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12-212-2Chords and Arcs1. PlanObjectives12To use congruent chords, arcs,and central anglesTo recognize properties oflines through the centerof a circleExamples123Using Theorem 12-4Using Theorem 12-5Using Diameters and ChordsGO for HelpWhat You’ll LearnCheck Skills You’ll Need To use congruent chords,Find the value of each variable. Leave your answer in simplest radical form.arcs, and central angles1. To recognize properties of11"22lines through the center ofa circle2. 53.c111445 . . . And Why45 aTo see how an archaeologistfinds the center and radiusof the rim of a jar, as inExercise 20Lesson 8-2b60 2825 New Vocabulary chordMath BackgroundTheorem 12-8 can be used toprove the theorem of analyticgeometry that states that anythree noncollinear pointsdetermine a unique circle. It alsocan be used to justify a method ofconstructing the circle. Constructthe perpendicular bisectors oftwo of the three possible segments.Construct a circle whose centeris the point of intersection ofthe perpendicular bisectors andwhose radius is the distancefrom the center to any of thethree points.1Using Congruent Chords, Arcs, and Central AnglesA segment whose endpoints are on a circle is called a chord.0The diagram shows the related chord and arc, PQ and PQ .PQOThe following theorem is about related central angles, chords,and arcs. It says, for example, that if two central angles in a circleare congruent, then so are the two chords and two arcs that theangles intercept.Key ConceptsTheorem 12-4Within a circle or in congruent circles(1) Congruent central angles have congruent chords.(2) Congruent chords have congruent arcs.More Math Background: p. 660C(3) Congruent arcs have congruent central angles.Lesson Planning andResourcesSee p. 660E for a list of theresources that support this lesson.For: Chords and Arcs ActivityUse: Interactive Textbook, 12-2You will prove Theorem 12-4 in Exercises 23, 24, and 35.1Using Theorem 12-40 0In the diagram, O P. Given that BC DF ,what can you conclude?PowerPointBell Ringer PracticeBy Theorem 12-4, &O &P andBC DF.Check Skills You’ll NeedFor intervention, direct students to:Quick CheckUsing 45 -45 -90 TrianglesLesson 8-2: Example 2Extra Skills, Word Problems, ProofPractice, Ch. 8EXAMPLE670BODCPF1 If you are instead given that BC DF, what can you conclude?0 0lO O lP; BC O DFChapter 12 CirclesUsing 30 -60 -90 TrianglesLesson 8-2: Example 4Extra Skills, Word Problems, ProofPractice, Ch. 8670Special NeedsBelow LevelL1Review with students why congruent arcs must be inthe same circle or in congruent circles. Also point outthat a chord is related to the minor arc it intercepts.learning style: verbalL2Have students draw diagrams for Theorems 12-6,12-7, and 12-8 that accurately represent thegiven information.learning style: visual

Theorem 12-5 shows a relationship between two chords and their distances fromthe center of a circle. You will prove part (2) in Exercise 38.Key ConceptsTheorem 12-52. TeachGuided InstructionWithin a circle or in congruent circlesVisual Learners(1) Chords equidistant from the center are congruent.On the board, copy the diagrambelow that summarizes Theorem12-4.(2) Congruent chords are equidistant from the center.ProofCongruentCentral AnglesProof of Theorem 12-5, Part Prove: AB CDSteel beams model congruentchords equidistant from thecenter to give the illusion ofa circle.EAGiven: O, OE OF,OE ' AB, OF ' CDFReasons1. OA OB OC OD1. Radii of a circle are congruent.CongruentChordsAsk: Can you conclude thatcongruent chords have congruentcentral angles? If so, how? yes;by the Law of Syllogism2. OE OF, OE ' AB, OF ' CD2. Given3. &AEO and &CFO are right angles.3. Def. of perpendicular segments4. #AEO #CFO4. HL Theorem5. &A &C5. CPCTC6. &B &A, &C &D6. Isosceles Triangle Theorem7. &B &D7. Transitive Property of Congruence8. &AOB &COD8. If two of a # are to two ofanother #, then the third are .9. AB CD9. central angles have chords.Alternative MethodAn alternate proof of part 1of Theorem 12-5 would usethe HL Theorem to prove AOE BOE COF DOFand then use CPCTC and theSegment Addition Postulate.This method also could be usedto prove Theorem 12PowerPointYou can use Theorem 12-5 to ﬁnd missing lengths in circles.Additional Examples2EXAMPLEUsing Theorem 12-5Multiple Choice What is the value of a in the circleat the ing Tip12.51825aA12.5 9PQ QR 12.5 GivenPQPQ QR PR Segment Addition PostulateIn a circle, the lengthof the perpendicularsegment from thecenter to a chord isthe distance from thecenter to the chord.1 In the diagram, radius OXbisects &AOB. What can youconclude?RO25 PR Substitute.Chords equidistant from the centera PR of a circle are congruent.a 25BSubstitute.The correct answer is D.Quick Check2 Find the value of x in the circle at the right.16lAOX O lBOX; AX O BX ;AX O BX1818162 Find AB.x36B4C4AQLesson 12-2 Chords and ArcsAdvanced LearnersX6717R7S14English Language Learners ELLL4Have students write a paragraph to explain whythe phrase that is not a diameter is necessary inTheorem 12-7.Ask: Is a diameter a chord? Explain. Yes; it is asegment with two endpoints on the circle.Is a radius a chord? Explain. No; it has only 1 pointon the circle.learning style: verballearning style: verbal671

