Geometry Student Text and Homework Helper...

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1

Odd-Numbered ANswersGeometry | Student Text and Homework Helper

Topic 1Lesson 1-1 pp. 7–91. 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.

hsm11gmte_0102_t02256.ai

X W

S

RQ

U V

T

15.


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.


A B C 23. always 25. sometimes27.


A

B

Location ofcell 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.


y

xO�4 4�2

4 33.


y

xO

�2

4

4�4

yes no

35. 14 37. F

Lesson 1-2 pp. 14–151. 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. 12

3 or -1 23 27. G

Lesson 1-3 pp. 19–211. ∠XYZ , ∠ZYX , ∠Y 3. ∠JKM, ∠MKJ, or ∠2 5. 90, right 7. Answers may vary. Sample:


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. 25–261. 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. 30–321. Answers may vary. Sample:


X Y

Z


X Y

Z

Find a segment on <XY

> so that you can construct <

YZ> as its perpendicular bisector.


PV

P B

VB

VP

P

V B

B1 1

1 1


PV

P B

VB

VP

P

V B

B1 1

1 1


2


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.


The three angle bisectors meet at a point. b.


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.


A

C

O

B

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

V WAB AB


15.


F 17.


Q

R

P

19.


Am �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. 33–341. 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

> is fixed. 5a. Ask your teacher to check

your work. b. no

Activity Lab 1-5 p. 361. 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. 37–391. 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. 3123.

73�

hsm11gmte_01cr_t05318

25.


L M


3


TEKS cumulative practice pp. 40–411. 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 = 1

2m∠AOC since OB is the angle bisector of ∠AOD. Thus, m∠BOC = 1

2 (4x + 12) = 2x + 6. b. x = 12. Sample answer: m∠COD = 1

2m∠AOD since OC is the angle bisector of ∠AOC. Thus, 4x + 12 = 1

2(120) = 60. If 4x + 12 = 60, then x = 12.

Topic 2Lesson 2-1 pp. 46–481. 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, 16. 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.

hsm11gmte_0201_t06678

hsm11gmte_0201_t06679

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.

hsm11gmte_0201_t06684

2004 2005 2006 2007 2008

40

60

80

100

Num

ber

of S

peci

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. 52–541. 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. 19.

hsm11gmte_0202_t06685

Obtuseangles

100�

21.

hsm11gmte_0202_t06687

People who want to helpothers Peace

Corpsvolunteers

23. If - y is positive, then y is negative; true. 25. If x2 7 0, then x 6 0; false: a counterexample is x = 1. 27. Conditional: If a person is a pianist, then that person is a musician. Converse: If a person is a musician, then that person is a pianist. Inverse: If a person is not a pianist, then that person is not a musician. Contrapositive: If a person is not a musician, then that person is not a pianist. The conditional and the contrapositive are true. The converse and the inverse are false; counterexample: a percussionist is a musician. 29. Conditional: If a number is an odd natural number less than 8, then the number is prime. Converse: If a number is prime, then it is an odd natural number less than 8. Inverse: If a number is not an odd natural number less than 8, then the number is not prime. Contrapositive: If a number is not prime, then it is not an odd natural number less than 8. All four statements are false; counterexamples: 1 and 11. 31. If you wear Snazzy sneakers, then you will look cool. 33. If two figures are congruent, then they have equal areas. 35. All integers divisible by 8 are divisible by 2. 37. Some musicians are students. 39. J

Lesson 2-3 pp. 57–591. Converse: If two segments are congruent, then they have the same length; true. Biconditional: Two segments have the same length if and only if they are congruent. 3. Converse: If it is Independence Day in the United States, then it is July 4; true. Biconditional: In the United States, it is July 4 if and only if it is Independence Day. 5. Yes; it uses clearly understood terms, is precise, and is reversible. 7. That statement, as a biconditional, is “an angle is a right angle if and only if it is greater than an acute angle.” Counterexamples to that statement are obtuse angles and straight angles. 9. A point is in Quadrant III if and only if it has two negative coordinates.


4


11. A number is a whole number if and only if it is a nonnegative integer. 13. If an integer is divisible by 100, then its last two digits are zeros. If the last two digits of an integer are zeros, then the integer is divisible by 100. 15. If x2 = 144, then x = 12 or x = -12. If x = 12 or x = -12, then x2 = 144. 17. good definition 19. If ∠A and ∠B are a linear pair, then ∠A and ∠B are supplementary. 21. If ∠A and ∠B are a linear pair, then ∠A and ∠B are adjacent and supplementary angles. 23. Answers may vary. Sample: A line is a circle on the sphere formed by the intersection of the sphere and a plane containing the center of the sphere. 25. B 27a. If you go to the store, then you want to buy milk; false. b. Answers may vary. Sample: A counterexample is going to the store because you want to buy juice.

Lesson 2-4 pp. 62–641. No conclusion is possible; the conclusion has been satisfied, but the hypothesis has not been satisfied. 3. No conclusion is possible; the same statement does not appear as the conclusion of one conditional and as the hypothesis of the other conditional. 5. $5.99; your family goes to your favorite restaurant on the night of your weekly game, which is on Tuesday. Chicken fingers are $5.99 on Tuesday. 7. Must be true; by E and A, it is breakfast time; by C, Curtis is drinking water. 9. Is not true; by E and A, it is breakfast time; by C, Curtis drinks water and nothing else. 11. Is not true; by A and E, it is breakfast time; by D, Julio is drinking juice and nothing else. 13. Alaska’s Mount McKinley is the highest mountain in the U.S. 15. From the table, if a quark has rest energy 540 MeV and a charge of - 1

3 e, then the flavor of the quark is strange; a given quark has rest energy 540 MeV and a charge of - 1

3 e. By the Law of Detachment, the flavor of the given quark is strange. 17. B

Lesson 2-5 pp. 68–701a. Mult. Prop. of Eq. b. Distr. Prop. c. Add. Prop. of Eq. 3a. Seg. Add. Post. b. Subst. Prop. c. Distr. Prop. d. Distr. Prop. e. Subtr. Prop. of Eq. f. Div. Prop. of Eq. 5. Since LR and RL are two ways to name the same segment and ∠CBA and ∠ABC are two ways to name the same ∠, then both statements are examples of saying that something is ≅ to itself. 7. KM = 35 (Given); KL + LM = KM (Seg. Add. Post.); (2x - 5) + 2x = 35 (Subst. Prop.); 4x - 5 = 35 (Distr. Prop.); 4x = 40 (Add. Prop. of Eq.); x = 10 (Div. Prop. of Eq.); KL = 2x - 5 (Given); KL = 2(10) - 5 (Subst. Prop.); KL = 15(Simplify). 9. The error is in the 5th step when both sides of the equation are divided by b - a, which is 0, and division by 0 is not defined. 11. Sym. Prop. of ≅ 13a. Given b. A midpt. divides a seg. into two ≅ segments. c. Substitution Prop. of Eq. d. 2x = 12 e. Div. Prop. of Eq. 15. ∠K 17. 3 19. Transitive only; A cannot be taller than A; if A is

taller than B, then B is not taller than A. 21. 58.5 23. 153.86 25. 58

Lesson 2-6 pp. 75–781. x = 15, x + 10 = 25, 4x - 35 = 25 3a. Vert. ∠s Thm. b. ∠1 ≅ ∠6 c. Vert. ∠s Thm. d. Trans. Prop. of ≅ 5. 55 7. ∠1 and ∠3 are given as vert. ∠s. Because ∠1 and ∠2 form a linear pair, ∠1 and ∠2 are suppl. Because ∠2 and ∠3 form a linear pair, ∠2 and ∠3 are suppl. So m∠1 + m∠2 = 180 and m∠2 + m∠3 = 180 by the def of suppl. ∠s. By the Trans. Prop. of Eq., m∠1 + m∠2 = m∠2 + m∠3. Subtract m∠2 from both sides. By the Subtr. Prop. of Eq., m∠1 = m∠3. ∠s with equal measure are ≅, so ∠1 ≅ ∠3. 9. 55°, 125°, 125° 11a. it is given b. m∠V c. 180 d. Division e. right 13. By Theorem 2-5: If two angles are ≅ and suppl., then each is a right angle. 15. m∠A = 30, m∠B = 60 17. m∠A = 90, m∠B = 90 19. Sample answer: The intersecting lines form two pairs of vertical angles. By the Vertical Angles Theorem, because one angle measures 90°, its vertical angle also measures 90°. Each of the remaining two angles forms a linear pair with a 90° angle, so each has a measure of 180° - 90°, or 90°. So all four angles are right angles. 21. theorem; This is an example of the Congruent Complements Theorem. 23a. ∠Y b. right ∠ c. m∠Y d. ∠X ≅ ∠Y 25. x = 30, y = 90; 60, 120, 60 27. x = 50, y = 20; 80, 100, 80 29. ∠5 _ ∠7 by the Vertical Angles Theorem. ∠5 _ ∠8 by the Transitive Property of Congruence. ∠8 _ ∠6 by the Vertical Angles Theorem. ∠5 _ ∠6 by the Transitive Property of Congruence. 31. 20

Topic Review pp. 79–821. conclusion 3. truth value 5. biconditional 7. Divide the previous term by 10; 1, 0.1 9. Subtract 7 from the previous term; 6, -1 11. Answers may vary. Sample: -1 # 2 = -2 and -2 is not greater than 2. 13. If a person is a motorcyclist, then that person wears a helmet. 15. If two angles form a linear pair, then the angles are supplementary. 17. Converse: If the measure of an angle is greater than 90 and less than 180, then the angle is obtuse. Inverse: If an angle is not obtuse, then it is not true that its measure is greater than 90 and less than 180. Contrapositive: If it is not true that the measure of angle is greater than 90 and less than 180, then the angle is not obtuse. All four statements are true. 19. Converse: If you play an instrument, then you play the tuba. Inverse: If you do not play the tuba, then you do not play an instrument. Contrapositive: If you do not play an instrument, then you do not play the tuba. The conditional and the contrapositive are true. The converse and inverse are false. 21. No; it is not reversible; a magazine is a counterexample. 23. No; it is not reversible; a line is a counterexample.


5


25. If two angles are complementary, then the sum of their measures is 90; if the sum of the measures of two angles is 90, then the angles are complementary. 27. m∠1 + m∠2 = 180 29. If your father buys new gardening gloves, then he will plant tomatoes. 31. BY 33. 18 35. 74 37. ∠1 is compl. to ∠2, ∠3 is compl. to ∠4, and ∠2 ≅ ∠4 are all given. m∠2 = m∠4 by the def. of ≅. ∠1 and ∠4 are compl. by the Subst. Prop. ∠1 ≅ ∠3 by the ≅ Compl. Thm.

TEKS cumulative practice pp. 83–851. D 3. C 5. B 7. D 9. B 11. C 13. C 15. C 17. D 19. 25 21. 121 23. 123454321 25. Converse: If you live in the United States, then you live in Oregon; false. Inverse: If you do not live in Oregon, then you do not live in the United States; false. Contrapositive: If you do not live in the United States, then you do not live in Oregon; true. 27a.

b. 9; the ones digit has four repeating digits (7, 9, 3, 1). Since 34 divided by 4 has a remainder of 2, the ones digit in 734 is the same as the ones digit in 72.

Topic 3Lesson 3-1 pp. 91–931. Plane JCD and plane ELH 3.

<GB

>, <JE>, <CL>, <FA>

5. <GB

>, <DH

>, <CL> 7. ∠7 and ∠6 (lines a and b with

transversal d); ∠2 and ∠5 (lines b and c with transversal e) 9. ∠5 and ∠6 (lines d and e with transversal b); ∠2 and ∠4 (lines b and e with transversal c); ∠1 and ∠3 (lines a and d with transversal c); ∠7 and ∠3 (lines a and c with transversal b); ∠7 and ∠1 (lines d and c with transversal a) 11. ∠1 and ∠2 are corresponding angles; ∠3 and ∠4 are alternate interior angles; ∠5 and ∠6 are corresponding angles 13. ∠1 and ∠2 are corresponding angles; ∠3 and ∠4 are same-side interior angles; ∠5 and ∠6 are alternate interior angles 15. Skew; answers may vary. Sample: Since the paths are not coplanar or parallel, they are skew. 17. 4 pairs 19. 4 pairs 21. False;

<ED

> and

<HG

> are

skew. 23. False; the planes intersect. 25. False; both lines are in plane ABC. 27. Answers may vary. Sample:

E illustrates corresponding angles, N illustrates alt. int. angles. 29. always 31. sometimes 33a. The lines of intersection are ||. b. Sample: the lines of intersection of a wall with the ceiling and floor (or the lines of intersection of any of the 6 planes with two different, opposite faces) 35. Yes; 37. B

AP

BC D

Lesson 3-2 pp. 98–1001. ∠1 (vertical angles), ∠7 (alternate interior angles), ∠4 (corresponding angles) 3. ∠3 (alternate interior angles), ∠1 (corresponding angles) 5. a } b; c } d; (Given) ∠1 ≅ ∠4 (Alternate interior angles are congruent.) ∠4 ≅ ∠3 (Corresponding angles are congruent.) ∠1 ≅ ∠3 (Transitive Property of Congruence) 7. ∠1 = 120 because corresponding angles are congruent; ∠2 = 60 because same-side interior angles are supplementary. 9. x = 115, x - 50 = 65 11. 20; 5x = 100, 4x = 80 13. x = 135, y = 45 15. 90; all the angles are congruent because each pair are vertical angles, corresponding angles, or supplementary angles. 17. a } b (Given); m∠1 + m∠2 = 180 and m∠3 + m∠4 = 180 (Same-side interior angles are supplementary.); ∠1 ≅ ∠4 (Given); ∠2 ≅ ∠3 (If two angles are supplementary to congruent angles, then the angles are congruent.) 19. Drawings will vary. Conjecture: The transversal is perpendicular to both parallel lines. 21. m∠1 = 48, m∠2 = 132 23. 65 25. 14

Lesson 3-3 pp. 104–1061.

<BE>}<CG

>; Converse of the Corresponding

Angles Theorem. 3. <CA

>}<HR

>; Converse of the

Corresponding Angles Theorem. 5a. Given b. ∠1 and ∠2 form a linear pair. c. Angles that form a linear pair are supplementary. d. ∠2 ≅ ∠3 e. If corresponding angles are congruent, then lines are parallel. 7. 59 9. a } b; if same-side interior angles are supplementary, then the lines are parallel. 11. a } b; if same-side interior angles are supplementary, then the lines are parallel. 13. none 15. a } b (Converse of the Alternate Exterior Angles Theorem) 17. none 19. ∠2 and ∠3 are supplementary (Linear Pair Postulate), so ∠3 ≅ ∠7 (Congruent Supplements Theorem). Therefore, / } m (Converse of the Corresponding Angles Theorem). 21. x = 10; m∠1 = m∠2 = 70 23. x = 2.5; m∠1 = m∠2 = 30 25. If alternate exterior angles are congruent, then the lines are parallel.


PowerOnes Place

Digit

71 7

72 9

73 3

74 1

75 7

76 9

77 3

78 1


6


27. Answers may vary. Samples given:

29. PL } NA; if same-side interior angles are supplementary, then the lines are parallel. 31. PN } LA; if same-side interior angles are supplementary, then the lines are parallel. 33. A 35. C

Lesson 3-4 pp. 109–1101a. ∠1 ≅ ∠2 ≅ ∠3 b. Yes; pieces B and C are parallel, and if a line is perpendicular to one of several parallel lines, it is perpendicular to all of the parallel lines. 3. Sample: Consider the edges of a rectangular prism. The top front edge and the bottom side edge are both perpendicular to a vertical edge of the prism. The top front edge and the bottom side edge are skew, so they cannot be parallel. 5. Measure any three interior angles to be right angles and opposite walls will be parallel because two walls perpendicular to the same wall are parallel. 7. The right triangles must have acute angles that measure 45. 9. The rungs are parallel to each other because they are all perpendicular to the same side. 11. The rungs are perpendicular to both sides. The rungs are perpendicular to one of two parallel sides, so they are perpendicular to both sides. This also means all of the rungs are parallel. 13. The rungs are parallel because they are all perpendicular to one side. 15. Reflexive: a # a; false; a line does not intersect itself at a right angle. Symmetric: If a#b, then b#a; true; lines a and b form right angles. Transitive: If a#b and b#c, then a#c; false; if a and c are both perpendicular to b, then a } c or a and c are skew. 17. a#d by Theorems 3-8 and 3-10 19. a#d by Theorem 3-10 21. a } d by Theorems 3-9 and 3-10 23. H

Lesson 3-5 pp. 114–1161. Sample answer:

S

1U T2 3

Y

X

By the Parallel Postulate, draw

<XY

> through T,

parallel to SU. Because angles that form a linear pair are supplementary, ∠1 and ∠UTX are supplementary. By the definition of supplementary angles, m∠1 + m∠UTX = 180. By the Angle Addition Postulate, m∠UTX = m∠2 + m∠3. By the Substitution Property, m∠1 + m∠2 + m∠3 = 180. ∠1 _ ∠U and ∠3 _ ∠S because if lines are parallel, alternate interior angles are congruent. Congruent angles have equal measure, so m∠1 = m∠U and m∠3 = m∠S. So, by the Substitution Property, m∠S + m∠2 + m∠U = 180.

3. x = y = 80 5. 123 7. m∠3 = 92, m∠4 = 88 9. 60; answers may vary. Sample: 180 , 3 = 60, so each angle is 60. 11. 30, 60 13. 12, 60 15. Answers may vary. Sample: m∠A + m∠B + m∠C = 180 by the Triangle Angle-Sum Theorem. It is given that m∠C = 90, so m∠A + m∠B + 90 = 180. By the Subtraction Property of Equality, m∠A + m∠B = 90. Thus ∠A and ∠B are complementary by the definition of complementary angles. 17. x = 7; m∠A = 35, m∠B = 55, m∠C = 90 19. x = 38; y = 36, z = 90; m∠ABC = 74 21. Sample answer:

A

B ED

C △ABC with right angle ACB (Given); Draw

<DE

>

through B, parallel to AC (Parallel Postulate); ∠DBC and ∠CBE are supplementary (Angles that form a linear pair are supplementary.); m∠DBC + m∠CBE = 180 (Definition of supplementary angles); m∠DBC = m∠DBA + m∠ABC (Angle Addition Postulate); m∠DBA + m∠ABC + m∠CBE = 180 (Substitution Property); ∠DBA _ ∠BAC and ∠CBE _ ∠ACB (If lines are parallel, then alternate interior angles are congruent.); m∠DBA = m∠BAC and m∠CBE = m∠ACB (Congruent angles have equal measure.); m∠BAC + m∠ABC + m∠ACB = 180 (Substitution Property); m∠ACB = 90 (Definition of right angle); m∠BAC + m∠ABC + 90 = 180 (Substitution Property); m∠BAC + m∠ABC = 90 (Subtraction Property of Equality); ∠BAC and ∠ABC are complementary (Def. of compl. angles)

j � k

∠4 suppl. ∠9

Given

Given

∠8 suppl. ∠9

If lines are �,then same-sideint. ⦞ are suppl.

∠4 ≅ ∠8

� � n

If corresp. ⦞are ≅, then linesare �.

≅ Suppl. Thm.



7


23. 40, 50 25. The bisector is parallel to the side common to the congruent angles. If the measure of each congruent angle is x, then the measure of the exterior angle is 2x so the bisector forms 2 angles of measure x. Alternate interior angles are congruent, so the bisector is parallel to the side common to the congruent interior angles. 27. H

Lesson 3-6 pp. 121–1221.


�

A

J

B3.


�

A

J

B

5. Constructions may vary. Sample using the following segments is given.

b

a

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12

b

a7.


P

�

9.


P

R S

�

11.


2

1

13. Construct a congruent alternate interior angle, then draw the parallel line.15.


b

a

17.


c c

b

b

19.


b

a

21–29. Constructions may vary. Samples are given.21.


D G

23.

a

ab

a



8


25.

b

a

hsm11gmte_0306_t11653

27.

a c

b

hsm11gmte_0306_t1165529. Not possible; if a = 2b = 2c, then b = c and a = b + c. So the shorter sides would meet at the midpoint of the longer side, forming a segment. 31. D 33. 28; if n is the number of points, thenn(n - 1)

2 is the number of segments.

