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Apply Properties of Chords

Apply Properties of Chords. Sunday, August 10, 2014. Essential Question: How do we use relationships of arcs and chords in a circle?. Lesson 6.3. M2 Unit 3: Day 3. Describe each figure as a minor arc, major arc or a semicircle. Find the arc measure. 1. BC. o. ANSWER. minor arc, 32.

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Apply Properties of Chords

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  1. Apply Properties of Chords Sunday, August 10, 2014 Essential Question: How do we use relationships of arcs and chords in a circle? Lesson 6.3 M2 Unit 3: Day 3

  2. Describe each figure as a minor arc, major arc or a semicircle. Find the arc measure. 1. BC o ANSWER minor arc, 32 Daily Homework Quiz Daily Homework Quiz

  3. CBE o ANSWER major arc, 212 Daily Homework Quiz Daily Homework Quiz Describe each figure as a minor arc, major arc or a semicircle. Find the arc measure. 2.

  4. BCE o ANSWER semicircle, 180 Daily Homework Quiz Daily Homework Quiz Describe each figure as a minor arc, major arc or a semicircle. Find the arc measure. 3.

  5. ~ ~ ~ = = = AE BC BC BC AE AE ANSWER mAFE = m BFC because the angles are vertical angles, so AFEBFC.Then arcs and are arcs that have the same measure in the same circle. By definition . Daily Homework Quiz Describe each figure as a minor arc, major arc or a semicircle. Find the arc measure. . 4. Explain why

  6. . o Two diameters ofPareABandCD.If m = 50 , find m and m . AC AD ACD o o ANSWER 310 ; 130 Daily Homework Quiz 5.

  7. Tell whether the segment is best described as a radius, chord, or diameter of C. 1.DC 4.AE 2.BD 3.DE Warm Ups radius diameter chord chord 5.Solve 4x = 8x – 12. 6.Solve 3x + 2 = 6x – 4. x = 2 x = 3

  8. Theorem 6.5 In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent.

  9. In the diagram, PQ, FGJK, and mJK= 80o. Find mFG Because FGand JKare congruent chords in congruent circles, the corresponding minor arcs FGand JKare congruent. So, mFG = mJK = 80o. Use congruent chords to find an arc measure SOLUTION

  10. Use the diagram of D. 1. If mAB = 110°, find mBC Because ABand BCare congruent chords in the same circle, the corresponding minor arcs ABand BCare congruent. So, mBC = mAB = 110o. SOLUTION

  11. Use the diagram of D. ANSWER 2. If mAC = 150°, find mAB And, mBC + mAB + mAC = 360° So, 2 mAB + mAC = 360° Because ABand BCare congruent chords in the same circle, the corresponding minor arcs ABand BCare congruent. So, mBC = mAB 2 mAB +150° = 360° 2 mAB = 360 – 150 2 mAB = 210 mAB = 105° mAB = 105° GUIDED PRACTICE Substitute Subtract Simplify

  12. Theorem 6.6 If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.

  13. Theorem 6.7 If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.

  14. Use the diagram of Eto find the length of AC. Tell what theorem you use. Diameter BDis perpendicular to AC. So, by Theorem 6.7, BDbisects AC, and CF = AF. Therefore, AC = 2 AF =2(7) = 14. Use a diameter SOLUTION

  15. Find the measure of the indicated arc in the diagram. 3. CD From the diagram Diameter BDis perpendicular to CE. So, by Theorem 6.7, BDbisects CE, Therefore mCD=mDE. So mCD=9x° =72° SOLUTION So 9x°=(80 – x)° So 10x°=80° x=8°

  16. Find the measure of the indicated arc in the diagram. 4. DE mCD=mDE. So mDE=72° 5. CE mCE=mDE + mCD So mCE=72°+ 72° = 144° SOLUTION SOLUTION

  17. Theorem 6.8 In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.

  18. In the diagram of C, QR = ST = 16. Find CU. Chords QRand STare congruent, so by Theorem 6.8 they are equidistant from C. Therefore, CU = CV. SOLUTION CU = CV Use Theorem 6.8 2x = 5x – 9 Substitute. Solve for x. x =3 So, CU = 2x = 2(3) = 6.

  19. Suppose ST = 32, and CU= CV = 12. Find the given length. 6. QR SOLUTION Since CU = CV. Therefore Chords QR and ST are equidistant from center and from theorem 6.8QR is congruent to ST QR = ST Use Theorem 6.8. QR= 32 Substitute.

  20. Suppose ST = 32, and CU= CV = 12. Find the given length. 1 1 SoQU = QR SoQU = (32) 2 2 7. QU SOLUTION Since CU is the line drawn from the center of the circle to the chord QR it will bisect the chord. Substitute. QU= 16

  21. Suppose ST = 32, and CU= CV = 12. Find the given length. 8. The radius of C Join the points Q and C. Now QUC is right angled triangle. Use the Pythagorean Theorem to find the QC which will represent the radius of the C SOLUTION

  22. 8. The radius of C ANSWER The radius of C = 20 Suppose ST = 32, and CU= CV = 12. Find the given length. SOLUTION So QC2 = QU2 + CU2 By Pythagoras Thm So QC2 = 162+ 122 Substitute So QC2 = 256 + 144 Square So QC2 = 400 Add So QC = 20 Simplify

  23. Homework Page 203-204 # 4 – 21 all.

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