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AUTHOR CK-12 Foundation

CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, webbased collaborative model termed the FlexBook®textbook, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2015 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/about/ terms-of-use. Printed: February 25, 2015

iii

Contents

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Contents 1

2

3

4

5

iv

Basics of Geometry, Answer Key 1.1 Geometry - Second Edition, Points, Lines, and Planes, Review Answers 1.2 Geometry - Second Edition, Segments and Distance, Review Answers . 1.3 Geometry - Second Edition, Angles and Measurement, Review Answers 1.4 Geometry - Second Edition, Midpoints and Bisectors, Review Answers . 1.5 Geometry - Second Edition, Angle Pairs, Review Answers . . . . . . . 1.6 Geometry - Second Edition, Classifying Polygons, Review Answers . . 1.7 Geometry - Second Edition, Chapter Review Answers . . . . . . . . . .

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1 2 5 7 10 13 14 17

Reasoning and Proof, Answer Key 2.1 Geometry - Second Edition, Inductive Reasoning, Review Answers . . . . . . . . . . . 2.2 Geometry - Second Edition, Conditional Statements, Review Answers . . . . . . . . . 2.3 Geometry - Second Edition, Deductive Reasoning, Review Answers . . . . . . . . . . 2.4 Geometry - Second Edition, Algebraic and Congruence Properties, Review Answers . . 2.5 Geometry - Second Edition, Proofs about Angle Pairs and Segments, Review Answers . 2.6 Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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18 19 21 23 26 29 33

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Parallel and Perpendicular Lines, Answer Key 3.1 Geometry - Second Edition, Lines and Angles, Review Answers . . . . . . . . . . . . 3.2 Geometry - Second Edition, Properties of Parallel Lines, Review Answers . . . . . . . 3.3 Geometry - Second Edition, Proving Lines Parallel, Review Answers . . . . . . . . . . 3.4 Geometry - Second Edition, Properties of Perpendicular Lines, Review Answers . . . . 3.5 Geometry - Second Edition, Parallel and Perpendicular Lines in the Coordinate Plane, Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Geometry - Second Edition, The Distance Formula, Review Answers . . . . . . . . . . 3.7 Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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34 35 37 40 42 43 47 49

Triangles and Congruence, Answer key 4.1 Geometry - Second Edition, Triangle Sums, Review Answers . . . . . . . . . . . . . . . . . . 4.2 Geometry - Second Edition, Congruent Figures, Review Answers . . . . . . . . . . . . . . . . 4.3 Geometry - Second Edition, Triangle Congruence using SSS and SAS, Review Answers . . . . 4.4 Geometry - Second Edition, Triangle Congruence using ASA, AAS, and HL, Review Answers 4.5 Geometry - Second Edition, Isosceles and Equilateral Triangles, Review Answers . . . . . . . 4.6 Chapter 4 Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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50 51 53 55 58 60 63

Relationships with Triangles, Answer Key 5.1 Geometry - Second Edition, Midsegments of a Triangle, Review Answers . . . . . 5.2 Geometry - Second Edition, Perpendicular Bisectors in Triangles, Review Answers 5.3 Geometry - Second Edition, Angle Bisectors in Triangles, Review Answers . . . . 5.4 Geometry - Second Edition, Medians and Altitudes in Triangles, Review Answers . 5.5 Geometry - Second Edition, Inequalities in Triangles, Review Answers . . . . . . . 5.6 Geometry - Second Edition, Extension: Indirect Proof, Review Answers . . . . . .

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64 65 67 71 73 75 76

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Contents

Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Polygons and Quadrilaterals, Answer Key 6.1 Geometry - Second Edition, Angles in Polygons, Review Answers . . . . . . . . . . . . 6.2 Geometry - Second Edition, Properties of Parallelograms, Review Answers . . . . . . . 6.3 Geometry - Second Edition, Proving Quadrilaterals are Parallelograms, Review Answers 6.4 Geometry - Second Edition, Rectangles, Rhombuses and Squares, Review Answers . . . 6.5 Geometry - Second Edition, Trapezoids and Kites, Review Answers . . . . . . . . . . . 6.6 Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Similarity, Answer Key 7.1 Geometry - Second Edition, Ratios and Proportions, Review Answers . . . 7.2 Geometry - Second Edition, Similar Polygons, Review Answers . . . . . . 7.3 Geometry - Second Edition, Similarity by AA, Review Answers . . . . . . 7.4 Geometry - Second Edition, Similarity by SSS and SAS, Review Answers . 7.5 Geometry - Second Edition, Proportionality Relationships, Review Answers 7.6 Geometry - Second Edition, Similarity Transformations, Review Answers . 7.7 Geometry - Second Edition, Extension: Self-Similarity, Review Answers . . 7.8 Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . .

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78 79 81 83 86 89 91

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92 93 94 95 96 97 99 101 103

Right Triangle Trigonometry, Answer Key 8.1 Geometry - Second Edition, The Pythagorean Theorem, Review Answers . . . . . . 8.2 Geometry - Second Edition, Converse of the Pythagorean Theorem, Review Answers 8.3 Geometry - Second Edition, Using Similar Right Triangles, Review Answers . . . . . 8.4 Geometry - Second Edition, Special Right Triangles, Review Answers . . . . . . . . 8.5 Geometry - Second Edition, Tangent, Sine and Cosine, Review Answers . . . . . . . 8.6 Geometry - Second Edition, Inverse Trigonometric Ratios, Review Answers . . . . . 8.7 Geometry - Second Edition, Extension: Laws of Sines and Cosines, Review Answers 8.8 Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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104 105 106 109 111 112 113 114 115

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Circles, Answer Key 9.1 Geometry - Second Edition, Parts of Circles and Tangent Lines, Review Answers . . . . . . . . 9.2 Geometry - Second Edition, Properties of Arcs, Review Answers . . . . . . . . . . . . . . . . . 9.3 Geometry - Second Edition, Properties of Chords, Review Answers . . . . . . . . . . . . . . . . 9.4 Geometry - Second Edition, Inscribed Angles, Review Answers . . . . . . . . . . . . . . . . . . 9.5 Geometry - Second Edition, Angles of Chords, Secants, and Tangents, Review Answers . . . . . 9.6 Geometry - Second Edition, Segments of Chords, Secants, and Tangents, Review Answers . . . 9.7 Geometry - Second Edition, Extension: Writing and Graphing the Equations of Circles, Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.8 Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10 Perimeter and Area, Answer Key 10.1 Geometry - Second Edition, Triangles and Parallelograms, Review Answers . . . . . . 10.2 Geometry - Second Edition, Trapezoids, Rhombi, and Kites, Review Answers . . . . . 10.3 Geometry - Second Edition, Areas of Similar Polygons, Review Answers . . . . . . . . 10.4 Geometry - Second Edition, Circumference and Arc Length, Review Answers . . . . . 10.5 Geometry - Second Edition, Areas of Circles and Sectors, Review Answers . . . . . . 10.6 Geometry - Second Edition, Area and Perimeter of Regular Polygons, Review Answers 10.7 Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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116 117 119 120 122 124 127 128 129 130 131 132 134 135 136 137 139

11 Surface Area and Volume, Answer Key 140 11.1 Geometry - Second Edition, Exploring Solids, Review Answers . . . . . . . . . . . . . . . . . . 141 v

Contents 11.2 11.3 11.4 11.5 11.6 11.7 11.8

www.ck12.org Geometry - Second Edition, Surface Area of Prisms and Cylinders, Review Answers Geometry - Second Edition, Surface Area of Pyramids and Cones, Review Answers . Geometry - Second Edition, Volume of Prisms and Cylinders, Review Answers . . . Geometry - Second Edition, Volume of Pyramids and Cones, Review Answers . . . . Geometry - Second Edition, Surface Area and Volume of Spheres, Review Answers . Geometry - Second Edition, Exploring Similar Solids, Review Answers . . . . . . . Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12 Rigid Transformations, Answer Key 12.1 Geometry - Second Edition, Exploring Symmetry, Review Answers . . . . . . . 12.2 Geometry - Second Edition, Translations and Vectors , Review Answers . . . . 12.3 Geometry - Second Edition, Reflections, Review Answers . . . . . . . . . . . . 12.4 Geometry - Second Edition, Rotations, Review Answers . . . . . . . . . . . . . 12.5 Geometry - Second Edition, Composition of Transformations, Review Answers 12.6 Geometry - Second Edition, Extension: Tessellations, Review Answers . . . . . 12.7 Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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152 153 155 156 158 162 163 164

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Chapter 1. Basics of Geometry, Answer Key

C HAPTER

1

Basics of Geometry, Answer Key

Chapter Outline 1.1

G EOMETRY - S ECOND E DITION , P OINTS , L INES , AND P LANES , R EVIEW A N SWERS

1.2

G EOMETRY - S ECOND E DITION , S EGMENTS AND D ISTANCE , R EVIEW A NSWERS

1.3

G EOMETRY - S ECOND E DITION , A NGLES AND M EASUREMENT, R EVIEW A N SWERS

1.4

G EOMETRY - S ECOND E DITION , M IDPOINTS AND B ISECTORS , R EVIEW A N SWERS

1.5

G EOMETRY - S ECOND E DITION , A NGLE PAIRS , R EVIEW A NSWERS

1.6

G EOMETRY - S ECOND E DITION , C LASSIFYING P OLYGONS , R EVIEW A NSWERS

1.7

G EOMETRY - S ECOND E DITION , C HAPTER R EVIEW A NSWERS

1

1.1. Geometry - Second Edition, Points, Lines, and Planes, Review Answers

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1.1 Geometry - Second Edition, Points, Lines, and Planes, Review Answers For 1-5, answers will vary. One possible answer for each is included.

1.

2.

3.

4.

←→ ←→ ← → ←→ 6. W X, YW , line m, XY and WY . 2

5.

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Chapter 1. Basics of Geometry, Answer Key

7. Plane V or plane RST . 8. In addition to the pictures to the right, three planes may not intersect at all and can be parallel.

9. A circle.

10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

26. 27. 28.

−→ ← → PQ intersects RS at point Q. − → AC and AB are coplanar and point D is not. −→ −→ Points E and H are coplanar, but their rays, EF and GH are non-coplanar. − → − → − → − → IJ , IK, IL, and IM with common endpoint I and J, K, L and M are non-collinear. Always Sometimes Sometimes Sometimes Never Always Sometimes Never Always Sometimes #18: By definition, a point does not take up any space, it is only a location. #21: The ray is never read “BA,” the endpoint is always stated first. To make #15 true, they must be three non-collinear points. For #16, the two rays must lie on the same line, which it does not state. For #20, four points could be coplanar, but you only need three points to make a plane, so the fourth point could be in another plane. For #23, theorems can also be proven true by definitions and previously proven theorems. The walls, ceiling and floor are all planes. When two of them intersect the intersection is a line (i.e. the ceiling and a wall). When two walls and either the ceiling or the floor intersect the intersection is a point. The spokes on a wheel are segments. They intersect at a point. Cities on a map are points and the distance between them can be found by measuring the segment connecting the points.

29-33. 3

1.1. Geometry - Second Edition, Points, Lines, and Planes, Review Answers

4

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Chapter 1. Basics of Geometry, Answer Key

1.2 Geometry - Second Edition, Segments and Distance, Review Answers 1. 2. 3. 4. 5. 6. 7. 8.

1.625 in 2.875 in 3.7 cm 8.2 cm 2.75 in 4.9 cm 4.625 in 8.7 cm

9. 10. O would be halfway between L and T , so that LO = OT = 8 cm 11. a. b. TA + AQ = T Q c. T Q = 15 in 12. a. b. HM + MA = HA c. AM = 11 cm 13. BC = 8 cm, BD = 25 cm, and CD = 17 cm

14. FE = 8 in, HG = 13 in, and FG = 17 in

15. a. b. c. d.

16. 17. 18. 19. 20. 21.

RS = 4 QS = 14 TS = 8 TV = 12

x = 3, HJ = 21, JK = 12, HK = 33 x = 11, HJ = 52, JK = 79, HK = 131 x = 1, HJ = 2 31 , JK = 5 32 , HK = 8 x = 17, HJ = 27, JK = 153, KH = 180 x = 16, HJ = 7, JK = 15, KH = 22 One possible answer. 5

1.2. Geometry - Second Edition, Segments and Distance, Review Answers

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|7 − (−6)|= 13 |−3 − 2|= 5 |0 − (−9)|= 9 |−4 − 1|= 5 Answers vary, but hopefully most students found their heights to be between 7 and 8 heads. Answers should include some reference to the idea that multiplying and dividing by ten (according to the prefixes) is much easier than keeping track of 12 inches in a ft, 3 ft in a yard, 5280 ft in a mile, etc. 28. Answers vary, but students should recognize that the pedometer is more likely to yield a false reading because a person’s stride length varies. One possible way to minimize this error would be to average a person’s stride length over a relatively long distance-i.e. count the number of steps taken in 100 m. 29. Answers vary. The cubit was the first recorded unit of measure and it was integral to the building of the Egyptian pyramids. 30. Students should comment on the “ideal” proportions found in the human face and how these correspond to our perception of beauty.

22. 23. 24. 25. 26. 27.

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Chapter 1. Basics of Geometry, Answer Key

1.3 Geometry - Second Edition, Angles and Measurement, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

False, two angles could be 5◦ and 30◦ . False, it is a straight angle. True True False, you use a compass. False, B is the vertex. True True True False, it is equal to the sum of the smaller angles within it. Acute

12. Obtuse

13. Obtuse

14. Acute

15. Obtuse

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1.3. Geometry - Second Edition, Angles and Measurement, Review Answers 16. Acute

17 & 18: Drawings should look exactly like 12 and 16, but with the appropriate arc marks.

19. 20. 21. 22. 23.

40◦ 122◦ 18◦ 87◦ AE = CD, ED = CB, m6 EDC = 90◦ , m6 EAC = m6 ABC

24. 25. An interior point would be (2, 0).

26. An interior point would be (2, 0). 8

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28. 29. 30. 31. 32.

33.

34.

35.

36.

Chapter 1. Basics of Geometry, Answer Key

27. m6 QOP = 100◦ m6 QOT = 130◦ m6 ROQ = 30◦ m6 SOP = 70◦ (x + 7)◦ + (2x + 19)◦ = 56◦ (3x + 26)◦ = 56◦ 3x◦ = 30◦ x = 10◦ (4x − 23)◦ + (4x − 23)◦ = 130◦ (8x − 46)◦ = 130◦ 8x◦ = 176◦ x = 22◦ (5x − 13)◦ + 90◦ = (16x − 55)◦ (5x + 77)◦ = (16x − 55)◦ 22◦ = 11x◦ x = 2◦ (x − 9)◦ + (5x + 1)◦ = (9x − 80)◦ (6x − 8)◦ = (9x − 80)◦ 72◦ = 3x◦ x = 24◦ Students should comment about the necessity to have a number of degrees in a line that is divisible by 30, 45, 60 and 90 degrees because these degree measures are prevalent in the study of geometrical figures. Basically, setting the measure of a straight line equal to 180 degrees allows us to have more whole number degree measures in common geometrical figures.

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1.4. Geometry - Second Edition, Midpoints and Bisectors, Review Answers

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1.4 Geometry - Second Edition, Midpoints and Bisectors, Review Answers

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

12 in 5 in 5 in 13 in 90◦ 10 in 24 in 90◦ 8 triangles PS QT ,V S 90◦ 45◦ bisector bisector PU is a segment bisector of QT 45◦ x = 9, y = 14◦ x = 14◦ x = 20◦ d = 13◦ x = 12 a = 22◦ , x = 12◦ 55◦ each

26. 37.5◦ each 10

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Chapter 1. Basics of Geometry, Answer Key

27. 3.5 cm each

28. 2 in each

29. You created a right, or 90◦ angle.

30. 31. 32. 33. 34.

(3, -5) (1.5, -6) (5, 5) (-4.5, 2) (7, 10) 11

1.4. Geometry - Second Edition, Midpoints and Bisectors, Review Answers

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35. (6, 9) 36. This is incorrect. She should have written AB = CD or AB ∼ = CD. 37. This formula will give the same answer.

x1 + x2 y1 + y2 , = (mx , my ) 2 2 y1 + y2 x1 + x2 = mx and = my 2 2 amp; x1 + x2 = 2mx and y1 + y2 = 2my amp; x1 = 2mx − x2

For#34,

and y1 = 2my − y2

x1 = 2(3) − (−1) = 7 y1 = 2(6) − 2 = 10

38. 39. A square or a rectangle.

40. Midpoint could be used to determine where you might want to make a stop halfway through a trip (if using a map the longitude and latitude could be used in the formula for midpoint). We often want to find the middle of something-the middle of a wall to hang a picture, the middle of a room to divide it in half, etc.

12

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Chapter 1. Basics of Geometry, Answer Key

1.5 Geometry - Second Edition, Angle Pairs, Review Answers 1. a. b. c. d.

45◦ 8◦ 71◦ (90 − z)◦

a. b. c. d.

135◦ 62◦ 148◦ (180 − x)◦

2.

3. 4. 5. 6. 7.

6 6 6 6

JNI and 6 MNL (or 6 INM and 6 JNL) INM and 6 MNL (or 6 INK and 6 KNL ) INJand 6 JNK INM and 6 MNL (or 6 INK and 6 KNL) a. b. c. d.

8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31.

117◦ 90◦ 63◦ 117◦

Always Sometimes Never Always Always Never Sometimes Always x = 7◦ x = 34◦ y = 13◦ x = 17◦ x = 15◦ y = 9◦ y = 8◦ x = 10.5◦ x = 4◦ y = 3◦ x = 67◦ , y = 40◦ x = 38◦ , y = 25◦ x = 15◦ , x = −4◦ x = 11◦ , √ x = −2◦ √ x = 1 + 102, x = 1 − 102 x = 11◦ , y = 7◦ 13

1.6. Geometry - Second Edition, Classifying Polygons, Review Answers

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1.6 Geometry - Second Edition, Classifying Polygons, Review Answers 1. 2. 3. 4. 5. 6. 7. 8.

10. 11. 12. 13. 14. 15. 16. 17. 18.

Acute scalene triangle Equilateral and equiangular triangle Right isosceles triangle. Obtuse scalene triangle Acute isosceles triangle Obtuse isosceles triangle No, there would be more than 180◦ in the triangle, which is impossible. No, same reason as #7.

9. All the angles in an equilateral triangle must be equal. So, an equilateral triangle is also an equiangular triangle. Concave pentagon Convex octagon Convex 17-gon Convex decagon Concave quadrilateral Concave hexagon A is not a polygon because the two sides do not meet at a vertex; B is not a polygon because one side is curved; C is not a polygon because it is not closed. 2 diagonals

19. 5 diagonals 14

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Chapter 1. Basics of Geometry, Answer Key

20. A dodecagon has twelve sides, so you can draw nine diagonals from one vertex. 21. The pattern is below

TABLE 1.1: Number of sides 3 4 5 6 7 8 9 10 11 12

Diagonals from one vertex 0 1 2 3 4 5 6 7 8 9

This shows us that the number diagonals from one vertex increase by one each time. So, for an n−gon, there are (n − 3) diagonals from one vertex. 22. Octagon has 20 total diagonals Nonagon has 27 total diagonals Decagon has 35 total diagonals Undecagon has 44 total diagonals Dodecagon has 54 total diagonals The pattern is 0, 2, 5, 9, 14, 20, 27, 35, 44, 54. To find the next term you would add one more than was added previously. For example, the next term you would add 11. The equation is n(n−3) 2 . 23. Sometimes 24. Always 25. Always 26. Never 27. Always 28. Sometimes, a square is ALWAYS a quadrilateral. 29. Sometimes, you can draw AT MOST n − 3 diagonals from one vertex. 30. Sometimes, a 5-point star is ALWAYS a decagon. For questions 31-34 answers will vary.

31. 15

1.6. Geometry - Second Edition, Classifying Polygons, Review Answers

32.

