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Chords and Arcs

For Exercises 1–5, use the diagram of L . 1. If YM and ZN are congruent chords, what can you conclude? 2. If YM and ZN are congruent chords, explain why you cannot conclude that LV = LC . 3. Suppose that YM has length 12 in., and its distance from

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Chords and Arcs

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  1. For Exercises 1–5, use the diagram of L. 1. If YM and ZN are congruent chords, what can you conclude? 2. If YM and ZN are congruent chords, explain why you cannot conclude that LV = LC. 3. Suppose that YM has length 12 in., and its distance from point L is 5 in. Find the radius of L to the nearest tenth. For Exercises 4 and 5, suppose that LVYM, YV = 11 cm, and L has a diameter of 26 cm. 4. Find YM. 5. Find LV to the nearest tenth. . . . YMZN; YLMZLN You do not know whether LV and LC are perpendicular to the chords. Chords and Arcs Lesson 12-2 Lesson Quiz 7.8 in. 22 cm 6.9 cm 12-3

  2. 1. A semicircle is a half circle. Answers will vary. Sample: STQ 2. A minor arc is less than a half circle. Answers will vary. Sample: ST 3. A major arc is more than a half circle. Answers will vary. Sample: RSQ 4. The vertex of a central angle is at the center of the circle. Answers may vary. Sample: TPQ 5. The measure of an arc is the measure of its central angle. m SPT = 86, so mST = 86. 6. The measure of an arc is the measure of its central angle. m SPQ = 180, so mSTQ = 180. Inscribed Angles Lesson 12-3 Check Skills You’ll Need Solutions 12-3

  3. Inscribed Angles Lesson 12-3 Check Skills You’ll Need Solutions (continued) 7. The measure of an arc is the measure of its central angle. m SPQ = 180 and m RPQ = 145. By the Angle Addition Postulate, m SPR + m RPQ = m SPQ, so m SPR + 145 = 180 and m SPR = 180 – 145 = 35. By the Angle Addition Postulate, m RPT = m RPS + m SPT = 35 + 86 = 121. Thus, mRST = 121. 8. The measure of an arc is the measure of its central angle. m SPQ = 180 and m SPT = 86. By the Angle Addition Postulate, m TPQ + m SPT = m SPQ, so m TPQ + 86 = 180 and m TPQ = 180 – 86 = 94. Thus, mTQ = 94. 12-3

  4. . . Inscribed Angles Lesson 12-3 Check Skills You’ll Need (For help, go to Lesson 10-6.) Identify the following in P at the right. 1. a semicircle 2. a minor arc 3. a major arc 4. a central angle Find the measure of each arc in P. 5. ST6. STQ 7. RST8. TQ Check Skills You’ll Need 12-3

  5. Inscribed Angles Lesson 12-3 Notes An inscribed angleis an angle whose vertex is on a circle and whose sides contain chords of the circle. An intercepted arcconsists of endpoints that lie on the sides of an inscribed angle and all the points of the circle between them. 12-3

  6. Inscribed Angles Lesson 12-3 Notes 12-3

  7. Inscribed Angles Lesson 12-3 Notes 12-3

  8. Inscribed Angles Lesson 12-3 Notes 12-3

  9. Inscribed Angles Lesson 12-3 Notes 12-3

  10. Find the values of x and y. 1 2 x = (80 + 70) Substitute. 1 2 x = mDEFInscribed Angle Theorem 1 2 x = (mDE + mEF) Arc Addition Postulate Because EFG is the intercepted arc of D, you need to find mFG in order to find mEFG. Inscribed Angles Lesson 12-3 Additional Examples Using the Inscribed Angle Theorem x = 75 Simplify. 12-3

  11. (continued) The arc measure of a circle is 360°, so mFG = 360 – 70 – 80 – 90 = 120. 1 2 y = mEFGInscribed Angle Theorem 1 2 y = (mEF + mFG) Arc Addition Postulate 1 2 y = (70 + 120) Substitute. Inscribed Angles Lesson 12-3 Additional Examples y = 95 Simplify. Quick Check 12-3

  12. . Find the values of a and b. The sum of the measures of the three angles of the triangle inscribed in O is 180. Inscribed Angles Lesson 12-3 Additional Examples Using Corollaries to Find Angle Measures By Corollary 2 to the Inscribed Angle Theorem, an angle inscribed in a semicircle is a right angle, so a = 90. Therefore, the angle whose intercepted arc has measure b must have measure 180 – 90 – 32, or 58. Because the inscribed angle has half the measure of the intercepted arc, the intercepted arc has twice the measure of the inscribed angle, so b = 2(58) = 116. Quick Check 12-3

  13. . . RS and TU are diameters of A. RB is tangent to A at point R. Find m BRT and m TRS. 1 2 m BRT = mRTThe measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc (Theorem 12-10). mRT = mURT – mURArc Addition Postulate 1 2 m BRT = (180 – 126) Substitute 180 for m and 126 for mUR. m BRT = 27 Simplify. Inscribed Angles Lesson 12-3 Additional Examples Using Theorem 12-10 12-3

  14. (continued) Use the properties of tangents to find m TRS. m BRS = 90 A tangent is perpendicular to the radius of a circle at its point of tangency. m BRS = m BRT + m TRSAngle Addition Postulate 90 = 27 + m TRSSubstitute. 63 = m TRSSolve. Inscribed Angles Lesson 12-3 Additional Examples Quick Check 12-3

  15. . m A = 100; m B = 75; m C = 80; m D = 105 m X = 80; m Y = 70; m Z = 90; m W = 120 . No; the diagonal would be a diameter of O and the inscribed angle would be a right angle, which was not found in Exercise 1 above. Inscribed Angles Lesson 12-3 Lesson Quiz In the diagram below, O circumscribes quadrilateral ABCD and is inscribed in quadrilateral XYZW. 1. Find the measure of each inscribed angle. 2. Find m  DCZ. 3. Are XAB and XBA congruent? Explain. 4. Find the angle measures in quadrilateral XYZW. 45 Yes; each is formed by a tangent and a chord, and they intercept the same arc. 5. Does a diagonal of quadrilateral ABCD intersect the center of the circle? Explain how you can tell. 12-3

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