PTOLEMY DAY 7 BUILDING A TABLE OF CHORDS AND ARCS: PART 1
|
|
- Natalie Elliott
- 6 years ago
- Views:
Transcription
1 PTOLEMY Y 7 UILING TLE OF HORS N RS: PRT 1 (1) WHY 360 N 120? Ptolemy divides his standard circle s diameter, R, into 120 parts, in keeping with his sexagesimal ways this way the radius of the circle is 60 of those 120 th -parts of the diameter, and each of those is called a R O part of the diameter s 120. nd he divides the arc of the circle (and hence the angles subtending them at the center of the circle) into 360 equal arcs, each one called a degree. s cumbersome as this is for us, there were some advantages to this for Ptolemy, who did not have decimal calculators at his disposal. (a) For one thing, 120 = 5! = , so 120 has lots of integral factors. 360 is triple that, so it also has lots of factors. This makes for more whole numbers of degrees when we cut the circumference of the circle into some whole number of parts; e.g. half a circle, a third of a circle, a quarter of a circle, a fifth of a circle, will all have whole number degree-values, as will a tenth, a twelfth, a fifteenth, and so on. (b) 360 : 120 = 3 : 1, a rough approximation of pi. In other words, one degree of arc is almost equal to one part of the diameter. (c) With this system, the equilateral triangle on the radius has 3 angles of 60 and 3 sides of 60 parts. That s rather pretty. (d) There are 365 days in a year; so now we have the Sun moving about 1 a day on its circle (just a little less, as we shall see). (2) HORS RE NOT S RS. To see the need for this table of chords and arcs, we need to see that chords do not have the same ratios as the arcs they subtend. We easily fall into the mistake of thinking that any two chords in this circle will have to each other the same ratio as the arcs of the circle they cut off. If you are tempted to think this, banish it from your thought! It is not true. nd it is easy to forget that this is not true, even for some veterans of trigonometry. One must burn this annoying fact into memory. If chords were as arcs, then it would be much easier to build up the Table! For instance, since the chord of 180 is 120 parts, it would follow that the chord of 90 must be 60 parts, and so on proportionally. las, it is not so. very quick proof of this: raw a square inscribed in our circle. Obviously the chord of 90 is the side of that square, and the chord of 180, the diameter, is the diagonal. ut the diagonal is not double the side, even though the arc is double the arc. 46
2 Or draw a regular HEXGON in a circle, letting,, be three consecutive sides. Then is the diameter of the circle, and is double. ut the arc (i.e. ) is not double the arc, but triple it! gain, M If M = M, and so arc M = arc M so arc M = ½ arc M is chord M = ½ chord? E No. E = ½ and M > E (hypotenuse) (3) PLN FOR UILING THE TLE: So we must find other inroads into filling out the entries on our table. Plan of attack: [1] FIN HORS OF 36 & 72. = items (4) (5) below. [2] FIN HORS OF 60, 90, 120 and note that the chord of the supplement to an angle whose chord is given, is given. = item (6). [3] FIN HORS OF 144, 108. = item (7). [4] PROVE THT HORS OF RS WHIH RE IFFERENES OF RS WITH GIVEN HORS RE GIVEN. = items (8) (9) [5] PROVE THT HORS OF RS WHIH RE HLVES OF RS WITH GIVEN HORS RE GIVEN. = item (10) [6] PROVE THT HORS OF RS WHIH RE SUMS OF RS WITH GIVEN HORS RE GIVEN. = item (12) [7] FIN HORS OF 1 N 1½. = items (13) (15) 47
3 [8] INTERPOLTION OF SIXTIETHS = item (16) [9] RELTION OF TLE OF HORS TO TLE OF SINES = item (17) NOTE: 36, 72, 60, 90 are all found directly. 120, 144, 108 are found as supplements. 12 is found by subtraction, i.e. by [4] above. 6, 3, 1½, ¾ are found by bisection, i.e. by [5] above. ll multiples of 1½ are found by addition, i.e. by [6] above. 1 and ½ are found by approximation. We will cover Steps [1] through [6] today, i.e. items (4) through (12), and we will cover Steps [7] through [9], i.e. items (13) through (17), in ay 8. (4) PRELIMINRY TO FINING THE HORS OF 36 & 72. First, Ptolemy shows that if we have a circle of diameter, perpendicular radius, and bisect radius at E, and join E, and draw a circle with center E, radius E, cutting radius at F, and join F, then F = side of regular inscribed pentagon F = side of regular inscribed decagon F E He uses Euclid s Elements, ook 13 Proposition 10, for this. So now we know that F is the chord of 72, and F is the chord of 36. This does not give us a numerical value for them yet, in terms of the 120 parts of diameter, but it will enable us to do that in the next step. 48
4 (5) FINING THE HORS OF 36 & 72. Now let s find numerical values, as precise as we like, for the chords of 36 and 72 : F E = 120 so E = ¼ = 30 and = ½ = 60 [given] so E = 2 E + 2 = = 4500 = so EF = E = so F = EF E = = i.e. hord 36 = so F 2 = but 2 = 3600 so F 2 = F = = so F = so hord 72 = NOTES: (a) These are decimal values, but are easily translated into sexagesimal values matching those on Ptolemy s Table. (b) Ptolemy speaks of getting or finding chords, or of chords being given. What he means is to find a way of determining a numerical value for their lengths, in units of one-hundred-twentieth parts of the diameter, to any desired degree of precision. We do this by beginning with chords whose exact values are known for geometrical reasons, then by showing how the sought chord is the result of a known operation on the known chords. 49