Guided Instruction321Lines Through the Center of a CircleError PreventionEXAMPLEThe Converse of the Perpendicular Bisector Theorem from Lesson 5-2 has specialapplications to a circle and its diameters, chords, and arcs.Because the figures in parts a andb do not show diameters, somestudents may not understand whyTheorems 12-6 and 12-7 apply.Have them reread the sectionabove Theorem 12-6 to reinforcethat the theorems apply to linesor segments that contain thecenter of the circle.Key ConceptsTheorem 12-6In a circle, a diameter that is perpendicular to a chord bisects the chordand its arcs.Theorem 12-7In a circle, a diameter that bisects a chord (that is not a diameter)is perpendicular to the chord.PowerPointTheorem 12-8Additional ExamplesIn a circle, the perpendicular bisector of a chord contains the centerof the circle.3 P and Q are points on O.The distance from O to PQ is 15in., and PQ 16 in. Find theradius of O. 17 in.ProofProof of Theorem 12-7L3XY and YZ are perpendicularchords within C that are alsoequidistant from center C. Whatis the most precise name forquadrilateral MYNC? Explain.M3SUsing Diameters and ChordsEXAMPLEAlgebra Find each missing length to the nearest tenth.a.LN 12(14) 7 A diameter ' to a chord bisectsthe chord.2 32 72rUse the Pythagorean Theorem.Kr3 cmr 7.6Find the square root of each side.NML14 cmCXReal-WorldConnectionThe center of the tire islocated on the perpendicularbisector of the flat part.672672RTYou will prove Theorems 12-6 and 12-8 in Exercises 25 and 36, respectively.ZSquare; congruent chords areequidistant from the center,and a diameter that bisects achord is # to the chord.VProof: TS TU because the radii of a circle are congruent. VS VU bythe deﬁnition of bisect. Thus, T and V are equidistant from S and U.By the Converse of the Perpendicular Bisector Theorem, T and V are on theperpendicular bisector of SU. Since two points determine one line, TV is theperpendicular bisector of SU. Another name for TV is QR. Thus, QR ' SU.ClosureNQProve: QR ' SU Daily Notetaking Guide 12-2—L1Adapted InstructionYUGiven: T with diameter QRbisecting SU at V.Resources Daily Notetaking Guide 12-2Chapter 12 Circlesb.A1511yC11BBC ' ACA diameter that bisects a chord that isnot a diameter is ' to the chord.y 2 112 15 2Use the Pythagorean Theorem.y2 104y 10.2Solve for y2.Find the square root of each side.

Quick Check3. Practice3 Use the circle at the right.a. Find the length of the chord. about 11b. Find the distance from the midpoint ofthe chord to the midpoint of its minor arc.2.86.8Assignment Guide4x1 A B 1-8, 17, 23, 24, 27,29-32, 352 A BEXERCISESFor more exercises, see Extra Skill, Word Problem, and Proof Practice.Practice and Problem SolvingPractice by ExampleExample 1GO forHelp1. B(page 670)XCAExample 2(page 671)0 0BC O YZ ; BC O YZ00 02. 0ET O GH O JN OML ; ET O GH O JN OML; lTFE O lHFG;lJKN O lMKLZ2. T1–2.See left. EYHFMLJG4. 23. 14To check students’ understandingof key skills and concepts, go overExercises 8, 12, 24, 29, 30.NO2x5Exercises 12, 14 Students mayfind it helpful to draw and labelthe third side of the triangle,using the fact that all radii arecongruent.10xx575.72.5 2.5757. 86. 5025188.1015x183.5x8x5Example 3Use the diagram at the right to complete Exercises 9 and 10.(page 672)9. Given that AB is a diameter of the circle and AB ' CD,then a. 9 b. 9 and c. 9 d. 9. See left.CEAc. lCEB12. 5.4x 16d. lDEAOName62215.9.9y42. 5, -1 3. 2, -6 4. 4, -4 5. 0, 0 6. -2, -4 ZO67. Z S Y8. V S W9. U S X10. Y S W11. U S Z12. W S V 2X4 x2 2 4VWUse matrices to ﬁnd the image of each ﬁgure under the given translation.13. translation 2, 4 14. translation -2, 1 44K2X 4 2 2W16.20.815. translation 5, -3 yy1518 x152U 4Find the vector that describes the given translation.xL3DateTranslations1. 2, -2 4xClassYx14.12.5L1Practice 12-213.8L2ReteachingPracticeD8.93.620 OBWhat is the image of Z under each translation?Algebra Find the value of x to the nearest tenth.L3L4Adapted Practice10. Given that AB is the perpendicular bisector of CD, thenAB contains 9. the center of the circle11. 6GPS Guided Problem SolvingEnrichmenta. CEb. DE42-4849-52Homework Quick CheckKFind the value of x.1.9. Answers may vary.Samples are given.Test PrepMixed ReviewIn Exercises 1 and 2, the circles are congruent. What can you conclude?Y242M 2N 4 2 0 2Q 42 4 J 2 2L 44 xZy424x4xPWrite a rule to describe each translation.16.17.yA' 4 212B' 24xB 418.y42A2J' 4 2 0 2L' 4QK6 x6Find a single translation that has the same effect as each compositionof translations.xyJL2 4K'19. 3, 5.2 followed by 1.2, 6 20. 4, -8 followed by 9, -5 21. 7, 11 followed by -7, -11 22. 1, 2 followed by 2, 1 42Q'RR' 4P 2 0 S 2 4S'P' 46 x Pearson Education, Inc. All rights reserved.A9-16, 18-22, 25, 26,28, 33, 34, 36, 37C Challenge38-4123. PNQ has vertices P(2, 5), N(-3, -1), and Q(4, 0).a. Determine the image of P under the translation -5, -6 .b. Use matrices to ﬁnd the image of PNQ under the translation -2, 3 .Lesson 12-2 Chords and Arcs673673