edu/rhumj Recommended Citation Davis, Noah and Jansens, Kyle (2017) "Squaring the Circle in Elliptic Geometry," Rose-Hulman Undergraduate Mathematics Journal: Vol. 18 : Iss. 2 , Article 1. Available at: https://scholar.rose-hulman.edu/rhumj/vol18/iss2/1 Rose- Hulman Undergraduate Mathematics Journal squaring the circle in elliptic geometry Noah Davis a Kyle Jansensb Volume 18, No. 2, Fall 2017 Sponsored by Rose-Hulman Institute of Technology Department of Mathematics Terre Haute, IN 47803 a [email protected] Aquinas College b scholar.rose-hulman.edu/rhumj Aquinas College Rose-Hulman Undergraduate Mathematics Journal Volume 18, No. 2, Fall 2017 squaring the circle in elliptic geometry Noah Davis Kyle Jansens Abstract. Constructing a regular quadrilateral (square) and circle of equal area was proved impossible in Euclidean geometry in 1882. Hyperbolic geometry, however, allows this construction. In this article, we complete the story, providing and proving a construction for squaring the circle in elliptic geometry. We also find the same additional requirements as the hyperbolic case: only certain angle sizes work for the squares and only certain radius sizes work for the circles; and the square and circle constructions do not rely on each other. Acknowledgements: We thank the Mohler-Thompson Program for supporting our work in summer 2014. Page 2 RHIT Undergrad. Math. J., Vol. 18, No. 2 1 Introduction In the Rose-Hulman Undergraduate Math Journal, 15 1 2014, Noah Davis demonstrated the construction of a hyperbolic circle and hyperbolic square in the Poincar´edisk [1]. -
Three Points on the Plane
THREE POINTS ON THE PLANE ALEXANDER GIVENTAL UC BERKELEY In this lecture, we examine some properties of a geometric figure formed by three points on the plane. Any three points not lying on the same line determine a triangle. Namely, the points are the vertices of the triangle, and the segments connecting them are the sides of it. We expect the reader to be familiar with a few basic notions and facts of geometry of triangles. References like “[K], §140” will point to specific sections in: Kiselev’s Geometry. Book I: Planimetry, Sumizdat, 2006, 248 pages. Everything we assume known, as well as the initial part of the present exposition, is contained in the first half of this textbook. Circumcenter. One says a circle is circumscribed about a triangle (or shorter, is the circumcircle of it), if the circle passes through all vertices of the triangle. Theorem 1. About every triangle, a circle can be circum- scribed, and such a circle is unique. Let △ABC be a triangle with vertices A, B, C (Figure 1). A point O equidistant from1 all the three vertices is the center of a circumcircle of radius OA = OB = OC. Thus we need to show that a point equidistant from the vertices exists and is unique. Indeed, the geometric locus 2 of points equidistant from A and B is the perpendicular bisector to the segment AB, i.e. it is the line (denoted by MN in Figure 1) perpendicular to the segment AB at its midpoint (see [K], §56). Likewise, the geometric locus of points equidistant from B and C is the perpendicular bisector P Q to the segment BC. -
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. -
Unit 6 Visualising Solid Shapes(Final)
• 3D shapes/objects are those which do not lie completely in a plane. • 3D objects have different views from different positions. • A solid is a polyhedron if it is made up of only polygonal faces, the faces meet at edges which are line segments and the edges meet at a point called vertex. • Euler’s formula for any polyhedron is, F + V – E = 2 Where F stands for number of faces, V for number of vertices and E for number of edges. • Types of polyhedrons: (a) Convex polyhedron A convex polyhedron is one in which all faces make it convex. e.g. (1) (2) (3) (4) 12/04/18 (1) and (2) are convex polyhedrons whereas (3) and (4) are non convex polyhedron. (b) Regular polyhedra or platonic solids: A polyhedron is regular if its faces are congruent regular polygons and the same number of faces meet at each vertex. For example, a cube is a platonic solid because all six of its faces are congruent squares. There are five such solids– tetrahedron, cube, octahedron, dodecahedron and icosahedron. e.g. • A prism is a polyhedron whose bottom and top faces (known as bases) are congruent polygons and faces known as lateral faces are parallelograms (when the side faces are rectangles, the shape is known as right prism). • A pyramid is a polyhedron whose base is a polygon and lateral faces are triangles. • A map depicts the location of a particular object/place in relation to other objects/places. The front, top and side of a figure are shown. Use centimetre cubes to build the figure. -
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 -
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.