5 (6) FINING THE HORS OF 60, 90, 120. hord 60 = Radius = 60 so 2 = 2 2 = = 7200 so = so hord 90 = (NOTE: I will stop putting the ellipsis in (...) just for simplicity. I will just truncate the expressions at an arbitrary place.) T If we now let T = 120, so that T must be 60, we know: T 2 = 2 T 2 = = = so T = = so hord 120 = NOTE: WE N FIN HORS OF SUPPLEMENTS. We can now find the chord for any angle which is the supplement of an angle whose chord is known. In the example above, we knew T, the chord of 60, and we knew, the diameter. That, together with the Pythagorean Theorem, was all we needed in order to compute T, the chord of the supplement of 60. There was nothing special about 120 and 60. So now, if we know the chord of any angle, we will also be able to compute the value of the chord of its supplement. (7) FINING THE HORS OF 144 & 108. So using the very same technique, we can find the chords of 144 and 108, since these are the supplements of 36 and 72 respectively, and the chords of those arcs are known. 50
6 (8) LEMM (FOR THE UPOMING PROOF THT HORS OF RS WHIH RE IFFERENES OF RS WITH GIVEN HORS RE GIVEN). This may be called the iagonal ross-product Theorem. The Theorem states that if is a cyclic quadrilateral (i.e. a quadrilateral inscribed in a circle), then = + or, put verbally: The rectangle contained by the diagonals is equal to the sums of the rectangles contained by the pairs of opposite sides. 6 E 5 7 gain, since this is just a matter of going through the steps, we will assume it is true and use it, but not bother proving it together. (9) PROOF THT HORS OF RS WHIH RE IFFERENES OF RS WITH GIVEN HORS RE GIVEN (i.e. calculable). Given: Prove: arcs & in degrees; chords & in 120 th parts of diameter chord is also given in 120 th parts of (i.e. can be calculated) Since & are given, thus is given [Euclid 1.47] 51
7 Since & are given, thus is given [1.47] Q.E.. ut = + [by the cross-product Lemma] and all terms in that equation are given except for. Hence is also given. NOTE: We are not just thinking of. (for example) as a rectangle, but as a product of two numbers. There s plenty of philosophically discussible matter there. EXMPLE of the use of this Theorem for our Table: If arc = 72 and arc = 60 then arc = 12. nd since the chords of 72 and 60 are already given, it follows by this Theorem that the chord of 12 is also given now, or calculable to any degree of accuracy we please. (10) PROOF THT HORS OF RS WHIH RE HLVES OF RS WITH GIVEN HORS RE GIVEN. Ptolemy next shows that if we know the chord of a known arc, then we can also calculate the value of the chord of half that arc. That will help fill in a whole lot of entries on the table! If we know arc in degrees (and its midpoint is ), and chord in 120 th parts of diameter, then we will also be able to calculate the value of chord in those units. E F 52
8 Given: Prove: Make: arc in degrees chord in 120 th parts of the diameter arc = arc is given (i.e. the chord of half the given arc ) E = F perpendicular to = E common = E [Euclid 3.27; they stand on equal arcs] so r r E so = E i.e. = E so r EF r F so F = EF i.e. F = ½ E so F = ½ [ E] F = ½ [ ] so F is given [since is given, is supplement of given] Now : = : F [r similar to r F; 6.8] so F = 2 so 2 is given [ & F are given] so is given Q.E.. EXMPLES: Since we had the chord of 12, now we have the chords of 6, 3, 1½, ¾ (11) RS WHOSE HORS RE NOW GIVEN: irect Geometry: Supplements: 36 (side of decagon) 144 = (radius, side of hexagon) 120 = (side of pentagon) 108 = (side of square) 180 (diameter) Subtraction: Supplements: 53
9 24 = = = = = = = = = = = = = = = = isection: Supplements: 42 = = = = = = Now we have all multiples of 6 isection: Supplements: 3 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Now we have all multiples of 3, and by isection: 1½ = 3 2 (12) HORS OF RS WHIH RE SUMS OF RS WITH GIVEN HORS RE GIVEN (this is like addition ) Ptolemy adds this Theorem that will help to fill in new entries on our Table: Given: rcs and are given in degrees hords and are given in diameter-parts Prove: is also given in diameter-parts F 54 E
10 Make: iameters FE and F E is given is given E is given [supplement of ] [supplement of ] [supplement of ] ut E = E + E [quadr. multiplication Lemma, item (11) above] so is given [all other terms given in there] so is given [supplement of ] Q.E.. PPLITION: We can add 1½ to itself thus, getting 1 out of every 3 terms on the table. Now we have 1½ 3 4½ 6 etc. NOTE: The use of given here means can be calculated to whatever degree of accuracy you please, e.g. to as many decimal or sexagesimal places you want. Sometimes we need numerical exercises to get the gist of this. For instance, we have seen that hord 36 = and hord 1½ = so now, using this Theorem, see if you can determine the numerical value for hord 37½. 55