33. 34. a rhombus or diamond

35. This triangle is to scale.

36. Use #9 to help you. It is the same construction, but do not draw the third side.

16

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Chapter 1. Basics of Geometry, Answer Key

1.7 Geometry - Second Edition, Chapter Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

E B L A H M F O J G I K D C N

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C HAPTER

2

Reasoning and Proof, Answer Key

Chapter Outline 2.1

G EOMETRY - S ECOND E DITION , I NDUCTIVE R EASONING , R EVIEW A NSWERS

2.2

G EOMETRY - S ECOND E DITION , C ONDITIONAL S TATEMENTS , R EVIEW A N SWERS

18

2.3

G EOMETRY - S ECOND E DITION , D EDUCTIVE R EASONING , R EVIEW A NSWERS

2.4

G EOMETRY - S ECOND E DITION , A LGEBRAIC AND C ONGRUENCE P ROPERTIES , R EVIEW A NSWERS

2.5

G EOMETRY - S ECOND E DITION , P ROOFS ABOUT A NGLE PAIRS AND S EG MENTS , R EVIEW A NSWERS

2.6

C HAPTER R EVIEW A NSWERS

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Chapter 2. Reasoning and Proof, Answer Key

2.1 Geometry - Second Edition, Inductive Reasoning, Review Answers 1. 9, 21 2. 20, 110 3.

a. b. there are two more points in each star than its figure number. c. n + 2 4. a. 10;

b. 48 c. 2n 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21.

20, 23; 107; 3n + 2 −19, −24; −164; −5n + 11 64, 128; 34, 359, 738, 368; 2n 12, 1; −307; −11n + 78 −12, 0; −93; odd terms: −3n + 12, even terms: −4n 6 7 35 n 7 , 8 ; 36 ; n+1 2n 12 14 70 23 , 27 ; 139 ; 4n−1 −13, 15; 71; (−1)n−1 (2n + 1) 21, −25; −137; (−1)n (4n − 3) 1 −1 −1 (−1)n 12 , 14 ; 70 ; 2n 8, 11; 73; odd terms 2n + 3, even terms −2n + 14 36, 49; 1225; n2 38, 51; the amount that is added is increasing by two with each term. 48, 63; the amount that is added is increasing by two with each term. 216, 343; the term number cubed, n3 . 8, 13; add the previous two terms together to get the current term. There is a good chance that Tommy will see a deer, but it is not definite. He is reasoning correctly, but there are other factors that might affect the outcome. 19

2.1. Geometry - Second Edition, Inductive Reasoning, Review Answers

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22. Maddie has experimented multiple times and recognized a pattern in her results. This is a good example of inductive reasoning. 23. Juan does not use inductive reasoning correctly. It is important that conclusions are based on multiple observations which establish a pattern of results. He only has one trial. 24. Answers vary-correct answers should include multiple experiments or trials which indicate a clear pattern for outcomes. 25. Answers vary. 26. n(n+3) 2 27. (n+1)(n+2) 2 28. n(n+1)(n+2) 2 29. Students should notice that the points are collinear. Thus, they could find the rule by finding the equation of the line using any two of the three points. The equation is y = 5x − 2. 30. The sequences in problems 5, 6 and 8 are of the same type. They can be modeled by linear equations because they have a constant “slope” or rate of change. In other words, the same value is added or subtracted each time to get the next term.

20

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Chapter 2. Reasoning and Proof, Answer Key

2.2 Geometry - Second Edition, Conditional Statements, Review Answers 1. 2. 3. 4. 5. 6. 7.

8.

9.

10. 11. 12. 13. 14. 15. 16. 17.

Hypothesis: 5 divides evenly into x. Conclusion: x ends in 0 or 5. Hypothesis: A triangle has three congruent sides. Conclusion: It is an equilateral triangle. Hypothesis: Three points lie in the same plane. Conclusion: The three points are coplanar. Hypothesis: x = 3. Conclusion: x2 = 9. Hypothesis: You take yoga. Conclusion: You are relaxed. Hypothesis: You are a baseball player. Conclusion: You wear a hat. Converse: If x ends in 0 or 5, then 5 divides evenly into x. True. Inverse: If 5 does not divide evenly into x, then x does not end in 0 or 5. True. Contrapositive: If x does not end in 0 or 5, then 5 does not divide evenly into it. True Converse: If you are relaxed, then you take yoga. False. You could have gone to a spa. Inverse: If you do not take yoga, then you are not relaxed. False. You can be relaxed without having had taking yoga. You could have gone to a spa. Contrapositive: If you are not relaxed, then you did not take yoga. True Converse: If you wear a hat, then you are a baseball player. False. You could be a cowboy or anyone else who wears a hat. Inverse: If you are not a baseball player, then you do not wear a hat. False. Again, you could be a cowboy. Contrapositive: If you do not wear a hat, then you are not a baseball player. True If a triangle is equilateral, then it has three congruent sides. True. A triangle has three congruent sides if and only if it is equilateral. If three points are coplanar, then they lie in the same plane. True. Three points lie in the same plane if and only if they are coplanar. If x2 = 9, then x = 3. False. x could also be -3. If B is the midpoint of AC, then AB = 5 and BC = 5. This is a true statement. If AB 6= 5 and BC 6= 5, then B is not the midpoint of AC. This is true. If B is noncollinear with A and C. If AB 6= 5 and BC 6= 5, then B is not the midpoint of AC. It is the same as #14. the original statement p→q ∼ p →∼ q ∼∼ p →∼∼ q p→q

18. the contrapositive p→q ∼ p →∼ q ∼ q →∼ p 19. the contrapositive p→q q→ p ∼ q →∼ p 21

2.2. Geometry - Second Edition, Conditional Statements, Review Answers

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20. the original statement p→q ∼ q →∼ p ∼∼ p →∼∼ q p→q 21. If a U.S. citizen can vote, then he or she is 18 or more years old. If a U.S. citizen is 18 or more years old, then he or she can vote. 22. If a whole number is prime, then it has exactly two distinct factors. If a whole number has exactly two distinct factors, then it is prime. 23. If points are collinear, then there is a line that contains the points. If there is a line that contains the points, then the points are collinear. 24. If 2x = 18, then x = 9. If x = 9, then 2x = 18. 25. a. b. c. d.

Yes. No, x could equal -4. No, again x could equal -4. Yes.

a. b. c. d.

Yes. Yes. Yes. Yes.

a. b. c. d.

Yes. Yes. Yes. Yes.

a. b. c. d.

Yes. No, 6 ABC could be any value between 0 and 90 degrees. No, again 6 ABC could be any value between 0 and 90 degrees. Yes.

26.

27.

28.

29. Answers vary. 30. Answers vary.

22

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Chapter 2. Reasoning and Proof, Answer Key

2.3 Geometry - Second Edition, Deductive Reasoning, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9.

10.

11.

12. 13.

14. 15. 16. 17.

18. 19. 20. 21.

I am a smart person. Law of Detachment No conclusion If a shape is a circle, then we don’t need to study it. Law of Syllogism. You don’t text while driving. Law of Contrapositive. It is sunny outside. Law of Detachment. You are not wearing sunglasses. Law of Contrapositive. My mom did not ask me to clean my room. Law of Contrapositive. If I go to the park, I will give my dog a bath. Law of Syllogism. This is a sound argument, but it doesn’t make sense because we know that circles exist. p → q q→r r→s s→t ∴ p→t p→q p ∴q p→q ∼q ∴∼ p If I need a new watch battery, then I go to the mall. If I go to the mall, then I will shop. If I shop, then I will buy shoes. Conclusion: If I need a new watch battery, then I will buy shoes. If Anna’s teacher gives notes, then Anna writes them down. If Anna writes down the notes, then she can do the homework. If Anna can do the homework, then she will get an A on the test. If Anna gets an A on the test, her parents will take her out for ice cream. Conclusion: If Anna’s teacher gives notes, then Anna’s parents will buy her ice cream. Inductive; a pattern of weather was observed. Deductive; Beth used a fact to determine what her sister would eat. Deductive; Jeff used a fact about Nolan Ryan. Either reasoning. Inductive; surfers observed patterns of weather and waves to determine when the best time to surf is. Deductive; surfers could take the given statement as a fact and use that to determine when the best time to surf is. Inductive; observed a pattern. Both-Inductive: Amani noticed a pattern of behavior. Deductive: Amani ruled out possible explanations until there was only one remaining. Deductive: The detectives narrowed their field of suspects by eliminating those who couldn’t have committed the crime. See the following table:

TABLE 2.1: p T F

∼p F T

p∧ ∼ p F F

22. See the following table: 23

2.3. Geometry - Second Edition, Deductive Reasoning, Review Answers

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TABLE 2.2: p T T F F

∼p F F T T

q T F T F

∼q F T F T

∼ p∨ ∼ q F T T T

q∨ ∼ q T T T T

p ∧ (p∨ ∼ q) T T F F

23. See the following table:

TABLE 2.3: p T T F F

∼q F T F T

q T F T F

24. See the following table:

TABLE 2.4: p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

∼r F T F T F T F T

p∧q T T F F F F F F

(p ∧ q)∨ ∼ r T T F T F T F T

∼ q∨r T F T T T F T T

p ∨ (∼ q ∨ r) T T T T T F T T

25. See the following table:

TABLE 2.5: p T T T T F F F F

q T T F F T T F F

26. See the following table:

24

r T F T F T F T F

∼q F F T T F F T T

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Chapter 2. Reasoning and Proof, Answer Key

TABLE 2.6: p T T T T F F F F 27. 28. 29. 30. 31. 32. 33. 34.

q T T F F T T F F

r T F T F T F T F

∼r F T F T F T F T

q∨ ∼ r T T F T T T F T

p ∧ (q∨ ∼ r) T T F T F F F F

There are two more T ’s in #24. We can conclude that parenthesis placement matters. p ∨ q ∨ r is always true except the one case when p, q, and r are all false. True; Law of Syllogism Not valid True; Law of Contrapositive Not valid True; Law of Detachment Not valid

25

2.4. Geometry - Second Edition, Algebraic and Congruence Properties, Review Answers

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2.4 Geometry - Second Edition, Algebraic and Congruence Properties, Review Answers 1. 3x + 11 = −16 3x = −27 Subtraction PoE x = −9 Division PoE 2. 7x − 3 = 3x − 35 4x − 3 = −35 Subtraction PoE 4x = −32 Addition PoE x = −8 Division PoE 2 3. 3 g + 1 = 19 2 Subtraction PoE 3 g = 18 g = 27 Multiplication PoE 4. 21 MN = 5 MN = 10 Multiplication PoE ◦ 6 5. 5m ABC = 540 m6 ABC = 108◦ Division PoE 6. 10b − 2(b + 3) = 5b 10b − 2b − 6 = 5b Distributive Property 8b − 6 = 5b Combine like terms − 6 = −3b Subtraction PoE 2=b Division PoE b=2 Symmetric PoE 7. 14 y + 56 = 31 3y + 10 = 4 Multiplication PoE (multiplied everything by 12) 3y = −6 Subtraction PoE y = −2 Division PoE 8. 14 AB + 31 AB = 12 + 21 AB 3AB + 4AB = 144 + 6AB Multiplication PoE (multiplied everything by 12) 7AB = 144 + 6AB Combine like terms AB = 144 Subtraction PoE 9. 3 = x 10. 12x − 32 11. x = 12 12. y + z = x + y 13. CD = 5 14. z + 4 = y − 7 15. Yes, they are collinear. 16 + 7 = 23 16. No, they are not collinear, 9 + 9 6= 16. I cannot be the midpoint. 17. 6 NOP must be an obtuse angle because it is supplementary with 56◦ , meaning that m6 NOP is 180◦ − 56◦ = 124◦ . 90◦ < 124◦ < 180◦ , so by definition 6 NOP is an obtuse angle. 18. 6 ABC ∼ = 6 DEF 6 GHI ∼ = 6 JKL; ∼ = 6 s have = measures; m6 ABC + m6 GHI = m6 DEF + m6 GHI; Substitution 19. M is the midpoint of AN, N is the midpoint MB; AM = MN, MN = NB; Transitive 20. 6 BFE or 6 BFG ← → ← → 21. EF⊥BF 22. Yes, EG = FH because EF = GH and EF + FG = EG and FG + GH = FH by the Segment Addition Postulate. FG = FG by the Reflexive Property and with substitution EF + FG = EG and FG + EF = FH. 26

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Chapter 2. Reasoning and Proof, Answer Key

Therefore, EG = FH by the Transitive Property. 23. Not necessarily, G could slide along EH. 24. See the following table:

TABLE 2.7: Statement 1. 6 EBF ∼ = 6 HCG 6 ABE ∼ 6 = DCH 2. m6 EBF = m6 HCG m 6 ABE = m6 DCH 3. m6 ABF = m6 EBF + m6 ABE m 6 DCG = m6 HCG + m6 DCH 4. m6 ABF = m6 EBF + m6 ABE m 6 DCG = m6 EBF + m6 ABE 5. m6 ABF = m6 DCG 6. 6 ABF ∼ = 6 DCG

Reason Given ∼ = angles have = measures Angle Addition Postulate Substitution PoE Transitive PoE ∼ = angles have = measures

25. See the following table:

TABLE 2.8: Statement 1. AB = CD 2. BC = BC 3. AB + BC = CD + BC 4. AC = BD 26. 27. 28. 29. 30. 31. 32.

Reason Given Reflexive PoE Addition PoE Segment Addition Postulate

No No Yes Yes No No See the following table:

TABLE 2.9: Statement 1. 6 DAB is a right angle 2. m6 DAB = 90◦ 3. AC bisects 6 DAB 4. m6 DAC = m6 BAC 5. m6 DAB = m6 DAC + m6 BAC 6. m6 DAB = m6 BAC + m6 BAC 7. m6 DAB = 2m6 BAC 8. 90◦ = 2m6 BAC 9. 45◦ = m6 BAC

Reason Given Definition of a right angle Given Definition of an angle bisector Angle Addition Postulate Substitution PoE Combine like terms Substitution PoE Division PoE

27

2.4. Geometry - Second Edition, Algebraic and Congruence Properties, Review Answers

33.

TABLE 2.10: Statement 1. 6 1 and 6 2 form a linear pair m6 1 = m6 2 2. 6 1 and 6 2 are supplementary 3. m6 1 + m6 2 = 180◦ 4. m6 1 + m6 1 = 180◦ 5. 2m6 1 = 180◦ 6. m6 1 = 90◦ 7. 6 1 is a right angle

28

Reason Given Linear Pair Postulate Definition of Supplementary Substitution Simplify Division PoE Definition of a right angle

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Chapter 2. Reasoning and Proof, Answer Key

2.5 Geometry - Second Edition, Proofs about Angle Pairs and Segments, Review Answers 1. See the following table:

TABLE 2.11: Statement 1. AC⊥BD, 6 1 ∼ =6 4 2. m6 1 = m6 4 3. 6 ACB and 6 ACD are right angles 4. m6 ACB = 90◦ m 6 ACD = 90◦ 5. m6 1 + m6 2 = m6 ACB m 6 3 + m6 4 = m6 ACD 6. m6 1 + m6 2 = 90◦ m 6 3 + m6 4 = 90◦ 7. m6 1 + m6 2 = m6 3 + m6 4 8. m6 1 + m6 2 = m6 3 + m6 1 9. m6 2 = m6 3 10. 6 2 ∼ =6 3

Reason Given ∼ = angles have = measures ⊥ lines create right angles Definition of right angles Angle Addition Postulate Substitution Substitution Substitution Subtraction PoE ∼ = angles have = measures

2. See the following table:

TABLE 2.12: Statement 1. 6 MLN ∼ = 6 OLP 2. m6 MLN = m6 OLP 3. m6 MLO = m6 MLN + m6 NLO m 6 NLP = m6 NLO + m6 OLP 4. m6 NLP = m6 NLO + m6 MLN 5. m6 NLP = m6 MLO 6. 6 NLP ∼ = 6 MLO

Reason Given ∼ = angles have = measures Angle Addition Postulate Substitution Substitution ∼ = angles have = measures

3. See the following table:

TABLE 2.13: Statement 1. AE⊥EC, BE⊥ED 2. 6 BED is a right angle 6 AEC is a right angle 3. m6 BED = 90◦ m 6 AEC = 90◦ 4. m6 BED = m6 2 + m6 3 m 6 AEC = m6 1 + m6 2

Reason Given ⊥ lines create right angles

Definition of a right angle Angle Addition Postulate 29

2.5. Geometry - Second Edition, Proofs about Angle Pairs and Segments, Review Answers

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TABLE 2.13: (continued) Statement 5. 90◦ = m6 2 + m6 3 90^◦ = m6 1 + m6 2 6. m6 2 + m6 3 = m6 1 + m6 2 7. m6 3 = m6 1 8. 6 3 ∼ =6 1

Reason Substitution Substitution Subtraction PoE ∼ = angles have = measures

4. See the following table:

TABLE 2.14: Statement 1. 6 L is supplementary to 6 M 6 P is supplementary to 6 O 6 L∼ =6 O 2. m6 L = m6 O 3. m6 L + m6 M = 180◦ m 6 P + m6 O = 180◦ 4. m6 L + m6 M = m6 P + m6 O 5. m6 L + m6 M = m6 P + m6 L 6. m6 M = m6 P 7. 6 M ∼ =6 P

Reason Given ∼ = angles have = measures Definition of supplementary angles Substitution Substitution Subtraction PoE ∼ = angles have = measures

5. See the following table:

TABLE 2.15: Statement 1. 6 1 ∼ =6 4 2. m6 1 = m6 4 3. 6 1 and 6 2 are a linear pair 6 3 and 6 4 are a linear pair 4. 6 1 and 6 2 are supplementary 6 3 and 6 4 are supplementary 5. m6 1 + m6 2 = 180◦ m 6 3 + m6 4 = 180◦ 6. m6 1 + m6 2 = m6 3 + m6 4 7. m6 1 + m6 2 = m6 3 + m6 1 8. m6 2 = m6 3 9. 6 2 ∼ =6 3

Reason Given ∼ = angles have = measures Given (by looking at the picture) could also be Definition of a Linear Pair Linear Pair Postulate

Definition of supplementary angles Substitution Substitution Subtraction PoE ∼ = angles have = measures

6. See the following table:

TABLE 2.16: Statement 1. 6 C and 6 F are right angles 2. m6 C = 90◦ , m6 F = 90◦ 30

Reason Given Definition of a right angle

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Chapter 2. Reasoning and Proof, Answer Key

TABLE 2.16: (continued) Statement 3. 90◦ + 90◦ = 180◦ 4. m6 C + m6 F = 180◦

Reason Addition of real numbers Substitution

7. See the following table:

TABLE 2.17: Statement 1. l⊥m 2. 6 1 and 6 2 are right angles 3. 6 1 ∼ =6 2

Reason Given ⊥ lines create right angles. Right Angles Theorem

8. See the following table:

TABLE 2.18: Statement 1. m6 1 = 90◦ 2. 6 1 and 6 2 are a linear pair 3. 6 1 and 6 2 are supplementary 4. m6 1 + m6 2 = 180◦ 5. 90◦ + m6 2 = 180◦ 6. m6 2 = 90◦

Reason Given Definition of a linear pair Linear Pair Postulate Definition of supplementary angles Substitution Subtraction PoE

9. See the following table:

TABLE 2.19: Statement 1. l⊥m 2. 6 1 and 6 2 make a right angle 3. m6 1 + m6 2 = 90◦ 4. 6 1 and 6 2 are complementary

Reason Given ⊥ lines create right angles Definition of a right angle Definition of complementary angles

10. See the following table:

TABLE 2.20: Statement 1. l⊥m, 6 2 ∼ =6 6 2. m6 2 = m6 6 3. 6 5 ∼ =6 2 6 4. m 5 = m6 2 5. m6 5 = m6 6

Reason Given ∼ = angles have = measures Vertical Angles Theorem ∼ = angles have = measures Transitive

31

2.5. Geometry - Second Edition, Proofs about Angle Pairs and Segments, Review Answers 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35.

