Properties or Parallel Lines

(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.

GEOMETRY LESSON 3-1GEOMETRY LESSON 3-1

Properties or Parallel LinesProperties or Parallel Lines

3-1

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.

Solutions



3-1

Use the diagram above. Identify which angle forms a pair of same-sideinterior 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.



3-1

One angle must be an interior angle, and the other must be an exterior angle.


(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.


3-1

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.



3-1

3 and 2 are adjacent angles that form a straight angle.

m 3 + m 2 = 180 because of the Angle Addition Postulate.


Which theorem or postulate gives the reason that m 3 + m 2 = 180?


3-1

1 and the 42° angle are corresponding angles. Because || m, m 1 = 42 by the Corresponding Angles Postulate.


In the diagram above, || m. Find m 1 and then m 2.

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.


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


In the diagram above, || m. Find the values of a, b, and c.


3-1

Properties or Parallel LinesProperties or Parallel LinesGEOMETRY LESSON 3-1GEOMETRY LESSON 3-1

Pages 118-121 Exercises

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

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.

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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

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3-1


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

11.m 1 = 75 because corr. of || lines are ; m 2 = 105 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.

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13.m 1 = 100 because same-side int. of || lines are suppl.; m 2 = 70 because alt. int. of || lines have = measure.

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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

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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.

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3-1


31.a. If two lines are || and cut by a transversal, then same-side ext. are suppl.

b. Given: a || bProve: 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

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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

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

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3-1

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.

Show that m 2 = m 6. Then show that m 6 + m 7 = 180, and substitute m 2 for m 6.


6

4

24

143


3-1

In the diagram below, m || n. Use the diagram for Exercises 1–5.

(For help, go to page 24 and Lesson 2-1.)


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.


Proving Lines ParallelProving Lines Parallel

3-2

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.

Solutions



3-2

Write the flow proof below of the Alternate Interior Angles Theorem as a paragraph proof.


If two lines and a transversal form alternate interior

angles that are congruent, then the two lines are parallel.

Prove: || m

Given: 1 2

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.


3-2

It is given that 3 and 2 are supplementary.

The diagram shows that 4 and 2 are supplementary.

Because 3 and 4 are congruent corresponding angles, EC || DK by the Converse of the Corresponding Angles Postulate.


Use the diagram above. Which lines, if any, must be parallel if 3 and 2are supplementary?

Because supplements of the same angle are congruent (Congruent Supplements Theorem), 3 4.


3-2

By the Vertical Angles Theorem, 2 is congruent to its vertical angle.

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.


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?

Because 1 2, 1 is congruent to the vertical angle of 2 by the Transitive Property of Congruence.


3-2

The labeled angles are alternate interior angles.

Write and solve the equation 5x – 66 = 14 + 3x.

5x – 66 = 14 + 3x

5x = 80 + 3x Add 66 to each side.

2x = 80 Subtract 3x from each side.

x = 40 Divide each side by 2.

If || m, the alternate interior angles are congruent, and their measures are equal.


Find the value of x for which || m.


3-2

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?

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

Proving Lines ParallelProving Lines ParallelGEOMETRY LESSON 3-2GEOMETRY LESSON 3-2


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 ||.

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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

s

s

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

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3-2


18. 30

19. 50

20. 59

21. 31

22. 5

23. 20

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.

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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.

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3-2


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.

s

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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, AB BH and AB BD, 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.

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3-2


38. Reflexive: a a; false; lines are two lines that intersect to form right .Symmetric: If a b, then b a; true; b and a intersect to form right .Transitive: If a b, and b c, then a c; 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.

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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.

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3-2


46. 47.45. 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.

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3-2


48. 49. 50. a. Answers may vary. Sample:

b. Given: a || b with transversal e,

c bisects AOB,

d bisects AXZ.

c. Prove: c || d

3-2


50. (continued)

d. To prove that c || d, show that 1 3.

1 3 if AOB

OXZ. AOB

OXZ by the Corr. Post.

s

e. 1. a || b (Given)

2. AOB AXZ (Corr. Post.)

3. m AOB = m AXZ (Def. of )

4. m AOB = m 1 + m 2; m AXZ = 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.)

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3-2


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.

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

=/

3-2

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.

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.

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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

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.

Suppose that m 1 = 3x + 10, m 2 = 3x + 14, and m 6 = x + 58 in the diagram above.

26


m || n by Converse of Corresponding Angles Post.

a || b by Converse of Same-Side Interior Angles Theorem.No lines must be parallel.

24


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.

(For help, go to Lesson 1-4.)

Classify each angle as acute, right, or obtuse.

1. 2. 3.


4. 30 + 90 + x = 180 5. 55 + x + 105 = 180

6. x + 58 = 90 7. 32 + x = 90


Parallel Lines and the Triangle Angle-Sum TheoremParallel Lines and the Triangle Angle-Sum Theorem

3-3

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

Solutions



3-3

Find m Z.

