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Lesson 6.2 Properties of Chords

Objective : Discover properties of chords of a circle. Lesson 6.2 Properties of Chords. Homework: Lesson 6.2/1-12. What is a chord? A chord is a segment with endpoints on a circle. Any chord divides the circle into two arcs. A diameter divides a circle into two semicircles.

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Lesson 6.2 Properties of Chords

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  1. Objective: Discover properties of chords of a circle Lesson 6.2 Properties of Chords Homework: Lesson 6.2/1-12

  2. What is a chord? A chord is a segment with endpoints on a circle. Any chord divides the circle into two arcs. A diameter divides a circle into two semicircles. Any other chord divides a circle into a minor arc and a major arc.

  3. Chord Arcs Conjecture In the same circle, two minor arcs are congruent if and only if their corresponding chords are congruent. IFF G and IFF and

  4. If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc. Perpendicular Bisector of a Chord Conjecture H

  5. is a diameter of the circle.

  6. If one chord is a perpendicular bisector of another chord, then the first chord passes through the center of the circle and is a diameter. Perpendicular Bisector to a Chord Conjecture is a diameter of the circle.

  7. ,  If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.

  8. Ex. 4: Using Chord Arcs Conjecture D (x + 40)° 2x° C A B 2x = x + 40 x = 40

  9. Perpendicular bisector to a chord can be used to locate a circle’s center as shown in the next few slides. Step 1: Draw any two chords that are not parallel to each other. Ex. 5: Finding the Center of a Circle

  10. Step 2: Draw the perpendicular bisector of each chord. These are the diameters. Ex. 5: Finding the Center of a Circle

  11. Step 3: The perpendicular bisectors intersect at the circle’s center. Ex. 5: Finding the Center of a Circle

  12. Chord Distance to the Center Conjecture

  13. AB  CD if and only if EF  EG.

  14. AB = 8; DE = 8, and CD = 5. Find CF. Ex. 7:

  15. (x + 40)° D 2x° 2x = x + 40 x = 40

  16. Ex.4: Solve for the missing sides. A 7m 3m C D BC = AB = AD ≈ 7m 14m 7.6m B

  17. Ex.6: QR = ST = 16. Find CU. x = 3

  18. Ex 7: AB = 8; DE = 8, and CD = 5. Find CF. CG = CF CG = 3 = CF

  19. Ex.8: Find the length of Tell what theorem you used. BF = 10 Diameter is the perpendicular bisector of the chord Therefore, DF = BF

  20. Ex.9: PV = PW, QR = 2x + 6, and ST = 3x – 1. Find QR. Congruent chords are equidistant from the center.

  21. Congruent chords intercept congruent arcs

  22. Ex.11: Congruent chords are equidistant from the center.

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