Geometry Student Text and Homework Helper ANSWERS · 2019. 6. 3. · Geometry | Student Text and...

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1

EvEn-numbErEd AnswErsGeometry | Student Text and Homework Helper

Topic 1Lesson 1-1 pp. 7–92. E, B, F 4. RS or SR, ST or TS, TW or WT , RT or TR, SW or WS, RW or WR 6.

<RS> 8. plane QUX, plane

QUV 10. plane XTQ, plane XTS 12.

hsm11gmte_0102_t02255.ai

X W

S

RQ

U V

T

14.


X W

S

RQ

U V

T

16. coplanar 18. Sample:


CG

F

DBE

20. Solution: Line is an undefined term. Opposite rays is a defined term. Opposite rays are two rays that share the same endpoint and form a line. 22. always 24. sometimes 26a. one b. one c. A line and a point not on that line are always coplanar; Postulate 1-4 says that through any three noncollinear points there is exactly one plane. Postulate 1-1 says that there is exactly one line that contains two of those three coplanar points. So through any line and a point not on that line, there is exactly one plane, so the line and the point are coplanar. 28. Answers may vary. Sample: 6:00 is the only “exact” time. Other times are about 1:38, 2:43, 3:49, 4:54, 7:05, 8:11, 9:16, 10:22, 11:27, and 12:33. 30a. Answers may vary. Sample: Since the plane is flat, the line would have to curve in order to contain the two points and not lie in the plane, but lines are straight, so the line must also be in plane P. b. One; points A, B, and C are noncollinear. By Postulate 1-4, they are coplanar. Thus, by part (a),

<AB

>

and <BC

> are coplanar.

32.


y

xO

�2 4

4

yes34. 1 36. D 38a. planes ABC, ACD, ABD, BCD b.

<BD

>, <CD

>, <AD

>

Lesson 1-2 pp. 14–152. 9 4. 25 6. 2 8. -2 or 8 10. ED = 10, DB = 10, EB = 20

12a. (2x + 3) - x + (4x - 3), or 5x b. GH = 9, JK = 15 14. -1.5 or - 3

2 16. 30 18. yes

20. no 22. 415 24. x = 1

5 26. D 28. A, B, C, and D

Lesson 1-3 pp. 19–212. ∠ABC, ∠CBA, ∠B, or ∠1 4. 70, acute 6. 110, obtuse 8. Answers may vary. Sample:


H

J

G

10. 75 12. 5.5 14. A 16. 150 18. 100 20. 22.5 22. about 60°; acute 24. about 88°; acute 26. x = 8; m∠AOB = 30, m∠BOC = 50, m∠COD = 30 28. m∠ABC = 45, m∠DBC = 34 30. A 32a. 4 b. 52

Lesson 1-4 pp. 25–262. No, the angles are not formed by opposite rays. 4. ∠DOC , ∠AOB 6. ∠EOC 8. 120; 60 10. 155 12a. 19.5 b. m∠RQS = 43, m∠TQS = 137 c. Answers may vary. Sample: 43 + 137 = 180 14. Both are correct; if you multiply both sides of the equation m∠ABX = 1

2m∠ABC by 2, you get 2m∠ABX = m∠ABC. 16a. x = 11, m∠FGH = 30 b. 30 c. 60 18. The four vertical angles formed by two intersecting lines are all right angles. 20. Yes; they are marked as ≅. 22. Yes; they form a linear pair. 24. y = x 26. J

Lesson 1-5 pp. 30–322.


P T

S Q

P T

QS

Find a segment on <SQ

> so that you can construct

<SP>

as its perpendicular bisector. Then bisect ∠PSQ. 4a. With P as center, draw an arc with radius slightly more than 12 PQ. Keeping that radius, draw an arc with Q as center. Those two arcs meet at two points; the line through those two points intersects PQ at its midpoint. b. Follow the steps in part (a) to find the midpoint C of PQ. Then repeat the process for segments PC and CQ.


2


6. possible


4 cm4 cm

5 cm

8.


B

AC

D

In the angle bisector construction, AB ≅ AC , BD ≅ CD, and AD ≅ AD. Using the statement that two triangles are ≅ if three pairs of sides are ≅, △ABD ≅ △ACD. Since the triangles are ≅, each angle of one △ is ≅ to an angle of the other △. So ∠BAD ≅ ∠CAD and AD

> is the angle bisector of ∠BAC .

10. #; contains the intersection of that line with the plane. 12.


X YAB

14.


D 16.


A B

18.


X

Q Z

20a.


X Y

b. The measure of each angle is 60°. c. Draw an angle congruent to one of the angles of the triangle from part (a) to get a 60° angle. Then construct its angle bisector to get two 30° angles. 22. yes 24. yes 26. J

Technology Lab 1-5 pp. 33–342. Yes. 4. The Construct tool fixes objects so that their relationship does not change. The Draw tool does not fix objects, so that the relationship between drawn objects can be changed. 6a–b. Ask your teacher to check your work.

c. yes d. Answers may vary. Sample: Adjust ∠JKM or ∠MKL until it is 45°.

Activity Lab 1-5 p. 362. no; no sides or vertices 4. no; intersecting sides 6. Sample: CLPK; sides are CL, LP, PK, KC; angles are ∠C, ∠L, ∠P, ∠K 8. octagon, concave 10. pentagon, concave

Topic Review pp. 37–392. perpendicular lines 4. Answers may vary: Sample: <QA

> and

<AB

> 6. Answers may vary. Sample: A, B, C

8. False; they have different endpoints. 10. 12 or 0.5 12. XY = 21, YZ = 29 14. right 16. 14 18. Answers may vary. Sample: ∠ADB and ∠BDF 20. Answers may vary. Sample: ∠ADC and ∠ADE 22. 1524.

hsm11gmte_01cr_t05319

60�

26.


B

TEKS cumulative practice pp. 40–412. H 4. G 6. H 8. F 10. J 12. 6 14. 4.5 16. - 0.25 18. 18 20. Answers may vary. Sample: If several angles have the same vertex, three letters are used to name each angle. 22a. x = 10 b. No; JH = 25 and HK = 23, so JH ≠ HK.

Topic 2Lesson 2-1 pp. 46–482. Find the next square; 36, 49. 4. Multiply the previous term by 12; 1

16, 132. 6. Subtract 3 from the

previous number; 3, 0. 8. the first letter of the months; J, J 10. the presidents of the U.S.; Madison, Monroe 12. state postal abbreviations in alphabetical order; CO, CT 14.

hsm11gmte_0201_t06676

P

B RPB

R

16.

Blue

18. H; the rest of the points form the shape of a heart. 20. star 22. 1 mi 24. Answers may vary. Sample: △ABC with m∠B = 90 26. Answers may vary. Sample: -2 and -3 28. Answers may vary. Sample: 1, 2, 3, 4, 5, . . . and 1, 2, 4, 8, 16, c 30. Answers may vary. Sample: Group the numbers in pairs that sum to 101, so the sum of the integers from 1 to 100 is 100 # 101

2 = 5,050; group the numbers in pairs that sum to n + 1, so the sum of the integers from 1 to n is

n(n + 1)2 . 32a. 1, 3, 6, 10, 15, 21


3


b. 5

15

6

21

n

n2 + n2

1

1

2

3

3

6

4

10

The values are the same.

c. n2 + n

2 = n(n + 1)2 ; since the diagram represents

n(n + 1), half of the diagram represents n2 + n

2 .

d.

hsm11gmte_0201_t06683

34. J

Lesson 2-2 pp. 52–542. Hypothesis: A figure is a rectangle. Conclusion: It has four sides. 4. If a group is half the people, then that group should make up half the Congress. 6. If an event has a probability of 1, then that event is certain to occur. 8. If an object or example is a counterexample for a conjecture, then the object or example shows that the conjecture is false. 10. If something is blue, then it has a color. 12. If something is wheat, then it is a grain. 14. false; softball 16. Answers may vary. Sample: If an angle is acute, its measure is less than 90; if the measure of an angle is 85, then it is acute. 18. Natalie is correct because a conditional statement and its contrapositive have the same truth value. 20.

new art.ai

Juniors Seniors

Captains

22. If 0 x 0 = 6, then x = -6; false: x = 6 is a counterexample. 24. If x3 6 0, then x 6 0; true. 26. Converse: If you play football, then you are a quarterback. Inverse: If you are not a quarterback, then you do not play football. Contrapositive: If you do not play football, then you are not a quarterback. The conditional and the contrapositive are true. The converse and the inverse are false: counterexample: a person who is a tailback plays football. 28. Converse: If x = 5, then 4x + 8 = 28. Inverse: If 4x + 8 ∙ 28, then x ∙ 5. Contrapositive: If x ∙ 5, then 4x + 8 ∙ 28. All four statements are true. 30. Conditional: If two lines lie in the same plane, then they are coplanar. Converse: If two lines are coplanar, then they lie in the same plane. Inverse: If two lines do not lie in the same plane, then they are not coplanar. Contrapositive: If two lines are not coplanar, then they do not lie in the same plane. All four statements are true. 32. If two lines intersect, then they meet in exactly one point. 34. If you identify any two (distinct) points, then exactly one line goes through those two points. 36. No squares are

triangles. 38. A 40. Answers may vary. Sample: Begin with 1 and 1 as the first two terms; after that, each term is the sum of the two preceding terms.

Lesson 2-3 pp. 57–592. Converse: If a number is even, then it is divisible by 20; false. 4. Converse: If ∼q ¡ ∼p is true, then p ¡ q is true; true. Biconditional: p ¡ q if and only if ∼q ¡ ∼p. 6. No; a straight angle has a measure greater than 90, but it is not an obtuse angle. 8. D 10. The sum of the digits of an integer is divisible by 9 if and only if the integer is divisible by 9. 12. If a line bisects a segment, then it intersects the segment only at its midpoint. If a line intersects a segment only at its midpoint, then the line bisects the segment. 14. If a polygon is a triangle, then it has exactly three sides. If a polygon has exactly three sides, then it is a triangle. 16. No; V could fit that description. 18. good definition 20. If ∠A and ∠B are a linear pair, then ∠A and ∠B are adjacent angles. 22. ∠A and ∠B are a linear pair if and only if ∠A and ∠B are adjacent and supplementary angles. 24a. If an integer is divisible by 10, then its last digit is 0. If the last digit of an integer is 0, then the integer is divisible by 10. b.

Integers withlast digit of 0

Integersdivisible

by 10

hsm11gmte_0203_t06695

c. Integers

divisible by 10

Integerswith lastdigit of

0

hsm11gmte_0203_t06696

d. Integers

divisible by 10

Integers withlast digit of 0

hsm11gmte_0203_t06697

e. Answers may vary. Sample: The two circles coincide. f. Answers may vary. Sample: A good definition can be written as a biconditional because either of the coinciding circles of its Venn diagram can be the hypothesis, and the other the conclusion. 26. J

Lesson 2-4 pp. 62–642. No conclusion is possible; the hypothesis has not been satisfied. 4. If a line intersects a segment at its midpoint, then it divides the segment into two congruent segments. 6. Must be true; by E and A, it is breakfast time; by D, Julio is drinking juice. 8. May be true; by E and A, it is breakfast time. You don’t know what Kira drinks at breakfast.


4


10. May be true; by E, Maria is drinking juice. You don’t know if she also drinks water. 12. Ask your teacher to check your work. 14. If you are studying botany, then you are studying a science. (Law of Syllogism only) No conclusion can be made about Shanti. 16a. The result is two more than the chosen integer. b. 3x + 6

3 = x + 2 c. The expression in part (b) is equivalent to the conjecture in part (a). In part (a) inductive reasoning was used to make a conjecture based on a pattern. In part (b) deductive reasoning was used in order to write and simplify an expression. 18a. Ben b. Andrea, Ben, and Claire; if Dion is also reading it, then all four would be reading it.

Lesson 2-5 pp. 68–702a. Distr. Prop. b. Subtr. Prop. of Eq. c. Div. Prop. of Eq. 4. Samples: ∠1 and ∠2 are a linear pair, m∠1 + m∠2 = 180, DB = EB 6. Domino C; Law of Syllogism 8. m∠GFI = 128 (Given); m∠GFE + m∠EFI = m∠GFI (Ang. Add. Post.); (9x - 2) + 4x = 128 (Subst. Prop.); 13x - 2 = 128(Distr. Prop.); 13x = 130 (Add. Prop. of Eq.); x = 10 (Div. Prop. of Eq.); m∠EFI = 4x (Given); m∠EFI = 4(10) (Subst. Prop.); m∠EFI = 40 (Simplify). 10. Div. Prop. of Eq. 12. Add. Prop. of Eq. 14. YU = AB 16. ∠POR 18. reflexive, symmetric, and transitive; because “has the same birthday as” satisfies all three properties 20. Symmetric only; A cannot live in a different state than A, and if A lives in a different state than B and B lives in a different state than C, then it is possible that A and C live in the same state. 22. 28 24. 82

Lesson 2-6 pp. 75–782. x = 14, y = 15; 3x + 8 = 50, 5x - 20 = 50, 5x + 4y = 130 4a. 90 b. 90 c. m∠3 d. ≅ 6. Answers may vary. Sample: scissors 8. Sample answer: ∠3 and ∠5 are vertical angles, so m∠3 = 138 because vertical angles are congruent. m∠1 + m∠2 + m∠3 = 180 and m∠2 = 21, so m∠1 + 21 + 138 = 180, so m∠1 = 21. 10. x = 50, y = 50 12. If two angles are suppl. to ≅ angles, then the two angles are ≅ to each other. 14. m∠A = 60, m∠B = 30 16. m∠A = 120,m∠B = 60 18. ∠1 and ∠2 are suppl., ∠3 and ∠4 are suppl. (Given); m∠1 + m∠2 = 180,m∠3 + m∠4 = 180 (If two angles are suppl., then the sum of their measures is 180.); m∠1 + m∠2 = m∠3 + m∠4 (Subst. Prop.); ∠2 ≅ ∠4 (Given); m∠2 = m∠4 (If two angles are ≅, their measures are = .); m∠1 = m∠3 (Subtr. Prop. of Eq.); ∠1 ≅ ∠3 (If two angles have the same measure, then they are ≅.) 20. conjecture; The counterexample of an obtuse angle that measures 101° shows this statement is incorrect, so it is a conjecture. 22. postulate; It is an accepted fact that two lines intersect in exactly one point. 24. Sample answer: Undefined terms can be used to define other terms. For example, the undefined terms point and line can be used in

the definition of a ray as part of a line that consists of one point, called the endpoint, and all the points of the line on one side of the endpoint. 26. x = 35, y = 70; 70, 110, 70 28. Sample:

14

32

Both ∠2 and ∠4 are supplementary to ∠3, because ∠2 and ∠3 form a linear pair, and ∠3 and ∠4 form a linear pair. By the definition of supplementary angles, m∠2 + m∠3 = 180 and m∠3 + m∠4 = 180. Then m∠2 + m∠3 = m∠3 + m∠4 by the Transitive Property of Equality. Subtract m∠3 from each side. By the Subtraction Property of Equality, m∠2 = m∠4. Angles with the same measure are congruent, so ∠2 _ ∠4.

30. 18 32. 60

Topic Review pp. 79–822. deductive reasoning 4. converse 6. theorem 8. Multiply the previous term by -1; 5, -5 10. Multiply the previous term by 4; 1536, 6144 12. Answers may vary. Sample: Portland, Maine 14. If two lines are nonparallel, then they intersect at one point. 16. If today is a certain holiday, then school is closed. 18. Converse: If a figure has four sides, then it is a square. Inverse: If a figure is not a square, then it does not have four sides. Contrapositive: If a figure does not have four sides, then it is not a square. The conditional and contrapositive are true. The converse and inverse are false. 20. Converse: If you are busy on Saturday night, then you baby-sit on Saturday night. Inverse: If you do not baby-sit on Saturday night, then you are not busy on Saturday night. Contrapositive: If you are not busy on Saturday night, then you do not baby-sit on Saturday night. The conditional and contrapositive are true. The converse and inverse are false. 22. yes 24. A phrase is an oxymoron if and only if it contains contradictory terms. 26. Colin will become a better player. 28. If two angles are vertical, then their measures are equal. 30a. Given b. Seg. Add. Post. c. Subst. Prop. d. Assoc. and Comm. Prop. e. Subtr. Prop. of Eq. f. Div. Prop. of Eq. 32. p - 2q 34. 74 36. 106

TEKS cumulative practice pp. 83–852. H 4. J 6. J 8. F 10. G 12. J 14. F 16. F 18. G 20. 34 22. 45 24. PQ = 8 - (-4) = 12; 34

# 12 = 9; 9 units from R to P is 8 - 9 = -1. The coordinate of Q is -1. 26. false; A perpendicular bisector of a line segment forms two pairs of vertical angles. 28. ∠1 and ∠2 are suppl. (Given); ∠1 ≅ ∠2 (Vert. Angles Thm.); ∠1 and ∠2 are right angles. (If two angles are ≅ and suppl., then each is a right angle.)


5


Topic 3Lesson 3-1 pp. 91–932.

<FG

> 4. Answers may vary. Sample:

<AB

>, <BH

>

6. Plane JCD 8. ∠2 and ∠3 (lines d and e with transversal c); ∠1 and ∠4 (lines a and b with transversal c); ∠5 and ∠4 (lines c and e with transversal b) 10. ∠6 and ∠8 (lines a and b with transversal d) 12. ∠1 and ∠2 are same-side interior angles; ∠3 and ∠4 are corresponding angles; ∠5 and ∠6 are corresponding angles 14. corresponding angles 16. 2 pairs 18. 2 pairs 20. true 22. true 24. true 26a. Lines may be intersecting, parallel, or skew. b. Answers may vary. Sample: In a classroom, two adjacent edges of the floor are intersecting, two opposite edges of the floor are parallel, and one edge of the floor is skew to each of the vertical edges of the opposite wall. 28. No; the floor and the wall intersect, so figures on those planes are not parallel. 30. sometimes 32. never 34. No; if two planes intersect, then their intersection is a single line, and the intersection of planes A and B is

<CD

>.

36. Answers may vary. Sample:


1

23

4

r

�

m

38.

Lesson 3-2 pp. 98–1002. ∠7 (vertical angles), ∠4 (alternate interior angles), ∠5 (corresponding angles) 4a. If two || lines are cut by a transversal, then the same-side interior angles are supplementary. b. If two || lines are cut by a transversal, then the same-side interior angles are supplementary. c. If two angles are supplementary to the same angle, then they are congruent. 6. ∠1 = 75 because corresponding angles are congruent; ∠2 = 105 because ∠2 forms a linear pair with the given angle. 8. ∠1 = 100 because same-side interior angles are supplementary; ∠2 = 70 because alternate interior angles are congruent. 10. 25; x + 40 = 65, 3x - 10 = 65 12. 32 14. x = 87, y = 31, w = 20, v = 42 16. / = m (Given); m∠2 + m∠3 = 180 (Angles that form a linear pair are supplementary.); m∠3 + m∠6 = 180 (Same-side interior angles are supplementary.); m∠2 + m∠3 = m∠3 + ∠6 (Substitution); m∠2 = m∠6 (Subtraction); ∠2 ≅ ∠6 (Definition of Congruence) 18a. 117 b. same-side interior angles 20a. In the diagram, ∠1 and ∠6 are one pair of same-side exterior angles. ∠4 and ∠7 are another pair of same-side exterior angles. Use a protractor to

find m∠1 + m∠6 = 180 and m∠4 + m∠7 = 180. b. Same-side exterior angles are supplementary. Explanations may vary. Sample: ∠1 is supplementary to ∠2 which is congruent to its corresponding angle ∠6. 22. Sample: The labels (60 - 2x)° and (2x - 60)° contain contradictory information because those angles are corresponding angles. If 60 - 2x = 2x - 60, then x = 30 and the measure of each angle is 0. 24. 60

Lesson 3-3 pp. 104–1062.