Definitions. (V.1). A magnitude is a part of a magnitude, the less of the greater, when it measures
hapter 8 Euclid s Elements ooks V 8.1 V.1-3 efinitions. (V.1). magnitude is a part of a magnitude, the less of the greater, when it measures the greater. (V.2). The greater is a multiple of the less when
More informationPlane geometry Circles: Problems with some Solutions
The University of Western ustralia SHL F MTHMTIS & STTISTIS UW MY FR YUNG MTHMTIINS Plane geometry ircles: Problems with some Solutions 1. Prove that for any triangle, the perpendicular bisectors of the
More informationIntroduction Circle Some terms related with a circle
141 ircle Introduction In our day-to-day life, we come across many objects which are round in shape, such as dials of many clocks, wheels of a vehicle, bangles, key rings, coins of denomination ` 1, `
More informationChapter 1. Some Basic Theorems. 1.1 The Pythagorean Theorem
hapter 1 Some asic Theorems 1.1 The ythagorean Theorem Theorem 1.1 (ythagoras). The lengths a b < c of the sides of a right triangle satisfy the relation a 2 + b 2 = c 2. roof. b a a 3 2 b 2 b 4 b a b
More informationCHAPTER (multiply by 10) 2 10 (double first line) 4 20 (double third line) 8 40 (double fourth line) (halve first line)
CHAPTER 1 1. The answers are given in the answer section of the text. For the Egyptian hieroglyphics, 375 is three hundreds, seven tens and five ones, while 4856 is four thousands, eight hundreds, five
More informationInteger (positive or negative whole numbers or zero) arithmetic
Integer (positive or negative whole numbers or zero) arithmetic The number line helps to visualize the process. The exercises below include the answers but see if you agree with them and if not try to
More informationPTOLEMY DAY 6 THE SEXAGESIMAL SYSTEM AND THE NEED FOR A TABLE OF CHORDS AND ARCS
PTOLEMY DAY 6 THE SEXAGESIMAL SYSTEM AND THE NEED FOR A TABLE OF CHORDS AND ARCS Before getting into any of the detailed versions of Ptolemy s models for the motions of the Sun and the planets, we need
More informationTheorem 1.2 (Converse of Pythagoras theorem). If the lengths of the sides of ABC satisfy a 2 + b 2 = c 2, then the triangle has a right angle at C.
hapter 1 Some asic Theorems 1.1 The ythagorean Theorem Theorem 1.1 (ythagoras). The lengths a b < c of the sides of a right triangle satisfy the relation a + b = c. roof. b a a 3 b b 4 b a b 4 1 a a 3
More information1 st Preparatory. Part (1)
Part (1) (1) omplete: 1) The square is a rectangle in which. 2) in a parallelogram in which m ( ) = 60, then m ( ) =. 3) The sum of measures of the angles of the quadrilateral equals. 4) The ray drawn
More information1. Draw and label a diagram to illustrate the property of a tangent to a circle.
Master 8.17 Extra Practice 1 Lesson 8.1 Properties of Tangents to a Circle 1. Draw and label a diagram to illustrate the property of a tangent to a circle. 2. Point O is the centre of the circle. Points
More informationSolve problems involving tangents to a circle. Solve problems involving chords of a circle
8UNIT ircle Geometry What You ll Learn How to Solve problems involving tangents to a circle Solve problems involving chords of a circle Solve problems involving the measures of angles in a circle Why Is
More informationArkansas Council of Teachers of Mathematics 2012 State Competition Geometry Exam. B. 28 (5x-41) 3 m (2x+25)
rkansas ouncil of Teachers of Mathematics 2012 State ompetition Geometry Exam For questions 1 through 25, mark your answer choice on the answer sheet provided. (Figures may not be drawn to scale.) fter
More informationFor math conventions used on the GRE, refer to this link:
GRE Review ISU Student Success Center Quantitative Workshop One Quantitative Section: Overview Your test will include either two or three 35-minute quantitative sections. There will be 20 questions in
More informationChapter (Circle) * Circle - circle is locus of such points which are at equidistant from a fixed point in
Chapter - 10 (Circle) Key Concept * Circle - circle is locus of such points which are at equidistant from a fixed point in a plane. * Concentric circle - Circle having same centre called concentric circle.
More informationMaths Assessment Framework Year 10 Higher
Success Criteria for all assessments: Higher Tier 90% 9 80% 8 70% 7 60% 6 50% 5 Please note the GCSE Mathematics is one of the first GCSEs which will be graded by number rather than A*, A, B, C etc. Roughly,
More informationChapter 19 Exercise 19.1
hapter 9 xercise 9... (i) n axiom is a statement that is accepted but cannot be proven, e.g. x + 0 = x. (ii) statement that can be proven logically: for example, ythagoras Theorem. (iii) The logical steps
More informationPark Forest Math Team. Meet #4. Geometry. Self-study Packet
Park Forest Math Team Meet #4 Self-study Packet Problem Categories for this Meet: 1. Mystery: Problem solving 2. : ngle measures in plane figures including supplements and complements 3. Number Theory:
More informationName: GEOMETRY: EXAM (A) A B C D E F G H D E. 1. How many non collinear points determine a plane?