32

6

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AHM, 6 PHE and 6 GHE, 6 AHC AM ∼ = MG,CP ∼ = PE, AH ∼ = HE, MH ∼ = HP, GH ∼ = HC 6 AMH, 6 HMG and 6 CPH, 6 HPE 6 AHC 6 MAH, 6 HAC and 6 MGH, 6 HGE GC AE, GC 6 AHM, 6 MHG 6 AGH ∼ = 6 HGE Given; ∼ = angles have = measures; m6 ACE = m6 ACH + m6 ECH; m6 ACE = m6 ACH + m6 ACH; Combine like terms; 21 m6 ACE = m6 ACH; AC is the angle bisector of 6 ACH; Definition of an angle bisector 90◦ 26◦ 154◦ 26◦ 64◦ 25◦ 75◦ 105◦ 90◦ 50◦ 40◦ 25◦ 130◦ 155◦ 130◦

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Chapter 2. Reasoning and Proof, Answer Key

2.6 Chapter Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

D F H B I C G A J E

33

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C HAPTER

3

Parallel and Perpendicular Lines, Answer Key

Chapter Outline 3.1

G EOMETRY - S ECOND E DITION , L INES AND A NGLES , R EVIEW A NSWERS

3.2

G EOMETRY - S ECOND E DITION , P ROPERTIES OF PARALLEL L INES , R EVIEW A NSWERS

3.3

G EOMETRY - S ECOND E DITION , P ROVING L INES PARALLEL , R EVIEW A N SWERS

34

3.4

G EOMETRY - S ECOND E DITION , P ROPERTIES OF P ERPENDICULAR L INES , R E VIEW A NSWERS

3.5

G EOMETRY - S ECOND E DITION , PARALLEL AND P ERPENDICULAR L INES IN THE C OORDINATE P LANE , R EVIEW A NSWERS

3.6

G EOMETRY - S ECOND E DITION , T HE D ISTANCE F ORMULA , R EVIEW A NSWERS

3.7

C HAPTER R EVIEW A NSWERS

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Chapter 3. Parallel and Perpendicular Lines, Answer Key

3.1 Geometry - Second Edition, Lines and Angles, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

AB and EZ, XY and BW , among others AB || VW , among others BC ⊥ BW , among others one, AV one, CD 6 6 6 3 6 2 6 1 6 8 6 8 6 5 m6 3 = 55◦ (vertical angles), m6 1 = 125◦ (linear pair), m6 4 = 125◦ (linear pair) m6 8 = 123◦ (vertical angles), m6 6 = 57◦ (linear pair), m6 7 = 57◦ (linear pair) No, we do not know anything about line m. No, even though they look parallel, we cannot assume it.

17.

19. 20. 21. 22.

18. Fold the paper so that the lines match up and the crease passes through the point you drew. Same as number 19. One way to do this is to use the edges of the ruler as guide lines. The sides of the ruler are parallel. Use the ruler to draw a line. Turn the ruler perpendicular to the first line (make sure it is perpendicular by matching up a marking on the ruler to the original line. Use the ruler edge to draw the perpendicular line. 35

3.1. Geometry - Second Edition, Lines and Angles, Review Answers

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23. Parallel lines are evident in the veins of the leaves of ferns and the markings on some animals and insects. Parallel planes are illustrated by the surface of a body of water and the bottom. 24. Trees are usually perpendicular to the ground. Each leaf of a fern is perpendicular to the stem. 25. Some branches of trees are skew. 26. Any two equations in the form y = b, where b is a constant. 27. Any two equations in the form x = b, where b is a constant. 28. These two lines are parallel to each other. ← → ← → 29. slope of AB equals slope of CD = − 56 ; these lines are parallel ← → ← → 30. slope of AB = − 53 , slope of CD = 53 ; these lines are perpendicular 31. It appears that the slopes of parallel lines are the same and the slopes of perpendicular lines are opposite reciprocals. 32. y = 2x − 11 33. y = − 53 x + 2 34. y = − 32 x + 6 35. y = 4x − 5

36

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Chapter 3. Parallel and Perpendicular Lines, Answer Key

3.2 Geometry - Second Edition, Properties of Parallel Lines, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.

Supplementary Congruent Congruent Supplementary Congruent Supplementary Supplementary Same Side Interior Alternate Interior None Same Side Interior Vertical Angles Corresponding Angles Alternate Exterior None 6 1, 6 3, 6 6, 6 9, 6 11, 6 14, and 6 16 x = 70◦ , y = 90◦ x = 15◦ , y = 40◦ x = 9◦ , y = 26◦ x = 21◦ , y = 17◦ x = 25◦ y = 18◦ x = 20◦ x = 31◦ y = 12◦ See the following table:

TABLE 3.1: Statement 1. l || m 2. 6 1 ∼ =6 5 3. m6 1 = m6 5 4. 6 1 and 6 3 are supplementary 5. m6 1 + m6 3 = 180◦ 6. m6 3 + m6 5 = 180◦ 7. 6 3 and 6 5 are supplementary

Reason Given Corresponding Angles Postulate ∼ = angles have = measures Linear Pair Postulate Definition of Supplementary Angles Substitution PoE Definition of Supplementary Angles

27. See the following table:

TABLE 3.2: Statement 1. l || m

Reason Given 37

3.2. Geometry - Second Edition, Properties of Parallel Lines, Review Answers

TABLE 3.2: (continued) Statement 2. 6 1 ∼ =6 5 6 3. 5 ∼ =6 8 4. 6 1 ∼ =6 8

Reason Corresponding Angles Postulate Vertical Angles Theorem Transitive PoC

28. See the following table:

TABLE 3.3: Statement 1. l || m 2. 6 4 and 6 6 are supplementary 3. m6 4 + m6 6 = 180◦ 4. 6 2 ∼ = 6 6, 6 4 ∼ =6 8 6 6 5. m 2 = m 6, m6 4 = m6 8 6. m6 2 + m6 8 = 180◦ 7. 6 2 and 6 8 are supplementary

Reason Given Same Side Interior Angles Theorem Definition of Supplementary Angles Corresponding Angles Postulate ∼ = angles have = measures Substitution PoE Definition of Supplementary Angles

29. See the following table:

TABLE 3.4: Statement 1. l || m, s || t 2. 6 4 ∼ = 6 12 6 3. 12 ∼ = 6 10 ∼ 4. 6 4 = 6 10

Reason Given Corresponding Angles Postulate Corresponding Angles Postulate Transitive PoC

30. See the following table:

TABLE 3.5: Statement 1. l || m, s || t 2. 6 2 ∼ = 6 13 3. 6 13 ∼ = 6 15 ∼ 6 4. 2 = 6 15

Reason Given Alternate Exterior Angles Theorem Corresponding Angles Postulate Transitive PoC

31. See the following table:

TABLE 3.6: Statement 1. l || m, s || t 2. 6 6 ∼ =6 9 6 3. 4 ∼ =6 7 4. 6 6 and 6 7 are supplementary 38

Reason Given Alternate Interior Angles Theorem Vertical Angles Theorem Same Side Interior Angles

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Chapter 3. Parallel and Perpendicular Lines, Answer Key

TABLE 3.6: (continued) Statement 5. 6 9 and 6 4 are supplementary 32. 33. 34. 35.

Reason Same Angle Supplements Theorem

m6 1 = 102◦ , m6 2 = 78◦ , m6 3 = 102◦ , m6 4 = 78◦ , m6 5 = 22◦ , m6 6 = 78◦ , m6 7 = 102◦ x = 15◦ , y = 21◦ x = 37◦ , y = 28◦ The Same Side Interior Angles Theorem says that two angles are supplementary, not congruent.

39

3.3. Geometry - Second Edition, Proving Lines Parallel, Review Answers

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3.3 Geometry - Second Edition, Proving Lines Parallel, Review Answers 1. Start by copying the same angle as in Investigation 3-1, but place the copy where the alternate interior angle would be.

2. This question could be considered a “trick question,” because you are still copying two congruent angles, not two supplementary ones, like asked. Indicate the consecutive interior angles with arc marks, but copy the adjacent angle to the one that was copied in # 14.

3. Given, 6 1 ∼ = 6 3, Given, 6 2 ∼ = 6 3, Corresponding Angles Theorem, Transitive Property ∼ 4. Given, 6 1 = 6 3, Given, 6 2 ∼ = 6 3, l || m 5. Give, Converse of the Alternate Interior Angles Theorem, Given, Converse of the Alternate Interior Angles Theorem, Parallel Lines Property 6. See the following table:

TABLE 3.7: Statement 1. m ⊥ l, n ⊥ l 2. m6 l = 90◦ , m6 2 = 90◦ 3. m6 1 = m6 2 4. m || n

Reason Given Definition of Perpendicular Lines Transitive Property Converse of Corresponding Angles Theorem

7. See the following table:

TABLE 3.8: Statement 1. 6 1 ∼ =6 3 2. m || n 40

Reason Given Converse of Alternate Interior Angles Theorem

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Chapter 3. Parallel and Perpendicular Lines, Answer Key

TABLE 3.8: (continued) Statement 3. m6 3 + m6 4 = 180◦ 4. m6 1 + m6 4 = 180◦ 5. 6 1 and 6 4 are supplementary

Reason Linear Pair Postulate Substitution Definition of Supplementary Angles

8. See the following table:

TABLE 3.9: Statement 1. 6 2 ∼ =6 4 2. m || n 3. 6 1 ∼ =6 3

Reason Given Converse of Corresponding Angles Theorem Alternate Interior Angles Theorem

9. See the following table:

TABLE 3.10: Statement 1. 6 2 ∼ =6 3 2. m || n 3. 6 1 ∼ =6 4

Reason Given Converse of Corresponding Angles Theorem Alternate Exterior Angles Theorem

10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

none yes, AK || LJ by Converse of Consecutive Interior Angles Theorem yes, LG || KD by Converse of Corresponding Angles Theorem none none yes, AD || GJ by Converse of Alternate Interior Angles Theorem 58◦ 73◦ 107◦ 58◦ 49◦ 107◦ 49◦ x = 30◦ x = 15◦ x = 12◦ x = 26◦ x = 5◦ Construction, the first and last lines are parallel. You might conjecture that two lines perpendicular to the same line are parallel to each other. 29. You could prove this using any of the converse theorems learned in this section because all four angles formed where the transversal intersects the two parallel lines are right angles. Thus, Alternate Interior Angles, Alternate Exterior Angles and Corresponding Angles are all congruent and the Same Side Interior Angles are supplementary. 30. These two angles should be supplementary if the lines are parallel. 41

3.4. Geometry - Second Edition, Properties of Perpendicular Lines, Review Answers

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3.4 Geometry - Second Edition, Properties of Perpendicular Lines, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.

90◦ 90◦ 45◦ 16◦ 72◦ 84◦ 41◦ 24◦ 78◦ 90◦ 126◦ 54◦ 180◦ ⊥ not ⊥ not ⊥ ⊥ 90◦ 34◦ 56◦ 90◦ 56◦ 134◦ 134◦ 34◦ See the following table:

TABLE 3.11: Statement 1. l ⊥ m, l ⊥ n 2. 6 1 and 6 2 are right angles 3. m6 1 = 90◦ , m6 2 = 90◦ 4. m6 1 = m6 2 5. 6 1 ∼ =6 2 6. m || n 27. 28. 29. 30. 31. 32.

42

x = 12◦ x = 9◦ x = 13.5◦ x = 8◦ x = 4◦ x = 30◦

Reason Given Definition of perpendicular lines Definition of right angles Transitive PoE ∼ = angles have = measures Converse of the Corresponding Angles Postulate

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Chapter 3. Parallel and Perpendicular Lines, Answer Key

3.5 Geometry - Second Edition, Parallel and Perpendicular Lines in the Coordinate Plane, Review Answers 1. 2. 3. 4. 5. 6. 7.

1 3

-1 2 7

-2 4 undefined Perpendicular

8. Parallel

9. Perpendicular 43

3.5. Geometry - Second Edition, Parallel and Perpendicular Lines in the Coordinate Plane, Review Answers www.ck12.org

10. Neither

11. Perpendicular

12. Parallel 44

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Chapter 3. Parallel and Perpendicular Lines, Answer Key

13. Neither

14. Parallel

15. y = −5x − 7 45

3.5. Geometry - Second Edition, Parallel and Perpendicular Lines in the Coordinate Plane, Review Answers www.ck12.org 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34.

46

y = 32 x − 5 y = 14 x + 2 y = − 32 x + 1 y = 2x + 1 y = x − 10 y = −x − 4 y = − 13 x − 4 y = − 25 x + 7 x = −1 y=8 y = −3x + 13 Perpendicular y = 23 x + 2 y = − 32 x − 4 Parallel y = − 15 x + 7 y = − 15 x − 3 Perpendicular y = x y = −x Neither y = −2x + 2 y = 2x − 3 ⊥: y = − 43 x − 1 ||: y = 43 x + 5 41 ⊥: y = 3x − 3 ||: y = − 13 x + 7 ⊥: y = 7 ||: x = −3 ⊥: y = x − 4 ||: y = −x + 8

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Chapter 3. Parallel and Perpendicular Lines, Answer Key

3.6 Geometry - Second Edition, The Distance Formula, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29.

17.09 units 19.20 units 5 units 17.80 units 22.20 units 14.21 units 6.40 units 9.22 units 6.32 units 6.71 units 12 units 7 units 4.12 units 18.03 units 2.83 units 7.81 units 4 units 9 units 5.66 units 9.49 units 4.12 units 4.47 units y = 21 x − 3 y = −3x + 5 y = − 23 x − 4 y = 52 x + 8 (9, -4) (8, -1) y = − 53 x − 6, (0, −6)

47

3.6. Geometry - Second Edition, The Distance Formula, Review Answers

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30. y = 2x + 1

31. There are 12 possible answers: (-27, 9), (23, 9), (-2, -16), (-2, 34), (-17, -11), (-17, 29), (13, -11), (13, 29), (-22, 24), (-22, -6), (18, 24), and (18, -6) 32. 1. Graph the two lines. 2. Determine the slope of a perpendicular line to the two lines. 3. Use the slope from #2 to count from one line to the next to find a point on each line that is also on a perpendicular line. 4. Determine coordinates of the points from #3. 5. Plug the points from #4 into the distance formula and solve.

48

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Chapter 3. Parallel and Perpendicular Lines, Answer Key

3.7 Chapter Review Answers m6 1 = 90◦

m6 2 = 118◦

m6 3 = 90◦

m6 4 = 98◦

m6 5 = 28◦

m6 6 = 118◦

m6 7 = 128◦

m6 8 = 52◦

m6 9 = 62◦

49

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C HAPTER

4

Triangles and Congruence, Answer key

Chapter Outline

50

4.1

G EOMETRY - S ECOND E DITION , T RIANGLE S UMS , R EVIEW A NSWERS

4.2

G EOMETRY - S ECOND E DITION , C ONGRUENT F IGURES , R EVIEW A NSWERS

4.3

G EOMETRY - S ECOND E DITION , T RIANGLE C ONGRUENCE USING SSS AND SAS, R EVIEW A NSWERS

4.4

G EOMETRY - S ECOND E DITION , T RIANGLE C ONGRUENCE USING ASA, AAS, AND HL, R EVIEW A NSWERS

4.5

G EOMETRY - S ECOND E DITION , I SOSCELES AND E QUILATERAL T RIANGLES , R EVIEW A NSWERS

4.6

C HAPTER 4 R EVIEW A NSWERS

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Chapter 4. Triangles and Congruence, Answer key

4.1 Geometry - Second Edition, Triangle Sums, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

43◦ 121◦ 41◦ 86◦ 61◦ 51◦ 13◦ 60◦ 70◦ 118◦ 68◦ 116◦ 161◦ 141◦ 135◦ a = 68◦ , b = 68◦ , c = 25◦ , d = 155◦ , e = 43.5◦ , f = 111.5◦ See the following table:

TABLE 4.1: Statement 1. Triangle with interior and exterior angles. 2. m6 1 + m6 2 + m6 3 = 180◦ 3. 6 3 and 6 4 are a linear pair, 6 2 and 6 5 are a linear pair, and 6 1 and 6 6 are a linear pair 4. 6 3 and 6 4 are supplementary, 6 2 and 6 5 are supplementary, and 6 1 and 6 6 are supplementary 5. m6 1 + m6 6 = 180◦ , m6 2 + m6 5 = 180◦ m 6 3 + m6 4 = 180◦ 6. m6 1 + m6 6 + m6 2 + m6 5 + m6 3 + m6 4 = 540◦ 7. m6 4 + m6 5 + m6 6 = 360◦

Reason Given Triangle Sum Theorem Definition of a linear pair Linear Pair Postulate

Definition of supplementary angles Combine the 3 equations from #5. Subtraction PoE

18. See the following table:

TABLE 4.2: Statement 1. 4ABC with right angle B 2. m6 B = 90◦ 3. m6 A + m6 B + m6 C = 180◦ 4. m6 A + 90◦ + m6 C = 180◦ 5. m6 A + m6 C = 90◦ 6. 6 A and 6 C are complementary

Reason Given Definition of a right angle Triangle Sum Theorem Substitution Subtraction PoE Definition of complementary angles

51

4.1. Geometry - Second Edition, Triangle Sums, Review Answers 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

52

x = 14◦ x = 9◦ x = 22◦ x = 17◦ x = 12◦ x = 30◦ x = 25◦ x = 7◦ x = ±8◦ x = 17◦ x = 11◦ x = 7◦

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Chapter 4. Triangles and Congruence, Answer key

4.2 Geometry - Second Edition, Congruent Figures, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23.

R∼ = 6 U, 6 A ∼ = 6 G, 6 T ∼ = 6 H, RA ∼ = UG, AT ∼ = GH, RT ∼ = UH ∼ ∼ ∼ ∼ ∼ ∼ 6 6 6 6 6 B = T, I = O, G = P, BI = T O, IG = OP, BG = T P Third Angle Theorem 90◦ , they are congruent supplements Reflexive, FG ∼ = FG Angle Bisector 4FGI ∼ = 4FGH 6 A∼ = 6 E and 6 B ∼ = 6 D by Alternate Interior Angles Theorem Vertical Angles Theorem No, we need to know if the other two sets of sides are congruent. AC ∼ = CE and BC ∼ = CD ∼ 4ABC = 4EDC Yes, 4FGH ∼ = 4KLM No Yes, 4ABE ∼ = 4DCE No 4BCD ∼ = 4ZY X CPCTC m6 A = m6 X = 86◦ , m6 B = m6 Z = 52◦ , m6 C = m6 Y = 42◦ m6 A = m6 C = m6 Y = m6 Z = 35◦ , m6 B = m6 X = 110◦ m6 A = m6 C = 28◦ , m6 ABE = m6 DBC = 90◦ , m6 D = m6 E = 62◦ m6 B = m6 D = 153◦ , m6 BAC = m6 ACD = 15◦ , m6 BCA = m6 CAD = 12◦ See the following table: 6

6

TABLE 4.3: Statement 1. 6 A ∼ = 6 D, 6 B ∼ =6 E 6 6 2. m A = m D, m6 B = m6 E 3. m6 A + m6 B + m6 C = 180◦ m 6 D + m6 E + m6 F = 180◦ 4. m6 A + m6 B + m6 C = m6 D + m6 E + m6 F 5. m6 A + m6 B + m6 C = m6 A + m6 B + m6 F 6. m6 C = m6 F 7. 6 C ∼ =6 F

24. 25. 26. 27. 28.

Reason Given ∼ = angles have = measures Triangle Sum Theorem Substitution PoE Substitution PoE Subtraction PoE ∼ = angles have = measures

Transitive PoC Reflexive PoC Symmetric PoC Reflexive PoC 4ABC is either isosceles or equiangular because the congruence statement tells us that 6 A ∼ = 6 B. 53

4.2. Geometry - Second Edition, Congruent Figures, Review Answers

29. ∼ ∼ 30. 4SMR ∼ 4SMT 4T MA 4AMR and 4SRA ∼ = = = = 4RAT ∼ = 4AT S ∼ = 4T SA

54

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Chapter 4. Triangles and Congruence, Answer key

4.3 Geometry - Second Edition, Triangle Congruence using SSS and SAS, Review Answers

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

Yes, 4DEF ∼ = 4IGH No, HJ and ED are not congruent because they have different tic marks No, the angles marked are not in the same place in the triangles. Yes, 4ABC ∼ = 4RSQ No, this is SSA, which is not a congruence postulate No, one triangle is SSS and the other is SAS. Yes, 4ABC ∼ = 4FED Yes, 4ABC ∼ = 4Y XZ ∼ AB = EF AB ∼ = HI 6 C∼ =6 G 6 C∼ =6 K AB ∼ = JL AB ∼ = ON See the following table:

TABLE 4.4: Statement 1. AB ∼ = DC, BE ∼ = CE ∼ 6 6 2. AEB = DEC 3. 4ABE ∼ = 4ACE

Reason Given Vertical Angles Theorem SAS

16. See the following table:

TABLE 4.5: Statement 1. AB ∼ = DC, AC ∼ = DB 2. BC ∼ = BC 3. 4ABC ∼ = 4DCB

Reason Given Reflexive PoC SSS

17. See the following table:

TABLE 4.6: Statement 1. B is a midpoint of DC, AB⊥DC 2. DB ∼ = BC 3. 6 ABD and 6 ABC are right angles 4. 6 ABD ∼ = 6 ABC ∼ 5. AB = AB 6. 4ABD ∼ = 4ABC

Reason Given Definition of a midpoint ⊥ lines create 4 right angles All right angles are ∼ = Reflexive PoC SAS 55

4.3. Geometry - Second Edition, Triangle Congruence using SSS and SAS, Review Answers 18. See the following table:

TABLE 4.7: Statement 1. AB is an angle bisector of 6 DAC, AD ∼ = AC 2. 6 DAB ∼ = 6 BAC ∼ 3. AB = AB 4. 4ABD ∼ = 4ABC

Reason Given Definition of an Angle Bisector Reflexive PoC SAS

19. See the following table:

TABLE 4.8: Statement 1. B is the midpoint of DC, AD ∼ = AC ∼ 2. DB = BC 3. AB ∼ = AB 4. 4ABD ∼ = 4ABC

Reason Given Definition of a Midpoint Reflexive PoC SSS

20. See the following table:

TABLE 4.9: Statement 1. B is the midpoint of DE and AC, 6 ABE is a right angle 2. DB ∼ = BE, AB ∼ = BC 3. m6 ABE = 90◦ 4. m6 ABE = m6 DBC 5. 4ABE ∼ = 4CBD

Reason Given Definition of a Midpoint Definition of a Right Angle Vertical Angle Theorem SAS

21. See the following table:

TABLE 4.10: Statement 1. DB is the angle bisector of 6 ADC, AD ∼ = DC 6 2. ADB ∼ = 6 BDC 3. DB ∼ = DB 4. 4ABD ∼ = 4CBD

22. 23. 24. 25. 56

Yes Yes No Yes

Reason Given Definition of an Angle Bisector Reflexive PoC SAS

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Chapter 4. Triangles and Congruence, Answer key

26. Check the measures of the three sides in your triangle with your ruler to make sure that they are 5cm, 3cm and 2cm. If you are having trouble, follow the directions in investigation 4-2 using these lengths. 27. Match up your construction with the original to see if they are the same. 28. Your triangle should look like this.

