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12.3 Arcs and Chords

12.3 Arcs and Chords. Geometry. Objectives/Assignment. Use properties of arcs of circles, as applied. Use properties of chords of circles. Reminder Quiz Tomorrow!!!!!! . Ex. 1: Finding Measures of Arcs. Find the measure of each arc of R. 80 °. Ex. 1: Finding Measures of Arcs.

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12.3 Arcs and Chords

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  1. 12.3 Arcs and Chords Geometry

  2. Objectives/Assignment • Use properties of arcs of circles, as applied. • Use properties of chords of circles. • Reminder Quiz Tomorrow!!!!!!

  3. Ex. 1: Finding Measures of Arcs • Find the measure of each arc of R. 80°

  4. Ex. 1: Finding Measures of Arcs • Find the measure of each arc of R. Solution: is a minor arc, so m = mMRN = 80° 80°

  5. Ex. 1: Finding Measures of Arcs • Find the measure of each arc of R. Solution: is a major arc, so m = 360° – 80° = 280° 80°

  6. Ex. 1: Finding Measures of Arcs • Find the measure of each arc of R. Solution: is a semicircle, so m = 180° 80°

  7. Ex. 2: Finding Measures of Arcs • Find the measure of each arc. m = m + m = 40° + 80° = 120° 40° 80° 110°

  8. Ex. 2: Finding Measures of Arcs • Find the measure of each arc. m = m + m = 120° + 110° = 230° 40° 80° 110°

  9. Ex. 2: Finding Measures of Arcs • Find the measure of each arc. m = 360° - m = 360° - 230° = 130° 40° 80° 110°

  10. if and only if  Theorem 12.6 • In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent.

  11. If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.  ,  Theorem 12.7

  12. If one chord is a perpendicular bisector of another chord, then the first chord is a diameter. Converse of Theorem 12.7 is a diameter of the circle.

  13. Because AD  DC, and  . So, m = m Ex. 4: Using Theorem 12.6 (x + 40)° • You can use Theorem 10.4 to find m . 2x° 2x = x + 40 Substitute Subtract x from each side. x = 40

  14. Theorem 12.7 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

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

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

  17. 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. Theorem 12.9

  18. AB = 8; DE = 8, and CD = 5. Find CF. Ex. 7: Using Theorem 12.9

  19. Ex. 7: Using Theorem 12.9 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.

  20. Ex. 7: Using Theorem 12.9 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

  21. Reminders: • Quiz after 12.3 • Last day for seniors is this Friday, make sure you return your books!

  22. Homework: Finish the worksheet 12.3 Last day for seniors is this Friday, make sure you return your books!

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