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Warm Up Section 4.3 Draw and label each of the following in a circle with center P.

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: . Answers to Warm Up Section 4.3. T. A. E. P. R. D. C. Properties of Chords

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Warm Up Section 4.3 Draw and label each of the following in a circle with center P.

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  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 T A E P R D C

  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. C B AB  CD if and only if _____  ______. CD AB D A

  5. 1. In the diagram, A  D, , and m EF = 125o. Find m BC. E B A D F C chords Because BC and EF are congruent ________ in congruent _______, the corresponding minor arcs BC and EF are __________ . So, m ______ = m ______ = ______o. 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. T S P Q R 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 ____  ____ . F E H G D GF GD

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

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

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

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

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

  13. In F, SP = 5, MP = 8, ST = SU,  and • NRQ is a right angle. Show that PTS  NRQ. N Step 1: Look at PTS !! Since MP = 8, and NQ bisects MP, we know MT= PT = 4. Since  , PTN is a right angle. 4 T 4 M P U 3 5 S R Q Use the Pythagorean Theorem to find TS. 42 + TS2 = 52 TS2 = 25 – 16 TS2 = 9 So, TS = 3.

  14. In F, SP = 5, MP = 8, ST = SU,  and • NRQ is a right angle. Show that PTS  NRQ. N 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. 8 T M P U 10 S R 6 Q Use the Pythagorean Theorem to find RQ. RQ2 + 82 = 102 RQ2 = 100 – 64 RQ2 = 36 So, RQ = 6.

  15. In F, SP = 5, MP = 8, ST = SU,  and • NRQ is a right angle. Show that PTS  NRQ. N 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: T M P U S R Q

  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. N N 24 T M P U 26 S R R 7 Q Q 242 + RQ2 = 262 RQ2 = 676 – 576 RQ2 = 100 So, RQ = 10.

  18. 7. In S, QN = 26, NR = 24, ST = SU,  and NRQ is a right angle. Show that PTS  NRQ. N 12 T P 5 T M P 13 U S S MP = RN = 24. So, PT = ½(24). SP is half of the diameter QN, so SP = ½(26) = 13. R Q 122 + ST2 = 132 ST2 = 169 –144 ST2 = 25 So, RQ = 5.

  19. 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: N T M P U S R Q Because the corresponding sides lengths are proportional (all have a ratio of ½), PTS  NRQ by SSS.

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