<PS>}<QT

>; Converse of the Corresponding

Angles Theorem. 4. <KR

>}<MT

>; Converse of the

Corresponding Angles Theorem. 6. Yes; ∠1 and ∠2 are alternate exterior angles, and if alternate exterior angles are congruent, then the lines are parallel. 8. 31 10. a } b; Converse of the Corresponding Angles Theorem. 12. none 14. / } m (Converse of the Corresponding Angles Theorem) 16. a } b (Converse of Corresponding Angles Theorem) 18. / } m (Converse of the Alternate Interior Angles Theorem) 20. The corresponding angles are congruent, so the oars are parallel by the Converse of the Corresponding Angles Theorem. 22. x = 5; m∠1 = m∠2 = 50 24. x = 1.25;m∠1 = m∠2 = 1026. Answers may vary. Samples given:

28. PL } NA; PN } LA; if same-side interior angles are supplementary, then the lines are parallel. 30. none32. Answers may vary. Sample:

1. / } m (Given) 2. ∠ABG ≅ ∠ACE (If lines are parallel, then corresponding angles are congruent.)


X Y

M N

� � n

∠2 ≅ ∠8

Given

Given

∠12 ≅ ∠8

If lines �, thenalt. int. ⦞ are ≅.

∠2 ≅ ∠12

j � k

If alt. ext. ⦞ are ≅,then lines are �.

Trans. Prop.of ≅



m

�K

B

C

n

A J

E

G12

43


6


3. BJ bisects ∠ABG, CK bisects ∠ACE (Given) 4. m∠1 = 1

2m∠ABG, m∠3 = 12m∠ACE (A bisector

divides an angle in half.) 5. 12m∠ABG = 12m∠ACE

(Multiplication Property of Equality) 6. m∠1 = m∠3 (Substitution) 7.

<CK

>}<BJ> (If

corresponding angles are congruent, then the lines are parallel.)

34. G 36. J

Lesson 3-4 pp. 109–1102a. Corresponding Angles b. ∠1 c. ∠3 d. Converse of Corresponding Angles Theorem 4. Since a and c are both perpendicular to b, a } c because, in a plane, if two lines are perpendicular to the same line, they are parallel. It is given that c } d, so a } d because if two lines are parallel to the same line, they are parallel. 6. In the diagram, a#b means the marked angle is a right angle. b } c means that the corresponding angle formed by a and c is a right angle, so a#c. 8. They are perpendicular; answers may vary. Sample: The yellow and blue lines are parallel because they are both perpendicular to the brown line. Since the pink line is perpendicular to one of those two parallel lines, it must also be perpendicular to the other. 10. All of the rungs are perpendicular to one side. The side is perpendicular to the top rung, and because all of the rungs are parallel to each other, the side is perpendicular to all of the rungs. 12. The sides are parallel because they are both perpendicular to one rung. 14. Reflexive: a } a; false; every line intersects itself. Symmetric: If a } b, then b } a; true; lines a and b are coplanar and do not meet. Transitive: If a } b and b } c, then a } c; true; that is Theorem 3-8. 16. a } d by Theorem 3-8 18. a#d by Theorem 3-10 20. a } d by Theorems 3-8 and 3-9 22. A

Lesson 3-5 pp. 114–1162. x = 70, y = 110, z = 30 4. c = 60 6. 115.5 8. a = 162, b = 18 10. 114 12. 60, 80 14. 102, 65, 13 16. x = 37; m∠P = 65, m∠Q = 78, m∠R = 37 18. a = 67, b = 58, c = 125, d = 23, e = 90 20. ∠1 is an exterior angle of the triangle. (Given); ∠1 and ∠4 are supplementary (Angles that form a straight angle are supplementary.); m∠1 + m∠4 = 180 (Definition of supplementary); m∠2 + m∠3 + m∠4 = 180 (Triangle Angle-Sum Theorem); m∠1 + m∠4 = m∠2 + m∠3 + m∠4 (Substitution Property); m∠1 = m∠2 + m∠3 (Subtraction Property of Equality) 22. 132; the smallest interior angle is adjacent to the largest exterior angle, so the largest exterior angle has measure 180 - 48 = 132. 24. 115 26. A 28a. 159; m∠A + m∠B + m∠C = 180, but m∠A = 21 and 180 - 21 = 159. b. 1 … m∠C … 68; ∠C is acute, so m∠B 7 90. m∠A + m∠C 6 90. Thus, 21 + m∠C 6 90, or m∠C 6 69. Also m∠C 7 0.

c. 91 … m∠B … 158; ∠B is obtuse, so m∠B 7 90. m∠B + m∠C = 159, so m∠B 6 159.

Lesson 3-6 pp. 121–1222.


�

J

A B

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

b

a

hsm11gmte_0306_t14506


b

a a6.


P

�

8.


P

R S

�

10.


P

R S

�


7


12.


p

p p

14a. Answers may vary. Sample:


c

cb. Sample: The sides are parallel and congruent. c. Ask your teacher to check your work.16.


b

m

18.


a a

a

20a–c.

d. Sample: The sides of the smaller triangle are half the lengths and parallel to the sides of the larger triangle. e. Ask your teacher to check your work.22. Answers may vary. Sample:


D G


b

a



a

b

hsm11gmte_0306_t11654


b c

a

hsm11gmte_0306_t11656

30. Not possible; the shorter sides would meet at a point on the longer side, forming a segment. 32. J

Lesson 3-7 pp. 127–1282. -5

6 4. 0

6.

O

y

x

hsm11gmte_0307_t11659

1

1

3

8.

hsm11gmte_0307_t011663

O 2 4

4

6

y

x

10. y + 6 = 23(x + 2) 12. y - 3 = 1

4(x + 5) or

y - 5 = 14(x - 3) 14. y = 3

5x + 5 16. The vertical change is y2 - y1. The horizontal change is x2 - x1. The slope m is the ratio of vertical change to horizontal change, so m should be equal to

y2 - y1x2 - x1

. 18. horizontal: y = -1; vertical: x = 0


b

c

a


8


20. (6, - 4)

hsm11gmte_0307_t11672

yO 4 8

x

2

�2 x � 6

y � �4

(6, �4)

22. ( - 1, 3)


2

4

(�1, 3)

x � �1

y � 3

2

y

xO�2�4

24. Undefined; the y-axis is a vertical line, and the slope of a vertical line is undefined; x = 0 26. y = 2x + 9 28. y = -3

2 x + 5 30. y = -2x + 4 32. Yes; the slope of the line through the first two points is 2, and the slope of the line through the last two points is 2, so the points lie on the same line. 34. Yes; the slope of the line through the first two points is -7

3, and the slope of the line through the last two points is -7

3, so the points lie on the same line. 36. G

Lesson 3-8 pp. 132–1332. No; the slope of /1 is 13, and the slope of /2 is 12. The slopes are not equal, so the lines are not parallel. 4. y = 1

3(x - 6) 6. y + 2 = -32(x - 6)

8. Yes; the slope of /1 is -1, and the slope of /2 is 1. Since the product of the slopes is -1, the lines are perpendicular. 10. y = -2(x - 4) 12. y = 4

3x 14. No; the slopes are 7 and -7, so the lines are not }. 16. Yes; both slopes are -2

5, so the lines are }. 18. Answers may vary. Samples: y = x + 1 and y = -x + 1; y = 2x + 1 and y = - 1

2x + 1 20. slope of AB = slope of CD = 2

3, AB } CD; slope of BC = slope of AD = -3, BC } AD 22. slope of AB = 1

2, slope of CD = 1

4, AB is not parallel to CD; slope of BC = -1, slope of AD = -1

2, BC is not parallel to AD 24. Yes; RS and VU both have slope 0; ST and WV both have slope -1; RW and TU both have slope 1. Therefore, opposite sides of the figure are }. 26. Yes; the slopes are -1 and 1, and their product is -1. 28. No; the slopes are 27 and -7

4, and their product is not -1. 30. No; the slope of GH is 35, the slope of HK is -5

8, and the slope of GK is -8

3. Since no pair of slopes have a product of -1, no two lines are perpendicular and the triangle is not a right triangle. 32. 12 34. 25

Lesson 3-9 pp. 137–1382. In Euclidian geometry, you can draw only one line through two points. 4. In Euclidian geometry, a

triangle can have at most one obtuse angle. 6a. At least one of the two curves is not a great circle, so it is not a line. b. The upper circle is not a line, so a piece of the upper circle cannot be a line segment. 8a. All lines of longitude are lines because they are all great circles. b. The equator is a line. The other lines of latitude are not great circles. c. infinitely many d. Only the equator is a line in spherical geometry. 10. In Euclidean geometry, there is exactly one line through point P that is parallel to line /. In spherical geometry, a line is a great circle. Every line that contains point P will intersect line /. So there is no line through point P that is parallel to line /. 12. A 14. C

Topic Review pp. 139–1432. great circle 4. ∠2 and ∠7, a and b, transversal d; ∠3 and ∠6, c and d, transversal e; ∠3 and ∠8, b and e, transversal c 6. ∠1 and ∠4, lines c and d, transversal b; ∠2 and ∠4, lines a and b, transversal d; ∠2 and ∠5, lines c and d, transversal a; ∠1 and ∠5, lines a and b, transversal c; ∠3 and ∠4, b and c, transversal e 8. corresponding angles 10. m∠1 = 120 because corresp. angles are ≅. m∠2 = 120 because ∠1 and ∠2 are vert. angles. 12. x = 118, y = 37 14. 20 16. none; ∠3 and ∠6 form a linear pair. 18. n } p; if alt. int. angles are ≅, then the lines are }. 20. a 22. x = 60, y = 60 24. 30 26. 328.

hsm11gmte_03cr_t11681.ai

a

b

30.


a

a

32. undefined 34. slope: -2; point: (-5, 3)


y

2

�2�4 �2

�4

xO

36. y + 9 = 3(x - 1) 38. neither 40. # 42. y - 2 = 8(x + 6) 44. Euclidean 46. spherical 48. neither 50. true 52. true

TEKS cumulative practice pp. 144–1452. H 4. G 6. J 8. G 10. G 12. 54.5 14. 32 16. The logic is not valid. You may have received the free month as a gift or through a coupon, without buying a one-year membership. 18. x = 40; 80; 64; 36


9


20. They are } ; planes A and B are#to the same line. 22a. No; answers may vary. Sample: The characteristics of a bleeble are that the outside shape is a noncircular oval and lines parallel to the longer axis divide the oval into sections, each of which contains a dot. The figure in question has lines#to the longer axis, so it is not a bleeble. b. Answers may vary. Sample: A bleeble is a noncircular oval, divided into sections by lines parallel to the longer axis of the oval, with one dot in each section.

Topic 4Lesson 4-1 pp. 150–1522. EF ≅ HI, FG ≅ IJ, EG ≅ HJ, ∠EFG ≅ ∠HIJ, ∠FGE ≅ ∠IJH, ∠FEG ≅ ∠IHJ 4. CM 6. ∠J 8. △CLM 10. ∠P ≅ ∠S, ∠O ≅ ∠I, ∠L ≅ ∠D, ∠Y ≅ ∠E 12. 45 ft 14. 52 16. 280 ft 18. 128 20. No; there are not three pairs of congruent corresponding sides. 22. Yes. The diagram shows that ∠CBD ≅ ∠ABD. By the Third Angles Theorem, ∠C ≅ ∠A. The diagram shows that CD ≅ AB and CB ≅ AD. BD ≅ BD by the Reflexive Property of Congruence. So △BCD ≅ △DAB by the definition of congruent triangles. 24. m∠B = m∠E = 12 26. AC = DF = 19 28. x = 15, t = 2 30. △NRZ, △ZRN (also △MLJ) 32. Answers may vary. Sample: If △PQR ≅ △XYZ , then PQ ≅ XY , QR ≅ YZ , PR ≅ XZ , ∠P ≅ ∠X , ∠Q ≅ ∠Y , and ∠R ≅ ∠Z . 34. Answers may vary. Sample: Sort congruent cards into three piles. 36a. 16 quadrilaterals, including the one given

hsm11_gmte_0401_t11227

b. 12 convex, 4 concave 38. 72 40. 28

Lesson 4-2 pp. 155–1572a. Answers will vary. Sample: I selected two pieces of string cut to the same length to indicate the same perimeter. b. Sample answer. The conjecture is incorrect. Using the same length of string, I was able to form triangles of different shape. New conjecture: If two triangles have the same perimeter and two

corresponding sides are congruent, then the triangles are congruent. 4. F is the midpoint of GI (Given), so IF ≅ GF because a midpoint divides a segment into two congruent segments. The other two pairs of sides are given as congruent, so △EFI ≅ △HFG by SSS. 6. You need to know LG ≅ MN; the diagram shows that LT ≅ MQ and ∠L ≅ ∠M. ∠L is included between LG and LT , and ∠M is included between MN and MQ. 8. If the 40° angle is always included between the two 5-in. sides, then all the triangles will be congruent by SAS. If the 40° angle is never included between the two 5-in. sides, then the angles of the triangle will be 40°, 40°, and 100°, with the 100° angle included between the 5-in. sides, so all the triangles will be congruent by SAS. But a triangle with the 40° angle included between the 5-in. sides will NOT be congruent to a triangle with the 40° angle not included between the 5-in. sides. 10. Not enough information; you need DY ≅ TK to show the triangles are congruent by SSS, or you need ∠H ≅ ∠P to show the triangles are congruent by SAS. 12. Yes; two sides of the original triangle are congruent, and finding successive midpoints of congruent segments results in pairs of (smaller) congruent segments. The bottom sides of the triangles are congruent because the length of each segment is half the length of the original triangle’s base. The base angles of the original isosceles triangles are congruent, so the triangles outlined in red are congruent by SAS. 14. Not necessarily; the congruent angles are not included between the pairs of congruent sides. 16. Given the # segments, ∠B ≅ ∠CMA because all rt. angles are ≅. M is the midpt. of AB (Given), so AM ≅ MB by the def. of midpt. Since DB ≅ CM (Given), then △ AMC ≅ △MBD by SAS. 18. H

Lesson 4-3 pp. 161–1622. △UST ≅ △RTS by ASA or AAS. 4. It is given that ∠FJG ≅ ∠HGJ and FG } JH. Then ∠FGJ ≅ ∠HJG because alternate interior angles are congruent. Since GJ ≅ GJ by the Reflexive Property of Congruence, △FGJ ≅ △HJG by ASA. 6. Answers may vary. Sample: Yes; ASA guarantees a unique triangle with vertices at the oak tree, the maple tree, and the time capsule. 8. No; the common side is included between the two congruent angles in one triangle, but it is not included between the congruent angles in the other triangle. 10. DH bisects ∠BDF (Given), so ∠BDH ≅ ∠FDH by the definition of angle bisector. ∠1 ≅ ∠2 (Given) and DH ≅ DH (Reflexive Property of Congruence), so △BDH ≅ △FDH by ASA. 12. Answers may vary. Sample:

A C D

F

B E

hsm11gmte_0403_t11231


10


14.

hsm11gmte_0403_t11230

L

J

M N

P

K

hsm11gmte_0403_t11230

L

J

M N

P

K

16.

18. J

Technology Lab 4-3 p. 1632. No, you cannot use SSA to prove △s ≅ because two sides of a △ and the nonincluded ∠ can be parts of two noncongruent △s . 4. If CE Ú CA there is exactly one △ACE for any m∠A. If CE 6 CA there is exactly one △ACE only when CE2 + EA2 = CA2; that is, when ∠A (and ∠C ) is an acute angle of a right triangle (with right ∠AEC ).

Lesson 4-4 pp. 166–1672. ∠ABD ≅ ∠CBD and ∠BDA ≅ ∠BDC (Given). BD ≅ BD by the Reflexive Property of Congruence. △ABD ≅ △CBD by ASA, so AB ≅ CB because corresponding parts of congruent triangles are congruent. 4a. △KRA b. ASA c. Corresponding parts of congruent triangles are congruent. 6. From the given information, △YCT ≅ △YRP by AAS. So CT ≅ RP because corresponding parts of congruent triangles are congruent. 8. From the definition of perpendicular bisector, PL ≅ QL and ∠PLK ≅ ∠QLK because all right angles are congruent. Since KL ≅ KL, then △PKL ≅ △QKL by SAS, and ∠P ≅ ∠Q because corresponding parts of congruent triangles are congruent. 10. The arcs with center P make PA ≅ PB, and the arcs with centers at A and B make CA ≅ CB. Since CP ≅ CP, △APC ≅ △BPC by SSS. ∠APC ≅ ∠BPC because corresponding parts of congruent triangles are congruent. ∠APC and ∠BPC are supplementary because they form a linear pair. So ∠APC is a right angle, which means its sides are perpendicular. 12. BE # AC (Given) and DF # AC (Given). ∠AEB and ∠CFD are right angles (Definition of perpendicular lines). ∠AEB ≅ ∠CFD (All right

angles are congruent.). BE ≅ DF (Given) and AF ≅ CE (Given), so AF = CE, AE + EF = CF + FE (Segment Addition Postulate), AE = CF (Subtraction Property of Equality) and AE ≅ CF . △AEB ≅ △CFD by SAS, so AB ≅ CD because corresponding parts of congruent triangles are congruent. 14. 36 16. 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. ∠KSA ≅ ∠BSA because corresponding parts of congruent triangles are congruent; the angles are also supplementary, so the measure of each is 90. Thus BK # AE by the definition of perpendicular lines. 18. 42 20. 10

Lesson 4-5 pp. 172–1732. UW ; Converse of Isosceles Triangle Theorem 4. Answers may vary. Sample: ∠VUY ; Isosceles Triangle Theorem 6. x = 80, y = 40 8. x = 40, y = 70 10. AC ≅ BC and ∠ACD ≅ ∠BCD (Given). Also, CD ≅ CD by the Refl. Prop. of ≅. So △CAD ≅ △CBD by SAS. Therefore AD ≅ BD and ∠ADC ≅ ∠BDC because corresp. parts of ≅ triangles are ≅. Since ∠ADC and ∠BDC are suppl., each must have measure 90. So CD # AB (Def. of #). 12. 2.5 14. 35 16a. RS b. RS; RS ≅ RS (Reflexive Property of Congruence) and ∠PRS ≅ ∠QRS (Definition of angle bisector). Also, ∠P ≅ ∠Q (Given). So △PRS ≅ △QRS by AAS. PR ≅ QR because corresponding parts of congruent triangles are congruent. 18. 6 20. ( - 5, 5), (0, 5), (0, 10), (5, 0), (5, -5), (10, 0) 22a–c. Ask your teacher to check your diagram. d. Answers will vary. Sample: If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle, then the triangles are congruent. 24. H

Lesson 4-6 pp. 176–1782. Answers may vary. Sample: ∠S ≅ ∠T by the Isosceles Triangle Theorem, so △PRS ≅ △PRT by AAS. 4. Since HV # GT (Given), △IGH and △ITV are right triangles. It is given that GH ≅ TV , and it is also given that point I is the midpoint of HV , so HI ≅ VI by the definition of midpoint. So △IGH ≅ △ITV by HL. 6. x = 3, y = 2 8. Yes; the two triangles are right triangles with congruent hypotenuses and congruent legs, so the two triangles are congruent by HL. Then RQ ≅ CB because corresponding parts of congruent triangles are congruent. 10. From the information about perpendicular segments, △MNL and △QPL are right triangles. It is given that ML ≅ QL and since NP is the base of isosceles △LNP (Given), LN ≅ LP (Definition of isosceles triangle). So △MNL ≅ △QPL by HL.

hsm11gmte_0403_t11232

W

95

30�

V

U


11


12.

hsm11gmte_0406_t11237

14.

hsm11gmte_0406_t11239

16. From the given perpendicular segments, ∠M ≅ ∠N because perpendicular lines form right angles, which are congruent. LO bisects ∠MLN (Given), so ∠MLO ≅ ∠NLO by the definition of angle bisector. The Reflexive Property of Congruence gives LO ≅ LO, so △LMO ≅ △LNO by AAS. 18. △AEB and △CEB are right triangles because the given information includes BE # EA and BE # EC . △ABC is equilateral (Given), so AB ≅ CB by the definition of equilateral. Also, BE ≅ BE by the Reflexive Property of Congruence. So △AEB ≅ △CEB by HL. 20. C

Lesson 4-7 pp. 181–1832. DF 4. P P

R

Q Q

S

hsm11gmte_0407_t11259

PQ is a common side.6. K K

M

L

J

L

hsm11gmte_0407_t11261

KL is a common side. 8. RS ≅ UT and RT ≅ US (Given), and ST ≅ ST (Reflexive Property of Congruence), so △RST ≅ △UTS by SSS. 10. ∠1 ≅ ∠2 and ∠3 ≅ ∠4 (Given), and QB ≅ QB by the Reflexive Property of Congruence. So △QTB ≅ △QUB by ASA. Thus QT ≅ QU (Corresponding parts of congruent triangles are congruent.). QE ≅ QE (Reflexive Property of Congruence), so △QET ≅ △QEU by SAS.

