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LESSON 11.2 CHORDS AND ARCS

LESSON 11.2 CHORDS AND ARCS. OBJECTIVE: To use chords, arcs and central angles to solve problems To recognize properties of lines through the center of a circle. Label each picture as a chord, arc or a central angle:. x . central angle. arc. chord. Theorem 11.4.

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LESSON 11.2 CHORDS AND ARCS

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  1. LESSON 11.2 CHORDS AND ARCS OBJECTIVE: To use chords, arcs and central angles to solve problems To recognize properties of lines through the center of a circle

  2. Label each picture as a chord, arc or a central angle: x central angle arc chord

  3. Theorem 11.4 Within one circle or within (two or more) congruent circles: central angles have  chords (1) chordshave  arcs (2)  arcs have  central angles (3)

  4. IFS AND THENS

  5. BC PF Example #1: In the diagram, circle O  circle D. Given that BC  PF, what can you conclude? And why (theorem)? P B F D O C Theorem:  arcs have  ’s O  D AND Theorem:  ’s have  chords

  6. Theorem 11.5 Within one circle or within (2 or more congruent circles): (Biconditional) If chords are equidistant from the (1) they are . center of a circle, then (2) If two or more chords are , then they are equidistant from the center.

  7. AB  CD A F EG  FG B AB  CD EG  FG G D C E IF THEN THEN IF

  8. Ex. #2 Find a. Give reason (theorem). a and PR are equidistant from center.  So, they are 25 un. Therefore, a = THEOREM If chordsareequidistant from the center of the Circle, then they are .

  9. Theorem 11.6 In a circle, if a diameter is perpendicular to a chord, then it bisects the chord and its arcs. THEN IF

  10. Theorem 11.7 In a circle, if a diameter bisects a chord (that is not another diameter) then it is perpendicular to the chord. THEN IF

  11. A A B B Theorem 11.8 In a circle, if a segment is the perpendicular bisector of a chord, then it contains the center of a circle THEN AB passes through the center of the circle. IF

  12. Ex. #3 Find r. State the reason (theorem). If KN were extended, it would be a diameter and it is  to LM. Therefore, it bisects LM. So, LN = 7. Why? r2 = 72 + 32 r2 = 49 + 9 If a diameter is  to a chord then it bisects the chord. r2 = 58 r = 58

  13. Ex. #4 Find y. State the reason (theorem) Is this a right triangle? Why? Yes. If a diameter bisects a chord then it is  to the chord. 152 = y2 + 112 225 = y2 + 121 104 = y2 2 26 = y

  14. ASSIGNMENT: Page 593 #1 – 16 Write out the theorem used for #3-16

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