48 + 67 + m Z = 180 Triangle Angle-Sum Theorem

115 + m Z = 180 Simplify.

m Z = 65 Subtract 115 from each side.



3-3

m ACB = 90 Definition of right angle

c + 70 = 90 Angle Addition Postulate

c = 20 Subtract 70 from each side.

Find c first, using the fact that ACB is a right angle.


In triangle ABC, ACB is a right angle, and CD AB.

Find the values of a, b, and c.


3-3

a + m ADC + c = 180 Triangle Angle-Sum Theorem

m ADC = 90 Definition of perpendicular lines

a + 90 + 20 = 180 Substitute 90 for m ADC and 20 for c.

a + 110 = 180 Simplify.

a = 70 Subtract 110 from each side.

70 + m CDB + b = 180 Triangle Angle-Sum Theorem

m CDB = 90 Definition of perpendicular lines

70 + 90 + b = 180 Substitute 90 for m CDB.

160 + b = 180 Simplify.

b = 20 Subtract 160 from each side.

To find b, use CDB.

To find a, use ADC.

(continued)



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.


Classify the triangle by its sides and its angles.


3-3

Find m 1.

m 1 + 90 = 125 Exterior Angle Theorem

m 1 = 35 Subtract 90 from each side.



3-3

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 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.

The angle formed by the back of the chair and the armrest increases as you lower the back of the lounge chair.



3-3

Parallel Lines and the Triangle Angle-Sum TheoremParallel Lines and the Triangle Angle-Sum TheoremGEOMETRY LESSON 3-3GEOMETRY LESSON 3-3

9. 70

10. 30

11. 60

12. acute, isosceles

13. acute, equiangular, equilateral

14. right, scalene

15. obtuse, isosceles


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

16.

17. Not possible; a right will always have one longest side opp. the right .

18.

3-3


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

s

27. 115.5

28. m 3 = 92; m 4 = 88

29. x = 147, y = 33

30. a = 162, b = 1831. x = 52.5;

52.5, 52.5, 75; acute

32. x = 7; 55, 35, 90; right

33. x = 37; 37, 65, 78; acute

3-3


39. eight

40. 32.5

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

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.

3-3


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

s

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46. 103

47. 32

48. a. 90

b. 180

c. 90

d. compl.

e. compl.

49. a. Add.

b. -Sum

c. Trans.

d. Subtr.

s

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

3-3


56. Greater than, because there are two with measure 90 where the meridians the equator.

57.

58.

59. 1

60.

61. 0

62. 115

s

13

17

119

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

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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.

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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


70. 53

71. 46

72. 7

73.

74.

3-3

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.


scalene right triangle

68

166

30

8

105


3-3

3-4

1. 2.

3.

(For help, go to Lesson 1-4 and 3-3.)


The Polygon Angle-Sum TheoremsThe Polygon Angle-Sum Theorems

Find the measure of each angle of quadrilateral ABCD.

Solutions


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


3-4

Name the polygon. Then identify its vertices, sides,

and angles.

The polygon can be named clockwise or counterclockwise, starting at any vertex.

Possible names are ABCDE and EDCBA.


Its vertices are A, B, C, D, and E.

Its angles are named by the vertices, A (or EAB or BAE), B (or ABC or CBA), C (or BCD or DCB), D (or CDE or EDC), and E (or DEA or AED).


3-4

Its sides are AB or BA, BC or CB, CD or DC, DE or ED, and EA or AE.


Starting with any side, count the number of sides clockwise around the figure. Because the polygon has 12 sides, it is a dodecagon.

Classify the polygon below by its sides. Identify it as convex or

concave.

Think of the polygon as a star. If you draw a diagonal connecting two points of the star that are next to each other,that diagonal lies outside the polygon, so the dodecagon is concave.


3-4

A decagon has 10 sides, so n = 10.

Sum = (n – 2)(180) Polygon Angle-Sum Theorem

= (10 – 2)(180) Substitute 10 for n.

= 8 • 180 Simplify.

= 1440

Find the sum of the measures of the angles of a decagon.



3-4

m X + m Y + m Z + m W = (4 – 2)(180) Polygon Angle-Sum Theorem

m X + m Y + 90 + 100 = 360 Substitute.

m X + m Y + 190 = 360 Simplify.

m X + m Y = 170 Subtract 190 from each side.

2m X = 170 Simplify.

m X = 85 Divide each side by 2.

m X + m X = 170 Substitute m X for m Y.


The figure has 4 sides, so n = 4.

Find m X in quadrilateral XYZW.


3-4

Because supplements of congruent angles are congruent, all the angles marked 1 have equal measures.

Sample: The hexagon is regular, so all its angles are congruent.

An exterior angle is the supplement of a polygon’s angle because they are adjacent angles that form a straight angle.