12. Since VT = VU + UT = UT + TS = US, VT ≅ US. Therefore, △QVT ≅ △PSU by SAS. 14a. m∠1 = 50,m∠2 = 50, m∠3 = 40, m∠4 = 90, m∠5 = 10, m∠6 = 40, m∠7 = 40, m∠8 = 80, m∠9 = 100 b. △ABC ≅ △FCG by ASA because ∠B ≅ ∠FCG (All right angles are congruent.), AB ≅ FC (Given), and ∠A ≅ ∠GFC (Their measures are equal.). 16. △PQT ≅ △RQT by SAS because QT ≅ QT (Reflexive Property of Congruence), ∠PQT ≅ ∠RQT (perpendicular lines form right angles, which are congruent), and PQ ≅ RQ (because QT bisects PR). Then ∠QTP ≅ ∠QTR because corresponding parts of congruent triangles are congruent. Also, ∠VQT ≅ ∠SQT (because QT bisects ∠VQS), so △VQT ≅ △SQT by ASA. So VQ ≅ SQ because corresponding parts of congruent triangles are congruent. 18. D 20. C

Topic Review pp. 184–1872. hypotenuse 4. congruent polygons 6. ∠U 8. ONMLK 10. 3 12. 35 14 145 16. MR 18. not enough information 20. AAS or ASA 22. △BEC ≅ △DEC by ASA so BE ≅ DE because corresp. parts of ≅ triangles are ≅. 24. If two parallel lines are cut by a transversal, their alt. int. angles are ≅, so ∠LKM ≅ ∠NMK . Then △LKM ≅ △NMK by SAS, and KN ≅ ML because corresp. parts of _triangles are ≅. 26. x = 55, y = 62.5 28. x = 7, y = 60 30. The given information about # segments means △PSQ and △RQS are rt. triangles. You know PQ ≅ RS (Given) and QS ≅ QS (Refl. Prop. of ≅). So △PSQ ≅ △RQS by HL. 32. △FIH ≅ △GHI by SAS

TEKS cumulative practice pp. 188–1892. G 4. H 6. H 8. 25 10. 3 12. AE _ DE and EB _ EC (Given), and ∠AEB _ ∠DEC (Vert. angles are ≅.) So △AEB _ △DEC by SAS. 14. 1) LN bisects ∠OLM and ∠ONM (Given); 2) ∠OLN _ ∠MLN and ∠ONL _ ∠MNL (Def. of Bisect); 3) LN _ LN (Reflexive); 4) △ONL _ △MNL (ASA); 5) ON _ MN (Corresponding Parts of Congruent Triangles are Congruent). 16a. The comet should appear in 2137. Here is the information from the passage: 1835 to 1910: 75 years; 1910 to 1986: 76 years; 1986 to 2061: 75 years. The time between appearances seems to alternate between 75 and 76 years. The period after 2061 would be a 76-year period, so it should appear in 2061 + 76 = 2137. b. Answers may vary. Sample: Fairly confident; to be more confident, you would need to know the exact dates of appearances and a greater number of appearance dates.


12


Topic 5Lesson 5-1 pp. 196–1982. (3, 1) 4. (6, 1) 6. (37

8, -3) 8a. Answers may vary. Sample: Distance Formula (Find KP, then divide it by 2.) b. Answers may vary. Sample: Distance Formula (If M is the given midpoint, find KM, and then multiply it by 2.) 10a. 10.8 b. (3, -4)12a.

OA

B

C

D

2

�2

�2

6

4

8

2 4 8x

y

The midpoints are the same, (5, 4). b. Answers may vary. Sample: The diagonals bisect each other. 14. 7 mi 16. 3.2 mi 18. 6 20. 8 22. 23.3 24. 165 units; flying T to V then to U is the shortest distance. 26. 5 28. 9.4 30. Yes; Sample: BC and AD lie on two horizontal lines. CD lies on a vertical line. A horizontal line and a vertical line are perpendicular. 32. Answers may vary. Sample: △ADM and △BCM are congruent by SAS. Since corresponding parts of congruent triangles are congruent, AM = MB. This means that M is the midpoint of AB. Therefore, since M lies halfway between C and D, the coordinates of the midpoint

M of AB is M = (x1 + x22 ,

y1 + y22 ) . 34. Answers may

vary. Sample: PR = 0 y2 - y1 0 and QR = 0 x2 - x1 0 .By the Pythagorean Theorem, PQ2 = QR2 + PR2.

Substitute the algebraic expressions for QR and PR: PQ2 = (x2 - x1)2 + (y2 - y1)2. Then take the square root of each side of the equation. Thus, PQ = 2(x2 - x1)2 + (y2 - y1)2. 36. D(5, 2) and E(7,3) 38. A(0, 0, 0), B(6, 0, 0), C(6, - 3, 0), D(0, - 3, 0), E(0, 0, 9), F(6, 0, 9), G(0, - 3, 9) 40. 11.7 units 42. F

Technology Lab 5-2 pp. 199–2002. yes 4. The areas are equal. 6a. The area of △ABC is 16 times the area of △GHI. b. The perimeter of △ABC is 4 times the perimeter of △GHI. c. The area of △ABC will be 64 times the area of the new △, and its perimeter will be 8 times the perimeter of the new △.

Lesson 5-2 pp. 203–2052. 437.5 ft 4. F B D

A C

E

hsm11gmte_0501_t10684

6. 18.5 8. 40 10. 160 12. 13 14. 6 16. △UTS;answers may vary. Sample: VS = SY = 1

2VY ,

VU = UZ = 12VZ and YT = TZ = 1

2YZ by the definition

of midpoint. Also ST = 12VZ , SU = 1

2YZ , and TU = 12VY

by the Triangle Midsegment Theorem. So △YST ≅△TUZ ≅ △SVU ≅ △UTS by SSS. 18. The midpoint of

RP is S(0 + (-4)2 ,

0 + (-8)2 ) = S(-2, -4). The midpoint

of RQ is T (0 + (-10)2 , 0 + 0

2 ) = T(-5, 0). The slope of

ST = 0 - (-4)-5 - (-2) = -4

3 and the slope of

PQ = -0 - (-8)-10 - (-4) = - 4

3 so ST } PQ.

ST = 2(-5 - (-2))2 + (0 - (-4))2 = 29 + 16 =225 = 5 and PQ = 2(-10 - (-4))2 + (0 - (-8))2 = 236 + 64 = 2100 = 10 so ST = 12PQ.

20. 100 22. C 24. 24 26. Draw CA>. Find P on

CA> such that CA = AP. Draw PD. Construct the

perpendicular bisector of PD. Label the intersection point B. Draw AB. This is a midsegment of △CPD. According to the Triangle Midsegment Theorem, AB } CD and AB = 1

2CD. 28. Answers may vary. Sample: You can verify that the Triangle Midsegment Theorem works for a particular triangle, whereas the proof of the Triangle Midsegment Theorem shows that it works for all triangles. 30. 8.4 32. 118

Lesson 5-3 pp. 210–2122a.

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P

Q RS


13


b. Answers may vary. Sample: Since P is on the perpendicular bisector of QR, it is equidistant from Q and R by the Perpendicular Bisector Theorem. Then PQ = PR, so △PQR is an isosceles triangle. 4a.

D

C

E

F

hsm11gmte_0106_t83046

b. You should verify by measuring that points on DF>

are equidistant from DC> and DE

>, and are thus on

the bisector of ∠CDE by the Converse of the Angle Bisector Theorem. 6. 3 8. Coleman School; it is on 6th Ave., which is (approximately) the perpendicular bisector of 14th St. between 8th Ave. and Union Square. 10. Draw HS and find its midpt., M. Through M, construct the line perpendicular to HS. Any point on this line will be equidistant from H and S. 12. At the point on XY that lies on the bisector of ∠GPL; the goalie does not know to which side of her the player will aim his shot, so she should keep herself equidistant from the sides of ∠GPL. Points on the bisector of ∠GPL are equidistant from PG and PL. If she moves to a point on the perpendicular bisector of GL, she will be closer to PL than to PG. 14. HL

> bisects

∠KHF ; point L is equidistant from the sides of the ∠, so L is on the bisector of ∠KHF by the Converse of the ∠ Bisector Thm. 16. 54; 54 18. 10 20. equidistant; RT; RZ 22. A point is on the perpendicular bisector of a segment if and only if it is equidistant from the endpoints of the segment.

24a. / : y = -34x + 25

2 ; m : x = 10 b. C(10, 5)

c. CA = 2(10 - 6)2 + (5 - 8)2 = 5;

CB = 2(10 - 10)2 + (5 - 0)2 = 5 d. C is equidistant from the sides of ∠AOB, so C is on the bisector of ∠AOB by the Converse of the ∠ Bisector Theorem. 26. PA = PB (Given); PA ≅ PB (Definition of congruence); ∠AMP ≅ ∠BMP (All right angles are congruent.);

<PM

>≅

<PM

> (Reflexive Property

of Congruence); △AMP ≅ △BMP (Hypotenuse-Leg Theorem); AM ≅ BM (Corresponding parts of congruent triangles are congruent.); PM is the perpendicular bisector of AB (Definition of perpendicular bisector). 28. C 30. Point A(2, 10) is 3 units above the line y = 7. Therefore point B must be 3 units below y = 7, so the y-coordinate of B must be 4. Since the coordinates of B are (2, k), k = 4.

Activity Lab 5-4 p. 2132. The # bisectors of the sides of a △ meet at a single point. 4. an equilateral △

Lesson 5-4 pp. 217–2182. (0, 0) 4. C 6. The perpendicular bisectors of a triangle intersect at a single point, even if that point is outside the triangle. 8. Answers may vary. Sample: For any triangle, the location of the point where the angle bisectors intersect is always inside the triangle. 10. An interpretation of the passage is that the treasure is equidistant from three Norway pines. To find the treasure, Ivars can draw a diagram of the situation and then find the circumcenter of the triangle whose vertices are the three pines. 12. P; the markings in the diagram show that P is the incenter of the triangular station and C is the circumcenter. If you stand at P, you will be equidistant from the three sides along which the buses are parked. If you move away from P, you will move closer to some of the buses. 14. False; if the points are collinear, then the perpendicular bisectors of the segments determined by the points will be parallel. Since the perpendicular bisectors are parallel, they will not intersect, so there is no point that is equidistant from all 3 points. 16. B

Technology Lab 5-5 p. 2192. Answers may vary. Sample: The three lines containing the altitudes of a △ are concurrent. The three medians of a △ are concurrent. 4. Answers may vary. Sample: For isosc. △s , these lines are the same: the # bisector of, altitude to, and median to the base, and the bisector of the vertex ∠. For equilateral △s , the ∠ bisector, median, altitude, and # bisector to each side are all the same line.

Lesson 5-5 pp. 222–2242. 1 : 2 4. BD 6. Answers may vary. Sample: The angle bisector of the vertex angle forms two triangles that are congruent by SAS. Therefore the 2 segments formed on the base are ≅ (so the angle bisector contains a median), and the two angles formed by the angle bisector and the base are right angles (so the angle bisector contains an altitude). Thus the median and the altitude are the same. 8.

hsm11gmte_0504_t10695

R T

S


14


10. Each of the three folds should be along a line containing a vertex, and should be perpendicular to the opposite side.12. an obtuse triangle

hsm11gmte_0504_t10696

14. ZY = 4.5, ZU = 13.5 16. Neither; it does not have a vertex of △ABC as an endpoint. 18. (4, 0) 20. ( - 2, 0) 22. A is the intersection of the perpendicular bisectors of the sides, so it is the circumcenter; B is the intersection of the medians, so it is the centroid; C is the intersection of the angle bisectors, so it is the incenter; D is the intersection of the altitudes, so it is the orthocenter. 24a. L(1, 3); M(5, 3); N(4, 0)

b. <AM

> : y = 3

5 x, <BN

> : y = -3x + 12;

<CL> : y = -3

7 x + 247

c. -37 (10

3 ) + 247 = -10

7 + 247 = 14

7 = 2

d. AM = 134, AP = 51369 = 2

3134;

CL = 158, CP = 52329 = 2

3158; BN = 140 = 2110,

BP = 51609 = 4

3110 26. H

Lesson 5-6 pp. 227–228

2. Assume temporarily that no ∠ is obtuse. 4. Assume temporarily that m∠2 … 90. 6. I and III 8. B 10. Assume temporarily that XB # AC . Then ∠BXA ≅ ∠BXC (All right angles are congruent.), ∠ABX ≅ ∠CBX (Given), and BX ≅ BX (Reflexive Property of Congruence), so △BXA = △BXC by ASA and BA = BC because corresponding parts of congruent triangles are congruent. But this contradicts the given statement that △ABC is scalene. Therefore the temporary assumption that XB # AC is wrong, and we can conclude that XB is not perpendicular to AC . 12. Assume temporarily that ∠A ≅ ∠B. Then BC = AC by the Converse of the Isosceles Triangle Theorem. But this contradicts the given statement that BC 7 AC . Therefore the temporary assumption is false, and we can conclude that ∠A is not congruent to ∠B. 14. The culprit entered the room through a hole in the roof; all the other possibilities were ruled out. 16. I and III D 18. H

Lesson 5-7 pp. 232–2342. PQ + QR 7 PR

Q

P

P Q R

R

4. Sample answer: The classmate’s statement is incorrect. By the Triangle Inequality Theorem, FG + GH must be greater than FH. If FG + GH = FH, then FG and GH would form a line segment with FH, not a triangle. 6. m∠2 = m∠4 because parallel lines form alternate interior angles that are congruent, and m∠1 7 m∠4 by the Corollary to the Triangle Exterior Angle Theorem. So m∠1 7 m∠2 by substitution. 8. 5

18 10. The sign, Topeka, and Wichita are either collinear or they determine the vertices of a triangle. If D is the distance between Topeka and Wichita, then 20 … D … 200. 12. Yes; 11 + 12 7 15, 11 + 15 7 12, and 12 + 15 7 11. 14. 11 in. 6 x 6 21 in. 16. 15 km6 x 6 55 km 18. ∠D, ∠C, ∠E 20. FH, GF, GH

22. AC, AB, BC 24a. m∠OTY b. m∠3 c. Isosceles Triangle Thm. d. Angle Addition Post. e. Comparison Prop. of Inequality f. Substitution g. Corollary to Triangle Exterior Angle Thm. h. Transitive Prop. of Inequality 26. By the Triangle Exterior Angle Theorem, m∠1 = m∠2 + m∠3. Since m∠2 7 0 and m∠3 7 0, you can apply the Comparison Property of Inequality and conclude that m∠1 7 m∠2 and m∠1 7 m∠3. 28. 129 30. 2

Lesson 5-8 pp. 238–2392. PR 6 RT 4. 6 6 x 6 38 6. The 40° opening; the lengths of the two sections of the robotic arm do not change as the arm moves. The included angle between the arm sections of the 60° opening is greater than the included angle of the 40° opening, so by the Hinge Thm., the tip of the arm is closer to the base for the 40° opening. 8. Ship A; the two triangles in the diagram have two pairs of ≅ corresp. sides. The included ∠ for Ship A measures 180 - 65 = 115 and the included ∠ for Ship B measures 180 - 70 = 110. Since 115 7 110, the side opposite 115 is longer than the side opposite 110, by the Hinge Thm. 10. D 12. Assume temporarily that there is a triangle with at least 2 obtuse angles. Then the sum of the measures of those two angles is > 180. This is impossible because the sum of the measures of all 3 angles is 180. Therefore the temporary assumption that the triangle could have at least 2 obtuse angles is false, and thus any triangle has at most one obtuse angle.


15


Topic Review pp. 240–2432. distance from a point to a line 4. 1.4 6. 14.4

8. 7.2 10. (1, 1) 12. 15 14. L(52, -1

2); M(72, 12);

slope of AB = 1 and slope of LM = 1, so LM } AB; AB = 212 and LM = 12, so LM = 1

2 AB.16. 40 18. 6 20. 33 22. (0, 0) 24. (4, 4) 26. 45 28. 25 30. AB is a median; it is a segment from a vertex to the midpt. of the opposite side. 32. (0, -1) 34. Assume temporarily that neither of the two numbers is even. That means each number is odd, so the product of the two numbers must be odd. That contradicts the statement that the product of the two numbers is even. Thus the temporary assumption is false, and we can conclude that at least one of the numbers must be even. 36. Assume temporarily that there is a triangle with two obtuse angles. Then the sum of the measures of those two angles is greater than 180, which contradicts the Triangle Angle-Sum Thm. Therefore the temporary assumption is false, and a triangle can have at most one obtuse angle. 38. Assume temporarily that each of the three integers is less than or equal to 3. Then the sum of the three integers must be less than or equal to 3 + 3 + 3, or 9. This contradicts the given statement that the sum of the three integers is greater than 9. Therefore the temporary assumption is false, and you can conclude that one of the integers must be greater than 3. 40. No; 5 + 8 is not greater than 15. 42. 1 ft. 6 x 6 25 ft. 44. 7

TEKS cumulative practice pp. 244–2452. H 4. G 6. H 8. G 10. 30 12. 3.5 14a.

hsm11gmte_05cu_t10736

b. Answers may vary. Sample: Draw a line. Use a compass to mark off a segment that is twice the length of the original base. Construct the # bis. of this segment. Use a compass to mark off a segment on the bisector equal in length to the original altitude. Draw a segment connecting the endpoints of the two legs. 16a. (7, 6); the circumcenter of a triangle is equidistant from the 3 vertices of the triangle, so the new tower should be located at the circumcenter of the triangle whose vertices are (3, 3) (for Westfield), (3, 9) (for Bayville), and (11, 3) (for Oxboro). Those three vertices determine a rt. triangle; the circumcenter of a rt. triangle is the midpt. of the

hypotenuse, so this will be (3 + 112 , 9 + 3

2 ) or (7, 6).

b. Using the Distance Formula, it is 5 mi. to Westfield.