29 and 30. These are the two triangles you should create in these two problems.

57

4.4. Geometry - Second Edition, Triangle Congruence using ASA, AAS, and HL, Review Answers www.ck12.org

4.4 Geometry - Second Edition, Triangle Congruence using ASA, AAS, and HL, Review Answers 1. Yes, AAS, 4ABC ∼ FDE 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

= Yes, ASA, 4ABC ∼ = 4IHG No No Yes, SAS, 4ABC ∼ = 4KLJ No Yes, SAS, 4RQP Yes, HL, 4ABC ∼ = 4QPR Yes, SAS, 4ABE ∼ = 4DBC No No Yes, ASA, 4KLM ∼ = 4MNO Yes, SSS, 4W ZY ∼ = 4Y XW Yes, AAS, 4W XY ∼ = 4QPO 6 DBC ∼ 6 = DBA because they are both right angles. 6 CDB ∼ = 6 ADB ∼ DB = DB See the following table:

TABLE 4.11: Statement 1. DB⊥AC, DB is the angle bisector of 6 CDA 2. 6 DBC and 6 ADB are right angles 3. 6 DBC ∼ = 6 ADB ∼ 4. 6 CDB = 6 ADB 5. DB ∼ = DB 6. 4CDB ∼ = 4ADB

Reason Given Definition of perpendicular All right angles are ∼ = Definition of an angle bisector Reflexive PoC ASA

19. CPCTC 20. 6 L ∼ = 6 O and 6 P ∼ = 6 N by the Alternate Interior Angles Theorem ∼ 6 6 21. LMP = NMO by the Vertical Angles Theorem 22. See the following table:

TABLE 4.12: Statement 1. LP || NO, LP ∼ = NO ∼ ∼ 6 6 6 2. L = O, P = 6 N 3. 4LMP ∼ = 6 OMN 23. CPCTC 24. See the following table: 58

Reason Given Alternate Interior Angles Theorem ASA

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Chapter 4. Triangles and Congruence, Answer key

TABLE 4.13: Statement 1. LP || NO, LP ∼ = NO ∼ ∼ 6 6 6 2. L = O, P = 6 N 3. 4LMP ∼ = 6 OMN 4. LM ∼ MO = 5. M is the midpoint of PN. 25. 26. 27. 28. 29. 30. 31.

Reason Given Alternate Interior Angles ASA CPCTC Definition of a midpoint

A∼ =6 N 6 C∼ =6 M PM ∼ = MN LM ∼ = MO or LP ∼ = NO UT ∼ = FG SU ∼ = FH See the following table: 6

TABLE 4.14: Statement 1. SV ⊥WU, T is the midpoint of SV and WU 2. 6 STW and 6 UTV are right angles 3. 6 STW ∼ = 6 UTV ∼ 4. ST = TV ,W T ∼ = TU 5. 4STW ∼ = 4UTV 6. W S ∼ = UV

Reason Given Definition of perpendicular All right angles are ∼ = Definition of a midpoint SAS CPCTC

32. See the following table:

TABLE 4.15: Statement 1. 6 K ∼ = 6 T, EI is the angle bisector of 6 KET 2. 6 KEI ∼ = 6 T EI 3. EI ∼ EI = 4. 4KEI ∼ = 4T EI 5. 6 KIE ∼ = 6 T IE 6. EI is the angle bisector of 6 KIT

Reason Given Definition of an angle bisector Reflexive PoC AAS CPCTC Definition of an angle bisector

59

4.5. Geometry - Second Edition, Isosceles and Equilateral Triangles, Review Answers

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4.5 Geometry - Second Edition, Isosceles and Equilateral Triangles, Review Answers All of the constructions are drawn to scale with the appropriate arc marks.

1.

2.

3.

4.

6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

17. 60

5. x = 10, y = 7 x = 14 x = 13◦ x = 16◦ x = 7◦ x=1 y=3 y = 11◦ , x = 4◦ x = 25◦ , y = 19◦ x = 3, y = 8 Yes, 4ABC is isosceles. 4ABD is congruent to 4CBD by ASA. Therefore segments AB and BC are congruent by CPCTC. Or, 6 A is congruent to 6 C by third angles theorem and thus the triangle is isosceles by the converse of the Base Angles Theorem.

www.ck12.org a. b. c. d. 18. 19. 20. 21. 22. 23. 24.

Chapter 4. Triangles and Congruence, Answer key

90◦ 30◦ 60◦ 2

Always Sometimes Sometimes Never Always a = 46◦ , b = 88◦ , c = 46◦ , d = 134◦ , e = 46◦ , f = 67◦ , g = 67◦ See the following table:

TABLE 4.16: Statement 1. Isosceles 4CIS, with base angles 6 C and 6 SIO is the angle bisector of 6 CIS 2. 6 C ∼ =6 S 6 3. CIO ∼ = 6 SIO 4. IO ∼ = IO 5. 4CIO ∼ = 4SIO 6. CO ∼ = OS 7. 6 IOC ∼ = 6 IOS 6 8. IOC and 6 IOS are supplementary 9. m6 IOC = m6 IOS = 90◦ 10. IO is the perpendicular bisector of CS

Reason Given Base Angles Theorem Definition of an Angle Bisector Reflexive PoC ASA CPCTC CPCTC Linear Pair Postulate Congruent Supplements Theorem Definition of a ⊥ bisector (Steps 6 and 9)

25. See the following table:

TABLE 4.17: Statement 1. Equilateral 4RST with RT ∼ = ST ∼ = RS 2. 6 R ∼ =6 S 3. 6 S ∼ =6 T 4. 6 R ∼ =6 T 5. 4RST is equilangular

Reason Given Base Angles Theorem Base Angles Theorem Transitive PoC Definition of an Equiangular 4

26. See the following table:

TABLE 4.18: Statement 1. Isosceles 4ICS with 6 C and 6 S, IO is the perpendicular bisector of CS 2. 6 C ∼ =6 S 3. CO ∼ = OS 4. m6 IOC = m6 IOS = 90◦ 5. 4CIO ∼ = 4SIO

Reason Given Base Angle Theorem Definition of a ⊥ bisector Definition of a ⊥ bisector ASA 61

4.5. Geometry - Second Edition, Isosceles and Equilateral Triangles, Review Answers

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TABLE 4.18: (continued) Statement 6. 6 CIO ∼ = 6 SIO 7. IO is the angle bisector of 6 CIS

Reason CPCTC Definition of an Angle Bisector

27. See the following table:

TABLE 4.19: Statement 1. Isosceles 4ABC with base angles 6 B and 6 C, Isosceles 4XY Z with base angles 6 Y and 6 Z, 6 C ∼ = 6 Z, BC ∼ = YZ 2. 6 B ∼ = 6 C, 6 Y ∼ =6 Z 6 3. 6 B ∼ Y = 4. 4ABC ∼ = 4XY Z

Reason Given

Base Angles Theorem Transitive PoC ASA

28. Bisect a 60◦ angle as shown.

29. Construct a 60◦ angle, then extend one side. The adjacent angle is 120◦ .

30. In investigations 3-2 and 3-3 you learned how to construct perpendiculars (i.e. 90◦ angles). You could make a 90◦ angle and copy your 30◦ onto it to make 120◦ . See investigation 1-2 for a review of copying an angle. 31. Method 1: Construct a 90◦ angle and bisect it. Method 2: Construct a 30◦ angle, bisect the 30◦ angle and copy the resulting 15◦ angle onto the original 30◦ to make a total of 45◦ .

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Chapter 4. Triangles and Congruence, Answer key

4.6 Chapter 4 Review Answers For 1-5, answers will vary. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

One leg and the hypotenuse from each are congruent, 4ABC ∼ = 4Y XZ ∼ Two angles and the side between them, 4ABC = EDC Two angles and a side that is NOT between them, 4ABC ∼ = 4SRT All three sides are congruent, 4ABC ∼ = 4CDA Two sides and the angle between them, 4ABF ∼ = 4ECD Linear Pair Postulate Base Angles Theorem Exterior Angles Theorem Property of Equilateral Triangles Triangle Sum Theorem Equilateral Triangle Theorem Property of an Isosceles Right Triangle

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C HAPTER

5

Relationships with Triangles, Answer Key

Chapter Outline 5.1

G EOMETRY - S ECOND E DITION , M IDSEGMENTS OF A T RIANGLE , R EVIEW A N SWERS

5.2

G EOMETRY - S ECOND E DITION , P ERPENDICULAR B ISECTORS IN T RIANGLES , R EVIEW A NSWERS

5.3

G EOMETRY - S ECOND E DITION , A NGLE B ISECTORS IN T RIANGLES , R EVIEW A NSWERS

5.4

G EOMETRY - S ECOND E DITION , M EDIANS AND A LTITUDES IN T RIANGLES , R EVIEW A NSWERS

5.5

G EOMETRY - S ECOND E DITION , I NEQUALITIES IN T RIANGLES , R EVIEW A N SWERS

5.6

G EOMETRY - S ECOND E DITION , E XTENSION : I NDIRECT P ROOF, R EVIEW A N SWERS

5.7

64

C HAPTER R EVIEW A NSWERS

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Chapter 5. Relationships with Triangles, Answer Key

5.1 Geometry - Second Edition, Midsegments of a Triangle, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

RS = TU = 6 TU = 8 x = 5, TU = 10 x=4 No, we cannot say that the triangles are congruent. We do not know any angle measures. y = 18 x = 12 x = 5.5 x=6 x = 14, y = 24 x = 6, z = 26 x = 5, y = 3 x = 1, z = 11 a. 53 b. 106 c. The perimeter of the larger triangle is double the perimeter of the midsegment triangle.

15. 16. 17. 18. 19.

(7, 1), (3, 6), (1, 3) (3, 6), (2, 2), (-5, -3) (2, 2), (1, -2), (-1, 1) (5, 0), (5, -4), (2, 0) GH = 13 , HI = 2, GI = − 12

20. 21. 22. 23. 24. 25.

(3, 4), (15, √ -2), (-3, -8) GH = 90 ≈ 9.49, Yes, GH is half of this side (0, 3), (0, -5) and (-4, -1) (-1, 4), (3, 4) and (5, -2) a. M(0, 3), N(−1, −2), O(−4, 0); b. slope of MN and AC = 5, slope of NO and AB = − 23 , and slope of MO and BC = 34 ; 65

5.1. Geometry - Second Edition, Midsegments of a Triangle, Review Answers c. MN =

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√ √ √ √ 26 and AC = 2 26; NO = 13 and AC = 2 13; OM = 5 and BC = 10.

26. a. M(1, 3), N(5, 2), O(2, 1); b. slope of MN and AC = − 41 , slope of NO and AB = 12 , and slope of MO and BC = −2; √ √ √ √ √ √ c. MN = 17 and AC = 2 17; NO = 10 and AC = 2 10; OM = 5 and BC = 2 5. 3 y1 +y3 2 y1 +y2 , M x1 +x 27. L x1 +x 2 , 2 2 , 2 28. slope of LM =

y1 +y3 y1 +y2 2 − 2 x1 +x3 x1 +x2 2 − 2

=

y3 −y2 x3 −x2

= slope of AT y1 + y3 y1 + y2 2 x1 + x3 x1 + x2 2 29. length of LM = + − − 2 2 2 2 s 2 2 x3 − x2 y3 − y2 = + 2 2 r 1 1 (x3 − x2 )2 + (y3 − y2 )2 = 4 q4 1 2 = 2 (x3 − x2 ) + (y3 − y2 )2 = 12 AT 30. We have just proven algebraically that the midsegment (or segment which connects midpoints of sides in a triangle) is parallel to and half the length of the third side. s

66

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Chapter 5. Relationships with Triangles, Answer Key

5.2 Geometry - Second Edition, Perpendicular Bisectors in Triangles, Review Answers

1.

2.

3. 4. Yes, but for #2, the circumcenter is not within the triangle. 5. For acute triangles, the circumcenter is inside the triangle. For right triangles, the circumcenter is on the hypotenuse. For obtuse triangles, the circumcenter is outside the triangle. 6. By the definition of a perpendicular bisector, all three sides are bisected and therefore each half is congruent and all six triangles are right triangles. Then, by the definition of a circumcenter, the distance from it to each vertex is congruent (the hypotenuses of each triangle). Therefore, all 6 triangles are congruent by HL.

7. 8. 9. 10. 11. 12. 13.

x = 16 x=8 x=5 x = 12 x = 31◦ x = 34 a. AE = EB, AD = DB 67

5.2. Geometry - Second Edition, Perpendicular Bisectors in Triangles, Review Answers

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b. No, AC 6= CB c. Yes, AD = DB 14. 15. 16. 17. 18. 19. 20.

No, not enough information No, we don’t know if T is the midpoint of XY . m = 12 (4, 2) y= √ −2x + 10 2 5 C is going to be on the perpendicular In the picture, it is √ above AB, but it also could be √ bisector of AB. √ below AB on y = −2x + 10. AB = 2 5, so AC is also 2 5. So, C will be 2 5 units above or below AB on y = −2x + 10.

21-25. drawing

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Chapter 5. Relationships with Triangles, Answer Key

26.

27. 28. The perpendicular bisector of one side in a triangle is the set of all points equidistant from the endpoints of that side. When we find the perpendicular bisector of a second side, we find all the points equidistant from the endpoints of the second side (one of which is an endpoint of the first side as well). This means that the intersection of these two lines is equidistant from all three vertices of the triangle. The segments connecting this point (the circumcenter) to each vertex would be the radius of the circumscribed circle. 29. Fill in the blanks: There is exactly one circle which contains any three points. 30. See the following table:

TABLE 5.1: Statement ← → 1. CD is the perpendicular bisector of AB 2. D is the midpoint of AB 3. AD ∼ = DB 4. 6 CDA and 6 CDB are right angles 5. 6 CDA ∼ = 6 CDB ∼ 6. CD = CD

Reason Given Definition of a perpendicular bisector Definition of a midpoint Definition of a perpendicular bisector Definition of right angles Reflexive PoC 69

5.2. Geometry - Second Edition, Perpendicular Bisectors in Triangles, Review Answers

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TABLE 5.1: (continued) Statement 7. 4CDA ∼ = 4CDB ∼ 8. AC = CB

Reason SAS CPCTC

31. See the following table:

TABLE 5.2: Statement 1. 4ABC is a right isosceles triangle and BD is the ⊥ bisector of AC 2. D is the midpoint of AC 3. AD ∼ = DC 4. AB ∼ = BC 5. BD ∼ = BD 6. 4ABD and 4CBD are congruent.

Reason Given Definition of a perpendicular bisector Definition of a midpoint Definition of Isosceles Triangle Reflexive Property of Congruence SSS

32. Since 6 ABC is a right angle and 6 ABD ∼ = 6 CBD (CPCTC), each must be 45◦ . Also, since 6 ABC is a right angle and 6 A ∼ = 6 C, by Base Angles Theorem, 6 A and 6 C = 45◦ . Therefore, by the converse of the Base Angles Theorem, 4ABD and 4CBD are isosceles.

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Chapter 5. Relationships with Triangles, Answer Key

5.3 Geometry - Second Edition, Angle Bisectors in Triangles, Review Answers 1-3. Construct the incenter using investigation 5-2. 4. Yes, by definition, angle bisectors are on the interior of the angle. So, the incenter will be on the interior of all three angles, or inside the triangle. 5. They will be the same point. 6. x = 6 7. x = 3 8. x = 8 9. x = 7 10. x = 9 11. x = 9 12. No, the line segment must be perpendicular to the sides of the angle also. 13. No, it doesn’t matter if the bisector is perpendicular to the interior ray. 14. Yes, the angles are marked congruent. 15. A is the incenter because it is on the angle bisectors. B is the circumcenter because it is equidistant to the vertices. 16. A is the circumcenter because it is equidistant to the vertices. B is the incenter because it is equidistant to the sides. 17. See the following table:

TABLE 5.3: Statement 1. AD ∼ = DC − → − → 2. BA⊥AD and BC⊥DC 3. 4. 5. 6. 7. 8.

6 6

DAB and 6 DCB are right angles DAB ∼ = 6 DCB BD ∼ = BD 4ABD ∼ = 4CBD 6 ABD ∼ = 6 DBC −→ BD bisects 6 ABC

18. 19. 20. 21.

Reason Given The shortest distance from a point to a line is perpendicular. Definition of perpendicular lines All right angles are congruent Reflexive PoC HL CPCTC Definition of an angle bisector

Incenter Circumcenter Circumcenter Incenter

22-25. In an equilateral triangle the circumcenter and the incenter are the same point. 71

5.3. Geometry - Second Edition, Angle Bisectors in Triangles, Review Answers

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26. See diagram for 29-31. − → − → 27. slope of BA is -2 and slope of BC is 21 . The rays are perpendicular because their slopes are opposite reciprocals. √ √ √ √ 28. AB = 20 = 2 5 and BC = 20 = 2 5. They are congruent. 29-31.

−→ 32. BD is the angle bisector of 6 ABC. Since AD⊥AB and CD⊥CB, 4DAB and 4DCB are right triangles. Since we have shown that AB ∼ = BC and we know BD ∼ = BD by the reflexive property, 4DAB ∼ = 4DCB by HL. Thus, −→ ∼ 6 ABD = 6 CBD by CPCTC. Now we can conclude that BD is the angle bisector of 6 ABC by definition of an angle bisector.

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Chapter 5. Relationships with Triangles, Answer Key

5.4 Geometry - Second Edition, Medians and Altitudes in Triangles, Review Answers 1-3. Use Investigation 5-3 to find the centroid. 4. The centroid will always be inside of a triangle because medians are always on the interior of a triangle. 5-7. Use Investigation 5-4 and 3-2 to find the orthocenter. For #6, the orthocenter will be outside of the triangle. 8. If a triangle is equilateral, then the incenter, circumcenter, orthocenter and centroid will all be the same point. This is because all of the sides are equal and all the angles are equal. 9. You only have to construct two lines for each point of concurrency. That is because any two lines intersect at one point. The fact that a third line intersects at this point does not change the location of the point. 10. y = 21 x + 2 11. y = −3x − 3 12. y = −x + 4 13. y = 31 x − 5 14. GE = 10 BE = 15 15. GF = 8 CF = 24 16. AG = 20 GD = 10 17. GC = 2x CF = 3x 18. x = 2, AD = 27 19. See the following table:

TABLE 5.4: Statement 1. 4ABC ∼ = 4DEF, AP and DO are altitudes ∼ 2. AB = DE 3. 6 P and 6 O are right angles 4. 6 P ∼ =6 O 6 5. ABC ∼ = 6 DEF 6. 6 ABC and 6 ABP are a linear pair 6 DEF and 6 DEO are a linear pair 7. 6 ABC and 6 ABP are supplementary 6 DEF and 6 DEO are supplementary 8. 6 ABP ∼ = 6 DEO 9. 4APB ∼ = 4DOE ∼ 10. AP = DO

Reason Given CPCTC Definition of an altitude All right angles are congruent CPCTC Definition of a linear pair Linear Pair Postulate Congruent Supplements Theorem AAS CPCTC

20. See the following table:

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5.4. Geometry - Second Edition, Medians and Altitudes in Triangles, Review Answers

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TABLE 5.5: Statement 1. Isosceles 4ABC with legs AB and AC BD⊥DC and CE⊥BE 2. 6 DBC ∼ = 6 ECB 3. 6 BEC and 6 CEB are right angles 4. 6 BEC ∼ = 6 CEB ∼ 5. BC = BC 6. 4BEC ∼ = 4CDB 7. BD ∼ = CE 21. 22. 23. 24. 25. 26. 27. 28.