GMTRY: XM () Name: 1. How many non collinear points determine a plane? ) none ) one ) two ) three 2. How many edges does a heagonal prism have? ) 6 ) 12 ) 18 ) 2. Name the intersection of planes Q and
More informationLEMMA. Trigonometriae Britannicae Chapter Eleven
Trigonometriae Britannicae -. hapter leven The ratio of other chords can be found with no less certainty, and for this purpose the following Lemma is useful. LMM. B If the line bisects the angle in the
More informationIndicate whether the statement is true or false.
PRACTICE EXAM IV Sections 6.1, 6.2, 8.1 8.4 Indicate whether the statement is true or false. 1. For a circle, the constant ratio of the circumference C to length of diameter d is represented by the number.
More informationYear 11 Mathematics: Specialist Course Outline
MATHEMATICS LEARNING AREA Year 11 Mathematics: Specialist Course Outline Text: Mathematics Specialist Units 1 and 2 A.J. Unit/time Topic/syllabus entry Resources Assessment 1 Preliminary work. 2 Representing
More informationSecondary School Certificate Examination Syllabus MATHEMATICS. Class X examination in 2011 and onwards. SSC Part-II (Class X)
Secondary School Certificate Examination Syllabus MATHEMATICS Class X examination in 2011 and onwards SSC Part-II (Class X) 15. Algebraic Manipulation: 15.1.1 Find highest common factor (H.C.F) and least
More informationHomework Assignments Math /02 Fall 2014
Homework Assignments Math 119-01/02 Fall 2014 Assignment 1 Due date : Friday, September 5 6th Edition Problem Set Section 6.1, Page 178: #1, 2, 3, 4, 5, 6. Section 6.2, Page 185: #1, 2, 3, 5, 6, 8, 10-14,
More informationIn this article we examine a famous and important result in
How to Prove It In this episode of How To Prove It, we prove a striking theorem first discovered by Ptolemy. We then discuss some nice applications of the theorem. lassroom In this article we examine a
More informationSolutions for practice questions: Chapter 7, Triangle Trigonometry If you find any errors, please let me know at
Solutions for practice questions: Chapter 7, Triangle Trigonometry If you find any errors, please let me know at mailto:peggy.frisbie@polk-fl.net.. The shortest distance from the chord [AB] to the center
More informationIntroduction to Trigonometry: Grade 9
Introduction to Trigonometry: Grade 9 Andy Soper October 6, 2013 This document was constructed and type-set using P C T E X (a dielect of L A T E X) 1 1 Before you start 1.1 About these notes. These notes
More informationCBSE Class IX Syllabus. Mathematics Class 9 Syllabus
Mathematics Class 9 Syllabus Course Structure First Term Units Unit Marks I Number System 17 II Algebra 25 III Geometry 37 IV Co-ordinate Geometry 6 V Mensuration 5 Total 90 Second Term Units Unit Marks
More informationCircle Theorems. Angles at the circumference are equal. The angle in a semi-circle is x The angle at the centre. Cyclic Quadrilateral
The angle in a semi-circle is 90 0 Angles at the circumference are equal. A B They must come from the same arc. Look out for a diameter. 2x Cyclic Quadrilateral Opposite angles add up to 180 0 A They must
More information10. Circles. Q 5 O is the centre of a circle of radius 5 cm. OP AB and OQ CD, AB CD, AB = 6 cm and CD = 8 cm. Determine PQ. Marks (2) Marks (2)
10. Circles Q 1 True or False: It is possible to draw two circles passing through three given non-collinear points. Mark (1) Q 2 State the following statement as true or false. Give reasons also.the perpendicular
More informationHomework Assignments Math /02 Fall 2017
Homework Assignments Math 119-01/02 Fall 2017 Assignment 1 Due date : Wednesday, August 30 Section 6.1, Page 178: #1, 2, 3, 4, 5, 6. Section 6.2, Page 185: #1, 2, 3, 5, 6, 8, 10-14, 16, 17, 18, 20, 22,
More informationMathematics Class 9 Syllabus. Course Structure. I Number System 17 II Algebra 25 III Geometry 37 IV Co-ordinate Geometry 6 V Mensuration 5 Total 90
Mathematics Class 9 Syllabus Course Structure First Term Units Unit Marks I Number System 17 II Algebra 25 III Geometry 37 IV Co-ordinate Geometry 6 V Mensuration 5 Total 90 Second Term Units Unit Marks
More informationChapter-wise questions
hapter-wise questions ircles 1. In the given figure, is circumscribing a circle. ind the length of. 3 15cm 5 2. In the given figure, is the center and. ind the radius of the circle if = 18 cm and = 3cm
More informationMATH II CCR MATH STANDARDS
RELATIONSHIPS BETWEEN QUANTITIES M.2HS.1 M.2HS.2 M.2HS.3 M.2HS.4 M.2HS.5 M.2HS.6 Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents
More informationMathematics 3210 Spring Semester, 2005 Homework notes, part 8 April 15, 2005
Mathematics 3210 Spring Semester, 2005 Homework notes, part 8 April 15, 2005 The underlying assumption for all problems is that all points, lines, etc., are taken within the Poincaré plane (or Poincaré
More informationREVISED vide circular No.63 on
Circular no. 63 COURSE STRUCTURE (FIRST TERM) CLASS -IX First Term Marks: 90 REVISED vide circular No.63 on 22.09.2015 UNIT I: NUMBER SYSTEMS 1. REAL NUMBERS (18 Periods) 1. Review of representation of
More informationYear 9 Term 3 Homework
Yimin Math Centre Year 9 Term 3 Homework Student Name: Grade: Date: Score: Table of contents 5 Year 9 Term 3 Week 5 Homework 1 5.1 Geometry (Review)................................... 1 5.1.1 Angle sum
More informationCircles in Neutral Geometry
Everything we do in this set of notes is Neutral. Definitions: 10.1 - Circles in Neutral Geometry circle is the set of points in a plane which lie at a positive, fixed distance r from some fixed point.