A regular hexagon is inscribed in a rectangle. Explain how you

know that all the angles labeled 1 have equal measures.


3-4

The Polygon Angle-Sum TheoremsThe Polygon Angle-Sum TheoremsGEOMETRY LESSON 3-4GEOMETRY LESSON 3-4

3-4

12. 1800

13. 1440

14. 3240

15. 180,000

16. 102

17. 103

18. 145

19. 37


1. yes

2. No; it has no sides.

3. No; it is not a plane figure.

4. No; two sides intersect between endpoints.

5. MWBFX; sides: MW, WB, BF, FX, XM ; : M, W, B, F, X

s

6. KCLP; sides: KC, CL, LP, PK; : K, C, L, P

7. HEPTAGN; sides: HE, EP, PT, TA, AG, GN, NH; : H, E, P, T, A, G, N

8. pentagon; convex

9. decagon; concave

10. pentagon; concave

11. 1080

s

s


31.

32. 3

33. 8

34. 13

35. 18

36. a. 3; 6

b. 4; 8

c. 5; 10

20. 60, 60, 120, 12021. 113, 119

22. 108; 72

23. 150; 30

24. 160; 20

25. 176.4; 3.6

26. 45, 45, 90

27.

28.

29.

30.

3-4


3-4

37. octagon; m 1 = 135; m 2 = 45

38. If you solve

= 130,

you get n = 7.2. This number is not an integer.

39. 20-80-80; 50-50-80

40. 108; 5

41. 144; 10

42. 162; 20

(n – 2) 180n

43. 150; 12

44. 180 – x;

45.

46. a. n • 180

b. (n – 2)180

c. 180n – 180(n – 2) = 360

d. Polygon Ext. -Sum Thm.

47. y = 103; z = 70; quad.

360x

45

48. w = 72, x = 59, y = 49, z = 121;

49. x = 36, 2x = 72, 3x = 108, 4x = 144; quad.

50–53. Answers may vary. Samples are given

50.

51.


b.

c. It is very close to 180

d. No, two sides cannot be collinear.

52.

53.

54. Yes; the sum of the measures of at the int. point is 360. The sum of the measures of all the is 180n. 180n – 360 = (n – 2)180

55. Answers may vary. Sample: The figure is a convex equilateral quad. The sum of its is 2 • 180 or 360.

56. octagon

57. a. (20, 162), (40, 171), (60, 174), (80,175.5), (100, 176.4), (120, 177),

(140, 177.4), (160, 177.75), (180, 178),

(200, 178.2)

s

s

s

3-4


58. a. [180(n – 2)] ÷ n = = 180 – .

b. As n gets larger, the size of the angles get closer to 180. The more sides it has, the closer the

polygon is to a circle.

59. 36

60–63. Answers may vary. Samples are given.

60.

61. Not possible; opp. sides would overlap.

62.

63. Not possible; opp. and adj. sides would overlap.

64. 4140

65. 20

66. 225

67. 360

68. 157.5

180n – 360n

360n

3-4


86. R69. 27

70. 10

71. 120, 25

72. 50, 4073. 104, 76, 35, 69

74. Distr. Prop.

75. Subst. Prop.

76. Reflexive Prop. of

77. Symm. Prop. of

78. Div. Prop.

79. Trans. Prop.

80. RT, RK

81. BRT, BRK

82. Answers may vary. Sample: BR and TK.

83. Answers may vary. Sample: BRM

84. TRM

85. TRK

3-4

1. 2.

3. Find the sum of the measures of the angles in an octagon.

4. A pentagon has two right angles, a 100° angle and a 120° angle. What is the measure of its fifth angle?

5. Find m ABC.

6. XBC is an exterior angle at vertex B. Find m XBC.

quadrilateral ABCD;

AB, BC, CD, DA

not a polygon becausetwo sides intersect at a point other than endpoints


1080

140

144

36


3-4

For Exercises 1 and 2, if the figure is a polygon, name it by its vertices and identify its sides. If the figure is not a polygon, explain why not.

ABCDEFGHIJ is a regular decagon.

Find the slope of the line that contains each pair of points.

1. A(–2, 2), B(4, –2) 2. P(3, 0), X(0, –5)

3. R(–3, –4), S(5, –4) 4. K(–3, 3), T(–3, 1)

5. C(0, 1), D(3, 3) 6. E(–1, 4), F(3, –2)

7. G(–8, –9), H(–3, –5) 8. L(7, –10), M(1, –4)

(For help, go to Page 151.)


Lines in the Coordinate PlaneLines in the Coordinate Plane

3-5

1. m = = = = –

2. m = = = =

3. m = = = = 0

4. m = = = , which is undefined; There is no slope.

