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Answers to Exercises CHAPTER 0 • CHAPTER 0 CHAPTER 0 • CHAPTER LESSON 0.1 1. possible answers: snowflakes and crystals; flowers and starfish 2. Answers might include shapes and patterns, perspective, proportions, or optical illusions. 3. Answers will vary. Bilateral symmetry. 4. A, B, C, and F 5. A, B, D, and E 6. 4 of diamonds; none 7. The line of reflection is a line along the surface of the lake. 8. One line of symmetry is a vertical line through the middle of the Taj Mahal, and the other is a horizontal line between the Taj Mahal and the reflecting pool. The pool reflects the building, giving the scene more reflectional symmetry than the building itself has. 9. Designs will vary, but all should have 2-fold rotational symmetry. 2 lines of reflectional symmetry It is impossible to have 2 lines of reflectional symmetry without rotational symmetry. 10. Answers will vary. 11. Answers will vary. Answers to Exercises ANSWERS TO EXERCISES 1 LESSON 0.2 1. compass and straightedge 2. Answer will be the Astrid or 8-Pointed Star, possibly with variations. 3. Answers will be a design from among the three choices. 4. The first design has 4-fold symmetry and four lines of reflectional symmetry. The triangle has 3-fold rotational symmetry and three lines of reflectional symmetry. The third design has 6-fold rotational symmetry if you ignore color and 2-fold rotational symmetry if you don’t. 5. possible answer: Answers to Exercises 2 lines of reflectional symmetry 2 ANSWERS TO EXERCISES 6. possible answer: 7. 3 lines of reflectional symmetry 3-fold rotational symmetry: rotated 120⬚, 240⬚, 360⬚ LESSON 0.3 4. possible answer: 1. Design with reflectional symmetry is drawn on this background. 2. Design should have rotational symmetry. Possible answer: 5. Drawing should resemble the hexagons in the given figure, except that each radius, and the side of each hexagon, should measure 1 inch. Answers to Exercises 3. possible answer: ANSWERS TO EXERCISES 3 LESSON 0.4 Answers to Exercises 1. Possible answer: The designs appear to go in and out of the page (they appear 3-D). The squares appear to spiral (although there are no curves in the drawing). The spiral appears to go down as if into a hole. Lines where curves meet look wavy. Bigger squares appear to bulge out. 2. Possible answer: When zebras group together, their stripes make it hard for predators to see individual zebras. 4 ANSWERS TO EXERCISES 3. Designs will vary. 4. Designs will vary. 5. Possible answers: 2-D: rectangle, triangle, trapezoid; 3-D: cylinder, cone, prism. The palace facade has one line of reflectional symmetry (bilateral symmetry). LESSON 0.5 5. 1. possible answers: Scotland, Nigeria 2. possible answer: 3. possible answer: 6. Answers will be a lusona from among the three choices. 7. It means to solve a problem boldly and decisively or in a creative way not considered by others. 8. The square knot has reflectional symmetry across a horizontal line. The less secure granny knot has 2-fold rotational symmetry. 9. The result is a regular pentagon. Answers to Exercises 4. possible answer: Cut the middle ring. ANSWERS TO EXERCISES 5 LESSON 0.6 Answers to Exercises 1. possible answers: Morocco, Iran, Spain, Malaysia 2. Alhambra 3. 2.1 cm 4. in terms of the original square tile: 4-fold in the center (center of orange) or corner (center of white); 2-fold in the midpoint of the edge of the tile (where two orange shapes meet) 6 ANSWERS TO EXERCISES 6. Designs will vary. 7. Designs will vary, but should contain a regular hexagon. 8. Tessellations will vary. CHAPTER 0 REVIEW 1. possible answer: Islamic, Hindu, Celtic 2. Sometimes the large hexagons appear as blocks with a corner removed, and sometimes they appear as corners with small cubes nestled in them. 3. Compass: A geometry tool used to construct circles. Straightedge: A geometry tool used to construct straight lines. 4. possible answer: 8. Wheel A has four lines of reflectional symmetry; Wheel C has five lines of reflectional symmetry. Wheels B and D do not have reflectional symmetry. 9. Wheels B and D have only rotational symmetry. Wheels A and B have 4-fold, Wheel C has 5-fold, and Wheel D has 3-fold rotational symmetry. 10, 11. Drawing should contain concentric circles and symmetry in some of the rings. 12. The mandala should contain all the required elements. 13a. The flag of Puerto Rico is not symmetric because of the star and the colors. 13b. The flag of Kenya does not have rotational symmetry because of the spearheads. 13c. possible answers: Japan 5. Answers to Exercises 6. possible answers: hexagon: honeycomb, snowflake; pentagon: starfish, flower 7. Answers will vary. Should be some form of an interweaving design. Nigeria ANSWERS TO EXERCISES 7 Answers to Exercises 1 CHAPTER 1 • CHAPTER CHAPTER 1 • CHAPTER LESSON 1.1 1. sample answers: point: balls, where lines cross; segment: lines of the parachute, court markings; collinear: points along a line of the court structure; coplanar: each boy’s hand, navel, and kneecap, points along two strings , TP 2. PT , RA , AT , TA , RT , TR 3. any two of the following: AR 4. any two of the following: MA,MS,AS,AM,SA,SM 5. 6. A B 7. K , PN 22. PM 24. A 25. 26. Y M N 27. C D B A 28. 10 29–31. y 5 3 L M 3 C x or CA or QP 8. AC 9. PQ or RT , RI or IR , and TI or IT 10. TR y 11. A 12. B S 11 O –5 –3 E A (4, 0) x D –5 B 32. Yes, P(6, 2), Q(5, 2), R(4, 6); the slope between any two of the points is 14. y R Q R P´ x x Q´ 14.3 cm 14. mCD 6.7 cm 13. mAB 15–17. Check each length to see if it is correct. Refer to text for measurements. . X is the midpoint of 18. R is the midpoint of PQ . No midpoints are WY . Y is the midpoint of XZ shown in ABC. 19. possible answers: or A B C D P R´ 33. possible answer: y Q N x P E CE AC BC CD 20. 34. P R A T G Y 35. Answers will vary. Possible answers: S M T 8 cm 8 cm 11 cm 11 cm Q 8 B X y D –3 Answers to Exercises , AC 21. AB , XZ 23. XY ANSWERS TO EXERCISES USING YOUR ALGEBRA SKILLS 1 1. (3, 4) 2. (9, 1.5) 3. (5.5, 5.5) 4. (6, 44) 5. Yes. The coordinates of the midpoint of a segment with endpoints (a, b) and (c, d ) are found ac by taking the average of the x-coordinates, 2 , bd and the average of the y-coordinates, 2. Thus the ac bd midpoint is 2 , 2 . 6. (3, 2) and (6, 4). To get the first point of trisection, sum the coordinates of points A and B to get (9, 6), then multiply those coordinates by 31 to get (3, 2). To get the second point of trisection, sum the coordinates of points A and B to get (9, 6), then multiply those coordinates by 32 to get (6, 4). This works because the coordinates of the first point are (0, 0). 7. Find the midpoint, then find the midpoint of each half. 8a. Midpoints for Figure 1 are (5.5, 6.5); for Figure 2, (16, 6.75); and for Figure 3, (29.75, 5.5). 8b. For these figures the midpoints of the two diagonals are the same point. Answers to Exercises ANSWERS TO EXERCISES 9 LESSON 1.2 1. TEN, NET, E; FOU, UOF, 1; ROU, UOR, 2 N 2. A 24. 135 D 67.5 22 22 A 67.5 D 26. no B G E 25. T 3. C C Y D I 4. N S A M 27. L Answers to Exercises 5. S, P, R, Q; none in the second figure 6. possible answer: B 28. C A D 7. 90° 8. 120° 9. 45° 10. 135° 11. 45° 12. 135° 13. 30° 14. 90° 15. Yes; mXQA mXQY 45° 90° 135°, which equals mAQY. 16. 69° 17. 110° 18. 40° 19. 125° 20. 55° 21. SML has the greater measure because mSML 30° and mBIG 20°. 22. A 30. One possibility is 4:00. H 31. 90 8 6 O T 32. T H R 33. S A T G 44 A 23. I 90 10 29. B ANSWERS TO EXERCISES N 34. W O B T 45. 180 km. The towns can be represented by three collinear points, P, S, and G. Because S is between P and G, PS SG PG. 12 cm 46. 6 cm 47. I 35. 37. 39. 41. 43. MY; CK; mI x 54° z 32° 242° no 36. 38. 40. 42. SEU; EUO; MO y 102° 288° They add to 360°. 6 cm 4.36 cm 2.18 cm 2.18 cm 48. MS DG means that the distance between M and S equals the distance between D and G. MS DG means that segment MS is congruent to segment DG. The first statement equates two numbers. The second statement concerns the congruence between two geometric figures. However, they convey the same information and are marked the same way on a diagram. 49. Answers will vary. Possible answer. C 70 4 cm Answers to Exercises 40 A B F 44. You will miss the target because the incoming angle is too big. 40 D 70 6 cm E Target Mirror Laser light source ANSWERS TO EXERCISES 11 LESSON 1.3 1–3. possible answers: D 1. 45 O 2. G T E R 3. 11. The measures of complementary angles sum to 90°, whereas the measures of supplementary angles sum to 180°. 12. No, supplementary angles can be unconnected, while a linear pair must share a vertex and a common side. 13a. an angle; measures less than 90° 13b. angles; have measures that add to 90° 13c. a point; divides a segment into two congruent segments 13d. a geometry tool; is used to measure the sizes of angles in degrees 14. E B P C B I G 4. Answers to Exercises A G and CD intersect at point P so that P is If AB between A and B, and P is between C and D, then APC and BPD are a pair of vertical angles. 15. true D S M 5. A 16. true P 6. E R D A 17. true B E C 7. 40 A 18. false 50 B 8. 19. false 140 40 C D 9. B is a Zoid. A Zoid is a creature that has in its interior a small triangle with a large black dot taking up most of its center. 10. A good definition places an object in a class and also differentiates it from other objects in that class. A good definition has no counterexamples. 12 D ANSWERS TO EXERCISES 20. false 28. One possible answer is A(8, 8), B(4.5, 6.5), C(11, 1). 50 y A B 50 80 6 S C 21. true –12 x –6 T 22. false; Though the converse is true, a counterexample is C T R 29. 12 cm 30. 36° 31. possible answer: A 23. false Answers to Exercises 32. 140 0 pt. C 1 pt. D 40 24. false 2 pt. A 3 pt. 33. C 1 pt. 2 pt. T 25. possible answers: (3, 0), (0, 2), or (6, 6) 26. possible answers: (2, 3) or (5, 1) 27. The reflected ray and the ray that passes through (called the refracted ray) are mirror images of each other. Or they form congruent angles with the mirror. A Reflected segment B Mirror 4 pt. 3 pt. 5 pt. 6 pt. Infinitely many points 120° 1 2 , 34a. 360° 3 3 left 60° 1 1 , 34b. 360° 6 6 missing 360° 34c. 9 40° ANSWERS TO EXERCISES 13 LESSON 1.4 1. FI and IVE ANC 18. PA 19. possible answer: 130 2. possible answers: 130 130 5 cm 20. possible answer: Answers to Exercises 3. 4. octagon 5. hexagon 6. heptagon 7. pentagon 8–10. One possible answer for each is shown. 8. pentagon FIVER 9. quadrilateral FOUR 10. equilateral quadrilateral BLOC 11a. a polygon; has eight sides 11b. a polygon; has at least one diagonal outside of the polygon 11c. a polygon; has 20 sides 11d. a polygon; has all sides of equal length 12. One possibility is C and Y are consecutive and YN are consecutive sides. angles; CY 13. 9; possible answer: , AD , BD , BE , CE 14. AC 15. TIN 16. WEN 17a. a 44, b 58, c 34 17b. mT 87° and mI 165° 14 ANSWERS TO EXERCISES 21. 84 in. 22. 5.25 cm 23. AB 14 m, CD 25 m 24. complementary angles: AOS and SOC; vertical angles: OCT and ECR or TCE and RCO 25. E A B C F D 26. possible answer: 5 20 27. All are possible except two points. 0 pts. 1 pt. 3 pts. 4 pts. 5 pts. 6 pts. 16. possible answers: LESSON 1.5 1. D 3. C 5. 2. A 4. B 6. C C C 14 cm 10 cm 40 A 7. T S M L A B A F 10 cm R 40 C A 8. possible answer: 2a 2a b – 2a b 6 cm B 40 6 cm A B 40 y C C (–3, 5) (2, 4) 10. possible answer: C (–4, 1) (1, 1) x 4 cm 22. (4,1) → (3,2) (1,1) → (2, 2) (2, 4) → (3, 1) (3, 5) → (2, 2) Yes, the quadrilaterals are congruent. 23. Find the midpoint of each rod. All the midpoints lie on the same line; place the edge of a ruler under this line. 24–26. Sample answers for 24–26 P P 24. 25. 80 A 4 cm B 11. possible answer: A 4 cm 120 P 4 cm Z 12. possible answers: (1, 1), (1, 0), or (4, 3) 13. possible answers: (3, 3), (3, 4), (11, 4), (11, 3), (0.5, 0.5), or (7.5, 0.5) 14. possible answers: (3, 1), (1, 9), (2, 2), (4, 8), (0, 1), or (1, 6) 15. possible answers: C E A N 26. E T A N Q U D A T C 7 cm 12 cm 45 A 9 cm B 45 A 9 cm B ANSWERS TO EXERCISES 15 Answers to Exercises 9. possible answer: A E 17. true 18. true 19. False, a diagonal connects nonconsecutive vertices. 20. False, an angle bisector divides an angle into two congruent angles. 21. true 3a 3a 10 cm LESSON 1.6 1. first figure: quadrilateral ABCD with two congruent sides and one right angle; second figure kite EFGH; third figure: trapezoid IJKL with two right angles; fourth figure parallelogram MNPQ 2. B 3. D 4. F 5. C 6. A, D, F I 7. D Z 8. 19b. Possible answer: Rotate one triangle so that a congruent pair of sides forms a diagonal of a parallelogram. 20. There are three possibilities: a rhombus, a concave kite, or a parallelogram. O N E B F Answers to Exercises 9. L A E 21a. right triangles 21b. isosceles right triangles 22. U Q 10. T H 23. I R 5 cm 5 cm G 11. 24. 12. square 13. possible answers: (2, 6), (2, 4); (6, 2), (2, 4); (3, 1), (1, 3) 14. 90 cm 15. (5, 6) 16. S(9, 0), I(4, 2) 17. S(3, 0), I(1, 5) 18. A(7, 6), N(5, 9) or A(5, 2), N(7, 1) 19a. Possible answer: Flip one triangle and align it so that a congruent pair of sides forms a diagonal of a kite. 16 ANSWERS TO EXERCISES 72 72 72 72 72 25. no y C (0, 5) B (4, 4) x A (5, 0) LESSON 1.7 1. Answers will vary. Sample answers:The green area in the irrigation photo is a circle,the water is a radius,and a path on the far side of the circle appears to be tangent to the circle.The wood bridge is an arc of a circle,and the railings are arcs of concentric circles. The horizontal support beam under the bridge is a chord. , BD , EC , EF 2. three of the following: AB 3. EC , EP , FP , BP , CP 4. AP , AE , AB , BC , CD , DF , 5. five of the following: EF EB , ED , FC , AC , DB , AF , AD , BF or EFC , EBC or EAC 6. EDC , FEC , DEC , . . . 7. two of the following: ECD , EDF B P Q A 17. equilateral; 3 to 1 s 18. 19. yes; yes Answers to Exercises , HB 8. FG 9. either F or B 10. possible answers: cars, trains, motorcycles; washing machines, dishwashers, vacuum cleaners; compact disc players, record players, car racing, Ferris wheel 110°; mPRQ 250° 11. mPQ 12. 16. Equilateral quadrilateral. (The figure is actually a rhombus, but students have not yet learned the properties needed to conclude that the sides are parallel.) y 5 5 –5 x 65 –5 215 20. yes; no 13. possible answers: concentric rings on cross sections of trees (annual rings),bull’s-eye or target, ripples from a rock falling into a pond. 14. y 5 –5 5 x –5 21. no; no 15. y 8 P Q 8 16 x ANSWERS TO EXERCISES 17 22. 80°. The slices of pizza do not overlap, so 60° x 140°, where x is the measure in degrees of the angle of the second slice. 23. 24. not possible 25. C 60 60 32. not possible 33. R G T 34. A B 27. about 0.986° 28. 15° 29. 4a + 2b 35. T 100 50 36. Answers to Exercises 120 B C 70 I P E A 37. 50 70 2p 2p F 30. 120 60 120 A 26. A 31. U 2p 55 2p Q E 120 T R K T I 18 ANSWERS TO EXERCISES 120 60 22. False. The two lines are not necessarily in the same plane, so they might be skew. LESSON 1.8 1. 2. 3. 4. 23. true 5. 6. 7. 8. 24. true 25. true 2 9. 60 boxes Answers to Exercises 3 4 10. 26. False. They divide space into seven or eight parts. 5 3 4 11. 12. 27. true 13. B, D 15. C 17. A 18. 19. 20. true 21. true 28. (3, 1) 14. B 16. D 29. perimeter 20.5 cm; m(largest angle) 100° 30. 8 cm 120 13 cm ANSWERS TO EXERCISES 19 18. pyramid with hexagonal base LESSON 1.9 1. Sample answer: Furniture movers might visualize how to rotate a couch to get it up a narrow staircase. 2. yes 3. W O M E N ; Nadine is ahead. 4. 28 posts 5. 28 days 6. 0 ft (The poles must be touching!) B 7. 20. pyramid with square base A B A 21. x 15, y 27 22. x 12, y 4 23. 8. Answers to Exercises 19. prism with hexagonal base A 9. A 3 cm B 3 cm 24. 3 cm A B 10. Polygons Quadrilaterals Trapezoids 11. 12. 13. 14. 15. 16. Triangles Obtuse Isosceles no C(2, 3), A(0, 0), B(0, 5), D(2, 1) Y(4, 1), C(3, 1), N(0, 3) (x, y) → (x 3, y 2) CP , EF GH ; i k, j k AB perimeter 34 cm 25. 