More informationExercise. and 13x. We know that, sum of angles of a quadrilateral = x = 360 x = (Common in both triangles) and AC = BD
9 Exercise 9.1 Question 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. Solution Given, the ratio of the angles of quadrilateral are 3 : 5 : 9
More informationProof of the Equivalent Area of a Circle and a Right Triangle with Leg Lengths of the Radius and Circumference
Proof of the Equivalent Area of a ircle and a Right Triangle with Leg Lengths of the Radius and ircumference Brennan ain July 22, 2018 Abstract In this paper I seek to prove Archimedes Theorem that a circle
More informationStepping stones for Number systems. 1) Concept of a number line : Marking using sticks on the floor. (1 stick length = 1 unit)
Quality for Equality Stepping stones for Number systems 1) Concept of a number line : Marking using sticks on the floor. (1 stick length = 1 unit) 2) Counting numbers: 1,2,3,... Natural numbers Represent
More informationCalifornia Common Core State Standards for Mathematics Standards Map Mathematics II
A Correlation of Pearson Integrated High School Mathematics Mathematics II Common Core, 2014 to the California Common Core State s for Mathematics s Map Mathematics II Copyright 2017 Pearson Education,
More informationProve that a + b = x + y. Join BD. In ABD, we have AOB = 180º AOB = 180º ( 1 + 2) AOB = 180º A
bhilasha lasses lass- IX ate: 03- -7 SLUTIN (hap 8,9,0) 50 ob no.-947967444. The sides and of a quadrilateral are produced as shown in fig. rove that a + b = x + y. Join. In, we have y a + + = 80º = 80º
More informationA. 180 B. 108 C. 360 D. 540
Part I - Multiple Choice - Circle your answer: 1. Find the area of the shaded sector. Q O 8 P A. 2 π B. 4 π C. 8 π D. 16 π 2. An octagon has sides. A. five B. six C. eight D. ten 3. The sum of the interior
More informationCOURSE STRUCTURE CLASS -IX
environment, observance of small family norms, removal of social barriers, elimination of gender biases; mathematical softwares. its beautiful structures and patterns, etc. COURSE STRUCTURE CLASS -IX Units
More informationThe focus of SECONDARY Mathematics II Critical Area 1: Critical Area 2: Critical Area 3: Critical Area 4: Critica l Area 5: Critical Area 6:
The focus of SECONDARY Mathematics II is on quadratic expressions, equations, and functions; comparing their characteristics and behavior to those of linear and exponential relationships from Secondary
More informationDavid Bressoud Macalester College, St. Paul, MN. NCTM Annual Mee,ng Washington, DC April 23, 2009
David Bressoud Macalester College, St. Paul, MN These slides are available at www.macalester.edu/~bressoud/talks NCTM Annual Mee,ng Washington, DC April 23, 2009 The task of the educator is to make the
More informationCOURSE STRUCTURE CLASS IX Maths
COURSE STRUCTURE CLASS IX Maths Units Unit Name Marks I NUMBER SYSTEMS 08 II ALGEBRA 17 III COORDINATE GEOMETRY 04 IV GEOMETRY 28 V MENSURATION 13 VI STATISTICS & PROBABILITY 10 Total 80 UNIT I: NUMBER
More informationGeometry Final Review. Chapter 1. Name: Per: Vocab. Example Problems
Geometry Final Review Name: Per: Vocab Word Acute angle Adjacent angles Angle bisector Collinear Line Linear pair Midpoint Obtuse angle Plane Pythagorean theorem Ray Right angle Supplementary angles Complementary
More informationC Given that angle BDC = 78 0 and DCA = Find angles BAC and DBA.