5. m = = =

6. m = = = = –

7. m = = =

8. m = = = = – 1

Solutions

y2 – y1

x2 – x1

y2 – y1

x2 – x1

y2 – y1

x2 – x1

y2 – y1

x2 – x1

y2 – y1

x2 – x1

y2 – y1

x2 – x1

y2 – y1

x2 – x1

y2 – y1

x2 – x1

–2 – 24 – (–2)

–5 – 0 0 – 3

–4 – (–4)5 – (–3)

1 – 3– 3 – (–3)

3 – 13 – 0–2 – 43 – (–1)

–5 – (–9)– 3 – (–8)

–4 – (–10) 1 – 7

–4 6

23

–5–3

53

08

–2 0

23

–6 4

32

6 –6

45



3-5

Use the slope and y-intercept to graph the line y = –2x + 9.

In the equation y = –2x + 9, the slope is –2 and the y-intercept is 9.

When an equation is written in the form y = mx + b, m is the slope and b is the y-intercept.



3-5

Use the x-intercept and y-intercept to graph 5x – 6y = 30.

To find the x-intercept, substitute0 for y and solve for x.

To find the y-intercept, substitute0 for x and solve for y.

5x – 6y = 305x – 6(0) = 30 5x – 0 = 30 5x = 30 x = 6

5x – 6y = 305(0) – 6y = 30 0 – 6y = 30 –6y = 30 y = –5

The x-intercept is 6.A point on the line is (6, 0).

The y-intercept is –5.A point on the line is (0, –5).



3-5

Plot (6, 0) and (0, –5). Draw the line containing the two points.

(continued)



3-5

Transform the equation –6x + 3y = 12 to slope-intercept form,

then graph the resulting equation.

Step 1: Transform the equation to slope-intercept form.

The y-intercept is 4 and the slope is 2.

–6x + 3y = 12 3y = 6x + 12 Add 6x to each side.

= + Divide each side by 3. y = 2x + 4

3y 3

6x 3

12 3

Step 2: Use the y-intercept and the slope to plot two points and draw the line containing them.



3-5

Write an equation in point-slope form of the line with slope –8

that contains P(3, –6).

y – y1 = m(x – x1) Use point-slope form.

y – (–6) = –8(x – 3) Substitute –8 for m and (3, –6) for (x1, y1).

y + 6 = –8(x – 3) Simplify.



3-5

Write an equation in point-slope form of the line that contains

the points G(4, –9) and H(–1, 1).

Step 1: Find the slope.

Use the formula for slope.

Substitute (4, –9) for (x1, y1) and (–1, 1) for (x2, y2).

Simplify.

m = –2

m =

m =

m =

y2 – y1

x2 – x1

1 – (–9)–1 – 4

10–5



3-5

Step 2: Select one of the points. Write the equation in point-slope form.

y – y1 = m(x – x1) Point-slope form

y – (–9) = –2(x – 4) Substitute –2 for m and (4, –9) for (x1, y1).

y + 9 = –2(x – 4) Simplify.


(continued)


3-5

Write equations for the horizontal line and the vertical line that

contain A(–7, –5).

Every point on the horizontal line through A(–7, –5) has the same y-coordinate, –5, as point A.

The equation of the line is y = –5.

It crosses the y-axis at (0, –5).

Every point on the vertical line through A(–7, –5) has the same x-coordinate, –7, as point A.

The equation of the line is x = –7.

It crosses the x-axis at (–7, 0).



3-5

Lines in the Coordinate PlaneLines in the Coordinate PlaneGEOMETRY LESSON 3-5GEOMETRY LESSON 3-5

7.

8.

1.

2.

3.

4.

5.

6.

3-5


13. y = –2x + 4

14. y = –2x + 4

9.

10.

11. y = 2x + 1

12. y = x + 1

3-5


15. y = – x + 1

16.

17. y – 3 = 2(x – 2)

18. y + 1 = 3(x – 4)

13

19. y – 5 = –1(x + 3)

20. y + 6 = –4(x + 2)

21. y – 1 = (x – 6)

22. y – 4 = 1(x – 0) or y – 4 = x

23–28. Equations may vary from the pt. chosen. Samples are given.

23. y – 5 = (x – 0)

24. y – 2 = – (x – 6)

12

35

12

25. y – 6 = 1(x – 2)26. y – 4 = 1(x + 4)

27. y – 0 = (x + 1)

28. y – 10 = (x – 8)

29. a. y = 7

b. x = 4

30. a. y = –2

b. x = 3

31. a. y = –1

b. x = 0

12

23

3-5


39. No; a line with no slope is a vertical line. 0 slope is a horizontal line.

40. a. m = 0; it is a horizontal line.

b. y = 0

41. a. Undefined; it is a vertical line.

b. x = 0

32. a. y = 4

b. x = 6

33.

34.

35.

36.

37.

38. a. 0.05

b. the cost per min

c. 4.95

d. the initial charge for a call

3-5


46.