Lesson 3-7 pp. 127–1281. 2 3. -3

2 5. undefined7. y

xO 2

2

�2

�4

hsm11gmte_0307_t11662

9.

hsm11gmte_0307_t11664

y2

�2x

O

11. y + 1 = -3(x - 4) 13. y - 6 = - (x + 2) or y - 3 = - (x - 1) 15. y - 2 = -1

2(x - 6) or y - 4 = -1

2(x - 2) 17. horizontal: y = 7; vertical: x = 4 19. Answers may vary. Sample: x = 5, y = 6, y = 6

5x, y - 6 = 65(x - 5)

21. (0, 0)

hsm11gmte_0307_t11673

y

x

2 x � 0y � 0

(0, 0)�2�2

O

23. 0; the x-axis is a horizontal line, and the slope of a horizontal line is 0; y = 0 25. Yes; if the ramp is 24 in. high and 72 in. long, the slope will be 24

72 = 0.3 whichis less than the maximum slope of 4

11 = 0.36. 27. y = -x + 2 29a. - 1

33; The slope represents the rate at which the pressure changes for each foot of a dive. b. 1; The y-intercept is the value of the pressurewhen the depth of the dive is 0. 31a. y = 5

2x

b. y - 5 = -52(x - 2) or y = -5

2x + 10 c. The abs. value of the slopes is the same, but one slope is pos. and the other is neg. One y-int. is 0 and the other is 10. 33. No; the slope of the line through the first two points is -1

3, and the slope of the line through the last two points is -1, so the points do not lie on the same

line. 35. B 37. Yes; if the sum of two numbers is 180 and one of them is less than 90, then the other must be greater than 90.

Lesson 3-8 pp. 132–133 1. Yes; the slope of /1 is -1

2, and the slope of /2 is -1

2, and two lines with the same slope are parallel.

3. y = -2x + 3 5. y - 4 = 12(x + 2) 7. No; the

slope of /1 is -1, and the slope of /2 is 45. Since the product of the slopes is not -1, the lines are not perpendicular. 9. y - 6 = -3

2(x - 6) 11. y - 4 = 1

2(x - 4) 13. Yes; both slopes are -1, so the lines are }. 15. No; the slope of the first line is -3

4, and the slope of the second line is -3. Since the slopes are not equal, the lines are not }. 17. -4 19. No; if two equations represent lines with the same slope and the same y-intercept, the equations must represent the same line. 21. slope of AB = slope of CD = -3

4, AB } CD; slope of BC = slope of AD = 1, BC } AD 23. slope of AB =slope of CD = 0, AB } CD; slope of BC = 3, slope of AD = 3

2, BC is not parallel to AD 25. A 27. Yes; the equations represent a horizontal line and a vertical line, and every horizontal line is perpendicular to every vertical line. 29a. y = -1

2x + 100 b. (100, 50) 31. y - 5 = 1

3(x - 4) 33. 7

Lesson 3-9 pp. 137–1381a. neither b. Euclidian c. spherical 3. All equiangular triangles in Euclidian geometry have three 60° angles.5a. False; Drawings may vary. Sample:

hsm11gmte_0305b_t11620.ai

b. False; Drawings may vary. Sample:


D

A B

�ADB � �ADC

C

c. False; Drawings may vary. Sample:


D

A B

�ADB � �ADC

C


9


7. Yes, vertical angles seem to be congruent in spherical geometry. Explanations may vary. Sample:


9. In spherical geometry, a line segment is an arc of a great circle. In Euclidian geometry, a line segment is a part of a line. 11. 2; Drawings may vary. Sample:

P

13. H 15. In both Euclidean and spherical geometries, triangles have three sides and three angles. In Euclidian geometry, triangles are flat and the sum of the measures of their angles is always 180. In spherical geometry, triangles have curved sides and the sum of the measures of their angles is always greater than 180.

Topic Review pp. 139–1431. remote interior angles 3. alternate interior angles 5. ∠5 and ∠8, lines a and b, transversal c; ∠2 and ∠6, a and e, transversal d 7. ∠1 and ∠7, lines c and d, transversal b 9. alt. int. angles 11. m∠1 = 75 because same-side int. angles are suppl; m∠2 = 105 because alt. int. angles are ≅. 13. 20 15. n } p; if corresp. angles are ≅, then the lines are }. 17. / } m; if same-side int. angles are suppl., then the lines are }. 19. } 21. 1st Street and 3rd Street are } because they are both#to Morris Avenue. Since 1st Street and 5th Street are both } to 3rd Street, 1st Street and 5th Street are } to each other. 23. x = 45, y = 45 25. 5527.

hsm11gmte_03cr_t11680.ai

m

Q

29.


a

b b

31. -1

33. slope: 2; y-intercept: -1

35. y = -12x + 12 37. y - 2 = 4(x - 4) or

y + 2 = 4(x - 3) 39. } 41. } 43. y + 3 = -6(x - 3) 45. spherical 47. neither 49. spherical 51. false 53. The equator is a great circle. Therefore, it is a line in spherical geometry.

TEKS cumulative practice pp. 144–1451. D 3. A 5. B 7. B 9. D 11. 42 13. 50 15. 317.

hsm11gmte_03cu_t11686.ai

NM

Q

P

19. x = 104, (x - 28) = 76, y = 35, (2y - 1) = 69 21. Converse: If a figure has at least two right angles, then it is a square (false); inverse: If a figure is not a square, then it does not have at least two right angles (false); contrapositive: If a figure does not have at least two right angles, then it is not a square (true). 23. It is given that / } m. ∠1 ≅ ∠2 by the Corresponding Angles Theorem. It is given that ∠2 ≅ ∠4. By the Transitive Property of Congruence, ∠1 ≅ ∠4. So n } p by the Converse of the Corresponding Angles Theorem.

Topic 4Lesson 4-1 pp. 150–1521. AB ≅ AB, BC ≅ BD, AC ≅ AD, ∠CAB ≅ ∠DAB, ∠C ≅ ∠D, ∠ABC ≅ ∠ABD 3. BK 5. ∠C 7. △KJB 9. PO ≅ SI, OL ≅ ID, LY ≅ DE, YP ≅ ES 11. 335 ft 13. 52 15. 45 ft 17. 128 19. Yes; two pairs of sides and two pairs of angles are marked as congruent; the third pair of sides are congruent by the Reflexive Property of Congruence, and the third pair of angles are congruent by the Third Angles Theorem. 21. ∠B ≅ ∠D (Given); it is also given that AB } DC , so ∠BAC ≅ ∠DCA because they are alternate interior angles. ∠BCA ≅ ∠DAC by the Third Angles Theorem. BC ≅ AD and AB ≅ DC (Given), and AC ≅ AC by the Reflexive Property of Congruence. So △ABC ≅ △CDA by the definition of congruent triangles. 23. m∠A = m∠D = 20 25. BC = EF = 8 27. C


O

2

2x

y4


10


29. x = 5 31. Two pairs of sides are given as congruent, and the third pair of sides are congruent by the Reflexive Property of Congruence. ∠A ≅ ∠C because perpendicular lines form right angles, and all right angles are congruent.∠ABD ≅ ∠CDB by the Alternate Interior Angles Theorem, and ∠ADB ≅ ∠CBD by the Third Angles Theorem. So ∠ABD ≅ ∠CDB by the definition of congruent angles. 33. Two pairs of congruent sides are given, and the third pair of sides are congruent because PQ bisects RT , so TS ≅ RS. PR } TQ, so ∠P ≅ ∠Q and ∠R ≅ ∠T because they are alternate interior angles; the third pair of angles are vertical angles, so they are congruent. Thus, △PRS ≅ △QTS by the definition of congruent triangles. 35. two; (3, 1) or (3, –7) 37. It is given that ∠A ≅ ∠D and ∠B ≅ ∠E, and so m∠A = m∠D and m∠B = m∠E by the definition of congruent angles. Thus m∠A + m∠B + m∠C = 180 and m∠D + m∠E + m∠F = 180 by the Triangle Angle-Sum Theorem. By the Substitution Property of Equality, we have m∠A + m∠B + m∠C = m∠D + m∠E + m∠F and m∠D + m∠E + m∠C = m∠D + m∠E + m∠F . This gives us m∠C = m∠E by the Subtraction Property of Equality, and so ∠C ≅ ∠E, by the definition of congruent angles. 39. 50

Lesson 4-2 pp. 155–1571a. Given b. Reflexive Property of Congruence c. △JKM d. △LMK 3. Three; answers will vary. Sample: If you know the lengths of two sides, you must know the angle measure in between them to form the third side that would make the triangles congruent. Knowing the measurements of two sides or two angles is not enough since the angle measures or side lengths could vary, respectively. 5. ZD ≅ ZD by the Reflexive Property of Congruence, and it is given that ZW ≅ ZS ≅ SD ≅ DW . So △WZD ≅ △SDZ by SSS. 7. You need to know RS ≅ WU; the diagram shows that ∠R ≅ ∠W and RT ≅ WV . ∠R is included between RT and RS , and ∠W is included between WV and WU . 9. △ANG ≅ △RWT by SAS. 11. △JEF ≅ △SFV (or △JEF ≅ △SFV ) by SSS. 13a. Answers may vary. Sample:

hsm11_gmte_0402_t11229

J

L

K

M

P

N

b. Answers may vary. Sample:

hsm11gmte_0402_t11228

J

L

K

M

P

N

15. FG } KL (Given), so ∠GFK ≅ ∠LKF because they are alternate interior angles. FG ≅ KL (Given) and KF ≅ KF (Reflexive Property of Congruence), so △FGK ≅ △KLF by SAS. 17. A 19. D

Lesson 4-3 pp. 161–1621. △PMO ≅ △NMO by ASA. 3. △VZY ≅ △VWY by AAS. 5. Given the perpendicular segments, ∠Q ≅ ∠S because perpendicular lines form right angles, and all right angles are congruent. It is given that T is the midpoint of PR, so PT ≅ RT by the definition of midpoint. ∠PTQ ≅ ∠RTS because vertical angles are congruent, so △PQT ≅ △RST by AAS. 7. Yes; use the Third Angles Theorem. Then you have three pairs of congruent angles and one pair of congruent sides, so you can use the ASA Postulate. 9. It is given that ∠N ≅ ∠P and MO ≅ QO. Also, ∠MON ≅ ∠QOP because vertical angles are congruent. So △MON ≅ △QOP by AAS. 11. Given the parallel lines, ∠BAC ≅ ∠DCA and ∠DAC ≅ ∠BCA because alternate interior angles are congruent. Also, AC ≅ AC by the Reflexive Property of Congruence. So △ABC ≅ △CDA by ASA. 13. △EAB ≅ △ECD, △EBC ≅ △EDA, △ABD ≅ △CDB, △ABC ≅ △CDA 15. It is given that ∠A ≅ ∠D and ∠B ≅ ∠E. By the Third Angles Theorem, ∠C ≅ ∠F . It is also given that AC ≅ DF . It follows that △ABC ≅ △DEF by ASA. 17. A 19. Converse: If you are too young to vote in the United States, then you are less than 18 years old; true.

Technology Lab 4-3 p. 1631. No, you cannot use AAA to prove △s ≅ because two △s can have 3 pairs of ≅ ∠s and not be the same size. 3. No; the circle intersects AB

> just once, so only

one △ is formed. If the ≅ ∠s are obtuse, then there could be an SSA congruency since a △ can have only one obtuse ∠.


11


Lesson 4-4 pp. 166–1671. △KLJ ≅ △OMN by SAS; KJ ≅ ON, ∠K ≅ ∠O, ∠J ≅ ∠N. 3. OM ≅ ER and ME ≅ RO (Given). OE ≅ OE by the Reflexive Property of Congruence. △MOE ≅ △REO by SSS, so ∠M ≅ ∠R because corresponding parts of congruent triangles are congruent. 5. A pair of congruent sides and a pair of congruent angles are given. Since PT ≅ PT (Reflexive Property of Congruence), △STP ≅ △OTP by SAS. ∠S ≅ ∠O because corresponding parts of congruent triangles are congruent. 7. KL bisects ∠PKQ, so ∠PKL ≅ ∠QKL. KL ≅ KL by the Reflexive Property of Congruence. △PKL ≅ △QKL by SAS, so ∠P ≅ ∠Q because corresponding parts of congruent triangles are congruent. 9. ∠PLK ≅ ∠QLK because perpendicular lines form right angles, and all right angles are congruent. From the angle bisector, ∠PKL ≅ ∠QKL. So with KL ≅ KL by the Reflexive Property of Congruence, △PLK ≅ △QLK by ASA and ∠P ≅ ∠Q because corresponding parts of congruent triangles are congruent. 11. The construction makes AC ≅ BE, AD ≅ BF, and CD ≅ EF. So △ACD ≅ △BEF by SSS. Thus ∠A ≅ ∠B because corresponding parts of congruent triangles are congruent. 13. It is given that JK } QP , so ∠K ≅ ∠Q and ∠J ≅ ∠P because they are alternate interior angles. With JK ≅ PQ (Given), △KJM ≅ △QPM by ASA and then JM ≅ PM because corresponding parts of congruent triangles are congruent. Thus M is the midpoint of JP by definition of midpoint. So KQ, which contains point M, bisects JP by the definition of segment bisector. 15. Using the given information and AE ≅ AE (Reflexive Property of Congruence), △AKE ≅ △ABE by SSS. Thus ∠KAS ≅ ∠BAS because corresponding parts of congruent triangles are congruent. In △KAS and △BAS, AK ≅ AB (Given) and AS ≅ AS (Reflexive Property of Congruence), so △KAS ≅ △BAS by SAS. Thus KS ≅ BS because corresponding parts of congruent triangles are congruent, and S is the midpoint of BK by the definition of midpoint. 17. 3.5 19. 38

Lesson 4-5 pp. 172–1731. VX ; Converse of Isosceles Triangle Theorem 3. VY ; Converse of Isosceles Triangle Theorem and Segment Addition Postulate 5. Sample answer: Yes, the builder is correct. AC ≅ AD is given, and by the definition of angle bisectors, ∠CAB ≅ ∠DAB. By the Reflexive Property of Congruence, AB ≅ AB. So by the SAS Postulate, △ACB ≅ △ADB. ∠ACB ≅ ∠ADB because corresponding parts of congruent triangles are congruent. 7. x = 38, y = 4 9. 52, 52 11. 64 13. 42 15. The triangle is an obtuse isosceles triangle; (x + 15) + (3x - 35) + 4x = 180, so the angle measures are 40, 40, and 100. 17a. isosceles triangle b. 900 ft; 1100 ft c. The tower is the perpendicular bisector of each base. 19. ( - 4, 0), (0, 0), (0, - 4), (4, 4), (4, 8), (8, 4)

21. ( - 1, 6), (2, 6), (2, 9), (5, 0), (5, 3), (8, 3) 23. B 25. Answers may vary. Sample: You need to know BC ≅ EF to prove the triangles congruent by SAS OR you need to know ∠A ≅ ∠D to prove the triangles congruent by ASA OR you need to know ∠C ≅ ∠F to prove the triangles congruent by AAS.

Lesson 4-6 pp. 176–1781a. If two angles are congruent and supplementary, they are right angles. b. definition of a right triangle c. Given d. Reflexive Property of Congruence e. HL 3a. △ABE and △DEB are right triangles. b. BE ≅ EB c. AB ≅ DE d. HL 5. From the given information about perpendicular segments, △PTM and △RMJ are right triangles. PM ≅ RJ (Given), and since M is the midpoint of TJ , TM ≅ JM. Thus △PTM ≅ △RMJ by HL. 7. x = -1, y = 3 9. Yes; two right triangles with congruent hypotenuses and a pair of congruent acute angles also have a pair of congruent right angles. So the two right triangles are congruent by AAS. 11.

hsm11gmte_0406_t11236

13.

hsm11gmte_0406_t11238

15. From the given information about the isosceles triangle, the right angles, and the midpoint, you can conclude that KG ≅ KE (Definition of isosceles triangle), △LKG and △DKE are right triangles (Definition of right triangles), and LK ≅ DK (Definition of midpoint). So △LKG ≅ △DKE by HL, and LG ≅ DE because corresponding parts of congruent triangles are congruent. 17. No, the triangles are not congruent. Explanations may vary. Sample: DF is the hypotenuse of △DEF , so it is the longest side of the triangle. Therefore, it is greater than 5 and greater than 13 because it is longer than either of the legs. So DF cannot be congruent to AC , which is the hypotenuse of △ABC and has length 13. 19. No; visualize “squeezing” at points A and C. That would change the distance AC but would not change any of the given information. Since you can change the length of AC , you cannot prove △ABC to be equilateral. 21. Answers may vary. Sample: △XRY ≅ △XRZ by HL, △YQX ≅ △YQZ by HL, △ZPX ≅ △ZPY by HL, and △XPS ≅ △YPS by SAS


12


Lesson 4-7 pp. 181–1831. ∠M 3. XY 5. A

B B

C R

P

hsm11gmte_0407_t11260

∠B is a common angle.7a. Given b. Reflexive Property of Congruence c. Given d. AAS e. Corresponding parts of congruent triangles are congruent. 9. QD ≅ UA and ∠QDA ≅ ∠UAD (Given), and DA ≅ DA (Reflexive Property of Congruence). So △QDA ≅ △UAD by SAS. 11. Answers may vary. Sample: AD ≅ ED and D is the midpoint of BF (Given). BD ≅ FD (Definition of midpoint), and ∠ADB ≅ ∠EDF (Vertical angles are congruent.). So △ADB ≅ △EDF by SAS, and ∠A ≅ ∠E because corresponding parts of congruent triangles are congruent. ∠ADC ≅ ∠EDG because vertical angles are congruent, so △ADC ≅ △EDG by ASA. 13. The overlapping triangles are △CAE and △CBD. It is given that AC ≅ BC and ∠A ≅ ∠B. Also, ∠C ≅ ∠C by the Reflexive Property of Congruence. So △CAE ≅ △CBD by ASA. 15. It is given that AC ≅ EC and CD ≅ CB, and ∠C ≅ ∠C by the Reflexive Property of Congruence. So △ACD ≅ △ECB by SAS, and ∠A ≅ ∠E because corresponding parts of congruent triangles are congruent.17. A

E

D

B

C

hsm11gmte_0407_t11266

a. AB ≅ DC, AD ≅ BC, AE ≅ EC, DE ≅ EB b. By showing that △ABC ≅ △CDA (by ASA), you can then prove that △AEB ≅ △CED and △AED ≅ △CEB (by ASA or AAS). The segments are congruent because corresponding parts of congruent triangles are congruent. 19. H 21a. J

L K

M

G

hsm11gmte_0407_t11264

b. It is given that L J } GK , so ∠J ≅ ∠K and ∠L ≅ ∠G because if lines are parallel, then alternate interior angles are congruent. Also, LM ≅ GM because M is the midpoint of LG (Given) and a midpoint

divides a segment into two congruent segments. So △LJM ≅ △GKM by AAS. c. If you use the congruent vertical angles JML and KMG, you can prove △LJM ≅ △GKM by ASA.

Topic Review pp. 184–1871. legs 3. corollary 5. ML 7. ST 9. 80 11. 5 13. 100 15. ∠D 17. not enough information 19. SAS 21. △TVY ≅ △YWX by AAS, so TV ≅ YW because corresp. parts of ≅ triangles are ≅. 23. △BEC ≅ △DEC by SSS, so ∠B ≅ ∠D because corresp. parts of _triangles are ≅. 25. x = 4, y = 65 27. x = 65, y = 90 29. LN # KM (Given), so △MLN and △KLN are rt. triangles. KL ≅ ML (Given) and LN ≅ LN (Refl. Prop. of ≅), so △KLN ≅ △MLN by HL. 31. △AEC ≅ △ABD by SAS, ASA, or AAS 33. △TAR ≅ △TSP by ASA

TEKS cumulative practice pp. 188–1891. B 3. C 5. A 7. B 9. 70 11. Yes, by the Converse of the Isosceles Triangle Theorem. 13. 1) AT _ SG and AT } SG (Given); 2) GT _ TG (Reflexive); 3) ∠GTA _ ∠TGS (Alternate Interior Angles Theorem); 4) △GAT _ △TSG (SAS) 15. They are congruent right triangles, because BD # AC. They share the leg BD, and have congruent hypotenuses AB and BC , so they are congruent by HL. 17. For LMNK, LM = KN = 4 and MN = LK = 2, while for PQRS, PQ = SR = 3 and PS = RQ = 2. The length of rectangle LMNK is 4 units, while the length of rectangle PQRS is 3 units, so the rectangles are not ≅. Answers may vary. Sample: To make the rectangles ≅, you could change the vertices of LMNK so that M has coordinates (1, 5) and N has coordinates (1, 3). OR you could change the vertices of PQRS so that R has coordinates (1, -4) and Q has coordinates (3, -4).

