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Properties of a Chord

Properties of a Chord. Circle Geometry. Homework: Lesson 6.2/1-12, 18 Quiz Friday Lessons 6.1 – 6.2 Ying Yang Project Due Friday. What is a chord?. A chord is a segment with endpoints on a circle. Any chord divides the circle into two arcs.

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Properties of a Chord

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  1. Properties of a Chord Circle Geometry Homework: Lesson 6.2/1-12, 18 Quiz Friday Lessons 6.1 – 6.2 Ying Yang Project Due Friday

  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. What is a chord? • The diameter of a circle is the longest chord of any circle since it passes through the center. • A diameter satisfies the definition of a chord, however, a chord is not necessarily a diameter. Therefore,every diameter is a chord, but not every chord is a diameter.

  4. Chord Property #1, #2 and #3 • A perpendicular line from the center of a chord to the center of a circle: #1: Makes a 90° angle with the chord #2: Creates two equal line segments RS and QR #3: Must pass through the center of the circle O

  5. is a diameter of the circle.

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

  7. AB  CD if and only if EF  EG.

  8. if and only if  Chord Arcs Conjecture • In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent.

  9. 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

  10. 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.

  11. Because AD = DC, and = . So, m = m Ex: Using Chord Arcs Conj. (x + 40)° 2x° 2x = x + 40 Substitute Subtract x from each side. x = 40

  12. Perpendicular Bisector to a Chord Conjecture 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: Finding the Center of a Circle

  13. Step 2: Draw the perpendicular bisector of each chord. These are the diameters. Ex: Finding the Center of a Circle con’t

  14. Step 3: The perpendicular bisectors intersect at the circle’s center. Ex: Finding the Center of a Circle con’t

  15. In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center. AB  CD if and only if EF  EG. Chord Distance to the Center Conjecture

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

  17. Ex: Find CF con’t Because AB and DE are congruent chords, they are equidistant from the center. So CF  CG. To find CG, first find DG. CG  DE, so CG bisects DE. Because DE = 8, DG = =4.

  18. Ex: Find CF con’t Then use DG to find CG. DG = 4 and CD = 5, so ∆CGD is a 3-4-5 right triangle. So CG = 3. Finally, use CG to find CF. Because CF  CG, CF = CG = 3

  19. An angle whose vertex is on the center of the circle and whose sides are radii of the circle An angle whose vertex is ON the circle and whose sides are chords of the circle

  20. Naming Arcs Arcs are defined by their endpoints Minor Arcs require the 2 endpoints Major Arcs require the 2 endpoints AND a point the arc passes through Semicircles also require the 2 endpoints (endpoints of the diameter) AND a point the arc passes through K

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

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

  23. What can you tell me about segment AC if you know it is the perpendicular bisectors of segments DB? D It’s the DIAMETER!!! A C B

  24. If one chord is a perpendicular bisector of another chord, then the first chord is a diameter. This 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. Finding the Center of a Circle

  25. Step 2: Draw the perpendicular bisector of each chord. These are diameters of the circle. Finding the Center of a Circle

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

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

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

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

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

  31. Congruent chords intercept congruent arcs

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

  33. Let’s get crazy… Find c. 3 8 c 12

  34. Let’s get crazy… Find c. 3 8 c 12 Step 1: What do we need to find? We need a radius to complete this big triangle.

  35. Let’s get crazy… Find c. 3 8 c 12 How do we find a radius? We can draw multiple radii (radiuses).

  36. Let’s get crazy… Find c. 5 4 3 8 c 12 How do we find a radius? Now what do we have and what will we do? We create this triangle, use the chord property to find on side & the Pythagorean theorem find the missing side.

  37. Let’s get crazy… Find c. 5 4 3 13 c 12 !!! Pythagorean Theorem !!! c2=a2+b2 c2=52+122 c2=169 c = 13.

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