26. 28. 30. 32. 34. point, line, plane AB vertex CD AB ABC congruent to 20 ANSWERS TO EXERCISES AB AB protractor CD AB 35. The distance is two times the radius. r r P Q PQ = 2r 36. They bisect each other and are perpendicular. A 17. Left photo: Three points determine a plane. Middle photo: Two intersecting lines determine a plane. Right photo: A line and a point not on the line determine a plane. 27. 29. 31. 33. P Q B CHAPTER 1 REVIEW 1. 2. 3. 4. 5. 6. 7. 8. true . False; it is written as QP true False; the vertex is point D. true true False; its measure is less than 90°. false; two possible counterexamples: T Y E A P 30. 31. A D A P R C D A 29. P P B C T APD and APC are a linear pair. APD and APC are the same angle. E Y A C B Answers to Exercises 9. true 10. true 11. False; they are supplementary. 12. true 13. true 14. true 15. False; it has five diagonals. 16. true 17. F 18. G 19. L 20. J 21. C 22. I 23. no match 24. A 25. no match T 26. 32. D 33. 34. 2 in. 5 in. 3 in. 35. 125 36. K 40 27. S 5 3 T 7 28. O P 37. A N R I E G ANSWERS TO EXERCISES 21 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 114° x 2, y 1 x 12, y 4 x 4, y 2.5 x 10, y 8 AB 16 cm 96° 105° 30° 51. He will get home at 5:46 (assuming he goes inside before he gets blown back again). 52. (2, 3) 53. Quadrilateral 54. Rectangle Square Rhombus 55. Answers to Exercises Trapezoid 48. 66 inches 49. 3 feet 50. The path taken by the midpoint of the ladder is an arc of a circle or a quarter-circle if the ladder slides all the way from the vertical to the horizontal. 57. Shed Path of flashlight 22 56. ANSWERS TO EXERCISES Answers to Exercises CHAPTER 2 • CHAPTER 2 CHAPTER 2 • CHAPTER LESSON 2.1 13. 14. 28. 29. possible answers: 30. sample answer: 31. 33. 35. 37. 39. 41. collinear dodecagon protractor diagonal 90° sample answer: 32. 34. 36. 38. 40. 42. Answers to Exercises 1. “All rocks sink.” Stony needs to find one rock that will not sink. 2. If two angles are formed by drawing a ray from a line, then their measures add up to 180°. 3. 10,000, 100,000, . . . . Each term is 10 times the previous term. 4. 65, 1, . . . . (Written with the common denominator 6, the pattern becomes 1 2 3 4 5 6 , , , , , , . . . .) 6 6 6 6 6 6 5. 17, 21 6. 28, 36 7. 21, 34 8. 49, 64 9. 10, 24 10. 64, 128 11. 12. 27. isosceles parallel radius regular perpendicular sample answer: I 15. 16. A 17. 1, 4, 7, 10, 13 18. 15, 21, 28, 36, 45 19. Answers will vary. 20. sample answers: 3, 6, 12, 24, 48, . . . and 4, 8, 12, 16, 20, . . . 21. Answers will vary. 22. 7th term: 56; 10th term: 110; 25th term: 650 23. Conjecture is false; 142 196 but 412 1681. 24. 11,111 11,111 123,454,321. For the sixth line, if 111,111 is multiplied by itself, then the product will be 12,345,654,321. But 1,111,111,111 1,111,111,111 1,234,567,900,987,654,321. 25. 26. G N T 43. sample answer: 44. possible answer: A C Clearly AC does not bisect A or C. 45. Methods will vary. It is not possible to draw a second triangle with the same angle measures and side length that is not congruent to the first. C 40 A 60 9 cm B ANSWERS TO EXERCISES 23 4. 5. 6. 7. LESSON 2.2 1. See table below. 2. See table below. 3. See table below. See table below. See table below. See table below. See table below. 1. (Lesson 2.2) n 1 2 3 4 5 6 … n … 20 f (n) 3 9 15 21 27 33 … 6n 3 … 117 … 20 … 56 … 20 … 148 Answers to Exercises 2. (Lesson 2.2) n 1 2 3 4 5 6 … f (n) 1 2 5 8 11 14 … n 3n 4 3. (Lesson 2.2) n f (n) 1 2 3 4 5 6 … 4 4 12 20 28 36 … n 8n 12 4. (Lesson 2.2) Number of sides 3 4 5 6 … n … 35 Number of triangles formed 1 2 3 4 … n2 … 33 5. (Lesson 2.2) Figure number 1 2 3 4 5 6 … n … 200 Number of tiles 8 16 24 32 40 48 … 8n … 1600 Figure number 1 2 3 4 5 6 … n … 200 Number of tiles 1 5 9 13 17 21 … 4n 3 … 797 6. (Lesson 2.2) 7. (Lesson 2.2) 24 Figure number 1 2 3 4 5 6 … Number of matchsticks 5 9 13 17 21 25 … Number of matchsticks in perimeter of figure 5 8 11 14 17 20 ANSWERS TO EXERCISES n … 200 4n 1 … 801 … 3n 2 … 602 8. 8n is steeper; the coefficient of n. 13. R E f (n) T 8n C 50 40 30 4n 1 4n 3 20 3n 2 14. 2 1 2 1 3 3 10 1 2 4 6 1 n 8 H H H H H H H H C C C C C C C C H H H H H H H H Octane (C8H18) 3 10. y 2x 3 11. 1 2 1 2 9. CnH2n2 H 1 1 L H 15. Márisol should respond by noticing that all the triangles José drew were isosceles, but it is possible to draw a triangle with no two sides congruent. In other words, she should show him a counterexample. 16. Methods will vary. It is not possible to draw another triangle with the given side lengths and angle measure that is not congruent to the first. C Answers to Exercises 9 cm T 45 A E 12. Q O M L 8 cm B 17. She could try eating one food at a time; inductive. 18. 2600 Y ANSWERS TO EXERCISES 25 n(n 1) 8. Use 2 to get 780 direct lines. Use a central hub with a line to each house to get 40 lines. The art shows the direct-line solution and the practical solution for six houses. LESSON 2.3 1. See table below. 2. n 1, 36 9. Use points for the 10 teams and segments connecting them to represent four games played n(n 1) between them. So, use 4 2 to get 180 games played. n(n 1) 10. If 2 66, then n(n 1) 132. What two consecutive numbers multiply to equal 132? 12 times 11. Thus, there were 12 people at the party. 11. true 12. true 13. False; an isosceles right triangle has two sides congruent. 14. false 3. You can draw a diagonal from one vertex to all the other vertices except three: the two adjacent vertices and the vertex itself. Answers to Exercises n(n 3) 4. 2, 560 n(n 1) 5. 2, 595 E B C A AED and BED are not a linear pair. n(n 1) 6. 2, 595 15. False; they are parallel. 16. true 17. False; a rectangle is a parallelogram with all of its angles congruent. 18. False; a diagonal is a segment in a polygon connecting any two nonconsecutive vertices. 19. true 20. 5049 7. Answers will include relationships between points in Exercises 5 and 6 and vertices of a polygon. The total number of segments connecting n random points is the number of diagonals of the n-sided n(n 3) polygon, 2, plus the number of sides, n. Thus, the total number of segments connecting n n(n 3) 2n n(n 3) 2n random points is 2 2 2 n2 3n 2n n2 n 2 2 D n(n 1) 2 . 1. (Lesson 2.3) 26 Lines 1 2 3 4 5 … n … 35 Regions 2 4 6 8 10 … 2n … 70 ANSWERS TO EXERCISES LESSON 2.4 1. inductive; deductive 2. mB 65°; deductive 3. inductive 25 18. 19. 36 4. DG 258 cm; deductive 5. 180°; 180°; The sum of the five angles is 180°; inductive. 20. 20 32 64 27 37 I E L 33. D B 140 O G 34. C T O D 35. K D N ANSWERS TO EXERCISES 27 Answers to Exercises 6. LNDA is a parallelogram; deductive. 7. Possible answer: AB CD, so AB BC CD BC. Because AB BC AC and CD BD . BC BD, AC BD, so AC 8. 3; 10; 7 9. Just over 45°; if m 90°, then 21m 45°. 10. 48°; 17°; 62°; mCPB 11. Possible answers: Because mAPC mBPD, add the same measure to both sides to get mAPC mCPD mBPD mCPD.By angle addition, mAPC mCPD mAPD and mBPD mCPD mCPB. Therefore, mAPD mCPB. 12. Answers will vary. 13. The pattern cannot be generalized because once the river is straight, it cannot get any shorter. 14. 900, 1080 15. 75, 91 4 16. 5, 12 17. 21. Sample answer: There were three sunny days in a row, so I assumed it would be sunny the fourth day, but it rained. 22. L 23. M 24. A 25. B 26. E 27. C 28. G 29. D 30. H 31. I W 32. 15. Answers to Exercises LESSON 2.5 1. a 60°, b 120°, c 120° 2. a 90°, b 90°, c 50° 3. a 77°, b 52°, c 77°, d 51° 4. a 60°, b c 120°, d f 115°, e 65°, g i 125°, h 55° 5. a 90°, b 163°, c 17°, d 110°, e 70° 6. The measures of the linear pair of angles add up to 170°, not 180°. 7. The angles at which he should cut measure 45°. 8. Greatest: 120°. Smallest: 60°. One possible explanation: The tree is perpendicular to the horizontal. The angle of the hill measures 30°. The smaller angle and the angle between the hill and the horizontal form a pair of complementary angles, so the smaller angle equals 90° 30° 60°. The smaller angle and larger angle form a linear pair, so the larger angle equals 180° 60° 120°. 9. The converse is not true. ; sample counterexample: A A O T 16. 17. 6 3 4 18. Possible answer: All the cards look exactly as they did, so it must be the 4 of diamonds, because it has rotational symmetry while the others do not. 19. 22.5° 20. 55 125 B 10. each must be a right angle 21. CnH2n H H 11. Let the measures of the congruent angles be x. They are supplementary, so x x 180°, 2x 180°, x 90°. Thus each angle is a right angle. 12. The ratio is 1. The ratio does not change as long as the lines don’t coincide. Because the demonstration does not explain why, it is not a proof. 13. P 14. 5 H H H H H C C C C C C C C H H H H H H H H 22. See table below. n(n 1) 80(79) 23. handshake problem: 2 ; 2 3160 pieces of string 24. 3160 intersections n(n 2) 25. 2 yields 760 handshakes. 26. 21; 252 n(n 3) 27. 2 560; there are 35 vertices. ; deductive. 28. M is the midpoint of AY 3 A H T 22. (Lesson 2.5) Rectangle 1 2 3 4 5 6 … n … 200 Perimeter of rectangle 10 14 18 22 26 30 … 4n 6 … 806 Number of squares 6 12 20 30 42 56 … … 40,602 (n 1)(n 2) 28 ANSWERS TO EXERCISES 13. No, tomorrow could be a holiday. Converse:“If tomorrow is a school day, then yesterday was part of the weekend;” false. 14. 42° 15. 20° 16. No, the lines are not parallel. 17. isosceles triangle 18. a parallelogram that is not also a rectangle or a rhombus 19. 18 cm 20. 39° 21. 3486 22. 30 squares (one 4-by-4, four 3-by-3, nine 2-by-2, and sixteen 1-by-1) 23. The triangle moved to the left 1 unit.Yes, congruent to original. LESSON 2.6 1. 63° 2. 90° 3. no 4. 57° 5. yes 6. 113° 7. a d 64°, b c 116°, e g i j k 108°, f h s 72°, m 105°, n 79°, p 90°, q 116°, t 119°; Possible explanation: Using the Vertical Angles Conjecture, n 79°. Using the Linear Pair Conjecture, p 90°. Using the Corresponding Angles Conjecture and b 116°, q 116°. 8. Possible answer: In the k diagram, lines and m are parallel 1 and intersected by transversal k. m Using the Corresponding Angles 2 3 Conjecture, 1 2. Using the Vertical Angles Conjecture, 2 3. Because 1 and 3 are both congruent to 2, they must be congruent to each other. So 1 3. Therefore,if two parallel lines are cut by a transversal, then alternate exterior angles are congruent. 9. 56° 114° 170° 180°. Thus, the lines marked as parallel cannot really be parallel. 10. Alternate interior angles measure 55°, but 55° 45° 180°. 11. The incoming and 45 90 outgoing angles measure 45 45°. Possible explanation: Yes, the alternate interior angles are congruent, and thus, by the Converse of the 45 Parallel Lines Conjecture, 90 the mirrors are parallel. 45 24. The quadrilateral was reflected across both axes or rotated 180° about the origin.Yes, congruent to original. y 4 x 4 25. The pentagon was reflected across the line x y.Yes, congruent to original. y N 5 O' L M' Q P x 5 O N' x E B A 6 E' M L' 26. See table below. 26. (Lesson 2.6) Figure number 1 2 3 4 5 6 … n … 35 Number of yellow squares 2 3 4 5 6 7 … n1 … 36 Number of blue squares 3 5 7 9 11 13 … 2n 1 … 71 Total number of squares 5 8 11 14 17 20 … 3n 2 … 107 ANSWERS TO EXERCISES 29 Answers to Exercises 12. Explanations will vary. Sample explanation:“I used the protractor to make corresponding angles congruent when I drew line PQ.” y 5 USING YOUR ALGEBRA SKILLS 2 Answers to Exercises 1. 2 12 2. 1 3 97 3. 4 6 2.1 4. y 23 5. x 4 6. Any point of the form (3p, 4p).Possible answer: (6,8),(9,12),and (12,16).To find another point go right 3 and down 4. 30 ANSWERS TO EXERCISES 7. 66.7 mi/h 8. At 6 m/s, Skater 1 is 4 m/s faster than Skater 2 at 2 m/s. 9. A 100% grade has a slope of 1. It has an inclination of 45°, so you probably could not drive up it.You might be able to walk up it. Grades higher than 100% are possible, but the angle of inclination would be greater than 45°. 10. The slope of the adobe house flat roof is approximately 0. The Connecticut roof is steeper, with a slope of about 2, and will shed the snow. 19d. possible answers: AFG and FGD or CFG and BGF 20. possible answers: GFC by the Vertical Angles Conjecture, BGF by the Corresponding Angles Conjecture, and DGH, using the Alternate Exterior Angles Conjecture 21. True. Converse:“If two polygons have the same number of sides, then the two polygons are congruent”; false. Counterexamples may vary; students might draw a concave quadrilateral and a convex quadrilateral. 22. CHAPTER 2 REVIEW 1. poor inductive reasoning, but Diana was probably just being funny 2. Answers will vary. 3. Answers will vary. 4. 19, 30 5. S, 36 6. 2, 5, 10, 17, 26, 37 7. 1, 2, 4, 8, 16, 32 8. 9. 56 124 7 cm 56 4.5 cm 4.5 cm 56 124 56 7 cm 900 930 See table below. See table below. n2, 302 900 n(n 1) 100(101) 15. 2, 2 5050 n(n 1) 16. 2 741; therefore, n 39 n(n 1) 17. 2 2926; therefore, n 77 18. n 2, 54 2 52 19a. possible answers: EFC and AFG, AFE and CFG, FGD and BGH, or BGF and HGD 19b. possible answers: AFE and EFC, AFG and CFG, BGF and FGD, or BGH and HGD (there are four other pairs) 19c. possible answers: EFC and FGD, AFE and BGF, CFG and DGH, or AFG and BGH RX and SU VX ; explanations will vary. 23. PV 24. The bisected angle measures 50° because of AIA. So each half measures 25°. The bisector is a transversal, so the measure of the other acute angle in the triangle is also 25° by AIA. However, this angle forms a linearpairwiththeanglemeasuring 165°,and 25° 165° 180°, which contradicts the Linear Pair Conjecture. 25. a 38°, b 38°, c 142°, d 38°, e 50°, f 65°, g 106°, h 74°; The angle with measure e forms a linear pair with an angle with measure 130° because of the CorrespondingAngles Conjecture.So e measures 50° because of the Linear Pair Conjecture. The angle with measure f is half of the angle with measure 130°, so f 65°.The angle with measure g is congruent to the angle with measure 106° by the CorrespondingAngles Conjecture,so g 106°. 12. (Chapter 2 Review) n 1 2 3 4 5 6 … f(n) 2 1 4 7 10 13 … n … 20 … 55 … 20 … 210 3n 5 13. (Chapter 2 Review) n 1 2 3 4 5 6 … f(n) 1 3 6 10 15 21 … n n(n 1) 2 ANSWERS TO EXERCISES 31 Answers to Exercises 10. 11. 12. 13. 14. Answers to Exercises CHAPTER 3 • CHAPTER 3 CHAPTER 3 • CHAPTER LESSON 3.1 1. B A D C F E 2. AB 9. For Exercise 7, trace the triangle. For Exercise 8, trace the segment onto three separate pieces of patty paper. Lay them on top of each other, and slide them around until the segments join at the endpoints and form a triangle. . Copy Q and 10. One method: Draw DU construct COY QUD. Duplicate DUA at point O. Construct OYP UDA. CD U O A AB ⫹ CD 3. AB EF EF Q AB ⫹ 2EF ⫺ CD CD Answers to Exercises D C Y 11. One construction method is to create congruent circles that pass through each other’s center. One side of the triangle is between the centers of the circles; the other sides meet where the circles intersect. 12. a 50°, b 130°, c 50°, d 130°, e 50°, f 50°, g 130°, h 130°, k 155°, l 90°, m 115°, n 65° 13. west 14. An isosceles triangle is a triangle that has at least one line of reflectional symmetry.Yes, all equilateral triangles are isosceles. 15. 4. 5. possible answer: L G P E copy 6. m3 m1 m2; possible answer: ⬔2 16. new coordinates: A(0, 0), Y(4, 0), D(0, 2) y 1 3 2 ⬔1 6 Y 7. possible answer: D' Y' C –6 AC D A' A 6 x BC –6 A B AB 8. 17. Methods will vary. It isn’t possible to draw a second triangle with the same side lengths that is not congruent to the first. 