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Transcript of Geometry Student Text and Homework Helper ... Plane < >, < >, < >, < > <...

1

Odd-Numbered ANswersGeometry | Student Text and Homework Helper

Topic 1 Lesson 1-1 pp. 7–9 1. Answers may vary. Sample: plane EBG, plane BFG 3. E, B, F, G 5. RS

> , SR

> , ST

> , TS

> , TW

> , WT

> , TR

> , RT

> ,

WR > , RW

> , WS

> , SW

> 7.

< VW

> 9. plane TSR, plane TSW

11. plane VWX, plane VWS 13.

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X W

S

RQ

U V

T

15.

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X W

S

RQ

U V

T

17. noncoplanar 19. Answers may vary. Sample: You can represent a plane by drawing a four-sided shape. However, planes do not have boundaries, so they extend past the drawn edges without end. Since a plane does not have an endpoint, the intersection of the two planes your friend drew cannot be a point. 21. Not always; AC

> contains BC

> , but they are not the

same ray.

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A B C 23. always 25. sometimes 27.

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A

B

Location of cell phone

By Postulate 1-1, the location of the cell phone and point A determine a line, and the location of the cell phone and point B determine a line. By Postulate 1-2, the two lines intersect at, or share, only one point. Since the cell phone signal is on both lines, its location must be at the intersection of the lines.

29. Answers may vary. Sample: now, think, exist 31.

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y

x O�4 4�2

4 33.

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y

xO

�2

4

4�4

yes no

35. 14 37. F

Lesson 1-2 pp. 14–15 1. 2 3. 24 5. about 1 h, 21 min 7a. 9 b. AY = 9, XY = 18 9. y = 15; AC = 24, DC = 12 11. Not always; the Segment Addition Postulate can be used only if P, Q, and R are collinear points.

13. 3 15. - 1.5 or -32 17. no 19. yes 21. The distance is @ 65 - 80 @ , or 15 mi. The driver added the values instead of subtracting them. 23. 2 25. 123 or -1 23 27. G

Lesson 1-3 pp. 19–21 1. ∠XYZ , ∠ZYX , ∠Y 3. ∠JKM, ∠MKJ, or ∠2 5. 90, right 7. Answers may vary. Sample:

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R S

T

9. ∠FHG 11. 130 13. yes;

15. 180 17. 30 19. 40 21. about 27.7 23. about 90°; right 25. Angle Addition Postulate 27. x = 18; m∠BOC = 52, m∠AOD = 108 29. m∠RQS = 43, m∠TQS = 137 31. J

Lesson 1-4 pp. 25–26 1. Yes, the angles share a common side and vertex, and have no interior points in common. 3. No, they are supplementary. 5. ∠EOC 7. Answers may vary. Sample: ∠AOB, ∠DOC 9. 35, 55 11. 115 13. No; they do not have a common vertex. 15. m∠EFG = 69, m∠GFH = 111 17a. ∠CBD; 41 b. 82 c. 49; 49 19. 30 21. No; JC and CD are not marked as ≅. 23. Yes; they are formed by

< JF > and

< ED

> . 25. C

Lesson 1-5 pp. 30–32 1. Answers may vary. Sample:

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X Y

Z

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X Y

Z

Find a segment on < XY

> so that you can construct <

YZ > as its perpendicular bisector.

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PV

P B

V B

V P

P

V B

B1 1

1 1

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PV

P B

V B

V P

P

V B

B1 1

1 1

2

Odd-Numbered ANswersGeometry | Student Text and Homework Helper

3. Answers may vary. Sample: Both constructions involve drawing arcs with the same radius from two different points, and using the point(s) of intersection of those arcs. Arcs must intersect at two points for the # bis., but only one point for the ∠ bis. 5a.

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The three angle bisectors meet at a point. b.

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c. For any triangle, the three angle bisectors meet at a point. 7. Not possible; the 2-cm sides meet on the 4-cm side, so they do not form a triangle. 9a. A segment has exactly one midpoint; using the Ruler Postulate (Post. 1-5), each point corresponds with exactly one number, and exactly one number represents half the length of a segment. b. A segment has infinitely many bisectors because infinitely many lines can be drawn through the midpoint. c. In the plane with the segment, there is one # bis. because only one line in that plane can be drawn through the midpoint so that it forms a right angle with the given segment. d. Consider the plane that is the # bis. of the segment. Any line in that plane that contains the midpoint of the segment is a # bis. of the segment, and there are infinitely many such lines. 11a–b.

