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Properties or Parallel Lines

Properties or Parallel Lines. GEOMETRY LESSON 3-1. (For help, go to page 24 or Skills Handbook, page 720.). Solve each equation. 1 . x + 2 x + 3 x = 180 2. ( w + 23) + (4 w + 7) = 180 3. 90 = 2 y – 30 4. 180 – 5 y = 135 Write an equation and solve the problem.

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Properties or Parallel Lines

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  1. Properties or Parallel Lines GEOMETRY LESSON 3-1 (For help, go to page 24 or Skills Handbook, page 720.) Solve each equation. 1. x + 2x + 3x = 180 2. (w + 23) + (4w + 7) = 180 3. 90 = 2y – 30 4. 180 – 5y = 135 Write an equation and solve the problem. 5. The sum of m 1 and twice its complement is 146. Find m 1. 6. The measures of two supplementary angles are in the ratio 2 : 3. Find their measures. 3-1

  2. Solutions 1. Combine like terms: 6x = 180; divide both sides by 6: x = 30 2. Combine like terms: 5w + 30 = 180; Subtract 30 from both sides: 5w = 150; divide both sides by 5: w = 30 3. Add 30 to both sides: 120 – 2y; divide both sides by 2: 60 = y 4. Subtract 180 from both sides: – 5y = –45; divide both sides by –5: y = 9 5. The complement of 1 is 90 – m 1; m 1 + 2(90 – m 1) = 146; distribute: m 1 + 180 – 2m = 146; combine like terms: 180 – m 1 = 146; subtract 180 from both sides: –m 1 = –34; multiply both sides by –1: m 1 = 34 6. Let 2x represent the measure of one of the angles and 3x represent the measure of the other angle. Then 2x + 3x = 180; combine like terms: 5x = 180; divide both sides by 5: x = 36. Then 2x = 72 and 3x = 108.0. Properties or Parallel Lines GEOMETRY LESSON 3-1 3-1

  3. Use the diagram above. Identify which angle forms a pair of same-side interior angles with 1. Identify which angle forms a pair of corresponding angles with 1. Same-side interior angles are on the same side of transversal t between lines p and q. 4, 8, and 5 are on the same side of the transversal as 1, but only 1 and 8 are interior. So 1 and 8 are same-side interior angles. Properties or Parallel Lines GEOMETRY LESSON 3-1 3-1

  4. (continued) Corresponding angles also lie on the same side of the transversal. The angle corresponding to 1 must lie in the same position relative to line q as 1 lies relative to line p. Because 1 is an interior angle, 1 and 5 are corresponding angles. Properties or Parallel Lines GEOMETRY LESSON 3-1 One angle must be an interior angle, and the other must be an exterior angle. 3-1

  5. Compare 2 and the vertical angle of 1. Classify the angles as alternate interior angles, same–side interior angles, or corresponding angles. The vertical angle of 1 is between the parallel runway segments. 2 is between the runway segments and on the opposite side of the transversal runway. Because alternate interior angles are not adjacent and lie between the lines on opposite sides of the transversal, 2 and the vertical angle of 1 are alternate interior angles. Properties or Parallel Lines GEOMETRY LESSON 3-1 3-1

  6. Which theorem or postulate gives the reason that m 3 + m 2 = 180? 3 and 2 are adjacent angles that form a straight angle. m 3 + m 2 = 180 because of the Angle Addition Postulate. Properties or Parallel Lines GEOMETRY LESSON 3-1 3-1

  7. In the diagram above, || m. Find m 1 and then m 2. 1 and the 42° angle are corresponding angles. Because || m, m 1 = 42 by the Corresponding Angles Postulate. Because 1 and 2 are adjacent angles that form a straight angle, m 1 + m 2 = 180 by the Angle Addition Postulate. If you substitute 42 for m 1, the equation becomes 42 + m 2 = 180. Subtract 42 from each side to find m 2 = 138. Properties or Parallel Lines GEOMETRY LESSON 3-1 3-1