11 cm 8 cm 10 cm 32 ANSWERS TO EXERCISES LESSON 3.2 1. A B 2. Q 3. D Edge of the paper Original segment 4. D 1 CD 2 1 CD 2 AB AB I 1 CD 2 2AB – 1 CD 2 5. A AB M L CD N MN = 1 (AB + CD) 2 8. The medians all intersect in one point. C N 6. Exercises 1–5 with patty paper: Exercise 1 This is the same as Investigation 1. Exercise 2 . Step 1 Draw a segment on patty paper. Label it QD Step 2 Fold your patty paper so that endpoints Q and D coincide. Crease along the fold. Step 3 Unfold and draw a line in the crease. Step 4 Label the point of intersection A. Step 5 Fold your patty paper so that endpoints Q and A coincide. Crease along the fold. Step 6 Unfold and draw a line in the crease. Step 7 Label the point of intersection B. Step 8 Fold your patty paper so that endpoints A and D coincide. Crease along the fold. Step 9 Unfold and draw a line in the crease. Step 10 Label the point of intersection C. A M L B appears to be parallel to EF , and its length 9. GH is half the length of EF . F G D H E ANSWERS TO EXERCISES 33 Answers to Exercises C Exercise 3 This is the same as Investigation 1. Exercise 4 Step 1 Do Investigation 1 to get 12CD. Step 2 On a second piece of patty paper, trace AB two times so that the two segments form a segment of length 2AB. Step 3 Lay the first piece of patty paper on top of the second so that the endpoints coincide and the shorter segment is on top of the longer segment. Step 4 Trace the rest of the longer segment with a different colored writing utensil. That will be the answer. Exercise 5 Step 1 Trace segments AB and CD so that the two segments form a segment of length AB CD. Step 2 Fold your patty paper so that points A and D coincide. Crease along the fold. Step 3 Unfold and draw a line in the crease. 7. The perpendicular bisectors all intersect in one point. 10. The quadrilateral appears to be a rhombus. V E O C R D 14. One way to balance it is along the median. The two halves weigh the same. sample figure: C S I 11. D Ness Station Umsar Station A Answers to Exercises Any point on the perpendicular bisector of the segment connecting the two offices would be equidistant from the two post offices. Therefore, any point on one side of the perpendicular bisector would be closer to the post office on that side. 12. It is a parallelogram. F L Area CDB = 158 in2 Area DAB = 158 in2 B Ruler 15. F 16. E 17. B 18. A 19. D 20. C 21. B, C, D, E, H, I, O, X (K in some fonts, though not this one) 22. Methods will vary. C T 10 cm A 40⬚ 13. The triangles are not necessarily congruent, but their areas are equal. A cardboard triangle would balance on its median. 34 ANSWERS TO EXERCISES A 70⬚ B It is not possible to draw a second triangle with the same angle and side measures that is not congruent to the first. LESSON 3.3 1. 6. The two folds are parallel. P B P Q I G The answer depends on the angle drawn and where P is placed. C 2. D A B 7. Fold the patty paper through the point so that two perpendiculars coincide to see the side closest to the point. Fold again using the perpendicular of the side closest to the point and the third perpendicular; compare those sides. 8. Draw a line. Mark two points on it, and label them A and C. Construct a perpendicular at C. congruent to CA . The altitude CD is Mark off CB also the angle bisector, median, and perpendicular bisector. A D C B 9. U O B T O T Answers to Exercises 3. B U 10. E Two altitudes fall outside the triangle, and one falls inside. 4. From the point, swing arcs on the line to construct a segment whose midpoint is that point. Then construct the perpendicular bisector of the segment. L A B 11. 12. Complement of ⬔A ⬔A 5. Construct perpendiculars from point Q and and RE congruent to QR . point R. Mark off QS Connect points S and E. Q R S E ANSWERS TO EXERCISES 35 13. See table below. 14. 18. not congruent 6 cm 6 cm 40⬚ 40⬚ 8 cm 15. F 8 cm 19. possible answer: Y T I 16. A E 20. possible answer: C D 5 cm F 9 cm B 5 cm Answers to Exercises 17. 9 cm 13. (Lesson 3.3) Rectangular pattern with triangles Rectangle 1 2 3 4 5 6 … Number of shaded triangles 2 9 20 35 54 77 … n … 35 … 2484 (2n 1)(n 1) 36 ANSWERS TO EXERCISES LESSON 3.4 1. 2. 3. 4. 5. 6. D F A C E 11. 12. The angle bisectors are perpendicular. The sum of the measures of the linear pair is 180°. The sum of half of each must be 90°. z z z 7. A B M R P 8. P A P E M S M M E U M O S 9a, b. 18. G 45⬚ 90⬚ 45⬚ N 9c. I A S L 135⬚ O T 45⬚ 10. Altitude Angle bisector Median ANSWERS TO EXERCISES 37 Answers to Exercises A R R 13. If a point is equally distant from the sides of an angle, then it is on the bisector of an angle. This is true for points in the same plane as the angle. Mosaic answers: Square pattern constructions: perpendiculars, equal segments, and midpoints; The triangles are not identical, as the downward ones have longer bases. 14. y 110° 15. mR 46° 16. Angle A is the largest; mA 66°, mB 64°, mC 50°. 17. STOP 19. 22a. A web of lines fills most of the plane, except a U-shaped region and a V-shaped region. (The U-shaped region is actually bounded by a section were of a parabola and straight lines. If AB extended to AB , the U would be a complete parabola.) B 5.6 cm 130⬚ A 3.5 cm C 20. A A B 6.5 cm C 21. No, the triangles don’t look the same. 60⬚ 40⬚ 60⬚ 40⬚ 8 cm Answers to Exercises 8 cm 38 ANSWERS TO EXERCISES D B C Parabola and half the 22b. a line segment parallel to AB length (The segment is actually the midsegment of ABD.) LESSON 3.5 1. 2. 3. Construct a segment with length z. Bisect the segment to get length 2z. Bisect again to get a segment with length 4z. Construct a square with each side of length 4z. 8. 1 S and 2 U by of the Alternate Interior Angles Conjecture 9. The ratios appear to be the same. 10. 1 3 and 2 4 by the Corresponding Angles Conjecture 11. A parallelogram 12. Using the Converse of the Parallel Lines Conjecture, the angle bisectors are parallel: BC . DAB ABC, so AD D B A 1z 4 13. Construct the perpendicular bisector of each of the three segments connecting the fire stations. 1z 2 z x A x x A Eliminate the rays beyond where the bisectors intersect. A point within any region will be closest to the fire station in that region. O B M 14. Z 15. x x 5. sample construction: R P T D A 16. I R R 60⬚ O E W T K R perpendicular to 6. Draw a line and construct ML it. Swing an arc from point M to point G so that MG RA. From point G, swing an arc to construct . Finish the parallelogram by swinging an arc of RG length RA from R and swinging an arc of length GR from M. There is only one possible parallelogram. M G L A R RC = KE = 8 cm C 17. a 72°, b 108°, c 108°, d 108°, e 72°, f 108°, g 108°, h 72°, j 90°, k 18°, l 90°, m 54°, n 62°, p 62°, q 59°, r 118°; Explanations will vary. Sample explanation: c is 108° because of the Corresponding Angles Conjecture. Using the Vertical Angles Conjecture, 2m 108°, so m 54°. p n because of the Corresponding Angles Conjecture. Using the Linear Pair Conjecture, n 62°, so p 62°. Using the Linear Pair Conjecture, r p 180°. Because p 62°, r 118°. 7. 1 S, 2 U ANSWERS TO EXERCISES 39 Answers to Exercises 4. C USING YOUR ALGEBRA SKILLS 3 Answers to Exercises 1. perpendicular 2. neither 3. perpendicular 4. parallel 5. possible answer: (2, 5) and (7, 11) 6. possible answer: (1, 5) and (2, 12) 7. Ordinary; no two slopes are the same, so no EM because their sides are parallel, although TE slopes are opposite reciprocals. 8. Ordinary, for the same reason as in Exercise 7— none of the sides are quite parallel. RO 9. trapezoid: KC 61; 10a. Slope HA slope ND slope NA 6. Quadrilateral HAND slope HD is a rectangle because opposite sides are parallel and adjacent sides are perpendicular. 40 ANSWERS TO EXERCISES midpoint AD 21, 3. The 10b. Midpoint HN diagonals of a rectangle bisect each other. 11a. Yes, the diagonals are perpendicular. 1; slope VR 1. Slope OE midpoint OE (2, 4). 11b. Midpoint VR The diagonals of OVER bisect each other. 11c. OVER appears to be a rhombus. Slope slope RE 51 and slope OR slope OV VE 5, so opposite sides are parallel. Also, all of the sides appear to have the same length. 12a. Both slopes equal 21. 12b. The segments are not parallel because they are coincident. 12c. distinct 13. (3, 6) LESSON 3.6 1. Sample description: Construct one of the segments, and mark arcs of the correct length from the endpoints. Draw sides to where those arcs meet. M angle at each end of that segment congruent to one of the angles in the book. Where they meet is the third vertex of the triangle. 5. C A A B C A M A S S A M S 2. O D B Sample description: Construct A and mark off the distance AB. From B swing an arc of length BC to intersect the other side of A at two points. Each gives a different triangle. C 6. A y ⫺x ____ 2 O O O T x T y T Sample description: Construct O. Mark off distances OD and OT on the sides of the angle. Connect D and T. I 3. Sample description: Mark the distance y, mark back the distance x, and bisect the remaining length of y x. Using an arc of that length, mark arcs on the ends of segment x. The point where they intersect is the vertex angle of the triangle. 7. I Y G Y I Y . Construct I at Sample description: Construct IY I and Y at Y. Label the intersection of the rays point G. 4. Yes, students’ constructions must be either larger or smaller than the triangle in the book. Sample description: Draw an angle. Mark off equal segments on the sides of the angle. Use a different compass setting to draw intersecting arcs from the ends of those segments. 8. Sample description: Draw an angle and mark off unequal distances on the sides. At the endpoint of the longer segment (not the angle vertex), swing an arc with the length of the shorter segment. From the endpoint of the shorter segment, swing an arc the length of the longer segment. Connect the endpoints of the segments to the intersection points of the arcs to form a quadrilateral. Sample description: Draw one side with a different length than the lengths in the book. Duplicate an ANSWERS TO EXERCISES 41 Answers to Exercises D y ⫺x ____ 2 x 12. new coordinates: E(4, 6), A(7, 0), T(1, 2) 9. y –5 Sample description: Draw a segment and draw an angle at one end of the segment. Mark off a distance equal to that segment on the other side of the angle. Draw an angle at that point and mark off the same distance. Connect that point to the other end of the original segment. 10. E T' A' –5 A –5 T E' 13. Reflectional symmetries Rotational symmetries Trapezoid 0 0 Kite 1 0 Parallelogram 0 2 Rhombus 2 2 Rectangle 2 2 Answers to Exercises Figure Sample description: Draw an angle and mark off equal lengths on the two sides. Use that length to determine another point that distance from the points on the sides. Connect that point with the two points on the side of the angle. 11. Answers will vary. The angle bisector lies between the median and the altitude. The order of the points is either M, R, S or S, R, M. One possible conjecture: In a scalene obtuse triangle the angle bisector is always between the median and the altitude. m⬔ABC = 111⬚ 14. half a cylinder 15. 503 16. 110⬚ E R B 3.2 cm R A S Altitude 42 Median M Angle bisector ANSWERS TO EXERCISES 110⬚ C x A 5.5 cm 110⬚ C LESSON 3.7 1. incenter Because the station needs to be equidistant from the paths, it will need to be on each of the angle bisectors. 2. circumcenter 3. incenter 8. Yes, any circle with a larger radius would not fit within the triangle. To get a circle with a larger radius tangent to two of the sides would force the circle to pass through the third side twice. 9. No, on an obtuse triangle the circle with the largest side of the triangle as the diameter of the circle creates the smallest circular region that contains the triangle. The circumscribed circle of an acute triangle does create the smallest circular region that contains the triangle. Stove Fridge Sink To find the point equidistant from three points, find the circumcenter of the triangle with those points as vertices. 5. Circumcenter. Find the perpendicular bisectors of two of the sides of the triangle formed by the classes. Locate the pie table where these two lines intersect. 6. 10. For an acute triangle, the circumcenter is inside the triangle; for an obtuse triangle, the circumcenter is outside the triangle. The circumcenter of a right triangle lies on the midpoint of the hypotenuse. 11. For an acute triangle, the orthocenter is inside the triangle; for an obtuse triangle, the orthocenter is outside the triangle. The orthocenter of a right triangle lies on the vertex of the right angle. 12. The midsegment appears parallel to side MA and half the length. A S M 7. H T 13. The base angles of the isosceles trapezoid appear congruent. A T O M ANSWERS TO EXERCISES 43 Answers to Exercises The center of the circular sink must be equidistant from the three counter edges, that is, the incenter of the triangle. 4. circumcenter 14. The measure of A is 90°. The angle inscribed in a semicircle appears to be a right angle. 21. Y 40⬚ 40⬚ A 4.8 cm 6.4 cm M T K 15. The two diagonals appear to be perpendicular bisectors of each other. A T E 22. construction of an angle bisector T 16. y 9 23. construction of a perpendicular line through a point on a line Answers to Exercises x+y=9 9 x 17. Construct the incenter by bisecting the two angles shown. Any other point on the angle bisector of the third angle must be equidistant from the two unfinished sides. From the incenter, make congruent arcs that intersect the unfinished sides. The intersection points are equidistant from the incenter. Use two congruent arcs to find another point that is equidistant from the two points you just constructed. The line that connects this point and the incenter is the angle bisector of the third angle. 18. Answers should describe the process of discovering that the midpoints of the altitudes are collinear for an isosceles right triangle. 19. a triangle M 20. 6.0 cm R 6.0 cm 60⬚ 6.0 cm 60⬚ O 60⬚ 60⬚ 6.0 cm 6.0 cm H 44 ANSWERS TO EXERCISES 24. construction of a line parallel to a given line through a point not on the line 25. construction of an equilateral triangle 26. construction of a perpendicular bisector LESSON 3.8 1. The center of gravity is the centroid. She needs to locate the incenter to create the largest circle within the triangle. 2. AM 20; SM 7; TM 14; UM 8 3. BG 24; IG 12 4. RH 42; TE 45 5. The points of concurrency are the same point for equilateral triangles because the segments are the same. 9. circumcenter 10. The shortest chord through P is a segment perpendicular to the diameter through P, which is the longest chord through P. O P 11. A B' C 6. B 12. rule: 2n 2, possible answer: H Centroid Incenter 7. ortho-/in-/centroid/circum-; the order changes when the angle becomes larger than 60°. The points become one when the triangle is equilateral. Orthocenter Centroid Circumcenter 8. Start by constructing a quadrilateral, then make a copy of it. Draw a diagonal in one, and draw a different diagonal in the second. Find the centroid of each of the four triangles. Construct a segment connecting the two centroids in each quadrilateral. Place the two quadrilaterals on top of each other matching the congruent segments and angles.Where the two segments connecting centroids intersect is the centroid of the quadrilateral. C D H C C H H H C C H H H C C C H H H H 14. a 128°, b 52°, c 128°, d 128°, e 52°, f 128°, g 52°, h 38°, k 52°, m 38°, n 71°, p 38° 15. Construct altitudes from the two accessible vertices to construct the orthocenter. Through the orthocenter, construct a line perpendicular to the southern boundary of the property. This method will divide the property equally only if the southern boundary is the base of an isosceles triangle. Altitude to missing vertex 16. 1580 greetings C⬘ D⬘ M4 M2 M3 M1 A H C 13. Orthocenter Incenter H B A⬘ B⬘ ANSWERS TO EXERCISES 45 Answers to Exercises Circumcenter CHAPTER 3 REVIEW 1. False; a geometric construction uses a straightedge and a compass. 2. False; a diagonal connects two non-consecutive vertices. 3. true 4. true 5. false B A 28. _1 z 2 y y Segment 29. 5x P C D R 3x 6. False; the lines can’t be a given distance from a segment because the segment has finite length and the lines are infinite. 7. false C x _1 z 2 4x Q 30. mA mD.You must first find B. mB 180° 2(mA). 2y B Answers to Exercises A D B 8. true 9. true 10. False; the orthocenter does not always lie inside the triangle. 