It is the same to the other two towns, since it is located at the circumcenter. c. Yes, Seabury and Medfield; using the Distance Formula, Seabury is 15 mi. from the tower and Medfield is 5 mi. away. Eastham is farther than 5 mi. away.

Topic 6Technology Lab 6-1 p. 2482. Answers may vary. Sample: The sum of the measures of all the ∠s around a point is 360.

Lesson 6-1 pp. 253–2542. 150 4. Answers may vary. Sample:


6. y = 103, z = 70 8. 36 10a. Ask your teacher to check your work. b. Adjacent angles are supplementary. Opposite angles are congruent. 12. octagon 14.


; 90

16a. Answers may vary. Sample: The sum of the interior angles measures = (n - 2)180. All angles of a regular n-gon are ≅. So each interior angle measure = 180(n - 2)

n , and 180(n - 2)

n = 180n - 360n = 180 - 360

n . b. As n gets larger, 360

n gets smaller. The interior angle measure gets closer to 180. The polygon becomes more like a circle. 18. 79

Lesson 6-2 pp. 259–2602. AB = CD = 13, BC = AD = 33 4. 3 6. 2.25 8. 4.5 10. 6.75 12. x = 6, y = 8 14. x = 7, y = 10 16. Suppose that RSTW and XYTZ are parallelograms. It follows from the def. of a ▱ that XY } TZ and TZ } RS Since two lines } to the same line are } to each other, XY } RS. 18. m∠1 = 71, m∠2 = 28, m∠3 = 81 20a. 2.5 ft b. 129 c. Answers may vary. Sample: As m∠E increases, m∠D decreases. ∠E and ∠D are suppl. 22. No; answers may vary. Sample: Corresponding sides ≅ does not prove parallelograms ≅. 24. Answers may vary. Sample: The lines of the paper are } and equally spaced. Place the right corner of the top edge of the card on the first line of the paper. Place the right corner of the bottom edge on the fourth line. Mark the points where the second and third lines intersect the card. The marks will be equally spaced because the edge of the card is a transversal for the equally spaced } lines of the paper. Repeat for the left side of the card. Connect the marks using a straightedge. 26. G 28. 60; sample explanation: A hexagon has 6 sides, so exterior angles measure =3606 = 60.


16


Lesson 6-3 pp. 267–2682. Answers may vary. Sample: 1. ∠M ≅ ∠P , ∠MNQ ≅ ∠PQN, ∠MQN ≅ ∠PNQ (Given) 2. m∠MNQ = m∠PQN, m∠MQN = m∠PNQ (Def. of ≅) 3. m∠MNQ + m∠PNQ = m∠PQN + m∠MQN (Add. Prop. of = ) 4. m∠MNP = m∠PQM (Angle Addition Postulate) 5. ∠MNP ≅ ∠PQM (Def. of ≅) 6. MNPQ is a ▱. (If both pairs of opp. angles of a quad. are ≅, then the quad. is a ▱.) 4. Answers may vary. Sample: 1. ∠A and ∠C are right angles, and AD ≅ CB. (Given) 2. △ABD and △CDB are right triangles. (Def. of right triangle) 3. DB ≅ BD (Refl. Prop. of ≅) 4. △ABD = △CDB (HL) 5. AB ≅ CD (Corresp. parts of ≅ △s are ≅.) 6. ABCD is a ▱. (If both pairs of opp. sides of a quad. are ≅, then the quad. is a ▱.) 6. x = 2, y = 6 8. x = 15, y = 25 10. 24 12. Yes; both pairs of opposite sides are congruent. 14. Yes; both pairs of opposite angles are congruent. 16. Answers may vary. Sample: 1. Draw BD. (Through any two points there is exactly one line.) 2. ∠CBD ≅ ∠ADB (Alt. int. angles are ≅.) 3. BC ≅ DA (Given) 4. BD ≅ BD (Refl. Prop. of ≅) 5. △BCD ≅ △DAB (SAS) 6. ∠BDC ≅ ∠DBA (Corresp. parts of ≅ triangles are ≅.) 7. AB } CD (If alt. int. angles are ≅, then lines are }.) 8. ABCD is a ▱. (Def. of ▱) 18. A 20. B

Lesson 6-4 pp. 272–2742. m∠1 = 60, m∠2 = 90, m∠3 = 30 4. x = 4; LN = MP = 4 6. x = 1; LN = MP = 4 8. x = 5

3; LN = MP = 29

3 10. Answers may vary. Sample: 1. ABCD is a rhombus. (Given) 2. AB ≅ AD and CB ≅ CD. (Def. of rhombus) 3. AC ≅ AC (Refl. Prop. of ≅) 4. △ABC ≅ △ADC (SSS) 5. ∠3 ≅ ∠4 and ∠2 ≅ ∠1. (Corresp. parts of ≅ triangles are ≅.) 6. AC bisects ∠BAD and ∠BCD. (Def. of ∠ bisector) 12. Rhombus; the parallelogram has 4 congruent sides and no right angles. 14a. Opp. sides are ≅ and }; consecutive angles are suppl.; opp. angles are ≅; diagonals bisect each other. b. All 4 sides are ≅; diagonals are # and bisect opp. angles. c. All 4 angles are rt. angles; diagonals are ≅. 16. Answers may vary. Sample:


18. AC = BD = 2 20. AC = BD = 1 22. x = 5, y = 4; all sides are 3. 24a. Paper folding is useful for observing patterns in angles. b. Each diagonal of a rhombus bisects a pair of opposite interior angles. 26. -1 28. J 30. Assume that triangle PQR is a right triangle.

Lesson 6-5 pp. 279–2802. 1 4. 11 6. 16 8. Answers may vary. Sample: Measure the lengths of the frame‘s diagonals. If they are ≅, then the frame has the shape of a rectangle, and therefore a parallelogram; measure the two

pairs of alt. int. angles formed by the turnbuckle (the transversal). If both pairs of angles are ≅, then both pairs of opposite sides of the frame are }. 10. Rhombus; one diagonal bisects a pair of opposite angles. 12. No; you only know that the diagonals bisect each other, which is true of all parallelograms. 14. Rhombus; answers may vary. Sample:


a b

16. Answers may vary. Sample: 1. AC bisects ∠ BAD and ∠ BCD. (Given) 2. ∠ 1 ≅ ∠ 2 and ∠ 3 ≅ ∠ 4. (Def. of bisect) 3. AC ≅ AC (Refl. Prop. of ≅) 4. △ABC ≅ △ADC (ASA) 5. AB ≅ AD and BC ≅ DC . (Corresp. parts of ≅ triangles are ≅.) 6. AB ≅ CD and BC ≅ AD. (Opp. sides of a ▱ are ≅.) 7. AB ≅ AD ≅ BC ≅ CD (Trans. Prop. of ≅) 8. ABCD is a rhombus. (Def. of rhombus) 18. Construct the midpt. of each diagonal. Copy the diagonals so the two midpts. coincide. Connect the endpoints of the diagonals. 20. Construct the midpts. of each diagonal. Construct two # lines, and mark off diagonal lengths on the # lines. Connect the endpoints of the diagonals. 22. Yes; ≅ diagonals in a ▱ mean it can be a rectangle with 2 opp. sides 2 cm long. 24. “If one diagonal of a ▱ bisects one angle, then the ▱ is a rhombus.” The new statement is true. If AC bisects ∠ BCD, then ∠ BCA ≅ ∠ DCA. (Def. of angle bisector) ABCD is a ▱, so ∠ B ≅ ∠ D. (Opp. angles of a ▱ are ≅.) AC ≅ AC (Reflexive Prop. of ≅), so △BCA ≅ △DCA (AAS) and BC ≅ DC . (Corresp. parts of ≅ triangles are ≅.) Since opp. sides of a ▱ are ≅, AB ≅ CD and BC ≅ DA. So AB ≅ BC ≅ CD ≅ DA, and ▱ABCD is a rhombus. (Def. of rhombus) 26. J 28. ( -7 + x

2 , 10 + y

2 ) =(-1, 4). -7 + x = -2, x = 5, and 10 + y = 8, y = -2, so Q(5, -2).

Lesson 6-6 pp. 286–2872. 1 4. x = 35, y = 30 6. Isosc. trapezoid; AB } DC (If alt. int. angles are congruent, then lines are parallel), and AD ≅ BC . (Corresp. parts of congruent triangles are congruent.) 8. 12 cm, 12 cm, 21 cm, 21 cm 10. Isosc. trapezoid; answers may vary. Sample:


12. Rectangle, square; answers may vary. Sample:


14. Kite, rhombus, square; answers may vary. Sample:



17


16. Answers may vary. Sample: 1. AB ≅ DC (Given) 2. ∠BAD ≅ ∠CDA (Base angles of an isosc. trapezoid are ≅.) 3. AD ≅ AD (Refl. Prop. of ≅) 4. △BAD ≅ △CDA (SAS) 5. BD ≅ CA (Corresp. parts of ≅ triangles are ≅.) 18. Answers may vary. Sample: 1. Draw TA and PR. (Construction) 2. TR ≅ PA (Given) 3. ∠TRA ≅ ∠PAR (Base angles of an isosc. trapezoid are ≅.) 4. RA ≅ RA (Refl. Prop. of ≅) 5. △TRA ≅ △PAR (SAS) 6. ∠RTA ≅ ∠APR (Corresp. parts of ≅ triangles are ≅.) 20. True; a square is a parallelogram with 4 rt. angles. 22. False; a rhombus has 4 congruent sides, and a kite does not. 24. False; counterexample: kites and trapezoids are not parallelograms. 26. Rulers are useful for measuring length, and protractors are useful for measuring angles. Conjecture: The diagonals of an isosceles trapezoid separate the trapezoid into four triangles: two triangles that are congruent and two that are not. 28. D is a point on

<BN

> such that

ND ≠ BN, and B and D are on opp. sides of N. 30. half the difference of the bases; Triangle Midsegment Theorem 32. H

Topic Review pp. 288–2912. equiangular polygon 4. trapezoid 6. 157.5, 22.5 8. 360, 360, 360 10. 69 12. m∠1 = 101, m∠2 = 79, m∠3 = 101 14. m∠1 = 45, m∠2 = 45, m∠3 = 45 16. x = 2, y = 5 18. yes 20. x = 4, y = 5 22. m∠1 = 124, m∠2 = 28, m∠3 = 62 24. always 26. sometimes 28. always 30. Yes, the parallelogram is a rhombus and a rectangle so it must be a square. 32. x = 4; a rectangle has ≅ diagonals that bisect each other. 34. m∠1 = 80, m∠2 = 100, m∠3 = 100 36. m∠1 = 56, m∠2 = 52

TEKS cumulative practice pp. 292–2932. H 4. G 6. J 8. G 10. H 12. 38 14. 108 16. The sum of the measures of the pentagon ∠ and the two hexagon ⦞ is 108 + 120 + 120 = 348. A circle has 360° so there is a gap of 12°. 18. Yes; if opp. sides are ≅ and }, then the quad. is a ▱. 20. Yes; if both pairs of opp. ∠s are ≅, then the quad. is a ▱.

Topic 7Lesson 7-1 pp. 299–3012. Scalene; side lengths are 215, 126, and 312.4. PQ = 2(-4 - 2)2 + (4 - 0)2

= 136 + 16 = 152 or 2113

QR = 2(2 - 0)2 + (0 - (-3))2 = 14 + 9 = 113

SR = 2(-6 - 0)2 + (1 - (-3))2 = 136 + 16 = 2113

SP = 2(-6 - (-4))2 + (1 - 4)2 = 14 + 9 = 113

Theorem 6-8 states if both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. PQ ≅ SR, and QR ≅ SP . Therefore PQRS is a parallelogram. By the definition of a parallelogram, PQ } SR and QR } SP .

6. M = (8 + 22 ,

4 + (-2)2 ) = (5, 1)

slope of AM = 1 - 35 - 3 = -1

slope of BC = -2 - 42 - 8 = 1

The product of the slopes is -1, so AM # BC . M is the midpoint of BC , so AM is the perpendicular bisector of BC . The Perpendicular Bisector Theorem (Theorem 5-2) states if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. Since point A is on the perpendicular bisector of BC , A is equidistant from the endpoints, B and C. Therefore, AB ≅ AC .

8. Rhombus; explanations may vary. Sample: All four sides are ≅ (with length 15), and diagonals are not ≅ (with lengths 2 and 4). 10. Square; explanations may vary. Sample: All sides are ≅ and consecutive sides are #. 12. Sample answer: Yes, the classmate is correct. If EFGH is a rhombus, it is also a parallelogram, and its opposite angles are congruent. If one pair of adjacent sides is perpendicular, then one angle and its opposite are both right angles. So the total measure for this pair of angles is 90 + 90 or 180. Since the total measure of interior angles of a quadrilateral is 360, subtract 180 from 360 to find that 180 is the total measure for the other two opposite angles. These opposite angles are also congruent, so the measure of each is 180 , 2 or 90, and they are also right angles. Therefore, EFGH is a rhombus with four right angles, or a square.14.

hsm11gmte_0607_t10950

x

�4 6O

y

�2

2

6

J

L

K

isosc.; rt. △


18


16.

hsm11gmte_0607_t10951

x

�2 2 4

H

B

F

O

y

�2

6

isosc.; rt. △18. Rectangle; opp. sides with = slopes makes it a ▱, and consecutive sides with slopes that are negative reciprocals are #. 20.

hsm11gmte_0607_t10953

2�2

4

4

V U

TSxO

y

parallelogram

22.

hsm11gmte_0607_t10955

4�4�8

8

4

�4

�8

�12

NM

QP

xO

y

square

24.

hsm11gmte_0607_t10957

NR

Q

PO 2�2�4�6

4

�2

x

y

parallelogram26. y

xO�4 4

2

�4

hsm11gmte_0607_t10960

K

H

J

I

rectangle

28a. To show that a quadrilateral is a square, show that consecutive sides are # and ≅, or that diagonals bisect each other, and are # and ≅. b. Answers may vary. Sample: Showing that the diagonals bisect each other and are ≅ and # uses only two pairs of coordinates. 30. G( - 4, 1), H(1, 3) 32. (0, 7.5), (3, 10), (6, 12.5) 34. ( - 1.8, 6), ( - 0.6, 7), (0.6, 8), (1.8, 9), (3, 10), (4.2, 11), (5.4, 12), (6.6, 13), (7.8, 14) 36. B 38. A

Lesson 7-2 pp. 304–3062. O(0, 0), S(0, a), T(a, a), W(a, 0) 4. Z(b, 0), W(b + c, 0), T(c, a), S(0, a) 6. S( - a, 0), Z(a, 0), W(b, c), T( - b, c). Another variable for b is acceptable. 8. Answers may vary. Sample:

y

xO

hsm11gmte_0608_t10963

A(0, 0)

D(d, e)C(b, c)

B(a, 0)

10a. T( - 2a, 0), R( - 2b, 2c), A(2b, 2c), P(2a, 0) b. Given: TRAP is an isosc. trapezoid; TR ≅ PA; D, E, F, and G are midpts. of sides. Prove: DEFG is a rhombus. c. Use the Midpoint Formula. d. Answers may vary. Sample: Use the Distance Formula to show

that DEFG is equilateral. 12a. W (a2, b2) , Z (c + e

2 , d2)

b. W(a, b), Z(c + e, d) c. W(2a, 2b), Z(2c + 2e, 2d) d. Answers may vary. Sample: Figure (c) avoids fractions. 14. Answers may vary. Sample: Place vertices at A(0, 0), B(a, 0), C(a + b, c), and D(b, c). Use the Distance Formula to find the lengths of opp. sides. 16. Answers may vary. Sample: Place vertices at A(0, 0), B(0, a), C(a, a), and D(a, 0). Use the fact that a horizontal line is # to a vertical line. 18. isosc. trapezoid 20. square 22. Answers may vary. Sample: A, C, H, F 24. Answers may vary. Sample: A, B, F, E 26. G 28. 1, 2, 3, 4

Technology Lab 7-3 p. 3072a. rhombus b. rectangle c. square 4. Answers may vary. Sample: 1. HG is a midsegment of △ACD, and EF is a midsegment of △ACB. (Construction) 2. HG } AC, EF } AC, HG = 1

2 AC, and EF = 12 AC .

(△ Midsegment Thm.) 3. HG } EF (If two lines are } to the same line, then they are } to each other.) 4. HG = EF (Transitive Prop. of Eq.) 5. EFGH is a ▱. (If one pair of opp. sides of a quad. is both ≅ and }, the quad. is a ▱.)

Lesson 7-3 pp. 309–3112. Yes; use Slope Formula. 4. Yes; use Midpoint Formula. 6. No; you need angle measures. 8. Yes; use Distance Formula.


19


10. Yes; show that they have same midpt. 12. Yes; find intersection point of segments.

14. M = (2a + 02 , 2b + 0

2 ) = (a, b)

OM = 2a2 + b2 FM = 2(2a - a)2 + (0 - b)2 = 2a2 + b2

EM = 2(0 - a)2 + (2b - b)2 = 2a2 + b2 Since EM = FM = OM, the midpoint of the hypotenuse is equidistant from the vertices of the right triangle.

16. Use the Midpoint Formula to find the coordinates of M and N. M = (2 + 2

2 , 4 + (-4)

2 ) = (2, 0);

N = (2 + (-6)2 , 4 + 4

2 ) = (-2, 4)

KN = 2(-2 - 2)2 + (4 - (-4))2 = 116 + 64 = 180 = 415

LN = 2(2 - (-6))2 + (0 - 4)2

= 164 + 16 = 180 = 415 The Distance Formula shows that KN and LM are the same length. So, by the definition of congruence, KN ≅ LM.

18. Answers may vary. Sample: y

x

hsm11gmte_0609_t10970

C(�b, 0) B(b, 0)

A(0, 2c)

O

Given: Isosc. △ABC with base BC and altitude AO. Prove: AO bisects BC . By the Distance Formula, CO = 2[0 - (-b)]2 + (0 - 0)2 = b and BO = 2(b - 0)2 + (0 - 0)2 = b. Since CO = BO, CO ≅ BO, so AO bisects BC by def. of seg. bisect.

20. Answers may vary. Sample: y

x

hsm11gmte_0609_t10973

K(�b, 0) T(b, 0)

I(0, a)

E(0, c)

Given: Kite KITE Prove: △KIE ≅ △TIE By the Distance Formula, KI = IT = 2a2 + b2, KE = TE = 2b2 + c2, and IE = 2(a - c)2. IE ≅ IE by the Reflexive Property of Congruence. So △KIE ≅ △TIE by SSS.

22. Answers may vary. Sample: Lines are # when the product of their slopes is -1; it is difficult to find the product without using coordinate methods. 24a. bc b. The point-slope formula for point (a, 0) and m = b

c is y - 0 = b

c (x - a) or y = bc (x - a). c. x = 0

d. The ordered pair (0, -abc ) satisfies

the equation of line q, x = 0. When

x = 0, y = bc (x - a) = b

c (0 - a) = -abc .