  8. In the diagram above, || m. Find the values of a, b, and c. Properties or Parallel Lines GEOMETRY LESSON 3-1 a = 65 Alternate Interior Angles Theorem c = 40 Alternate Interior Angles Theorem a + b + c = 180 Angle Addition Postulate 65 + b + 40 = 180 Substitution Property of Equality b = 75 Subtraction Property of Equality 3-1

  9. Pages 118-121 Exercises 5. 1 and 2: corr. 3 and 4: alt. int. 5 and 6: corr. 6. 1 and 2: same-side int. 3 and 4: corr. 5 and 6: corr. 7. 1 and 2: corr. 3 and 4: same-side int. 5 and 6: alt. int. 8. alt. int. 9.a. 2 b. 1 c. corr. 10.a. Def. of b. Def. of right c. Corr. of i lines are . d. Subst. e. Def. of right f. Def. of 1.PQ and SR with transversal SQ; alt. int. 2.PS and QR with transversal SQ; alt. int. 3.PS and QR with transversal PQ; same-side int. 4.PS and QR with transversal SR; corr. s s s s s s s s s s s s s s s Properties or Parallel Lines GEOMETRY LESSON 3-1 3-1

  10. 11. m 1 = 75 because corr. of || lines are ; m 2 = 105 because same-side int. of || lines are suppl. 12.m 1 = 120 because corr. of || lines are ; m 2 = 60 because same-side int. of || lines are suppl. 13.m 1 = 100 because same-side int. of || lines are suppl.; m 2 = 70 because alt. int. of || lines have = measure. 11. m 1 = 75 because corr. of || lines are ; m 2 = 105 because same-side int. of || lines are suppl. 12.m 1 = 120 because corr. of || lines are ; m 2 = 60 because same-side int. of || lines are suppl. 13.m 1 = 100 because same-side int. of || lines are suppl.; m 2 = 70 because alt. int. of || lines have = measure. 14. 70; 70, 110 15. 25; 65 16. 20; 100, 80 17.m 1 = m 3 = m 6 = m 8 = m 9 = m 11 = m 13 = m 15 = 52; m 2 = m 4 = m 5 = m 7 = m 10 = m 12 = m 14 = 128 18. You must find the measure of one . All that are vert., corr., or alt. int. to that will have that measure. All other will be the suppl. of that measure. 19. two 20. four 21. two 22. four 23. 32 s s s s s s s s s s s s s s Properties or Parallel Lines GEOMETRY LESSON 3-1 3-1

  11. s s s s s s s s s Properties or Parallel Lines GEOMETRY LESSON 3-1 28. Answers may vary. Sample: E illustrates corr. ( 1 and 3, 2 and 4) and same-side int. ( 1 and 2, 3 and 4); I illustrates alt. int. ( 1 and 4, 2 and 3) and same-side int. ( 1 and 3, 2 and 4). 29. a. alt. int. b. He knew that alt. int. of || lines are . 30.a. 57 b. same-side int. 24.x = 76, y = 37, v = 42, w = 25 25.x = 135, y = 45 26. The labeled are corr. and should be . If you solve 2x – 60 = 60 – 2x, you get x = 30. This would be impossible since 2x – 60 and 60 – 2x would equal 0. 27.Trans means across or over. A transversal cuts across other lines. 3-1

  12. s s s s Properties or Parallel Lines GEOMETRY LESSON 3-1 31.a. If two lines are || and cut by a transversal, then same-side ext. are suppl. b.Given: a || b Prove: 4 and 5 are suppl. 1.a || b (Given) 2.m 5 + m 6 = 180 ( Add. Post.) 3. 4 6 (Corr. are ) 4.m 5 + m 4 = 180 (Subst.) 5. 4 and 5 are suppl. (Def. of suppl.) 32.1.a || b (Given) 2. 1 2 (Vert. are .) 3. 2 3 (Corr. are .) 4. 1 3 (Trans. Prop.) 33. Never; the two planes do not intersect. 34. Sometimes; if they are ||. 35. Sometimes; they may be skew. 36. Sometimes; they may be ||. 37. D 38. G 39. D 40. I 3-1