11. A 12. B or K 13. I 14. H 15. G 16. D 17. J 18. C 19. 20. 2y ⬔D ⬔B ⬔A A 31. 21. 4x A Copy B y 22. y D F 32. 23. Construct a 90° angle and bisect it twice. 24. I y 25. incenter 26. Dakota Davis should locate the circumcenter of the triangular region formed by the three stones, which is the location equidistant from the stones. B 27. A 46 z ANSWERS TO EXERCISES C R x T 33. rotational symmetry 34. neither 35. both 36. reflectional symmetry 37. D 38. A 39. C 40. B 41. False; an isosceles triangle has two congruent sides. 42. true 43. False; any non-acute triangle is a counterexample. 44. False; possible explanation: The orthocenter is the point of intersection of the three altitudes. 45. true 46. False; any linear pair of angles is a counterexample. 47. False; each side is adjacent to one congruent side and one noncongruent side, so two consecutive sides may not be congruent. 48. false; 58c. no 59. Q Q' 49. False; the measure of an arc is equal to the measure of its central angle. 50. false; TD 2DR 51. False; a radius is not a chord. 52. true 53. False; inductive reasoning is the process of observing data, recognizing patterns, and making generalizations about those patterns. 54. paradox 55a. 2 and 6 or 3 and 5 55b. 1 and 5 55c. 138° 56. 55 57. possible answer: 58a. yes 58b. If the month has 31 days, then the month is October. 60. (Chapter 3 Review) n f(n) 1 2 3 4 5 6 … 1 2 5 8 11 14 … n … 20 … 56 f(n) 3n 4 61. (Chapter 3 Review) n 1 2 3 4 5 6 … f(n) 0 3 8 15 24 35 … n f(n) n2 … 20 … 399 1 ANSWERS TO EXERCISES 47 Answers to Exercises 60. See table below. 61. See table below. 62. a 38°, b 38°, c 142°, d 38°, e 50°, f 65°, g 106°, h 74°. Possible explanation: The angle with measure c is congruent to an angle with measure 142° because of the Corresponding Angles Conjecture, so c 142°. The angle with measure 130° is congruent to the bisected angle by the Corresponding Angles Conjecture. The angle with measure f has half the measure of the bisected angle, so f 65°. 63. Triangles will vary. Check that the triangle is scalene and that at least two angle bisectors have been constructed. 64. mFAD 30° so mADC 30°, but its vertical angle has measure 26°. This is a contradiction. 65. minimum: 101 regions by 100 parallel lines; maximum: 5051 regions by 100 intersecting, noncurrent lines Answers to Exercises CHAPTER 4 • CHAPTER 4 CHAPTER 4 • CHAPTER angles P and D can be drawn at each endpoint using the protractor. LESSON 4.1 Q 1. The angle measures change, but the sum remains 180°. 2. 73° 3. 60° 4. 110° 5. 24° 6. 3 360° 180° 900° 7. 3 180° 180° 360° 8. 69°; 47°; 116°; 93°; 86° 9. 30°; 50°; 82°; 28°; 32°; 78°; 118°; 50° 55⬚ P 10. Answers to Exercises ⬔R ⬔M 11. ⬔A ⬔G ⬔L 12. First construct E, using the method used in Exercise 10. 85⬚ 40⬚ 7 cm 17. The third angles of the triangles also have the same measures; are equal in measure 18. You know from the Triangle Sum Conjecture that mA mB mC 180°, and mD mE mF 180°. By the transitive property, mA mB mC mD mE mF. You also know that mA mD, and mB mE. You can substitute for mD and mE in the longer equation to get mA mB mC mA mB mF. Subtracting equal terms from both sides, you are left with mC mF. 19. For any triangle, the sum of the angle measures is 180°, by the Triangle Sum Conjecture. Since the triangle is equiangular, each angle has the same measure, say x. So x x x 180°, and x 60°. 20. false E R A 13. Fold ⬔M ⬔A 21. false E ⬔R ⬔G ⬔L A R 22. false 14. From the Triangle Sum Conjecture mA mS mM 180°. Because M is a right angle, mM 90°. By substitution, mA mS 90° 180°. By subtraction, mA mS 90°. So two wrongs make a right! 15. Answers will vary. See the proof on page 202. To prove the Triangle Sum Conjecture, the Linear Pair Conjecture and the Alternate Interior Angles Conjecture must be accepted as true. 16. It is easier to draw PDQ if the Triangle Sum Conjecture is used to find that the measure of can be drawn to be 7 cm, and D is 85°. Then PD 48 ANSWERS TO EXERCISES D 23. false 24. true 25. eight; 100 LESSON 4.2 F D E C A B 16. P M N K G H 17. possible answer: Fold 1 Fold 3 Fold 2 Fold 4 105⬚ 60⬚ 45⬚ 18. 19. 20. 21. 22. 23. 0 perpendicular parallel parallel neither parallelogram 40 8 16 24 32 40 24. New: (6, 3), (2, 5), (3, 0). Triangles are congruent. 25. New: (3, 3), (3, 1), (1, 5). Triangles are congruent. ANSWERS TO EXERCISES 49 Answers to Exercises 1. 79° 2. 54° 3. 107.5° 4. 44°; 35 cm 5. 76°; 3.5 cm 6. 72°; 36°; 8.6 cm 7. 78°; 93 cm 8. 75 m; 81° 9. 160 in.; 126° 10. a 124°, b 56°, c 56°, d 38°, e 38°, f 76°, g 66°, h 104°, k 76°, n 86°, p 38°; Possible explanation: The angles with measures 66° and d form a triangle with the angle with measure e and its adjacent angle. Because d, e, and the adjacent angle are all congruent, 3d 66° 180°. Solve to get d 38°. This is also the measure of one of the base angles of the isosceles triangle with vertex angle measure h. Using the Isosceles Triangle Conjecture, the other base angle measures d, so 2d h 180°, or 76° h 180°. Therefore, h 104°. 11. a 36°, b 36°, c 72°, d 108°, e 36°; none 12a. Yes. Two sides are radii of a circle. Radii must be congruent; therefore, each triangle must be isosceles. 12b. 60° 13. NCA 14. IEC 15. USING YOUR ALGEBRA SKILLS 4 1. 3. 5. 7. false not a solution not a solution y 4 true solution x7 x 8 1 10. n 2 9. x 4.2 Answers to Exercises 2. 4. 6. 8. 11. x 2 12. t 18 9 2 14a. x 4 13. n 5 14b. x 94; The two methods produce identical results. Multiplying by the lowest common denominator (which is comprised of the factors of both denominators) and then reducing common factors (which clears the denominators on either side) is the same as simply multiplying each numerator by the opposite denominator (or cross multiplying). Algebraically you could show that the two methods are equivalent as follows: c a b d c a bd b bd d abd bcd b d ad bc The method of “clearing fractions” results in the method of “cross multiplying.” 50 ANSWERS TO EXERCISES 15. You get an equation that is always false, so there is no solution to the equation. 16. Camella is not correct. Because the equation 0 0 is always true, the truth of the equation does not depend on the value of x. Therefore, x can be any real number. Camella’s answer, x 0, is only one of infinitely many solutions. 17. 2x x 2x If x equals the measure of the vertex angle, then the base angles each measure 2x. Applying the Triangle Sum Conjecture results in the following equation: x 2x 2x 180° 5x 180° x 36° The measure of the vertex angle is 36° and the measure of each base angle is 72°. LESSON 4.3 1. yes 2. no 4 5 9 3. no 5 6 12 Answers to Exercises 4. yes 5. a, b, c 6. c, b, a 7. b, a, c 8. a, c, b 9. a, b, c 10. v, z, y, w, x 11. 6 length 102 12. By the Triangle Inequality Conjecture, the sum of 11 cm and 25 cm should be greater than 48 cm. 13. b 55°, but 55° 130° 180°, which is impossible by the Triangle Sum Conjecture. 14. 135° 15. 72° 16. 72° 17. a b c 180° and x c 180°. Subtract c from both sides of both equations to get x 180 c and a b 180 c. Substitute a b for 180 c in the first equation to get x a b. 18. 45° 19. a 52°, b 38°, c 110°, d 35° 20. a 90°, b 68°, c 112°, d 112°, e 68°, f 56°, g 124°, h 124° 21. By the Triangle Sum Conjecture, the third angle must measure 36° in the small triangle, but it measures 32° in the large triangle. These are the same angle, so they can’t have different measures. 22. ABE 23. FNK 24. cannot be determined ANSWERS TO EXERCISES 51 13. Cannot be determined. SSA is not a congruence conjecture. 14. AIN by SSS or SAS 15. Cannot be determined. Parts do not correspond. 16. SAO by SAS 17. Cannot be determined. Parts do not correspond. 18. RAY by SAS and PR is (0, 0). 19. The midpoint of SD Therefore, DRO SPO by SAS. 20. Because the LEV is marking out two triangles that are congruent by SAS, measuring the length of the segment leading to the finish will also approximate the distance across the crater. 21. 22. LESSON 4.4 1. Answers will vary. Possible answer: If three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent (all corresponding angles are also congruent). 2. Answers will vary. Possible answer: The picture statement means that if two sides of one triangle are congruent to two sides of another triangle, and the angles between those sides are also congruent, then the two triangles are congruent. If you know this: then you also know this: Answers to Exercises 3. Answers will vary. Possible answer: 23. a 37°, b 143°, c 37°, d 58°, e 37°, f 53°, g 48°, h 84°, k 96°, m 26°, p 69°, r 111°, s 69°; Possible explanation: The angle with measure h is the vertex angle of an isosceles triangle with base angles measuring 48°, so h 2(48) 180°, and h 84°. The angle with measure s and the angle with measure p are corresponding angles formed by parallel lines, so s p 69°. 24. 3 cm third side 19 cm 25. See table below. 13 3 26a. y 6 b. y c. y 4x 2 3 27. (5, 3) 4. SAS 5. SSS 6. cannot be determined 7. SSS 8. SAS 9. SSS (and the Converse of the Isosceles Triangle Conjecture) 10. yes, ABC ADE by SAS 11. Possible answer: Boards nailed diagonally in the corners of the gate form triangles in those corners. Triangles are rigid, so the triangles in the gate’s corners will increase the stability of those corners and keep them from changing shape. 12. FLE by SSS 25. (Lesson 4.4) Side length 1 2 3 4 5 Elbows 4 4 4 4 4 T’s 0 4 8 12 16 Crosses 0 1 4 9 16 … 4n 4 52 ANSWERS TO EXERCISES n … 4 20 4 76 361 (n 1)2 LESSON 4.5 1. If two angles and the included side of one triangle are congruent to the corresponding side and angles of another triangle, then the triangles are congruent. 2. If two angles and a non-included side of one triangle are congruent to the corresponding side and angles of another triangle, then the triangles are congruent. If you know this: then you also know this: 3. Answers will vary. Possible answer: L K M 27. 28a. about 100 km southeast of San Francisco 28b. Yes. No, two towns would narrow it down to two locations. The third circle narrows it down to one. ORE GO N Boise IDAH Eureka Sacramento San Francisco Reno NEV O Elko ADA U TA H LI FO RN Las Vegas IA Los Angeles Miles 0 50 100 0 100 Kilometers 200 ARIZ ONA 400 ANSWERS TO EXERCISES 53 Answers to Exercises 22. Construction will show a similar but larger (or smaller) triangle constructed from a drawn triangle by duplicating two angles on either end of a new side that is not congruent to the corresponding side. 23. Draw a line segment. Construct a perpendicular. Bisect the right angle. Construct a triangle with two congruent sides and with a vertex that measures 135°. 24. 125 25. False. One possible counterexample is a kite. 26. None. One triangle is determined by SAS. CA 4. ASA 5. cannot be determined 6. SAA 7. cannot be determined 8. ASA 9. cannot be determined 10. FED by SSS 11. WTA by ASA or SAA 12. SAT by SAS 13. PRN by ASA or SAS; SRE by ASA 14. Cannot be determined. Parts do not correspond. 15. MRA by SAS 16. Cannot be determined.AAA does not guarantee congruence. 17. WKL by ASA 18. Yes, ABC ADE by SAA or ASA. slope CD 3 and slope 19. Slope AB BC 1 3, so AB BC , CD DA , and slope DA DA . ABC CDA by SAA. BC 20. 21. The construction is the same as the construction using ASA once you find the third angle, which is used here. (Finding the third angle is not shown.) 13. cannot be determined 14. KEI by ASA 15. UTE by SAS 16. Answers to Exercises LESSON 4.6 BD (same segment), A C 1. Yes. BD (given), and ABD CBD (given), so DBA CB by CPCTC. DBC by SAA. AB WN and C W (given), and 2. Yes. CN RNC ONW (vertical angles), CNR ON by CPCTC. WON by ASA. RN 3. Cannot be determined. The congruent parts lead to the ambiguous case SSA. IT 4. Yes. S I, G A (given), and TS (definition of midpoint), so TIA TSG by IA by CPCTC. SAA. SG FR and UO UR (given), and UF 5. Yes. FO (same segment), so FOU FRU by SSS. UF O R by CPCTC. MA and ME MR (given), and 6. Yes. MN M M (same angle), so EMA RMN by SAS. E R by CPCTC. EU and BU ET (given), and 7. Yes. BT UT UT (same segment), so TUB UTE by SSS. B E by CPCTC. 8. Cannot be determined. HLF LHA by and HF are not corresponding sides. ASA, but HA 9. Cannot be determined. AAA does not guarantee congruence. 10. Yes. The triangles are congruent by SAS. 11. Yes. The triangles are congruent by SAS, and the angles are congruent by CPCTC. and DF to form ABC and DEF. 12. Draw AC because all were drawn with AB CB DE FE DF for the same reason. the same radius. AC ABC DEF by SSS. Therefore, B E by CPCTC. 17. 18. a 112°, b 68°, c 44°, d 44°, e 136°, f 68°, g 68°, h 56°, k 68°, l 56°, m 124°; Possible explanation: f and g are measures of base angles of an isosceles triangle, so f g. The vertex angle measure is 44°, so subtract 44° from 180° and divide by 2 to get f 68°. The angle with measure m is the exterior angle of a triangle. Add the remote interior angle measures 56° and 68° to get m 124°. 19. ASA. The “long segment in the sand” is a shared side of both triangles 20. (4, 1) 21. See table below. 22. Value C is always decreasing. 23. x 3, y 10 21. (Lesson 4.6) 54 Number of sides 3 4 5 6 7 … 12 … n Number of struts needed to make polygon rigid 0 1 2 3 4 … 9 … n3 ANSWERS TO EXERCISES 3. See flowchart below. 4. See flowchart below. LESSON 4.7 1. See flowchart below. 2. See flowchart below. 1. (Lesson 4.7) 1 SE SU }? 2 ESM USO 4 E U }? 3 Given Given } } 5 ? ? MS SO }? CPCTC ASA Congruence Conjecture 1 2 }? Vertical Angles Conjecture 2. (Lesson 4.7) 1 3 CI IM Definition of midpoint Given 2 Answers to Exercises I is midpoint of CM 4 I is midpoint of BL 6 IL IB of }? Definition midpoint }? Given } } ? ? MB }? CL 7 }? CIL MIB CPCTC by SAS 5 1 2 }? Vertical Angles Conjecture 3. (Lesson 4.7) 1 3 NS is a median Given 2 Same segment 4 S is a midpoint Definition of median 6 WS SE ESN WSN ? } } ? SSS Definition of midpoint 5 7 E W ? } ? } CPCTC WN NE }? Given 4. (Lesson 4.7) 1 NS is an NS NS 2 angle bisector } 1 ? } 2 Definition of ? angle bisector Given 3 W E } ? Given 4 5 WNS ENS ? ? } } ? SAA } 6 WN NE } ? CPCTC 7 NEW is isosceles of }? Definition isosceles triangle NS NS }? Same segment ANSWERS TO EXERCISES 55 5. See flowchart below. BC , CD AD 6. Given: ABC with BA is the angle bisector of ABC. Show: BD 12. ACK by SSS 13. a 72°, b 36°, c 144°, d 36°, e 144°, f 18°, g 162°, h 144°, j 36°, k 54°, m 126° 14. The circumcenter is equidistant from all three vertices because it is on the perpendicular bisector of each side. Every point on the perpendicular bisector of a segment is equidistant from the endpoints. Similarly, the incenter is equidistant from all three sides because it is on the angle bisector of each angle, and every point on an angle bisector is equidistant from the sides of the angle. 15. ASA. The fishing pole forms the side. “Perpendicular to the ground” forms one angle. “Same angle on her line of sight” forms the other angle. 2 16. 7 17. A D 1 2 B C BA ⬵ BC CD ⬵ AD BD ⬵ BD Given Given Same segment 䉭ABD ⬵ 䉭CBD SSS ⬔1 ⬵ ⬔2 CPCTC Answers to Exercises → BD bisects ⬔ABC Definition of angle bisector 7. The angle bisector does not go to the midpoint of the opposite side in every triangle, only in an isosceles triangle. , because it is across from the smallest angle 8. NE , which is across in NAE. It is shorter than AE from the smallest angle in LAE. 9. The triangles are congruent by SSS, so the two central angles cannot have different measures. 10. PRN by ASA; SRE by ASA 11. Cannot be determined. Parts do not correspond. 18. y 4 Y X B O B' Y'4 O' x X' 5. (Lesson 4.7) 1 SA NE 3 ? Given ᎏ 2 SE NA 3 4 AIA Conjecture ESN ANS 4 ? 1 2 ᎏ ? AIA Conjecture ᎏ ? Given ᎏ 5 SN SN Same segment 56 ANSWERS TO EXERCISES 6 ? ᎏ ? ᎏ ? ASA ᎏ 7 ? SA NE ᎏ ? CPCTC ᎏ This proof shows that in a parallelogram, opposite sides are congruent. 7. LESSON 4.8 1. 2. 3. 4. 5. 6. 6 90°; 18° 45° See flowchart below. See flowchart below. 