Base Angles Theorem Definition of perpendicular lines All right angles are congruent Reflexive PoC AAS CPCTC

M(2, 5) y = 2x + 1 N(1, −3) y = −4x + 1 intersection (0, 1) Centroid (1, -1) (1, 3)

29. Midpoint of one side is x1 +x2 +x3 y1 +y2 +y3 . , 3 3 30. (1, -5)

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Reason Given

x1 +x2 y1 +y2 , 2 , 2

using the third vertex, the centroid is

x +x2

x3 +2( 1 2 3

)

y +y2

,

y3 +2( 1 2 3

)

=

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Chapter 5. Relationships with Triangles, Answer Key

5.5 Geometry - Second Edition, Inequalities in Triangles, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

AB, BC, AC BC, AB, AC AC, BC, AB 6 B, 6 A, 6 C 6 B, 6 C, 6 A 6 C, 6 B, 6 A No, 6 + 6 < 13 No, 1 + 2 = 3 Yes Yes No, 23 + 56 < 85 Yes 1 < 3rd side <17 11 < 3rd side <19 12 < 3rd side <52 Both legs must be longer than 12 0 < x < 10.3 m6 1 > m6 2 because 7 > 6 IJ, IG, GJ, GH, JH m6 1 < m6 2, m6 3 > m6 4 a=b a>b a

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5.6. Geometry - Second Edition, Extension: Indirect Proof, Review Answers

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5.6 Geometry - Second Edition, Extension: Indirect Proof, Review Answers Answers will vary. Here are some hints. Assume n is odd, therefore n = 2a + 1. Use the definition of an equilateral triangle to lead you towards a contradiction. Remember the square root of a number can be negative or positive. Use the definition of an isosceles triangle to lead you towards a contradiction. If x + y is even, then x + y = 2n, where n is any integer. Use the Triangle Sum Theorem to lead you towards a contradiction. With the assumption of the opposite of AB + BC = AC, these three lengths could make a triangle, thus making A, B, and C non-collinear. 8. If we assume that we have an even number of nickels, then the value of the coin collection must be a multiple of ten and we have a contradiction. 9. Assume that the last answer on the quiz is false. This implies that the fourth answer is true. If the fourth answer is true, then the one before it (the third answer) is false. However, this contradicts the fact that the third answer is true. 10. None. To prove this by contradiction, select each statement as the “true” statement and you will see that at least one of the other statements will also be true. If Charlie is right, then Rebecca is also right. If Larry is right, then Rebecca is right. If Rebecca is right, then Larry is right. 1. 2. 3. 4. 5. 6. 7.

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Chapter 5. Relationships with Triangles, Answer Key

5.7 Chapter Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

BE AE AH CE AG The point of concurrency is the circumcenter and use Investigation 5-1 to help you. The circle should pass through all the vertices of the triangle (inscribed triangle). The point of concurrency is the incenter and use Investigation 5-2 to help you. The circle should touch all the sides of the triangle (inscribed circle). The point of concurrency is the centroid and it is two-thirds of the median’s length from the vertex (among other true ratios). It is also the balancing point of a triangle. The point of concurrency is called the orthocenter. The circumcenter and the orthocenter can lie outside a triangle when the triangle is obtuse. x − 7 < third side < 3x + 5

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C HAPTER

6

Polygons and Quadrilaterals, Answer Key

Chapter Outline

78

6.1

G EOMETRY - S ECOND E DITION , A NGLES IN P OLYGONS , R EVIEW A NSWERS

6.2

G EOMETRY - S ECOND E DITION , P ROPERTIES OF PARALLELOGRAMS , R EVIEW A NSWERS

6.3

G EOMETRY - S ECOND E DITION , P ROVING Q UADRILATERALS ARE PARALLEL OGRAMS , R EVIEW A NSWERS

6.4

G EOMETRY - S ECOND E DITION , R ECTANGLES , R HOMBUSES AND S QUARES , R EVIEW A NSWERS

6.5

G EOMETRY - S ECOND E DITION , T RAPEZOIDS AND K ITES , R EVIEW A NSWERS

6.6

C HAPTER R EVIEW A NSWERS

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Chapter 6. Polygons and Quadrilaterals, Answer Key

6.1 Geometry - Second Edition, Angles in Polygons, Review Answers 1. See the following table:

TABLE 6.1: # of sides 3 4 5 6 7 8 9 10 11 12 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29.

# of 4s from one vertex 1 2 3 4 5 6 7 8 9 10

4s × 180◦ (sum) 180◦ 360◦ 540◦ 720◦ 900◦ 1080◦ 1260◦ 1440◦ 1620◦ 1800◦

Each angle in a regular n−gon 60◦ 90◦ 108◦ 120◦ 128.57◦ 135◦ 140◦ 144◦ 147.27◦ 150◦

Sum of the exterior angles 360◦ 360◦ 360◦ 360◦ 360◦ 360◦ 360◦ 360◦ 360◦ 360◦

2340◦ 3780◦ 26 20 157.5◦ 165◦ 30◦ 10◦ 360◦ 18 30 17 24 10 11 x = 60◦ x = 90◦ , y = 20◦ x = 35◦ y = 115◦ x = 105◦ x = 51◦ , y = 108◦ x = 70◦ , y = 70◦ , z = 90◦ x = 72.5◦ , y = 107.5◦ x = 90◦ , y = 64◦ x = 52◦ , y = 128◦ , z = 123◦ larger angles are 135◦ smallest angle is 36◦ x = 117.5◦ 79

6.1. Geometry - Second Edition, Angles in Polygons, Review Answers ◦

◦

30. 180◦ − (n−2)180 = 360 n n ◦ ◦ 360◦ 180 n−180 n+360 = n ◦ n 360◦ 360 = n n 31. a = 120◦ , b = 60◦ , c = 48◦ , d = 60◦ , e = 48◦ , f = 84◦ , g = 120◦ , h = 108◦ , j = 96◦

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Chapter 6. Polygons and Quadrilaterals, Answer Key

6.2 Geometry - Second Edition, Properties of Parallelograms, Review Answers m6 A = 108◦ , m6 C = 108◦ , m6 D = 72◦ m6 P = 37◦ , m6 Q = 143◦ , m6 D = 37◦ all angles are 90◦ m6 E = m6 G = (180 − x)◦ , m6 H = x◦ a = b = 53◦ c=6 d = 10, e = 14 f = 5, g = 3 h = 25◦ , j = 11◦ , k = 8◦ m = 25◦ , n = 19◦ p = 8, q = 3 r = 1, s = 2 t = 3, u = 4 96◦ 85◦ 43◦ 42◦ 12 2 64◦ 42◦ (2, 1), Find the midpoint of one of the diagonals since the midpoints are the same for both slope of EF = slope of GH = 41 ; slope of EH = slope of FG = − 52 ; Slopes of opposite sides are the same, therefore opposite √ sides are parallel. √ 24. EF = HG = 17; FG = EH = 29; lengths of opposite sides are the same (congruent). 25. A quadrilateral in the coordinate plane can be show to be a parallelogram by showing any one of the three properties of parallelograms shown in questions 22-24. 26. See the following table: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23.

TABLE 6.2: Statement 1. ABCD is a parallelogram with diagonal BD 2. AB || DC, AD || BC 3. 6 ABD ∼ = 6 BDC, 6 ADB ∼ = 6 DBC ∼ 4. DB = DB 5. 4ABD ∼ = 4CDB 6 6. 6 A ∼ C =

Reason Given Definition of a parallelogram Alternate Interior Angles Theorem Reflexive PoC ASA CPCTC

27. See the following table:

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6.2. Geometry - Second Edition, Properties of Parallelograms, Review Answers

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TABLE 6.3: Statement 1. ABCD is a parallelogram with diagonals BD and AC 2. AB || DC, AD || BC 3. 6 ABD ∼ = 6 BDC, 6 CAB ∼ = 6 ACD DC 4. AB ∼ = 5. 4DEC ∼ = 4BEA 6. AE ∼ = EC, DE ∼ = EB

Reason Given Definition of a parallelogram Alternate Interior Angles Theorem Opposite Sides Theorem ASA CPCTC

28. See the following table:

TABLE 6.4: Statements 1. ABCD is a parallelogram 2. m6 1 = m6 3 and m6 2 = m6 4 3. m6 1 + m6 2 + m6 3 + m6 4 = 360◦ 4. m6 1 + m6 2 + m6 1 + m6 2 = 360◦ 5. 2(m6 1 + m6 2) = 360◦ 6. m6 1 + m6 2 = 180◦ 29. 30. 31. 32.

82

w = 135◦ x = 16 y = 105◦ z = 60◦

Reasons Given Opposite angles congruent in parallelogram Sum of angles in quadrilateral is 360◦ Substitution Simplification Division POE

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Chapter 6. Polygons and Quadrilaterals, Answer Key

6.3 Geometry - Second Edition, Proving Quadrilaterals are Parallelograms, Review Answers 1. No 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.

Yes Yes Yes No No Yes No Yes Yes No No x=5 x = 8◦ , y = 10◦ x = 4, y = 3 Yes Yes No See the following table:

TABLE 6.5: Statement 1. 6 A ∼ = 6 C, 6 D ∼ =6 B 2. m6 A = m6 C, m6 D = m6 B 3. m6 A + m6 B + m6 C + m6 D = 360◦ 4. m6 A + m6 A + m6 B + m6 B = 360◦ 5. 2m6 A + 2m6 B = 360◦ 2m 6 A + 2m6 D = 360◦ 6. m6 A + m6 B = 180◦ m 6 A + m6 D = 180◦ 7. 6 A and 6 B are supplementary 6 A and 6 D are supplementary 8. AD || BC, AB || DC 9. ABCD is a parallelogram

Reason Given ∼ = angles have = measures Definition of a quadrilateral Substitution PoE Combine Like Terms Division PoE Definition of Supplementary Angles Consecutive Interior Angles Converse Definition of a Parallelogram

20. See the following table:

TABLE 6.6: Statement 1. AE ∼ = EC, DE ∼ = EB 6 2. AED ∼ = 6 BEC 6 DEC ∼ = 6 AEB

Reason Given Vertical Angles Theorem 83

6.3. Geometry - Second Edition, Proving Quadrilaterals are Parallelograms, Review Answers

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TABLE 6.6: (continued) Statement 3. 4AED ∼ = 4CEB ∼ 4AEB = 4CED 4. AB ∼ = DC, AD ∼ = BC 5. ABCD is a parallelogram

Reason SAS CPCTC Opposite Sides Converse

21. See the following table:

TABLE 6.7: Statement 1. 6 ADB ∼ = 6 CBD, AD ∼ = BC 2. AD || BC 3. ABCD is a parallelogram

22. 23. 24. 25. 26.

see graph -2√ 3 5 see graph The triangle is formed by the midsegments of the triangle formed when the parallelograms overlap. Four congruent triangles are formed within this center triangle, which is also congruent to the three outer triangles.

27. see graph 84

Reason Given Alternate Interior Angles Converse Theorem 5-10

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28. 29. 30. 31.

Chapter 6. Polygons and Quadrilaterals, Answer Key

parallelogram slope of W X = slope of Y Z = 3; slope of XY = slope of ZW = − 12 opposite sides parallel midpoint of diagonal YW is (1.5, 3.5); midpoint of diagonal XZ is (1.5, 3.5); midpoints bisect each other Each side of the parallelogram is parallel to the diagonal. For example, XY || DU || ZW , so opposite sides are parallel. They are also half the length of the diagonal so opposite sides are congruent. Either proves that W XY Z is a parallelogram.

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6.4. Geometry - Second Edition, Rectangles, Rhombuses and Squares, Review Answers

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6.4 Geometry - Second Edition, Rectangles, Rhombuses and Squares, Review Answers 1. a. b. c. d. e.

13 26 24 10 90◦

a. b. c. d. e.

12 21.4 11 54◦ 90◦

a. b. c. d.

90◦ 90◦ 45◦ 45◦

2.

3.

4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23.

24. 25. 26. 27. 86

Rectangle, the diagonals bisect each other and are congruent. Rhombus, all sides are congruent and the diagonals are perpendicular. None Parallelogram, the diagonals bisect each other. Square, the diagonals bisect each other, are congruent, and perpendicular. Rectangle, all angles are right angles. None Square, all the angles and sides are congruent. Parallelogram, one set of sides are parallel and congruent. Sometimes, with the figure is a square. Always Sometimes, when it is a square. Always Sometimes, when it is a square. Never Square Rhombus Rectangle Parallelogram Answers will vary. One possibility: Another way to determine if a quadrilateral is a square would be to find the length of all the sides using the distance formula. All sides must be equal. Then, find the slopes of each side. If the adjacent sides have perpendicular slopes, then the angles are all 90◦ and thus congruent. ◦ , z = 37◦ x = 10, w = 53◦ , y = 37√ ◦ ◦ x = 45 , y = 90 , z = 2 2 x = y = 13, w = z = 25◦ See the following table:

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Chapter 6. Polygons and Quadrilaterals, Answer Key

TABLE 6.8: Statements 1. ABCD is a rectangle 2. BW ∼ = WC, AY ∼ = Y D, BX ∼ = XA, CZ ∼ = ZD 3. BD = AC 4. XY is a midsegment in 4ABD ZY is a midsegment in 4ACD XW is a midsegment in 4ABC W Z is a midsegment in 4BCD 5. XY = 21 BD = W Z and XW = 12 AC = Y Z 6. 12 BD = 12 AC 7. XY = W Z = Y Z = XW 8. W XY Z is a rhombus

Reasons 1. Given 2. Definition of a midpoint 3. Diagonals are congruent in a rectangle 4. Definition of a midsegment in a triangle

5. Midsegment in a triangle is half the length of the parallel side. 6. Division POE 7. Substitution 8. Definition of a rhombus

28. Answers may vary. The quadrilateral inscribed in the rhombus will always be a rectangle because the diagonals of a rhombus are perpendicular and the opposite sides of the inscribed quadrilateral will be parallel to the diagonals and thus perpendicular to one another. 29. Answers may vary. First, the square is a rhombus, the inscribed quadrilateral will be a rectangle (see problem 28). Second, the diagonals of the square are congruent so the sides of the inscribed quadrilateral will be congruent (see problem 27). Since the sides of the inscribed quadrilateral are perpendicular and congruent the parallelogram is a square.

30. Start by drawing a segment 2 inches long. Construct the perpendicular bisector of this segment. Mark off points on the perpendicular bisector .75 inches from the point of intersection. Connect these points to the endpoint of your original segment.

31. There are an infinite number of rectangles with diagonals of length 3 inches. The picture to the left shows three possible rectangles. Start by drawing a segment 3 inches long. Construct the perpendicular bisector of the segment to find the midpoint. Anchor your compass at the midpoint of the segment and construct a circle which contains the endpoints of your segment (radius 1.5 inches). Now you can draw a second diameter to 87

6.4. Geometry - Second Edition, Rectangles, Rhombuses and Squares, Review Answers your circle and connect the endpoints to form a rectangle with diagonal length 3 inches.

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Chapter 6. Polygons and Quadrilaterals, Answer Key

6.5 Geometry - Second Edition, Trapezoids and Kites, Review Answers 1. a. b. c. d.

55◦ 125◦ 90◦ 110◦

a. b. c. d. e.

50◦ 50◦ 90◦ 25◦ 115◦

2.

3. No, if the parallel sides were congruent, then it would be a parallelogram. By the definition of a trapezoid, it can never be a parallelogram (exactly one pair of parallel sides). 4. Yes, the diagonals do not have to bisect each other.

5. Construct two perpendicular lines to make the diagonals. One diagonal is bisected, so measure an equal length on either side of the point of intersection on one diagonal. Mark this as two vertices. The other two vertices are on the other diagonal. Place them anywhere on this diagonal and connect the four points to create the kite. Answers will vary. 6. 33 7. 28 8. 8 9. 11 10. 37 11. 5 12. x = 4 √ 13. x = 5, y = 73 14. x = 11, y = 17 15. y = 5◦ 16. y = 45◦ 17. x = 12◦ , y = 8◦ 18. parallelogram 19. square 20. kite 21. trapezoid 22. None 23. isosceles trapezoid 24. rectangle 89

6.5. Geometry - Second Edition, Trapezoids and Kites, Review Answers

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25. rhombus 26. See the following table:

TABLE 6.9: Statement 1. KE ∼ = T E and KI ∼ = TI ∼ 2. EI = EI 3. 4EKI ∼ = 4ET I 4. 6 KES ∼ = 6 T ES and 6 KIS ∼ = 6 T IS 5. EI is the angle bisector of 6 KET and 6 KIT

Reason Given Reflexive PoC SSS CPCTC Definition of an angle bisector

27. See the following table:

TABLE 6.10: Statement 1. KE ∼ = T E and KI ∼ = TI 2. 4KET and 4KIT are isosceles triangles 3. EI is the angle bisector of 6 KET and 6 KIT 4. EI is the perpendicular bisector of KT 5. KT ⊥ EI

Reason Given Definition of isosceles triangles Theorem 6-22 Isosceles Triangle Theorem

28. See the following table:

TABLE 6.11: Statement 1. T RAP is an isosceles trapezoid with T R || AP 2. T P ∼ = RA ∼ 3. AP = AP 4. 6 T PA ∼ = 6 RAP 5. 4T PA ∼ = 4RAP ∼ 6. TA = RP

Reason Given Definition of isosceles trapezoid Reflexive PoC Base angles congruent in isosceles trapezoid SAS CPCTC

29. The sides of the parallelogram inscribed inside a kite will be parallel to the diagonals because they are triangle midsegments. Since the diagonals in a kite are perpendicular, the sides of the parallelogram will be perpendicular as well. The diagonals in a kite are not congruent so only opposite sides of the parallelogram will be congruent and thus preventing the parallelogram from being a square. 30. Since the diagonals are congruent and the sides of the inscribed parallelogram are half the length of the diagonals they are parallel to (because they are triangle midsegments), they are all congruent. This makes the inscribed parallelogram a rhombus.

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Chapter 6. Polygons and Quadrilaterals, Answer Key

6.6 Chapter Review Answers

1. 2. 3. 4. 5. 6. 7. 8.

Never Always Always Sometimes Sometimes Never Always Sometimes

TABLE 6.12: Opposite sides ||

Diagonals ⊥

Opposite sides ∼ =

Opposite angles ∼ =

Diagonals ∼ =

No No

No No, base 6 s∼ = Non-vertex 6 s Yes All 6 s ∼ = Yes All 6 s ∼ =

No Yes

Trapezoid Isosceles Trapezoid Kite

One set One set

Diagonals bisect each other No No

No

No

Yes

No Non-parallel sides No

Parallelogram Rectangle Rhombus Square

Both sets Both sets Both sets Both sets

Yes Yes Yes Yes

No No Yes Yes

Yes Yes All sides ∼ = All sides ∼ =

No No Yes No Yes

a = 64◦ , b = 118◦ , c = 82◦ , d = 99◦ , e = 106◦ , f = 88◦ , g = 150◦ , h = 56◦ , j = 74◦ , k = 136◦

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C HAPTER

7

Similarity, Answer Key

Chapter Outline 7.1

G EOMETRY - S ECOND E DITION , R ATIOS AND P ROPORTIONS , R EVIEW A N SWERS

7.2

G EOMETRY - S ECOND E DITION , S IMILAR P OLYGONS , R EVIEW A NSWERS

7.3

G EOMETRY - S ECOND E DITION , S IMILARITY BY AA, R EVIEW A NSWERS

7.4

G EOMETRY - S ECOND E DITION , S IMILARITY BY SSS AND SAS, R EVIEW A N SWERS

7.5

G EOMETRY - S ECOND E DITION , P ROPORTIONALITY R ELATIONSHIPS , R EVIEW A NSWERS

7.6

G EOMETRY - S ECOND E DITION , S IMILARITY T RANSFORMATIONS , R EVIEW A N SWERS

7.7

G EOMETRY - S ECOND E DITION , E XTENSION : S ELF -S IMILARITY, R EVIEW A N SWERS

7.8

92

C HAPTER R EVIEW A NSWERS

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Chapter 7. Similarity, Answer Key

7.1 Geometry - Second Edition, Ratios and Proportions, Review Answers 1. a. b. c. d. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

14.

