1. Warm Up Section 4.3 Draw and label each of the following in a circle with center P. (1). Radius: (2). Diameter: (3). Chord that is NOT a diameter: (4). Secant: (5). Tangent:

2. Answers to Warm Up Section 4.3 P A C E D R T

3. Properties of Chords Section 4.3 Standard: MM2G3 ad Essential Question: Can I understand and use properties of chords to solve problems?

4. In this section you will learn to use relationships of arcs and chords in a circle. In the same circle, or congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent. A C D B AB  CD if and only if _____  ______. AB CD

5. 1. In the diagram, A  D,, and m EF = 125 o. Find m BC. A C D B F E Because BC and EF are congruent ________ in congruent _______, the corresponding minor arcs BC and EF are __________. So, m ______ = m ______ = ______ o. chords circles congruent 125 BC EF

6. If one chord is a perpendicular bisector of another chord, then the first chord is a diameter. If QS is a perpendicular bisector of TR, then ____ is a diameter of the circle. R P S Q T QS

7. If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc. If EG is a diameter and TR  DF, then HD  HF and ____  ____. D H E G F GD GF

8. 2. If m TV = 121 o, find m RS. R S V T 6 6 If the chords are congruent, then the arcs are congruent. So, m RS = 121 o

9. 3. Find the measure of CB, BE, and CE. E DB C (80 – x) o Since BD is a diameter, it bisects, the chord and the arcs. 4(16) = 64 so mCB = m BE = 64 o mCE = 2(64 o ) = 128 o A 4xo4xo 4x = 80 – x 5x = 80 x = 16

10. In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center. ED A C B F G if and only if _____  ____ FE GE

11. 4. In the diagram of F, AB = CD = 12. Find EF. E B D A C 7x – 8 G 3x3x F Chords and are congruent, so they are equidistant from F. Therefore EF = 6 7x – 8 = 3x 4x = 8 x = 2 So, EF = 3x = 3(2) = 6

12. 5.In the diagram of F, suppose AB = 27 and EF = GF = 7. Find CD. E B D A C G F Since and are both 7 units from the center, they are congurent. So, AB = CD = 27.

13. 6.In F, SP = 5, MP = 8, ST = SU,  and  NRQ is a right angle. Show that  PTS   NRQ. S N T MP Q U Step 1: Look at  PTS !! Since MP = 8, and NQ bisects MP, we know MT= PT = 4. Since ,  PTN is a right angle. R 5 44 Use the Pythagorean Theorem to find TS. 4 2 + TS 2 = 5 2 TS 2 = 25 – 16 TS 2 = 9 So, TS = 3. 3

14. 6.In F, SP = 5, MP = 8, ST = SU,  and  NRQ is a right angle. Show that  PTS   NRQ. S N T MP Q U Step 2: Now, look at  NRQ! Since the radius of the circle is 5, QN = 10. Since ST = SU, MP and RN are equidistant from the center. Hence, MP = RN = 8. R 10 Use the Pythagorean Theorem to find RQ. RQ 2 + 8 2 = 10 2 RQ 2 = 100 – 64 RQ 2 = 36 So, RQ = 6. 8 6

15. 6.In F, SP = 5, MP = 8, ST = SU,  and  NRQ is a right angle. Show that  PTS   NRQ. S N T MP Q U R Step 3: Identify ratios of corresponding sides. In  PTS, PT = 4, TS = 3, and PS = 5. In  NRQ, NR = 8, RQ = 6, and QN = 10. Find the corresponding ratios:

16. Because the corresponding sides lengths are proportional (all have a ratio of ½),  PTS   NRQ by SSS.

17. 7. In S, QN = 26, NR = 24, ST = SU,  and  NRQ is a right angle. Show that  PTS   NRQ. S N T MP Q U R 24 2 + RQ 2 = 26 2 RQ 2 = 676 – 576 RQ 2 = 100 So, RQ = 10. N Q R 26 24 7

18. 7. In S, QN = 26, NR = 24, ST = SU,  and  NRQ is a right angle. Show that  PTS   NRQ. S N T MP Q U R 12 2 + ST 2 = 13 2 ST 2 = 169 –144 ST 2 = 25 So, RQ = 5. S T P 12 MP = RN = 24. So, PT = ½(24). SP is half of the diameter QN, so SP = ½(26) = 13. 13 5

19. S N T MP Q U R Step 3: Identify ratios of corresponding sides. In  PTS, PT = 12, TS = 5, and PS = 13, In  NRQ, NR = 24, RQ = 10, and QN = 26. Find the corresponding ratios: Because the corresponding sides lengths are proportional (all have a ratio of ½),  PTS   NRQ by SSS.