12-2 Chords And Arcs
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
Error Prevention!BExercises 17–19 Students may0Find m AB . (Hint: You will need to use trigonometry in Exercise 19.)about 123.99017. 10818.19.C108 BOO16O3017ABDABA16Apply Your Skillsconfuse the measure of an arcwith arc length. Remind themthat the letter m signals themeasure of the arc.Technology TipExercise 19 Review with studentshow to use the sin-1 calculatorfunction key.20. Archaeology An archaeologist found several jar fragments including a largepiece of the circular rim. How can she find the center and radius of the rim tohelp her reconstruct the jar? See margin.Tactile Learners21. Geometry in 3 Dimensions In the figure at the right,sphere O with radius 13 cm is intersected by a plane5 cm from center O. Find the radius of cross section A.5 cm O 13 cm12 cm22. Geometry in 3 Dimensions A plane intersects a sphereAthat has radius 10 in. forming cross section B with radius8 in. How far is the plane from the center of the sphere? 6 in.Exercise 20 Have students usea compass and straightedge andthe concepts in this exercise tofind the center of a circle circumscribed about three noncollinearpoints. This method also can beused in Exercise 30.Real-WorldConnectionCareers Field archaeologistsanalyze artifacts to provideglimpses of life in the past.23. Complete this proof of Theorem 12-4, Part (1).Given: P with &KPM &LPNExercises 30–32 Discuss theseexercises as a class. Have studentssuggest different solution methods,such as using the PythagoreanTheorem or showing that ACBDis a rhombus.Prove: KM LN a–c. PL; PM; All radii of a circle are O.Proof: KP a. 9 b. 9 NP because c. 9.#KPM d. 9 by e. 9. KM LN by f. 9.d–f. kLPN; SAS; CPCTC24. Complete this proof of Theorem 12-4, Part (2).Exercise 33 Show students howb. AB O CDRecall that in a circlecongruent centralangles interceptcongruent arcs.36. X is equidist. from Wand Y, since XW and XYare radii. So X is on the# bis. of WY by theConv. of the # Bis. Thm.But is the # bis. ofWY , so contains X.674b. 9c. 9Proof AEB CEDd. 9NCEAD(Hint: Begin by drawing OA and OB.)GO for HelpFor a guide to solvingExercise 26, see p. 677.27. He doesn’t knowthat the chords areequidistant from thecenter.674e. 9CPCTCAB CDf. 925. Prove Theorem 12-6. See back of book.Given: O with diameter ED ' AB at C0 0Prove: AC BC and AD BDd. SSSf. O central ' have Oarcs.EA EC ED EBa. 9Problem Solving Hintc. Givene. lAEB O lCEDMB20. She can draw 2 chords,and their # bisectors, ofthe partial circle. Theintersection pt. of the #bisectors will be thecenter and she can thenmeasure the radius.24. a. All radii of a circle areO.LPGiven: E with congruent chords AB and CD0 0Prove: AB CD See margin.they can substitute x 2 intox2 y2 25 to find the positiveand negative y-coordinates forthe chord.KAECODB26. Two concentric circles have radii of 4 cm and 8 cm. A segment tangent to thesmaller circle is a chord of the larger circle. What is the length of the segment?about 13.9 cmP27. Error Analysis Scott looks at this figure and concludesthat ST PR. What is wrong with Scott’s conclusion?RSee left.Q28. Open-Ended Use a circular object such as a can or aSTsaucer to draw a circle. Construct the center of the circle.Check students’ work.29. Writing Theorems 12-4 and 12-5 both begin with the phrase “Within a circle orin congruent circles.” Explain why “congruent” is essential for both theorems.Circles can have O chords or O central ' without having both.Chapter 12 Circles38. All radii of (0 are O, sokAOB O kCOD by SSS.lA O lC by CPCTC.Also, lOEA O lOFCsince both are rt. '.Thus, kOEA O kOFC byAAS, and OE O OF byCPCTC.39. 1. (A with CE # BD(Given) 2. CF O CF(Refl. Prop. of O)3. BF O FD (diameter #to a chord bisects thechord.) 4. lCFB andlCFD are rt. ' (Def. of#). 5. kCFB O kCFD(SAS) 6. BC0O CD 0(CPCTC) 7. BC O DC(O chords have O arcs.)