  13. 41.[2] a.  First show that 1 7. Then show that 7 5. Finally, show that 1 5 (OR other valid solution plan). b.  1 7 because vert. are . 7 5 because corr. of || lines are . Finally, by the Trans. Prop. of , 1 5. [1] incorrect sequence of steps OR incorrect logical argument s s Properties or Parallel Lines GEOMETRY LESSON 3-1 42. 121 43. 59 44. 29.5 45. (0.5, 7) 46. (–0.5, 3.5) 47. (3, 3) 48. add 4; 20, 24 49. multiply by –2; 16, –32 50. subtract 7; –5, –12 3-1

  14. 1. Complete: and 4 are alternate interior angles. 2. Complete: and 8 are corresponding angles. 3. Suppose that m 3 = 37. Find m 6. 4. Suppose that m 1 = x + 12 and m 5 = 3x – 36. Find x. 5. If a transversal intersects two parallel lines, then same-side exterior angles are supplementary. Write a Plan for Proof. Given: m || n Prove: 2 and 7 are supplementary. 6 4 Show that m 2 = m 6. Then show that m 6 + m 7 = 180, and substitute m 2 for m 6. Properties or Parallel Lines GEOMETRY LESSON 3-1 In the diagram below, m || n. Use the diagram for Exercises 1–5. 143 24 3-1

  15. Proving Lines Parallel GEOMETRY LESSON 3-2 (For help, go to page 24 and Lesson 2-1.) Solve each equation. 1. 2x + 5 = 27 2. 8a – 12 = 20 3.x – 30 + 4x + 80 = 180 4. 9x – 7 = 3x + 29 Write the converse of each conditional statement. Determine the truth value of the converse. 5. If a triangle is a right triangle, then it has a 90° angle. 6. If two angles are vertical angles, then they are congruent. 7. If two angles are same-side interior angles, then they are supplementary. 3-2

  16. Proving Lines Parallel GEOMETRY LESSON 3-2 Solutions 1. Subtract 5 from both sides: 2x = 22; divide both sides by 2: x = 11 2. Add 12 to both sides: 8a = 32; divide both sides by 8: a = 4 3. Combine like terms: 5x + 50 = 180; subtract 50 from both sides: 5x = 130; divide both sides by 5: x = 26 4. Add –3x + 7 to both sides: 6x = 36; divide both sides by 6: x = 6 5. Reverse the hypothesis and conclusion: If a triangle has a 90° angle, then it is a right triangle. By definition of a right triangle, it is true. 6. Reverse the hypothesis and conclusion: If two angles are congruent, they are vertical angles. A counterexample is the congruent base angles of an isosceles triangle, which are not vertical angles. The converse is false. 7. Reverse the hypothesis and conclusion: If two angles are supplementary, then they are the same-side interior angles of parallel lines. A counterexample is two angles that form a straight angle. The converse is false. 3-2

  17. If two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel. Given:  1 2 Prove:  || m By the Vertical Angles Theorem, 3 1. 1 2, so 3 2 by the Transitive Property of Congruence. Because 3 and 2 are corresponding angles, || m by the Converse of the Corresponding Angles Postulate. Proving Lines Parallel GEOMETRY LESSON 3-2 Write the flow proof below of the Alternate Interior Angles Theorem as a paragraph proof. 3-2

  18. Use the diagram above. Which lines, if any, must be parallel if 3 and 2 are supplementary? It is given that 3 and 2 are supplementary. The diagram shows that 4 and 2 are supplementary. Because supplements of the same angle are congruent (Congruent Supplements Theorem), 3 4. Because 3 and 4 are congruent corresponding angles, EC || DK by the Converse of the Corresponding Angles Postulate. Proving Lines Parallel GEOMETRY LESSON 3-2 3-2

  19. Use the diagram above. Which angle would you use with 1 to prove the theorem In a plane, if two lines are perpendicular to the same line, then they are parallel to each other (Theorem 3-6) using the Converse of the Alternate Interior Angles Theorem instead of the Converse of the Corresponding Angles Postulate? By the Vertical Angles Theorem, 2 is congruent to its vertical angle. Because 1 2, 1 is congruent to the vertical angle of 2 by the Transitive Property of Congruence. Because alternate interior angles are congruent, you can use the vertical angle of 2 and the Converse of the Alternate Interior Angles Theorem to prove that the lines are parallel. Proving Lines Parallel GEOMETRY LESSON 3-2 3-2