1 Isosceles 䉭ABC with AC ⬵ BC and CD bisects ⬔C 2 AD ⬵ BD AC ⬵ BC 2 Given Same segment D is midpoint of AB ⬔ACD ⬵ ⬔DCB CPCTC 䉭ADC ⬵ 䉭BDC AD ⬵ BD Def. of midpoint 7 CD is angle bisector of ⬔ACB Def. of angle bisector 8 CD is altitude of 䉭ABC Conjecture B (Exercise 5) CPCTC 4 䉭ADC ⬵ 䉭BDC SSS 6 Conjecture A (Exercise 4) 4 3 CD ⬵ CD Given 5 Given 3 1 9 CD ⬜ AB Def. of altitude 8. Yes. First show that the three exterior triangles are congruent by SAS. CD is a median Def. of median Answers to Exercises 4. (Lesson 4.8) 1 } 2 ? CD is the bisector of C Given 1 2 Definition of angle bisector 3 ABC is isosceles with AC BC 5 }? SAS Given 4 ADC BDC CD CD Same segment 5. (Lesson 4.8) 1 ABC is isosceles with AC BC, and CD is the bisector of C 2 ADC BDC 4 }? CPCTC Conjecture A Given 3 1 and 2 form a linear pair Definition of linear pair 1 2 5 6 Congruent supplementary angles are 90 1 and 2 are supplementary Linear Pair Conjecture 1 and 2 are right angles 7 AB CD } Definition of ? perpendicular } is an altitude 8 ? CD Definition of altitude ANSWERS TO EXERCISES 57 9. C A B D Drawing the vertex angle bisector as an auxiliary segment, we have two triangles. We can show them to be congruent by SAS, as we did in Exercise 4. Then, A B, by CPCTC. Therefore, base angles of an isosceles triangle are congruent. 10. The proof is similar to the one on page 245, but in reverse, and using the Converse of the Isosceles Triangle Conjecture. 11. 5 14. y 3x 16 15. 120 16. (4, 6) or (4, 0) or any point at which the x-coordinate is either 1 or 7 and the y-coordinate does not equal 3 17. Hugo and Duane can locate the site of the fireworks by creating a diagram using SSS. Fireworks 340 m/s • 3 s = 1.02 km 340 m/s • 5 s = 1.7 km Hugo 1.5 km Duane 18. Cn H2n H H H 30⬚ H Answers to Exercises H 12. a 128°, b 128°, c 52°, d 76°, e 104°, f 104°, g 76°, h 52°, j 70°, k 70°, l 40°, m 110°, n 58° 13. between 16 and 17 minutes 58 ANSWERS TO EXERCISES H H C C C C C H H C C H H H H H CHAPTER 4 REVIEW P ⬔L A y ⬔A ⬔P L 34. Construct P. Mark off the length PB on one ray. From point B, mark off the two segments that intersect the other ray of P at distance x. S2 S1 x x P z B ANSWERS TO EXERCISES 59 Answers to Exercises 1. Their rigidity gives strength. 2. The Triangle Sum Conjecture states that the sum of the measures of the angles in every triangle is 180°. Possible answers: It applies to all triangles; many other conjectures rely on it. 3. The angle bisector of the vertex angle is also the median and the altitude. 4. The distance between A and B is along the segment connecting them. The distance from A to C to B can’t be shorter than the distance from A to B. Therefore, AC CB AB. Points A, B, and C form a triangle. Therefore, the sum of the lengths of any two sides is greater than the length of the third side. 5. SSS, SAS, ASA, or SAA 6. In some cases, two different triangles can be constructed using the same two sides and non-included angle. 7. cannot be determined 8. ZAP by SAA 9. OSU by SSS 10. cannot be determined 11. APR by SAS 12. NGI by SAS 13. cannot be determined 14. DCE by SAA or ASA 15. RBO or OBR by SAS UT by 16. AMD UMT by SAS, AD CPCTC 17. cannot be determined 18. cannot be determined AL by 19. TRI ALS by SAA, TR CPCTC KV by 20. SVE NIK by SSS, EL overlapping segments property 21. cannot be determined 22. cannot be determined 23. LAZ IAR by ASA, LRI IZL by ASA, and LRD IZD by ASA 24. yes. PTS TPO by ASA or SAA 25. ANG is isosceles, so A G. However, the sum of mA mN mG 188°. The measures of the three angles of a triangle must sum to 180°. 26. ROW NOG by ASA, implying that OG . However, the two segments shown are OW not equal in measure. 27. a g e d b f c. Thus, c is the longest segment, and a and g are the shortest. 28. x 20° 29. Yes. TRE SAE by SAA, so sides are congruent by CPCTC. 30. Yes. FRM RFA by SAA. RFM FRA by CPCTC. Because base angles are congruent, FRD is isosceles. 31. x 48° 32. The legs form two triangles that are congruent by SAS. Because alternate interior angles are congruent by CPCTC, the seat must be parallel to the floor. 33. Construct P and A to be adjacent. The angle that forms a linear pair with the conjunction of P and A is L. Construct A. Mark off the length AL on one ray. Construct L. Extend the unconnected sides of the angles until they meet. Label the point of intersection P. 37. Possible method: Construct an equilateral triangle and bisect one angle to obtain 30°. Adjacent to that angle, construct a right angle and bisect it to obtain 45°. 38. d, a b, c, e, f Answers to Exercises 35. See flowchart below. 36. Given three sides, only one triangle is possible; therefore, the shelves on the right hold their shape. The shelves on the left have no triangles and move freely as a parallelogram. 35. (Chapter 4 Review) TMI RME 2 } } } ? ? Vertical angles ? ME TMI EMR TM 3 ? 5 M is midpoint ? ? ? and IR of TE ? Definition ? SAS 1 ? of midpoint ? Given 4 ? ? MR IM ? Definition of midpoint } } 60 } } } } } } ANSWERS TO EXERCISES } } } T E or R I 6 ? ? } } ? } CPCTC RE TI 7 }? }? of }? Converse AIA Conjecture Answers to Exercises CHAPTER 5 • CHAPTER 5 13. 17 14. 15 15. the twelfth century 16. The angles of the trapezoid measure 67.5° and 112.5°; 67.5° is half the value of each angle of a regular octagon, and 112.5° is half the value of 360° 135°. CHAPTER 5 • CHAPTER LESSON 5.1 1. See table below. 2. See table below. 3. 122° 4. 136° 5. 108°; 36° 6. 108°; 106° 7. 105°; 82° 8. 120°; 38° 9. The sum of the interior angle measures of the quadrilateral is 358°. It should be 360°. 10. The measures of the interior angles shown sum to 554°. However, the figure is a pentagon, so the measures of its interior angles should sum to 540°. 11. 18 12. a 116°, b 64°, c 90°, d 82°, e 99°, f 88°, g 150°, h 56°, j 106°, k 74°, m 136°, n 118°, p 99°; Possible explanation: The sum of the angles of a quadrilateral is 360°, so a b 98° d 360°. Substituting 116° for a and 64° for b gives d 82°. Using the larger quadrilateral, e p 64° 98° 360°. Substituting e for p, the equation simplifies to 2e 198°, so e 99°. The sum of the angles of a pentagon is 540°, so e p f 138° 116° 540°. Substituting 99° for e and p gives f 88°. 135⬚ 67.5⬚ 1. (Lesson 5.1) Number of sides of polygon Sum of measures of angles 7 8 9 10 11 20 55 100 900° 1080° 1260° 1440° 1620° 3240° 9540° 17640° 2. (Lesson 5.1) Number of sides of equiangular polygon Measure of each angle of equiangular polygon 5 6 7 8 9 10 12 16 100 108° 120° 12874° 135° 140° 144° 150° 1571° 1762° 2 5 ANSWERS TO EXERCISES 61 Answers to Exercises 17. Answers will vary; see the answer for Developing Proof on page 259. Using the Triangle Sum Conjecture, a b j c d k e f l g h i 4(180°), or 720°. The four angles in the center sum to 360°, so j k l i 360°. Subtract to get a b c d e f g h 360°. 18. x 120° 19. The segments joining the opposite midpoints of a quadrilateral always bisect each other. 20. D 21. Counterexample: The base angles of an isosceles right triangle measure 45°; thus they are complementary. LESSON 5.2 360° 72°; 60° 15 43 a 108° 1 6. b 453° 3 5 7. c 517°, d 1157° 8. e 72°, f 45°, g 117°, h 126° 9. a 30°, b 30°, c 106°, d 136° 10. a 162°, b 83°, c 102°, d 39°, e 129°, f 51°, g 55°, h 97°, k 83° 11. See flowchart below. 12. Yes. The maximum is three. The minimum is zero. A polygon might have no acute interior angles. Answers to Exercises 1. 2. 3. 4. 5. 13. Answers will vary. Possible proof using the diagram on the left: a b i 180°, c d h 180°, and e f g 180° by the Triangle Sum Conjecture. a b c d e f g h i 540° by the addition property of equality. Therefore, the sum of the measures of the angles of a pentagon is 540°. To use the other diagram, students must remember to subtract 360° to account for angle measures k through o. 14. regular polygons: equilateral triangle and regular dodecagon; angle measures: 60°, 150°, and 150° 15. regular polygons: square, regular hexagon, and regular dodecagon; angle measures: 90°, 120°, and 150° CR by 16. Yes. RAC DCA by SAS. AD CPCTC. 17. Yes. DAT RAT by SSS. D R by CPCTC. 11. (Lesson 5.2) 1 a b 180 ? Linear Pair Conjecture 2 c d 180 ? Linear Pair Conjecture 3 e f 180 ? Linear Pair Conjecture 62 ANSWERS TO EXERCISES 4 540° ? a+b+c+d+e+f= Addition property of equality 180° 5 ? a+c+e= ? Triangle Sum Conjecture 6 360° ? b+d+f= Subtraction property of equality 13. possible answer: LESSON 5.3 W 1. 64 cm 2. 21°; 146° 3. 52°; 128° 4. 15 cm 5. 72°; 61° 6. 99°; 38 cm 7. w 120°, x 45°, y 30° 8. w 1.6 cm, x 48°, y 42° 9. See flowchart below. 10. Answers may vary. This proof uses the Kite Angle Bisector Conjecture. S 2 1 N X H is the other base. S and H are a pair of OW base angles. O and W are a pair of base angles. HO . SW 14. Only one kite is possible F because three sides determine N B a triangle. E 15. S H Y B O I E Given: Kite BENY with vertex angles B and N is the perpendicular bisector Show: Diagonal BN . of diagonal YE BY . From the From the definition of kite, BE Kite Angle Bisector Conjecture, 1 2. BX because they are the same segment. By SAS, BX XE . BXY BXE. So by CPCTC, XY Because YXB and EXB form a linear pair, they are supplementary, so mYXB mEXB 180°. By CPCTC, YXB EXB, or mYXB mEXB, so by substitution, 2mYXB 180°, or mYXB 90°. So mYXB mEXB 90°. XE and YXB and EXB are right Because XY is the perpendicular bisector of YE . angles, BN I 11. possible answer: E I 16. infinitely many, possible construction: B E Z Q N 17. 80°, 80°, 100°, 100° 18. Because ABCD is an isosceles trapezoid, A BH B. AGF BHE by SAA. Thus, AG by CPCTC. 19. a 80°, b 20°, c 160°, d 20°, e 80°, f 80°, g 110°, h 70°, m 110°, n 100°; Possible explanation: Because d forms a linear pair with e and its congruent adjacent angle,d 2e 180°.Substituting d 20° gives 2e 160°,so e 80°. Using theVerticalAngles Conjecture and d 20°, the unlabeled angle in the small right triangle measures 20°, which means h 70°. Because g and h are a linear pair, they are supplementary, so g 110°. K E 12. possible answer: O I U . Q and U are a pair of base The other base is ZI angles. Z and I are a pair of base angles. 9. (Lesson 5.3) 1 BE BY Given 2 EN YN Given BN BN 3 ? ᎏ ? ᎏ Same segment BYN 4 ? BEN ᎏ ? Congruence ᎏ shortcut SSS 2 5 1 ? and ᎏ ? 4 3 ᎏ ? CPCTC ᎏ bisects B, BN bisects N BN 6 ? and ᎏ ? ᎏ Definition of angle bisector ANSWERS TO EXERCISES 63 Answers to Exercises T W Answers to Exercises LESSON 5.4 1. three; one 2. 28 3. 60°; 140° 4. 65° 5. 23 6. 129°; 73°; 42 cm 7. 35 8. See flowchart below. 9. Parallelogram. Draw a diagonal of the original quadrilateral. The diagonal forms two tri-angles. Each of the two midsegments is parallel to the diagonal, and thus the midsegments are parallel to each other. Now draw the other diagonal of the original quadrilateral. By the same reasoning, the second pair of midsegments is parallel. Therefore, the quadrilateral formed by joining the midpoints is a parallelogram. 10. The length of the edge of the top base measures 30 m. We know this by the Trapezoid Midsegment Conjecture. 11. Ladie drives a stake into the ground to create a triangle for which the trees are the other two vertices. She finds the midpoint from the stake to each tree. The distance between these midpoints is half the distance between the trees. 13. If a quadrilateral is a kite, then exactly one diagonal bisects a pair of opposite angles. Both the original and converse statements are true. 14. a 54°, b 72°, c 108°, d 72°, e 162°, f 18°, g 81°, h 49.5°, i 130.5°, k 49.5°, m 162°, n 99°; Possible explanation: The third angle of the triangle containing f and g measures 81°, so using the Vertical Angles Conjecture, the vertex angle of the triangle containing h also measures 81°. Subtract 81° from 180° and divide by 2 to get h 49.5°. The other base angle must also measure 49.5°. By the Corresponding Angles Conjecture, k 49.5°. 15. (3, 8) 16. (0, 8) 17. coordinates: E(2, 3.5), Z(6, 5); the slope of 83, and the slope of YT 83 EZ 18. R F N Cabin K There is only one kite, but more than one way to construct it. 12. Explanations will vary. 80 60 cm 40 60 8. (Lesson 5.4) 1 FOA with 3 midsegment LN Given midsegment RD 64 ANSWERS TO EXERCISES } ? Triangle Midsegment Conjecture 2 IOA with Given LN OA RD OA 4 } ? Triangle Midsegment Conjecture RD LN 5 }? Two lines parallel to the same line are parallel 10. LESSON 5.5 1. 2. 3. 4. 5. 6. 7. 34 cm; 27 cm 132°; 48° 16 in.; 14 in. 63 m 80 63°; 78° Vb T Vc S L A 11. (b a, c) 12. possible answer: 8. R a O D b c d P P 9. R 13. See flowchart below. 14. The parallelogram linkage is used for the sewing box so that the drawers remain parallel to each other (and to the ground) so that the contents cannot fall out. Vh 15. a 135°, b 90° 16. a 120°, b 108°, c 90°, d 42°, e 69° Vw 13. (Lesson 5.5) 1 LEAN is a parallelogram ? Given ᎏ 2 EA LN ? Definition of ᎏ parallelogram 3 AEN LNE AIA Conjecture EAL NLA 4 ? ᎏ AIA Conjecture LN AE 5 ? ᎏ 7 AET LNT 6 ET NT and LA bisect EN each other CPCTC 9 ? ᎏ ASA LT AT 8 ? ᎏ CPCTC ? ᎏ Definition of segment bisector Opposite sides congruent ANSWERS TO EXERCISES 65 Answers to Exercises b d c a 17. x 104°, y 98°. The quadrilaterals on the left and right sides are kites. Nonvertex angles are congruent. The quadrilateral at the bottom is an isosceles trapezoid. Base angles are congruent, and consecutive angles between the bases are supplementary. 18. a 84°, b 96° 19. No. The congruent angles and side do not correspond. 20. Answers to Exercises 21. Parallelogram. Because the triangles are congruent by SAS, 1 2. So, the segments are 66 ANSWERS TO EXERCISES parallel. Use a similar argument to show that the other pair of opposite sides is parallel. 1 2 22. Kite or dart. Radii of the same circle are congruent. If the circles have equal radii, a rhombus is formed. USING YOUR ALGEBRA SKILLS 5 1. y (0, 1) (1, –1) 2. x y (3, 8) (0, 4) x 3. y 4. y x 2 74 6 5. y 1 3 x 13 6. y x 1 7. y 3x 5 2 8 8. y 5x 5 9. y 80 4x 10. y 3x 26 1 11. y 4x 3 6 12. y 5x 13. y x 1 2 43 14. y 9x 9 (2, 9) (0, 6) Answers to Exercises x ANSWERS TO EXERCISES 67 LESSON 5.6 17. E V L O 1. Sometimes true; it is true only if the parallelogram is a rectangle. false true 18. Constructions will vary. 2. Always true; by the definition of rectangle, all the angles are congruent. By the Quadrilateral Sum Conjecture and division, they all measure 90°, so any two angles in a rectangle, including consecutive angles, are supplementary. 3. Always true by the Rectangle Diagonals Conjecture. 4. Sometimes true; it is true only if the rectangle is a square. A B B E false Answers to Exercises 5. Always true by the Square Diagonals Conjecture. 6. Sometimes true; it is true only if the rhombus is equiangular. false 7. Always true; all squares fit the definition of rectangle. 8. Always true; all sides of a square are congruent and form right angles, so the sides become the legs of the isosceles right triangle and the diagonal is the hypotenuse. 9. Always true by the Parallelogram Opposite Angles Conjecture. 10. Sometimes true; it is true only if the parallelogram is a rectangle. Consecutive angles of a parallelogram are always supplementary, but are congruent only if they are right angles. 11. 20 12. 37° 13. 45°, 90° 14. DIAM is not a rhombus because it is not equilateral and opposite sides are not parallel. 15. BOXY is a rectangle because its adjacent sides are perpendicular. 16. Yes. TILE is a rhombus, and a rhombus is a parallelogram. 68 ANSWERS TO EXERCISES E E I S 20. Converse: If the diagonals of a quadrilateral are congruent and bisect each other, then the quadrilateral is a rectangle. Given: Quadrilateral ABCD with diagonals BD . AC and BD bisect each other AC Show: ABCD is a rectangle A 8 7 true K 19. one possible construction: P true A K D 1 2 E 6 5 B 3 4 C Because the diagonals are congruent and bisect BE DE EC . Using the each other, AE Vertical Angles Conjecture, AEB CED and BEC DEA. So AEB CED and AED CEB by SAS. Using the Isosceles Triangle Conjecture and CPCTC, 1 2 5 6, and 3 4 7 8. Each angle of the quadrilateral is the sum of two angles, one from each set, so for example, mDAB m1 m8. By the addition property of equality, m1 m8 m2 m3 m5 m4 m6 m7. So mDAB mABC mBCD mCDA. So the quadrilateral is equiangular. Using CD . 