The slopes are all different and the y-intercepts are the same.


48.


42. The eq. is in standard form; change to slope-intercept form, because it is easy to graph the eq. from that form.

43. The eq. is in slope-int. form; use slope-int. form, because the eq. is already in that form.

44. The eq. is in point-slope form; use point-slope form, because the eq. is already in that form.

45.

The slopes are the same and the y-intercepts are different.

3-5


49.

50.

51.

52. = 0.3, = 0.083; > ;

it is possible only if the ramp zigzags.

53. The y-intercepts are the same, and the lines have the same steepness. One line rises from left to right while the other falls from left to right.

54. Answers may vary. Sample: x = 5, y – 6 = 2(x – 5), y = x + 1

55. (2, 0), (0, 4); m = = = –2

y – 0 = –2(x – 2), 2x + y = 4 or y = –2x + 4

56. a. y – 0 = (x – 0) or y = x

b. y – 5 = – (x – 2) or y = – x + 10

310

112

310

112

0 – 42 – 0

–42

52

52

52

52

3-5


3-5

67. C

68. C

56. (continued)c. The abs. value of the slopes is the same, but one

slope is pos. and the other is neg. One y-int. is at

(0, 0) and the other is at (0, 10).

57. Yes; the slope of AB = the slope of BC.

58. No; the slope of DE the slope of EF.

59. Yes; the slope of GH = the slope of HI.

60. Yes; the slope of JK = the slope of KL.

61. y – 2 = 3(x + 2); 3x – y = –8

62. y – 5 = (x – 5); x – 2y = –5

63. y – 6 = (x – 2); 2x – 3y = –14

64. D

65. G

66. B

12

23

=/


72. 1620

73. 2160

74. yes

75. No; || lines never intersect, but they are not skew.

76. No; all obtuse have two acute .

77. a = 5; m MPR = 30

78. a = 4; m MPR = 21

69. [2] a. Line a: y = x + 12 OR equivalent equation;

Line b: 3y + 4x = 19 OR equivalent equation

b.

point of intersection: (22, 9)

[1] at least one correct eq. or graph

70. 1260

71. 540

32

s

s

3-5


79. a = 2; m QPR = 8

80. a = 6; m MPQ = 21

3-5

1. Find the x-intercept and the y-intercept of the line 5x + 4y = –80.

2. Write an equation in point-slope form of the line with slope –1 that contains point C.

3. Write an equation in point-slope form of the line that contains points A and B.

4. Write an equation of the line that contains B and C.

5. Graph and label the equations of the lines in Exercises 2–4 above.


x-intercept: –16, y-intercept: –20

y + 4 = –1(x – 6)

y – 5 = 3(x – 1) or y + 4 = 3(x + 2)

y = –4


3-5

Three points are on a coordinate plane: A(1, 5), B(–2, –4), and C(6, –4).

(For help, go to page 151 and Lesson 3-5.)

1. F(2, 5), B(–2, 3) 2. H(0, –5), D(2, 0)

3. E(1, 1), F(2, –4)

4. y = 2x – 5 5. x + y = 20

6. 2x – 3y = 6 7. x = y

8. y = 7 9. y = x + 723


Slopes of Parallel and Perpendicular Lines Slopes of Parallel and Perpendicular Lines

3-6

Find the slope of the line through each pair of points.

Find the slope of each line.

1. m = = = = y2 – y1

x2 – x1

3 – 5– 2 – 2

– 2– 4

12

2. m = = =y2 – y1

x2 – x1

0 – (–5) 2 – 0

52

3. m = = = = –5 y2 – y1

x2 – x1

– 4 – 1 2 –1

– 5 1

4. The equation is in slope-intercept form and the slope is the coefficient of x, which is 2.

5. Rewrite the equation in slope-intercept form: y = – x + 20; the slope is the coefficient of x, which is – 1.

Solutions


Slopes of Parallel and Perpendicular LinesSlopes of Parallel and Perpendicular Lines

3-6

6. Rewrite the equation in slope-intercept form: y = x – 2; the slope

is the coefficient of x, which is .

7. The equation in slope-intercept form is y = x + 0; the slope is the coefficient of x, which is 1.

8. The equation in slope-intercept form is y = 0x + 7; the slope is the coefficient of x, which is 0.

9. The equation is in slope-intercept form and the slope is the coefficient

of x, which is .

23

23

Solutions (continued)


23


3-6

Find and compare the slopes of the lines.

Each line has slope –1.

The y-intercepts are 3 and –7.

The lines have the same slope and different y-intercepts, so they are parallel.

Slope of line 1 = = = = –1 y2 – y1

x2 – x1

5 – 3–2 – 0

2–2

Slope of line 2 = = = = –1 y2 – y1

x2 – x1

–10 –(–7) 3 – 0

–3 3



3-6

Line 1 contains P(0, 3) and Q(–2, 5). Line 2 contains R(0, –7)

and S(3, –10). Are lines 1 and 2 parallel? Explain.