  20. Find the value of x for which || m. If || m, the alternate interior angles are congruent, and their measures are equal. Proving Lines Parallel GEOMETRY LESSON 3-2 The labeled angles are alternate interior angles. Write and solve the equation 5x – 66 = 14 + 3x. 5x – 66 = 14 + 3x 5x = 80 + 3xAdd 66 to each side. 2x = 80 Subtract 3x from each side. x = 40 Divide each side by 2. 3-2

  21. Suppose that the top and bottom pieces of a picture frame are cut to make 60° angles with the exterior sides of the frame. At what angle should the two sides be cut to ensure that opposite sides of the frame will be parallel? Proving Lines Parallel GEOMETRY LESSON 3-2 In order for the opposite sides of the frame to be parallel, same-side interior angles must be supplementary. Two 90° angles are supplementary, so find an adjacent angle that, together with 60°, will form a 90° angle: 90° – 60° = 30°. 3-2

  22. Pages 125-129 Exercises 8.a || b; Conv. of Corr. Post. 9. none 10.a || b; Conv. of Alt. Int. Thm. 11. || m; Conv. of Corr. Post. 12. none 13.a || b; Conv. of Corr. Post. 14. none 4.a || b; if two lines and a transversal form same-side int. that are suppl., then the lines are ||. 5.a || b; if two lines and a transversal form same-side int. that are suppl., then the lines are ||. 6. none 7. none 1.BE || CG; Conv. of Corr. Post. 2.CA || HR; Conv. of Corr. Post. 3.JO || LM; if two lines and a transversal form same-side int. that are suppl., then the lines are ||. s s s s s s s s s Proving Lines Parallel GEOMETRY LESSON 3-2 3-2

  23. 24. When the frame is put together, each of the frame is a right . Two right are suppl. By the Conv. of the Same-Side Int. Thm., opp. sides of the frame are ||. 25. The corr. are , so the lines are || by the Conv. of Corr. Post. 26. a. Corr. b–c. 1, 3 (any order) d. Conv. of Corr. 15. || m; Conv. of Alt. Int. Thm. 16.a. Def. of b. Given c. All right are . d. Conv. of Corr. Post. 17.a. 1 b. 1 c. 2 d. 3 e. Conv. of Corr. s s s s s s s s s s Proving Lines Parallel GEOMETRY LESSON 3-2 18. 30 19. 50 20. 59 21. 31 22. 5 23. 20 3-2

  24. 27. 10; m 1 = m 2 = 70 28. 5; m 1 = m 2 = 50 29. 2.5; m 1 = m 2 = 30 30. 1.25; m 1 = m 2 = 10 31. The corr. he draws are . 32. PL || NA and PN || LA by Conv. of Same-Side Int. Thm. 33.PL || NA by Conv. of Same-Side Int. Thm. 34. none 35.PN || LA by Conv. of Same-Side Int. Thm. 36. Answers may vary. Sample: In the diagram, ABBH and ABBD, but BH || BD. They intersect. 37. Reflexive: a || a; false; any line intersects itself. Symmetric: If a || b, then b || a; true; b and a are coplanar and do not intersect. Transitive: In general, if a || b, and b || c, then a || c; true; however, when a || b, and b || a, it does not follow that a || a. s s s s Proving Lines Parallel GEOMETRY LESSON 3-2 3-2

  25. 42. Answers may vary. Sample: 3 11; || m by Conv. of the Alt. Int. Thm. and j || k by Conv. of Corr. Post. 43. Answers may vary. Sample: 3 and 12 are suppl.; j || k by the Conv. of Corr. Post. 44. Vert. Thm. and Conv. of Corr. Post. 38. Reflexive: aa; false; lines are two lines that intersect to form right . Symmetric: If ab, then ba; true; b and a intersect to form right . Transitive: If ab, and bc, then ac; false; in a plane, two lines to the same line are ||. 39. The corr. are , and the oars are || by the Conv. of Corr. Post. 40. Answers may vary. Sample: 3 9; j || k by Conv. of the Alt. Int. Thm. 41. Answers may vary. Sample: 3 9; j || k by Conv. of the Alt. Int. Thm. and || m by Conv. of Same-Side Int. Thm. s s s s s s s s s s s s Proving Lines Parallel GEOMETRY LESSON 3-2 3-2