1 5 and the Converse of AIA, AB Using 3 7 and the Converse of AIA, AD . Therefore ABCD is an equiangular BC parallelogram, so it is a rectangle. 21. If the diagonals are congruent and bisect each other, then the room is rectangular (converse of the Rectangle Diagonals Conjecture). 90°. Because 4 and 5 form a linear pair, m4 m5 180°. Substitute 90° for m4 and solve to get m5 90°. By definition of congruent angles, 5 3, and 5 and 3 are alternate BC by the Converse of the interior angles, so AD Parallel Lines Conjecture. Similarly, 1 and 5 are CD by the congruent corresponding angles, so AB Converse of the Parallel Lines Conjecture. Thus, ABCD is a parallelogram by the definition of parallelogram. Because it is an equiangular parallelogram, ABCD is a rectangle. 28. a 54°, b 36°, c 72°, d 108°, e 36°, f 144°, g 18°, h 48°, j 48°, k 84° 29. possible answers: (1, 0); (0, 1); (1, 2); (2, 3) 8 86 30. y 9x 9 or 8x 9y 86 7 12 31. y 1x 0 5 or 7x 10y 24 32. velocity 1.8 mi/h; angle of path 106.1° clockwise from the north Answers to Exercises 22. The platform stays parallel to the floor because opposite sides of a rectangle are parallel (a rectangle is a parallelogram). 23. The crosswalks form a parallelogram: The streets are of different widths, so the crosswalks are of different lengths. The streets would have to meet at right angles for the crosswalks to form a rectangle. The corners would have to be right angles and the streets would also have to be of the same width for the crosswalk to form a square. 24. Place one side of the ruler along one side of the angle. Draw a line with the other side of the ruler. Repeat with the other side of the angle. Draw a line from the vertex of the angle to the point where the two lines meet. 25. Rotate your ruler so that each endpoint of the segment barely shows on each side of the ruler. Draw the parallel lines on each side of your ruler. Now rotate your ruler the other way and repeat the process to get a rhombus. The original segment is one diagonal of the rhombus. The other diagonal will be the perpendicular bisector of the original segment. 26. See flowchart below. 27. Yes, it is true for rectangles. Given: 1 2 3 4 Show: ABCD is a rectangle By the Quadrilateral Sum Conjecture, m1 m2 m3 m4 360°. It is given that all four angles are congruent, so each angle measures 2 mi/h 60 1.5 mi/h 26. (Lesson 5.6) 1 QU AD Given 2 QD AU ? Given 3 4 QUD ADU ? SSS 5 1 2 3 4 ? CPCTC DU DU Same segment 8 QU UA AD DQ Given 6 7 QUAD is a QU AD QD AU parallelogram Converse of the Parallel Lines Conjecture Definition of parallelogram 9 QUAD is a rhombus ? Definition of rhombus ANSWERS TO EXERCISES 69 5. parallelogram 6. sample flowchart proof: LESSON 5.7 1. See flowchart below. 2. See flowchart below. 3. See flowchart below. 2 5 4. 1 SP OA SOP APO SP OA Given Given 3 2 IY GO Opposite sides of rectangle are congruent 1 2 3 YOG OYI Definition of rectangle 4 SAS 6 AIA Conjecture 4 1 YO OY Same segment YOG OYI SAS 3 4 CPCTC PO PO 7 Same segment 5 PA SO CPCTC Converse of AIA Conjecture 8 YG IO SOAP is a parallelogram Definition of parallelogram 1. (Lesson 5.7) 2 SO KA 4 3 4 AIA Answers to Exercises Definition of parallelogram SK 1 3 SOAK is a parallelogram Given 5 ? OA ᎏ 1 2 ? Definition of ? AIA ᎏ ᎏ parallelogram 6 7 ? SOA ᎏ AKS ? ASA ᎏ ? SA SA ᎏ ? Same segment ᎏ 2. (Lesson 5.7) 1 Parallelogram BATH with diagonal BT Given 4 Parallelogram BATH with diagonal HA ? Given ᎏ 2 BAT THB 3 BAT THB CPCTC Conjecture proved in Exercise 1 ATH THA 5 6 ? ? BAH = ᎏ HBA ᎏ ? ? CPCTC ᎏ ᎏ Conjecture proved in Exercise 1 3. (Lesson 5.7) RT WA 1 ? ᎏ ? ᎏ ? Given ᎏ AT WR 2 ? ? ᎏ ᎏ WRT TAW 4 ? Given ᎏ WT WT 3 70 ? ᎏ ? ᎏ ? Same segment ᎏ ANSWERS TO EXERCISES ? ᎏ ? ᎏ ? SSS ᎏ 2 5 1 ? ᎏ WA RT 6 ? CPCTC ᎏ 3 7 4 ? ᎏ ? CPCTC ᎏ ? ? ᎏ ᎏ ? ᎏ TA RW 8 by Converse of the Parallel Lines Conjecture 9 WATR is a parallelogram ? Definition of ? ? ᎏ ᎏ ᎏ parallelogram ? Converse of the ᎏ Parallel Lines Conjecture 7. 1 BEAR is a parallelogram Given 2 AB RE 3 BR EA Given 4 AR EB Parallelogram Opposite Sides Conjecture 5 Parallelogram Opposite Sides Conjecture EBR ARB RAE BEA SSS 6 EBR ARB RAE BEA CPCTC 7 BEAR is a rectangle Definition of rectangle 9 6 12 3 6 inches Answers to Exercises is parallel to ZT , corresponding 8. Because AR 3 and 2 are congruent. Opposite sides of parallelogram ZART are congruent so AR TZ. Because the trapezoid is isosceles, AR PT, and substituting gives ZT PT making PTZ isosceles and 1 and 2 congruent. By substitution, 1 and 3 are congruent. 9. See sample flowchart below. 10. If the fabric is pulled along the warp or the weft, nothing happens. However, if the fabric is pulled along the bias, it can be stretched because the rectangles are pulled into parallelograms. 11. 30° angles in 4-pointed star, 30° angles in 6-pointed star; yes 12. He should measure the alternate interior angles to see whether they’re congruent.If they are, the edges are parallel. : y 2x 3; QI : y 21x 2 13. ES 14. (12, 7) 1 15. 3 16. 9. (Lesson 5.7) 2 TH GR Given 1 Isosceles trapezoid GTHR 3 GT GT Same segment Given 4 5 RGT HTG SAS 6 TR GH CPCTC RGT HTG Isosceles Trapezoid Conjecture ANSWERS TO EXERCISES 71 distance between the two points is twice the length of the midsegment. 7. x 10°, y 40° 8. x 60 cm 9. a 116°, c 64° 10. 100 11. x 38 cm 12. y 34 cm, z 51 cm 13. See table below. 14. a 72°, b 108° 15. a 120°, b 60°, c 60°, d 120°, e 60°, f 30°, g 108°, m 24°, p 84°; Possible explanation: Because c 60°, the angle that forms a linear pair with e and its congruent adjacent angle measures 60°. So 60° 2e 180°, and e 60°. The triangle containing f has a 60° angle. The other angle is a right angle because it forms a linear pair with a right angle. So f 30° by the Triangle Sum Conjecture. Because g is an interior angle in an equiangular pentagon, divide 540° by 5 to get g 108°. 16. 15 stones 17. (1, 0) 18. When the swing is motionless, the seat, the bar at the top, and the chains form a rectangle. When you swing left to right, the rectangle changes to a parallelogram. The opposite sides stay equal in length, so they stay parallel. The seat and the bar at the top are also parallel to the ground. 19. a = 60°, b = 120° CHAPTER 5 REVIEW 1. 360° divided by the number of sides 2. Sample answers: Using an interior angle, set the interior angle measure formula equal to the angle and solve for n. Using an exterior angle, divide into 360°. Or find the interior angle measure and go from there. 3. Trace both sides of the ruler as shown at right. Answers to Exercises 4. Make a rhombus using the double-edged straightedge, and draw a diagonal connecting the angle vertex to the opposite vertex. 5. Sample answer: Measure the diagonals with string to see if they are congruent and bisect each other. 6. Sample answer: Draw a third point and connect it with each of the two points to form two sides of a triangle. Find the midpoints of the two sides and connect them to construct the midsegment. The 13. (Chapter 5 Review) Opposite sides are parallel Opposite sides are congruent Opposite angles are congruent Diagonals bisect each other Diagonals are perpendicular Diagonals are congruent Exactly one line of symmetry Exactly two lines of symmetry 72 ANSWERS TO EXERCISES Kite Isosceles trapezoid Parallelogram Rhombus Rectangle Square No No Yes Yes Yes Yes No No Yes Yes Yes Yes No No Yes Yes Yes Yes No No Yes Yes Yes Yes Yes No No Yes No Yes No Yes No No Yes Yes Yes Yes No No No No No No No Yes Yes No 20. 24. possible answers: Resultant vector R x F Y 900 km/h x F z R Y D z 50 km/h D Speed: 901.4 km/h. Direction: slightly west of north. Figure is approximate. E 21. R S 25. 26. 27. 28. 1x 2 1x 2 Q 20 sides 12 cm See flowchart below. possible answer: A B 4 22. possible answers: R R x y C Given: Parallelogram ABCD CD and AD CB Show: AB See flowchart below. L 23. N Answers to Exercises F L 2 D Y Y F 1 3 E x L z P 27. (Chapter 5 Review) 2 Definition of rhombus ? ᎏ 1 DENI is a rhombus ? Given ᎏ DI ? DE ᎏ 3 NI ? NE ᎏ ? Definition of ᎏ rhombus DN 4 ? DN ᎏ 2, 4 DEN DIN 5 ? ? ᎏ ᎏ ? SSS ᎏ 6 1 ? ᎏ ? 3 ᎏ ? CPCTC ᎏ 7 DN bisects IDE and INE ? Definition of ᎏ angle bisector ? Same segment ᎏ 28. (Chapter 5 Review) 2 AD CB 4 1 3 5 1 ABCD is a parallelogram Given BD BD Same segment 3 AB CD Definition of parallelogram 8 6 2 4 AIA AB CD CPCTC AIA Definition of parallelogram 7 ABD CDB ASA 9 AD CB CPCTC ANSWERS TO EXERCISES 73 Answers to Exercises CHAPTER 6 • CHAPTER 6 CHAPTER 6 • CHAPTER LESSON 6.1 1. 3. 5. 6. 50° 30° 76 in. possible answer: 2. 55° 4. 105° 11. Construct a line and label a point T on it. On the line, mark off two points, L and M, each on the same side of T at distances r and t from T, respectively. Construct circle L with radius r. Construct circle M with radius t. r L M T t Answers to Exercises Target 7. Possible answer: The perpendicular to the tangent passes through the center of the circle. Use the T-square to find two diameters of the Frisbee. The intersection of these two lines is the center. . Construct a line through point T 8. Construct OT . perpendicular to OT r T O , OY , and OZ . Construct tangents 9. Construct OX through points X, Y, and Z. Z O t Y X 10. Construct tangent circles M and N. Construct equilateral MNP. Construct circle P with radius s. M s s N P 74 ANSWERS TO EXERCISES 12. Start with the construction from Exercise 11. , mark off length TK so that TK s On line LM and K is on the opposite side of T from L and M. Construct circle K with radius s. r L M T t K s 13. Sample answer: If the three points do not lie on the same semicircle, the tangents form a circumscribed triangle. If two of the three points are on opposite sides of a diameter, the tangent lines to those two points are parallel. If the points lie on the same semicircle, they form a triangle outside the circle, with one side touching (called an exscribed triangle). 14. sample answer: internally tangent: wheels on a roller-coaster car in a loop, one bubble inside another; externally tangent: touching coins, a snowman, a computer mouse ball and its roller balls 15. Constructions will vary. 16. 360° 30° 90° 90° 150° and 150° % 41.6 360° 17. Angles A and B must be right angles, but this would make the sum of the angle measures in the quadrilateral shown greater than 360°. 18a. rhombus 18b. rectangle 18c. kite 18d. parallelogram 19. 78° 20. 9.7 km 21. x 55° 55° 180° and 40° y y 180°, so x y 70° 11 22. 2 1 LESSON 6.2 5 cm 5 cm 15. M(4, 3), N(4, 3), O(4, 3) 16. The center of the circle is the circumcenter of the triangle. Possible construction: 17. possible construction: O 18. 13.8 cm 19. They can draw two chords and locate the intersection of their perpendicular bisectors. The radius is just over 5 km. 20. See flowchart below. 20. (Lesson 6.2) 1 AB CD ? Given ᎏ 2 AO CO All radii of a circle are congruent 3 AOB COD 4 ? ᎏ ? ᎏ ? SSS Congruence ᎏ Conjecture 5 COD ? AOB ᎏ ? CPCTC ᎏ BO DO ? All radii of a circle are congruent ᎏ ANSWERS TO EXERCISES 75 Answers to Exercises 1. 165°, definition of measure of an arc 2. 84°, Chord Arcs Conj. 3. 70°, Chord Central Angles Conj. 4. 8 cm, Chord Distance to Center Conj. 68°; mB 34° (Because OBC is 5. mAC isosceles, mB mC, mB mC 68°, and therefore mB 34°.) 6. w 115°, x 115°, y 65°; Chord Arcs Conjecture 7. 20 cm, Perpendicular to a Chord Conj. 8. w 110°, x 48°, y 82°, z 120°; definition of arc measure 9. x 96°, Chord Arcs Conjecture; y 96°, Chord Central Angles Conjecture; z 42°, Isosceles Triangle Conjecture and Triangle Sum Conjecture. 10. x 66°, y 48°, z 66°; Corresponding Angles Conjecture, Isosceles Triangle Conjecture, Linear Pair Conjecture 11. The length of the chord is greater than the length of the diameter. 12. The perpendicular bisector of the segment does not pass through the center of the circle. 13. The longer chord is closer to the center; the longest chord, which is the diameter, passes through the center. 14. The central angle of the smaller circle is larger, because the chord is closer to the center. 1 21. y 7 x; (0, 0) is a point on this line. 22a. true C A X B Answers to Exercises D is the perpendicular Possible explanation: If AB , then every point on AB is bisector of CD equidistant from endpoints C and D. Therefore AD and BC BD . Because CD is not the AC , C is not equidistant perpendicular bisector of AB from A and B. Likewise, D is not equidistant from and BC are not congruent, and A and B. So, AC and BD are not congruent. Thus ACBD AD has exactly two pairs of consecutive congruent sides, so it is a kite. 22b. false, isosceles trapezoid 22c. false, rectangle 23. 140°, 160°, 60°; 180° minus the measure of the intercepted arc 76 ANSWERS TO EXERCISES 24. Using the Tangent Conjecture, 2 and 4 are right angles, so m2 m4 180°. According to the Quadrilateral Sum Conjecture, m1 m2 m3 m4 360°, so by the Subtraction Property of Equality, m1 m3 180°, or m1 180° m3. The measure of a central angle equals the measure of . its arc, so by substitution, m1 180° mAB 25. the circumcenter of the triangle formed by the three light switches 26. 7 27. Station Beta is closer. Downed aircraft N 4.6 mi 48 7.2 mi 72 38 Alpha 28. D Beta 312 8.2 mi LESSON 6.3 22. a 108°; b 72°; c 36°; d 108°; e 108°; f 72°; g 108°; h 90°; l 36°; m 18°; n 54°; p 36° 23. (2, 1); Possible method: Plot the three points. Construct the midpoint and the perpendicular bisector of the segments connecting two different pairs of points. The center is the point of intersection of the two lines. To check, construct the circle through the three given points. 24. See flowchart below. 25. Start with an equilateral triangle whose vertices are the centers of the three congruent circles. Then locate the incenter/circumcenter/orthocenter/ centroid (all the same point because the triangle is equilateral) to find the center of the larger circle. To find the radius, construct a segment from the incenter of the triangle through the vertex of the triangle to a point on the circle. It works on acute and right triangles. 20. The camera can be placed anywhere on the major arc (measuring 268°) of a circle such that the row of students is a chord intersecting the circle to form a minor arc measuring 92°. This illustrates the conjecture that inscribed angles that intercept the same arc are congruent (Inscribed Angles Intercepting Arcs Conjecture). 21. two congruent externally tangent circles with half the diameter of the original circle 26. mA 60°, mB 36°, mC 90°; 60° 36° 90° 180° 24. (Lesson 6.3) 1 OR CD 2 ORC and ORD are right angles ? Definition of ᎏ perpendicular lines 3 6 OCD is isosceles 7 C ? D ᎏ ? Given ᎏ 5 OC OD ? ᎏ All radii of a circle are congruent ? Definition of ᎏ isosceles triangle mORC ⫽ 90 mORD ⫽ 90 ? Definition of ᎏ right angle ? ᎏ 7. Isosceles Triangle Conjecture 4 mORC ⫽ mORD (ORC ORD) Substitution ? ᎏ property 8 OCR ? ᎏ ? ᎏ ODR SAA 9 CR ? DR ᎏ ? ᎏ CPCTC 10 OR bisects CD ? Definition of bisect ᎏ ANSWERS TO EXERCISES 77 Answers to Exercises 1. 65° 2. 30° 3. 70° 4. 50° 5. 140°, 42° 6. 90°, 100° 7. 50° 8. 148° 9. 44° 10. 142° 11. 120°, 60° 12. 140°, 111° 13. 71°, 41° 14. 180° 15. 75° 16. The two inscribed angles intercept the same arc, so they should be congruent. 17. BFE DFA (Vertical Angles Conjecture). BGD FHD (all right angles congruent). Therefore, B D (Third Angle Conjecture). , mD 1 mEC , AC EC mB 21 mAC 2 18. Possible answer: Place the corner so that it is an inscribed angle. Trace the inscribed angle. Use the side of the paper to construct the hypotenuse of the right triangle (which is the diameter). Repeat the process. The place where the two diameters intersect is the center. 