Are the lines y = –5x + 4 and x = –5y + 4 parallel? Explain.

The equation y = –5x + 4 is in slope-intercept form. Write the equation x = –5y + 4 in slope-intercept form.

The line y = –5x + 4 has slope –5.

The lines are not parallel because their slopes are not equal.

x = –5y + 4

x – 4 = –5y Subtract 4 from each side.

– x + = y Divide each side by –5.

y = – x +

15

45

15

45

The line x = –5y + 4 has slope – .15



3-6

Write an equation in point-slope form for the line

parallel to 6x – 3y = 9 that contains (–5, –8).

Step 1: To find the slope of the line, rewrite the equation in slope-intercept form.

6x – 3y = 9 –3y = –6x + 9 Subtract 6x from each side. y = 2x – 3 Divide each side by –3.

The line 6x – 3y = 9 has slope 2.

Step 2: Use point-slope form to write an equation for the new line.

y – y1 = m(x – x1)y – (–8) = 2(x – (–5)) Substitute 2 for m and (–5, –8) for (x1, y1). y + 8 = 2(x + 5) Simplify.



3-6

Step 2: Find the product of the slopes.

m1 • m2 = – • – = 127

72

Line 1 contains M(0, 8) and N(4, –6). Line 2 contains P(–2, 9)

and Q(5, 7). Are lines 1 and 2 perpendicular? Explain.

m1 = slope of line 1 = = = = – y2 – y1

x2 – x1

–6 – 8 4 – 0

–14 4

72

m2 = slope of line 2 = = = = – y2 – y1

x2 – x1

7 – 95 – (–2)

–2 7

27

Lines 1 and 2 are not perpendicular because the product of their slopes is not –1.


Step 1: Find the slope of each line.


3-6

Write an equation for a line perpendicular to 5x + 2y = 1 that contains (10, 0).

Step 1: To find the slope of the given line, rewrite the equation in slope-intercept form. 5x + 2y = 1

2y = –5x + 1 Subtract 5x from each side.

y = – x + Divide each side by 2.52

12

The line 5x + 2y = 1 has slope – .52

Step 2: Find the slope of a line perpendicular to 5x + 2y = 1. Let m be the slope of the perpendicular line.

– m = –1 The product of the slopes of perpendicular lines is –1.

m = –1 • ( – ) Multiply each side by – .

m = Simplify.

52 2

5252

5



3-6

Step 3: Use point-slope form, y – y1 = m(x – x1), to write an equation for the new line.

(continued)


y – 0 = (x – 10) Substitute for m and (10, 0) for (x1, y1).

y = (x – 10) Simplify.

25

25

25


3-6

The equation for a line containing a lead strip is y = x – 9.

Write an equation for a line perpendicular to it that contains (1, 7).

12

Step 1: Identify the slope of the given line.

slope

y = x – 912

Step 2: Find the slope of the line perpendicular to the given line. Let m be the slope of the perpendicular line.

m = –1 The product of the slopes of perpendicular lines is –1.

m = –2 Multiply each side by 2.

12



3-6

y – y1 = m(x – x1)

y – 7 = –2(x – 1) Substitute –2 for m and (1, 7) for (x1, y1).

(continued)

Step 3: Use point-slope form to write an equation for the new line.

Step 4: Write the equation in slope-intercept form.

y – 7 = –2(x – 1)

y – 7 = –2x + 2 Use the Distributive Property.

y = –2x + 9 Add 7 to each side.



3-6

Slopes of Parallel and Perpendicular LinesSlopes of Parallel and Perpendicular LinesGEOMETRY LESSON 3-6GEOMETRY LESSON 3-6

10. No; one slope = – and the other slope = –3.

11. Yes; the lines both have a slope of – but different y-intercepts.

12. y – 3 = –2(x – 0) or y – 3 = –2x

13. y – 0 = (x – 6) or

y = (x – 6)

6. Yes; the lines both have a slope of 2 but different y-intercepts.

7. Yes; the lines both

have a slope of but

different y-intercepts.

8. Yes; the lines both have a slope of –1 but different y-intercepts.

9. No; one slope = 7 and the other slope = –7.


1. Yes; both slopes = – .

2. No; the slope of 1 = ,

and the slope of 2 = .

3. No; the slope of 1 = ,

and the slope of 2 = 2.

4. Yes; both slopes = 4.

5. Yes; both slopes = 0.

12

1312

32

34

34

25

13

13

3-6


3-6

25. No; • 2 –1.

26. Yes; 1 • (–1) = –1.

27. Yes; one is vertical and the other is horizontal.

28. No; – • (–3) –1.

29. Yes; – • = –1.

30. No; • – –1.

14. y – 4 = (x + 2)15. y + 2 = – (x – 6)

16. Yes; the slope of 1 =

– , and the slope of

2 = 2; – • 2 = –1.