UNERSTNING IRLE THEREMS-PRT NE. ommon terms: (a) R- ny portion of a circumference of a circle. (b) HR- line that crosses a circle from one point to another. If this chord passes through the centre then
More information10.5 Areas of Circles and Sectors
10.5. Areas of Circles and Sectors www.ck12.org 10.5 Areas of Circles and Sectors Learning Objectives Find the area of circles, sectors, and segments. Review Queue Find the area of the shaded region in
More information( ) ( ) Geometry Team Solutions FAMAT Regional February = 5. = 24p.
. A 6 6 The semi perimeter is so the perimeter is 6. The third side of the triangle is 7. Using Heron s formula to find the area ( )( )( ) 4 6 = 6 6. 5. B Draw the altitude from Q to RP. This forms a 454590
More information2. A diagonal of a parallelogram divides it into two congruent triangles. 5. Diagonals of a rectangle bisect each other and are equal and vice-versa.
QURILTERLS 1. Sum of the angles of a quadrilateral is 360. 2. diagonal of a parallelogram divides it into two congruent triangles. 3. In a parallelogram, (i) opposite sides are equal (ii) opposite angles
More informationReteaching , or 37.5% 360. Geometric Probability. Name Date Class
Name ate lass Reteaching Geometric Probability INV 6 You have calculated probabilities of events that occur when coins are tossed and number cubes are rolled. Now you will learn about geometric probability.
More informationMATHEMATICS (IX-X) (CODE NO. 041) Session
MATHEMATICS (IX-X) (CODE NO. 041) Session 2018-19 The Syllabus in the subject of Mathematics has undergone changes from time to time in accordance with growth of the subject and emerging needs of the society.
More informationExhaustion: From Eudoxus to Archimedes
Exhaustion: From Eudoxus to Archimedes Franz Lemmermeyer April 22, 2005 Abstract Disclaimer: Eventually, I plan to polish this and use my own diagrams; so far, most of it is lifted from the web. Exhaustion
More informationAnswer Explanations for: ACT June 2012, Form 70C
Answer Explanations for: ACT June 2012, Form 70C Mathematics 1) C) A mean is a regular average and can be found using the following formula: (average of set) = (sum of items in set)/(number of items in
More informationMid-Chapter Quiz: Lessons 10-1 through Refer to. 1. Name the circle. SOLUTION: The center of the circle is A. Therefore, the circle is ANSWER:
Refer to. 1. Name the circle. The center of the circle is A. Therefore, the circle is 2. Name a diameter. ; since is a chord that passes through the center, it is a diameter. 3. Name a chord that is not
More informationSSC EXAMINATION GEOMETRY (SET-A)
GRND TEST SS EXMINTION GEOMETRY (SET-) SOLUTION Q. Solve any five sub-questions: [5M] ns. ns. 60 & D have equal height ( ) ( D) D D ( ) ( D) Slope of the line ns. 60 cos D [/M] [/M] tan tan 60 cos cos
More informationDirections: Examine the Unit Circle on the Cartesian Plane (Unit Circle: Circle centered at the origin whose radius is of length 1)
Name: Period: Discovering the Unit Circle Activity Secondary III For this activity, you will be investigating the Unit Circle. You will examine the degree and radian measures of angles. Note: 180 radians.
More informationChapter 7 Sect. 2. A pythagorean triple is a set of three nonzero whole numbers a, b, and c, that satisfy the equation a 2 + b 2 = c 2.
Chapter 7 Sect. 2 The well-known right triangle relationship called the Pythagorean Theorem is named for Pythagoras, a Greek mathematician who lived in the sixth century b.c. We now know that the Babylonians,
More informationDESK Secondary Math II
Mathematical Practices The Standards for Mathematical Practice in Secondary Mathematics I describe mathematical habits of mind that teachers should seek to develop in their students. Students become mathematically
More informationGEOMETRY ADDITIONAL PRACTICE ITEMS
GEOMETRY ADDITIONAL PRACTICE ITEMS Geometry Additional Practice Items This section has two parts. The first part is a set of 4 sample items for Geometry. The second part contains a table that shows for
More informationThe Golden Section, the Pentagon and the Dodecahedron
The Golden Section, the Pentagon and the Dodecahedron C. Godsalve email:seagods@hotmail.com July, 009 Contents Introduction The Golden Ratio 3 The Pentagon 3 4 The Dodecahedron 8 A few more details 4 Introduction
More informationIntegrated Mathematics I, II, III 2016 Scope and Sequence
Mathematics I, II, III 2016 Scope and Sequence I Big Ideas Math 2016 Mathematics I, II, and III Scope and Sequence Number and Quantity The Real Number System (N-RN) Properties of exponents to rational
More information1. (25 points) Consider the region bounded by the curves x 2 = y 3 and y = 1. (a) Sketch both curves and shade in the region. x 2 = y 3.
Test Solutions. (5 points) Consider the region bounded by the curves x = y 3 and y =. (a) Sketch both curves and shade in the region. x = y 3 y = (b) Find the area of the region above. Solution: Observing
More information02)
GRE / GMATmath,! abscissa, scalene, intercept, markup, such that, break even. abscissa. (4, 2) 4abscissa, 2ordinate absolute value acre add adjacent angles altitude ; angle () acute angle (90 ) right angle
More informationCh 10 Review. Multiple Choice Identify the choice that best completes the statement or answers the question.