  26. 45. It is given that || m, so 4 8 by Corr. Post. It is also given that 12 8, so 4 12 by Trans. Prop. of . So, j || k by the Conv. of Corr. Post. s s Proving Lines Parallel GEOMETRY LESSON 3-2 47. 46. 3-2

  27. 50.a. Answers may vary. Sample: b. Given: a || b with transversal e, c bisects AOB, d bisects AXZ. c. Prove: c || d Proving Lines Parallel GEOMETRY LESSON 3-2 48. 49. 3-2

  28. 50. (continued) d. To prove that c || d, show that 1 3. 1 3 if AOB OXZ. AOB OXZ by the Corr. Post. e.1.a || b (Given) 2.AOBAXZ (Corr. Post.) 3.mAOB = mAXZ (Def. of ) 4.mAOB = m 1 + m 2; mAXZ = m 3 + m 4 ( Add. Post.) 5.c bisects AOB; d bisects AXZ. (Given) 6.m 1 = m 2; m 3 = m 4 (Def. of bisector) 7.m 1 + m 2 = m 3 + m 4 (Trans. Prop. of ) 8.m 1 + m 1 = m 3 + m 3 (Subst.) 9. 2m 1 = 2m 3 (Add. Prop.) 10.m 1 = m 3 (Div. Prop.) 11.c || d (Conv. of Corr. Post.) s s s s Proving Lines Parallel GEOMETRY LESSON 3-2 3-2

  29. 51. C 52. F 53. B 54.[2]a. 136 + (x + 21) = 180 so x = 23 (OR equivalent equation resulting in x = 23). 55. (continued) b.m 1 + m 3 = 180. If 2x – 38 + 6x + 18 = 180, then x = 25. The measures are 2x – 38 = 12 and 25, but 12 25. So a can’t be || to b. [3]  appropriate methods, but with one computational error [2]  incorrect diagram solved correctly OR correct diagram solved incorrectly 54. (continued) b.x + 21 = 2x so x = 21. Lines c and d are not || because x cannot = both 21 and 23 (OR equivalent explanation). [1] incorrect equations OR incorrect solutions 55.[4]a. = / Proving Lines Parallel GEOMETRY LESSON 3-2 3-2

  30. s s s s Proving Lines Parallel GEOMETRY LESSON 3-2 55. (continued) [1] correct answer (lines a and b are not ||), without work shown 56.m 1 = 70 since it is a suppl. of the 110° . m 2 = 110 since same-side int. are suppl. 57.m 1 = 66 because alt. int. are . m 2 = 180 – 94 = 86 because same-side int. are suppl. 61. If you form the past tense of a verb, then you add ed to the verb. Original statement is false, converse is false. 62. If there are clouds in the sky, then it is raining. Original statement is true, converse is false. 63. 201.1 in.2 64. 28.3 cm2 58. If you are west of the Mississippi River, then you are in Nebraska. Original statement is true, converse is false. 59. If a circle has a radius of 4 cm, then it has a diameter of 8 cm. Both are true. 60. If same-side int. are suppl., then a line intersects a pair of || lines. Both are true. 3-2

  31. Proving Lines Parallel GEOMETRY LESSON 3-2 65. 63.6 ft2 66. 78.5 in.2 67. 6.2 m2 68. 4.5 m2 69. 17.7 ft2 70. 0.3 m2 3-2