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A

C

O

B

c. O is the center of the circle. 13.

V W AB AB

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15.

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F 17.

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Q

R

P

19.

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A m �A1

4

21. A 23. no 25. D 27. x2 - 2 = x

x2 - x - 2 = 0 (x - 2)(x + 1) = 0 x = 2 or x = -1 (not possible) x = 2

Technology Lab 1-5 pp. 33–34 1. Yes. It is possible to make m∠HGF = 90 and EG = GF. When these conditions are met,

< HG

> is a

perpendicular bisector. 3. The position of EF relative to

< GH

> can change, whereas the position of AB

relative to < DC

Activity Lab 1-5 p. 36 1. yes 3. no; not a plane figure 5. Sample: FBWMX; sides are FB, BW, WM, MX, XF ; angles are ∠F, ∠B, ∠W, ∠M, ∠X 7. Sample: AGNHEPT; sides are AG, GN, NH, HE, EP, PT , TA; angles are ∠A, ∠G, ∠N, ∠H, ∠E, ∠P, ∠T 9. nonagon or enneagon, convex

Topic Review pp. 37–39 1. angle bisector 3. construction 5.

< QR

> 7. True;

Postulate 1-1 states, “Through any two points, there is exactly one line.” 9. -7, 3 11. 15 13. acute 15. 36 17. Answers may vary. Sample: ∠ADB and ∠BDC 19. Answers may vary. Sample: ∠ADC and ∠EDF 21. 31 23.

73�

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25.

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L M

3

Odd-Numbered ANswersGeometry | Student Text and Homework Helper

TEKS cumulative practice pp. 40–41 1. D 3. C 5. A 7. C 9. D 11. B 13. 45 15. 35 17. 7 19. No, is P, Q, and R are not collinear, then Q is not on PR, so it cannot be the midpoint of PR. 21a. 2x + 6; Sample answer: m∠AOC = m∠COD = 4x + 12 since OC is the angle bisector of ∠AOD. m∠BOC = 12m∠AOC since OB is the angle bisector of ∠AOD. Thus, m∠BOC = 12 (4x + 12) = 2x + 6. b. x = 12. Sample answer: m∠COD = 12m∠AOD since OC is the angle bisector of ∠AOC. Thus, 4x + 12 = 12(120) = 60. If 4x + 12 = 60, then x = 12.

Topic 2 Lesson 2-1 pp. 46–48 1. Double the previous term; 80, 160. 3. Add -2, +3, -4, +5, c; -3, 4. 5. The numerator is 1, and the denominator is the next whole number; 15,

1 6.

7. the first letters of the counting numbers; N, T 9. Multiply the previous number by 2, by 3, by 4, by 5, c; 720, 5040. 11. U.S. coins of descending value; dime, nickel 13. zodiac signs; Gemini, Cancer 15.

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17. 102 cm 19. blue 21. 75°F 23. and 25. Answers may vary. Samples are given. 23. two right angles 25. -2 and -3 27a. sì-shí-sān; lìu-shí-qī; bā-shí-sì b. Yes; the second part of the number repeats each ten numbers. 29a.

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2004 2005 2006 2007 2008

40

60

80

100

N um

be r

of S

pe ci

es

Bird Count

Year

b. Answers may vary. Sample: Using just the data from 2005 to 2008, the gain is 7 species in 3 years, or between 2 and 3 species each year. The year 2015 is 7 years after 2008, so the number of new species will be between 14 and 21 more than 90; an estimate is 90 + 17 or 107 species. 31. 1 * 1 : 64 squares; 2 * 2 : 49 squares;

3 * 3 : 36 squares; 4 * 4 : 25 squares; 5 * 5 : 16 squares; 6 * 6 : 9 squares; 7 * 7 : 4 squares; 8 * 8 : 1 square;

total number of squares: 204 33. C 35. 2

Lesson 2-2 pp. 52–54 1. Hypothesis: You are an American citizen. Conclusion: You have the right to vote. 3. Hypothesis:

You want to be healthy. Conclusion: You should eat vegetables. 5. If you have never made a mistake, then you have never tried anything new. 7. Yes, he is correct; both are true, because a conditional and its contrapositive have the same truth value. 9. If a point is in the first quadrant of a coordinate plane, then both coordinates of that point are positive. 11. If a number is a whole number, then it is an integer. 13. false; Mexico 15. true 17. Answers may vary. Sample: If a person is a pitcher, then that person is a baseball player. If a person is a baseball player, then that person is an athlete. If a person is a pitcher, then that person is an athlete.