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Warm Up Section 3.2 Draw and label each of the following in a circle with center P.

Warm Up Section 3.2 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: . Warm Up Section 3.2 answers

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Warm Up Section 3.2 Draw and label each of the following in a circle with center P.

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  1. Warm Up Section 3.2 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. Warm Up Section 3.2 answers 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: T A E P R D C

  3. Properties of Chords Section 3.2 Standard: MCC9-12.G.C.2 Essential Question: Can I understand and use properties of chords to solve problems?

  4. Parts of circles are called Arcs. If the part of the circle is less than half the circle it is called a minor arc. If the part of the circle is more than half the circle it is called a major arc. And if it is exactlyhalf the circle it is called a semicircle.

  5. The circle below has center P. Draw a diameter on your circle. Label the endpoints of the diameter H and K. Put another point on your circle and label it T. There are two semicircles pictured in your drawing. One can be symbolized HK and the other HTK . K T P H

  6. K T P Name two minor arcs(you only need two letters to name a minor arc because you always travel the shortest distance unless told otherwise.) _______ , _______ Name two major arcs(you must use three letters to name a major arc because you always travel the shortest distance unless told otherwise.) _______ , _______ H HT TK KHT HKT

  7. The circle below has center M. Mark two points on your circle and label them R and S. Now draw the following rays: and Color the inside of RMS. Name the arc that is inside the colored part of the angle. This is called the intercepted arc. Arc: ________ The angle is called acentral anglebecause its vertex is the center of the circle. S R M RS

  8. S R M Arcs are measured in degrees. The measure of the intercepted arc is equal to the measure of the central angle that forms that arc. So, if mRMS = 60o, then m RS = 60o.

  9. 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  DC if and only if _____  ______. DC AB D A

  10. 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

  11. STICKY NOTE PROBLEM: In the diagram, A  D, arc BC  FE, BC = 10 cm. What is the measure of segment EF Sticky Note Problem E B A D F C If the arcs are congruent, the chords will be congruent. BC=EF=10 cm.

  12. 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

  13. 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 EG  DF, then HD  HF and ____  ____ . F E H G D GF GD

  14. 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

  15. 3. Find the measure of CB, BE, and CE. C Since BD is a perpendicular 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

  16. 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

  17. 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

  18. 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.

  19. In S, 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.

  20. 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 equidistantfrom 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.

  21. 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

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

  23. 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 10 Q Q 242 + RQ2 = 262 RQ2 = 676 – 576 RQ2 = 100 So, RQ = 10.

  24. 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 Cords equidistant from center  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, ST = 5.

  25. 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  Thrm.

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