  32. Suppose that m 1 = 3x + 10, m 2 = 3x + 14, and m 6 = x + 58 in the diagram above. Proving Lines Parallel GEOMETRY LESSON 3-2 Use the diagram and the given information to determine which lines, if any, are parallel. Justify your answer with a theorem or postulate. 1. 6 3 2. 1 and 4 are supplementary. 3. 2 4 4. Find the value of x for which a || b. 5. Find the value of x for which m || n. m || n by Converse of Corresponding Angles Post. a || b by Converse of Same-Side Interior Angles Theorem. No lines must be parallel. 24 26 3-2

  33. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 (For help, go to Lesson 1-4.) Classify each angle as acute, right, or obtuse. 1.2.3. Solve each equation. 4. 30 + 90 + x = 180 5. 55 + x + 105 = 180 6.x + 58 = 90 7. 32 + x = 90 3-3

  34. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 Solutions 1. The measure of the angle is 90°, so it is a right angle. 2. The measure of the angle is between 0° and 90°, so it is an acute angle. 3. The measure of the angle is between 0° and 90°, so it is an acute angle. 4. Combine like terms: 120 + x = 180; subtract 120 from both sides: x = 60 5. Combine like terms: x + 160 = 180; subtract 160 from both sides: x = 20 6. Subtract 58 from both sides: x = 32 7. Subtract 32 from both sides: x = 58 3-3

  35. Find mZ. 48 + 67 + mZ = 180 Triangle Angle-Sum Theorem 115 + mZ = 180 Simplify. mZ = 65 Subtract 115 from each side. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 3-3

  36. In triangle ABC, ACB is a right angle, and CDAB. Find the values of a, b, and c. Find c first, using the fact that ACB is a right angle. mACB = 90 Definition of right angle c + 70 = 90 Angle Addition Postulate c = 20 Subtract 70 from each side. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 3-3

  37. (continued) To find a, use ADC. a + mADC + c = 180 Triangle Angle-Sum Theorem mADC = 90 Definition of perpendicular lines a + 90 + 20 = 180 Substitute 90 for mADC and 20 for c. a + 110 = 180 Simplify. a = 70 Subtract 110 from each side. To find b, use CDB. 70 + m CDB +b = 180 Triangle Angle-Sum Theorem mCDB = 90 Definition of perpendicular lines 70 + 90 + b = 180 Substitute 90 for mCDB. 160 + b = 180 Simplify. b = 20 Subtract 160 from each side. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 3-3

  38. Classify the triangle by its sides and its angles. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 The three sides of the triangle have three different lengths, so the triangle is scalene. One angle has a measure greater than 90, so the triangle is obtuse. The triangle is an obtuse scalene triangle. 3-3

  39. Find m 1. m 1 + 90 = 125 Exterior Angle Theorem m 1 = 35 Subtract 90 from each side. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 3-3

  40. Explain what happens to the angle formed by the back of the chair and the armrest as you lower the back of the lounge chair. The angle formed by the back of the chair and the armrest increases as you lower the back of the lounge chair. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 The exterior angle and the angle formed by the back of the chair and the armrest are adjacent angles, which together form a straight angle. As one measure increases, the other measure decreases. 3-3

  41. 16. 17. Not possible; a right will always have one longest side opp. the right . 18. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 Pages 135-139 Exercises 1. 30 2. 83.1 3. 90 4. 71 5. 90 6.x = 70; y = 110; z = 30 7.t = 60; w = 60 8.x = 80; y = 80 9. 70 10. 30 11. 60 12. acute, isosceles 13. acute, equiangular, equilateral 14. right, scalene 15. obtuse, isosceles 3-3

  42. 19. 20. 21. 22. 23. 24. a. 5, 6, 8 b. 1 and 3 for 5 1 and 2 for 6 1 and 2 for 8 c. They are vert. . 25. a. 2 b. 6 26. 123 27. 115.5 28.m 3 = 92; m 4 = 88 29.x = 147, y = 33 30.a = 162, b = 18 31.x = 52.5; 52.5, 52.5, 75; acute 32.x = 7; 55, 35, 90; right 33.x = 37; 37, 65, 78; acute s Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 3-3