Ch 10 Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. In the diagram shown, the measure of ADC is a. 55 b. 70 c. 90 d. 180 2. What is the measure
More informationBrunswick School Department Honors Geometry Unit 6: Right Triangles and Trigonometry
Understandings Questions Knowledge Vocabulary Skills Right triangles have many real-world applications. What is a right triangle? How to find the geometric mean of two numbers? What is the Pythagorean
More informationVertical Progression FOR THE NC STANDARD COURSE OF STUDY IN MATHEMATICS. N C D e p a r t m e n t o f P u b l i c I n s t r u c t i o n
Vertical Progression FOR THE NC STANDARD COURSE OF STUDY IN MATHEMATICS N C D e p a r t m e n t o f P u b l i c I n s t r u c t i o n Table of Contents DOMAIN GRADES / COURSE PAGE K 1 2 3 4 5 6 7 8 M1
More informationRMT 2014 Geometry Test Solutions February 15, 2014
RMT 014 Geometry Test Solutions February 15, 014 1. The coordinates of three vertices of a parallelogram are A(1, 1), B(, 4), and C( 5, 1). Compute the area of the parallelogram. Answer: 18 Solution: Note
More informationHigher Geometry Problems
Higher Geometry Problems (1) Look up Eucidean Geometry on Wikipedia, and write down the English translation given of each of the first four postulates of Euclid. Rewrite each postulate as a clear statement
More informationEpisodes from The History of Trigonometry David Bressoud Macalester College, St. Paul, MN
Episodes from The History of Trigonometry David Bressoud Macalester College, St. Paul, MN Winona State University Winona, MN October 8, 2013 A pdf file of these slides is available at www.macalester.edu/~bressoud/talks
More informationGeometry Facts Circles & Cyclic Quadrilaterals
Geometry Facts Circles & Cyclic Quadrilaterals Circles, chords, secants and tangents combine to give us many relationships that are useful in solving problems. Power of a Point Theorem: The simplest of
More informationMeritPath.com. Problems and Solutions, INMO-2011
Problems and Solutions, INMO-011 1. Let,, be points on the sides,, respectively of a triangle such that and. Prove that is equilateral. Solution 1: c ka kc b kb a Let ;. Note that +, and hence. Similarly,
More informationChapter 18 Exercise 18.1
hapter 18 Eercise 18.1 Q. 1. (i) 180 37 = 143 ( = 143 ) (ii) 180 117 = 63 ( = 63 ) 180 90 = 90 (y = 90 ) (iii) + + 3 + 45 = 180 4.5 = 135 (iv) 180 90 = y 90 = y = 30 45 = y 66 + ( + y) + 47 = 180 + y =
More informationWritten test, 25 problems / 90 minutes
Sponsored by: UGA Math Department and UGA Math Club Written test, 5 problems / 90 minutes October, 06 WITH SOLUTIONS Problem. Let a represent a digit from to 9. Which a gives a! aa + a = 06? Here aa indicates
More information~ 1 ~ Geometry 2 nd Semester Review Find the value for the variable for each of the following situations
Geometry nd Semester Review 018 Find the value for the variable for each of the following situations. 7. 400 m 1. 7 8. y. 8.9 cm 0 0 9.. 19 6 60 1 11 10. 45 4. 58 5 11. 5. 11 6. 18 1 slide 4.1 meters long
More informationCBSE OSWAAL BOOKS LEARNING MADE SIMPLE. Published by : 1/11, Sahitya Kunj, M.G. Road, Agra , UP (India) Ph.: ,
OSWAAL BOOKS LEARNING MADE SIMPLE CBSE SOLVED PAPER 2018 MATHEMATICS CLASS 9 Published by : OSWAAL BOOKS 1/11, Sahitya Kunj, M.G. Road, Agra - 282002, UP (India) Ph.: 0562 2857671, 2527781 email: contact@oswaalbooks.com
More informationLesson 12.1 Skills Practice
Lesson 12.1 Skills Practice Introduction to ircles ircle, Radius, and iameter Vocabulary efine each term in your own words. 1. circle circle is a collection of points on the same plane equidistant from
More informationHigher Geometry Problems
Higher Geometry Problems (1 Look up Eucidean Geometry on Wikipedia, and write down the English translation given of each of the first four postulates of Euclid. Rewrite each postulate as a clear statement
More information4 The Trigonometric Functions
Mathematics Learning Centre, University of Sydney 8 The Trigonometric Functions The definitions in the previous section apply to between 0 and, since the angles in a right angle triangle can never be greater
More information. Then 2 3 the circumference is 8π 25 in. 2(1) both of which violate the original assumptions. Therefore, we have no solution.