So p and q intersect at (0, -abc ) . e. The ordered

pair (0, -abc ) satisfies the equation of line q, x = 0.

When x = 0, y = ac (x - b) = a

c (0 - b) = -abc . So q and r

intersect at (0, -abc ) . f. (0, -ab

c ) 26. 10 28. 19

Topic Review pp. 312–3132. coordinate geometry 4. isosceles 6. kite 8. isosc. trapezoid 10. (a - b, c)

TEKS cumulative practice pp. 314–3152. J 4. J 6. J 8. -3

2 10. 109° 12. trapezoid 14. parallelogram 16. S(0, 0), T(b + c, d); (b + c

2 , d2) , d

b + c 18. ABCD is a square. Let the length of each side be a. By the Distance Formula, AC = a12 and BD = a12. AC = BD, so AC _ BD. 20. If L and N are opp. vertices, then the fourth vertex is (-4, 2). If L and M are opp. vertices, then the fourth vertex is (0, 8). If M and N are opp. vertices, then the fourth vertex is (6, -2).

Topic 8Lesson 8-1 pp. 322–3242. Yes; distances between corresponding pairs of points are equal. 4. Ask your teacher to check your work. 6a. Answers may vary. Sample: ∠ Q S ∠ Q′ b. QR and Q′R′; RS and R′S′; SP and S′P′; QP and Q′P′ 8a. Answers may vary. Sample: G S M b. GW and MR; WP and RT; PN and TX; NB and XS; BG and SM 10. U′(1, 16), G′(2, 12) 12. (x, y) S (x - 5, y - 7) 14. (-3, 7), (-2, 5), (-1, 7), (0, 5), (1, 7) 16. Translate a line segment in some other direction than along the segment. Then connect the endpoints of the line segment and its image to form a parallelogram. 18. T6-2, 147 (x, y) 20.


O�2

1

�4�8x

y

22. (x, y) S (x + 1, y - 3) 24. A 26a. C(-5, 2) b. Yes; answers may vary. Sample: The slope of DB = 1 and the slope of AC = -1, so DB # AC . Since ABCD is a parallelogram with perpendicular diagonals, ABCD is a rhombus.


20


Activity Lab 8-2 p. 3252. 90° 4. Open the compass to the length of the segment from the line of reflection to the point you want to reflect. Using the same compass setting, draw an arc that intersects the perpendicular on the opposite side of the line of reflection.

Lesson 8-2 pp. 329–3312. J′(-1, 4), A′(-3, 5), G′(-2, 1)

4. J′(1, 6), A′(3, 5), G′(2, 9)

6. J′(3, 4), A′(1, 5), G′(2, 1)

8. Answers may vary. Sample: scissors, airplanes, mirrors 10.

12a. -1 b. B′(0, 2); C′( -3, 3) c.

d. The coordinates of P′ will be (b, a); the x- and y-coordinates will switch.

14.

16a.

b. Answers may vary. Sample: For mirror writing, he would write from right to left. Because da Vinci was left-handed, his writing hand would not have covered up the words he had just written. 18. No; each point moves a distance equal to twice the point’s distance from the line of reflection. 20. ( - 1, - 4) 22. ( - 3, 2) 24. (1, - 1) 26. Yes; follow the steps of Exercise 25 using one leg of an isosc. △ to first form a ▱. Then reflect the original △ across the # bis. of the base of the second △ to form an isosc. trapezoid. 28. Yes; reflect an isosc. △ across its base. 30. Yes; reflect an isosc. rt. △ across its hyp. 32a. (3, 1) b. (-1, -3) c. (-3, -1) d. (1, 3) e. They are the same point. 34. H

Lesson 8-3 pp. 335–3372.

H

J

G

F

y4

O 2 42

4

x

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4. about the origin: L′(1, 2), M′(2, 6), N′(22, 4) about L: L′(2, 21), M′(3, 3), N′(21, 1) about M: L′(5, 26), M′(6, 22), N′(2, 24) about N: L′(7, 0), M′(8, 4), N′(4, 2) 6. 168.75° 8. any two rotations of a° and b° if a 7 0, b 7 0, and a + b = 360 10. Answers may vary. Sample: 200° rotation about point O 12.

hsm11gmte_0903_t15042

A

DB

D

A

P

B


yA

G

JJA

G xO 4

2

4


O 4

2

8A A

G

JJ

Gy

x

y

xO 4

4A¿

G = G¿J¿J

A


y

y x

x

52

2

5

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A

A

CB

B

C

y

xO 63

2

5

7

hsm11gmte_0902_t15047

H

G

H

G

F F

hsm11gmte_0902_t15044

Leonardo da Vinciwas left-handed.


21


14.

hsm11gmte_0903_t15034

DDB

R

P

R

B

16. V(2, 3), W(1, 25), X(4, 0), Y(0, 2) 18. Answers may vary. Sample: Because M is the midpoint of the diagonals, VM = QM and TM = NM. Since V and Q are equidistant from M, and m∠VMQ = 180, you know that r(180°, M) = Q. Similarly, r(180°, M) (T ) = N. Every point on VT can be rotated in this same way, so r(180°, T ) (VT ) = QN. Also, TQ can be mapped to NV , so r(180°, M ) (TQ ) = NV . Because rotations are rigid motions and preserve distance, TV = NQ and TQ = VN. 20. a 180° rotation 22. Ask your teacher to check your work. 24. F

Lesson 8-4 pp. 341–3432. line; rotational: 180°

hsm11gmte_0904_t15016

4. rotational: 90° 6. rotational: 180° 8. no symmetry 10. rotational: 60° 12. rotational: 180° 14. You could use pencil and paper to sketch each quadrilateral and its line of symmetry; 2

16. You could use pencil and paper to sketch each quadrilateral and its line of symmetry; 4

18. Answers may vary. Sample: TOMATO, HOAX 20. Line 22. Yes; explanations may vary. Sample: The angle bis. divides the angle into two ≅ angles. By the def. of a line of symmetry, the angle bis. is a line of symmetry. 24. Answers may vary. Sample:

30 , 10 = 3; ∙ 8 - 1 ∙ = ∙ 1 - 8 ∙ , 80 + 3 6 88; 80

80 = 3333

26. (3, -4)

28. reflectional symmetry across the y-axis

xO

y

2 2

4

2

30. point symmetry about the origin

x

y

2

O2 2

2

32. Answers may vary. Samples are given.

34. Yes; all regular polygons with n sides have rotational symmetry with angle of rotation 360°

n and center of rotation at the center of the polygon. 36a. 2 b.

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38. rotational: 180°; point 40. none 42. 60 44. 2

Lesson 8-5 pp. 347–3492. reflection; x = -1

2 4. rotation; center (3, 0), angle of rotation 180° 6. rotation; center (0, 2), angle of rotation 180° 8. 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.


M

MM

m

10. A rotation; the center of rotation is C and the angle of rotation is 190°.


LL

L

m

C


22


12.


Ox

y

4

2

4

6

BN

PP

N

B

14.

x

42

2

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N

P

BB

PN

4 2

y

16. y4

2

O 42-2-2

-4

-4xF″

D″

E″

F

D

E

18. Rx = 12

∘ T60, 27 20. Answers may vary. Sample: Translate the black R so that one point moves to its corresponding point on the blue R. Then reflect across a line passing through that point and the point halfway between two other corresponding points.

22. y

x

2

4

2

hsm11gmte_0906_t15250

2

2B

AA B

B

A12

a 180° rotation about (0, 0) 24. No; answers may vary. Sample: The diagram shows a counterexample. A” is the image of A for a 40° rotation about P(A’), followed by a reflection across line /, while A’’’ is the image of A for a reflection across line / (A’’’), followed by a 40° rotation about P.

hsm11gmte_0906_t15214

A

A

A

A

A

P40

40

26. G 28. Yes; in triangles BLC and ITG, ∠BLC ≅ ∠ITG because both are right angles, and ∠C ≅ ∠G and BC ≅ IG because corresponding parts of corresponding triangle are congruent. So △BLC ≅ △ITG by AAS, and BL ≅ IT because corresponding parts of corresponding triangle are congruent.

Lesson 8-6 pp. 353–3542. GC ≅ FD; Sample: Reflect GC over the y-axis; then translate 1 unit right and 2 units down. 4a. rotations, translations, and glide reflections b. translations and glide reflections 6. Sample: Reflect triangle LMN over the x-axis; then translate 2 units to the left. 8. Rotate, translate, and reflect △CIQ so that CQ and NZ coincide and points I and V are on the same side of the line containing CQ. Since rigid transformations preserve angle measures, point I is on the same line as point V. Suppose temporarily that I and V do not coincide. Since ∠CIQ and ∠NVZ are congruent corresponding angles, IQ and VZ are parallel. But Q and Z coincide, which contradicts conclusion that IQ } VZ . Therefore, I and V coincide, and △NVZ ≅ △CIQ. 10. Answers may vary. Sample: Translate the top triangle down 6 units; reflect across the x-axis; rotate the bottom triangle 180º about the point ( - 3, 0), then reflect across the line x = -3; reflect the bottom triangle across the line x = -4, then rotate 180° about the point ( - 4, 0). 12. Sample: reflections, translations, rotations, glide reflections 14. 31.5

Activity Lab 8-7 p. 3552. y

2 6 10�2

2

6

10

x

R�

S�

O

4. y

2 6 10�2

2

6

10

x

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R�

S�

O


23


6. R′S′ = 2RS

8. R″S″ = 5 (1 - 12)2 + ( -1

2 - 2)2

= 514 + 25

4 = 5264 = 1

2126

10. Answers may vary. Sample: The length of a line segment after a dilation of scale n is equal to n times the length of the original segment. 12. Dilations of a line through the origin all coincide with the original line.

Lesson 8-7 pp. 361–3632. enlargement; 1.5 4. enlargement; 2 6. enlargement; 1.5 8.

hsm11gmte_0905_t14987

y

xOP

R

Q

34

94

32

, D34

(x, y) = (34 x, 34 y)

10. 1.2 cm 12. 0.28 cm 14. N′(-4, 7) 16. Q′(-300, 400), R′(-200, -100), T′(300, 100), W′(300, 500) 18. Q′(-3, 4), R′(-4, 9), T′(-9, 7), W′(-9, 3) 20. S(-1, -5), U(2, 1), J(5, -5)22.

hsm11gmte_0905_t14976

y

x

6 6

6

10

NM

P Q

N

M

P

Q24. Let / be given by the equation y = ax for some real number a. If C = (c1, c2) is on /, then c2 = ac1, and kc2 = k (ac1) = a (kc1), so Dk(C ) = (kc1, kc2) is also on /. 26. Connect corresp. points A and A′, and B and B′. Extend AA′ and BB′ until they intersect. The intersection point is the center of dilation. The scale factor is the length of A′B′ divided by the length of AB. 28. Answers may vary. Sample: The image is an equilateral triangle with sides 10 in. long. For two of the pairs of corresp. sides, the corresp. sides lie on the same line. The third pair of corresp. sides are }.

30a. y

xO 2 4�2R�

Q�

P�

R

Q

P

hsm11gmte_0905_t14972

�4

4

2

b. Answers may vary. Sample: The origin is the midpt. of each segment joining an image point to its preimage. A figure has point symmetry if there is a point P through which the figure reflects onto itself. 32.

2 4 6

y

xO

2

�2

T′

R′

I′

34. C 36. C

Lesson 8-8 pp. 368–3702. H(21, 4) S H′(22, 2)

J(1, 4) S J′(2, 2) K(1, 24) S K′(2, 22) L(–1, 24) S L′(22, 22);

4a. (x, y) S (3x, 4y); Stretching ABCD horizontally by a factor of 3 and vertically by a factor of 4 results in a square with side length 24. b. (x, y) S (1

4x, 13 y) ; Compressing ABCD horizontally by a factor of 14 and vertically by a factor of 13 results in a square with side length 2. 6. L(0, 0) S L′(0, 0)

M(0, 4) S M′(0, 2) N(4, 0) S N′(1, 0)

8. A″(0, 3) S A(0, 1) B″(0, 0) S B(0, 0) C″(22, 0) S C(21, 0)

10. You can reverse the transformations by horizontally compressing the image by a scale factor of 12, then dilating by a scale factor of 13. 12. Answers may vary. Sample: Translate EFGH 2 units to the left, then dilate the image by a scale factor of 2 centered at (0, 4) to get E′F′G′H′. 14. a translation 2 units to the right: (x, y) S (x 1 2, y) 16. The rigid transformation could be a translation, a reflection, or a rotation. The non-rigid transformation must be a dilation because other non-rigid transformations do not preserve angle measure. 18. F 20. Answers may vary. Sample: Identify two points on each line. Use the Slope Formula to find the slope of each line. If the slopes are negative reciprocals, then the lines are perpendicular.


24


Topic Review pp. 371–3752. glide reflection 4. rigid transformation 6. R′(-4, 3), S′(-6, 6), T′(-10, 8)

8. T6-1, 77( x, y ) 10. A′(2, 4), B′(10, 1), C′(3, 0)

12.

14. P′(4, -1) 16. reflectional; rotational: 180°; point

18. reflectional; rotational: 120°

20. line, rotational and point 22. same; rotation 24. opposite; rotation and glide reflection 26. (r(90°, O) ° Rx@axis)(△XYZ) 28. enlargement; 230.

32. P′(-4, 2), Q′(4, 2), R′(4, -1), S′(-4, -1) 34. P′(-1, 4), Q′(1, 4), R′(1, -2), S′(-1, -2) 36. E(-3, 6), F(3, 6), G(3, -3), H(-3, -3)

TEKS cumulative practice pp. 376–3772. F 4. J 6. H 8. H 10. J 12. 95 14. 62 16. Quadrants I and IV

x

y

�3

3

6

B�

E�

D�

7

D

E

B

270 �


18. Yes; the coordinates of the vertices are A(1, 4), B(2, -2), and C (-4, -3). The slope of AB is 4 - (-2)

1 - 2 = - 61 and the slope of BC is -2 - (-3)2 - (-4) = 1

6.

Since the product of their slopes is -1, AB # BC, so ∠ABC is a rt. ∠. Thus △ABC is a rt. △ OR other appropriate method.

Topic 9Lesson 9-1 pp. 383–3852. ∠C ≅ ∠W, ∠A ≅ ∠V, ∠B ≅ ∠T; CAWV = AB

VT = BCTW 4. △ABC ∙ △DEF (in any order); scale

factor is 3 : 5 6. △JKL ∙ △PQR; scale factor is 2 : 1 8. x = 8, y = 9, z = 5.25 10. 120 pixels wide by 90 pixels high 12. 5 : 3 14. 25 16. The distance on the map is about 2.8 cm, so the actual distance is (2.8)(112), or about 314 km. 18. Answers may vary. Sample: If two figures are congruent, then corresponding angles are congruent and corresponding sides are congruent. Therefore, the ratio of the lengths of each pair of corresponding sides is 1, so the sides are proportional with a scale factor of 1 : 1. 20. All angles in any rectangle are right angles, so all corresponding angles are congruent. The ratio of two pair of consecutive sides for each rectangle is the same. Since opposite sides of a parallelogram are equal, the other two pair of sides will also have the same ratio. So corresponding sides form equal ratios and are proportional. So BCEG ∙ LJAW. 22a. The slopes of AB, CD, AE, and FG are all -2. The slope of BC , AD, EF , and AG is 12. For each pair of consecutive sides of ABCD, the slopes are negative reciprocals, so ABCD has four right angles. Similarly, AEFG has four right angles. The measures of ∠A, ∠ABC , ∠BCD, ∠CDA, ∠E, ∠F , and ∠G, are all 90. b. By the Distance Formula, AB = BC = CD = AD = 15 and AE = EF = FG = AG = 215.


y

x

�2�4�6�8�10 O

2

4

6

8T�

S�

R�


y

xO

A

BC

4

8

A�

B�C�

x � 4

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y

4

4

�4

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xG� T�

B�

G

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H � H�

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O x

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25


c. All the angles of AEFG and ABCD are congruent. ABAE = BC

EF = CDFG = AD

AB = 15215

= 12. The corresponding

sides are proportional, so AEFG ∙ ABCD. 24. Ask your teacher to check your work. 26. x = 4.4; 4 : 3 28a. 24 in., 32 in. b. Ratio of perimeters: 54

72 = 34, scale

factor is 1216 = 3

4; they are the same. 30. J

Lesson 9-2 pp. 389–3912.

4. D0.5 ∘ r(90°,O) 6. D0.5 ∘ Ry@axis 8. Answers may vary. Sample: There is no similarity transformation. The figures are not similar because corresponding sides are not proportional. 10a. Yes, the triangles are similar because there is a similarity transformation between them: a rotation of 180° followed by a dilation by a factor of 2.5. b. 330 m 12. Yes, the triangles are similar because there is a similarity transformation that maps △AJL to △ABC . 14. Answers may vary. Sample: No, there is not a similarity transformation. To create △NOP , △ABC is reflected across the x-axis. Then the x-coordinates are scaled by a factor of 5 and the y-coordinates are scaled by a factor of 4. Because the reflected triangle is 5 times as wide as △ABC but only 4 times as tall, the figures are not similar. Therefore, there is no similarity transformation between them. 16a. true b. true c. false 18. The figures are not similar. 20. Yes; Answers may vary. Sample: A dilation with scale factor 2 maps ABCDE to STUVW, so the pentagons are similar. Therefore, their corresponding sides are proportional. STAB = TU

BC = UVCD = VW

DE = WSEA 22. sometimes 24. always

26. No, the photos will not be similar. There is no similarity transformation that maps the smaller photo to the larger photo. 28a. 104 inches by 24 inches; b. Enlarging the banner is a dilation, which is an example of a similarity transformation. 30. 6 32. 13

Lesson 9-3 pp. 395–3972. Not similar; using the sides that contain the right angles, the ratio of the shorter sides is 1 : 1, while the ratio of the longer sides is 5 : 4. 4. ∠A ≅ ∠A (Reflexive Property of Congruence), and ∠ABC ≅ ∠ACD (Given), so △ABC ∙ △ACD by AA ∙.

6. In △PQR and △STV , ∠Q ≅ ∠T because perpendicular lines form right angles, which are congruent. The sides that contain the angles are proportional (given). So △PQR ∙ ∠STV by SAS ∙, and ∠KRV ≅ ∠KVR because corresponding angles of similar triangles are congruent. Thus △VKR is isosceles by the Converse of Isosceles Triangle Theorem. 8. A pair of congruent angles are given and the two right angles are congruent, so the triangles are similar by AA ∙; 13.75 ft or 13 ft 9 in. 10a. No; the ratios of the sides that form the vertex angles are equal, but the vertex angles may not be congruent. b. Yes; Sample explanation: An isosceles right triangle has two angles 45°, so any two isosceles right triangles are similar by AA ∙. 12. Yes; the two parallel lines and the two sides determine two pairs of congruent corresponding angles, so the two triangles are similar by AA ∙. 14. 4 : 3; sample explanation: Since ∠P ≅ ∠S and ∠PQM ≅ ∠STR, △PQM ∙ △STR by AA ∙. So the ratio MQ

RT = PMSR = the ratio of

corresponding sides in △PMN and △SRW ; namely, 4 : 3. 16. 6 18. 10 20. BC

AC = EFDF , EF # AF , BC # AF

(Given); ∠ACB and ∠DFE are right angles. (Definition of perpendicular); ∠ACB ≅ ∠DFE (All right angles are congruent.); △ABC ∙ △DEF (SAS ∙); ∠BAC ≅ ∠EDF (Definition of similar); /1 } /2 (If corresponding angles are congruent, then lines are parallel.) 22.