  43. 34.x = 38, y = 36, z = 90; ABD: 36, 90, 54; right; BCD: 90, 52, 38; right; ABC: 74, 52, 54; acute 35.a = 67, b = 58, c = 125, d = 23, e = 90; FGH: 58, 67, 55; acute; FEH: 125, 32, 23; obtuse; EFG: 67, 23, 90; right 36.x = 32, y = 62, z = 32, w = 118; ILK: 118, 32, 30; obtuse Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 37. 60; 180  3 = 60 38. Yes, an equilateral is isosc. Because if three sides of a are , then at least two sides are . No, the third side of an isosc. does not need to be to the other two. 39. eight 40. 32.5 3-3

  44. 41. Check students’ work. Answers may vary. Sample: The two ext. Formed at vertex A are vert. and thus have the same measure. 42. 30 and 60 43. a. 40, 60, 80 b. acute 44. 160 45. 100 46. 103 47. 32 48. a. 90 b. 180 c. 90 d. compl. e. compl. 49. a. Add. b. -Sum c. Trans. d. Subtr. 50. 132; since the missing is 68, the largest ext. Is 180 – 48 = 132. 51. Check students’ work. 52. a. 81 b. 45, 63, 72 c. acute 53. 120 or 60 54. 135 or 45 55. 90 s s s Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 3-3

  45. 56. Greater than, because there are two with measure 90 where the meridians the equator. 57. 58. 59. 1 60. 61. 0 62. 115 63. Answers may vary. Sample: The measure of the ext. is = to the sum of the measures of the two remote int. . Since these are , the formed by the bisector of the ext. are to each of them. Therefore, the bisector is || to the included side of the remote int. by the Conv. of the Alt Int. Thm. 64. B 65. G 66. B 67. H 1 3 1 7 s s s s s s s 1 19 Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 3-3

  46. s s Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 68. [2]a. b. The correct equation is 2x = (2x – 40) + (x – 15) with the sol. x = 55. The three int. measure 70, 40, and 70. [1] incorrect sketch, equation, OR solution 69.[2]a. 159; the sum of the three of the is 180, so m Y + m M + m F = 180. Since m F = 21, m Y + m M + 21 = 180. Subtr. 21 from both sides results in m Y + m M = 159. 69. (continued) b. 1 to 68 ; since Y is obtuse, its whole number range is from 91 to 158, allowing the measure of 1 for m M when m Y = 158. When m Y = 91, then m M = 68. [1] incorrect answer to part (a) or (b) OR incorrect computation in either part 3-3

  47. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 70. 53 71. 46 72. 7 73. 74. 3-3

  48. 1. A triangle with a 90° angle has sides that are 3 cm, 4 cm, and 5 cm long. Classify the triangle by its sides and angles. Use the diagram for Exercises 2–6. 2. Find m 3 if m 2 = 70 and m 4 = 42. 3. Find m 5 if m 2 = 76 and m 3 = 90. 4. Find x if m 1 = 4x, m 3 = 2x + 28, and m 4 = 32. 5. Find x if m 2 = 10x, m 3 = 5x + 40, and m 4 = 3x – 4. 6. Find m 3 if m 1 = 125 and m 5 = 160. Parallel Lines and the Triangle Angle-Sum Theorem GEOMETRY LESSON 3-3 scalene right triangle 68 166 30 8 105 3-3

  49. The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 (For help, go to Lesson 1-4 and 3-3.) Find the measure of each angle of quadrilateral ABCD. 1.2. 3. 3-4

  50. 1.m DAB = 32 + 45 = 77; m B = 65; m BCD = 70 + 61 = 131; m D = 87 2.m DAC = m ACD = m D and m CAB = m B = m BCA; by the Triangle Angle-Sum Theorem, the sum of the measures of the angles is 180, so each angle measures , or 60. So, m DAB = 60 + 60 = 120, m B = 60, m BCD = 60 + 60 = 120, and m D = 60. 3. By the Triangle Angle-Sum Theorem m A + 55 + 55 = 180, so m A = 70. m ABC = 55 + 30 = 85; by the Triangle Angle-Sum Theorem, m C + 30 + 25 = 180, so m C = 125; m ADC = 55 + 25 = 80 180 3 The Polygon Angle-Sum Theorems GEOMETRY LESSON 3-4 Solutions 3-4

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