15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32.

4:3 5:8 6:19 6:8:5

2:1 1:3 2:1 1:1 5:4:3 x = 18◦ , angles are 54◦ , 54◦ , 72◦ x = 3; 9, 12, 15 x = 4; 12, 20 x = 16; 64, 112 X = 4; 20, 36 x = 4; 12, 44 a+b c+d b = d d(a + b) = b(c + d) ad + bd = bc + bd ad = bc a−b c−d b = d d(a − b) = b(c − d) ad − bd = bc − bd ad = bc x = 12 x = −5 y = 16 x = 12, −12 y = −21 z = 3.75 x = 13.9 gallons The president makes $800,000, vice president makes $600,000 and the financial officer makes $400,000. 1 32 cups water 60 marshmallows; 6 cups miniatures False True True False 28 18 7 24

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7.2. Geometry - Second Edition, Similar Polygons, Review Answers

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7.2 Geometry - Second Edition, Similar Polygons, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31.

94

True False False False True True False True IG BG BI 6 B∼ = AT = HT = 6 H, 6 I ∼ = 6 A, 6 G ∼ = 6 T, HA 3 5 5 or 3 HT = 35 IG = 27 57, 95, 35 or 53 m6 E = 113◦ , m6 Q = 112◦ 3 2 3 or 2 12 21 6 18 No, 32 26 6= 12 Yes, 4ABC ∼ 4NML Yes, ABCD ∼ STUV Yes, 4EFG ∼ 4LMN x = 12, y = 15 31 x = 20, y = 7 ≈ 14.6 a ≈ 7.4, b = 9.6 X = 6, y = 10.5 121 1:3 30u2 , 270u2 , 1 : 9, this is the ratio of the lengths squared or

1 2 3 .

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Chapter 7. Similarity, Answer Key

7.3 Geometry - Second Edition, Similarity by AA, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.

4T RI T R, T I, AM 12 6 6, 12 4ABE ∼ 4CDE because 6 BAE ∼ = 6 DCE and 6 ABE ∼ = 6 CDE by the Alternate Interior Angles Theorem. AE BE Answers will vary. One possibility: CE = DE One possibility: 4AED and 4BEC AC = 22.4 Only two angles are needed because of the 3rd Angle Theorem. Congruent triangles have the same shape AND size. Similar triangles only have the same shape. Also, congruent triangles are always similar, but similar triangles are not always congruent. Yes, right angles are congruent and solving for the missing angle in each triangle, we find that the other two angles are congruent as well. FE = 43 k k = 16 right, right, similar Yes, 4DEG ∼ 4FDG ∼ 4FED Yes, 4HLI ∼ 4HKJ No only vertical angles are congruent Yes, they are ⊥ to the same line. Yes, the two right angles are congruent and 6 OEC and 6 NEA are vertical angles. x = 48 f t. Yes, we can use the Pythagorean Theorem to find EA. EA = 93.3 f t. 70 ft 29 ft 2 in 24 ft Answers will vary. Check your answer by considering whether or not it is reasonable.

27. 28. m6 1 + m6 2 = 90◦ , therefore m6 GDF = m6 2 and m6 EDF = 6 1. This shows that the three angles in each triangle are congruent to the three corresponding angles in each of the other triangles. Thus, they are all similar. 29. DF 30. GD 31. FE

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7.4. Geometry - Second Edition, Similarity by SSS and SAS, Review Answers

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7.4 Geometry - Second Edition, Similarity by SSS and SAS, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31.

96

Yes, SSS. The side lengths are proportional. No. One is much larger than the other. There are 2.2 cm in an inch, so that is the scale factor. There is no need. With the A and A parts of ASA we have triangles with two congruent angles. The triangles are similar by AA. 4DFE DF, EF, DF DH = 7.5 4DBE SAS 27 AB, BE, AC 7 8 Yes, 21 = 24 . This proportion will be valid as long as AC || DE. Yes, 4ABC ∼ 4DFE, SAS No, the angle is not between the given sides Yes, 4ABC ∼ 4DFE, SSS Yes, 4ABE ∼ 4DBC, SAS 15 No, 10 20 6= 25 24 No, 32 6= 16 20 x=3 x = 6, y = 3.5 The building is 10 ft tall. The child’s shadow is 105 inches long. The side lengths are 15, 36, 39 The radio tower √ is 55 ft. √ AB = BC = 11.25, AC = 3, DE = EF = 5, DF = 2 AB BC AC 3 DE = EF = DF = 2 Yes, 4ABC ∼ 4DEF by SSS similarity. slope of CA = slope of LO = undefined (vertical); slope of AR = slope of OT = 0 (horizontal). 90◦ , vertical and horizontal lines are perpendicular. T O = 6, OL = 8,CA = 4 and AR = 3; LO : CA = OT : AR = 2 : 1 Yes, by SAS similarity.

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Chapter 7. Similarity, Answer Key

7.5 Geometry - Second Edition, Proportionality Relationships, Review Answers 4ECF ∼ 4BCD DF CD FE DF, DB 14.4 21.6 16.8 45 The parallel sides are in the same ratio as the sides of the similar triangles, not the segments of the sides. yes no yes no x=9 y = 10 y = 16 z=4 x=8 x = 2.5 a = 4.8, b = 9.6 a = 4.5, b = 4, c = 10 a = 1.8, b = 37 x = 5, y = 7 3 2 b or 1.5b 16 5 a or 3.2a Casey mistakenly used the length of the angle bisector in the proportion rather than the other side length. The correct proportion is a5 = 75 , thus a = 25 7. 28. The path will intersect the third side 2.25 m from the 3 m side and 3.75 m from the 5 m side. 29. a = 42m and b = 56m 30. Blanks are in red.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27.

TABLE 7.1: Statement − → 1. AC is the angle bisector of 6 BADX, A, B are collinear − → ← → and AC || XD 2. 6 BAC ∼ = 6 CAD ∼ 6 6 3. X = BAC 4. 6 CAD ∼ = 6 ADX ∼ 6 6 5. X = ADX 6. 4XAD is isosceles 7. AX ∼ = AD 8. AX = AD BA BC 9. AX = CD

Reason Given Definition of an angle bisector Corresponding Angles Postulate Alternate Interior Angles Theorem Transitive PoC Base Angles Converse Definition of an Isosceles Triangle Congruent segments are also equal Theorem 7-7 97

7.5. Geometry - Second Edition, Proportionality Relationships, Review Answers

TABLE 7.1: (continued) Statement BC BA = CD 10. AD

98

Reason Substitution PoE

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Chapter 7. Similarity, Answer Key

7.6 Geometry - Second Edition, Similarity Transformations, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

(2, 6) (-8, 12) (4.5, -6.5) k = 23 k=9 k = 21 20, 26, 34 2 23 , 3, 5 k = 52 14 k = 11 original: 20, dilation: 80, ratio: 4:1 If k = 1, then the dilation is congruent to the original figure. A0 (6, 12), B0 (−9, 21),C0 (−3, −6) A0 (9, 6), B0 (−3, −12),C0 (0, −7.5)

15. 16. k = 2 17. A00 (4, 8), B00 (48, 16),C00 (40, 40) 18. k = 2 19. √ a. √5 b. √ 5 c. 3 √5 d. 2 √5 e. 4 5 20.

√ a. 5 √ 5 b. 10 √5 c. 20 5 99

7.6. Geometry - Second Edition, Similarity Transformations, Review Answers

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21. a. OA : OA0 = 1 : 2 AB : A0 B0 = 1 : 2 b. OA : OA00 = 1 : 4 AB : A00 B00 = 1 : 4

23. 24. 25. 26. 27.

22. x=3 y = 2x + 1 (3, 7) This point is the center of the dilation. The scale factor is 3.

28. 0 O0 0 G0 0 D0 29. DDO = 3, OOG = 3 and GGD = 3. 30. 3 31. To dilate the original figure by a scale factor of 4 make one additional tick mark with your compass. 32. To dilate the original figure by a scale factor of 12 construct the perpendicular bisectors of CG,CO and CD to find the midpoints of the segments which will be your G0 , O0 and D0 respectively.

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Chapter 7. Similarity, Answer Key

7.7 Geometry - Second Edition, Extension: Self-Similarity, Review Answers 1. 2. See the following table:

TABLE 7.2:

Stage 0 Stage 1 Stage 2 Stage 3 Stage 4 Stage 5

Number of Segments

Length of each Segment

1 2 4 8 16 32

1

Total Length of the Segments 1

1 3 1 9 1 27 1 81 1 243

2 3 4 9 8 27 16 81 32 243

3. There will be 2n segments. 4. The length of each segment will be

1 3n

units.

5. 1 6. Number of edges: 192 Edge length: 27 Perimeter:

192 27

7. 8. See the following table:

TABLE 7.3: Color No Color

Stage 0 0 1

Stage 1 1 8

Stage 2 9 64

Stage 3 73 512 101

7.7. Geometry - Second Edition, Extension: Self-Similarity, Review Answers

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9. Answers will vary. Many different flowers (roses) and vegetables (broccoli, cauliflower, and artichokes) are examples of fractals in nature. 10. Answers will vary.

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Chapter 7. Similarity, Answer Key

7.8 Chapter Review Answers 1. a. x = 12 b. x = 14.5 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

x = 10◦ ; 50◦ , 60◦ , 70◦ 3.75 gallons yes no yes, AA yes, SSS no no A0 (10.5, 3), B0 (6, 13.5),C0 (−1.5, 6) x = 19 3 x=1 z=6 a = 5, b = 7.5

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C HAPTER

8

Right Triangle Trigonometry, Answer Key

Chapter Outline 8.1

G EOMETRY - S ECOND E DITION , T HE P YTHAGOREAN T HEOREM , R EVIEW A N SWERS

8.2

G EOMETRY - S ECOND E DITION , C ONVERSE OF THE P YTHAGOREAN T HEO REM , R EVIEW A NSWERS

8.3

G EOMETRY - S ECOND E DITION , U SING S IMILAR R IGHT T RIANGLES , R EVIEW A NSWERS

8.4

G EOMETRY - S ECOND E DITION , S PECIAL R IGHT T RIANGLES , R EVIEW A N SWERS

8.5

G EOMETRY - S ECOND E DITION , TANGENT, S INE AND C OSINE , R EVIEW A N SWERS

104

8.6

G EOMETRY - S ECOND E DITION , I NVERSE T RIGONOMETRIC R ATIOS , R EVIEW A NSWERS

8.7

G EOMETRY - S ECOND E DITION , E XTENSION : L AWS OF S INES AND C OSINES , R EVIEW A NSWERS

8.8

C HAPTER R EVIEW A NSWERS

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Chapter 8. Right Triangle Trigonometry, Answer Key

8.1 Geometry - Second Edition, The Pythagorean Theorem, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32.

√ 505 √ 9√ 5 799 12 10 √ 10 14 26√ 3p 41 2 2 √x + y 9 2 yes no no yes yes no √ 20 √39 14√ 429 17 287 √4 4√ 5 493 √ 5 10 36.6 × 20.6 33.6 √ × 25.2 3 2 4 √s 16 3 a2 + 2ab + b2 c2 + 4 21 ab = c2 + 2ab a2 + 2ab + b2 = c2 + 2ab, which simplifies to a2 + b2 = c2 1 1 2 2 2 (a + b)(a 1+ b) = 2 (a1 + 2ab + b ) 1 2 2 ab + 2 c = ab + 2 c 1 2 1 2 2 2 2 2 2 2 2 (a + 2ab + b ) = ab + 2 c ⇒ a + 2ab + b = 2ab + c , which simplifies to a + b = c

105

8.2. Geometry - Second Edition, Converse of the Pythagorean Theorem, Review Answers

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8.2 Geometry - Second Edition, Converse of the Pythagorean Theorem, Review Answers 1. a. c = 15 b. 12 < c < 5 c. 15 < c < 21 2. a. a = 7 b. 7 < a < 24 c. 1 < c < 7 3. It is a right triangle because 8, 15, 17 is a Pythagorean triple. The “x” indicates that this set is a multiple of 8, 15, 17. 4. right 5. no 6. right 7. acute 8. right 9. obtuse 10. right 11. acute 12. acute 13. right 14. obtuse 15. obtuse 16. acute

17. obtuse 106

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Chapter 8. Right Triangle Trigonometry, Answer Key

18. One way is to use the distance formula to find the distances of all three sides and then use the converse of the Pythagorean Theorem. The second way would be to find the slope of all three sides and determine if two sides are perpendicular. 19. c = 13 √ 20. d = 194 21. The sides of 4ABC are a multiple of 3, 4, 5 which is a right triangle. 6 A is opposite the largest side, which is the hypotenuse, making it 90◦ . 22. See the following table:

TABLE 8.1: Statement 1. In 4ABC, a2 + b2 < c2 , and c is the longest side. In 4LMN, 6 N is a right angle. 2. a2 + b2 = h2 3. c2 > h2 4. c > h 5. 6 C is the largest angle in 4ABC. 6. m6 N = 90◦ 7. m6 C > m6 N 8. m6 C > 90◦ 9. 6 C is an obtuse angle. 10. 4ABC is an obtuse triangle.

23. 24. 25. 26.

Reason Given Pythagorean Theorem Transitive PoE Take the square root of both sides The largest angle is opposite the longest side. Definition of a right angle SSS Inequality Theorem Transitive PoE Definition of an obtuse angle. Definition of an obtuse triangle.

right obtuse acute (1, 5), (-2, -3)

27 and 28. answers vary, you can check your answer by plotting the points on graph paper and measuring with a protractor or using the distance formula to verify the appropriate inequality. 29 and 30. While your diagram may be different because your angle at A may be different, the construction should look something like this: 107

8.2. Geometry - Second Edition, Converse of the Pythagorean Theorem, Review Answers

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31. The sum of the angles in a triangle must be 180◦ , if 6 C is 90◦ , then both 6 A and 6 B are acute. 32. You could construct a line perpendicular to AB through 6 B (you will need to extend the segment beyond B to do the construction). Next, select any point on this perpendicular segment and call it C. By connecting A and C you will make 4ABC.

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Chapter 8. Right Triangle Trigonometry, Answer Key

8.3 Geometry - Second Edition, Using Similar Right Triangles, Review Answers 4KML ∼ √4JML ∼ 4JKL KM = 6√ 3 JK = 6 √7 KL√= 3 21 16 √2 15√ 7 2 √ 35 14 √6 20√ 10 2 102√ x = 12√ 5 y = 5 √5 z=9 2 x = 4√ y = 465 √ z = 14 5 √ √ 8 41 17. x = 32 5 , y = √ 5 , z = 2 41 18. x = 9,√ y = 3 34 19. x = 9 20481 , y = 81 40 , z = 40 20. See the following table: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

TABLE 8.2: Statement 1. 4ABD with AC⊥DB and 6 DAB is a right angle. 2. 6 DCA and 6 ACB are right angles 3. 6 DAB ∼ = 6 DCA ∼ = 6 ACB ∼ 6 6 4. D = D 5. 4CAD ∼ = 4ABD ∼ 6 6 6. B = B 7. 4CBA ∼ = 4ABD 8. 4CAD ∼ = 4CBA

Reason Given Definition of perpendicular lines. All right angles are congruent. Reflexive PoC AA Similarity Postulate Reflexive PoC AA Similarity Postulate Transitive PoC

21. See the following table:

TABLE 8.3: Statement 1. 4ABD with AC⊥DB and 6 DAB is a right angle. 2. 4ABD ∼ 4CBA ∼ 4CAD AB 3. BC AB = DB

Reason Given Theorem 8-5 Corresponding sides of similar triangles are proportional.

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8.3. Geometry - Second Edition, Using Similar Right Triangles, Review Answers 22. 23. 24. 25. 26. 27. 28. 29. 30.

6.1% 10.4% 9.4% ratios are 31 and 93 , which both reduce to the common ratio 3. Yes, this is true for the next pair of terms since 27 9 also reduces to 3. geometric mean; geometric mean 10 20 1 See the following table:

TABLE 8.4: Statement a b 1. ae = d+e and db = d+e 2. a2 = e(d + e) and b2 = d(d + e) 3. a2 + b2 = e(d + e) + d(d + e) 4. a2 + b2 = (e + d)(d + e) 5. c = d + e 6. a2 + b2 = c2

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Reason Theorem 8-7 Cross-Multiplication Property Combine equations from #2. Distributive Property Segment Addition Postulate Substitution PoE

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Chapter 8. Right Triangle Trigonometry, Answer Key

8.4 Geometry - Second Edition, Special Right Triangles, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

√ x √2 x √ 3, 2x 15 √2 11 2 8 √ 90 2√or 127.3 ft. a = 2 √ 2, b = 2 c = 6 2, d √ = 12 e = f =√13 2 g = 10 3, h√= 20 k = 8, j√= 8 3 √ x = 11 3, y = 22 3 m = 9, n√= 18 √ q = 14 6, p√= 28 3 s = 9,t = 3√ 3 x = w√ =9 2 √ a = 9 √3, b = 18 √3 p = 6 15, q = 6 5 Yes, it’s a 30-60-90 triangle. No, it is√not even a right triangle. 16 + 6√ 3 8 + 8√ 3 x√ :x 3 4 √2 in 3 3 in 2 √ 25 2 4 √3 f t 27 2 3 in 2 12 3960 √ ft s 3 2

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8.5. Geometry - Second Edition, Tangent, Sine and Cosine, Review Answers

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8.5 Geometry - Second Edition, Tangent, Sine and Cosine, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

112

d f f e f d d e d e f e

D, D equal, complement reciprocals 0.4067 0.7071 28.6363 0.6820 sin A = 45 , cos A = 53 , tan A = 43 √ √ 2 sin A = 2 , cos A = 2 2 , tan A = 1 √ √ sin A = 31 , cos A = 2 3 2 , tan A = 4 2 x = 9.37, y = 12.72 x = 14.12, y = 19.42 x = 20.84, y = 22.32 x = 19.32, y = 5.18 x = 5.85, y = 12.46 x = 20.89, y = 13.43 x = 435.86 f t. x = 56 m 25.3 ft 42.9 ft 94.6 ft 49 ft 14 miles The hypotenuse is the longest side in a right triangle. Since the sine and cosine ratios are each a leg divided by the hypotenuse, the denominator is always going to be greater than the numerator. This ensures a ratio that is less than 1.

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Chapter 8. Right Triangle Trigonometry, Answer Key

8.6 Geometry - Second Edition, Inverse Trigonometric Ratios, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

23. 24.

25. 26. 27. 28. 29. 30. 31. 32.

33.7◦ 31.0◦ 44.7◦ 39.4◦ 46.6◦ 36.9◦ 34.6◦ 82.9◦ 70.2◦ m6 A = √ 38◦ , BC = 9.38, AC = 15.23 AB = 4√ 10, m6 A = 18.4◦ , m6 B = 71.6◦ ◦ BC = 51, m6 A = 45.6, m6 C = √ 44.4 ◦ m6 A = 60 √ , BC = 12, AC = 12 3 CB = 7 5, m6 A = 48.2◦ , m6 B = 41.8◦ m6 B = 50◦ , AC = 38.14, AB = 49.78 You would use a trig ratio when given a side and an angle and the Pythagorean Theorem if you are given two sides and no angles. 47.6◦ 1.6◦ 44.0◦ 192 ◦ 11 f t ≈ 17 f t 5 in; 54 51◦ For problem 20: since the earth tilts on its axis, the position of the sun in the sky varies throughout the year for most places on earth. Thus, the angle at which the sun hits a particular object will vary at different times of the year. For problem 21: the water pressure in the hose will affect the path of the water, the more pressure, the longer the water will travel in a straight path before gravity causes the path of the water to arc and come back down towards the ground. Tommy used OA instead of OA for his tangent ratio. Tommy used the correct ratio in his equation here, but he used the incorrect angle measure he found previously which caused his answer to be incorrect. This illustrates the benefit of using given information whenever possible. Tommy could have used Pythagorean Theorem to find the hypotenuse instead of a trigonometric ratio. cos 50◦ sin 20◦ As the angle measures increase, the sine value increases. As the angle measures increase, the cosine value decreases. The sine and cosine values are between 0 and 1. tan 85◦ = 11.43, tan 89◦ = 57.29, and tan 89.5◦ = 114.59. As the tangent values get closer to 90◦ , they get larger and larger. There is no maximum, the values approach infinity. The sine and cosine ratios will always be less than one because the denominator of the ratios is the hypotenuse which is always longer than either leg. Thus, the numerator is always less than the denominator in these ratios resulting in a value less than one.