Topic 5Lesson 5-1 pp. 196–1981. (4, 2) 3. (3.5, 1) or (7

2, 1) 5. ( - 2.25, 2.1) 7. (0, - 1); found 25 of the x-distance A to B, which is 2, and added it to the x-coordinate of A; found 25 of the y-distance from A to B, which is also 2, and added it to the y-coordinate of A. 9a. 19.2 b. ( -3

2, 0) or

( - 1.5, 0) 11a. 5.4 b. (-1, 0.5) 13. 6.7 mi 15. 8.9 mi 17. 3.2 mi 19. 18 21. 9 23. 10 25. 8.2 27. 8.5 29. Everett, Charleston, Brookline, Fairfield, Davenport 31. Yes; Sample: CD divides the horizontal distance between A and B into two equal segments, AD and BC . CD also divides the vertical distance between A and B into two equal segments, DM and CM. CD is perpendicular to AD and to CD. Therefore, by SAS, △ADM ≅ △BCM.


13


33. Answers may vary. Sample: the formula

G(x1 + 35 (x2 - x1), y1 + 3

5 (y2 - y1))gives G(-7 + 3

5 (8), 7 + 35 (-4)) or G(-11

5 , 235 ) .

Number sense can help you add and subtract the

fractions given. The coordinates are G(- 115 ,23

5 ) .

35. Answers may vary. Sample: Since

ST = 2(x2 - x1)2 + (y2 - y1)2, substituting

(y1 + n) for y1 and (y2 + n) for y2 means

ST = 2(x2 - x1)2 + (( y2 + n) - ( y1 + n))2.

And ((y2 + n) - (y1 + n))2 is simplified as (y2 - y1)2.

So PQ = ST . 37. Terminal A: (3, 254 ) ; Terminal B:

( -2, - 52) ; Terminal C: (2, 92) 39. 6.5 units

41. D 43a. (-8, 9) b. 17.9 units

Technology Lab 5-2 pp. 199–2001. The length of a midsegment of a △ is half the length of the third side of the △, and its slope is the same as the slope of the third side. 3a. DE ≅ BF ≅ FC , EF ≅ AD ≅ DB, DF ≅ AE ≅ EC b. A triangle’s three midsegments divide the △ into four ≅ △s . 5a. The area of △ABC is 4 times the area of each small △. b. The perimeter of △ABC is 2 times the perimeter of each small △.

Lesson 5-2 pp. 203–2051. 114 ft 9 in.; because the red segments divide the legs into four congruent parts, the white segment divides each leg into two congruent parts. The white segment is a midsegment of the triangular face of the building, so its length is one half the length of the base. 3. 40; ST is a midsegment of △PQR, so by the Triangle Midsegment Theorem, ST } PR. Then m∠QPR = m∠QST because corresponding angles of

parallel lines are congruent. 5. Slope of HJ = 2 - 04 - 2 = 1

and slope of EF = 6 - 25 - 1 = 1. The slopes are equal,

so HJ } EF . HJ = 2(2 - 4)2 + (0 - 2)2 = 28 = 222

and EF = 2(1 - 5)2 + (2 - 6)2 = 232 = 422,

so HJ = 12EF . 7. 37 9. 50 11. 80 13. 4 15. G(4, 4);

H(0, 2); J(8, 0) 17. The midpoint of FG is

J( -6 + 42 , 4 + 8

2 ) = J(-1, 6). The midpoint

of FH is K( -6 + 22 ,

4 + (-2)2 ) = K(-2, 1).

The slope of JK = 1 - 6-2 - (-1) = 5 and the

slope of GH = -2 - 82 - 4 = 5, so JK } GH.

JK = 2(-2 - (-1))2 + (1 - 6)2 = 21 + 25 = 226

and GH = 2(2 - 4)2 + (-2 - 8)2 = 24 + 100 =2104 = 2226, so JK = 12GH. 19. 60 21. 55

23. 52 25. 26 27. 30 cm 29. 1.8 31. 80

Lesson 5-3 pp. 210–2121. Answers may vary. Sample answer: Conjecture: Quadrilateral ABCD is a square. 3.

Z

M

X

Y

△XMZ ≅ △YMZ. Conjecture: the two right triangles formed by the midpoint of a side of an equilateral triangle and two of its vertices are congruent.

5. MB is the perpendicular bisector of JK . 7. 9 9. No; it is not on the perpendicular bisector of the street connecting the two schools. 11. PM ≅ PM (Reflexive Property of Congruence); AP ≅ BP (Given); M is the midpoint of AB (Given); AM ≅ BM (Definition of a midpoint); △AMP ≅ △BMP (SSS Postulate); ∠AMP ≅ ∠BMP (Corresponding parts of congruent triangles are congruent.); m∠AMP + m∠BMP = 180 (Definition of a linear pair); m∠AMP = m∠BMP = 90 (∠AMP and ∠BMP are congruent.); PM is the perpendicular bisector of AB (Definition of perpendicular bisector). 13. 27; 27 15. 9 17. 5 19. 10 21. M is on the perpendicular bisectors of AB and BC , so MA = MB and MB = MC by the Perpendicular Bisector Theorem. So MA = MB = MC by the Transitive Property of Equality. ∠AME ≅ ∠BME ≅ ∠CME because if a line is perpendicular to a plane, then it is perpendicular to every line in the plane that contains the intersection of the plane and the line (and all right angles are congruent). EM ≅ EM by the Reflexive Property of Congruence. So △EAM ≅ △EBM ≅ △ECM by SAS. Therefore, EA ≅ EB ≅ EC , since corresponding parts of congruent triangles are congruent. So EA = EB = EC . 23. Line / through the midpoints of two sides of △ABC is equidistant from A, B, and C. This is because △1 ≅ △2 and △3 ≅ △4 by AAS. AD ≅ BE and BE ≅ CF because corresponding parts of congruent triangles are congruent. By the Transitive Property of Congruence, AD ≅ BE ≅ CF . By the definition of congruence, AD = BE = CF , so points A, B, and C are equidistant from line /.

4

3

2

1

A

C

B

D

E F

�

hsm11gmte_0502_t10689


14


25. By the definition of a bisector, AM ≅ BM, and ∠PMA and ∠PMB are right angles. Because all right angles are ≅, ∠PMA ≅ ∠PMB. By the Reflexive Property of Congruence, PM ≅ PM. So △APM ≅ △BPM by SAS. Corresponding parts of congruent triangles are congruent, so AP ≅ BP and AP = BP . 27. In right △SPQ and right △SRQ, ∠PQS ≅ ∠RQS (Given), ∠QPS ≅ ∠QRS (All right angles are congruent.), and QS ≅ QS (Reflexive Property of Congruence). So △SPQ ≅ △SRQ by AAS, and SP ≅ SR (or SP = SR) because corresponding parts of congruent triangles are congruent. 29. H

Activity Lab 5-4 p. 2131. The bisectors of the ∠s of a △ meet at a single point. 3a. The # bisectors meet at a single point, but that point is outside the △. b. Answers may vary. Ask your teacher to check your work.

Lesson 5-4 pp. 217–2181. ( - 2, - 3) 3. The circumcenter of the triangle formed by the lifeguard chair, the snack bar, and the volleyball court is equidistant from the vertices of the triangle. Place the recycling barrel at the intersection pt. of two of the triangle’s perpendicular bisectors.

S

V

P

L

hsm11gmte_0503_t13232

5. Z 7. an equilateral triangle 9. m∠DGE = 130, m∠DGF = 120, m∠EGF = 110 11. From the given information, ∠XBF ≅ ∠XBE and ∠XAE ≅ ∠XAD; also, ∠XFB ≅ ∠XEB ≅ ∠XEA ≅ ∠XDA because perpendicular lines form right angles, which are congruent. XB ≅ XB and XA ≅ XA by the Reflexive Property of Congruence, so △XFB ≅ △XEB and △XEA ≅ △XDA by AAS. Thus XF = XE and XE = XD because corresponding parts of congruent triangles are congruent, and XF = XD by the Transitive Property of Equality. Thus X is on the bisector of ∠BCA by the Converse of the Angle Bisector Theorem. Since n is the bisector of ∠BCA (Given), n contains X. 13. true 15. Method 1: MK ≅ MR and ∠MKV and ∠MRV are right angles. So M is equidistant from the sides of ∠KVR. It follows that M is on the angle bisector. Thus VM is the bisector of ∠KVR. Method 2: ∠MKV and ∠MRV are right angles, so △MKV ≅ △MRV are right triangles. MK ≅ MR (Given) and MV ≅ MV (Refl. Prop. of ≅), so △MKV ≅ △MRV (HL). Then ∠KVM ≅ ∠RVM since they are corresponding parts of congruent triangles, so VM bisects ∠KVR.

Technology Lab 5-5 p. 2191. They appear to be concurrent; yes. 3. acute △ : inside, inside, inside, inside; rt. △ : on, inside, on, inside; obtuse △ : outside, inside, outside, inside

Lesson 5-5 pp. 222–2241. 125 3. AE 5. DE 7.

hsm11gmte_0504_t10694

M

L N

9. Ask your teacher to check your work. The folds should show the perpendicular bisectors of the sides to identify the midpoint of each side. They should also go through each vertex and the midpoint of the opposite side. 11. C 13. TY = 18, TW = 27 15. VY = 6, YX = 3 17. Altitude; it extends from a vertex of △ABC and is perpendicular to the opposite side. 19. (6, 4) 21. A is the intersection of the altitudes, so it is the orthocenter; B is the intersection of the angle bisectors, so it is the incenter; C is the intersection of the medians, so it is the centroid; D is the intersection of the perpendicular bisectors of the sides, so it is the circumcenter. 23. incenter 25. B27.

Lesson 5-6 pp. 227–2281. Assume temporarily that it is not raining outside. 3. Assume temporarily that XY and AB are not congruent. 5. I and II 7a. right angle b. right angles c. 90 d. 180 e. 90 f. 90 g. 0 h. more than one right angle i. at most one right angle 9. Assume temporarily that / } p. Then ∠1 = ∠2 because if lines are parallel, then corresponding angles are congruent. But this contradicts the given statement that ∠1 R ∠2. Therefore the temporary assumption is false, and we can conclude that / is not parallel to p.

hsm11gmte_0504_t10699

A

P

C

BCircumcenter: outside

hsm11gmte_0504_t10701

A

P

C

BIncenter: inside

hsm11gmte_0504_t10707

A

P

C

BCentroid: inside


15


11. B

A

D

X C

hsm11gmte_0505_t10718

Assume temporarily that AX ≅ XC . Since the triangle is scalene, we may suppose that BA 6 BC . Find D on BA

T so that BD = BC . We have

∠DBX ≅ ∠CBX (Given) and BX ≅ BX (Reflexive Property of Congruence). So △DBX ≅ △CBX by SAS. Then DX ≅ CX (Corresponding parts of congruent triangles are congruent.), so DX ≅ AX (Transitive Property of Congruence). Let m∠BDX = q. Then m∠BCX = q (Corresponding parts of congruent triangles are congruent.) and m∠XAD = q (Isosceles Triangle Theorem). Therefore, m∠BAX = 180 - q. Now the sum of the angle measures in △ABC is 180 - q + 72 + q = 252. This contradicts the Triangle Angle-Sum Theorem. Therefore, the temporary assumption that AX ≅ XC is incorrect. So AX R XC .

13. Assume temporarily that at least one base angle is a right angle. Then both base angles must be right angles, by the Isosceles Triangle Theorem. But this contradicts the fact that a triangle is formed, because in a plane two lines perpendicular to the same line are parallel. Therefore the temporary assumption is false that at least one base angle is a right angle, and we can conclude that neither base angle is a right angle. 15. Assume that lines / and m intersect at point P. Given the angle measures in the diagram, the measure of ∠PQR is 180 - x (measures of angles in a linear pair are supplementary). Then the sum of measures in △RQP is 180 - x + x + m∠RPQ, which is greater than 180 since m∠RPQ must be greater than zero. This contradicts the Triangle Angle-Sum Theorem, so therefore lines / and m don’t intersect and they are parallel. 17. D 19. All acute triangles; the centroid and incenter are always inside. The circumcenter and orthocenter are on the triangle if it is a rt. triangle; they are outside the triangle if it is obtuse.

Lesson 5-7 pp. 232–2341. XZ + ZY 7 XY

X

XZ

YZ

Y

3. Sample answer: You can verify that the Triangle Inequality Theorem works for a particular triangle, while the proof of the Triangle Inequality Theorem works for all given triangles. 5. This is true by the

Corollary to the Triangle Exterior Angle Theorem. 7. (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (3, 7), (4, 3), (4, 4), (4, 5), (4, 6), (4, 7), (4, 8) 9. The dashed red line and the courtyard walkway determine three sides of a triangle, so by the Triangle Inequality Theorem, the path that follows the dashed red line is longer than the courtyard walkway. 11. No; 2 + 3 is not greater than 6. 13. Yes; 2 + 9 7 10, 9 + 10 7 2, and 2 + 10 7 9. 15. 5 m 6 x 6 41 m 17. Place the computer at the corner that forms a right angle; place the bookshelf along the wall opposite the right angle. In a right triangle the right angle is the largest angle, and the longest side of a triangle is opposite the largest angle. 19. ∠G, ∠H, ∠J 21. TU, UV, TV 23. YZ, XZ, XY 25. Answers may vary. Sample: The sum of the interior angle measures of a triangle is 180, so m∠T + m∠P + m∠A = 180. Since m∠T = 90, m∠P + m∠A = 90, and so m∠T 7 m∠A (Comparison Prop. of Inequality). Therefore PA 7 PT by Theorem 5-11. 27. AB 29. 5

Lesson 5-8 pp. 238–2391. AB 6 AD 3a. Converse of the Isosceles Triangle Theorem b. Given c. Def. of midpt. d. BC = CD e. Given f. Hinge Theorem 5. 6 6 x 6 24 7. △ABE ≅ △CBD (Given) so △ABE and △CBD are isosc. with AB = EB = DB = CB. Since m∠EBD 7 m∠ABE (Given), ED 7 AE by the Hinge Thm. 9. In △AOB and △AOC , AO = AO = 7, OB = OC = 15, AB = 182, and AC = 168. Since AB 7 AC , m∠AOB 7 m∠AOC by the Converse of the Hinge Thm. 11. J 13. D

Topic Review pp. 240–2431. median 3. incenter 5. 7.6 7. (0, 0) 9. (6, -2) 11. (-6, -7) 13. 11 15. Let point S be second base and point T be third base. Find the midpt. M of ST and then through M construct the line / # to ST . Points of the baseball field that are on line / are equidistant from second and third base. 17. 40 19. 11 21. 33 23. (3, 2) 25. (5, 1) 27. 40 29. AB is an altitude; it is a segment from a vertex that is # to the opposite side. 31. QZ = 8, QM = 12 33. (2, -3) 35. Assume temporarily that the third line intersects neither of the first two. Then it is }to both of them. Since the first two lines are } to the same line, they are } to each other. This contradicts the given information. Therefore the temporary assumption is false, and the third line must intersect at least one of the two others. 37. Assume temporarily that an equilateral triangle has an obtuse angle. Since all the angles are ≅ in an equilateral triangle, then all three angles must be obtuse. But we showed in Ex. 36 that a triangle can have at most one obtuse angle. Therefore the temporary assumption is false, and an equilateral triangle cannot have an obtuse angle. 39. RS, ST, RT 41. Yes; 10 + 12 7 20, 10 + 20 7 12, and 12 + 20 7 10. 43. 6 45. 6


16


TEKS cumulative practice pp. 244–2451. A 3. C 5. A 7. C 9. 1.5 11. 4 13a. / and m do not have any points in common. b. Answers may vary. Sample: Use the Law of Detachment to conclude that / and m do not intersect. Then use the Law of Detachment again to conclude that / and m do not have any points in common. 15. Assume temporarily that △ABC is a rt. triangle, as well as obtuse (Given). Then one angle measure is 90 (Def. of rt. triangle), and one angle measure is 7 90 (Def. of obtuse triangle). The sum of these two measures is 7 180. This contradicts the Triangle Angle-Sum Thm., so the temporary assumption is false. Hence, △ABC is not a rt. triangle.

Topic 6Technology Lab 6-1 p. 2481. Answers may vary. Sample: ∠s form a circle, which has 360 degrees. 3a. Ask your teacher to check your work. b. regular △, regular hexagon c. 3 sides: 120; 4 sides: 90; 5 sides: 72; 6 sides: 60; 8 sides: 45 d. Answers may vary. Sample: To tile a plane, the measure of ext. ∠ must be a factor of 360.

Lesson 6-1 pp. 253–2541. 135 3. 108 5. 20, 80, 80; 80, 50, 50 7. w = 72, x = 59, y = 49, z = 121 9a. 180n ; (n - 2) # 180 b. 180n - [(n - 2) # 180] = 360 ; Polygon Ext. Angle-Sum Theorem 11. Yes; answers may vary. Sample: The sum of the measures of interior angles = (n - 2) # 180. 13. 45, 45, 90 15a. You should select geometry software, because they can use it to make regular polygons and measure angles. b. Conjecture 1: The sum of the measures of the exterior angles of a regular polygon, one at each vertex, is 360. Conjecture 2: The measures of the exterior angles of a regular polygon are equal. 17. 225 19. 81

Lesson 6-2 pp. 259–2601. x = 15, y = 45 3. 3 5. 9 7. 2.25 9. 4.5 11a. definition of a parallelogram b. If lines are parallel, then alternate interior angles are congruent. c. Opposite sides of a parallelogram are congruent. d. △ABE ≅ △CDE e. Corresponding parts of congruent triangles are congruent. f. AC and BD bisect each other at E. 13. x = 5, y = 7 15. Answers may vary. Sample: Suppose RSTW and XYTZ are parallelograms. It follows that ∠R ≅ ∠T and ∠T ≅ ∠X , since opp. angles of a ▱ are ≅. By the Transitive Prop. of ≅, ∠R ≅ ∠X . 17. m∠1 = 38, m∠2 = 32, m∠3 = 110 19. m∠1 = 95, m∠2 = 37, m∠3 = 37 21. Since ▱ABCD is a parallelogram, it follows from the def. of a ▱ that AB } CD and BC } DA. Since same-side int. angles are suppl. ∠A is suppl. to ∠B and ∠A is suppl. to ∠D.

23. Answers may vary. Sample: It is given that AB } CD and CD } EF . By constructing BG } AC and DH } CE, it follows from the Def. of ▱ that ABGC and CDHE are parallelograms. AC ≅ BG and CE ≅ DH since opp. sides of a ▱ are ≅. It is given that AC ≅ CE, so by the Trans. Prop. of ≅, BG ≅ DH. Also, BG } DH because if two lines are } to the same line, then they are } to each other. Also, ∠3 ≅ ∠6 and ∠GBD ≅ ∠HDF , and if lines are }, corresp. angles are ≅. So, by AAS, △GBD ≅ △HDF . Finally, BD ≅ DF since corresp. parts of ≅ triangles are ≅. 25. D 27. C

Lesson 6-3 pp. 267–2681. Answers may vary. Sample: 1. AB ≅ CD, DE ≅ FC , EA ≅ BF (Given) 2. DE = FC, EA = BF (Def. of ≅) 3. DE + EA = FC + BF (Add. Prop. of = ) 4. DA = BC (Segment Addition Postulate) 5. DA ≅ BC (Def. of ≅) 6. ABCD is a ▱. (If both pairs of opposite. sides of a quad. are ≅, then the quad. is a ▱.) 3. Answers may vary. Sample: 1. M is the midpoint of HK and JL. (Given) 2. HM ≅ KM, JM ≅ LM (Def. of midpoint) 3. HK and JL bisect each other. (Def. of bisect) 4. HJKL is a ▱. (If the diagonals of a quad. bisect each other, then the quad. is a ▱.) 5. x = 21, y = 39 7. 5 9. x = 3, y = 11 11. Answers may vary. Sample:

A B

E F

GHCD13. No; only one diagonal is bisected. 15. The connecting pieces AD and BC are congruent, and the distances AB and CD between where the two pieces attach are equal. The side lengths of ABCD do not change as the tackle box opens and closes. Since both pairs of opposite sides are congruent, ABCD is always a parallelogram. By the definition of a parallelogram, AB is parallel to CD, so the shelves are always parallel to each other. 17. Answers may vary. Sample: 1. ∠A is suppl. to ∠B. (Given) 2. BC } AD (If same-side int. angles are supplementary, then lines are }.) 3. ∠A is suppl. to ∠D. (Given) 4. AB } DC (If same-side int. angles are supplementary, then lines are }.) 5. ABCD is a ▱. (Def. of ▱) 19. F 21. Answers may vary. Sample: 1. △NRJ ≅ △CPT (Given) 2. NJ ≅ CT (Corresp. parts of ≅ △s are ≅.) 3. JN } CT (Given) 4. JNTC is a ▱. (If one pair of opp. sides of a quad. is both ≅ and }, then the quad. is a ▱.)