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 ABQR = AC

QS = BCRS to get QX

QR = ACQS = BC

RS . So

BC = XY and AC = QY. Then △ABC ≅ △QXY by SSS. ∠B ≅ ∠QXY and ∠C ≅ ∠QXY (Corresponding parts of congruent triangles are congruent.). Since XY } RS, ∠QXY ≅ ∠R and ∠QYX ≅ ∠S (If lines are parallel, then corresponding angles are congruent.). ∠B ≅ ∠R and ∠C ≅ ∠S (Transitive Property). Therefore, △ABC ≅ △QRS by AA ∙.

24. G

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y

4

6

8

2

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O

M� T�

A�


26




xO

y

2

2

4

6

4 6

A

B

C

D

Slope of AB = 1, slope of BC = -1, slope of CD = 1, slope of AD = -1; for each pair of consecutive sides, the slopes are negative reciprocals, so the figure has 4 right angles. By the Distance Formula, AB = BC = CD = AD = 18, so all 4 sides are congruent. Since ABCD is a quadrilateral with 4 right angles and 4 congruent sides, it is a square.

Lesson 9-4 pp. 403–4052. Answers may vary. Sample: △QPR ∙ △SPQ ∙ △SQR 4. Answers may vary. Sample: 140 or 2110 6. 25 8. x = 20, y = 1015 10. x = 317, y = 12 12a. 4 cm b.


2 cm

4 cm

8 cmc. Answers may vary. Sample: Draw a 10-cm segment. Construct a perpendicular of length 4 cm that is 2 cm from one endpoint; connect to form a triangle. 14. 2.5 16. 1 18. Yes; the proportion a1ab

= 1abb

is true by the Cross Products Property and satisfies the definition of the geometric mean. 20. 1212 units 22. 1) ∠DCE is a right triangle. (Given); 2) FC is an altitude of ∠DCE. (Given); 3) △FCE ∙ △FDC (Theorem 9-3); 4) DF

FC = FCEF (Corresponding sides of

similar triangles are proportional.) 24. WY and YX; WZWY = WY

WX and ZXYX = YX

WX 26. 3 28. 6 30. △ACD ∙ △CBD by Theorem 9-3, so AD

CD = CDBD

because corresponding sides of similar triangles are proportional. 32. Right △ABC with altitude to the hypotenuse AB (given); ab = b

c (Corollary 1 to Theorem

9-3); Slope of <AC

>= b

a and slope of <BC

>= - b

c . Since ab = b

c , - ab = - b

c . Therefore the product of the slopes ba # - a

b = -1. 34. 8.50 m 36. /1 = 12, /2 = 12, a = 1, s2 = 1 38. /2 = 213, h = 4, a = 13, s1 = 1 40a. x

a = ac ,

yb = b

c b. c2 = a2 + b2 c. The square of the hypotenuse equals the sum of the squares of the legs. 42. B 44. Yes; they can both be correct. The three possibilities for the measures of the four angles are 38, 124, 124, 74; 38, 124, 38, 160; and 38, 124, 99, 99.

Activity Lab 9-4 pp. 406–4072. Using the quadratic formula, x = -b { 2b2 - 4ac

2a with a = 1, b = -1, c = -1 gives

x = - (-1) { 2(-1)2 - 4(1)(-1)2(1) = 1 { 21 + 4

2 ; since

1 - 15 6 0, x = 1 + 252 . 4. 13, 21, 34, 55, 89, 144,

233, 377, 610 6. For terms in the Fibonacci sequence, the ratio of each term to the previous term gets closer and closer to the golden ratio. 8a.


D

D

Q D

Q

Q

Q D Q D Q D Q D Q D

Q

Q Q

Q D Q D Q D Q DQ DQ DQ DQ DQ Q Q Q

Q

Q

Q

QQ D Q D Q D

D DQ

Q

Q

b. The number of ancestors in each generation is a Fibonacci number. 10. If AD = 1.618 and DB = 1, then AB = 2.618. So 1x = x

2.618 and x2 = 2.618. So x = 22.618 = 1.618.

Lesson 9-5 pp. 412–4142. 7.5 4. 3.6 6. 8.25 mm 8. 9.6 10. Isosceles; AC : BC is 1 : 1 by the Triangle-Angle-Bisector Theorem. 12. JP by the Triangle Proportionality Theorem 14. MP by the Corollary to the Triangle Proportionality Theorem 16. 10

3 18. ABBC = WP

PC by the

Triangle Proportionality Theorem, and WPPC = WX

XY . Therefore AB

BC = WXXY by the Transitive Property

of Equality. 20. By the Triangle Proportionality Theorem, CD

DB = CAAF . By the Corresponding Angles

Postulate, ∠3 ≅ ∠1. Since AD bisects ∠CAB, ∠1 ≅ ∠2. By the Alternate Interior Angles Theorem, ∠2 ≅ ∠4. So ∠3 ≅ ∠4 by the Transitive Property of Congruence. By the Converse of the Isosceles Triangle Theorem, BA = AF . Substituting BA for AF, CD

DB = CABA .

22. 750 ft 24. Use the diagram with Ex. 20, with AD } EB. It is given that CD

DB = CABA , and you want to

prove that ∠1 ≅ ∠2. By the Triangle Proportionality Theorem, CA

AF = CDDB . So CA

BA = CAAF , by the Transitive

Property of Equality, and BA = AF . Therefore, ∠3 ≅ ∠4 by the Isosceles Triangle Theorem. Using properties of parallel lines, ∠1 ≅ ∠3 and ∠2 ≅ ∠4. So ∠1 ≅ ∠2 by the Transitive Property of Congruence, and AD bisects ∠CAB by the definition of angle bisector. 26a. 90 units b. 14 units 28. Use the Triangle Proportionality Theorem to write the proportion AB

BD = ACCE , then find the values of BD,

AC, and CE to calculate the unknown length AB. 30. 20 32. 118


27


Topic Review pp. 415–4172. proportion 4. means, extremes (in either order) 6. △PQR ∼ △XYZ ; 3 : 2 (or △XYZ ∼ △PQR; 2 : 3) 8. x = 63, y = 8 10. No; the corresponding side lengths are not proportional. 12. Answers may vary. Sample: The figures are similar because a composition of a translation, a rotation, and a dilation maps p onto d. 14. 45 ft 16. If lines are 7 , the corresp. ⦞are ≅, so △RPT ∼ △SGT by AA∼. 18. 2215 20. 235

22. x = 12, y = 425 24. 3.6 26. 12 28. 77

TEKS cumulative practice pp. 418–4192. H 4. J 6. G 8. F 10. G 12. 138.72 14. 135 16. 3.75 18.

hsm11gmte_07cu_t10196.ai

Am

Draw arcs from A intersecting line m at two points. Open the compass wider and draw an arc above m from each of the two points. Draw a line from A to the intersection of arcs above A.

Topic 10Activity Lab 10-1 p. 4222a. a2 + b2 = c2 b. Yes 4. No; c2 7 a2 + b2 for an obtuse △, and c2 6 a2 + b2 for an acute △.

Lesson 10-1 pp. 427–429

2. 34 4. eb = bc and da = a

c by Corollary 2 of Theorem 9-3. By the Cross Products Property, b2 = ec and a2 = dc. Thus, b2 + a2 = ec + dc by the Addition Property of Equality. By factoring, b2 + a2 = c(e + d). By the Segment Addition Postulate, c = e + d. So b2 + a2 = c(c) by the Substitution Property. Simplify to get b2 + a2 = c2. 6. 1105 8. 513 10. No; 82 + 242 ≠ 252 12. 4.2 in. 14. Use the Distance Formula to find the length of

each side of △ABC . AC = 2(-6 - 1)2 + (6 - (-1))2

= 149 + 49 = 198 BC = 2(-2 - 1)2 + (-4 - (-1))2

= 19 + 9 = 118 AB = 2(-6 - (-2))2 + (6 - (-4))2

= 116 + 100 = 1116

(198)2 + (118)2 = (1116)2, so by the Converse of the Pythagorean Theorem, △ABC is a right triangle. By the Hinge Theorem, the longest side

(with length 1116) lies opposite the largest angle. So ∠C must be the right angle. Therefore, AC ⊥ BC .

16. 815 18. 29 20. 84 22. 17 m 24. right 26. 2830 km 28. 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 7 a2 + b2. c2 7 x2 and c 7 x by the Prop. of Inequalities. If c 7 x, then m∠C 7 m∠E by the Converse of the Hinge Thm. An angle with measure 7 90 is obtuse, so △ABC is an obtuse △. 30. 4 32. 61

Lesson 10-2 pp. 433–4342. 512 4. 14.1 cm 6. x = 20, y = 2013 8. x = 5, y = 513 10. x = 4, y = 2 12. 50 ft 14. a = 6, b = 612, c = 213, d = 6 16. a = 4, b = 4 18. a = 14, b = 612 20. 424 ft 22. Answers may vary. Sample: A ramp up to a door is 12 ft long. The ramp forms a 30° angle with the ground. How high off the ground is the door? Answer: 6 ft 24a. 13 units b. 213 units c. S13 units 26. J

Technology Lab 10-3 p. 4352a. The ratio becomes larger as m∠A increases. b. 0; 1 4. It does not change; the ratio becomes smaller as m∠A increases; 1, 0; yes; cosine; it does not change; the ratio becomes larger as ∠A increases; 0; very large; yes; tangent.

Lesson 10-3 pp. 438–4402. 412

9 ; 79; 4127 4. 11.5 6. 14.4 8. 106.5 10. 39 in.

12. 21 14. 46 16. 24 18. about 17 ft 8 in. 20. 49 22. 36 24. w ≈ 6.7, x ≈ 8.1 26. 52 m 28. Answers may vary. Samples are given. a. sin A = opposite

hypotenuse,

and the hypotenuse of a right triangle is always the longest side, so sin A is a proper fraction, and sin

A 6 1. b. cos A = adjacenthypotenuse, and the hypotenuse of a

right triangle is always the longest side, so cos A is a proper fraction, and cos A 6 1. 30a.

30

60

2a

a

a 3

hsm11gmte_0803_t12433

Using the ratio of sides 1 : 13 : 2 for a 30°@60°@90° triangle, tan 60° = 13

1 = 13.

b. Answers may vary. Sample: sin 60° = 13 # cos 60° 32. A 34. 11, 10, 9, 8, or 7; BC 7 AC , so m∠A 7 m∠B by the Converse of the Hinge Thm.; m∠A + m∠B = 12 by the Triangle Angle-Sum Thm.


28


Lesson 10-4 pp. 443–4452. angle of depression from boat to sub 4. angle of depression from tree to boat 6. angle of elevation from Maya to top of waterfall 8. angle of depression from top of waterfall to Maya 10. 502.4 m 12. 263.3 yd 14. 769 ft 16. 72, 72 18. 27, 27 20a. Length of any guy wire = distance on the ground from the tower to the guy wire div. by the cosine of the angle formed by the guy wire and the ground. b. Height of attachment = distance on the ground from the tower to the guy wire times the tangent of the angle formed by the guy wire and the ground. 22. 193 m 24. about 2.8 26. about 27.7 ft 28. C

Topic Review pp. 446–4472. ∠ of elevation 4. 21113 6. 1212 8. x = 7, y = 712 10. x = 613, y = 12

12. 70.7 ft 14. 1620 or 45; 12

20 or 35; 1612 or 43 16. 33.1

TEKS cumulative practice pp. 448–4492. H 4. H 6. G 8. H 10. 18.7 12. 28 14. 77 16. 31 18. Explanations may vary. Sample: AD ≅ CD by the definitions of median and midpt, and BD ≅ BD by the Reflexive Prop. of ≅. Since the △ was constructed with BC 7 BA, by the Converse of the Hinge Theorem, the larger included angle is opposite the longer third side. This means that m∠BDC 7 m∠BDA. Since those two angles are adjacent and form a straight angle, then m∠BDC 7 90, which means the angle is obtuse. 20a. ED

CB = DFBA, 20

25 = DF40, DF = 32 in. b. Yes, since

△EDF ∙ △CBA, ∠DEF ≅ ∠BCA. Since complements of ≅ ⦞ are ≅, ∠DFE ≅ ∠ACF. Therefore, EF } AC because if ≅ corresp. ⦞, lines are }.

Topic 11Lesson 11-1 pp. 456–4582. BDF¬, CDB¬, DEB¬, EFC¬, EFD¬, FBD¬, FBE¬, CFD¬ 4. 128 6. 218 8. 52 10. 20p cm 12. 8.4p m 14. 99 ft 16. 19 in. 18. 8p ft 20. 33p in. 22. 3p m 24. 180 26. 55 28. 290 30a. 6 b. 30 c. 120 32. 40 34. The circumference is doubled; explanations may vary. Sample: Since C = 2pr , doubling the radius results in 2p(2r) = 2(2pr) = 2C . 36. Since AR ≅ RW and AR + RW = AW by the Segment Addition Postulate, AW = 2 # AR. So the radius of the outer circle is twice the radius of the inner circle. Because ∠QAR and ∠SAU are vertical angles, and m∠SAT = 1

2 m∠SAU, m∠QAR = 2 # m∠SAT . The length of

ST¬ = m∠SAT360 # 2p(2 # AR) = m∠SAT

90 # p(AR) and

the length of QR¬ = m∠QAR360

# 2p(AR) = 2 # m∠SAT360

#

2p(AR) = m∠SAT90

# p(AR). Therefore the length of

ST¬ = the length of QR¬ by the Transitive Property of Equality. 38. 325.7 yd, 333.5 yd, 341.4 yd, 349.2 yd, 357.1 yd, 364.9 yd, 372.8 yd, 380.6 yd 40. G

Activity Lab 11-2 pp. 459–4602. Ask your teacher to check your work. 4. Ask your teacher to check your work. 6. 2p; a circle has an angle measure of 360°. Dividing the circumference, 2pr , by the length of a radius r, we find that there are 2p radians in 360°. 8. p18 radians 10. p2 radians

12. 3p2 radians 14. 5p18 radians 16. 17p18 radians

18. 130° 20. 240° 22. 150°

Lesson 11-2 pp. 464–4652. 4p3 4. p3 , 1.05 6. p9 , 0.35 8. 198° 10. 172° 12. 270° 14. 10.5 m 16. ≈196 ft 18. ≈42.2 in .; The distance the tip of the wiper travels is the length of the arc of a circle with radius 22 in. and central angle 110°, and 110

360(p)(22) = 121p9 ≈ 42.2.

20. Ask your teacher to check your work. 22. The student forgot to include parentheses around 2 * p. 24. ≈4008.7 mi 26. C 28. Using the Distance Formula, AB = CD = 3, BC = AD = 5 and RS = TV = 6, ST = RV = 10. The slopes of AB and CD = 0 and the slopes of BC and AD are undefined. So both AB and CD are perpendicular to BC and AD. Therefore ABCD is a rectangle and angles A, B, C, and D are all right angles. The slopes of RS and TV are undefined, and the slopes of ST and RV = 0. So RSTV is a rectangle and angles R, S, T, and V are all right angles. Since all right angles are congruent, the pairs of corresponding angles are congruent. The short sides of the two rectangles are 3 and 6, and the long sides are 5 and 10. Since 36 = 5

10 = 12, the corresponding sides

are proportional. Therefore ABCD ∙ RSTV by the definition of similar polygons.

Activity Lab 11-3 p. 4662. b ≈ 1

2C 4. A = pr # r = pr2

6. A = b # h = 2r # 12pr = pr2; yes

Lesson 11-3 pp. 472–4732. 0.7225p ft2 4. 22.1 cm2 6. 3.3 m2

8. 40.5p yd2 10. 169p6 m2 12. (54p + 20.2513) cm2

14. (120p + 3613) m2 16. 25p4 m2 18. 116 mm2

20. Yes; ∠AOC ≅ ∠BOD (Vertical angles are ≅.), so the two sectors are ≅ and will have = areas.


29


22a.

hsm11gmte_1007_t14626

10 ft

10 ft

8 ft

2 ft10 ft

8 ft

Boat path

b. 239 ft2 24. G 26a. A = pr2 = 81p, so r2 = 81 and r = 9 yd. C = 2pr = 2p(9) = 18p yd b. C = 18p yd, so the length of a 45° arc is 45

36018p = 1818p = 9p

4 ≈ 7.1 yd.

Lesson 11-4 pp. 477–4782. x2 + (y - 3)2 = 49 4. x2 + y2 = 16 6. (x + 9)2 + (y + 4)2 = 5 8. center (3, -8); radius 10

10. (x - 4)2 + (y + 4)2 = 4 12. (x - 1)2 + (y - 2)2 = 17 14. (x + 10)2 + (y + 5)2 = 125 16. position (-4, 9); range 12 18. (x - 2)2 + (y - 2)2 = 16 20. (x - 4)2 + (y - 3)2 = 25 22. (x - 3)2 + (y - 3)2 = 8 24. No; the x and y terms are not squared. 26. circumference 16p units; area 64p units2 28. The circles are concentric with the center (-6, -5). 30.

(3, 2), (2, 3) 32.

(3, 5) 34. about 11.5, 11.5, 49.8, and 49.8 units2 36a. x2 + y2 = 15,681,600 b. 69.1 mi c. about 32 days 38. J

Topic Review pp. 479–4812. radian 4. 30º 6. 330 8. 22p

9 in 10. 25p9 m

12. p3 14. p 16. 150º 18. 26.2 ft 20. 49p4 ft2

22. 18.3 m2 24. x2 + (y + 2)2 = 9 26. (x + 3)2 + (y + 4)2 = 25 28. center (7, -5); radius 6

TEKS cumulative practice pp. 482–4832. H 4. G 6. G 8. G 10. G 12. 50 14. 310 16. 9.4 18. 25.73 20. 3.93 22. 0.10 24. 86 26. (x - 2)2 + (y - 4)2 = 428.


arc

radius

central angle

sector

segment

diameter

center

30. (x - 3)2 + y2 = 41 32a. 6 * 9p = 169.65 cm2. b. No; the diameter of each can is 6 cm, so three cans across need 3 * 6 = 18 cm. Since the box is only 16 cm across, the cans will not fit.