19. possible answer: Answers to Exercises LESSON 6.4 mABD by the 1. Proof: mACD 21mAD Inscribed Angle Conjecture. ACD ABD. 2. Proof: By the Inscribed Angle Conjecture, mACB 21mAD B 21(180°) 90°. ACB is a right angle. 3. Proof: By the Inscribed Angle Conjecture, and mL 1mYCI 1(360° mC 21mYLI 2 2 1 mYLI ) 180° 2mYLI 180° mC. L and C are supplementary. (A similar proof can be used to show that I and Y are supplementary.) 2m2 4. Proof: 1 2 by AIA. mBC 2m1 mAD by the Inscribed Angle Conjecture. AD . BC 5. True. Opposite angles of a parallelogram are congruent. If it is inscribed in a circle, the opposite angles are also supplementary. So they are right angles, and the parallelogram is indeed equiangular, or a rectangle. and DB . Diagonal RG 6. Construct diagonals RG is the perpendicular bisector of BD by the Kite Diagonal Bisector Conjecture. Because kite BRDG and RG are chords of is inscribed in the circle, BD the circle. By the Perpendicular Bisector of a Chord Conjecture, the perpendicular bisector of BD passes through the center of the circle. Because RG is a chord of the circle that passes through the is a center, by the definition of diameter, RG diameter. , , and AR . Because they 7. Draw in radii GR ER , TR TR . are radii of the same circle, GR ER and AR By the Parallel Lines Intercepted Arcs Conjecture, ET . Their central angles must also be GA congruent, so GRA ERT. Thus GRA ET by CPCTC. GATE is ERT by SAS, so GA an isosceles trapezoid. 8. x 65°, y 40°, z 148°; half the measure of the intercepted arc 9. Because the radii of a circle are congruent, OB , so OAB is isosceles. By the Isosceles OA Triangle Conjecture, 2 3, so m2 m3. By the Triangle Sum Conjecture, m1 m2 m3 180°, so by substitution, m1 2(m2) 180°, or m1 180° 2(m2). 78 ANSWERS TO EXERCISES , so BC By the Tangent Conjecture, OB mOBC 90°. By Angle Addition, m2 m4 mOBC, so m2 m4 90°, or m2 90° m4. Substitute 90° m4 for m2 in the earlier equation: m1 180° 2(90° m4). Simplifying gives m1 2(m4). m1, substitution gives mAB Because mAB . 2(m4), or m4 21mAB 10a. S; An equilateral triangle is equiangular, but a rhombus is not equiangular. 10b. A 10c. N; Only one diagonal is the perpendicular bisector of the other. 10d. A 10e. S; An equilateral triangle has rotational symmetry and three lines of symmetry; a parallelogram has rotational symmetry but no line of symmetry. 11. L 12. J 13. K 14. D 15. E 16. B 17. H 18. G 19. N 20. a b b a 180°, so a b 90° 2 21. 2 1 RS . OP OQ OR 22. It is given that PQ because all radii in a circle are congruent. OS Therefore, OPQ ORS by SSS and 2 1 by CPCTC. Now we move on to the smaller right triangles inside OPQ and ORS. It is given that PQ and OV RS . Therefore, OTQ and OT OVS are right angles by the definition of perpendicular lines and OTQ OVS because all right angles are congruent. Thus, OTQ OV by CPCTC. OVS by SAA and OT LESSON 6.5 1. d 5 cm 2. C 10 cm 12 3. r m 11 4. C 5.5 or 2 18. (Lesson 6.5) 1 2 ? MA MT ᎏ Given ? SA ST ᎏ Given 3 MAST is a kite ? Definition of kite ᎏ is the perpendicular bisector of AT MS 4 ? ᎏ ? Kite Diagonal Bisector Conjecture ᎏ ANSWERS TO EXERCISES 79 Answers to Exercises 5. C 12 cm 6. d 46 m 7. C 15.7 cm 8. C 25.1 cm 9. r 7.0 m 10. C 84.8 in. 11. 565 ft 12. C 6 cm 13. 16 in. 14. Trees grow more in years with more rain; 244 yr 15. 1399 tiles 16. g 40°, n 30°, x 70°; y 142°; z 110°; Conjecture: The measure of the angle formed by two intersecting chords is one-half the sum of the measures of the two intercepted arcs. In the mLG . diagrams, mAEN 21mAN and mLNG 1mLG 17. mAGN 21mAN 2 by the Inscribed Angle Conjecture. mAEN mAGN mLNG by the Triangle Exterior mLG Angle Conjecture. So, mAEN 21mAN by substitution and the distributive property. 18. See flowchart below. 19. b 90°, c 42°, d 70°, e 48°, f 132°, g 52° 5 150° 20. 360° or 12 21. The base angles of the isosceles triangle have a measure of 39°. Because the corresponding angles are congruent, m is parallel to n. 22. 10x 2y LESSON 6.6 Answers to Exercises 1. 4398 km/h 2. 11 m/s 3. 37,000,000 revolutions 4. 637 revolutions 5. Mama; C 50 in. 6. 168 cm 7. d 7.6 ft. The table will fit, but the chairs may be a little tight in a 12-by-14 ft room. 12 chairs 192 in., 12 spaces 96 in., C 288 in., d 91.7 in. 7.6 ft. 8. 0.35 ft/s 9. a 37.5°, s 17.5°, x 20°; y 80°; z 61°; Conjecture: The measure of an angle formed by two secants that intersect outside a circle is one-half the difference of the larger arc measure and the smaller arc measure. 80 ANSWERS TO EXERCISES and mSAN 1mSN by 10. mESA 21mEA 2 the Inscribed Angle Conjecture. mSAN mESA mECA by the Triangle Exterior Angle 1mEA mECA by Conjecture. So, 21mSN 2 substitution. With a little more algebra, mECA 1 mSN mEA . 2 11. Both triangles are isosceles, so the base angles in each triangle are congruent. But one of each base angle is part of a vertical pair. So, a b by the Vertical Angles Conjecture and transitivity. 12. C 13. 38° 14. 48° 15. 30 cm 400 rev 2 . 2 6 ft 1 min 16. 1 min 1 rev 60 s 1089 ft/s 17. 12 cm third side 60 cm. This is based on the Triangle Inequality Conjecture. USING YOUR ALGEBRA SKILLS 6 50 40 Cost ($) 1 1. 2, 3 2. (3, 3) 2 3. 5, 3 4. (7, 2) 5. infinitely many solutions 6. no solution 7. 4. The lines intersect at the solution point, (7, 2). The circumcenter is 272 , 178 . Only two perpendicular bisectors are needed because the third bisector intersects at the same point. 9a. Plan A: y 4x 20; Plan B: y 7x; x 6 h 40 min, y $46.67 y 9b. y y 2x 16 5 30 Plan B 20 (0, 20) 10 2 5 ( 7, 2) x 10 4 6 Hours x The point of intersection shows when both plans cost the same, the answer for 9a. 9c. Plan B; 721 hours, with Plan A 10. (3, 1), (0, 3), (3, 1) y 12 x 32 –5 –10 y y 13 x 2 x 6. The lines are parallel (the slopes are the same, but the y-intercepts are different). y There is no solution. y = –2x + 9 5 x –5 –5 y = –2x – 1 –10 8. Possible solution using the equations y 21x 1 and y 32x 134 : 1 y 2x 1 2 14 1 3x 3 2x 1 4x 28 3x 6 22 7x 22 x 7 1 22 y 2 7 1 11a. y 43x 23 y x 12 11b. (6, 6) 11c. (6, 6); (6, 6) 11d. The diagonals intersect at their midpoints, which supports the conjecture that the diagonals of a parallelogram bisect each other. 2 13 12. y 3x 3 13. (4, 7) 3 15. 3, 2 69 6 14. 1, 4 7 16. The circumcenter is the midpoint of the hypotenuse. For a right triangle, the perpendicular bisectors of the legs are two of the midsegments of the triangle. As with any pair of triangle midsegments, they intersect at the midpoint of the third side. 18 y 7 ANSWERS TO EXERCISES 81 Answers to Exercises 5. The lines are the same. There are infinitely many solutions. 4 6 _23 , 46 _23 Plan A Answers to Exercises LESSON 6.7 4 1. 3 in. 2. 8 m 3. 14 cm 4. 9 m 5. 6 ft 6. 4 m 7. 27 in. 8. 100 cm 9. 217 m/min 10. 4200 mi 11. Desks are about 17 meters from the center. About four desks will fit because an arc with one-half the radius and the same central angle will be one-half as long as the outer arc. 12. The measure of the central angle is 7.2° because of the Corresponding Angles Conjecture. Therefore, 500 376.20 C, so C 25,000 mi. 13. 18°/s. No, the angular velocity is measured in degrees per second not in distance per second, so it is the same at every point on the carousel. 14. Outer horse 2.5 m/s, inner horse 1.9 m/s. One horse has traveled farther in the same amount of time (tangential velocity), but both horses have rotated the same number of times (angular velocity). 15. a 50°, b 75°, x 25°; y 45°; z 35°; Conjecture: The measure of the angle formed by an intersecting tangent and secant to a circle is one-half the difference of the larger intercepted arc measure and the smaller intercepted arc measure. 16. Let z represent the measure of the exterior . and tangent PB angle of PBA formed by AB 1 By the Tangent Chord Conjecture, z 2mAB . By . the Inscribed Angle Conjecture, mBAP 21mBC By the Triangle Exterior Angle Conjecture, z mBPA mBAP, or mBPA z mBAP. 82 ANSWERS TO EXERCISES 1mBC . By substitution, mBPA 21mAB 2 Using the distributive property, mBPA 1 (mAB mBC ). 2 17. a 70°, b 110°, c 110°, d 70°, e 20°, f 20°, g 90°, h 70°, k 20°, m 20°, n 20°, p 140°, r 80°, s 100°, t 80°, u 120° 18. possible answer: 19. 170° 8 289 20. y 1x 5 15 21. 45° 22. The sum of the lengths is 8 cm for Case 1, Case 2, Case 3, and Case 10. 23. 6 24. Yes, as long as the three points are noncollinear; possible answer: connect the points with segments, then find the point of concurrency of the perpendicular bisectors (same as circumcenter construction). CHAPTER 6 REVIEW 23. Sample answer: Construct a right angle and and RT with any label the vertex R. Mark off RE lengths. From point E, swing an arc with radius RT. From point T, swing an arc with radius RE. Label the intersection of the arcs as C. Construct the and RC . Their intersection is the diagonals ET center of the circumscribed circle. The circle’s radius is the distance from the center to a vertex. It is not possible to construct an inscribed circle in a rectangle unless it is a square. E C R T 24. Sample answer: Construct acute angle R. Mark off equal lengths RM and RH. From points M and H, swing arcs of lengths equal to RM. Label the intersection of the arcs as O. Construct RHOM. The intersection of the diagonals is the center of the inscribed circle. Construct a perpendicular to a side to find the radius. It is not possible to construct a circumscribed circle unless the rhombus is a square. H O R M 4 32 25. 4x 3y 32, or y 3x 3 26. (3, 2) 27. d 0.318 m 28. Melanie: 151 m/min or 9 km/h; Melody: 94 m/min or 6 km/h. 2(6357) 2(6378) 29. 1.849 1.852 1.855 360 60 360 60 30. Possible location 7700 ft 5280 ft 5500 ft D T Possible location ANSWERS TO EXERCISES 83 Answers to Exercises 1. Answers will vary. 2. Draw two nonparallel chords. The intersection of their perpendicular bisectors is the center of the circle. Fold the paper so that two semicircles coincide. Repeat with two different semicircles. The center is the intersection of the two folds. Place the outside or inside corner of the L in the circle so that it is an inscribed right angle. Trace the sides of the corner. Draw the hypotenuse of the right triangle (which is the diameter of the circle). Repeat. The center is the intersection of the two diameters. 3. The velocity vector is always perpendicular to the radius at the point of tangency to the object’s circular path. 4. Sample answer: An arc measure is between 0° and 360°.An arc length is proportional to arc measure and depends on the radius of the circle. 5. 55° 6. 65° 7. 128° 8. 118° 9. 91° 10. 66° 11. 125.7 cm 12. 42.0 cm 13. 15 cm 14. 14 ft 15. 2 57° 2 35° 180° 16. 84° 56° 56° 158° 360° 1 (180° 108°) 36° 17. mEKL 21 mEL 2 YL by Converse of the Parallel mKLY. KE Lines Conjecture. 360° 56° 152° 152° mMI . 18. mJI 1 mMI mMJI. By the mJMI 21 mJI 2 Converse of the Isosceles Triangle Conjecture, JIM is isosceles. 2mKEM 140°. mKI 19. mKIM 140° 70° 70° mMI . mIKM 1 1 mMI mKI mIMK. By the Converse of the 2 2 Isosceles Triangle Conjecture, KIM is isosceles. 20. Ertha can trace the incomplete circle on paper. She can lay the corner of the pad on the circle to trace an inscribed right angle. Then Ertha should mark the endpoints of the intercepted arc and use the pad to construct the hypotenuse of the right triangle, which is the diameter of the circle. 21. Sample answer: Construct perpendicular bisectors of two sides of the triangle. The point at which they intersect (the circumcenter) is the center of the circle. The distance from the circumcenter to each vertex is the radius. 22. Sample answer: Construct the incenter (from the angle bisectors) of the triangle. From the incenter, which is the center of the circle, construct a perpendicular to a side. The distance from the incenter to the foot of the perpendicular is the radius. 200 31. ft 63.7 ft 32. 8 m 25.1 m 8 33. The circumference is 346 0 2(45) 12; the diameter is 12 cm. 34. False. 20° 20° 140° 180°. An angle with measure 140° is obtuse. 35. true D 36. false C Answers to Exercises A 53. False. The ratio of the circumference to the diameter is . 24 cm 24 cm 54. false; 24 24 48 48 96 48 cm 55. true 56. This is a paradox. 57. a 58°, b 61°, c 58°, d 122°, e 58°, f 64°, g 116°, h 52°, i 64°, k 64°, l 105°, m 105°, n 105°, p 75°, q 116°, r 90°, s 58°, t 122°, u 105°, v 75°, w 61°, x 29°, y 151° 58. TAR YRA by SAS, TAE YR E by SAA 59. FTO YTO by SAA, SAS, or SSS; FLO YLO by SAA, SAS, or SSS; FTL YTL by SSS, SAS, or ASA 60. PTR ART by SSS or SAS; TPA RAP by SSS, SAS, SAA, or ASA; TLP RLA by SAA or ASA 61. ASA 84°, length of AC 11.2 35.2 in. 62. mAC 63. x 63°, y 27°, w 126° 64. sample answer: B 37. true 38. true 39. true 40. true 41. False. (7 2) 180° 900°. It could have seven sides. 42. False. The sum of the measures of any triangle is 180°. 43. False. The sum of the measures of one set of exterior angles for any polygon is 360°. The sum of the measures of the interior angles of a triangle is 180° and of a quadrilateral is 360°. Neither is greater than 360°, so these are two counterexamples. 44. False. The consecutive angles between the bases are supplementary. 45. False. 48° 48° 132° 180° 46. False. Inscribed angles that intercept the same arc are congruent. 47. False. The measure of an inscribed angle is half the measure of the arc. 48. true and BD bisect each other, but AC is 49. False. AC . not perpendicular to BD D C A B 48 cm 30 150 65. See table below. 66a. The circle with its contents has 3-fold rotational symmetry, the entire tile does not. 66b. No, it does not have reflectional symmetry. 67. 68. 9.375 cm 50. False. It could be isosceles. 51. False. 100° 100° 100° 60° 360° CD 52. false; AB C A 69. 90 90 70. D B 65. (Chapter 6 Review) 84 n 1 2 3 4 5 6 f (n) 5 1 3 7 11 15 ANSWERS TO EXERCISES ... n ... 20 . . . 9 4n . . . 71 Answers to Exercises CHAPTER 7 • CHAPTER 7 11. CHAPTER 7 • CHAPTER LESSON 7.1 1. Rigid; reflected, but the size and shape do not change. 2. Nonrigid; the shape changes. 3. Nonrigid; the size changes. 4. 5. P P ᐉ 6. , 7. possible answer: a boat moving across the water 8. possible answer: a Ferris wheel 9a. Sample answer: Fold the paper so that the images coincide, and crease. or 18. 19. P(a, b), Q(a, b), R(a, b) 20. possible construction: P 9b. Construct a segment that connects two corresponding points. Construct the perpendicular bisector of that segment. 10a. Extend the three horizontal segments onto the other side of the reflection line. Use your compass to measure lengths of segments and distances from the reflection line. 10b. 21. 50th figure: 154 (50 shaded, 104 unshaded); nth figure: 3n 4 (n shaded, 2(n 2) unshaded) 22. 46 23. It is given that 1 2, and 2 3 because of the Vertical Angles Conjecture, so 1 3. Segment DC is congruent to itself. DCE and DCB are both right angles, so they are congruent. Therefore, DCB DCE by CE by CPCTC. ASA, and BC 16. (Lesson 7.1) Number of sides of regular polygon 3 4 5 6 7 8 ... n Number of reflectional symmetries 3 4 5 6 7 8 ... n Number of rotational symmetries ( 360°) 3 4 5 6 7 8 ... n ANSWERS TO EXERCISES 85 Answers to Exercises 12. reflectional symmetry 13. 4-fold rotational and reflectional symmetry 14. reflectional symmetry 15. 7-fold symmetry: possible answers are F or J. 9-fold symmetry: possible answers are E or H. Basket K has 3-fold rotational symmetry but not reflectional symmetry. 16. See table below; n, n 17. LESSON 7.2 1. y translation 5 x 5 6. Rules that involve x or y changing signs, or switching places, produce reflections. If both x and y change signs, the rule produces a rotation. Rules that produce translations involve a constant being added to the x and/or y terms. 5, 0 is the translation vector for Exercise 1. 7. (x, y) → (x, y) 8. (x, y) → (x, y) 9. N 2. reflection y Cue ball 8 W E 8 ball x –5 Answers to Exercises S 10. There are two possible points, one on the N wall and one on the W wall. –8 3. reflection y N 5 H W –4 x 7 T y S 11. –5 4. E H'' reflection N 5 W E T x 5 H 12. by the Minimal Path Conjecture Proposed freeway 5. rotation y 5 5 Mason x Perry 13. 