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8.7. Geometry - Second Edition, Extension: Laws of Sines and Cosines, Review Answers

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8.7 Geometry - Second Edition, Extension: Laws of Sines and Cosines, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

114

m6 B = 84◦ , a = 10.9, b = 13.4 m6 B = 47◦ , a = 16.4, c = 11.8 m6 A = 38.8◦ , m6 C = 39.2◦ , c = 16.2 b = 8.5, m6 A = 96.1◦ , m6 C = 55.9◦ m6 A = 25.7◦ , m6 B = 36.6, m6 C = 117.7◦ m6 A = 81◦ , m6 B = 55.4◦ , m6 C = 43.6◦ b = 11.8, m6 A = 42◦ , m6 C = 57◦ b = 8.0, m6 B = 25.2◦ , m6 C = 39.8◦ m6 A = 33.6◦ , m6 B = 50.7◦ , m6 C = 95.7◦ m6 C = 95◦ , AC = 3.2, AB = 16.6 BC = 33.7, m6 C = 39.3◦ , m6 B = 76.7◦ m6 A = 42◦ , BC = 34.9, AC = 22.0 m6 B = 105◦ , m6 C = 55◦ , AC = 14.1 m6 B = 35◦ , AB = 12, BC = 5 Yes, BC would still be 5 units (see isosceles triangle below); the measures of 6 C are supplementary as shown below.

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Chapter 8. Right Triangle Trigonometry, Answer Key

8.8 Chapter Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

◦ 6 BC = 4.4, √ AC = 10.0, m A = 26 AB = 5 √10, m6 A = 18.4◦ , m6 B = 71.6◦ ◦ ◦ 6 BC = 6 7, m6 A = 41.4 √ , m C = 48.6 ◦ m6 A = 30 √ , AC = 25 3, BC = 25 BC = 7 13, m6 A = 31◦ , m6 B √ = 59◦ ◦ m6 B = 45 , AC = 32, AB = 32 2 m6 B = 63◦ , BC = 19.1, AB = 8.7 m6 C = √ 19◦ , AC = 22.7, AB = 7.8 BC = 4 13, m6 B = 33.7◦ , m6 C = 56.3◦ acute right, Pythagorean triple obtuse right acute obtuse x = 2√ x = 2 √110 x=6 7 2576.5 ft. x = 29.2◦ AC = 16.1, m6 A = 41.6◦ , m6 C = 63.4◦ m6 A = 123.7◦ , m6 B = 26.3◦ , m6 C = 30◦

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C HAPTER

9

Circles, Answer Key

Chapter Outline

116

9.1

G EOMETRY - S ECOND E DITION , PARTS OF C IRCLES AND TANGENT L INES , R EVIEW A NSWERS

9.2

G EOMETRY - S ECOND E DITION , P ROPERTIES OF A RCS , R EVIEW A NSWERS

9.3

G EOMETRY - S ECOND E DITION , P ROPERTIES OF C HORDS , R EVIEW A NSWERS

9.4

G EOMETRY - S ECOND E DITION , I NSCRIBED A NGLES , R EVIEW A NSWERS

9.5

G EOMETRY - S ECOND E DITION , A NGLES OF C HORDS , S ECANTS , AND TAN GENTS , R EVIEW A NSWERS

9.6

G EOMETRY - S ECOND E DITION , S EGMENTS OF C HORDS , S ECANTS , AND TAN GENTS , R EVIEW A NSWERS

9.7

G EOMETRY - S ECOND E DITION , E XTENSION : W RITING AND G RAPHING THE E QUATIONS OF C IRCLES , R EVIEW A NSWERS

9.8

C HAPTER R EVIEW A NSWERS

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Chapter 9. Circles, Answer Key

9.1 Geometry - Second Edition, Parts of Circles and Tangent Lines, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

diameter secant chord point of tangency common external tangent common internal tangent center radius the diameter 4 lines

11. 3 lines

12. none

13. 14. 15. 16. 17. 18. 19.

J

J

J

radius of B = 4, radius of C = 5, radius of D = 2, radius of J ∼J D = E because they have the same radius length. 2 common tangents CE = 7 y = x−2 yes no

J

E =2

117

9.1. Geometry - Second Edition, Parts of Circles and Tangent Lines, Review Answers 20. 21. 22. 23. 24. 25. 26. 27.

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yes √ 4 √10 4 11 x=9 x=3 x = 5√ x=8 2 a. b. c. d.

Yes, by AA. m6 CAE = m6 DBE = 90◦ and 6 AEC ∼ = 6 BED by vertical angles. BC = 37 AD = 35 m6 C = 53.1◦

28. See the following table:

TABLE 9.1: Statement 1. AB and CB with points of tangency at A and C. AD and DC are radii. 2. AD ∼ = DC 3. DA⊥AB and DC⊥CB 4. m6 BAD = 90◦ and m6 BCD = 90◦ 5. Draw BD. 6. 4ADB and 4DCB are right triangles 7. DB ∼ = DB 8. 4ABD ∼ = 4CBD 9. AB ∼ = CB

Reason Given All radii are congruent. Tangent to a Circle Theorem Definition of perpendicular lines Connecting two existing points Definition of right triangles (Step 4) Reflexive PoC HL CPCTC

29. a. kite b. center, bisects 30. AT ∼ = BT ∼ = CT ∼ = DT by theorem 10-2 and the transitive property. 31. 9.23 8 32. √8 ; √ 3 3 3 ← → ←→ 33. Since AW and W B both share point W and are perpendicular to VW because a tangent is perpendicular to the radius of the circle. Therefore A, B and W are collinear. V T ∼ = VW because they are tangent segments to circle A from the same point, V , outside the circle. Similarly, VW ∼ = VU because they are tangent segments to circle B from V . By the transitive property of congruence, V T ∼ = VU. Therefore, all three segments are congruent.

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Chapter 9. Circles, Answer Key

9.2 Geometry - Second Edition, Properties of Arcs, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36.

minor major semicircle major minor semicircle c ∼ c yes, CD = DE ◦ 66 228◦ yes, they are in the same circle with equal central angles yes, the central angles are vertical angles, so they are equal, making the arcs equal no, we don’t know the measure of the corresponding central angles. 90◦ 49◦ 82◦ 16◦ 188◦ 172◦ 196◦ 270◦ x = 54◦ x = 47◦ x = 25◦ J ∼J A= B ◦ 62 77◦ 139◦ 118◦ 257◦ 319◦ 75◦ 105◦ 68◦ 105◦ 255◦ 217◦

119

9.3. Geometry - Second Edition, Properties of Chords, Review Answers

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9.3 Geometry - Second Edition, Properties of Chords, Review Answers 1. No, see picture. The two chords can be congruent and perpendicular, but will not bisect each other.

2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29.

120

AC c DF c JF DE 6 HGC 6 AGC AG, HG,CG, FG, JG, DG 107◦ 8◦ 118◦ 133◦ 140◦ 120◦ x = 64◦ , y = 4 x = 8,√ y = 10 x = 3 √26, y ≈ 12.3 x=9 5 x = 9, y = 4 x = 4.5 x=3 x = 7√ x = 4 11 c = 121.3◦ mAB c = 112.9◦ mAB c∼ c by Theorem 10-5. BF ∼ = FD and BF = FD ∼ CA = AF by Theorem 10-6. QS is a diameter by Theorem 10-4. a-c shown in the diagram below; d. it is the center; e. shown in the diagram; this construction is not done to scale and your chords might be in different places but this should give you an idea of what it should look like.

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Chapter 9. Circles, Answer Key

30. for AB: a. b. c. d.

(1, 5) m = 0, ⊥m is undefined x=1 for BC: a. 92 , 23 b. m = 7, ⊥m = − 71 c. y = − 17 x + 15 7 e. Point of intersection (center of the circle) is (1, 2). f. radius is 5 units

31. a. 120◦ b. 60◦

121

9.4. Geometry - Second Edition, Inscribed Angles, Review Answers

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9.4 Geometry - Second Edition, Inscribed Angles, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29.

48◦ 120◦ 54◦ 45◦ 87◦ 27◦ 100.5◦ 95.5◦ 76.5◦ 84.5◦ 51◦ 46◦ x = 180◦ , y = 21◦ x = 60◦ , y = 49◦ x = 30◦ , y = 60◦ x = 72◦ , y = 92◦ x = 200◦ , y = 100◦ x = 68◦ , y = 99◦ x = 93◦ , y = 97◦ x = 10◦ x = 24◦ x = 74◦ , y = 106◦ x = 35◦ , y = 35◦ 55◦ 70◦ 110◦ 90◦ 20◦ 90◦

TABLE 9.2: Statement 1. Inscribed 6 ABC and diameter BD m6 ABE = x◦ and m6 CBE = y◦ 2. x◦ + y◦ = m6 ABC 3. AE ∼ = EB and EB ∼ = EC 4. 4AEB and 4EBC are isosceles 5. m6 EAB = x◦ and m6 ECB = y◦ 6. m6 AED = 2x◦ and m6 CED = 2y◦ c = 2x◦ and mDC c = 2y◦ 7. mAD c + mDC c = mAC c 8. mAD c = 2x◦ + 2y◦ 9. mAC c = 2(x◦ + y◦ ) 10. mAC c = 2m6 ABC 11. mAC 122

Reason Given Angle Addition Postulate All radii are congruent Definition of an isosceles triangle Isosceles Triangle Theorem Exterior Angle Theorem The measure of an arc is the same as its central angle. Arc Addition Postulate Substitution Distributive PoE Subsitution

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Chapter 9. Circles, Answer Key

TABLE 9.2: (continued) Statement c 12. m6 ABC = 12 mAC

Reason Division PoE

TABLE 9.3: Statement c 1. 6 ACB and 6 ADB intercept AB c 2. m6 ACB = 12 mAB 1 c m 6 ADB = 2 mAB 6 6 3. m ACB = m ADB 4. 6 ACB ∼ = 6 ADB

Reason 1. Given 2. Inscribed Angle Theorem 3. Transitive Property 4. Definition of Congruence

← → 32. Since AC || OD, m6 CAB = m6 DOB by Corresponding Angles Postulate.

c so mDB c This makes D the midpoint of CB. c c and m6 CAB = 1 mCB, c = 1 mCB. Also, m6 DOB = mDB 2 2

123

9.5. Geometry - Second Edition, Angles of Chords, Secants, and Tangents, Review Answers

1.

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9.5 Geometry - Second Edition, Angles of Chords, Secants, and Tangents, Review Answers

a.

b.

c. 2. No, by definition a tangent line cannot pass through a circle, so it can never intersect with any line inside of one. 3.

a.

b. 4. 5. 6. 7. 8. 9. 10. 11. 124

center, equal inside, intercepted on, half outside, half x = 103◦ x = 25◦ x = 100◦ x = 44◦

www.ck12.org 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31.

Chapter 9. Circles, Answer Key

x = 38◦ x = 54.5◦ x = 63◦ , y = 243◦ x = 216◦ x = 42◦ x = 150◦ x = 66◦ x = 113◦ x = 60, y = 40◦ , z = 80◦ x = 60◦ , y = 25◦ x = 35◦ , y = 55◦ x = 75◦ x = 45◦ x = 35◦ , y = 35◦ x = 60◦ x = 47◦ , y = 78◦ x = 84◦ , y = 156◦ x = 10◦ x = 3◦ See the following table:

TABLE 9.4: Statement 1. Intersecting chords AC and BD. 2. Draw BC

3. 4. 5. 6.

m6 m6 m6 m6

c DBC = 12 mDC 1 c ACB = 2 mAB a = m6 DBC + m6 ACB c + 1 mAB c a = 1 mDC 2

2

Reason Given Construction

Inscribed Angle Theorem Inscribed Angle Theorem Exterior Angle Theorem Substitution

32. See the following table:

TABLE 9.5: Statement − → − → 1. Intersecting secants AB and AC. 2. Draw BE.

Reason Given Construction

c 3. m6 BEC = 12 mBC

Inscribed Angle Theorem 125

9.5. Geometry - Second Edition, Angles of Chords, Secants, and Tangents, Review Answers

TABLE 9.5: (continued) Statement c 4. m6 DBE = 12 mDE 6 6 5. m a + m DBE = m6 BEC 6. m6 a = m6 BEC − m6 DBE c 1 c 7. m6 a = 12 m BC − 2 mDE c − mDE c 8. m6 a = 1 mBC 2

126

Reason Inscribed Angle Theorem Exterior Angle Theorem Subtraction PoE Substitution Distributive Property

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Chapter 9. Circles, Answer Key

9.6 Geometry - Second Edition, Segments of Chords, Secants, and Tangents, Review Answers 1. x = 12 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

x = 1.5 x = 12 x = 7.5 √ x=6 2 x = 10 x = 10 x=8 x=9 x = 22.4 x = 11 x = 20 x = 120 7√ ≈ 17.14 x = 4 66 x = 6√ x= √ 231 x = 4 42 x = 10 The error is in the set up. It should be 10 · 10 = y · (15 + y). The correct answer is y = 5. 10 inches x=7 x=5 x=3 x=3 x=8 x=6 x=2 x=8 x=2 x = 12, y = 3

127

9.7. Geometry - Second Edition, Extension: Writing and Graphing the Equations of Circles, Reviewwww.ck12.org Answers

9.7 Geometry - Second Edition, Extension: Writing and Graphing the Equations of Circles, Review Answers 1. center: (-5, 3), radius = 4 2. 3. 4. 5. 6. 7. 8. 9.

center: (0, -8), radius = 2 √ center: (7, 10), radius = 2√ 5 center: (-2, 0), radius = 2 2 (x − 4)2 + (y + 2)2 = 16 (x + 1)2 + (y − 2)2 = 7 (x − 2)2 + (y − 2)2 = 4 (x + 4)2 + (y + 3)2 = 25 a. yes b. no c. yes

(x − 2)2 + (y − 3)2 = 52 (x − 10)2 + y2 = 29 (x + 3)2 + (y − 8)2 = 200 (x − 6)2 + (y + 6)2 = 325 7 37 a-d. ⊥ bisector of AB is y = − 24 x + 24 , ⊥ bisector of BC is y = x + 8 (e) center of circle (-5, 3) (f) radius 25 2 2 (g) (x + 5) + (y − 3) = 625 15. (x − 2)2 + (y − 2)2 = 25 16. (x + 3)2 + (y − 1)2 = 289 10. 11. 12. 13. 14.

128

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Chapter 9. Circles, Answer Key

9.8 Chapter Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

I A D G C B H E J F

129

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C HAPTER

10Perimeter and Area, Answer Key

Chapter Outline 10.1

G EOMETRY - S ECOND E DITION , T RIANGLES AND PARALLELOGRAMS , R EVIEW A NSWERS

10.2

G EOMETRY - S ECOND E DITION , T RAPEZOIDS , R HOMBI , AND K ITES , R EVIEW A NSWERS

10.3

G EOMETRY - S ECOND E DITION , A REAS OF S IMILAR P OLYGONS , R EVIEW A N SWERS

130

10.4

G EOMETRY - S ECOND E DITION , C IRCUMFERENCE AND A RC L ENGTH , R EVIEW A NSWERS

10.5

G EOMETRY - S ECOND E DITION , A REAS OF C IRCLES AND S ECTORS , R EVIEW A NSWERS

10.6

G EOMETRY - S ECOND E DITION , A REA AND P ERIMETER OF R EGULAR P OLY GONS , R EVIEW A NSWERS

10.7

C HAPTER R EVIEW A NSWERS

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Chapter 10. Perimeter and Area, Answer Key

10.1 Geometry - Second Edition, Triangles and Parallelograms, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35.

A = 144 in2 , P = 48 in A = 144 cm2 , P = 50 cm A = 360 m2 A = 112 u2 , P = 44 u A = 324 f t 2 , P = 72 f t P = 36 f t A = 36 in2 A = 210 cm2 6m Possible answers: 10 × 6, 12 × 4 Possible answers: 9 × 10, 3 × 30 If the areas are congruent, then the figures are congruent. We know this statement is false, #11 would be a counterexample. √ 8 2 cm P ≈ 54.9 √cm A = 96 2 ≈ 135.8 cm2 15 in P ≈ 74.3 in A = 180 in2 315 units2 90 units2 14 units2 407.5 units2 560 units2 30 units2 814 units2 72 units2 72 units2 24 acres 6×4 12 × 24 √ √ 3 h = 3 √3, A = 9 √ h = 5 √3, A = 25 √ 3 2 h = 2x 3, A = x4 3 x = 20 f t, y = 60 f t Perimeter is 16 units, Area is 15 square units

131

10.2. Geometry - Second Edition, Trapezoids, Rhombi, and Kites, Review Answers

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10.2 Geometry - Second Edition, Trapezoids, Rhombi, and Kites, Review Answers 1. If a kite and a rhombus have the same diagonal lengths the areas will be the same. This is because both formulas are dependent upon the diagonals. If they are the same, the areas will be the same too. This does not mean the two shapes are congruent, however. 2. h(b1 ) + 24s 1 h(b1 ) + 2 12 · h · b2 −b 2 hb1 + h(b22−b1 )

2hb1 +hb2 −hb1 2 hb1 +hb2 = h2 (b1 + b2 ) 2

3. 44s 4 · 21 12 d1 · 12 d2 4 8 d1 · d2 1 2 d1 d2 4. 24s + 24s 2 12 · 12 d1 · x + 2 12 · 12 d1 (d2 − x) 1 1 1 2 d1 · x + 2 d1 d2 − 2 d1 x 1 2 d1 d2 5. 160 units2 6. 315 units2 7. 96 units2 2 8. 77 units √ 9. 100 3 units2 10. 84 units2 11. 1000 units2 12. 63 units2 13. 62.5 units2 14. A = 480 units2 P = 104 units√ 15. A = 36 1 + 3 units2 √ P = 12 2 + 2 units 2 16. A = 108 units√ P = 12 3 + 2 units √ √ 17. A = 5 3 5 + 77 units2 P = 52 units √ 18. A = 396 3 units2 P = 116 √ units 19. A = 256 5 units2 P = 96 units 20. A = 12 units2 21. 24 units2 22. Any two numbers with a product of 64 would work. 23. Any two numbers with a product of 108 would work. 24. 90 units2 25. kite, 24 units2

132

www.ck12.org 26. 27. 28. 29. 30. 31.

Chapter 10. Perimeter and Area, Answer Key

2 2 Trapezoid, 47.5 √ units units rhombus, 12 5 units2 8, 14 9, 12 192 units2

a. 200 f t 2 b. 400 f t 2 c. 21 32. a. 300 f t 2 b. 900 f t 2 c. 31

133

10.3. Geometry - Second Edition, Areas of Similar Polygons, Review Answers

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10.3 Geometry - Second Edition, Areas of Similar Polygons, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

134

9 25 1 16 49 4 36 121 1 6 2 9 7 3 5 12 1 4 1 2

5 units2 24 units 100 cm 468.75 cm2 96 units2 198 f t 2 54 in 32 units 4 9 2 3

√ √ Diagonals are 12 and 16. The length of the sides are 12 2 and 16 2. Because the diagonals of these rhombi are congruent, the rhombi are actually squares. √ 25 2 2.34 inches 1 Scale: 192 , length of model 5.44 inches 27.5 by 20 cm, yes because the drawing is 10.8 by 7.87 inches 9 by 6 inches 10 by 14 inches Baby Bella $0.05, Mama Mia $0.046, Big Daddy $0.046, the Mama Mia or Big Daddy are the best deals. 1.5 bottles, so she’ll need to buy 2 bottles.