Lesson 6-4 pp. 272–2741. m∠1 = 55, m∠2 = 35, m∠3 = 55, m∠4 = 90 3. m∠1 = 90, m∠2 = 55, m∠3 = 90 5. x = 3; LN = MP = 7 7. x = 9; LN = MP = 67 9. x = 2.5; LN = MP = 12.5 11. Rectangle; the parallelogram has 4 right angles and does not have 4 congruent sides. 13a. given b. def. of rectangle c. Refl. Prop. of ≅ d. def. of rectangle e. AB ≅ DC f. △ABC ≅ △DCB g. All rt. angles are ≅. h. Corresp.


17


parts of ≅ triangles are ≅. 15. m∠H = m∠J = 58, m∠K = m∠G = 122, HK = KJ = JG = GH = 6 17. AC = BD = 16 19. AC = BD = 1 21. x = 3, y = 5; all sides are 15. 23. Answers may vary. Sample: 1. PLAN is a rectangle. (Given) 2. PA ≅ LN (Diagonals of a rectangle are ≅.) 3. TP = TL = TN = TA (Diagonals of a ▱ bisect each other.) 4. ∠LTP ≅ ∠ATN (Vert. angles are ≅.) 5. △LTP ≅ △ATN (SAS) 25. 2 27. B 29. A

Lesson 6-5 pp. 279–2801. 12 3. 10 5. 7 7a. AE ≅ CE and DE ≅ BE; No. The diagonals bisect each other, so by Theorem 6-11, the quadrilateral is a parallelogram. However, you do not know if the diagonals are congruent, so you do not have enough information to show that the quadrilateral is a rectangle. b. AD ≅ BC , AB ≅ DC , and m∠DAB = 90; Yes. By Theorem 6-8, since opposite sides are congruent, the quadrilateral is a parallelogram. Because m∠DAB = 90, the quadrilateral is a rectangle. c. AB } CD, AD } BC , and AC ≅ DB; Yes. The quadrilateral is a rectangle by Theorem 6-18. d. AE ≅ CE ≅ DE ≅ BE; Yes. The diagonals bisect each other, so by Theorem 6-11, the quadrilateral is a parallelogram. The diagonals are congruent, so by Theorem 6-18, the quadrilateral is a rectangle. 9. Square; since it is a rhombus and a rectangle, it must be a square. 11. Rhombus; diagonals are perpendicular. 13. Rectangle, square; answers may vary. Sample:


a

15. Rectangle, rhombus, square; answers may vary. Sample:

a b

a b

b

b b

b


17. Answers may vary. Sample: 1. AC ≅ BD, ▱ABCD (Given) 2. AD ≅ BC (Opposite sides of a ▱ are ≅.) 3. DC ≅ DC (Refl. Prop. of ≅) 4. △ADC ≅ △BCD (SSS) 5. ∠ ADC ≅ ∠BCD (Corresp. parts of ≅ triangles are ≅.) 6. m∠ ADC + m∠ BCD = 180 (Same-side int. angles are suppl.) 7. ∠ ADC and ∠ BCD are rt. angles. (Def. of rt. angle) 8. ABCD is a rectangle. (Def. of rectangle) 19. Construct the midpt. of ≅ diagonals. Copy the diagonals so the two midpts. coincide. Connect the endpoints of the diagonals. 21. No; if the diagonals of a ▱ are ≅, then it would have to be a rectangle and have rt. angles. 23. No; in a ▱, consecutive angles must be suppl., so all angles must be rt. This would make it a rectangle. 25. B 27. B

Lesson 6-6 pp. 286–2871. Answers may vary. Sample: 1. Draw AE } DC . (Construction) 2. AECD is a parallelogram. (Def. of parallelogram) 3. AE ≅ DC (Opp. sides of a parallelogram are congruent.) 4. ∠1 ≅ ∠C (If lines are parallel, corresp. angles are congruent.) 5. ∠B ≅ ∠1 (Isosc. Triangle Thm.) 6. ∠B ≅ ∠C (Transitive Prop. of Congruence) 7. ∠D and ∠C are suppl. (Same-Side Int. Ang. Post.) 8. ∠BAD and ∠B are suppl. (Same-Side Int. Ang. Post.) 9. ∠BAD ≅ ∠D (Angles suppl. to congruent angles are congruent.) 3. 28 5. No; explanations may vary. Sample: Assume KM bisects both angles. Then ∠MKL ≅ ∠MKN ≅ ∠KML ≅ ∠KMN. Both pairs of sides of KLMN would be parallel, and KLMN would be a parallelogram. It is impossible for an isosceles trapezoid to also be a parallelogram, so KM cannot bisect ∠LMN and ∠LKN. 7. m∠A = m∠B = 60, m∠CDA = 120 9. Answers may vary. Sample: Given: Trapezoid ABCD with BC } AD, ∠A ≅ ∠D Prove: ABCD is an isosc. trapezoid. There are two cases to consider, whether ∠A and ∠D are acute or obtuse. ∠A and ∠D cannot be right angles; if they were, then ABCD would be a rectangle, not a trapezoid. Case I: ∠A and ∠D are acute. 1.

<AB

> is not } to

<DC

>.

(Def. of trapezoid) 2. Extend <AB

> and

<DC

> to meet

at point T. (Construction) 3. ∠A ≅ ∠D (Given) 4. AT ≅ DT (Converse of Isosc. Thm.) 5. ∠TBC ≅ ∠A and ∠TCB ≅ ∠D. (If } lines, then corresp. angles are ≅.) 6. ∠TBC ≅ ∠TCB (Transitive Prop. of ≅) 7. TB ≅ TC (Converse of Isosc. △ Thm.) 8. TB + AB = TA, TC + DC = TD (Seg. Add. Post.) 9. AB = TA - TB, DC = TD - TC (Subtr. Prop. of Eq.) 10. AB = DC (Subst. Prop. of Eq.) 11. ABCD is an isosc. trapezoid. (Def. of isosc. trapezoid) Case II: ∠A and ∠D are obtuse. 1. ∠B is supplementary to ∠A and ∠C is supplementary to ∠D (Same-Side Int. Angles Post.) 2. ∠B ≅ ∠C (Congruent Suppl. Thm.) 3. ∠B and ∠C are acute (suppl. to obtuse angles) 4. ABCD is an isosc. trapezoid. (Converse of Thm. 6–20, Case I)


18


11. ▱, rhombus, rectangle, square; answers may vary. Sample:

hsm11gmte_0606_t10939.aihsm11gmte_0606_t10940.ai

13. Kite, isosc. trapezoid, rhombus, square; answers may vary. Sample:


15. Rectangle, isosc. trapezoid, kite, square; answers may vary. Sample:



17. Given: Trapezoid ABCD with BC } AD, BD ≅ AC Prove: AB ≅ DC 1. Draw BP # AD and CQ # AD. (Construction) 2. BP ≅ CQ (Opp. sides of a rectangle are ≅.) 3. BD ≅ AC (Given) 4. △BPD ≅ △CQA (HL) 5. ∠BDP ≅ ∠CAQ (Corresp. parts of ≅ triangles are ≅.) 6. AD ≅ DA (Refl. Prop. of ≅) 7. △BAD ≅ △CDA (SAS) 8. AB ≅ DC (Corresp. parts of ≅ triangles are ≅.) 19. 1. kite ABCD with AB ≅ AD, BC ≅ CD (Given) 2. Draw AC . (Construction) 3. △ABC ≅ △ADC (SSS) 4. ∠B ≅ ∠D (Corresp. parts of ≅ triangles are ≅.) 21. False; a trapezoid has exactly one pair of parallel sides. 23. True; a square is a parallelogram that has four congruent sides and four rt. angles. 25. False; a rhombus without 4 rt. angles is not a square. 27. Answers may vary. Sample: 1. BI # TP (Perp. Transversal Thm.) 2. Draw RI and AI (Construction) 3. ∠RBI and ∠ABI are right angles (def. of perp.) 4. ∠RBI ≅ ∠ABI (all rt. angles are congruent) 5. RB ≅ AB (def. of bisector) 6. IB ≅ IB (Ref. Prop. of Congruence) 7. △RBI ≅ △ABI (SAS) 8. ∠IRB ≅ ∠IAB and RI ≅ AI (CPCTC) 9. ∠TRB ≅ ∠PAB (Thm. 6–20) 10. ∠TRI ≅ ∠PAI (Angle Addition Postulate, def. of congruent angles) 11. △TRI ≅ △PAI (SAS) 12. TI ≅ PI (CPCTC) 13. BI is the perp. bis. of TP (Def. of perp. bisector) 29. half the sum of the bases; Trapezoid Midsegment Thm. 31. B 33. 1. DE ≅ FG and EF ≅ GD. (Given) 2. DF ≅ FD (Reflection Property of Congruence) 3. △EDF ≅ △GFD (SSS) 4. ∠E ≅ ∠G (Corresponding parts of congruent triangles are congruent.)

Topic Review pp. 288–2911. rhombus 3. consecutive angles 5. 120, 60 7. 108, 72 9. 159 11. m∠1 = 38, m∠2 = 43, m∠3 = 99

13. m∠1 = 37, m∠2 = 26, m∠3 = 26 15. x = 3, y = 7 17. no 19. x = 29, y = 28 21. m∠1 = 58, m∠2 = 32, m∠3 = 90 23. sometimes 25. sometimes 27. sometimes 29. No; two sides are } in all parallelograms. 31. x = 18; a diagonal bisects a pair of angles in a rhombus. 33. m∠1 = 135, m∠2 = 135, m∠3 = 45 35. m∠1 = 90, m∠2 = 25 37. 2

TEKS cumulative practice pp. 292–2931. D 3. B 5. A 7. A 9. A 11. 52 13. 68 15. The lengths must satisfy the △ Inequality Thm.: n + 1 + 2n 7 5n - 4 (so 3n + 1 7 5n - 4, 5 7 2n, n 6 2.5), n + 1 + 5n - 4 7 2n (so 6n - 3 7 2n, 4n 7 3, n 7 0.75), and 2n + 5n - 4 7 n + 1 (so 7n - 4 7 n + 1, 6n 7 5, n 7 5

6). So 56 6 n 6 2.5. 17. No; both diagonals must bisect each other. 19. Yes; if both pairs of opp. sides are ≅, then the quad. is a ▱. 21. Using C(5, 7) and D(10, -5), the coordinates of the midpoint are

(5 + 102 ,

7 + (-5)2 ) = (7.5, 1). The length CD is 125 + 144 = 1169 = 13.

Topic 7Lesson 7-1 pp. 299–3011. Scalene; side lengths are 4, 5, and 117. 3. Isosceles; side lengths are 212, 134, and 134.

5. midpoint of PR = (8 + (-1)2 ,

5 + (-2)2 ) = (7

2 , 32)

midpoint of QS = (5 + 22 , -4 + 7

2 ) = (72 , 32)

Theorem 6-11 states if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. The midpoints of PR and QS are the same point, and the diagonals bisect each other. Therefore, PQRS is a parallelogram. By the definition of a parallelogram, PQ } SR and QR } SP . 7. None; explanations may vary. Sample: It has no right angles or consecutive sides ≅. 9. Rectangle; explanations may vary. Sample: Consecutive sides are # and not ≅. 11. PS = QR = 5 and PQ = RS = 134. The slopes of PS and QR equal 34, and the slopes of PQ and RS equal -5

3. Since consecutive sides are not perpendicular, PQRS is not a rectangle.13.

hsm11gmte_0607_t10949

O

y

x

2

2

4 6 8

4

6

8 R

I

T

scalene; not rt. △


19


15.

hsm11gmte_0607_t10987

x

�4 4

B

A

C

O

y

4

8

�8

scalene; not a rt. △17. Yes; PR = SW = 4, PQ = ST = 110, and QR = TW = 312, so △PQR ≅ △STW by SSS.19.

hsm11gmte_0607_t10952

x�2�4 2 4P

Q R

SO

y

4

isosc. trapezoid21.

hsm11gmte_0607_t10954

GH

IFO 2 4 6 8

4

6

2

x

y

kite23.

hsm11gmte_0607_t10956

AB

CD

O 2 4 6 8

4

6

2

x

y

rhombus25. y

xO

6

2

4 8

4

�2

hsm11gmte_0607_t10958

K

J

M

L

quadrilateral

27. slope of DE = 2; slope of AB = 2; DE = 1

215; AB = 15. So DE } AB and DE = 12 AB.

29. Answers may vary. Sample: Chairs are not at vertices of a ▱. Move the right-most chair down by 1 grid unit. 31a. rectangle b. rectangle c. Yes; corresp. sides are ≅ and corresp. angles are ≅ (rt. angles),

so ABCD ≅ EFGH. 33. ( -1, 6 23) , (1, 8 13) , (3, 10),

(5, 1123) , (7, 13 13) 35. ( -3 + a (12

n ), 5 + a (10n ) ) for

a = 1, 2, 3, c , n - 1 37. J 39. No; the slope of AC is 0, the slope of AB is -3

2, and the slope of BC is 1. No slopes have a product of -1, so the sides are not perpendicular.

Lesson 7-2 pp. 304–3061. O(0, 0), S(0, h), T(b, h), W(b, 0) 3. S(-b

2, -b2) ,

T (-b2, b2) , W (b

2, b2) , Z (b2, -b

2) 5. W(r, 0), T(0, t),

S( - r, 0), Z(0, - t) 7. Yes, ABCD is a rhombus. The slope of AC is - 1, and the slope of BD is 1, so the diagonals are #. 9. P( - r, s) 11a. Answers may vary. Sample:

y

x

hsm11gmte_0608_t10965

A(0, 0)

C(b, 2c)

B(2b, 0)

b. Answers may vary. Sample: y

xO

hsm11gmte_0608_t10966

A(�b, 0)

C(0, 2c)

B(b, 0)

c. 2b2 + 4c2, 2b2 + 4c2 d. 2b2 + 4c2, 2b2 + 4c2 e. The results are the same. 13. The diagonals of a rhombus are #. 15. Answers may vary. Sample: Place vertices at A( - a, - b), B( - a, b), C(a, b), and D(a, - b). Show that each diagonal has (0, 0) as its midpt. 17. parallelogram 19. kite 21. Answers may vary. Sample:

y

x

hsm11gmte_0609_t10967

(�b, 0)

(0, �b)

(b, 0)

(0, b)

O

23. Answers may vary. Sample: B, D, H, F 25. C 27. C

Technology Lab 7-3 p. 3071a. The quad. is a ▱. b. yes c. no 3. Ratios of lengths of sides = ratios of lengths of perimeters = 1 : 2; ratio of areas = 1 : 4. The sides are }. 5a. ▱; opp. sides are } and ≅. b. Rhombus; joining midpts. produces 4 ≅ rt. △s , so hypotenuses are ≅.


20


c. Rectangle; diagonals of a rhombus are #, so the sides of new figure are #. d. Square; sides are ≅ and #. e. ▱; opp. sides are }. f. Rhombus; all sides are ≅. g. Rectangle; sides are #.

Lesson 7-3 pp. 309–3111. Yes; use Distance Formula. 3. Yes; use Slope Formula and property of # lines. 5. Yes; show two points on AB is equidistant from the sides of ∠CAD. 7. Yes; use Slope Formula and property of # lines. 9. Yes; answers may vary. Sample: Show four sides have the same length or show diagonals #. 11. No; you need angle measures. 13. The Distance Formula shows that EG and FH are the same length.

15. M = ( -2a + 02 , 0 + 2b

2 ) = (-a, b);

N = (2c + 02 , 0 + 2b

2 ) = (c, b)

PN = 2(c - (-2a))2 + (b - 0)2 = 2c2 + 4ac + 4a2 + b2 RM = 2(2c - (-a))2 + (0 - b)2 = 24c2 + 4ac + a2 + b2 Since PN ≅ RM, PN = RM so 2c2 + 4ac + 4a2 + b2 = 24c2 + 4ac + a2 + b2. By squaring each side and combining like terms, 3a2 = 3c2. By dividing each side by 3 and taking square roots, a = {c, but by definition both a and c are both positive, so a = c. Since R = (2a, 0), PQ = 2(-2a - 0)2 + (0 - 2b)2 = 24a2 + 4b2 and RQ = 2(2a - 0)2 + (0 - 2b)2 = 24a2 + 4b2 so PQ ≅ RQ and △PQR is isosceles.

17. Answers may vary. Sample:y

xD(0, 0) G(a, 0)

F(b, c)E(b – a, c)

Given: DEFG is a rhombus. Prove: GE # DF . Since DEFG is a rhombus, FG = DG = a. So FG = 2(a - b)2 + (0 - c)2

= 2a2 - 2ab + b2 + c2 = a. By squaring each side and solving for c2, c2 = 2ab - b2. The slope of DF = c - 0

b - 0 = cb and

the slope of GE = 0 - ca - (b - a) = - c

2a - b.

The product of the slopes of the diagonals is

- c2a - b

# cb = - c2

2ab - b2 = -c2

c2 = -1, so GE # DF .

19. Answers may vary. Sample:y

x

hsm11gmte_0609_t10971

T

R(2b, 2c) A(2d, 2c)

P (2a, 0)

LM

K

N

Given: Trapezoid TRAP; M, L, N, and K are midpoints of its sides. Prove: MLNK is a ▱. By the Midpoint Formula, the coordinates of the midpoints are M(b, c), L(b + d, 2c), N(a + d, c), and K(a, 0). By the Slope Formula, the slope of ML = c

d , the slope of LN = cb - a, the slope of

NK = cd , and the slope of KM = c

b - a. Since slopes

are = , ML } NK and LN } KM. Therefore, MLNK is a ▱ by def. of ▱.

21. Answers may vary. Sample:

y

xO

hsm11gmte_0609_t10968

M(0, 2b)

P (2a, 0)

N(2a, 2b)W

U

T V

Given: MNPO is a rectangle. T, W, V, U are midpoints of its sides. Prove: TWVU is a rhombus. By the Midpoint Formula, the coordinates of the midpoints are T(0, b), W(a, 2b), V(2a, b), and U(a, 0). By the Slope Formula, slope of TW = 2b - b

a - 0 = ba , slope of

WV = 2b - ba - 2a = - ba , slope of VU = b - 0

2a - a = ba , slope

of UT = b - 00 - a = - ba . So TW } VU and WV } UT .

Therefore, TWVU is a ▱. By the Slope Formula, slope of TV = 0, and slope of WU is undefined. TV # WU because horiz. and vert, lines are #. Since the diagonals of ▱TWVU are #, it must be a rhombus.

23a. L(3q, 3r), M(3p + 3q, 3r), N(3p, 0) b. equation of

<AM

>: y = r

p + q x

equation of <BN

>: y = 2r

2q - p (x - 3p)

equation of <CL>: y = r

q - 2p (x - 6p)

c. P (2p + 2q, 2r) d. The coordinates of P satisfy the equation for

<CL>:

y = rq - 2p (x - 6p)

2r = rq - 2p (2p + 2q - 6p)

2r = rq - 2p (2q - 4p)

2r = 2r


21


e. AM = 2(3p + 3q - 0)2 + (3r - 0)2

= 2(3p + 3q)2 + (3r)2;

23 AM = 2

32(3p + 3q)2 + (3r)2

= 549[(3p + 3q)2 + (3r)2]

= 549[(3p + 3q)2] + [4

9 (3r)2]

= 549[(9p2 + 18pq + 9q)2] + [4

9 (9r2)]

= 2(4p2 + 8pq + 4q2) + (4r2)

= 2(2p + 2q)2 + (2r)2;

AP = 2(2p + 2q - 0)2 + (2r - 0)2

= 2(2p + 2q)2 + (2r)2

So AP = 23 AM. You can find the other two distances

similarly. 25a. (a, 0) b. D(-b, 0), B(-b, a) c. By the Slope Formula, the slope of /1 is ba and the slope of /2 is a

-b.

So the product of the slopes is (ba ) ( a

-b) = ab-ab = -1.

27. 50 29. 15

Topic Review pp. 312–3131. coordinate proof 3. scalene 5. parallelogram 7. rhombus 9. F (0, 2b), L(a, 0), P(0, -2b), S(-a, 0) 11. Answers may vary. Sample:

y

x


MN

LKE(0, 2a)

F(2b, 2c)

G(0, 0)

D(�2b, 2c)

Given: Kite DEFG; K, L, M, N are midpoints of sides Prove: KLMN is a rectangle. By the Midpoint Formula, coordinates of midpoints are K(-b, a + c), L(b, a + c), M(b, c), and N(-b, c). By the Slope Formula, slope of KL = slope of NM = 0, and slope of KN and slope of LM are undefined. KL } NM and KN } LM so KLMN is a ▱. KL # LM, LM # NM, KN # NM, and KN # KL so KLMN is a rectangle.

