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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.

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Presentation on theme: "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."— Presentation transcript:

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x arcchord central angle

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

4 IFS AND THENS

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

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

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

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

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

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. IF THEN

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

12 Ex. #3 Find r. State the reason (theorem). If KN were extended, it would be a diameter and it is  to LM. Therefore, r 2 = 7 2 + 3 2 r 2 = 49 + 9 r 2 = 58 r = 58 If a diameter is  to a chord then it bisects the chord. it bisects LM. So, LN = 7. Why?

13 Ex. #4 Find y. State the reason (theorem) Is this a right triangle? Yes. 15 2 = y 2 + 11 2 225 = y 2 + 121 104 = y 2 2 26 = y If a diameter bisects a chord then it is  to the chord. Why?

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


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