WYSE MATH SECTIONAL 04 SOLUTIONS Ans B: (a) and (d) are equivalent Therefore since the determinant of the product of matrices is the product of their determinants, it too cannot be zero If three square
More informationGeometry Note Cards EXAMPLE:
Geometry Note Cards EXAMPLE: Lined Side Word and Explanation Blank Side Picture with Statements Sections 12-4 through 12-5 1) Theorem 12-3 (p. 790) 2) Theorem 12-14 (p. 790) 3) Theorem 12-15 (p. 793) 4)
More informationName Geometry Common Core Regents Review Packet - 3. Topic 1 : Equation of a circle
Name Geometry Common Core Regents Review Packet - 3 Topic 1 : Equation of a circle Equation with center (0,0) and radius r Equation with center (h,k) and radius r ( ) ( ) 1. The endpoints of a diameter
More informationhmhco.com Adaptive. Intuitive. Transformative. AGA Scope and Sequence
hmhco.com Adaptive. Intuitive. Transformative. AGA Algebra 1 Geometry Algebra 2 Scope and Sequence Number and Quantity The Real Number System (N-RN) Properties of exponents to rational exponents Properties
More informationMASSACHUSETTS ASSOCIATION OF MATHEMATICS LEAGUES STATE PLAYOFFS Arithmetic and Number Theory 1.
STTE PLYOFFS 004 Round 1 rithmetic and Number Theory 1.. 3. 1. How many integers have a reciprocal that is greater than 1 and less than 1 50. 1 π?. Let 9 b,10 b, and 11 b be numbers in base b. In what
More informationBENCHMARKS GRADE LEVEL INDICATORS STRATEGIES/RESOURCES
GRADE OHIO ACADEMIC CONTENT STANDARDS MATHEMATICS CURRICULUM GUIDE Tenth Grade Number, Number Sense and Operations Standard Students demonstrate number sense, including an understanding of number systems
More informationCorrelation of 2012 Texas Essential Knowledge and Skills (TEKS) for Algebra I and Geometry to Moving with Math SUMS Moving with Math SUMS Algebra 1
Correlation of 2012 Texas Essential Knowledge and Skills (TEKS) for Algebra I and Geometry to Moving with Math SUMS Moving with Math SUMS Algebra 1 ALGEBRA I A.1 Mathematical process standards. The student
More informationContent Descriptions Based on the state-mandated content standards. Analytic Geometry
Content Descriptions Based on the state-mandated content standards Analytic Geometry Introduction The State Board of Education is required by Georgia law (A+ Educational Reform Act of 2000, O.C.G.A. 20-2-281)
More informationCircle geometry investigation: Student worksheet
Circle geometry investigation: Student worksheet http://topdrawer.aamt.edu.au/geometric-reasoning/good-teaching/exploringcircles/explore-predict-confirm/circle-geometry-investigations About these activities
More informationPage 1
Pacing Chart Unit Week Day CCSS Standards Objective I Can Statements 121 CCSS.MATH.CONTENT.HSG.C.A.1 Prove that all circles are similar. Prove that all circles are similar. I can prove that all circles
More informationCLASS IX GEOMETRY MOCK TEST PAPER
Total time:3hrs darsha vidyalay hunashyal P. M.M=80 STION- 10 1=10 1) Name the point in a triangle that touches all sides of given triangle. Write its symbol of representation. 2) Where is thocenter of
More informationLecture 1: Axioms and Models
Lecture 1: Axioms and Models 1.1 Geometry Although the study of geometry dates back at least to the early Babylonian and Egyptian societies, our modern systematic approach to the subject originates in
More informationPythagoras Theorem and Its Applications
Lecture 10 Pythagoras Theorem and Its pplications Theorem I (Pythagoras Theorem) or a right-angled triangle with two legs a, b and hypotenuse c, the sum of squares of legs is equal to the square of its
More informationChemical Applications of Symmetry and Group Theory Prof. Manabendra Chandra Department of Chemistry Indian Institute of Technology, Kanpur.
Chemical Applications of Symmetry and Group Theory Prof. Manabendra Chandra Department of Chemistry Indian Institute of Technology, Kanpur Lecture 08 Hello and welcome to the day 3 of the second week of
More informationPOINTS, LINES, DISTANCES
POINTS, LINES, DISTANCES NIKOS APOSTOLAKIS Examples/Exercises: (1) Find the equation of the line that passes through (4, 5), (4, ) () Find the equation of the line that passes through the points (1, ),
More informationMA 460 Supplement: Analytic geometry
M 460 Supplement: nalytic geometry Donu rapura In the 1600 s Descartes introduced cartesian coordinates which changed the way we now do geometry. This also paved for subsequent developments such as calculus.
More informationSolutions to Exercises in Chapter 1
Solutions to Exercises in hapter 1 1.6.1 heck that the formula 1 a c b d works for rectangles but not for 4 parallelograms. b a c a d d b c FIGURE S1.1: Exercise 1.6.1. rectangle and a parallelogram For
More informationLesson Plan by: Stephanie Miller
Lesson: Pythagorean Theorem and Distance Formula Length: 45 minutes Grade: Geometry Academic Standards: MA.G.1.1 2000 Find the lengths and midpoints of line segments in one- or two-dimensional coordinate
More information3. MATHEMATICS (CODE NO. 041) The Syllabus in the subject of Mathematics has undergone changes from time to time in accordance with growth of the subject and emerging needs of the society. The present
More information