TEKS cumulative practice pp. 314–3151. C 3. D 5. A 7. 53 9. 48

11. Given: G(-2, -4), H(1, 2), I(7, 5), J(4, -1) Prove: GHIJ is a rhombus. Proof: Using distance formula: GH = 2(1 - (-2))2 + (2 - (-4))2

= 29 + 36 = 245 HI = 2(7 - 1)2 + (5 - 2)2

= 236 + 9 = 245 IJ = 2(4 - 7)2 + (-1 - 5)2

= 29 + 36 = 245

JG = 2((4 - (-2))2 + (-1 - (-4))2 = 236 + 9 = 245

GH = HI = IJ = JG All sides are congruent. By definition of a parallelogram, GHIJ is a parallelogram because opposite sides are congruent. By definition of a rhombus, GHIJ is a rhombus because all parallelograms with congruent sides are rhombuses.

13. kite 15. S( - a, - b), T( - a, b); ( - a, 0), no slope 17. Ask your teacher to check your work. 19. Answers may vary. Samples: a. (0, 0), (3, 0), (3, 3), (0, 3) b. (0, 0), (3, 0), (4, 2), (1, 2) c. (0, 0), (3, 0), (3, 5), (0, 5) d. (0, 0), (3, 0), (2, 2), (1, 2)

Topic 8Lesson 8-1 pp. 322–3241. Yes; distances between corresponding pairs of points are equal. 3. No; distances between corresponding pairs of points are not equal. 5. 24 mi east and 81 mi south 7a. Answers may vary. Sample: ∠ R S ∠ R′ b. RP and R′P′; PT and P′T′; RT and R′T′ 9. (x, y) S (x - 3, y + 1) 11. The vertices of P′L′A′T′ are P′(0, -3), L′(1, -2), A′(2, -2), and T′(1, -3). Slope of PP′ = slope of LL′ = slope of AA′ = slope of TT′ = -3

2, so PP′ } LL′ } AA′ } TT′. 13. at least 5 ft east and 10 ft north 15. The midpts. of AB, BC, and AC are (-3, 2), (-1, -2), and (0, 1), respectively. The translation that maps (x, y) onto (x + 4, y + 2) translates those midpts. to (1, 4), (3, 0), and (4, 3), respectively. The same translation moves A, B, and C to A′(2, 7), B′(0, 1), and C′(6, -1), so the midpts. of A′B′, B′C′, and A′C′ are (1, 4), (3, 0), and (4, 3), respectively. 17. Answers may vary. Sample:

T64, -17 (△JKL) T62, -17 (△JKL) T64, -47 (△JKL)

19. T613, -2.57 (x, y)

21.


Ox

y

2

�2�6

4

6

23. (x, y) S (x - 4, y - 3) 25. G


22


Activity Lab 8-2 p. 3251. The distances are equal. 3. The results are the same; the reflection line is the # bis. of the seg. joining a pt. and its image. 5. Ask your teacher to check your work.

Lesson 8-2 pp. 329–3311. J′(1, -4), A′(3, -5), G′(2, -1)

3. J′(1, 0), A′(3, -1), G′(2, 3)

5. J′(-3, 4), A′(-5, 5), G′(-4, 1)

7a. Figure 3 = Rj (Figure 1) because line j is the perpendicular bisector of the line segments between corresponding vertices of Figures 1 and 3. b. Figure 2 = Rn (Figure 4) because line n is the perpendicular bisector of the line segments between corresponding vertices of Figures 2 and 4. c. Figure 4 = Rn (Figure 2) because line n is the perpendicular bisector of the line segments between corresponding vertices of Figures 4 and 2. 9a.

geom12_te_c09_l02_t0003.ai

P

P

N

L M

N

b. square; Since RLM (M) = M, RLM (N) = N′, and reflections preserve distance, RLM (MN) = MN′. So MN = MN′ and NN′ = 2MN. Since LM = 2MN and NN′ = 2MN, by substitution LM = NN′. Therefore, in the new figure PNN′P′ the length equals the width, so the figure is a square. 11.

13. y = x + 2; (x, y) S (y - 2, x + 2)

15.

17. Reflect point D across the mirrored wall to D′. Aim the camera at the point P where CD′ intersects the mirrored wall.

To show that a ray of light traveling from D to P will bounce off P and into the camera at C, show that ∠1 ≅ ∠3. By the definition of reflection across a line, the mirrored wall is the # bis. of DD′ so MD _ MD′ and ∠DMP _ ∠D′MP . PM _ PM, so △DMP _ △D′MP by SAS, and ∠1 _ ∠2 because corresp. parts of _ triangles are _. Thus, since ∠2 _ ∠3 (vert. angles), ∠1 _ ∠3 by the Trans. Prop. of _. 19. ( - 1, - 2) 21. ( - 3, 2) 23. ( - 5, - 3) 25. Yes; reflect a △ across any side and then reflect the image across the # bis. of that side. The combination of the original △ and the second image △ forms a ▱. 27. Yes; reflect an acute scalene △ across any side, an obtuse scalene △ across its longest side, a nonright isosc. (but not equilateral) △ across either leg, or a nonisosc. rt. △ across its hyp. 29. Yes; follow the steps of Exercise 25 using a rt. △ and the hyp. as the first reflection line. 31. The slope of AB is a - bb - a = a - b

-1(a - b) = -1. The slope of y = x is 1. Since (1)(-1) = -1, the lines are #. The midpt. of AB is

(b + a2 , a + b

2 ) , which is a point on the line y = x. 33. D 35. Answers may vary. Sample: In triangle ABC with right angle C, assume temporarily that the hypotenuse, AB, is not the longest side. One of the legs—say BC—is longer than AB. Therefore BC 7 AB, so m∠A 7 m∠C because in a triangle, the larger angle is opposite the longer side. Since angle C is a right angle, m∠A 7 90. Then m∠A + m∠C 7 90 + m∠C = 90 + 90 = 180. The statement m∠A + m∠C 7 180 contradicts the Triangle Angle-Sum Theorem. The assumption that AB is not the longest side of right triangle ABC must be false. Therefore, the hypotenuse of a right triangle must be its longest side.

O

4

4

y

x

J A

GG

J A



yA

G

J

J A

Gx

O 6

4

2

y

xO4 4

hsm11gmte_0902_t15054

A J

G G

AJ


hsm11gmte_0902_t15201

O

y

x

6 2

6

C

BA C

A

B

y x

C D

P12

3M

D

hsm11gmte_0902_t15203


23


Lesson 8-3 pp. 335–3371.

geom12_te_c09_l03_t0003.ai

AH

J

G

F

y

2

4

O 44

2

x

3. (x, y) S (y, -x); P′(2, 3), R′(5, 22), S′(0, 0) 5. Draw two segments connecting preimage points A and B to image points A′ and B′. If AA′ and BB′ are not collinear, construct the # bis. of AA′ and BB′ to find C, the center of rotation. m∠ACA′ is the ∠ of rotation. If AA′ and BB′ are collinear, then the midpoint of AA′ is the center of rotation, and 180° is the angle of rotation. 7. Yes; the angle of rotation of a composition of two rotations is the sum of the two angles of rotations. Since x + y = y + x, the two compositions give the same image. 9. Square; all sides are ≅ and all angles are 90°.

B

CD

A

x

y

O 26

6

2

hsm11gmte_0903_t15022

11. The image of ED is BA, not AB.13.

hsm11gmte_0903_t15043

E C

TR

R E

T

P

C15. Answers may vary. Sample: a reflection followed by a 270° rotation 17. (0, 230) 19. No. Rotations can only map each diagonal onto itself. You can use the properties of reflections to show that the lengths of diagonals are equal. 21. Sample answer: Your classmate is not correct; the puzzle piece will fit into Location A with a 90° rotation followed by a translation, but it requires a reflection to fit into Location B. 23. A 25a. Converse: If two lines do not intersect, then they are parallel; Inverse: If two lines are not parallel, then they intersect; Contrapositive: If two lines intersect, then they are not parallel. b. Converse: False; a counterexample is two skew lines; Inverse: False; a counterexample is two skew lines; Contrapositive: True.

Lesson 8-4 pp. 341–3431a.

Type of Symmetry

LanguageHorizontal

LineVertical

Line Point

English B, C, D, E,H, I, K, O, X

A, H, I, M,O, T, U, V,W, X, Y

H, I, N, OS, X, Z

Greek B, E, H, I, K, , O,

, , X

A, , H,, I, ,

M, , O,, T, ,, X, ,

Z, H, , I,N, , O,

, X

Alphabet Symmetry

b. Greek; explanations may vary. Sample: The Greek alphabet has more letters with at least one kind of symmetry and more letters with multiple symmetries. 3. line

hsm11gmte_0904_t15017

5. line; rotational: 60°

hsm11gmte_0904_t15018

7. no symmetry 9. line: any line passing through the center; rotational: any angle 11. line

13. line; rotational: 90°

15. You could use pencil and paper to sketch each quadrilateral and its line of symmetry; 1

17. You could use pencil and paper to sketch each quadrilateral and its line of symmetry; none

19. The other two vertices are (-1, -5) and (2, 3). The slopes of two opposite sides are -2 and the slopes of the other two opposite sides are 83, so the quadrilateral has two pairs of opposite sides parallel. 21. Line; ask your teacher to check your sketch; rotational: 90°; point 23. Not necessarily; the two other angles of the △ would need to be ≅. 25. (-3, 4) 27. (4, 3)


24


29. reflectional symmetry across x = -2

x

O

y

4 2

4

31. reflectional symmetry across the y-axis

y

xO 22

2

33. Answers may vary. Samples are given.

35. No; Sample: An equilateral triangle does not have point symmetry. 37. No; Sample: A parallelogram only has reflectional symmetry if all sides have equal lengths or if all angles are right angles. 39. line 41. line 43. 2.5 45. 6

Lesson 8-5 pp. 347–3491. rotation; center C, angle of rotation 180° 3. glide reflection; (Ry = 0 ∘ T<11, 0>)(x, y) 5. glide reflection; (Rx = 4 ∘ T<0, 4>)(x, y) 7. R( - 6, 3), O( - 2, - 5), and S(5, 6)9. A translation; the arrow in the diagram shows the direction, determined by a line perpendicular to / and m. The distance is twice the distance between / and m.


TT

T

m

11. A rotation; the center of rotation is C and the angle of rotation is 150°.


N

N

N

m

C

13. y

x

4

6

2

hsm11gmte_0906_t15244

N

P

B

OP

NB

4

15. (-2, -3)

17. y4

2

O 42-2

-2

-4

-4x

R″R

P″ P

Q″Q

19. r(180°, O)

21. y

x

2

4

4

2

hsm11gmte_0906_t15248

O

2

2B

A

B

A

A

B1

2

a 180° rotation about (0, 0) 23. y

x

2 8

2

hsm11gmte_0906_t15252

O

4

2

B BB

AA A

2 1

a translation 4 units left 25. A 27. C

Lesson 8-6 pp. 353–3541. △BVQ ≅ △ETJ; Sample: Rotate triangle BVQ 180° about the origin; then translate 2 units right and 2 units up. 3. FAKE ≅ CMDI; Sample: Rotate FAKE 180° about the origin. 5. Sample: Rotate triangle LMN 180° about the origin. 7. Rotate and translate △HLA so that HL coincides with EK. Since rigid transformations preserve angle measures, point A lies along the same line as ES, and point A also lies along the same line as KS. These two lines must intersect at a single point, so S coincides with A. Therefore there is a congruence transformation that maps △HLA to △EKS and △HLA ≅ △EKS.


25


9. congruent 11a. Congruence transformations preserve distances and angle measures. b. Use SAS, proven in Problem 4. 13. Sample: Draw and label the midpoint of GH as point M. Draw FM. Using the identity mapping, FM ≅ FM. It is given that FG ≅ FH, and GM ≅ MH by the definition of midpoint. Therefore, if triangle FHM is reflected across FM, it will overlap triangle FGM. Because there is a rigid transformation mapping triangle FHM onto triangle FGM, △FHM ≅ △FGM. Therefore, ∠G ≅ ∠H because corresp. parts of ≅ triangles are ≅. 15. Sample: For a figure to have x° rotational symmetry, x must divide into 360 with no remainder. Since 50 leaves a remainder when divided into 360, a figure cannot have 50° rotational symmetry.

Activity Lab 8-7 p. 3551. RS = 2(2 - 1)2 + (-1 - 4)2 = 21 + 25 = 226 3. R′(2, 8), S′(5, - 1) 5. R′S′ = 2(4 - 2)2 + (-2 - 8)2 = 24 + 100 = 2104 = 22267. y

2 4�2

2

4

x

geom12_te_ccs_c09_l06_t0002.ai

R��

S��O

9. R″S″ = 12 RS 11. The slopes of the dilated lines are

all equal to the slope of the original line.

Lesson 8-7 pp. 361–3631. enlargement; 1.5 3. reduction; 13 5. reduction; 0.5 7.

hsm11gmte_0905_t14986

y

xO40

30

20

P

R

Q , D10 (x, y) = (10x, 10y)

9. D(3, Q) (x, y) = (3x - 4, 3y - 8)

4 8

y

xO

4

�4

�8

P′ R′

Q′

11. 0.2 cm 13. L′(-15, 0) 15. A′(-9, 3) 17. Q′(-1.5, 0), R′(-1, -2.5), T′(1.5, -1.5), W′(1.5, 0.5) 19. Q′(3, 31), R′(-7, -19), T′(-57, 1), W′(-57, 41)

21.

hsm11gmte_0905_t14974

y

xO814

8

6

N

P Q

M

MN

PQ

23. Ask your teacher to check your work. 25a. Since <AB

> does not pass through the origin, it has x- and

y-intercepts (a, 0) and (0, b) for some a, b ≠ 0. So Dk(a, 0) = (ka, 0) and Dk(0, b) = (0, kb). So

<A′B′

>

has different x- and y-intercepts than <AB

> and thus <

AB>≠

<A′B′

>. b. The slope of

<AB

> is

b2 - a2b1 - a1

. Since

A′ = (ka1, ka2) and B′ = (kb1, kb2), the slope of <A′B′

>

is kb2 - ka2kb1 - ka1

. c. If a1 = b1 then <AB

> is vertical. Also,

a1 = b1 implies ka1 = kb1, so <A′B′

> is vertical.

27. x = 3, y = 60; the ratio of the corresp. sides is the same as the scale factor of the dilation, which is 4 : 2, or 2 : 1. To find x, solve the proportion x + 3

x = 21.

y = 2y - 60 and so y = 60 because dilations do not affect the measure of an angle. 29. y

xO

48 4

12

R

Q

P

QP

R

hsm11gmte_0905_t14969

8

2

4

31. 1 ft 33. A dilation with scale factor 1 will preserve congruence. 35. G

Lesson 8-8 pp. 368–3701. A(21, 1) S A′(22, 4)

B(1, 1) S B′(2, 4) C(1, 21) S C′(2, 24) D(21, 21) S D′(22, 24);

3. Q(24, 3) S Q′(22, 1) R(4, 3) S R′(2, 1) S(4, 23) S S′(2, 21) T(24, 23) S T′(22, 21);

5a. (x, y) S (rx, ry), where r 7 1 b. (x, y) S (x, by), where 0 6 b 6 1 7. L(0, 0) S L′(22, 0)

M(0, 24) S M′(22, 26) N(24, 0) S N′(24, 0)


26


9. A″(0, 23) S A(4, 21) B″(0, 0) S B(4, 0) C″(3, 0) S C(6, 0)

11. Answers may vary. Sample: Dilate EFGH by a scale factor of 2, then stretch the image vertically by a scale factor of 2 to get E′F′G′H′. 13. Answers may vary. Sample: Stretch the rectangle horizontally by a scale factor of 3. Rotate the rectangle 270° about point (12, 0). Translate the rectangle 2 units to the right and 2 units up. 15. Answers may vary. Sample: a horizontal stretch with factor 2 followed by a reflection over the line y = 4 followed by a translation 1 unit right; No; Sample answer: Since a horizontal stretch does not preserve congruence, the figures are not congruent. 17. D 19. B

Topic Review pp. 371–3751. transformation 3. translation 5a. No; the distances between corresponding points in the image and preimage are not the same. b. LA, W 7. T6-5, 107( x, y )9. A′(6, -4), B′(-2, -1), C′(5, 0)

11. A′(4, 6), B′(1, -2), C′(0, 5)

13.

15. W′(-1, -3), X′(2, -5), Y′(8, 0), and Z′(-1, -2). 17. rotational: 180°; point 19. one 21.

E is translated right, twice the distance between / and m.

23. same; translation 25. △T′A′M′ with vertices T′(-4, -9), A′(0, -5), M′(-1, -10) 27. Answers may vary. Sample: Yes, the letters are congruent. The p can be mapped to the d with a composition of a translation followed by a rotation. 29. M′(-15, 20), A′(-30, -5), T′(0, 0), H′(15, 10) 31. L′N′ = 6.5 ft, M′N′ = 11.25 ft 33. P′(-2, 1), Q′(2, 1), R′(2, -1

2), S′( -2, -12)

35. P′(-2, 3), Q′(0, 3), R′(0, 32), S′( -2, 32) 37. Answers may vary. Sample answer: a stretch transformation (x, y) S (3x, y) followed by a translation 1 unit down

TEKS cumulative practice pp. 376–3771. D 3. C 5. D 7. C 9. A 11. B 13. 29 15. 30 17. AB ≅ CB (Given); ∠A ≅ ∠C (Isosc. △ Thm.); BD # AC (Given); ∠ADB and ∠CDB are rt. ⦞ (Def. of #); ∠ADB ≅ ∠CDB (All rt. ⦞ are ≅.); △ABD ≅ △CDB (AAS) OR other correct proof 19. Yes, LMNO ≅ RSTV because (Rx@axis ° T65, -17)(LMNO) = RSTV

Topic 9Lesson 9-1 pp. 383–3851. ∠R ≅ ∠D, ∠S ≅ ∠E, ∠T ≅ ∠F, ∠V ≅ ∠G; RSDE = ST

EF = TVFG = VR

GD 3. ∠K ≅ ∠H, ∠L ≅ ∠G,∠M ≅ ∠F, ∠N ≅ ∠D, ∠P ≅ ∠C;KLHG = LM

GF = MNFD = NP

DC = PKCH 5. ABDC ∙ FEDG

(or ABDC ∙ FGDE, ABDC ∙ DEFG, ABDC ∙ DGFE) 7. x = 4, y = 3 9. x = 12.75, y = 18.5 11. 51 13. 16.5 15. 5 in. 17. No; for polygons with more than 3 sides, you also need to know that corresponding angles are congruent in order to state that the polygons are similar. 19. 1 : 3 21. 60 ft 23. Ask your teacher to check your work. 25. x = 10; 2 : 1 27. 14 ft by 21 ft 29. A 31. C

Lesson 9-2 pp. 389–3911.

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5. D1.5 ∘ r(180°,O) 7. Answers may vary. Sample: △AVS is similar to △RGI. Translate △RGI so that points R and A coincide. Rotate by 180°. Then dilate with center A and scale factor 1.5. 9. Answers may vary. Sample: The similarity transformation is a rotation about point A followed by a dilation with respect to point A. △SCA is similar to △ELA. 11. Answers may vary. Sample: Use the Distance Formula to find the lengths of the sides. Then verify that the ratios of corresponding sides are proportional. AB = 4, BC = 3, AC = 5, DE = 8, EF = 6, and DF = 10. Since AB

DE = BCFE = AC

DF = 0.5, the ratios of corresponding sides are proportional, and the figures are similar. 13. Distance is preserved by rigid transformations, but not by similarity transformations. 15. Yes, the image of the transparency is rotated in space and dilated. The image on the wall is similar to the preimage on the transparency. 17. A dilation with scale factor 2 maps △EFG to △KJL, so the figures are similar; EF

KJ = FGJL = EG

KL and ∠E ≅ ∠K , ∠F ≅ ∠J, and ∠G ≅ ∠L 19. A dilation with scale factor 32 maps QRTD to MSNP, so

the figures are similar; QRMS = RT

SN = TDNP = QD

MP and ∠Q ≅ ∠M, ∠R ≅ ∠S, ∠T ≅ ∠N, and ∠D ≅ ∠P . 21a–c. Ask your teacher to check your work. d. Conjecture: Two triangles with two pairs of congruent corresponding angles are similar. 23. always 25. never 27. Answers may vary. Sample: Yes, a rigid transformation is a similarity transformation with a scale factor of 1. The preimage and image of a rigid transformation are congruent, so they are also similar. 29. 2 31. 22

Lesson 9-3 pp. 395–3971. △FGH ∙ △KJH by AA ∙. 3. △RST ∙ △PSQ by SAS ∙. 5. ∠MPN ≅ ∠QPR (Vertical angles are congruent.),

and the given information tells us PRPN = PQ

PM = 21. So

△MNP ∙ △QRP by SAS ∙. 7. There are a pair of congruent vertical angles and a pair of congruent right angles, so the triangles are similar by AA ∙; 180 ft 9. about 169.2 m 11. 180 ft 13.


