Geometry Student Text and Homework Helper ANSWERS · 2019. 6. 3. · 1 ODD-NUMBERED Geometry |...

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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. 123 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 = 12m∠AOC since OB is the angle bisector of ∠AOD. Thus, m∠BOC = 12 (4x + 12) = 2x + 6. b. x = 12. Sample answer: m∠COD = 12m∠AOD since OC is the angle bisector of ∠AOC. Thus, 4x + 12 = 12(120) = 60. If 4x + 12 = 60, then x = 12.

Topic 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 - 13 e, then the flavor of the quark is strange; a given quark has rest energy 540 MeV and a charge of - 13 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. -32 5. undefined7. y

xO 2

2

�2

�4

hsm11gmte_0307_t11662

9.

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y2

�2x

O

11. y + 1 = -3(x - 4) 13. y - 6 = - (x + 2) or y - 3 = - (x - 1) 15. y - 2 = -12(x - 6) or y - 4 = -12(x - 2) 17. horizontal: y = 7; vertical: x = 4 19. Answers may vary. Sample: x = 5, y = 6, y = 65x, 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 2472 = 0.3 whichis less than the maximum slope of 411 = 0.36. 27. y = -x + 2 29a. - 133; 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 = 52x b. y - 5 = -52(x - 2) or y = -

52x + 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 -13, 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 -

12, and the slope of /2 is

-12, 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 = -32(x - 6) 11. y - 4 = 12(x - 4) 13. Yes; both slopes are -1, so the lines are }. 15. No; the slope of the first line is -34, 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 = -34, 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 = 32, 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 = -12x + 100 b. (100, 50) 31. y - 5 = 13(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) ory + 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 (72, 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. ( -32, 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 + 35 (y2 - y1))gives G(-7 + 35 (8), 7 + 35 (-4)) or G(-115 , 235 ) .Number sense can help you add and subtract the

fractions given. The coordinates are G(- 115 ,235 ) .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 units41. 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 ofparallel lines are congruent. 5. Slope of HJ = 2 - 04 - 2 = 1and slope of EF = 6 - 25 - 1 = 1. The slopes are equal,so HJ } EF . HJ = 2(2 - 4)2 + (0 - 2)2 = 28 = 222and 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 + 82 ) = 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 56). 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 ) = (72 , 32)

midpoint of QS = (5 + 22 , -4 + 72 ) = (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 -53. 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 = 1215; 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 (12n ), 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 -32, 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(-b2, -b2) , T (-b2, b2) , W (b2, b2) , Z (b2, -b2) 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 + 2b2 ) = (-a, b); N = (2c + 02 , 0 + 2b2 ) = (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 - 0b - 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 = - c22ab - b2 = -c2c2 = -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 = cd , 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 - ba - 0 =

ba , slope of

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

b - 02a - 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 = rp + q x

equation of <BN

>: y = 2r2q - p (x - 3p)

equation of <CL>: y = rq - 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 =232(3p + 3q)2 + (3r)2

= 549[(3p + 3q)2 + (3r)2] = 549[(3p + 3q)2] + [49 (3r)2] = 549[(9p2 + 18pq + 9q)2] + [49 (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′ = -32, 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

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