17. Yes; the slope of 1 =

– , and the slope of

2 = ; – • = –1.

18. No; the slope of 1 =

–1, and the slope of

2 = ; –1 • –1.

12

32

12 1

2

32 2

332

23

45

45

19. Yes; the slope of 1 = –1, and the slope of 2 = 1; –1 • 1 = –1.


20. y – 6 = – (x – 6)

21. y = –2(x – 4)

22. y – 4 = (x – 4)

23. y = x

24. y = – x

32

12

45

32

12

1223

32

27

74

=/

=/

=/

=/


31. slope of AB = slope of

CD = ; AB || CD

slope of BC = slope of

AD = –3; BC || AD


CD = – ; AB || CD

slope of BC = slope of

AD = 1; BC || AD

33. slope of AB = ; slope

of CD = ; AB || CD

slope of BC = –1;

slope of AD 5 – ;

BC || AD

23

34

121

4

12


CD = 0; AB || CD

slope of BC = 3 and

slope of AD = ;

BC || AD

35. Answers may vary.

Sample: y = x + 5,

y = – x + 5

36. No; two II lines with the same y-intercept are the same line.

32

455

4

37. RS and VU are horizontal with slope = 0; RS II VU; slope of RW = slope of UT = 1; RW || UT; slope of WV = slope of ST = –1; WV || ST

38. No; because no pairs of slopes have a product of –1.

39. The lines will have the same slope.

3-6


44. neither

45.

46.

47. AC: d = (7 – 9)2 + (11 – 1)2 = 104

BD: d = (13 – 3)2 + (7 – 5)2 = 104

AC BD

48. slope of AC = –5; slope of BD = ; since –5 • = –1,

AC BD; midpoint AC = (8, 6); midpoint BD = (8, 6);

since the midpoints are the same, the diagonals

bisect each other.

40. When lines are , the product of their slopes is –1. So, two lines to the same line must have the same slope.

41. a. y + 20 = (x – 35)

b. because you are given a point and can quickly find the slope

42. ||

43.

34

15

15

3-6


49. a–b. Answers may vary.

Sample:

c. The other possible locations for S are

(–2, 3) and (8, 7).

50. y – 5 = (x – 4)

51. B

52. I

53. C

54. [2] a. slope of line c:

b. 0

[1] at least one correct slope

55. y – 3 = – (x – 0) or y – 3 = – x

56. y – 2 = (x + 4)13

1 – (–2)–4 – 2

3–6=

12= – ; slope of line to c: 2

12

12

53

3-6


57. y + 2 = (x – 3)

58. Refl. Prop. of 59. Mult. Prop. of =

60. Dist. Prop.

61. Symm. Prop. Of

62. If you are in geometry class, then you are at school.

63. If you travel to Switzerland, then you have a passport.

34

3-6

1. Are lines 1 and 2 parallel? Explain.

2. Are the lines x + 4y = 8 and 2x + 6y = 16 parallel? Explain.

3. Write an equation in point-slope form for the line parallel to –18x + 2y = 7 that contains (3, 1).

4. Are the lines y = x + 5 and 3x + 2y = 10 perpendicular? Explain.

5. Write an equation in point-slope form for the line

perpendicular to y = – x – 2 that contains (–5, –8).16


23

Yes; the lines have the same slope and different y-intercepts.

No; their slopes are not equal.

y – 1 = 9(x – 3)

Yes; the product of their slopes is –1.

y + 8 = 6(x + 5)


3-6

(For help, go to Lesson 1-5.)

1. a segment 2. an obtuse angle 3. an acute angle

4. a segment 5. an acute angle 6. an obtuse angle


Constructing Parallel and Perpendicular LinesConstructing Parallel and Perpendicular Lines

3-7

Use a straightedge to draw each figure. Then use a straightedge and compass to construct a figure congruent to it.

Use a straightedge to draw each figure. Then use a straightedge and compass to bisect it.



Solutions

1. 2.

3. 4.

5. 6.

3-7

Use the method learned for constructing congruent angles.

Step 2: With the same compass setting, put the compass point on point N. Draw an arc.

Step 4: Use a straightedge to draw line m through the point you located and point N.

Step 1: With the compass point on point H, draw an arc that intersects the sides of H.

Step 3: Put the compass point below point N where the arc intersects HN.Open the compass to the length where the arc intersects line .Keeping the same compass setting, put the compass point above point N where the arc intersects HN. Draw an arc to locate a point.



Examine the diagram below. Explain how to

construct 1 congruent to H.

3-7

Construct a quadrilateral with both pairs of sides parallel.

Step 1: Draw point A and two rays with endpoints at A. Label point B on one ray and point C on the other ray.