A

C B J

L

K

30�

Draw a line segment. JK of any length. At J, construct an angle congruent to ∠C . At K, construct an angle congruent to ∠B. Extend the sides of the angles until they intersect. Label the intersection point L. △LKJ ∙ △ABC by AA ∙.

15. Use the Distance Formula: AB = AC = 215 and BC = 212, while RS = RT = 15 and ST = 12. △ABC ∙ △RST (SSS ∙) because AB

RS = BCST = AC

RT = 2. 17. 20 19. It is given that /1 } /2, so ∠BAC ≅ ∠EDF because if lines are parallel, then corresponding angles are congruent. The given perpendicular lines mean ∠ACB ≅ ∠DFE because parallel lines form right angles, which are congruent. So △ABC ∙ △DEF by AA ∙, and BC

EF = ACDF because corresponding sides of

similar triangles are proportional. Then Property of Proportions (2) lets us conclude that BC

AC = EFDF .

21.


A

B C

Q

R

X Y

S

Choose point X on QR so that QX = AB. Then draw XY } RS (Through a point not on a line, there is exactly one line parallel to the given line.) ∠A ≅ ∠Q (Given) and ∠QXY ≅ ∠R (If two lines are parallel, then corresponding angles are congruent.), so △QXY ∙ △QRS by AA ∙.

Therefore, QXQR = XY

RS = QYQS because corresponding

sides of similar triangles are proportional. Since QX = AB, substitute QX for AB in the given proportion AB

QR = ACQS to get QX

QR = ACQS . Therefore,

QXQR = QY

QS = ACQS , and QY = AC. So △ABC ≅ △QXY

by SAS. ∠B ≅ ∠QXY (Corresponding parts of congruent triangles are congruent.) and ∠B ≅ ∠R by the Transitive Property of Congruence. Therefore, △ABC ≅ △QRS by AA ∙.

23. C 25. C

Lesson 9-4 pp. 403–4051. Answers may vary. Sample: △KJL ∙ △NJK ∙ △NKL 3. Answers may vary. Sample: △OMN ∙ △PMO ∙ △PON 5. 12 7. x = 613, y = 313 9. x = 10, y = 2121 11. 5 ft 13. 10110 15. 2 17. 114 19. △PTQ ∙ △PQR, △QTR ∙ △PQR, △PTQ ∙ △QTR, △QST ∙ △QTR, △TSR ∙ △QTR, △QST ∙ △TSR, △QST ∙ △PQR, △TSR ∙ △PQR, △QST ∙ △PTQ, △TSR ∙ △PTQ 21. It is given that CDE is a right triangle with altitude FC and right ∠DCE. Since FC is an altitude, FC # DE, which means that ∠CFE and ∠DFC are right angles. ∠CFE ≅ ∠DCE ≅ ∠DFC because all right angles are congruent; ∠E ≅ ∠E and ∠D ≅ ∠D by reflexivity. Thus △FCE ∙ △CDE and △FDC ∙ △CDE by AA∙ . Thus, ∠EFC ≅ ∠D because corresponding angles of similar triangles are congruent. And ∠CFE ≅ ∠DFC since they are both right angles. So △FCE ∙ △FDC ∙ △CDE. 23. YZ; WZ

YZ = YZZX 25. No; Answers may vary. Sample:

The classmate assumes that triangle similarity has already been proven, but a proof of Theorem 9-3 must include a proof of similarity. 27. 4 29. 5 31. △ABC ∙ △ACD and △ABC ∙ △CBD by Theorem 9-3. Thus, AB

AC = ACAD and AB

BC = BCBD because

corresponding sides of similar triangles are proportional.33a.


A D B

C

Given: AC # BC , AB # CD Prove: AC # BC = AB # CD


28


b. The conjecture is true. You can express the area of △ABC as 12(AC )(BC ) or as 12(AB )(CD), so AC # BC = AB # CD. 35. (22, 6), (10, 6) 37. /1 = 612, /2 = 612, h = 12, s2 = 6 39. /1 = 6, h = 12, a = 313, s2 = 9 41. Answers may vary. Sample: △ABC ∙ △DEC (AA ∙ Postulate), so AC

DC = BCEC (Corresponding sides of similar triangles

are in proportion). By the Subtraction Property of Equality, AC - DC

DC = BC - ECEC , or AD

DC = BEEC . 43. H

Activity Lab 9-4 pp. 406–4071. x + 1

x = x1; find the cross products to write x2 = x + 1

or x2 - x - 1 = 0. 3. 1.6180 5. 1.625, 1.615, 1.619, 1.618, 1.618, 1.618, 1.618, 1.618, 1.618 7. Answers may vary. Rectangle 4 has dimensions that are close to the golden ratio. 9. The number of petals is a Fibonacci number.

Lesson 9-5 pp. 412–4141. XR

RQ = YSSQ (Given); XR + RQ

RQ = YS + SQSQ (Property of

Proportions (3)); XQ = XR + RQ, YQ = YS + SQ

(Segment Addition Postulate); XQRQ = YQ

SQ (Substitution); ∠Q ≅ ∠Q (Reflexive Property of Congruence); △XQY ∙ △RQS (SAS ∙ Postulate); ∠1 ≅ ∠2 (Corresp. ∠s of ∙ △s are ≅.); RS } XY (If corresp. ∠s are ≅, the lines are } .) 3. 35 5. 8 mm 7. 3 5

13 9. Answers may vary. Sample: 10 and 15, 8 and 12, or any two sides in the ratio 2 : 3 where the shorter side is greater than 6 cm and is less than 15 cm 11. KS by the Triangle-Angle-Bisector Theorem 13. KM by the Triangle-Angle-Bisector Theorem 15. 20 17. 5.2 19a. Answers may vary. Sample: A midsegment of a parallelogram connects the midpoints of two opposite sides of the parallelogram. b.


D

P

A B

Q

C

Given: PQ is a midsegment of parallelogram ABCD. Prove: PQ } AB, PQ } DC AD ≅ BC and AD } BC (properties of

parallelograms), so PD = 12(AD) = 1

2(BC) = QC, and

PA = 12(AD) = 1

2(BC) = BQ, both by the definition of midpoint. Therefore, ABQP and PQCD are parallelograms because each has a pair of opposite sides that are congruent and parallel. So PQ } AB and PQ } DC because opposite sides of a parallelogram are parallel.

c.


D

P

AT

B

Q

C

Given: Parallelogram ABCD and midsegment PQ Prove: PQ bisects BD.

From part (b) of this exercise, AB } PQ } DC . Since AP ≅ PD by the definition of midsegment, DT ≅ TB because if parallel lines cut off congruent segments on one transversal, they cut off congruent segments on every transversal (Corrolary to Thm. 9-4). Since PQ contains the midpoint of BD, PQ bisects BD by the definition of bisector. Also, point T is the midpoint of both diagonals (because the diagonals of a parallelogram have the same midpoint), so it bisects both diagonals of the parallelogram.

21. 575 ft 23. 12.5 cm or 4.5 cm 25. DE } AC is given. If lines are parallel, then corresponding angles are congruent, so ∠BAC ≅ ∠BDE and ∠BCA ≅ ∠BED. So, by the AA ∙ Postulate, △ABC ≅ △BDE. Because corresponding sides of similar triangles are proportional, AB

DB = CBEB . By

the Segment Addition Postulate, AB = AD + DB and CB = CE + EB. So AD + DB

DB = CE + EBEB

and ADDB + 1 = CE

EB + 1 by the Substitution

Property. Therefore ADDB = CE

EB by the Subtraction Property of Equality.27. Answers may vary. Sample: You might select a compass and straightedge to perform the measurements manually.

X ZM

Y

Conjecture: The bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.29. 5-cm side: 2.4 cm, 2.6 cm 12-cm; side: 31

3 cm, 82

3 cm; 13-cm side: about 3.8 cm, about 9.2 cm 31. 52 33. 66

Topic Review pp. 415–4171. similar 3. scale factor 5. JEHN ∼ JKLP ; 3 : 4 (or JKLP ∼ JEHN; 4 : 3) 7. 31

2 ft. tall by 514 ft. wide

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11. No; because all of the dimensions of the airplane must dilate by the same scale factor for the figures to be similar. 13. 120 ft 15. The ratio of each pair of corresp. sides is 2 : 1, so △AMY ∼ △ECD by SSS∼. 17. 12 19. x = 622, y = 626 21. x = 2221; y = 423 23. 7.5 25. 22.5 27. 17.5


29


TEKS cumulative practice pp. 418–4191. C 3. D 5. A 7. C 9. B 11. C 13. 30 15. 80 17. x = 8; the diagonals of a rectangle are ≅. 19. Answers may vary. Sample: No; the model would be more than 10 ft. tall.

Topic 10Activity Lab 10-1 p. 4221a. The areas are equal. b. a2 + b2 + 2ab, c2 + 2ab c. The sum of the areas of the two smaller squares = the area of the larger square. d. The same relationship occurs. 3. The sum of the squares of the lengths of the two legs = the square of the length of the hypotenuse.

Lesson 10-1 pp. 427–4291. 10 3. 20 5. 14.1 ft 7. 412 9. No; 192 + 202 ≠ 282 11. Yes; 332 + 562 = 652. 13. Yes; 72 + 242 = 252; so ∠S is a right angle. 15. 10 17. 212 19. 50 21. 35 23. obtuse 25. acute 27. Draw right △FDE with legs DE of length a and EF of length b, and hypotenuse of length x. Then a2 + b2 = x2 by the Pythagorean Thm. You are given △ABC with sides of length a, b, and c, and a2 + b2 = c2. By subst., c2 = x2, so c = x. Since all side lengths of △ABC and △FDE are the same, △ABC ≅ △FDE by SSS. ∠C ≅ ∠E because corresp. parts of ≅ triangles are ≅, so m∠C = 90. Therefore, △ABC is a right △. 29. Draw right △FDE with legs DE of length a and EF of length b, and hypotenuse of length x. By the Pythagorean Thm., a2 + b2 = x2. △ABC has sides of length a, b, and c, where c2 6 a2 + b2. c2 6 x2 and c 6 x by the Prop. of Inequalities. If c 6 x, then m∠C 6 m∠E by the Converse of the Hinge Thm. Thus m∠C is less than 90, and since C is the largest angle in △ABC by Thm. 5-10, ∠A and ∠B have measures less than 90 as well. An angle with measure 6 90 is acute, so △ABC is an acute △. 31. 23

Lesson 10-2 pp. 433–4341. x = 15, y = 15 3. 110 5. 25.5 ft 7. x = 13, y = 3 9. x = 24, y = 1213 11. x = 9, y = 18 13. a = 7, b = 14, c = 7, d = 713 15. a = 1013, b = 513, c = 15, d = 5 17. a = 3, b = 7 19. 14.4 s 21a. 8.5 m b. 3.1 m 23. Answers may vary. Samples using the following segment are given.

a

hsm11gmte_0802_t12428

a.

a

a

aa

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b.

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25. C 27. AC = 6; 3AC = AC

12, AC2 = 36

Technology Lab 10-3 p. 4351. The ratio does not change. 3. yes; sine 5. Sample: When the ratios are equal, the ⦞ are complements.

Lesson 10-3 pp. 438–4401. 7

25 ; 2425; 7

24 3. 132 ; 12; 13 5. 8.3 7. 17.0

9. 21.4 11. 1085 ft 13. 58 15. 59 17. 66 19. 44 and 136 21. 74 23. w = 3, x ≈ 41 25. w ≈ 68.3, x ≈ 151.6 27a. 6.25 ft b. about 53 29a. They are equal; yes; sine and cosine of compl. angles are equal. b. ∠B; ∠A c. Sample: The cosine is the complement’s sine. 31a. 1.5 AU b. 5.2 AU 33. G

Lesson 10-4 pp. 443–4451. angle of elevation from sub to boat 3. angle of elevation from boat to tree 5. angle of elevation from Max to top of waterfall 7. angle of depression from top of waterfall to Max 9. 34.2 ft 11. 986 m 13. 0.6 km 15. 64° 17. 46, 46 19. 20, 20 21. 3300 m 23. 5 25. 0.5; about 85 27. Ask your teacher to check your work. 29a. rhombus and square b. rhombus; No information is given about a right angle. c. Answers may vary. Sample:

Parallelogram

Square

Rhombus Rectangle

hsm11gmte_0804_t12434

Topic Review pp. 446–4471. Trigonometric ratios 3. Pythagorean triple 5. 17

7. 913 9. 512 11. x = 7, y = 713 13. 211920 or 119

10 ; 1820 or 9

10; 211918 or 119

9 15. 16.5 17. 38.2 ft


30


TEKS cumulative practice pp. 448–4491. B 3. B 5. A 7. B 9. C 11. 4.6 13. 112 15. 2340 17a. BD = AC; 2x - 8 + x - 4 = x + 2; 3x - 12 = x + 2; 2x = 14; x = 7 b. 9 19. No; the distance from (1, 7) to (4, 3) is 5, but the distance from (1, 7) to (6.5, 11) is about 6.8. A regular polygon must have all sides ≅.

Topic 11Lesson 11-1 pp. 456–4581. BC¬, BD¬, CD¬, CE¬, DE¬, DF¬, EF¬, FB¬ 3. BCE¬, BFE¬, CBF¬, CDF¬ 5. 180 7. 270 9. 308 11. 6p ft

13. 14p in. 15. B 17. 7p2 cm 19. 27p m 21. 2.6p in. 23. 70 25. 110 27. 235 29. Find the measure of the major arc, then use the definition of arc length; or find the length of the minor arc using the definition, then subtract that length from the circumference of the circle. 31. 38 33. 31 m 35. 3 : 4 37. AP ≅ BP (Radii of a circle are congruent.); △APB is isosceles (definition of an isosceles triangle); ∠A ≅ ∠B (Isosceles Triangle Theorem); AB } PC (Given); ∠B ≅ ∠BPC (Alternate Interior Angles Theorem); ∠A ≅ ∠CPD (Corresponding Angles Postulate); ∠BPC ≅ ∠CPD (Transitive Property of Congruence); m∠BPC = mBC¬ and m∠CPD = mCD¬ (The measure of a minor arc is equal to the measure of its corresponding central angle.); mBC¬ = mCD¬ (Transitive Property of Equality). 39. A 41. B

Activity Lab 11-2 pp. 459–4601. Ask your teacher to check your work. 3. Ask your teacher to check your work. 5. Ask your teacher to check your work. 7. about 57.30°; The numbers are very close. 9. p4 radians 11. 2p3 radians

13. 31p18 radians 15. 4p9 radians 17. p12 radians

19. 67.5° 21. 315°

Lesson 11-2 pp. 464–4651. 3 3. 1 5. 8p9 , 2.79 7. 360° 9. 120° 11. 90° 13. 3.1 cm 15. 51.8 ft 17. ≈32 ft 19a. ≈11,048 km, ≈33,144 km, ≈27,620 km, ≈276,198 km b. ≈18.1 h 21. ≈11 radians 23. ≈6.3 cm 25. us = 2p

2pr ; us = 1

r ; ur = s; s = ru 27. G

Activity Lab 11-3 p. 4661. They are equal. 3. b ≈ 1

2C = 12(2pr) = pr

5. b ≈ 2r, h ≈ 12pr

Lesson 11-3 pp. 472–4731. 30.25p cm2 3. p9 in.2 5. 18.3 ft2 7. about 282,743 ft2 9. 64p cm2 11. 56p cm2 13. (4 - p) ft2 15. (784 - 196p) in.2 17. 24p in.2 19. Ask your teacher to check your work. 21. 84.2 ft2; Ask your teacher to check your work. 23. B 25. B

Lesson 11-4 pp. 477–4781. (x - 2)2 + (y + 8)2 = 81 3. (x - 0.2)2 + (y - 1.1)2 = 0.16 5. (x + 6)2 + (y - 3)2 = 64 7. center (-7, 5); radius 4


�4�8

(�7, 5)

O

8y

4

x

9. (x + 4)2 + (y - 2)2 = 16 11. (x + 2)2 + (y - 6)2 = 16 13. (x - 7)2 + (y + 2)2 = 52 15. position (5, 7); range 9 17. x2 + y2 = 4 19. x2 + (y - 3)2 = 4 21. (x - 5)2 + (y - 3)2 = 13 23. No; the x term is not squared. 25. Yes; it is a circle with center (1, -2) and radius 3. 27. The graph is the point (0, 0). 29. 41231.

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�2

4y

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(2, 2), (-2, 2)

33. Let (x, y) be any point on the circle. Then the radius r is the distance from (0, 0) to (x, y). d = 2(x2 - x1)2 + (y2 - y1)2 Distance Formula

r = 2(x - 0)2 + (y - 0)2 Substitute r for d, (0, 0) for (x1, y1) and (x, y) for (x2, y2)

r = 2x2 + y2 Simplify. r2 = x2 + y2 Square both sides. The equation of a circle with radius r and center

(0, 0) is x2 + y2 = r2.35. (x + 1)2 + (y - 3)2 + (z - 2)2 = 6 37. D 39. No; the second statement is the converse of the first statement, and a conditional and its converse are not equivalent statements.

Topic Review pp. 479–4811. sector 3. concentric circles 5. 120 7. 120 9. p mm 11. 4p m 13. p8 15. 360º 17. 135º 19. 144p in.2 21. 41.0 cm2 23. 36.2 cm2 25. (x - 3)2 + (y - 2)2 = 4 27. (x - 1)2 + (y - 4)2 = 9

TEKS cumulative practice pp. 482–4831. B 3. C 5. A 7. B 9. B 11. 40 13. 270 15. 31.4 17. 31.42 19. 84.8 21. 2.09 23. 150 25. 46 27. x2 + (y + 1)2 = 4


31


29.

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y

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31. the graph is the point (0, 0) 33a. Answers may vary. Sample: Subtract the minor arc segment from the area of the circle; or add the areas of the major sector and the triangle that is part of the minor arc sector. b. (25p - 50) unit2; (75p + 50) unit2

Topic 12Lesson 12-1 pp. 490–4911. 120 3. 30 5. 113.1 km 7. 4.8 9. 8 in. 11. 14.2 in. 13a. Rectangle; AB is tangent to }D and }E; DB # AB and AE # AB (A line tangent to a } is # to the radius.); BC } AE (Two coplanar lines # to the same line are }.) So ABCE is a ▱ with two rt. angles. Therefore, ABCE is a rectangle. b. 35 in. c. 35.5 in. 15. Answers may vary. Sample: One square is inscribed in the circle, and the other square circumscribes the circle. If the circle has radius a, each side of the smaller square has length a12 and the area of the square is 2a2. Each side of the larger square has length 2a, and the area of the square is 4a2. So the larger square has double the area of the smaller square. 17. It is given that BC is tangent to } A at D and DB ≅ DC . It follows that AD # BC since lines tangent to a } are # to the radius at the point of tangency. By def. of #, ∠ ADB and ∠ ADC are rt. angles. Also ∠ ADB ≅ ∠ ADC since rt. angles are ≅. By the Refl. Prop. of ≅, AD ≅ AD. It follows that △ADB ≅ △ADC by SAS. Thus, AB ≅ AC since corresp. parts of ≅ triangles are ≅. 19. Assume

<AB

> is not

tangent to }O. Then either <AB

> does not intersect }O

or <AB

> intersects }O at two pts. If

<AB

> does not

intersect }O, then P is not on }O, which contradicts OP being a radius. If

<AB

> intersects }O at two pts., P

and Q, then OP ≅ OQ (≅ radii), △OPQ is isosc., and ∠OPQ ≅ ∠OQP. But ∠ OPQ is a rt. ∠ since AB # OP, and △OPQ has two rt. angles. This is a contradiction also, so

<AB

> is tangent to }O. 21. 22

Lesson 12-2 pp. 496–4981. BC¬ ≅ YZ¬, BC ≅ YZ 3. 14 5. 10 7. Since ∠AOB ≅ ∠COD, it follows that m∠AOB = m∠COD. Now m∠AOB = mAB¬ and m∠COD = mCD¬ (Definition of arc measure). So mAB¬ = mCD¬ (Substitution). Therefore, AB¬ ≅ CD¬ (Definition of ≅ arcs). 9. } O with AB ≅ CD (given); AO ≅ BO ≅ CO ≅ DO (All radii of a } are ≅.); △AOB ≅ △COD (SSS); ∠AOB ≅ ∠COD (Corresp. parts of ≅ triangles are ≅.); AB¬ ≅ CD¬ (≅ central angles have ≅ arcs.). 11. XW ≅ XY (All radii of a circle are ≅.); X is on the # bis. of WY (Converse of # Bis. Thm.); / is the # bis. of WY (Given); X is on

/, so / contains the center of } X . 13a. Ask your teacher to check your work. b. Conjecture: If two perpendicular chords of a circle have a common endpoint, the segment that joins the other two endpoints is a diameter of the circle. 15. 5 in. 17. 10 ft 19. 6 in. 21.