A and B are congruent. CD is a chord of both circles.GOnlineBAHomework Help31. AB 24 cm, radius 13 cm. How long is CD? 10 cmVisit: PHSchool.comWeb Code: aue-120232. radius 13 ft, CD 24 ft. How long is AB? 10 ftD33. Multiple Choice In the diagram at the right, theendpoints of the chord are the points where theline x 2 intersects the circle x2 1 y2 5 25. Whatis the length of the chord? C4.68.09.210.0y0 035. 1. (P with QS0O RT(Given) 2. m QS 0mlQPS and m RT mlRPT (Arc measure central0 lmeasure.)03. m QS m RT(Def. of O) 4. lQPS OlRPT (Subst.)Proof35. Prove Theorem 12-4, Part (3).0 0Given: P with QS RTLesson QuizFor Exercises 1–5, use thediagram of L below.x 23Mx-3OProve: &QPS &RPTY-3Prove: / contains the center of X.See margin p. 674.RQXTW/ZYProof37. PQ and PR are chords of C. Prove that if C is on the bisector of &QPR, thenPQ 5 PR. See back of book.Proof38. Prove Theorem 12-5, Part (2).Given: O with AB CD39. Given: A with CE ' BD0 0Prove: BC DCProve: OE OF 38-39. See margin.AECFEConnectionThe cylinders used on milktank trucks lie on the lateralsurface and have a verticalbase at each end.Proof41. If two circles are concentric and a chord of the larger circle is tangent to thesmaller circle, prove that the point of tangency is the midpoint of the chord.See margin.Lesson 12-2 Chords and Arcslesson quiz, PHSchool.com, Web Code: aua-120241. Let O be the center ofthe circles, and P be thept. of tangency of thelarger circle’s chord tothe smaller circle. ThenOP is # to the chord,and therefore bisects it.2. If YM and ZN are congruentchords, explain why youcannot conclude thatLV LC. You do not knowwhether LV and LC areperpendicular to the chords.3. Suppose that YM has length 12in., and its distance from pointL is 5 in. Find the radius of Lto the nearest tenth. 7.8 in.D40. Dairy The diameter of the base of a59 in.cylindrical milk tank is 59 in. The lengthof the tank is 470 in. You estimate that20 in.the depth of the milk in the tank is 20 in.470 in.Find the number of gallons of milk in thenot to scaletank to the nearest gallon. (1 gal 231 in.3) 1661 galReal-World1. If YM and ZN are congruentchords, what can youconclude?00YM O ZN ; lYLM O lZLN5. Find LV to the nearest tenth.6.9 cmAFCZ4. Find YM. 22 cmBBLFor Exercises 4 and 5, supposethat LV # YM, YV 11 cm, and L has a diameter of 26 cm.DONCGiven: / is the ' bisector of WY.See left.V336. Prove Theorem 12-8.SChallengePowerPoint34. Critical Thinking The diameter of a circle is 20 cm. Two chords parallel to thediameter are 6 cm and 16 cm long. What are the possible distances between thechords to the nearest tenth of a centimeter? 3.5 cm, 15.5 cmPC4. Assess & ReteachCGPS 30. AB 8 in., CD 6 in. How long is a radius? 5 in.Alternative AssessmentHave students use only picturesand mathematical symbols toexpress each theorem in thislesson. Then have each studentexchange their work with apartner who writes a paragraphbelow the picture evaluating thepresentation for accuracy andclarity. Use the drawings andparagraphs to assess students’understanding.675So P is the midpt. ofthe chord.675
Test PrepTest PrepResourcesMultiple ChoiceFor additional practice with avariety of test item formats: Standardized Test Prep, p. 711 Test-Taking Strategies, p. 706 Test-Taking Strategies withTransparencies42. The diameter of a circle is 25 cm and a chord of the same circle is 16 cm.To the nearest tenth, what is the distance of the chord from the center ofthe circle? BA. 9.0 cmB. 9.6 cmC. 18.0 cmD. 19.2 cm43. In the figure at the right, what is the valueof x to the nearest tenth? GF. 3.0G. 6.2H. 6.8J. 9.05x8 O44. A 9-cm chord is 11 cm from the center of a circle.What is the radius of the circle? BA. 9.0 cmB. 11.9 cmC. 13.0 cmD. 14.2 cm45. The radius of a circle is 10.8 ft. The length of a chord is 12 ft. What is theapproximate distance of the chord from the center of the circle? HF. 1.2 ftG. 4.7 ftH. 9.0 ftJ. 12.4 ftc46. What can you NOT conclude from the diagramdat the right? DA. c dC. c2 e2 b2b eOaB. a bD. e d47. A and B intersect at points C and D. Each circle has radius 6 in. andAB 8 in. What is CD? JF. 4.5 in.G. 6 in.H. 8 in.J. 8.9 in.Short Response48. Circles M and N are congruent with radiimeasuring 13 cm. PQ is a chord of both circlesand PQ 18 cm. To the nearest tenth,find MN. Justify your answer. See margin.PMXNQMixed ReviewGO forHelpLesson 12-1Assume that lines that appear to be tangent are tangent. O is the center of eachcircle. Find the value of x to the nearest tenth.49.50. 