Step 2: Construct a ray parallel to AC through point B.



3-7

(continued)

Step 3: Construct a ray parallel to AC through point C.

Step 4: Label point D where the ray parallel

to AC intersects the ray parallel to AB.

Quadrilateral ABDC has both pairs of

opposite sides parallel.



3-7

In constructing a perpendicular to line at point P, why must

you open the compass wider to make the second arc?

With the compass tip on A and B, a smaller compass setting would make arcs that do not intersect at all. Once again, without another point, you could not draw a unique line.


With the compass tip on A and B, the same compass setting would make arcs that intersect at point P on line . Without another point, you could not draw a unique line.


3-7

Point R is the same distance from point E as it is from point F because the arc was made with one compass opening.

Point G is the same distance from point E as it is from point F because both arcs were made with the same compass opening.


Examine the construction. At what special point does RG

meet line ?

This means that RG intersects line at the midpoint of EF, and RG is the perpendicular bisector of EF.


3-7

Constructing Parallel and Perpendicular LinesConstructing Parallel and Perpendicular LinesGEOMETRY LESSON 3-7GEOMETRY LESSON 3-7

5–7. Constructions may vary. Samples using the following segments are shown.

5.

6.


1.

|| AB

2.

|| AB

3.

|| AB

4.

|| AB

3-7


11.

RS

12.

RS

7.

8.

AB

9.

AB

10.

RS

3-7


17.

18.

13.

RS

14.

1 2

15. Construct a alt. int. ; then draw the || line.

16. a–b

17–25. Constructions may vary. Samples are given.

3-7


23.

24. a.

19.

20.

21. a.

b. The sides are || and .

c. Check students’ work.

22.

3-7


26. a–b. Check students’ work.

c. p || m; in a plane, two lines to a

third line are ||.


27.

The quad. is a rectangle.

24. (continued)b. Answers may vary. Sample:

c. One; if the lengths for the 3 sides are given, only one is possible; many different quad. are possible because the formed by the sides can vary.

25. a–c.

d. The sides of the smaller are

half the length of the sides of the larger that they are || to.

e. Check students’ work.

s

3-7


32.

33.

28.

The quad. is a square.

29.

30.

31.

3-7


36. Not possible; the smaller sides would meet at the midpoint of the longer side, forming a segment.

37. A

38. I

39. [2] a.

34.

35. Not possible; if a = 2b = 2c, then 2a = 2b + 2c or a = b + c. The smaller sides would meet at the midpoint of the longer side, forming a segment.

39. (continued)b. All points on the arc

with center G are the same dist. from G, so GR = GT. The is isosc. because an isosc. must have at least two sides of the same length.

[1] incorrect sketch OR incorrect classification

3-7


40. [2] a. I, IV, III, I

b. (III): location of compass at

points C and G; (I): same

as c and the intersection points of

with arcs drawn in (III)

[1] incorrect sequence OR incorrect location of compass point

41. No; the slopes are different.

42. No; the slopes are different.

43. Yes; the slopes are both – .

44. 10

45. 8.9

46. EB

47. DF

13

3-7

1. Construct a line through D that is parallel to XY.

2. Construct a quadrilateral with one pair of parallel sides of lengths p and q.

3. Construct the line perpendicular 4. Construct the perpendicular to line m at point Z. to line n through point O.


Answers may vary. Sample given:


3-7

Draw a figure similar to the one given. Then complete the construction.

Answers may vary. Sample given:

3-A

Parallel and Perpendicular LinesParallel and Perpendicular LinesGEOMETRY CHAPTER 3GEOMETRY CHAPTER 3

Page 176

1. acute isosceles2. obtuse scalene

3. m 1 = 65 because corr. are ; m 2 = 65 because vert. are .

4. m 1 = 85 because alt. int. are ; m 2 = 110 because same-side int. are suppl.

s

s

s

s

5. m 1 = 85 because corr. are ; m 2 = 95 because same-side int. are suppl.

6. m 1 = 70 because corr. are ; m 2 = 110 because same-side int. are suppl.

7. yes

8. yes

9. no

10. no

s

s

s

s

11. 5

12. 25

13. 6

14. 75

15.

Parallel and Perpendicular LinesParallel and Perpendicular LinesGEOMETRY CHAPTER 3GEOMETRY CHAPTER 3

20. y + 1 = –5(x – 3) or y = –5x + 14

21. Draw segments connecting nonconsecutive vertices. If no points of the segments are outside the polygon, then it is convex. Otherwise, the polygon is concave.

22. 109

23. x = 85; y = 100; z = 100

24.

25. ||

26. neither

27.

28. 30

16.

17. Answers may vary. Sample: Z


19. a. Given

b. Corr. are .

c. Given

d. Trans. Prop.

e. If corr.

are , then the lines

are ||.

s

s

3-A

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