23. The length of a chord or an arc is determined not only by the measure of the central angles, but also by the radius of the }. 25.

Given: } O with AB ≅ CD Prove: ∠ AOB ≅ ∠COD Proof: In circle O, AO = BO = CO = DO (radii of a } are ≅.) and AB ≅ CD (Given). So △ AOB ≅ △COD (SSS) and ∠ AOB ≅ ∠COD (Corresp. parts of ≅ △s are ≅.).

27.

Given: }O with AB ≅ CD, OE # AB, OF # CD Prove: OE ≅ OF Proof: All radii of }O are ≅, and it is given that AB ≅ CD, so △AOB ≅ △COD by SSS. ∠A ≅ ∠C (Corresp. parts of ≅ △s are ≅.). ∠OEA and ∠OFC are rt. ⦞ (# lines form rt. ⦞.). So ∠OEA ≅ ∠OFC (Rt. ⦞ are ≅.). Thus, △OEA ≅ △OFC by AAS, and OE ≅ OF (Corresp. parts of ≅ △s are ≅.).

29. CE = ED, BC¬ ≅ BD¬ 31. F

Lesson 12-3 pp. 501–5031. 58 3. a = 50, b = 90, c = 90 5. 123 7. e = 65, f = 130 9. }O with inscribed ∠ABC (given); m∠ABO = 1

2mAP¬ and

m∠OBC = 12m PC¬ (Inscribed Angle Theorem, Case I);

m∠ABO + m∠OBC = m∠ABC (Angle Addition Postulate); 12mAP¬ + 1

2m PC¬ = m∠ABC (Substitution

Property); 12(mAP¬ + m PC¬) = m∠ABC (Distributive

Property); 12mAC¬ = m∠ABC (Arc Addition Postulate)



A

O

D

C

B

O

A

E F

D

CB

hsm11gmte_1202_t15295


32


11. Answers may vary. Sample: a. If the cameras’ lenses open at congruent angles, then in the positions shown they share the same arc of the scene. b. No; the distances from each position of the scene to each camera affect the look of the scene. 13. a = 30, b = 60, c = 62, d = 124, e = 60 15. }O with ∠CAB inscribed in a semicircle (Given); m∠CAB = 1

2m BDC¬ (Inscribed Angle Theorem);

m BDC¬ = 180 (A semicircle has a measure of 180.); m∠CAB = 90 (Substitution Property); ∠CAB is a right angle (Definition of right angle). 17. GH and tangent / intersecting } E at H (Given); draw <HE

> intersecting } E at D so HD is a diameter (Two

points determine a line.); ∠DHI is a right angle (A tangent line is perpendicular to the radius at the point of tangency.); m∠DHI = 90 (Definition of right angle); mDGH¬ = 180 (A semicircle has a measure of 180.); m∠DHG + m∠GHI = m∠DHI (Angle Addition Postulate); mDG¬ + m GFH¬ = mDGH¬ (Arc Addition Postulate); m∠DHG + m∠GHI = 90 (Substitution Property); mDG¬ + m GFH¬ = 180 (Substitution Property); 12(mDG¬ + m GFH¬) = 90 (Multiplication Property of Equality); m∠DHG + m∠GHI = 1

2(mDG¬ + m GFH¬) (Substitution

Property); m∠DHG = 12mDG¬ (Inscribed Angle

Theorem); 12mDG¬ + m∠GHI = 12mDG¬ + 1

2m GFH¬ (Substitution Property and Distributive Property); m∠GHI = 1

2m GFH¬ (Subtraction Property of Equality). 19. J

Technology Lab 12-4 p. 5041. Ask your teacher to check your work. 3. BF # FC = EF # FD 5. Ask your teacher to check your work. 7. DG # DF = DE # DB 9. Ask your teacher to check your work. 11. (DG)2 = DE # DB

Lesson 12-4 pp. 508–5101. 11.5 3. x = 25.8, y ≈ 12.4 5. 180 - x 7. 46 9. x = 72, y = 36 11a. (8 - 413) in. b. 4

2 + 13 in.;

42 + 13

# 2 - 132 - 13

= 8 - 4134 - 3 = 8 - 413

13.


X

ZY

VW

Given: A } with secant segments XV and ZV Prove: XV # WV = ZV # YV . Proof: Draw X Y and Z W . (2 pts. determine a line.); ∠X V Y ≅ ∠Z V W (Refl. Prop. of ≅); ∠V X Y ≅ ∠W Z V (2 inscribed ⦞ that intercept the same arc are ≅.); △X V Y ∙ △Z V W (AA∙ ); X VZ V = Y V

W V (In similar figures, corresp. sides are proportional.); X V # W V = ZV # YV (Prop. of Proportion)

15a. △ACD b. tan A = DCAC = DC

1 = DC = length

of tangent seg. c. secant A = ADAC = AD

1 = AD =

length of secant seg. 17. m∠1 = 12m QRP¬ - 1

2mPQ¬

and m∠2 = 12m RQP¬ - 1

2m RP¬ (vertex outside }, m∠ = half differences of intercepted arcs); m∠1 + m∠2 = 1

2m QRP¬ + 12m RQP¬ - 1

2mPQ¬ - 12m RP¬

(Addition Prop. of = ); m∠1 + m∠2 = 1

2mQR¬ + 12m RP¬ + 1

2mQR¬ + 12mPQ¬ - 1

2mPQ¬ - 12m RP¬ (Arc Add. Postulate and Distr. Prop.); m∠1 + m∠2 = mQR¬ (Distr. Prop.) 19. c = b - a 21. x ≈ 10.7, y = 10 23. x ≈ 10.9, y ≈ 2.325.


A

O C

B

D

E

Given: < BA

> is tangent to } O at A;

<BC

> is tangent

to } O at C . Prove: m∠ABC = 1

2 (mADC¬

- m AC¬

) Proof: 1. Draw AC (2 pts. determine a line.) 2. m∠ACE = 12 m A D C¬ (The measure of an ∠ formed by a tangent and a chord is half the measure of the intercepted arc.) 3. m∠ACE = m∠ABC + m∠BAC (Ext. ∠ Thm.) 4. m∠BAC = 12 mAC

¬ (The measure of an ∠

formed by a tangent and a chord is half the measure of the intercepted arc.) 5. 12 mA D C¬ = m∠ABC + m∠BAC (Subst.

Prop. of = ) 6. 12 mA D C¬ = m∠ABC + 12 mAC ¬

(Subst. Prop. of = ) 7. m∠ABC = 12 mA D C¬

- 12 mAC ¬

(Subtr. Prop. of = )

8. m∠ABC = 12 (mADC¬

- mAC ¬

) (Distr. Prop.)


A

O

C

D

B

Given: } O with tangent < BA

> and secant

< BC

>

Prove: m∠ABC = 12 (m AC¬- mDA

¬)

Proof: 1. Draw AD. (2 pts. determine a line.)

2. m∠DAB = 12 m AD¬

(The measure of an ∠ formed by a tangent and a chord is half the measure of the intercepted arc.) 3. m∠ADC =12 m AC¬

(The measure of an inscribed ∠ is half the measure of its intercepted arc.) 4. m∠ABC =m∠ADC - m∠DAB (Subtr. Prop. of = and Ext.

∠ Thm.) 5. m∠ABC = 12 m AC¬- 1

2 m AD¬

(Subst.

Prop. of = ) 6. m∠B = 12 (m AC¬ - m AD

¬) (Distr. Prop.)


33


27.


X

A

Y

B

ZC

O

Given: Equilateral △ABC is inscribed in }O; XY, YZ, and XZ are tangents to }O. Prove: △X Y Z is equilateral. Proof: m AB

¬= mBC¬= m AC

¬= 120, since chords AB, BC, and CA are all ≅. So the measures of ∠X, ∠Y, and ∠Z are 12 (240 - 120) = 60, and △X Y Z is equiangular, so it is also equilateral.

29. 112

Topic Review pp. 511–5131. a secant of 3. tangents to 5. 20 7. 120 9. 2∶1 or 21 11. 4.5 13. a = 80, b = 40, c = 40, d = 100 15. a = 118, b = 49, c = 144, d = 98 17. 37 19. 6.5

TEKS cumulative practice pp. 514–5151. C 3. C 5. D 7. C 9. 1.24 11. 22.5 13. 7.5 15. DE = 2(1 + 2)2 + (1 - 4)2 ≈ 4.2 EF = 2(-2 - 4)2 + (4 - 7)2 ≈ 6.7 FD = 2(4 - 1)2 + (7 - 1)2 ≈ 6.7 Perimeter is about 17.7 units. 17. 30 19a.

BH = BK , CK = CJ, DJ = DI, and AH = AI by Theorem 12-3. Also AH = HB since HI¬ = HK¬, and likewise BK = BC , etc. Therefore ABCD is a rhombus, since all its sides are congruent. Finally, since mHK¬ = 90 and m∠B = 90, ABCD is a square. Therefore the new figure is a square. b. 2 c. Yes, because the tangent lines at the vertices of the inscribed polygon form a new polygon with equivalent side lengths.

Topic 13Activity Lab 13-1 and 13-2 pp. 518–5191. 9; 5 3. A = b # h; Explanations may vary. Sample: A ▱ can be transformed into a rectangle with the same base and height. 5. The bases are the same; the

height of the ▱ is half the height of the △; 12h 7. a rotation of 180° around point M or point N 9. 12h 11. Yes; no; explanations may vary. Sample: If you rotate the small △ about point A, the image of the small △ and the bottom part of the original △ form a ▱. 13a. Rotate the △ by 180° about point N. b. Area = 1

2(b1 + b2)h

Lesson 13-1 pp. 523–5251. 11.2 units 3. 16 8

13 units 5. 6 units2 7. 12 units2 9. 3 units2 11a. 1390 ft2 b. Find the entire area and subtract the areas for flowers. c. (50)(31) - 2[1

2 (10)(16)] = 1550 - 160 = 1390 ft2

13. 4 in. 15. 18 in.; 12 in. 17. 525 cm2 19. You can use mental math to find the area of each rectangle and add the areas together: 136 yd2.

21a.

hsm11gmte_1001_t14614

y

x

x 2

y 0O

2

4

4y x 31

2

b. 16 units2 23. 28 units2 25. 20 units2 27. The area does not change; the height and base length AB do not change. 29. 126 m2 31. B 33. No; the sum of the two shorter legs is 6 + 4. By the Triangle Inequality Thm., that sum must be greater than the length of the third side of the triangle. Since 6 + 4 6 11, a triangle with sides 6, 4, and 11 is not possible.

Lesson 13-2 pp. 529–5311. 15 units2 3. 18 units2 5. 30 ft2 7. 72 m2 9. 18 m2 11. 1200 ft2 13. 24 m2 15. 144.5 cm2 17. 150 cm2 19. Analyze the given information: The figure is composed of a trapezoid and an isosceles triangle. Formulate a plan: You can find the height of the triangle using the Pythagorean theorem, which will help you find the area of the triangle. Then you can find the area of the trapezoid and add the two areas together. Determine and justify the solution: The height of the triangle is 5 ft, because 52 + 122 = 132. The area of the triangle is 12

# 5 # 24 = 60. The area of

the trapezoid is 12 (24 + 18) # 26 = 546. The area of

the composite figure is 60 ft 2 + 546 ft 2 = 606 ft 2. Evaluate the problem-solving process: Divide the figure in a different way and confirm that you get the same total area. For example, extend the altitude of the isosceles triangle until it meets the other base of the trapezoid, dividing the composite figure into a trapezoid with bases 26 ft and 31 ft and height 12 ft, a trapezoid with bases 6 ft and 12 ft and height 26 ft, and one right triangle with base 12 ft and height 5 ft. The area of the figure is 342 ft2 + 234 ft2 + 30 ft2, or 606 ft2. 21. height; 18 cm; bases: 12 cm and 24 cm

geom12_gm_c12_csr_t0013.ai

H

B

A

D

C

K

JI


34


23. About 35.4 cm2 25. A27.

Lesson 13-3 pp. 536–5381. m∠1 = 120, m∠2 = 60, m∠3 = 30 3. m∠7 = 60, m∠8 = 30, m∠9 = 60 5. 12,080 in.2 7. 2192.4 cm2 9. 27.7 in.2 11. 210 in.2 13. 38413 in.2 15. 9.7 ft 17. 60013 m2 or about 1039.2 m2 19. m∠1 = 36, m∠2 = 18, m∠3 = 72 21. 11213 ft2 23. (980 + 14713 ) m2 25. 19.0; 26.0 27. The apothem is # to a side of the pentagon. Two right triangles are formed with the radii of the pentagon. The triangles are ≅ by HL. So the angles formed by the apothem and radii are ≅ because corresp. parts of ≅ triangles are ≅. Therefore, the apothem bisects the vertex angle. 29a. (2.8, 2.8) b. 5.6 units2 c. 45 units2 31. F 33. (2, 213) and (2, -213), or equivalent decimal approximations (2, 3.464) and (2, - 3.464); The length of each side of the triangle is 4 units. The third vertex must lie on the altitude of the triangle, which is a point on the line x = 2 that has x-coordinate 2. Using the Distance Formula, 2(2 - 0)2 + (y - 0)2 = 4; 4 + y2 = 16; y2 = 12; y = {112; y = {213.

Lesson 13-4 pp. 543–5451. 1 : 2; 1 : 4 3. 4 : 3; 16 : 9 5. 7 : 3; 7 : 3 7. 1 : 10;

1 : 10 9. x = 212 cm, y = 312 cm 11. x = 8133 cm,

y = 413 cm 13. x = 8 cm, y = 12 cm 15. 54 m2 17. 309 m2 19. 252 m2 21. 1280.18 ft2 23. The perimeter increases by 4s to be 6s + 2t, and the area is multiplied by 3 to be 3st. 25a–c. Answers may vary. Samples are given.a.

b. 96 mm; 336 mm 2 c. 457 yd; 7619 yd 2 27a. 613 cm 2 b. 54 13 cm 2; 13.5 13 cm 2; 96 13 cm 2 29. 91.2 ft 31. 26

3 33. 19.5

Lesson 13-5 pp. 548–5501. 173.8 cm2 3. 21.4 ft2 5. 1200 cm2 7. 408.3 mm2 9. 27.7 m2 11. 7554.0 m2 13. 128.1 mm2 15a. 72 b. 36 c. 8.1 in. d. 11.8 in. e. 58.8 in. f. 237.8 in.2 17. 5523 yd2 19. 17.6 ft, 21.4 ft2 21. 6.2 mi, 3 mi2 23. about 925.8 cm2 25. area of Hexagon A = area of Hexagon B 27. area of Decagon A = 0.01 ~ (area of Decagon B) 29. 16213 ft2 or about 280.6 ft2 31. 24 in.2 33. 320 ft 35. 17 37. 47.2

Topic Review pp. 551–5531. composite figure 3. radius 5. 10 m2 7. 30 ft2 9. 30 ft2 11. 96 ft2 13. 256 ft2 15. 913 in.2 17. 240013 cm2

19.

20.8 in.2

21.

127.3 cm2

23. 9∶4 25. 4∶1 27. 73.5 ft2 29. 124.7 in.2 31. 331.4 ft2 33. 100.8 cm2

TEKS cumulative practice pp. 554–5551. A 3. B 5. B 7. B 9. B 11. 363.3 13. 9891 15. 1

20 17. 6 19. 516

21.

23. Yes; m∠C = 90 and sin 90° = 1, so 12 bh = 14 cm2 and 12ab (sin C ) = 14 cm2.

25. Yes; by the Distance Formula, AB = CD = 10, BC = 4, AD = 16, AF = GH = 5, FG = 2, and AH = 8. Since AB

AF = 105 = 2

1, BCFG = 4

2 = 21,

CDGH = 10

5 = 21, AD

AH = 168 = 2

1, corresp. sides are proportional. By the Refl. Prop. of ≅, ∠A ≅ ∠A. The slopes of AD, BC, and FG = 0 and the slopes of CD and GH = - 43, so AD } BC } FG and CD } GH because } lines have the same

hsm11gmte_1002_t14618

hsm11gmte_1002_t14619

hsm11gmte_1004_t14621

33 mm21 mm

16 mm

42 mm


4 in.

hsm11gmte_1007_t14637

7 cm

hsm11gmte_10cu_t14638

hsm11gmte_10cu_t14639


35


slope. If lines are } , then corresp. ⦞ are ≅, so ∠AFG ≅ ∠B and ∠D ≅ ∠GHA. Also, ∠C ≅ ∠G by the Polygon-∠-Sum Thm. Since corresp. sides are proportional and pairs of corresp. ⦞ are ≅, ABCD ∙ AFGH.

Topic 14Lesson 14-1 pp. 563–5641. 8 3. 12 5. 5 7. two concentric circles 9. rectangle 11. Answers may vary. Sample: a pentagon and a rectangle 13a.


b.


15.


17.


rectangle triangle19. sphere 21. 60 23. A circle 25.


27a. A(3, 0, 0), B(3, 5, 0), C(0, 5, 0), D(0, 5, 4) b. 5 units c. 3 units d. 4 units 29. J 31. F

Lesson 14-2 pp. 569–5711. 1726 cm2

3. (80 + 3212) in.2, or about 125.3 in.2


4 in.

4 in.

8 in.

4 V2 in.

4 V2 in.

5. 220 ft2 7. 1121 cm2 9. 170 m2 11. 101.5p in.2 13. 4080 mm2 15a. 94 units2 b. 376 units2 c. 4∶1 d. 438 units2; 1752 units2; 4∶1 e. The surface area is multiplied by 4. 17. 20 cm 19a. A(3, 0, 0), B(3, 5, 0), C(0, 5, 0), D(0, 5, 4) b. 5 units c. 3 units d. 4 units e. 94 units2 21a. The lateral area is doubled. b. The

surface area is more than doubled. c. If r doubles, S.A. = 2p(2r)2 + 2p(2r)h = 8pr2 + 4prh = 2(4pr2 + 2prh). So the surface area 2pr2 + 2prh is more than doubled. 23. (182p + 232) cm2 25. 6.1 27. 7.

Lesson 14-3 pp. 577–5781. 31 m2 3. 47 cm2 5. 33p ft2 7. PT is a leg in each of rt. triangles PTA, PTB, PTC, and PTD. Since PA, PB, PC , and PD are each the hypotenuse in those rt. triangles, PT must be shorter than PA, PB, PC , and PD. 9. 4 in. 11. 58 m2 13. 45 m2 15. Slant height; slant height; the slant height is shorter because it is one leg of a rt. △ with the lateral edge as the hypotenuse, and it is steeper because it rises the same vertical distance for a shorter horizontal distance. 17. Answers may vary. Sample: 740 cm2

10 cm

8 cm

12 cm

19. L.A. = 10015 cm2 S.A. = (10015 + 100) cm2

21. A 23. B 25. A dilation with center (0, 0) takes P (x, y) to P′(nx, ny). If A(3, 4) S A′(15

2 , 10), then n # 4 = 10, so n = 2.5. The other points will be B′(17.5, 12.5), C′(15, 2.5), and D′(-5, -10).

