1. Arcs and Chords Section 11.4

2. Goal  Use properties of chords of circles.

3. Theorems  11.4 Diameter Perpendicular to a Chord  11.5 Converse 11.4  11.6 Congruent Arcs

4. Arcs and Chords  A chord is a line segment with endpoints on the circle.  The endpoints of chords also create arcs.  If a chord is not a diameter, then its endpoints divide a circle into a major and a minor arc. O B A Chord C Arc

5. Congruent Arcs and Chords  In a circle or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent.

6. TV  WS. Find mWS. 9n – 11 = 7n + 11 2n = 22 n = 11 = 88°  chords have  arcs. Def. of  arcs Substitute the given measures. Subtract 7n and add 11 to both sides. Divide both sides by 2. Substitute 11 for n. Simplify. mTV = mWS mWS = 7(11) + 11 TV  WS Example 1

7. PT bisects RPS. Find RT. 6x = 20 – 4x 10x = 20 x = 2 RT = 6(2) RT = 12 Add 4x to both sides. Divide both sides by 10. Substitute 2 for x. Simplify. RPT  SPT RT = TS mRT  mTS Your Turn:

8. Jewelry A circular piece of jade is hung from a chain by two wires around the stone. JM  KL and = 90. Find. Example 2

10. Your Turn: A.42.5 B.85 C.127.5 D.170

11. Example 3

12. WX= YZDefinition of congruent segments 7x – 2= 5x + 6Substitution 2x= 8Add 2 to each side. x= 4Divide each side by 2. So, WX = 7x – 2 = 7(4) – 2 or 26. Answer: WX = 26

13. Your Turn: A.6 B.8 C.9 D.13

14. Theorem 11.4 Diameter Perpendicular to a Chord  If a diameter (or radius)of a circle is perpendicular to a chord, then it bisects the chord and its arc. o

15. Theorem 11.5 Diameter Perpendicular to a Chord Converse  The perpendicular bisector of a chord is a diameter (or radius) of the circle.  Converse of Theorem 10.3 O

16. Example 4 ANSWER The length of is 10. BD BD = BE + ED Segment Addition Postulate = 5 + 5 Substitute 5 for BE and ED. = 10 Simplify. In C the diameter AF is perpendicular to BD. Use the diagram to find the length of BD. SOLUTION Because AF is a diameter that is perpendicular to BD, you can use Theorem 11.4 to conclude that AF bisects BD. So, BE = ED = 5.

17. Your Turn Find the Length of a Segment 1.Find the length of JM. 2. Find the length of SR. ANSWER 12 ANSWER 30

18. Finding the Center of a Circle  Theorem 11.5 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.

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

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

21. Example 4 Find the value of x. a.b. Because, it follows that. So, x = DE = 3. ABDE  AB  SOLUTION a. So,, and x = 60. mQPmRS==60° Because, it follows that QP RS  QPRS .

22. Your Turn Find Measures of Angles and Chords ANSWER 4 3 1. 2. 3. ANSWER 30 Find the value of x.

23. Example 5 Answer:

24. Your Turn: A.14 B.80 C.160 D.180

25. Solving Problems  Draw a circle with a chord that is 15 inches long and 8 inches from the center of the circle.  Draw a radius so that it forms a right triangle.  How could you find the length of the radius? 8cm 15cm O A B D ∆ODB is a right triangle andSolution: x

26. Example 6 CERAMIC TILE In the ceramic stepping stone below, diameter AB is 18 inches long and chord EF is 8 inches long. Find CD.

27. Example 6 Step 1Draw radius CE. This forms right ΔCDE.

28. Example 6 Step 2Find CE and DE. Since AB = 18 inches, CB = 9 inches. All radii of a circle are congruent, so CE = 9 inches. Since diameter AB is perpendicular to EF, AB bisects chord EF by Theorem 10.3. So, DE =1/2(8) or 4 inches. 9 4

29. Example 6 Step 3Use the Pythagorean Theorem to find CD. CD 2 + DE 2 = CE 2 Pythagorean Theorem CD 2 + 4 2 = 9 2 Substitution CD 2 + 16= 81Simplify. CD 2 = 65Subtract 16 from each side. Take the positive square root. Answer: 9 4

30. Your Turn: A.3.87 B.4.25 C.4.90 D.5.32 In the circle below, diameter QS is 14 inches long and chord RT is 10 inches long. Find VU.

31.  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 11.6 Congruent Arcs

32. Example 7 Since chords EF and GH are congruent, they are equidistant from P. So, PQ = PR.

33. Example 7 PQ= PR 4x – 3= 2x + 3Substitution x= 3Simplify. So, PQ = 4(3) – 3 or 9 Answer: PQ = 9

34. Your Turn: A.7 B.10 C.13 D.15

35. Example 8 AB = 8; DE = 8, and CD = 5. Find CF.

36. Example 8 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.

37. Example 8 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

38. Assignment  Pg. 610 – 612;#1 - 19 odd, 23 – 31 odd.