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Chapter 10. Perimeter and Area, Answer Key

10.4 Geometry - Second Edition, Circumference and Arc Length, Review Answers TABLE 10.1: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

diameter 15 8 6 84 18 25 2 36

radius 7.5 4 3 42 9 12.5 1 18

Circumference 15π 8π 6π 84π 18π 25π 2π 36π

r = 44 π in C = 20 cm 16 The√diameter is the same length as the diagonals of the square. 32 2 16π 9π 80π 15π r = 108 r = 30 r = 72 120◦ 162◦ 15◦ 40π ≈ 125.7 in. a. 26π ≈ 81.7 in b. 775 complete rotations

26. The Little Cheese, 3.59:1; The Big Cheese, 3.49:1; The Cheese Monster, 3.14:1; Michael should buy The Little Cheese 27. 31 gumdrops 28. 18 in 29. 93 in 30. 30 ft

135

10.5. Geometry - Second Edition, Areas of Circles and Sectors, Review Answers

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10.5 Geometry - Second Edition, Areas of Circles and Sectors, Review Answers TABLE 10.2: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31.

136

radius 2 4 5 12 9√ 3 10 17.5

Area 4π 16π 25π 144π 81π 90π 306.25π

7 π 30 π √6

49 π 900 π

π

36

54π 1.0416π 189π √ 2.6π − 4 3 33π 20.25π √ − 40.5 8 3 2 15 120◦ 10◦ 198◦ 123.61 292.25 1033.58 13.73 21.21 54.4 Square ≈ 10, 000 f t 2 ; Circle ≈ 12, 732 f t 2 ; the circle has more area. 18 units 40◦

circumference 4π 8π 10π 24π 18π √ 6 10π 35π 14 60 √ 12 π

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Chapter 10. Perimeter and Area, Answer Key

10.6 Geometry - Second Edition, Area and Perimeter of Regular Polygons, Review Answers 1. radius 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

apothem 6 equilateral 10√cm 5 3 cm 60 cm √ 150 3 √ A = 384 3 P = 96√ A=8 2 P = 6.12 A = 68.26 A = 72 A = 688.19 P = 100 A = 73.47 P = 15.45 A = 68.26 P = 63 6.5 12 a = 11.01 a = 14.49 93.86, 94.15 30π ≈ 94.25 The perimeter of the 40-gon is closer to the circumference because it is closer in shape to the circle. The more sides a polygon has, the closer it is to a circle. 695.29, 703.96 225π ≈ 706.86 The area of the 40-gon is closer to the area of the circle because it is closer in shape to the circle than the 20-gon. Start with 21 asn. n = 6, so all the internal triangles are equilateral triangles with sides s.Therefore the apothem √ √ 3 3 s (s)(6). Reducing this 1 is s from the 30-60-90 ratio. Plugging this in for n and a, we have A = 2

2

√ we end up with A = 3 2 3 s2 .

2

26. a. b. c. d. e.

◦ ◦ sin x2 = 2rs ; cos x2 = ar s = 2r sin 2x ◦ a = r cos x2 1 x◦ x◦ 2r sin r cos = r2 sin 2 2 2 x◦ x◦ 2 nr sin 2 cos 2

x◦ 2

cos

x◦ 2

27. 421.21 cm2 28. 77.25 in2 137

10.6. Geometry - Second Edition, Area and Perimeter of Regular Polygons, Review Answers

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29. 195.23 cm2 30. 153.44 in2 31. polygon with 30 sides: 254.30 in2 ; circle 254.47 in2 ; They are very close, the more sides a regular polygon has the closer to a circle it becomes. 2 x s s 32. First, take s = 2r sin 2 and solve for r to get = 2 sin x◦ . Next, replace r in the formula to get n 2 sin x◦ sin (2) (2) ◦ ns2 cos( x ) We can reduce this to 4 sin x◦2 . (2) 2 33. 16055.49 cm 34. 4478.46 in2

138

x◦ 2

cos

x◦ 2

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Chapter 10. Perimeter and Area, Answer Key

10.7 Chapter Review Answers 1. A = 225 P = 60 2. A = 198 P = 58 3. A = 124.71 P = 48 4. A = 139.36 P = 45 5. A = 3000 P = 232 6. A = 403.06 P = 72 7. 72 8. 154 √ 9. 162 3 10. C = 34π A = 289π 11. C = 30π A = 225π 12. 54 units2 13. 1070.12 14. 1220.39 15. 70.06

139

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C HAPTER

11

Surface Area and Volume, Answer Key

Chapter Outline 11.1

G EOMETRY - S ECOND E DITION , E XPLORING S OLIDS , R EVIEW A NSWERS

11.2

G EOMETRY - S ECOND E DITION , S URFACE A REA OF P RISMS AND C YLINDERS , R EVIEW A NSWERS

11.3

G EOMETRY - S ECOND E DITION , S URFACE A REA OF P YRAMIDS AND C ONES , R EVIEW A NSWERS

11.4

G EOMETRY - S ECOND E DITION , VOLUME OF P RISMS AND C YLINDERS , R E VIEW A NSWERS

11.5

G EOMETRY - S ECOND E DITION , VOLUME OF P YRAMIDS AND C ONES , R EVIEW A NSWERS

11.6

G EOMETRY - S ECOND E DITION , S URFACE A REA AND VOLUME OF S PHERES , R EVIEW A NSWERS

11.7

G EOMETRY - S ECOND E DITION , E XPLORING S IMILAR S OLIDS , R EVIEW A N SWERS

11.8

140

C HAPTER R EVIEW A NSWERS

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Chapter 11. Surface Area and Volume, Answer Key

11.1 Geometry - Second Edition, Exploring Solids, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

V =8 F =9 E = 30 F =6 E =6 V =6 F =9 V =6 Yes, hexagonal pyramid. F = 7,V = 7, E = 12 No, a cone has a curved face. Yes, hexagonal prism. F = 8,V = 12, E = 18 No a hemisphere has a face. Yes, trapezoidal prism. F = 6,V = 8, E = 12 Yes, concave decagonal prism. F = 10,V = 16, E = 24 Rectangle Circle Trapezoid

18.

19.

20. 21. 22. 23. 24.

Regular Icosahedron Decagonal Pyramid Trapezoidal Prism All 11 nets 141

11.1. Geometry - Second Edition, Exploring Solids, Review Answers

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25. The truncated icosahedron has 60 vertices, by Euler’s Theorem. F +V = E + 2 32 +V = 90 + 2 V = 60 26. regular tetrahedron 27. Use the construction directions from problem 26 to make an equilateral triangle with midsegments. Using one of the midpoints of the equilateral triangle as a vertex, construct another adjacent equilateral triangle with midsegments. Your result should look like the picture below.

28. regular dodecahedron, 31 29. 19 30. 1 red face, 8 yellow faces, 7 blue faces and 4 green faces

142

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Chapter 11. Surface Area and Volume, Answer Key

11.2 Geometry - Second Edition, Surface Area of Prisms and Cylinders, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9.

9 f t.2 10, 000 cm2 triangles, A = 6 The rectangles are 3 × 6, 4 × 6, and 6 × 5. Their areas are 18, 24, and 30. 72 84 Lateral surface area is the area of all the sides, total surface area includes the bases. rectangle, 2πrh a. 96 in2 b. 192 in2

10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

350π cm2 1606.4 390.2 486π 182 34π 2808 x=8 x = 40 x = 25 60π in2 4100π cm2 The height could be 1, 3, 5, or 15. 4060 f t 2 2940 f t 2 5320 f t 2 22 gal $341 5 in by 4π + 1 in, 20π + 5 in2 ≈ 67.83 in2 x2 − 16 in2 , x = 25 in 5 2 2 x π, x = 8

143

11.3. Geometry - Second Edition, Surface Area of Pyramids and Cones, Review Answers

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11.3 Geometry - Second Edition, Surface Area of Pyramids and Cones, Review Answers 1. 2. 3. 4. 5. 6.

vertex y lateral edge w z t

7. √ 8. 5 10 cm 9. 15 in 10. To find the slant√ height, we need to find the distance from the center of the edge of the equilateral triangle. This distance is 3.

11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 144

This is a picture of the base. The slant height is 62 + 671 135 64 1413.72 360 422.35 1847.26 896 1507.96 3, the √ lateral faces 36√ 3 s2 3 576; 321.53 1159.25 1152.23 1473.76 100.8◦ 7

√ 2 √ 3 = l 2 → l = 39

www.ck12.org 29. 30. 31. 32. 33.

Chapter 11. Surface Area and Volume, Answer Key

24 175π 10 in 13 in 360 in2

145

11.4. Geometry - Second Edition, Volume of Prisms and Cylinders, Review Answers

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11.4 Geometry - Second Edition, Volume of Prisms and Cylinders, Review Answers 1. No, the volumes do not have to be the same. One cylinder could have a height of 8 and a radius of 4, while another could have a height of 22 and a radius of 2. Both have a surface area of 96π, but the volumes are not the same. 2. 960 cubes, yes this is the same as the volume. 3. 280 in3 4. 4π in3 5. 6 in 6. r = 9 7. 5 8. 36 units3 9. a. 64 in3 b. 128 in3 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

146

882π cm3 3960 902.54 4580.44 147 50.27 7776 x=7 x = 24 x = 32 294π in3 24000π cm3 75π m3 330, 000 f t.3 165, 000 f t.3 495, 000 f t.3 36891.56 cm3 15901.92 cm3 r = 3 cm, h = 12 cm 11 cm 300.44 in3

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Chapter 11. Surface Area and Volume, Answer Key

11.5 Geometry - Second Edition, Volume of Pyramids and Cones, Review Answers Unless otherwise specified, all units are units3 . 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

9680 1280 778.71 3392.92 400 396.55 5277.88 128 1884.96 100.53 113.10 188.50 42 200 1066.67 √ 9 √3 2 √ 6 18 √ 2 1 3 s 2 12 Find √ the volume of one square pyramid then multiply it by 2. 72 √3 1 3 2 3s h = 13.5 in h = 3.6 cm r = 3 cm 112 in3 190.87 cm3 471.24 cm3 h = 9 m, r = 6 m 15 ft

147

11.6. Geometry - Second Edition, Surface Area and Volume of Spheres, Review Answers

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11.6 Geometry - Second Edition, Surface Area and Volume of Spheres, Review Answers 1. No, all the cross sections must be circles because there are no edges. 2. SA = 256π in2 3 V = 2048 3 π in 3. SA = 324π cm2 V = 972π cm3 4. SA = 1600π f t 2 3 V = 32000 3 π ft 2 5. SA = 16π m 2 V = 32 3πm 6. SA = 900π f t 2 V = 4500π f t 3 7. SA = 1024π in2 3 V = 16384 3 π in 8. SA = 676π cm2 3 V = 8788 3 π cm 9. SA = 2500π yd 2 3 V = 62500 3 π yd 10. r = 5.5 in 11. r = 33 m 12. V = 43 π f t 3 13. SA = 36π mi2 14. r = 4.31 cm 15. r = 7.5 f t. 16. 2025π cm2 17. 1900π units2 18. 4680 f t 2 19. 91.875π units2 20. 381703.51 cm3 21. 7120.94 units3 22. 191134.50 f t 3 23. 121.86 units3 350π 2 24. h = 20 3 cm, SA = 3 cm 25. 21.21 in3 26. 12π cm3 , 19 minutes 27. a. b. c. d.

28. 148

SA = 2πr2 + 2πrh SA = 4πr2 SA = 4πr2 They are the same. Think back to the explanation for the formula for the surface area of a sphere using the baseball-it is really the sum of the area of four circles. For the cylinder, the SA is the sum of the areas of the two circular bases and the lateral area. The lateral area is 2πrh, when we replace h with r this part of the formula becomes the area of two more circles. That makes the total surface area of the cylinder equal to the area of four circles, just like the sphere.

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Chapter 11. Surface Area and Volume, Answer Key

a. 24429 in3 b. 732.87 lbs c. 50 in 29. 25,132.74 miles 30. 201 million square miles 31. 268 billion cubic miles

149

11.7. Geometry - Second Edition, Exploring Similar Solids, Review Answers

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11.7 Geometry - Second Edition, Exploring Similar Solids, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32.

150

42 No, 14 10 6= 35 Yes, the scale factor is 4:3. Yes, the scale factor is 3:5. No, the top base is not in the same proportion as the rest of the given lengths. Yes, cubes have the same length for each side. So, comparing two cubes, the scale factor is just the ratio of the sides. 1:16 8:343 125:729 8:11 5:12 87.48π The volume would be 43 or 64 times larger. 4:9 60 cm 91125 m3 2:3 4:9 y = 8, x√= h = 12√ w = 4 5, z = 6 5 Vs = 170.67,V √ l = 576 √ LAs = 16 5, LAl = 36 5 Yes, just like the cubes spheres and hemispheres only have a radius to compare. So, all spheres and hemispheres are similar. 49:144, 343:1728 98π, 288π The ratio of the lateral areas is 49:144, which is the same as the ratio of the total surface area. 9:25, about 2.78 times as strong 27:125 Animal A, Animal B’s weight is about 4.63 times the weight of animal A but his bones are only 2.78 times as strong. 81 sq in small $0.216, large $0.486 8:27 The larger can for $2.50 is a better deal. Using the cost of the canning material and the ratio of the volume of beans, the “equivalent” cost of producing the larger can is $2.62. If we just use the volume of bean ratio (as a consumer would) the cost should be $2.87. Both of these are higher than the $2.50 price.

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Chapter 11. Surface Area and Volume, Answer Key

11.8 Chapter Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

F K G A E D J B L C H I H G A B D J I E F C

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C HAPTER

12

Rigid Transformations, Answer Key

Chapter Outline 12.1

G EOMETRY - S ECOND E DITION , E XPLORING S YMMETRY, R EVIEW A NSWERS

12.2

G EOMETRY - S ECOND E DITION , T RANSLATIONS AND V ECTORS , R EVIEW A N SWERS

12.3

G EOMETRY - S ECOND E DITION , R EFLECTIONS , R EVIEW A NSWERS

12.4

G EOMETRY - S ECOND E DITION , R OTATIONS , R EVIEW A NSWERS

12.5

G EOMETRY - S ECOND E DITION , C OMPOSITION OF T RANSFORMATIONS , R E VIEW A NSWERS

12.6

G EOMETRY - S ECOND E DITION , E XTENSION : T ESSELLATIONS , R EVIEW A N SWERS

12.7

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C HAPTER R EVIEW A NSWERS

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Chapter 12. Rigid Transformations, Answer Key

12.1 Geometry - Second Edition, Exploring Symmetry, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

sometimes always always never sometimes never never always always sometimes a kite that is not a rhombus a circle an isosceles trapezoid n

15.

16.

17. 18. none

20. 21. 22. 23. 24. 25.

19. H is the only one with rotational symmetry, 180◦ . line symmetry rotational symmetry line symmetry line symmetry (horizontal) rotational symmetry 153

12.1. Geometry - Second Edition, Exploring Symmetry, Review Answers 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37.

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2 lines 6 lines 4 lines 180◦ 60◦ , 120◦ , 180◦ , 240◦ , 300◦ 90◦ , 180◦ , 270◦ none 120◦ , 240◦ 40◦ , 80◦ , 120◦ , 160◦ , 200◦ , 240◦ , 280◦ , 320◦ 8 lines of symmetry; angles of rotation: 45◦ , 90◦ , 135◦ , 180◦ , 225◦ , 270◦ , and 315◦ 3 line of symmetry; angles of rotation: 120◦ , 240◦ 1 line of symmetry; no rotational symmetry

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Chapter 12. Rigid Transformations, Answer Key

12.2 Geometry - Second Edition, Translations and Vectors , Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

A vector has direction and size, a ray is part of a line, so it has direction, but no size. A0 (−1, −6) B0 (9, −1) C(0, 6) A00 (4, −15) D(7, 16) A000 (9, −24) All four points are collinear. A0 (−8, −14), B0 (−5, −17),C0 (−7, −5) A0 (5, −3), B0 (8, −6),C0 (6, 6) A0 (−6, −10), B0 (−3, −13),C0 (−5, −1) A0 (−11, 1), B0 (−8, −2),C0 (−10, 10) (x, y) → (x − 6, y + 2) (x, y) → (x + 9, y − 7) (x, y) → (x − 3, y − 5) (x, y) → (x + 8, y + 4) √ √ √ Using the distance formula, AB = A0 B0 = 5, BC = B0C0 = 3 5, and AC = A0C0 = 5 2. (x, y) → (x − 8, y − 4) *

19. GH= h6, 3i *

20. KJ= h−2, 4i *

21. LM= h3, −1i

22.

23. 25. 26. 27. 28. 29. 30. 31. 32.

24. D0 (9, −9), E 0 (12, 7), F 0 (10, 14) Q0 (−9, −6),U 0 (−6, 0), A0 (1, −9), D0 (−2, −15) h−3, 8i h9, −12i h0, −7i (x, y) → (x − 7, y + 2) (x, y) → (x + 11, y + 25) (x, y) → (x + 15, y − 9) 155

12.3. Geometry - Second Edition, Reflections, Review Answers

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12.3 Geometry - Second Edition, Reflections, Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

d p (-3, 2), (-8, 4), (-6, 7), (-4, 7) (-6, 4), (-2, 6), (-8, 8) (2, 2), (8, 3), (6, -3) (2, 6), (-6, 2), (4, -2) (2, -2), (8, -6) (2, -4), (-4, 2), (-2, -6) (2, 3), (4, 8), (7, 6), (7, 4) (4, 6), (6, 2), (8, 8) (2, 4), (-4, 3), (-2, 9) (-4, -14), (4, -10), (-6, -6) (-2, -2), (-6, -8) (-4, 2), (2, -4), (-6, -2) y = −2 y−axis y=x

18-20.

21. It is the same as a translation of 8 units down. 22-24.

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Chapter 12. Rigid Transformations, Answer Key

25. It is the same as a translation of 12 units to the left. 26-28.

29. A rotation of 180◦ .

30. 31. Perpendicular Bisector

32.

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12.4. Geometry - Second Edition, Rotations, Review Answers

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12.4 Geometry - Second Edition, Rotations, Review Answers

1.

2.

4. 5. 6. 7. 8. 9. 10. 11. 158

3. d d, they are the same because the direction of the rotation does not matter. 270◦ 90◦ Not rotating the figure at all; 0◦ (-6, -2) (-6, -4) (2, -2) and (6, 4)

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Chapter 12. Rigid Transformations, Answer Key

12.

13.

14.

15.

159

12.4. Geometry - Second Edition, Rotations, Review Answers

16.

17. 18. 19. 20. 21. 22. 23.

x=3 x = 4.5 x = 21 90◦ 180◦ 180◦

24-26.

27. A rotation of 180◦ . 28-30. 160

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Chapter 12. Rigid Transformations, Answer Key

31. Angle of rotation is double the angle between the lines.

161

12.5. Geometry - Second Edition, Composition of Transformations, Review Answers

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12.5 Geometry - Second Edition, Composition of Transformations, Review Answers 1. Every isometry produces a congruent figure to the original. If you compose transformations, each image will still be congruent to the original. 2. a translation 3. a rotation 4. (2, 2), (-2, -4), (0, -8), (4, -6) 5. (x, y) → (x + 6, −y) 6. (x, y) → (x − 6, −y) 7. No, because order does not matter. 8. (-2, -3), (-4, 2), (-9, -3) 9. (x, y) → (−x, y − 5) 10. (x, y) → (−x, y + 5) 11. (2, -10), (10, -6), (8, -4) 12. A translation of 12 units down. 13. (x, y) → (x, y + 12) 14. This image is 12 units above the original. 15. #11 → (x, y) → (x, y − 12), #14 → (x, y) → (x, y + 12), the 12’s are in the opposite direction. 16. (-8, 2), (-6, 10), (-2, 8), (-3, 4) 17. A rotation of 270◦ 18. A rotation of 90◦ 19. It is in the 4th quadrant and are 180◦ apart. 20. #16 → (x, y) → (y, −x), #19 → (x, y) → (−y, x), the values have the opposite sign. 21. 14 units 22. 14 units 23. rotation, 180◦ 24. the origin 25. 166◦ 26. 122◦ 27. 315◦ 28. 31 units 29. 2(b − a), right 30. 2(b − a), left

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Chapter 12. Rigid Transformations, Answer Key

12.6 Geometry - Second Edition, Extension: Tessellations, Review Answers 1-7. Yes, all quadrilaterals will tessellate. 8. Equilateral triangle, square, and regular hexagon. 9. Here is one possibility.

10. The figure is an equilateral concave hexagon.

11.

12. 13. Answers will vary.

163

12.7. Chapter Review Answers

12.7 Chapter Review Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

164

C E A F J B H D I G

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