Topic 12Lesson 12-1 pp. 490–4912. 47 4. 253.0 km 6. 178.9 km 8. 3.6 cm 10. 78 cm 12. It is given that BA and BC are tangent to } O at A and C. It follows that AB # OA and BC # OC since lines tangent to a } are # to the radius at the point of tangency. By the def. of rt. △, △ BAO and △ BCO are rt. triangles. Also, AO ≅ OC , since radii of a circle are ≅, and by the Refl. Prop. of ≅, BO ≅ BO. By HL, △ BAO ≅ △BCO. Thus, it follows that BA ≅ BC , as corresp. parts of ≅ triangles are ≅. 14. All 4 are ≅; the two tangents to each coin from A are ≅, so by the Transitive Prop. of ≅, all the tangents are ≅. 16. At each vertex, let the radius of a circle be the distance from the vertex to either point of tangency of the inscribed circle. 18. } A and } B are given with common tangents DF and CE. It follows that GD = GC and GE = GF since two tan. segments from a pt. to a } are ≅. By the Div. Prop. of = , GDGC = 1 and GF

GE = 1. Also by the Trans. Prop. of = , GDGC = GF

GE. ∠ DGC ≅ ∠ EGF since vert. angles are ≅. Thus, by the SAS ∙ Thm., △ GDC ∙ △GFE. 20. 66 22. 390

Lesson 12-2 pp. 496–4982. Answers may vary. Sample: ET¬ ≅ GH¬ ≅ JN¬ ≅ ML¬; ET ≅ GH ≅ JN ≅ ML; ∠ TFE ≅ ∠HFG; ∠ JKN ≅ ∠MKL 4. 8

O

-8

4

4

y

x

(3, −8)

hsm11gmte_1205_t12855

O�2

�2

2

4

2

y

x

hsm11gmte_1205_t12867

x

y

O

8

6�2


30


6. The center is at the intersection of GH and KM, because if a chord is the # bis. of another chord, then the first chord is a diameter; two diameters intersect at the center of a circle. 8. } O with ∠ AOB ≅ ∠COD (Given); AO ≅ BO ≅ CO ≅ DO (All radii of a } are ≅.). △ AOB ≅ △COD (SAS); AB ≅ CD (Corresp. parts of ≅ triangles are ≅.). 10. } O with diameter ED # AB at C (Given). Draw OA and OB (2 pts. determine a line.). ∠ ACO and ∠ BCO are rt. angles (# lines form rt. angles.). △ ACO and △ BCO are rt. triangles (Def. of a rt. △). OA ≅ OB (All radii of a }are ≅.); OC ≅ OC (Refl. Prop. of ≅); △ ACO ≅ △BCO (HL);AC ≅ BC (Corresp. parts of ≅ triangles are ≅.);∠ AOC ≅ ∠ BOC (Corresp. parts of ≅ triangles are ≅.); AD¬ ≅ BD¬ (≅ central angles have ≅ arcs.). 12. } A with CE # BD (Given); CF ≅ CF (Refl. Prop. of ≅); BF ≅ DF (A diameter # to a chord bisects the chord.); ∠ BFC and ∠ DFC are rt. angles(# lines form rt. angles.); ∠ BFC ≅ ∠DFC (Rt. angles are ≅.); △ BFC ≅ △DFC (SAS); BC ≅ CD (Corresp. parts of ≅ triangles are ≅.); BC¬ ≅ DC¬ (≅ chords have ≅ arcs.). 14. Answers may vary. Sample: Use a carpenter’s square to locate two chords and the perpendicular bisectors of each. The two perpendicular bisectors will intersect at the center of the circle. 16. 10 cm 18. 12 cm 20. 13.9 cm 22. 9.2 units 24.


A

O

D

C

B

Given: } O with AB¬ ≅ CD¬ Prove:∠ AOB ≅ ∠COD Proof: m∠AOB = mAB¬ and m∠COD = mCD¬ (definition of arc measure) AB¬ ≅ CD¬ (given), so mAB¬ = mCD¬ (Def. of ≅ arcs). Therefore, m∠AOB = m∠COD (Substitution). Hence ∠ AOB ≅ ∠COD (Def. of ≅ ⦞).

26.


A

O

D

C

B

Given: } O with AB¬ ≅ CD¬ Prove: AB ≅ CD Proof: It is given that AB¬ ≅ CD¬, so ∠ AOB ≅ ∠COD (If arcs are ≅, then their central ⦞ are ≅.). Also, AO = BO = CO = DO (Radii of a } are ≅.), so △ AOB ≅ △COD (SAS), and AB ≅ CD (Corresp. parts of ≅ △s are ≅.).

28.


O

D

C

B

Given: Concentric circles, BC is tangent to the smaller circle at D Prove: D is the midpt. of BC Proof: It is given that BC is tangent to the smaller circle, so BC # OD (a tangent is # to a radius at the point of tangency). OD is part of a diameter of the larger circle, so BD ≅ CD (If a diameter is # to a chord, it bisects the chord.). D is the midpt. of BC (Def. of midpt.)

30. B 32. During one revolution the bicycle moves C = pd = p(17) = 53.4 in., or about 4.45 ft. So the number of revolutions needed to travel 800 ft. is 8004.45 ≈ 180 revolutions.

Lesson 12-3 pp. 501–5032. a = 85, b = 47.5, c = 90 4. p = 90, q = 122 6. x = 65, y = 130 8. No; since opposite angles of a quadrilateral inscribed in a circle must be supplementary, the only rhombus that meets the criteria is a square. 10. } S with inscribed ∠ PQR (given); m∠PQT = 1

2m PT¬ (Inscribed Angle Theorem,

Case I); m∠RQT = 12m RT¬ (Inscribed Angle Theorem,

Case I); m PR¬ = m PT¬ - m RT¬ (Arc Addition Postulate); m∠PQR = m∠PQT - m∠RQT (Angle Addition Postulate); m∠PQR = 1

2m PT¬ - 12m RT¬

(Substitution Property); m∠PQR = 12m PR¬ (Substitution

Property) 12. a = 22, b = 78, c = 156 14. }O, ∠A intercepts BC¬, and ∠D intercepts BC¬ (Given); m∠A = 1

2mBC¬ and m∠D = 12mBC¬ (Inscribed

Angle Theorem); m∠A = m∠D (Substitution Property); ∠A ≅ ∠D (Def. of Congruent angles) 16. Quadrilateral ABCD inscribed in }O (Given); m∠A = 1

2m BCD¬ and m∠C = 12m BAD¬ (Inscribed

Angle Theorem); m∠A + m∠C = 12m BCD¬ + 1

2m BAD¬

(Addition Property); m BCD¬ + m BAD¬ = 360

(The arc measure of a circle is 360.); 12m BCD¬ + 1

2m BAD¬ = 180 (Multiplication Prop.); m∠A + m∠C = 180 (Substitution Property); ∠A and ∠C are supplementary (Definition of supplementary); m∠B = 1

2mADC¬ and m∠D = 12m ABC¬ (Inscribed

Angle Theorem); m∠B + m∠D = 12mADC¬ + 1

2m ABC¬

(Addition Property); mADC¬ + m ABC¬ = 360 (Arc measure of circle is 360.); 12mADC¬ + 1

2m ABC¬ = 180 (Multiplication Prop.); m∠B + m∠D = 180 (Substitution Property); ∠B and ∠D are supplementary (Definition of supplementary angles).


31


18. A 20. No; answers may vary. Sample: Although BD bisects ∠ABC , you cannot assume that it also bisects ∠ADC . Proof: From the given information, ∠A ≅ ∠C and BD bisects ∠ABC . By the def. of angle bisector, ∠ABD ≅ ∠CBD. BD ≅ BD by the Refl. Prop. of ≅. △ADB ≅ △CDB by AAS. So ∠ADB ≅ ∠CDB because corresp. parts of ≅ triangles are ≅.

Technology Lab 12-4 p. 5042. Ask your teacher to check your work. 4. The products of the lengths of the segments of the chords are equal. 6. Ask your teacher to check your work. 8. The product of the lengths of one secant seg. and its external seg. = the product of the lengths of the other secant seg. and its external seg. 10. Ask your teacher to check your work. 12. The square of the length of the tangent seg. = the product of the lengths of the secant seg. and its external seg.

Lesson 12-4 pp. 508–5102. 13.2 4. 360 - x 6. 180 - y 8. x = 115, y = 74 10. 140 12. 95, 104, 86, 7514.


X YV

T

Given: A circle with tangent T V and secant X V Prove: X V # YV = (T V)2. Proof: 1. Draw T X and T Y . (2 pts. determine a line.) 2. m∠TX V = 1

2 m T Y¬ (The measure of an inscribed ∠ is half the measure of the intercepted arc.) 3. m∠V T Y = 1

2 m T Y¬ (The measure of an ∠ formed by a chord and a tangent is half the measure of the intercepted arc.) 4. m∠T X V = m∠V T Y (Trans. Prop. of = ) 5. ∠T V Y ≅ ∠T V X (Reflexive Prop. of ≅) 6. △T V Y ∙ △X V T (AA ∙ ) 7. Y V

T V = T VX V (In

similar figures, corresp. sides are proportional.) 8. X V # Y V = (T V ) 2 (Prop. of Proportion)

16. m∠1 = 12m QRP¬ - 1

2mPQ¬ (vertex outside } ,

m∠ = 12 difference of intercepted arcs);

m∠1 + mPQ¬ = 12m QRP¬ + 1

2mPQ¬

(Add. Prop. of = ); m∠1 + mPQ¬ = 1

2 (m QRP¬ + mPQ¬) (Distr. Prop.);

m∠1 + mPQ¬ = 12(360) (arc measure of } is 360);

m∠1 + mPQ¬ = 180 (Simplify.)18a. Answers may vary. Sample: Geometry software is useful for quickly finding the measures of multiple arcs. b. Conjecture: A right angle formed by two perpendicular tangents and the minor arc the angle intercepts have the same measure; the measure of each is 90. 20. mPQ¬ = 120, mQR¬ = 140, m PR¬ = 100 22. x ≈ 8.9, y = 2 24. 1. }O with secants

<CA

> and

<CE> (Given) 2. Draw BE

(2 pts. determine a line.) 3. m∠BEC = 12mBD¬ and

m∠ABE = 12mAE¬ (The measure of an inscribed angle

is half the measure of its intercepted arc.)

4. m∠BEC + m∠BCE = m∠ABE (Ext. Angle Thm.) 5. 12mBD¬ + m∠BCE = 1

2mAE¬ (Subst. Prop. of = )

6. m∠BCE = 12mAE¬ - 1

2mBD¬ (Subtr. Prop. of = )

7. m∠BCE = 12(mAE¬ - mBD¬) (Distr. Prop.)

8. ∠BCE ≅ ∠ACE (Refl. Prop. of ≅) 9. m∠ACE = 1

2(mAE¬ - mBD¬) (Subst. Prop. of = )

26. 1. (PQ)2 = (QS)(QR) (Square of the tangent equals the product of the secant times the external segment.) 2. b2 = (c + a)(c - a) (Substitution) 3. b2 = c2 - a2 (Distributive Prop.) 4. b2 + a2 = c2 (Addition Prop. of Equality) 28. 10

Topic Review pp. 511–5132. chord 4. inscribed ∠ 6. 13 8. 90 10. AB is a

diameter of the circle. 12. 11812 ≈ 6.7 14. a = 40,

b = 140, c = 90 16. a = 90, b = 90, c = 70, d = 65 18. a = 95, b = 85 20. 4

TEKS cumulative practice pp. 514–5152. F 4. J 6. G 8. G 10. 5 12. 130 14. 34 16. 11.2 18. slope of HK = 7 - 1

-3 - 6 = 6-9 = - 23 ;

slope of MN = -8 - 10-5 - 7 = -18

-12 = 32 ;

HK and MN are perpendicular because the slopes are opposite reciprocals.

Topic 13Activity Lab 13-1 and 13-2 pp. 518–5192. They are the same. 4. 8; 3 6. A = b # h; A = 1

2b # h 8. b1 + b2 10. 12(b1 + b2)h 12. Rotate the entire △ 180° about the midpt. of any side to form a ▱ that has the same base b and height h as the △. Since two △s form the ▱, its area is twice the area of a △. That is, bh = Area of ▱ = 2 # Area of △. So Area of △ = 1

2 bh. 14. d1 = 9 units, d2 = 4 units; the area of each △ is half the area of the kite.

Area of kite = 2 # Area of △ = 2 # (12 bh) = bh, where

b = d1 and h = 12 d2. So, Area of kite = 1

2d1d2.

Lesson 13-1 pp. 523–5252. 0.24 units 4. 15 units2 6. 6 units2 8. 27 units2 10. 21 units2 12. B 14. 14 cm 16. 312.5 ft2 18. 12,800 m2 20. You can use mental math to find the area of the triangle and the area of the parallelogram and add the areas together: 57 m2.22a.

hsm11gmte_1001_t14615

y

xO 1 3

1

y 2x 4

y x 234

b. 6 units2 24. 60 units2 26. 88 units2 28. 34 in.2 30a. 54 in.2 b. 54 in.2 32. J


32


Lesson 13-2 pp. 529–5312. 15 units2 4. 42,900 mi2 6. 5213 ft2 8. 80 in.2 10. 24 ft2 12. 96 in.2 14. 472 in.2 16. 108 ft2 18. 56 ft2 20. 100 + 5013 or about 186.6 in.2 22. 1.5 m2

24a.

b. trapezoid c. 42 units2 26. H

Lesson 13-3 pp. 536–5382. m∠4 = 90, m∠5 = 45, m∠6 = 45 4. 2144.475 cm2 6. 2475 in2 8. 841.8 ft2 10. 93.5 m2 12. 72 cm2 14. 1213 in2 16. 73 cm2 18a. 9.1 in. b. 6 in. c. 3.7 in. d. Answers may vary. Sample: About 4 in.; the length of a side of a pentagon should be between 3.7 in. and 6 in. 20. The apothem is one leg of a rt. △ that has the radius as the hypotenuse. 22. 2,90413 in2 24. 17.0; 18

26a. b = s, h = 132 s; A = 1

2bh = 12s # 13

2 s = s2234

b. a = s136 ; A = 1

2ap = 12(s13

6 )(3s) = s2234 28. For

regular n-gon ABCDE . . . , let P be the intersection of the bisectors of ∠ ABC and ∠ BCD. BC ≅ DC , ∠BCP ≅ ∠DCP, and CP ≅ CP, so △BCP ≅ △DCP, and ∠CBP ≅ ∠CDP because corresp. parts of ≅ triangles are ≅. Since ∠ BCP is half the size of ∠ ABC and ∠ABC ≅ ∠CDE, ∠CDP is half the size of ∠CDE. By a similar argument, P is on the bisector of each angle around the polygon. The smaller angles formed by each of the angle bisectors are all ≅. By the Converse of the Isosc. Triangle Thm., each of △APB, △BPC, △CPD, etc., are isosc. with AP ≅ BP ≅ CP, etc. Thus, P is equidistant from the polygon’s vertices. So P is the center of the polygon, and the angle bisectors are radii. 30. B 32. B

Lesson 13-4 pp. 543–5452. 3 : 5; 9 : 25 4. 1 : 2; 1 : 2 6. 4 : 1; 4 : 1 8. x = 2 cm, y = 3 cm 10. x = 4 cm, y = 6 cm 12. x = 412 cm, y = 612 cm 14. 24 in.2 16. 59 ft2 18. +63.20 20. 0.3 cm2 22. The perimeter increases by 8 feet; 64 ft 24. While the ratio of lengths is 2 : 1, the ratio of areas is 4 : 1. 26. The area is halved. 28. Answers may vary. Sample: The proposed playground is more than adequate. The number of students has approximately doubled. The proposed playground would be four times larger than the original playground. 30a. 24 ft; 33.75 ft2 b. The perimeter doubles; the area quadruples. c. The perimeter increases by 15; the area doubles. 32. 225 34. 155

Lesson 13-5 pp. 548–5502. 123.1 yd2 4. 141.7 in2 6. 12.4 mm2 8. 2540.5 cm2 10. 18.0 ft2 12. 311.3 km2 14. 0.8 ft2 16. 1,459,000 ft2 18. 20.8 m, 20.8 m2 20. 61.2 m, 282.8 m2 22. about 29.7 cm2 24. area of Pentagon A ≈ 1.53 ~ (area of Pentagon B) 26. area of Octagon B ≈ 1.17 ~ (area of Octagon A) 28. D 30. about 48.2 cm2 32. 0.65 34. 140.3 36. 62 38. 12

Topic Review pp. 551–5532. apothem 4. height of a parallelogram 6. 90 in.2 8. 160 ft2 10. 9613 mm2 12. 117 cm2 14. 54 m2 16. 28 m2 18. 112.5 m2

20.

128 mm2

22. 4∶9 24. 1∶4 26. 212∶5 28. 232.5 cm2 30. 8 m2 32. 24.6 ft2 34. 70.4 m2

TEKS cumulative practice pp. 554–5552. G 4. H 6. J 8. G 10. 55 12. 27 14. 17,000 16. 900 18. 13.5 20. No; using the Distance Formula, AB = 2(10 - 2)2 + (9 - 3)2 = 10; BC = 2(10 - 10)2 + (-3 - 9)2 = 12; AC = 2(10 - 2)2 + (-3 - 3)2 = 10; △ABC is isosc., but it is not equilateral. 22. A rectangle; using the Distance Formula, PR = 2(-2 - 1)2 + (5 - (-1))2 =315 and QS = 2(1 - (-2))2 + (5 - (-1))2 = 315. So PR ≅ QS. If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. 24. No; the apothem is a leg of a rt. △ and the radius is the hypotenuse of that △. 26. By the Pythagorean Theorem, x = 40. Because the two △s are ∙ , 40

30 = 30y

and y = 22.5. Total distance = 40 + 22.5 = 62.5 yd.

Topic 14Lesson 14-1 pp. 563–5642. 12 4. 8 6. 9 8. triangle 10. Answers may vary. Sample: an isosceles triangle and a rectangle 12. Answers may vary. Sample: a triangle and a trapezoid 14. No; if F = V , then F + V = 2F , so F + V is even. So E ≠ 9 because E + 2 must be even. 16.


rectangle

x 6

y 0

y x

4

x 0

y

xO

hsm11gmte_1002_t14617

12

8

8


8 mm


33


18. cone 20. a cylinder attached to a cone 22. Yes; the intersection is a line segment when the plane is tangent to the curved surface of the cone.

24.


26.


28. B

30. D 32. (5, 2), (-1, 0), OR (3, -4); The fourth vertex lies on a line parallel to an opposite side such that the length of the side is equal to the length of the opposite side.

Lesson 14-2 pp. 569–5712. 216 cm2


6 ft

6 ft

6 ft

4a. hexagonal prism b. 240 cm2 c. 4813 ≈ 83.1 cm2 d. 240 + 4813 ≈ 323.1 cm2 6. 108 in.2 8. 82 in.2 10. 40p cm2 12. 236 in.2 14. Answers may vary. Sample:


3 cm4 cm

14 cm

4 cm

16. (220.5p + 222) mm2 or about 914.7 mm2 18. A cylinder and a prism both have two bases that are } and ≅. The bases of a cylinder are circles, and the bases of a prism are polygons. 20a. r ≈ 1.2 in. h = 6 in. b. about 54 in.2 22. 110 in.2 24. (220 - 8p) in.2 26. 13 28. 68

Lesson 14-3 pp. 577–5782. 51 m2 4. 179 in.2 6. 138 m2 8. Answers may vary. Sample:


6 cm

4 cm4 cm

10. 8 ft 12. 471 ft2 14. 1580.6 ft2 16. 4 18. 129.6

20. L.A. = 25p15 cm2 S.A. = (25p15 + 25p) cm2

22. H 24. Yes; if the legs of each isosceles triangle are consecutive sides of the parallelogram, then the parallelogram is a rhombus.

Lesson 14-4 pp. 582–5832. 37.5p m3; 117.8 m3 4. 14 cm3 6. 144 cm3 8. 40 cm 10. 6 ft 12. from 28 to 42 pots of flowers 14. Answers may vary. Sample: If two plane figures have the same height and the same width at every level, then they have the same area. 16. Volume is 27 times greater. Using V = B # h / # w # h for a rectangular prism,(3/) # (3w) # (3h) = 27 # / # w # h = 27 # V . 18. 125.7 cm3 20. cylinder with r = 2 and h = 4; 16p units3 22. cylinder with r = 2 and h = 4; 16p units3 24. A 26. C

Activity Lab 14-5 p. 584

2. The heights are about equal. 4. 13 6. The areas are = . 8. about 3 cones 10. V = 1

3Bh; the volume of a cylinder is Bh. The volume of a cone is 13 the volume of the cylinder, or 13Bh.