5.5O 140 x 6.840676xO2Lesson 8-551. From the top of a building you look down at an object on the ground. If youreyes are 50 feet above the ground and the angle of depression is 50 , how far isthe object on the ground from the base of the building? about 42 ftLesson 7-552. The legs of a right triangle are 10 in. and 24 in. long. Find the lengths, to thenearest tenth, of the segments into which the bisector of the right angle dividesthe hypotenuse. 18.4 in. and 7.6 in.Chapter 12 Circles48. [2] PN is a radius of (N.Thus, PN 13 cm.PX 12 PQ 9,kPNX is a rt. k, so6763.5NX N 9.38 cm. Thus,MN 18.8 cm, to thenearest tenth.[1] incorrect length ORincorrect explanation
Use: Interactive Textbook, 12-2 670 12-2 1. Plan Objectives 1 To use congruent chords, arcs, and central angles 2 To recognize properties of lines through the center of a circle Examples 1 Using Theorem 12-4 2 Using Theorem 12-5 3 Using Diameters and Chords Math Background Theorem 12-8 can be used to prove the theorem of analytic geometry that .File Size: 619KBPage Count: 7Explore furtherDownload Free Practice 12 2 Chords And Arcs Answer Key .www.findanswerkey.com12-2 Practice Chords and Arcs - studyres.comstudyres.comCircles - Arcs and chords Worksheetswww.math-worksheet.orgUnit Circle Worksheet with Answers. Find angle based on .www.mathwarehouse.comCircles worksheet day #1 - Ms. Sullivan's Geometry Websitesullivangeometry.weebly.comRecommended to you based on what's popular Feedback
Holt McDougal Geometry Arcs and Chords Example 3A: Applying Congruent Angles, Arcs, and Chords TV WS. Find mWS. 9n – 11 7n 11 2n 22 n 11 88 chords have arcs. Def. of arcs Substitute the given measures. Subtract 7n and add 11 to both sides. Divide both sides by 2. Substitute 11 for n. Simplify. mTV mWS mWS 7(11) 11
Holt McDougal Geometry Arcs and Chords Example 1a: Applying Congruent Angles, Arcs, and Chords TV WS. Find mWS. 9n – 11 7n 11 2n 22 n 11 88 chords have arcs. Def. of arcs Substitute the given measures. Subtract 7n and add 11 to both sides. Divide both sides by 2. Substitute 11 for n. Simplify. mTV mWS mWS 7(11) 11 TV WS
The L-Acoustics ARCS stand can hold up to four L-Acoustics ARCS Wide, L-Acoustics ARCS Focus, or a combination of the two, when used vertically. When deployed vertically, the L-Acoustics ARCS mounting table can be titled downwards at angles of 15 to 30 , 45 or 60 from the original angle. L-Acoustics ARCS Wide / Focus Color: RAL 7035 BP1000
traveling from any town to any other town within the country. Problem 6. (a) A. circle is divided into 10 equal arcs. are joined in pairs by 5 chords. Is. it. these chords are of equal length? (b) A. circle is divided into 20 equal arcs. are joined in pairs by 10 chords. Is it these chords ar
The following lessons introduce some of the basics of jazz guitar chords. Topics include: 1. Some of the most common chord voicings used 2. Intervals 3. Triads 4. Seventh Chords, Drop 2 and Drop 3 5. Extensions and Sus chords 6. Quartal Chords 7. Freddie Green Style 8. Sample Comping Rhythms 9. Bossa Nova Style 10. Chords with Bass Line
The green octave: Low string chords (C2-B2) Major & Minor chords triggered by velocity The blue octave: Low, Mid, High chords played together (C1-B1) Major & Minor chords triggered by velocity Dynamics: The mod wheel gives you further possibilities to adjust the dynamics and enables you to mix sustain and tremolo chords.
BARRE CHORDS FOR THE GUITAR by DON COMANDA FOREWORD Following are over 1,700 chords. Barre chords are shown throughout except in cases where a Barre chord would give too “thick” a sound. A quick reference section is included showing major, minor, and dominant seventh chords. Similarly fingered chords appear with different names.
AAMI HE75, Human factors engineering – Design of medical devices, Clause 9, Usability Testing, provides an excellent guide to the types of formative evaluations that are useful in early device UI development such as cognitive walkthroughs, heuristic evaluations, and walk-through-talk-through usability tests. Annex D of IEC 62366 also provides descriptions of these formative techniques .