Lesson 14-4 pp. 582–5831. about 280.6 cm3 3. 180 m3 5. 22.5 in.3 7. 3445 in.3 9. 5 in. 11. 96 ft3 13. 3 cm 15a. 88,757 ft3 b. 663,901 gal 17a. 1044.1 cm2 b. 553.6 cm2 c. 1481.3 cm3 19. 98.2 in.3 21. cylinder with r = 4 and h = 2; 32p units3 23. cylinder with r = 5 and h = 2, and a hole of radius 1; 48p units3 25. H

Activity Lab 14-5 p. 5841. The areas are equal. 3. about 3 pyramids 5. V = 1

3Bh; the volume of a cube is Bh. The volume of the pyramid is 13 the volume of the cube, or 13Bh. 7. The heights are = . 9. 13

Lesson 14-5 pp. 587–589

1. 443.7 cm3 3. 562.9 ft3 5. 163 p ft3; 17 ft3

7. 4p m3; 13 m3 9. 123 in.3 11. 10,368 ft3 13a. 79,000 m3 b. 202

3 m, or about 20.7 m 15a. 120p ft3 b. 60p ft3 c. 240p ft3 17a. The frustum has a volume that is the difference between the volumes of the entire cone and the small cone. The frustum has volume V = 1

3pR2H - 13pr2h or


29 cm

19 cm

6.5 cm

6.5 cm


36


13p(R2H - r2h). b. about 784.6 in.3. 19. cone with r = 4 and h = 3; 16p 21. cylinder with r = 4, h = 3, with a cone with r = 4, h = 3 removed from it; 32p 23a. 47.1 m b. 176.7 m2 c. 389.6 m3 25. G

Lesson 14-6 pp. 593–5951. 900p m2 3. 1024p mm2 5. 500

3 p ft3; 524 ft3

7. 11252 p in.3; 1767 in.3 9. 441p cm2 11. 62 cm2

13. 451 in.2 15. 130 cm2 17. Answers may vary. Sample: sphere with r = 3 in., cylinder with r = 3 in. and h = 4 in. 19. 0.9 in. 21. 1.7 lb 23. An infinite number of planes pass through the center of a sphere, so there are an infinite number of great circles. 25. 288p cm3 27. 1125

2 p mi3 29. 38,792.4 ft3

31. 22p cm2; 463 p cm3 33. D 35. B 37. C

Technology Lab 14-7 p. 5961. The ratio of volumes is the scale factor cubed. 3. The ratio of volumes is the scale factor cubed. 5. The ratio of surface areas is the scale factor squared.

Lesson 14-7 pp. 601–6031. S.A. is multiplied by 100, V is multiplied by 1000. 3. S.A. is multiplied by 6.25, V is multiplied by 15.625. 5. yes; 3 : 2 7. 5 : 6 9. 240 in.3 11. 175 in.2 13. 6000 toothpicks 15. Yes; explanations may vary. Sample: The ratio of the radii of the spheres equals the ratios of all other corresponding linear dimensions. 17. about 1000 cm3 19a. 100 times b. 1000 times c. His weight is 1000 times the weight of an average person, but his bones can support only 600 times the weight of an average person. 21. 7500 23. 18 25. 59

Topic Review pp. 604–6071. sphere 3. cross section 5. Answers may vary. Sample:

7. 8 9. a circle 11. 36 cm3 13. 208 in.2 15. 32.5p cm2 17. about 185.6 ft2 19. about 50.3 in.2 21. 84 m3 23. 410.5 yd3 25. S.A. = 314.2 in.2; V = 523.6 in.3 27. S.A. = 50.3 ft2; V = 33.5 ft3 29. 904.78 cm3 31. 8.6 in.2 33. 27∶64

TEKS cumulative practice pp. 608–6091. B 3. B 5. D 7. D 9. C 11. 124 13. 77 15. 484

17. Answers may vary. Sample:

19. 1 gal

21. C = 2pr so 3p = 2pr and r = 32. Using

V = 43 pr 3, then V = 4

3 p(32)3 = 9

2 p cm3.23. Answers may vary. Sample: 6 in. by 7 in. by 8 in.; cylinder: V = 108p in.3 ≈ 339.3 in.3; rectangular prism: V = 336 in.3.

Topic 15Lesson 15-1 pp. 614–6161a. about 0.297 or 29.7% b. about 59 3. 1

11

5. 411 7. 10

11 9. 711 11a. 5

30 or 16 b. 2530 or 56 13. 30%

15. 1265 17. 55

65 or 1113

19. yellow, round: about 1133 yellow, wrinkled: about 378 green, round: about 378 green, wrinkled: about 126

21. and 23. Ask your teacher to check your work. 25. 3 : 7 27. H

Lesson 15-2 pp. 619–6211. 12 3. 35 5. 1 7. 14, or 25% 9. 13, or about 33%

11. 1625, or 64% 13. p4 , or about 78.5% 15. 16

49, or

about 33% 17. 1, or 100% 19. 12 units 21. 35; The

probability that it is in any one of the 5 sectors is 15, and 3 sectors are not shaded. 23. about 5.6% 25. Ask your teacher to check your work. 27a. 1.4% b. About 72 coins; the probability of winning on each toss is 0.014 and (71)(0.014) 6 1 6 (72)(0.014). 29a. yes b. no c. 23 31. G 33. G

Lesson 15-3 pp. 626–6271. 800 3. 720 5. 151,200 7a. 12,144 b. 5,100,480 9. 21 11. 2

28 13. 24 ways 15. 2,496,144 17. 1792

19. yes; for example, when r = 0 or 1 21. (n - 1)! 23. H 25. 96

Lesson 15-4 pp. 631–6321. independent 3. independent 5. 1

40 7. 320 9. 25

11. 1315 13. 11

20 15. 34 17. 164 19. 18 21a. 3

10 b. 29


4 in.

5 in.5 in.

5 in.

5 in.

8 in.

hsm11gmte_11ct_t14466


37


c. 115 23. Yes; events A and B are mutually exclusive

when P(A and B) = 0. 25. A 27. 40p cm

Lesson 15-5 pp. 635–6361. 0.14 3. 0.16 5. 7

23 7. The sample space is limited to cats only. 9. 12 11. 18 13. 14 15a. 7

16 b. 716 c. Since

P(B) and P(B 0 A) are the same, the occurrence of event A has no effect on the probability of event B occurring. A and B are independent events. d. Since P(D) and P(D 0 C) are the same, the occurrence of event C has no effect on the probability of event D occurring. D and C are independent events. e. Since P(F) and P(F 0 C) are not equal, the occurrence of event C does have an effect on the probability of event F occurring. F and C are dependent events. 17a. 6%, 9%, 31%, 40%, 85% b. 1

10 c. 940 19. 6 21. 7.5

Lesson 15-6 pp. 640–6411. 19

31 3. 481 5. 19 7. 4

27 9. 136 11. 1

36 13. 112 15. 0

17. 20% 19. 24% 21. 37 23. 1142 25. C

Topic Review pp. 642–6451. permutation 3. dependent 5. 3

10 7. 0 9. 512

11. 38, or 37.5, 13. 12, or 50, 15. 120 17. 840

19. 6720 21. 36 23. 78,624,000 25. 0.115

27. 716 29. 7

11 31. 1926 33. 57 or about 71.4,

TEKS cumulative practice pp. 646–6471. B 3. A 5. B 7. A 9. C 11. D 13. B 15. C

AddiTionAL pRAcTicETopic 1 pp. 648–6491. true 3. true 5. true 7. 10 sets 9. 12 11. 3 13. 27

15. 6 17. 60, 120 19. 31 21. ∠JQN, ∠KQP 23. ∠KQP 25. ∠PQM, ∠LQN 27. 50, 130 29. 31.

33.

Topic 2 pp. 650–6511. 37, 42 3. 36, 49 5.

7. Answers may vary. Sample: Multiplying a number by 1 does not result in a larger number. 9. Hypothesis: You can predict the future. Conclusion: You can control the future. 11. If you can control the future, then you can predict the future. If Dan needs glasses, then Dan is nearsighted. 13. Conditional: If two numbers are even, then their product is even; true. Converse: If the product of two numbers is even, then the two numbers are even; false. Inverse: If two numbers are not both even, then their product is not even; false. Contrapositive: If the product of two numbers is not even, then the two numbers are not both even; true. 15. yes 17. yes 19. If a person can be the president of the United States, then the person is a citizen of the United States; true. If a person is a citizen of the United States, then the person can be president of the United States; false. 21. Linda’s band will win $500. 23. You will have a nightmare; Law of Detachment. 25. Symmetric Property of Congruence 27. Transitive Property of Congruence 29. x = 15 31. counterexample:

Topic 3 pp. 652–6531. ∠5 and ∠7; d, e, and b 3. ∠7 and ∠8; b, c, and d 5. m∠1 = 125; Corresponding Angles Postulate, m∠2 = 55; Same-Side Interior Angles Theorem 7. m∠1 = 82; Same-Side Interior Angles Theorem, m∠2 = 82; Alternate Exterior Angles Theorem 9. x = 62; (3x - 44) = 142, (x - 24) = 38 11a. Vertical Angles Theorem b. Same-Side Interior Angles Postulate c. Same-Side Interior Angles Postulate d. Congruent Supplements Theorem e. Transitive Property of Congruence f. Vertical Angles Theorem g. Transitive Property of Congruence 13. a # / and a } b means that / # b since a line # to one of two } lines is # to the other (Thm. 3–10). Since / # b and / } m, this means that b # m for the same reason. 15. x = 65 17.


38


19. y - 2 = -52 (x - 4) or y + 3 = -5

2 (x - 6)

21. y + 5 = -1(x - 3) or y - 3 = -1(x + 5)

23. y = x + 2 25. y - 4 = x or y = x + 4

27. y + 8 = -15 (x - 5) or y = -1

5 x - 7

29. y = 12(x - 4) or y = 1

2 x - 2 31. neither

33. Euclidean geometry

Topic 4 pp. 654–6551. ∠G 3. ∠T 5. △ATS 7. △ATS 9. Yes; corresponding sides and corresponding angles are congruent. 11. Yes; corresponding sides and corresponding angles are congruent. 13. ∠T ≅ ∠S, ∠Y ≅ ∠W, TY ≅ SW; ASA 15. Not enough information; You need PA ≅ RL. 17. ZP bisects XY means that XM ≅ YM. It is given that PX ≅ PY and PM ≅ PM by the Reflexive Property of Congruence. Thus, △PXM ≅ △PYM by SSS and ∠XPM ≅ ∠YPM by definition of Congruence, PZ ≅ PZ by the Reflexive Property of Congruence, so △PXZ ≅ △PYZ by SAS. 19. ∠1 ≅ ∠2 is given. ∠ABP ≅ ∠DCP by the Congruent Supplements Theorem. ∠3 ≅ ∠4 and AP ≅ DP are given. △ABP ≅ △DCP by AAS. 21.

<LO

>}<MN

>, so

∠OLN ≅ ∠MNL. LN ≅ LN by the Reflexive Property of Congruence. Since LO ≅ MN, △MLN ≅ △ONL by SAS, and ∠MLN ≅ ∠ONL by CPCTC (Corresponding Parts of Congruent Triangles are Congruent). 23. ∠1 ≅ ∠2, ∠3 ≅ ∠4 (both given), and SQ ≅ SQ (Reflexive Property of Congruence), so that △PQS ≅ △RQS by AAS and PQ ≅ RQ by CPCTC. Point M is the midpoint of PR, so PM ≅ RM. Also, QM ≅ QM by the Reflexive Property of Congruence, so △PQM ≅ △RQM by SSS. ∠PMQ ≅ ∠RMQ by CPCTC. 25. 25 27. x = 45; y = 90; z = 45 29. AP ≅ BP and PC ≅ PD (both given). ∠APC ≅ ∠BPD as vertical angles, so △APC ≅ △BPD by SAS. ∠ACP ≅ ∠BDP because CPCTC. ∠PCD ≅ ∠PDC by the Isosceles Triangle Theorem. m∠ACP + m∠PCD = m∠BDP + m∠PDC by the Addition Property of Equality, so m∠ACD = m∠BDC by the Angle Addition Postulate and substitution. QD ≅ QC by the Converse of the Isosceles Triangle Theorem, and △QCD is isosceles by the definition of isosceles triangle. 31. △RQM ≅ △QRS; SSS 33. ∠1 ≅ ∠2 (given), so MC ≅ MD by the Converse of the Isosceles Triangle Theorem. AM ≅ BM by the definition of midpoint. MC # AC means that ∠ACM is a right angle and △ACM is a right triangle. MD # BD means that ∠BDM is a right angle and △BDM is a right triangle. △ACM ≅ △BDM by HL.

Topic 5 pp. 656–6571a. 1.4 b. (5

2 , 12) 3a. 19.1 b. (-12 , 3) 5. 32

7. QR } MN, QS } LN, RS } LM 9. 37 11. 5

13. 52 15. It is given that A, B, and C are equidistant from P and Q. By the Converse of the Bisector Theorem, A, B, and C are on the same line, namely the perpendicular bisector of PQ. 17. RZ 19. (2, 2)

21. 5, 10 23. Median; the line intersects A and bisects a side at B. 25. Perpendicular bisector; the segment is perpendicular to and bisects a side at B, but it does not intersect a vertex. 27. (17

4 , 2) 29. Assume △ABC is not a right triangle. 31. I and III 33. ∠B is not an acute angle. 35. Yes; 9 + 11 7 15 37. PQ, QR, PR 39. 11 6 x 6 29

Topic 6 pp. 658–6591. 2520 3. x = 110; y = 102, z = 82 5. interior: 162; exterior: 18 7. x = 26; y = 11 9. 22 11. Since PQRS is a parallelogram, PS } QR and PQ } SR. So PA } SB. Since QDCA is a parallelogram, AB } QR. Thus, PS } AB because two lines parallel to the same line are parallel. PABS is a parallelogram by the definition of parallelogram. AP ≅ BS since opposite sides of a parallelogram are congruent. 13. DG = 3.2, EG = 2.05 15. x = 2, y = 30 17. parallelogram; m∠1 = 45; m∠2 = 45; m∠3 = 80; m∠4 = 55 19. By the Converse of the Isosceles Triangle Theorem and given that PA = PB, it follows that PD = PA = PB = PC . Thus, the diagonals of ABCD bisect each other (so ABCD is a parallelogram) and are congruent (by the Segment Addition Postulate), so ABCD is a rectangle. 21. x = 16 23. x = 8 25. m∠1 = 45; m∠2 = 74

Topic 7 pp. 660–6611. Use the Distance Formula: AB = 2(3 - (-3))2 + (2 - 1)2 = 136 + 1 = 137 BC = 2(5 - 3)2 + (-1 - 2)2 = 14 + 9 = 113 CD = 2(5 - (-1))2 + (-1 - (-2))2 = 136 + 1 = 137 DA = 2(-1 - (-3))2 + (-2 - 1)2 = 14 + 9 = 113 ABCD is a parallelogram because its opposites sides are congruent. 3. isosceles;

5. scalene;


39


7. rectangle;

9. trapezoid;

11. Yes, the triangles are congruent. Use the Distance Formula to find the lengths of the sides of each triangle. Triangle in Quadrant 2:

s1 = 2(-4 - (-3))2 + (4 - 1)2 = 11 + 9 = 110

s2 = 2(-4 - (-1))2 + (4 - 2)2 = 19 + 4 = 113

s3 = 2(-3 - (-1))2 + (1 - 2)2 = 14 + 1 = 15

Triangle in Quadrant 4:

s1 = 2(1 - 2)2 + (-4 - (-1))2 = 11 + 9 = 110

s2 = 2(2 - 4)2 + (-1 - (-4))2 = 14 + 9 = 113

s3 = 2(2 - 4)2 + (-1 - (-2))2 = 14 + 1 = 15

Because the corresponding sides of the triangles are equal in measure, the triangles are congruent. 13. (6, 3) and (3, 6) 15. (0, 0), (0, m), ( - m, m), ( - m, 0) 17. D(0, b); S( - a, 0) 19. D( - a, c), S(b, 0) 21. Use the Midpoint Formula to find coordinates of M and N. M(2a + 0

2 , 2b + 02 ) = (a, b)

N(2c + 2d2 , 2b + 0

2 ) = (c + d, b)

Use the Distance Formula:

RV = 2(2d - 0)2 + (0 - 0)2 = 24d2 = 2d

ST = 2(2c - 2a)2 + (2b - 2b)2

= 2(2c - 2a)2 = 2(c - a)

MN = 2(c + d - a)2 + (b - b)2

= 2(c + d - a)2 = c + d - a

Find 12(RV + ST ): 12 (RV + ST ) = 1

2(2d + 2(c - a)) = d + c - a = MN

Topic 8 pp. 662–6631. E 3. (x, y) S (x + 4, y - 2) 5. P′(-15, -11),Q′(-11, -6), R′(-4, 1) 7. (3, 0)

9.

11. rectangle 13.

15.

17. 180° rotational symmetry; no reflectional symmetry19. 120° rotational symmetry; reflectional symmetry:

21. rotation of 180° about (1, 0) 23. (-6, 0)

25. (87, 2) 27. (6

5, 4135 )

29. A′(6, 2), B′(-4, 8), C′(-2, -4)

Topic 9 pp. 664–665

1. x = 803 ; y = 6; z = 16

3 3. x = 30; y = 4 5. No. A square and a rhombus are not necessarily similar, since a square must have four congruent angles, but a rhombus only needs two pairs of congruent angles. 7. Answers may vary. Sample: a translation right 12 units followed by a dilation with scale factor 12 and center T′. 9. Yes; △QCT ∙ △MCP by SAS∙ . 11. No 13. 24 feet 15. 20 17. x = 15; y = 2; z = 215 19. Place △ABC in the coordinate plane with A(-a, 0) B(b, 0), C(0, 1ab), and D(0, 0). Slope of

AC = 1aba . Slope of BC = - 1ab

b . The product 2aba # - 2ab

b = -abab = -1, so AC # BC . 21. x = 117

10

23. 4.5 mi

Topic 10 pp. 666–6671. 15 3. 315 5. 2121 7. 188 ft 9. acute 11. obtuse 13. acute 15. 612 cm, 612 cm 17. 813 in., 16 in. 19. 413 units, 813 units


40


21. x = 29 23. x = 7.2 25. x = 62 27. 53.2 ft 29. 653 ft 31. 78 ft 33. 2000 ft

Topic 11 pp. 668–6691a. 6p cm b. 2p cm 3a. 18p cm b. 92p cm 5a. 15p in. b. 45

8 p in. 7. 482 rotations

9. 3p2 , 4.71 11. 35p18 , 6.11 13. p18, 0.17

15. 115° 17. 315° 19. 52° 21a. 7.5 min or 7 min 30 s b. 1.5 h or 1 h 30 min 23. (12p - 913) in.2 25. (4p - 8) m2 27. (16p

3 - 413 ) mm2

29. x2 + y2 = 16 31. (x - 9)2 + (y + 3)2 = 49 33. (-1, -3), r = 1;

35a. (x - 113)2 + (y - 215)2 = 852 b. (x - 113)2 + (y - 215)2 = 1702

Topic 12 pp. 670–6711. x = 65 3. x = 6 5. no; 92 + 142 ∙ 172 7. Tangents to a circle from a point outside the circle are congruent, so AS = AP, BP = BQ, CQ = CR, and DR = DS. By the Segment Addition Postulate and various Properties of Equality, AB + DC = AP + BP +DR + CR = AS + BQ + DS + CQ = BQ + CQ + AS +DS = BC + AD. 9. x = 5.2 11. yes; Each side of the polygon is a chord of the circle, and the perpendicular bisector of any chord contains the center of the circle. 13. a = 154; b = 76 15. a = 105; b = 100 17. a = 120; b = 60; c = 75 19. ∠A ≅ ∠D since they both intercept BC¬. ∠BPA ≅ ∠CPD since they are vertical angles. △APB ∙ △DPC by AA ∙. 21. x = 70 23. x = 5.6 25. x ≈ 5.6; y ≈ 11.9 27. 18 ft

Topic 13 pp. 672–6731. 80 cm2 3. 143 ft2 5. 120 yd2 7. 72 cm2

9. 254 13 mm2 11. 48 cm2 13. 3500 ft2 15. 7513

128 in.2

17. 3 : 4; 9 : 16 19. 560 in.2 21. 30.1 ft2 23. 43.2 cm2 25. 65.4 mm2 27. 29.5 ft2

Topic 14 pp. 674–6751a. 6; A, B, C, D, E, F b. 9; AB, AC, AD, BC, BE, CF, DE, DF, EF c. 5; △ABC, △DEF, ▱ABED, ▱ACFD, ▱BCFE 3. equilateral triangle; 7 + 7 = 12 + 2 5. 28p cm2; 36p cm2 7. 360 m2; (360 + 4813) m2 9. 55.0 ft2; 83.2 ft2 11. 19.3 cm2; 21.3 cm2 13. 16 mm3 15. 540 cm3 17. 49.3 in.3 19. 5.7 in.3 21. 36p ft3,

113 ft3; 36p ft2, 113 ft2 23. 4p3 ft3, 4 ft3; 4p ft2,

13 ft2 25. 243p2 m3, 382 m3; 81p m2, 254 m2

27. 32p3 in.3 29. 9.62 ft2; 2.81 ft3 31. yes; 3 : 4

33. ≈ 166.4 cm3

Topic 15 pp. 676–677

1. 16 3. 112 5. 16 7. 34 9. 19

30 11. 14 13. 1 - p4

15. 38 17. 1136 19. 12 21. 120 23. combination;

125,970 25. dependent 27. 1924 29. 25 31. 0.375, or

37.5% 33. 0.175, or 17.5% 35. 0.45, or 45%

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