Lesson 14-5 pp. 587–5892. 363.6 m3 4. about 4.7 cm3 6. 22

3 p in.3; 23 in.3 8. The volume of the cylinder is 3 times the volume of the cone. (V of cylinder = Bh, V of cone = 1

3Bh). 10. 312 cm3 12. They are = ; both volumes are 13pr 2h. 14. 73 cm3 16. 6,312,000 ft3 18. Volume is halved;

V = 13Bh, so if h is replaced with h2, then the volume is

13B(h

2) = 12[1

3Bh] . 20. cone with r = 3 and h = 4; 12p 22. cone with r = 4, h = 51

3, with a cone with r = 1, h = 11

3 cut off the top, and a cylinder with r = 1, h = 4 cut out of its center, 24p 24. A26.

hsm11gmte_1105_t14505

The centroidis inside.

The incenteris inside.

The circumcenteris outside.


34


Lesson 14-6 pp. 593–5952. 400p in.2 4. 40,000p yd2 6. 288p cm3; 905 cm3 8. 4624p mm2 10. 1089

256 p in.2 12. 232 in.2 14. 1006 m2 16. S.A. ≈ 108 cm2; V ≈ 108 cm3 18a. sphere with r = 4 b. 256

3 p units3 c. 64p units2 20. C 22. 8 in. sphere; the volume of the three spheres is 3(4.5p), or 13.5p, units3, and that of the large sphere is 851

3p units3. 24. 1256 p yd3

26. 3436 p m3 28. Yes; the volume of frozen

yogurt is 2563 p cm3, and the volume of the cone is

64p cm3. 30. about 148,250,000 km2 32. 26pcm2; 62

3 p cm3 34. B 36. G

Technology Lab 14-7 p. 5962. The ratio of surface areas is the scale factor squared. 4. The ratio of lateral areas is the scale factor squared. 6. The ratio of volumes is the scale factor cubed.

Lesson 14-7 pp. 601–6032. S.A. is multiplied by 1

16, V is multiplied by 164. 4. no

6. no 8. 6 : 7 10. 24 ft3 12. 325 yd2 14. In this case, the surface area of the enlarged cylinder is 864p cm2 and the volume is 3456p cm3. So when the height of the cylinder is multiplied by 2 and the radius is multiplied by 3, the surface area is multiplied by 6.75 and the volume is multiplied by 18. 16. yes;

60 : 8060 = 40

30 = 6045 = 4

3 18. No; the same increase to all

the dimensions does not result in proportional ratios unless the original prism is a cube. 20a. 384 cm3 b. 16 : 1 c. A: 384 cm2; B: 24 cm2 22. 10 24. 116

Topic Review pp. 604–6072. pyramid 4. Answers may vary. Sample:

6. 8 8. 5 10.

12. 66p m3 14. 36p cm2 16. B = S.A. - L.A.2

18. 576 m2 20. about 391.6 in.2 22. 24.5 ft3 24. 13.9 m3 26. S.A. = 153.9 cm2; V = 179.6 cm3 28. S.A. = 8.0 ft2; V = 2.1 ft3 30. 314 m2


34. 64∶27

TEKS cumulative practice pp. 608–6092. G 4. F 6. F 8. F 10. 128 12. 484 14. 72 16. 15 18. C, A, E, D, B 20. 234.25 22. The surface area equals 12 the surface area of the

log plus the area of the rectangular surface of the cut through the center. Using inches: • 12 (S.A. of log) =

12[2p(9)(36) + 2p(92)] = 12 (810p) = 405p

• area of rectangular face = 18 # 36 = 648 • 12 (S.A. of log) + area of rectangular

face = 405p + 648 ≈ 1920.345025The surface area of one of the halves is about 1920 in.2.

Topic 15Lesson 15-1 pp. 614–6162. about 0.937 or 93.7%; about 8432 people 4. 4

11

6. 211 8. 7

11 10. 911 12 26 or 13 14. about 133 16. 9

65

18. 5865 20–22. Answers will vary. Ask your teacher to

check your work. 24a. 16 b. 0.156 or 15.6% 26. D 28. C

Lesson 15-2 pp. 619–6212. 1

10 4. 25 6. 25 8. 25, or 40% 10. 29, or 22% 12. 59,

or 56% 14. 149, or about 2% 16. 24

49, or about

49% 18. 310, or 30% 20. 3

20, or 15% 22. 919, or

about 47% 24. 14; mAB¬ = 90, so the length of

AB¬ = 90360

# 2pr = 14# 2pr . The ratio of the length

of AB¬ to the circumference is 14. 26. p20, or about 16% 28. 36 s 30. A 32. D 34. P = 4s = 24, so s is 6 ft. The diagonal of a square is the hypotenuse of a 45°@45°@90° triangle with leg s, and its length is s12. So the diagonal of the square is 612 ft.

Lesson 15-3 pp. 626–6272. 6760 4. 10,897,286,400 6. 210 8a. 4060 b. 142,506 10. 1320 12. 1

15,890,700 14. 56 16. 1

1365 18. The friend used a permutation instead of a combination. There are 56 ways. 20a. 1

42 b. Sample: The number of possible codes is actually a permutation. 22. A 24. 135° 26. 150



5

55

44

4



35


Lesson 15-4 pp. 631–6322. dependent 4. dependent 6. 14 8. 35 10. 8

15

12. 1340 14. 58 16. 34 18. 80% 20. Mutually exclusive

events cannot happen at the same time, but overlapping events can occur at the same time. 22. No; answers may vary. Sample: 12 + 1

3 ∙ 23

24. dependent events; If one event occurs, then the other event cannot occur because they are mutually exclusive. So the outcome of the second event depends on the outcome of the first event. 26. 1

36

Lesson 15-5 pp. 635–6362. 0.2 4. 7

13 6. about 0.44 8. about 0.54 10. 15

12. 58 14. 35 16. Answers may vary. Sample: The probability is 1. The sum of the relative frequencies is also 1 because they represent the experimental probabilities of each possible outcome. 18. 0.56 20. - 4

Lesson 15-6 pp. 640–6412. 12 4. 2

27 6. 1681 8. 18 10. 1

12 12. 112 14. 1

12 16. 18.75% 18. 30% 20. 36% 22. 0.07 24. 69.05% 26. H

Topic Review pp. 642–6452. experimental probability 4. geometric probability

6. 910 8. 35 10. 12, or 50, 12. 16, or about 16.7,

14. 12, or 50, 16. 40,320 18. 220 20. 90 22. 1140 24. 2300 26. 0.595 28. Answers may vary. Sample: If the events are mutually exclusive, the P(A and B) =0 and the formula becomes P(A or B) = P(A) + P(B). Otherwise, the formula is used as stated. 30. 0.95 32. 22

145 34. 5574 or about 74.3,

TEKS cumulative practice pp. 646–6472. J 4. H 6. F 8. H 10. J 12. J 14. J 16. Answers may vary. Sample: Draw a line, mark off length AB on that line, and label the endpoints of the segment as D and F. Using D as center and AC as radius, draw an arc. Using F as center and BC as radius, draw an arc that intersects the arc with center D. Label the intersection of the two arcs as point E. [This construction uses SSS ≅; other valid constructions use SAS or ASA ≅.]

AddiTionAL pRAcTicETopic 1 pp. 648–6492. false 4. true 6. false

8.

10. Sample answer: D has a coordinate of 3 and F has a coordinate of 7. 12. 6 blocks 14. BD, AF 16. 8 18. 40 20. 7, 5, 12 22. ∠LQN 24. ∠MQK, ∠JQL 26. 36, 54 28. 5630. 32.

34.

Topic 2 pp. 650–6512. 8, 85 4.

6. 8 cm 8. a rectangle 10. Hypothesis: Dan is nearsighted. Conclusion: Dan needs glasses. 12. Converse: If x = 9, then 3x - 7 = 20; true. Inverse: If 3x - 7 ≠ 20, then x ≠ 9; true. Contrapositive: If x ≠ 9, then 3x - 7 ≠ 20; true. 14. no; to be even the real number must be an integer. 16. no; a segment with endpoints on the circle that does not pass through the center of the circle is not a diameter. 18. If a figure is a square then it has four sides; true. If a figure has four sides, then it is a square; false. 20. Shauna will gain a promotion. 22. If you enjoy all foods, then you eat bread; Law of Syllogism. 24b. Multiplication Property of Equality; Distributive Property c. Division Property of Equality d. 4c - a = 4b e. Symmetric Property of Equality 26. Division Property of Equality 28. x = 24 30. x = 25

Topic 3 pp. 652–6532. ∠1 and ∠3; b, c, and a, ∠2 and ∠4; d, e, and a 4. ∠6 and ∠7; d, e, and b 6. m∠1 = 58; Alternate Interior Angles Theorem, m∠2 = 122; Same-Side Interior Angles Theorem 8. x = 45; x + 15 = 2x - 30 = 60 10. x = 17, y = 35; 5x = (3x + 34) = 85, 2y = (3y - 35) = 70 12. x = 54 14. x = 25; y = 19 16. x = 73; y = 13


36


18.

20. y + 1 = 2(x + 1) or y - 3 = 2(x - 1)

22. y = -127 x + 4

7 24. y = -15 x + 1

26. y - 3 = 12 (x - 6) or y = 1

2 x 28. y - 2 = -1(x - 3) or y = -x + 5 30. y + 1 = -3(x - 5) or y = -3x + 14 32. spherical geometry

Topic 4 pp. 654–6552. SA 4. RE 6. TS 8. ∠A 10. No; the only known corresponding part that is congruent is UV . 12. Yes; corresponding sides and corresponding angles are congruent. 14. EA ≅ PL, AL ≅ LA, ∠EAL ≅ ∠PLA; SAS 16. CN ≅ AO, NA ≅ OC, AC ≅ CA; SSS 18. It is given that m∠1 = m∠2 and m∠3 = m∠4. By the Angle Addition Postulate, m∠DPA = m∠CPB. Since P is the midpoint of AB, PA ≅ PB. It is given that PD ≅ PC , so △ADP ≅ △BCP by SAS. 20. MP } NS and RS } PQ mean that ∠1 ≅ ∠4 and ∠2 ≅ ∠3, respectively, by the Alternate Interior Angles Theorem. It is given that MR = NQ,so MR + RQ = NQ + RQ, or MQ = NR, by the Addition Property of Equality. Therefore, △MQP ≅ △NRS by ASA. 22. OS ≅ OS by the Reflexive Property of Congruence. Since ∠T ≅ ∠E and ∠TSO ≅ ∠EOS, △TSO ≅ △EOS by AAS, and TO ≅ ES by CPCTC. 24. ∠1 ≅ ∠2, so ∠APO ≅ ∠BQO because supplements of congruent angles are congruent. PO ≅ QO (given) and ∠AOP ≅ ∠BOQ as vertical angles, so △AOP ≅ △BOQ by ASA. ∠A ≅ ∠B by CPCTC. 26. 65 28. PX ≅ PY (given) means that ∠1 ≅ ∠2 by the Isosceles Triangle Theorem. ∠3 ≅ ∠4 by the Congruent Supplements Theorem. ∠5 ≅ ∠6 (given), so △PXA ≅ △PYB by ASA. PA ≅ PB by CPCTC and △PAB is isosceles by the definition of isosceles triangle. 30. △ARO ≅ △RAF; HL 32. △AON ≅ △MOP; AAS

Topic 5 pp. 656–657

2a. 7.3 b. (32 , 3) 4. 25

7 6. 7 8. 20 10. 140 ft

12. 7 14. 152 16. C 18. (1, 5) 20. 60, 90

22. 12, 36 24. Altitude; the segment intersects A and is perpendicular to a side at B. 26. (0, 34 )

28. (2911, 9

11) 30. Assume points J, K, and L are not collinear. 32. I and III 34. No; 2 + 3 7 5 36. No; 8 + 9 7 18 38. AC 6 XZ

Topic 6 pp. 658–6592. x = 100; (x + 5) = 105 4. x = 122; (x - 6) = 116 6. x = 30; y = 55 8. x = 12; y = 4 10. 18 12. The diagonals of a parallelogram bisect each other, so P is the midpoint of AC . P and M are midpoints of two sides of △ACD, so PM } AD by the Triangle Midsegment Theorem. 14. y = 16 16. rhombus; m∠1 = 50; m∠2 = 90; m∠3 = 40; m∠4 = 40 18. rhombus; m∠1 = 90; m∠2 = 27; m∠3 = 63; m∠4 = 63 20. ∠1 and ∠2 are complementary. The diagonals of a rhombus are perpendicular so, by definition of perpendicular, ∠BKC is a right angle. So △BKC is a right triangle by definition. ∠CBK and ∠1 are complementary because they are acute angles in a right triangle. By definition of a rhombus, CD ≅ CB. So by the Converse of the Isosceles Triangle Theorem, ∠CBK ≅ ∠2. By the Transitive Property of Congruence, ∠1 and ∠2 are complementary. 22. x = 7 24. m∠1 = 110; m∠2 = 25 26. m∠1 = 74; m∠2 = 106

Topic 7 pp. 660–6612. Use the Slope Formula:

slope of AB = 3 - 52 - (-1) = -2

3

slope of BC = 0 - 32 - 2 = -3

0 ; undefined

slope of CD = -2 - 0-1-2 = 2

3

slope of DA = -2 - 5-1 - (-1) = -7

0 ; undefined

Sides BC and DA are parallel, and sides AB and CD have different slopes. Use the Distance Formula: AB = 2(2 - (-1))2 + (3 - 5)2 = 19 + 4 = 113 CD = 2(-1-2)2 + (-2 - 0)2 = 19 + 4 = 113 Sides AB and CD are congruent.

4. equilateral;

6. isosceles;


37


8. square;

10. rhombus;

12. D(-3, -2) 14. (0, 0), (0, h), (b, 0) 16. (-b2, 0), (-b2, h), (0, h), (b1 - b2, 0) 18. D( - c, 0); S(0, - b) 20. Given: Square DRSQ with K, L, M, N midpoints of DR, RS, SQ, and QD, respectively. Prove: KLMN is a square. K(a, a), L(a, - a), M( - a, - a), and N( - a, a) are midpoints of the sides of the square. KL = LM = MN = NK = 2a. The slopes of KL and MN are undefined. The slopes of LM and NK are 0, so adjacent sides are perpendicular to each other. Since all angles are right angles, the quadrilateral is a rectangle. A rectangle with all congruent sides is a square. 22. The line through R(2a, 0) and M( - a, b) is y = b

-3a (x - 2a). The line through Q( - 2a, 0) and

N(a, b) is y = b3a (x + 2a). For each line, when

x = 0, y = 2b3 , so the three medians all contain

point H(0, 2b3 ) .

Topic 8 pp. 662–6632. C 4. (x, y) S (x - 8, y ) 6. A translation that moves all points of y = x - 1 up 4 units maps y = x - 1 onto y = x + 3.8.

10.

12.

14.

16. 180° rotational symmetry; reflectional symmetry:

18. neither20.

22. (6, 6) 24. (0.5,-1) 26. (54, 1)

28.


38


30. A′(134 , 2), B′( -1

2, -1), C′(14, 5)

Topic 9 pp. 664–6652. x = 9

5; y = 92 4. x = 12; y = 8 6. Answers may vary.

Sample: a translation right 14 units and up 4 units followed by a dilation with scale factor 12 and center B′. 8. They are similar. Sample answer: There is a scale factor k such that k # FH = XY . And since ∠F ≅ ∠X and ∠H ≅ ∠Z, DK (△FGH) ≅ △XYZ by ASA. Therefore, △FGH ∙ △XYZ . 10. Yes; △XYZ ∙ △EWN by AA∙ . 12. △CAB ∙ △CED by SSS ∙. ∠A ≅ ∠E as corresponding angles of similar triangles, so AB } ED by the converse of the Alternate Interior Angles Theorem. 14. 12 16. 16 18. x = 170; y = 121;z = 130 20. x = 72

5 22. x = 20

Topic 10 pp. 666–6672. 513 4. 312 6. 413 8. obtuse 10. right 12. right 14. 4 in., 412 in. 16. 13 cm, 1313 cm 18. 12 units, 12 units, 2 units 20. x = 5.6 22. x = 11.0 24. x = 49 26. No; sample answer: In the plan, change 100 ft to 114.5 ft. 28. 4.6 ft 30. 139 ft 32. 90 ft; 143 ft

Topic 11 pp. 668–669

2a. 20p ft b. 53p ft 4a. 10p in. b. 254 p in.

6a. 8p in. b. 245 p in. 8. 5p9 , 1.75 10. p4 , 0.79

12. 25p36 , 2.18 14. 270° 16. 150° 18. 234°

20a. 2:00, 10:00 b. 4:00, 8:00 22. 49p3 ft2

24. 81p8 cm2 26. 36p yd2 28. 26 in.2

30. x2 + (y - 5)2 = 9 32. (x + 4)2 + y2 = 37

34. (1, -2), r = 15;

Topic 12 pp. 670–6712. x = 10 4. yes; 152 + 202 = 252 6. yes; 102 + 202 = (1015)2 8. x = 14.8 10. x = 5.3 12. 1.82 units 14. a = 38; b = 52; c = 104; d = 90 16. a = 55; b = 72; c = 178; d = 89 18. a = 43; b = 90; c = 47; d = 94 20. x = 193; y = 60.5 22. x = 112.5; y = 67.5 24. x = 11.5 26. x = 90; y = 150

Topic 13 pp. 672–6732. 10.825 ft2 4. 102 m2 6. The sides are not marked parallel, so you cannot determine the area. 8. 15 in.2 10. 3213 ft2 12. 18 yd2 14. 2.625 in.2 16. 5 : 8; 25 : 64 18. 5 : 16; 25 : 256 20. 48.4 cm 22. 78.0 in.2 24. 20.0 m2 26. 109.2 yd2 28. 70.7 ft2

Topic 14 pp. 674–6752. rectangle; 7 + 10 = 15 + 2 4. hexagon; 7 + 10 = 15 + 2 6. 10813 in.2; 14413 in.2 8. 37.5 cm2; 47.9 cm2 10. 112 cm2; 176 cm2 12. 339.3 cm2; 439.8 cm2 14. 47.1 m3 16. 400 in.3 18. 228.8 m3

20. 500p3 cm3, 524 cm3; 100p cm2, 314 cm2

22. 256p3 in.3, 268 in.3; 64p in.2, 201 in.2 24. p6 in.3,

1 in.3; p in.2, 3 in.2 26. 256p3 m3 28. 343p

6 ft3

30. yes; 2 : 1 32. 307.2 ft3 34. ≈ 6.7 yd3

Topic 15 pp. 676–677

2. 112 4. 14 6. 1

12 8. 56 10. 70 12. 13 14. 724 16. 59

18. 72 20. 36 22. permutation; 336 24. independent 26. independent 28. 13

61 30. 41120 32. 0.225, or

22.5% 34. 0.15, or 15% 36. 0.1875, or 18.75%

Geometry Student Text and Homework Helper ANSWERS · 2019. 6. 3. · Geometry | Student Text and...

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Transcript of Geometry Student Text and Homework Helper ANSWERS · 2019. 6. 3. · Geometry | Student Text and...