14. 86 H' S ANSWERS TO EXERCISES , 15. possible answer: HIKED 16. one, unless it is equilateral, in which case it has three 17. two, unless it is a square, in which case it has four 18. 19. sample construction: 20. sample construction: 21. false; possible counterexample: trapezoid with two right angles 22. false; possible counterexample: isosceles trapezoid Answers to Exercises ANSWERS TO EXERCISES 87 LESSON 7.3 1. 10, 10 2. A 180° rotation. If the centers of rotation differ, rotate 180° and add a translation. 3a. 20 cm 3b. 20 cm, but in the opposite direction 4a. 80° counterclockwise 4b. 80° clockwise 5. 180° 6. 3 cm 7. possible answer: 11. Answers may vary. Possible answer: reflection across the figure’s horizontal axis and 60° clockwise rotation. 12. 13. , 14. Sample answer: Draw a figure on an overhead transparency and then project the image onto a screen. 15. possible answers: rotational: playing card, ceiling fan, propeller blade; reflectional: human body, backpack 16. one: yes; two: no; three: yes ′O ′A N ′′ A H ′ A′ H ′′ O N ′′ ′H O Answers to Exercises ′N 8. possible answer: Center of rotation 17. possible answer: A 9. 18a. 18b. a b d e 10. Two reflections across intersecting lines yield a rotation. The measure of the angle of rotation is twice the measure of the angle between the lines of reflection, or twice 90°, or 180°. 88 ANSWERS TO EXERCISES O 5 12 B 9 0 12 7 14 11 ? ? 11 0 13 ? ? 20 c a f d 2a 2b d–e 3c 0 3b 4c ? ? ? ? ? ? 0 d f LESSON 7.4 1. Answers will vary. 2. Answers will vary. 3. 33.42 4. 34.6 5. 32.4.3.4 6. 3.4.6.43.42.6 7. 33.4232.4.3.4 8. 3632.4.12 9a. The dual of a square tessellation is a square tessellation. 9b. The dual of a hexagon tessellation is a triangle tessellation. 9c. If tessellation A is the dual of tessellation B, then tessellation B is the dual of tessellation A. 10. The dual is a 34 38 tessellation of isosceles right triangles. 11. 15. Answers will vary. 1 16. y 2x 4 y 4 –3 5 x –6 17. possible answer: TOT 18. 12. Answers to Exercises N 8-ball W E Cue ball 13. A ring of ten pentagons fits around a decagon, and another decagon can fit into any two of the pentagons. But another ring of pentagons around the second decagon doesn’t leave room for a third decagon. 14. S ANSWERS TO EXERCISES 89 LESSON 7.5 1. Answers will vary. 2. The dual is a 5354 tessellation. By the Triangle Sum Conjecture, a b c 180°. Around each point, we have 2(a b c) 2 180° 360°. Therefore, a triangle will fill the plane edge to edge without gaps or overlaps. Thus, a triangle can be used to create a monohedral tiling. 6. three ways 7. 8. y 2x 3 3. y Answers to Exercises 8 4. Yes. The four angles of the quadrilateral will be around each point of intersection in the tessellation. a a c c 5. b b b a c c a b a 90 b a c b c b a a c b c b a ANSWERS TO EXERCISES c a c b 5 –2 x LESSON 7.6 1. 2. 3. 4. 5. 6. 7. Answers will vary. Answers will vary. Answers will vary. regular hexagons squares or parallelograms squares or parallelograms 2 12. y 3x 3; the slope is the opposite sign. y 5 –10 10 x 13. 3.4.6.4 4.6.12 440 rev 2 28 ft 1 min 1290 ft/s 14. 1 min 60 s 1 rev 8. 9. Answers will vary. 10. Answers will vary. 11. B E A S ANSWERS TO EXERCISES 91 Answers to Exercises 15. Possible explanations: 15a. true; The kite diagonal between vertex angles is the perpendicular bisector of the other diagonal; in a square, diagonals would bisect each other 15b. False; it could be an isosceles trapezoid. 15c. False; it could be a rectangle. 15d. true; Parallel lines cut off congruent arcs of a circle, so inscribed angles (the base angles of the trapezoid) are congruent. LESSON 7.7 1. equilateral triangles. 2. regular hexagons. 3. Answers to Exercises 4. 9. true 10. true 11. False; it could be a kite or an isosceles trapezoid. 12. The path would be 14 of Earth’s circumference, approximately 6280 miles, which will take 126 hours, or around 5 14 days. 13a. Using the Reflection Line Conjecture, the line of reflection is the perpendicular bisector of and BB . Because these segments are both AA perpendicular to the reflection line, they are is parallel to parallel to each other. Note that if AB the reflection line, quadrilateral AABB will be a rectangle instead of a trapezoid. 13b. Yes; it has reflectional symmetry, so legs and base angles are congruent. 13c. greatest: near each of the acute vertices; least: at the intersection of the diagonals (where A, C, and B become collinear and A, C, and B become collinear) 14a. 5. Answers will vary. 6. Answers will vary. 7. sample design: 8. False; they must bisect each other in a parallelogram. 92 ANSWERS TO EXERCISES 14b. 31 5 6 0 4 128 108 28 15 0 ? ? 13? 3 5 9 8 7 ? ? 6 0 ? ? 9 2 2 ? 1 10 29 30 50 LESSON 7.8 1. parallelograms 2. parallelograms 3. 5. Answers will vary. 6. Answers will vary. 7. Circumcenter is (3, 4); orthocenter is (10, 8). 8. 9. 4. 10. Answers to Exercises ANSWERS TO EXERCISES 93 USING YOUR ALGEBRA SKILLS 7 1 1. y 6x Answers to Exercises 2. y 2x 2 3. Centroid is 2, 23; orthocenter is (0, 5). 94 ANSWERS TO EXERCISES 4. Centroid is (4, 0); orthocenter is (3, 0). 4 5. 1, 3 6. (1, 1) 7. (5, 8) 22. CHAPTER 7 REVIEW T H 23. Use a grid of squares. Tessellate by translation. 24. Use a grid of equilateral triangles. Tessellate by rotation. 25. Use a grid of parallelograms. Tessellate by glide reflection. 26. Yes. It is a glide reflection for one pair of sides and midpoint rotation for the other two sides. Answers to Exercises 1. true 2. true 3. true 4. true 5. true 6. true 7. False; a regular pentagon does not create a monohedral tessellation and a regular hexagon does. 8. true 9. true 10. False; two counterexamples are given in Lesson 7.5. 11. False; any hexagon with one pair of opposite sides parallel and congruent will create a monohedral tessellation. 12. This statement can be both true and false. 13. 6-fold rotational symmetry 14. translational symmetry 15. Reflectional; color arrangements will vary, but the white candle must be in the middle. 16. The two towers are not the reflection (or even the translation) of each other. Each tower individually has bilateral symmetry. The center portion has bilateral symmetry. 17. Answers will vary. 18. Answers will vary. 19. 3632.4.3.4; 2-uniform 20. 4.82; semiregular 1 21. y 2x 27. No.Because the shape is suitable for glide reflection,the rows of parallelograms should alternate the direction in which they lean (row 1 leans right,row 2 leans left,row 3 leans right,and so on). 28. y x ANSWERS TO EXERCISES 95 Answers to Exercises CHAPTER 8 • CHAPTER 8 CHAPTER 8 • CHAPTER LESSON 8.1 1. 228 m 2 2. 41.85 cm 2 3. 8 yd 4. 21 cm 5. 91 ft 2 6. 182 m 2 7. 96 in2 8. 210 cm 9. A 42 ft 2 10. sample answer: 6 cm 12 cm 48 cm2 Answers to Exercises 4 cm 8 cm 11. 12. 13. 14. 48 cm2 3 square units 10 square units 712 square units sample answers: 8 cm 16 cm 64 cm2 23. 500 cm2 24a. smallest: 191.88 cm2; largest: 194.68 cm2 24b. Answers will vary. Sample answer: about 193 cm2. 24c. Answers will vary. The smallest and largest area values differ at the ones place, so the digits after the decimal point are insignificant compared to the effect of the limit of precision in the measurements. 25a. In one Ohio Star block, the sum of the red patches is 36 in2, the sum of the blue patches is 72 in2, and the yellow patch is 36 in2. 25b. 42 25c. About 1814 in2 of red fabric, about 3629 in2 of blue fabric, and about 1814 in2 of yellow fabric. The border requires 5580 in2 (if it does not need the extra 20%). 26. 100; 36 64. The area of the square on the longer side is the same as the sum of the areas on the other two legs. 27. a 76°,b 52°,c 104°,d 52°,e 76°, f 47°, g 90°, h 43°, k 104°, m 86°. Explanations will vary. 28. sample construction: 64 cm2 4 cm 8 cm 15. possible answer: 16 cm 16 cm 4 cm A 30° M 30° 16 cm 16 cm 16. 23.1 m2 17. 2(4)(3) 2(5.5)(3) 57 m2 18. For a constant perimeter, area is maximized by a square. 100 m 4 25 m per side; A 625 m 2. 1 19. 530 20. 112 21. 96 square units 22. 32 square units 96 ANSWERS TO EXERCISES 29a. 29c. 29b. LESSON 8.2 1. 20 cm2 2. 49.5 m2 3. 300 square units 4. 60 cm2 5. 6 cm 6. 9 ft 7. 30 ft 8. 5 cm 9. 16 m 10. 168 cm 11. 12 cm 12. 3.6 ft; 10.8 ft 13. sample answer: .) 17. 12(To see why, draw altitude PQ 18. more than half, because the top card completely covers one corner of the bottom card 19a. 86 in. of balsa wood and 960 in2 of Mylar 19b. 56 in. (or less, if he tilts the kite) 20. 3600 shingles (to cover an area of 900 ft2) 21. The isosceles triangle is a right triangle because the angles on either side of the right angle are complementary. If you use the trapezoid area formula, the area of the trapezoid is 1 (a b)(a b). If you add the areas of the 2 three triangles, the area of the trapezoid is 1 c 2 ab. 2 22. A b1 B h D 9 cm 12 cm 9 cm 14. sample answer: 4 cm 7 cm 8 cm 7 cm 10 cm 9 cm 15. sample answer: 46 cm 12 cm 45 cm 35 cm 12 cm 56 cm 16. The length of the base of the triangle equals the sum of the lengths of both bases of the trapezoid. C Given: trapezoid ABCD with height h. area of ABD 21 hb1; area of BCD 21hb2; area of trapezoid sum of areas of two triangles 21hb1 b2 23. 1141 square units 24. 7 square units 25. 70 m 26. 144 cm2 27. 828 ft 2; 144 ft 28. 1440 cm2; 220 cm 29a. incenter 29b. orthocenter 29c. centroid 30. a 34°, b 68°, c 68°, d 56°, e 56°, f 90°, g 34°, h 56°, m 56°, n 90°, p 34°. Possible explanation: Let O be the center 112° by the Inscribed Angle of the circle. mBC by the Central Conjecture, and d e mBC Angle Conjecture. OBA is congruent to OCA is a semicircle, so by SSS, so d e 56°. DEC mDE 68°. By the Inscribed Angle Conjecture, p 34°. Using OEC and the Triangle Sum Conjecture, n 90°. 31. 32.623.6.3.6 2 cm 3 cm 7 cm 3 cm 9 cm ANSWERS TO EXERCISES 97 Answers to Exercises 12 cm b2 LESSON 8.3 Answers to Exercises 1a. 121,952 ft 2 1b. 244 gal of base paint and 488 gal of finishing paint 2. He should buy at least four rolls of wallpaper. (The area of each roll is 125 ft2. The total surface area to be papered is 480 ft2.) If paper cut off at the corners is wasted, he’ll need 5 rolls. 3. 1552 ft2; 776 ft 2 more surface area 4. 21 5. 336 ft2; $1780 98 ANSWERS TO EXERCISES 6. $760 7. 220 terra cotta tiles, 1107 blue tiles; $1598.15 8. 72 cm 2 9. AB 16.5 cm, BD 15.3 cm 10. 60 cm2 by either method 11. Because AOB is isosceles, mA 20° and 82°. mAOB 140°. mA A B 140° and mCD mBD because parallel lines intercept mAC 360° 140° 82° 69°. congruent arcs on a circle. 2 12. E 11. a2 2ab b2 USING YOUR ALGEBRA SKILLS 8 1. x 2 6x 5 a b a a2 ab a b ab b2 b 2. 2x 2 7x x5 x x 12. a2 b2 b a2 ab ab b 2 5 x x2 x1 x a 13. (x 15)(x 4) 1 2x 7 x x 15 x x2 15x 4 4x 60 14. (x 12)(x 2) x 12 x x2 12x 2 2x 24 7 15. (x 5)(x 4) 3. 6x 2 19x 10 3x 2 x x x x 5 x2 5x 4x 20 2 x x 4 2x 5 16. (x 3)2 (x 3)(x 3) x x 4. (3)(2x 1) 5. (x 5)(x 3) x5 3 x 3 3 x2 3x 3x 9 Answers to Exercises 5 x 17. (x 6)(x 6) 5 x 6 x2 6x 6x 36 x 2x 1 x3 x2 x x 3 x 6 18. (2x 7)(2x 7) 1 6. (2x 3)(x 4) x x4 2x 3 x 3 x 4 7. x 2 26x 165 x 15 x2 15x 11x 165 x 11 9. x 2 8x 16 x 4 x x2 4x 4 4x 16 8. 12x 2 13x 35 3x 7 4x 12x 2 28x 5 15x 35 10. 4x 2 25 2x 5 2x 4x 2 10x 5 10x 25 2x 7 2x 4x 2 14x 7 14x 49 19. x 4 or x 1 20. x 10 or x 3 21. x 3 or x 8 1 22. x 2 or x 4 h 23a and b. h h4 1 23c. 2[h (h 4)]h 48 23d. h 8 or h 6. The height cannot be negative, so the only valid solution is h 6. The height is 6 feet, one base is 6 feet, and the other base is 10 feet. ANSWERS TO EXERCISES 99 LESSON 8.4 2092 cm2 2. 74 cm 256 cm 4. 33 cm2 63 cm 6. 490 cm2 57.6 m 8. 25 ft 42 cm2 10. 58 cm2 1 1 1 1 11. a 2s; A 2asn 2 2s s 4 s 2 12. It is impossible to increase its area, because a regular pentagon maximizes the area. Any dragging of the vertices decreases the area. (Subsequent dragging to space them out more evenly can increase the area again, but never beyond that of the regular pentagon.) 13. 996 cm2 14. 497 cm2 15. total surface area 13,680 in2 95 ft2; cost $8075 16. Area is 20 square units. Answers to Exercises 1. 3. 5. 7. 9. 17. Area is 36 square units. y y – _4 x 12 3 y – _1 x 6 3 (6, 4) x 18. Conjecture: The three medians of a triangle divide the triangle into six triangles of equal area. Argument: Triangles 1 and 2 have equal area because they have equal bases and the same height. Because the centroid divides each median into thirds, you can show that the height of triangles 1 and 2 is 31 the height of the whole triangle. Each has an area 61 the area of the whole triangle. By the same argument, the other small triangles also have areas 61 the area of the whole triangle. y y _12x 5 1 (2, 6) y 2x 10 x 100 ANSWERS TO EXERCISES h 2 19. nw ny 2x 20. 504 cm2 21. 840 cm2 LESSON 8.5 1. 9 in2 2. 49 cm2 3. 0.8 m2 4. 3 cm 5. 3 in. 6. 0.5 m 7. 36 in2 8. 7846 m2 9. 25 48, or about 30.5 square units 10. 100 128, or about 186 square units 11. 16. A r 2 because the 100-gon almost completely fills the circle. 17. 456 cm2 18. 36 ft2 19. The triangles have equal area when the point is at the intersection of the two diagonals. There is no other location at which all four triangles have equal area. 2 24° 48° 20. x mDE 21. 90° 38° 28° 28° 180° 22. 24 cm r 18 cm 12 cm 804 m2 11,310 km2 154 m2 4 times 18 cm Answers to Exercises 12. 13. 14. 15. 6 cm ANSWERS TO EXERCISES 101 LESSON 8.6 Answers to Exercises 1. 6 cm2 64 2. 3 cm2 3. 192 cm2 4. ( 2) cm2 5. (48 32) cm2 6. 33 cm2 7. 21 cm2 105 8. 2 cm2 9. 6 cm 10. 7 cm 11. 75 12. 100 13. 42 14. $448 15a. 15b. 15c. 102 ANSWERS TO EXERCISES 15d. 16. sample answer: 17a. (144 36) cm2; 78.54% 17b. (144 36) cm2; 78.54% 17c. (144 36) cm2; 78.54% 17d. (144 36) cm2; 78.54% 18. 480 m2 19. AB 17.0 cm, AG 6.6 cm 90 20. True. If 24 360 2r, then r 48 cm. 360 21. True. If n 24, then n 15. 22. False. It could be a rhombus. 23. true; Triangle Inequality Conjecture LESSON 8.7 1. 150 cm2 2. 4070 cm2 3. 216 cm2 4. 340 cm2 5. 103.7 cm2 6. 1187.5 cm2 7. 1604.4 cm2 8. 1040 cm2 2 9. 414.7 cm 10. 329.1 cm2 11. area of square 4 area of trapezoid 4 area of triangle 12. $1570 13. sample answer: 15. a 75°, b 75°, c 30°, d 60°, e 150°, f 30° 16. About 23 days. Each sector is about 1.767 km2. 17. a 50°, b 50°, c 80°, d 100°, e 80°, f 100°, g 80°, h 80°, k 80°, m 20°, n 80°. Explanations will vary. Sample explanation: The angle with measure d corresponds to the angle forming a linear pair with g. Because d 100°, by the Parallel Lines Conjecture, the angle adjacent to g measures 100°, and by the Linear Pair Conjecture, g 80°. The angle with measure f corresponds to the angle measuring 100°, so f 100°. The angles measuring g and k are the base angles of an isosceles triangle, so by the Isosceles Triangle Conjecture, k 80°. 18. 398 square units Answers to Exercises 14. sample tiling 33.42/32.4.3.4/44 ANSWERS TO EXERCISES 103 CHAPTER 8 REVIEW 1. B (parallelogram) 3. C (trapezoid) 5. F (regular polygon) 7. J (sector) 9. G (cylinder) 11. 2. A (triangle) 4. E (kite) 6. D (circle) 8. I (annulus) 10. H (cone) 12. 20. 23. 26. 29. 32. 32 cm 21. 32 cm 81 cm2 24. 48 cm 153.9 cm2 27. 72 cm2 300 cm2 30. 940 cm2 Area is 112 square units. 22. 25. 28. 31. 15 cm 40° 30.9 cm2 1356 cm2 y Apothem D (6, 8) A (0, 0) C (20, 8) B (14, 0) x 33. Area is 81 square units. 13. y R (4, 15) U (9, 5) Answers to Exercises F (0, 0) 14. Sample answer: Construct an altitude from the vertex of an obtuse angle to the base. Cut off the right triangle and move it to the opposite side, forming a rectangle. Because the parallelogram’s area hasn’t changed, its area equals the area of the rectangle. Because the area of the rectangle is given by the formula A bh, the area of the parallelogram is also given by A bh. b O (4, –3) 34. 6 cm 35. 172.5 cm2 36. sample answers: b h h 15. Sample answer: Make a copy of the trapezoid and put the two copies together to form a parallelogram with base b1 b2 and height h. Thus the area of one trapezoid is given by the formula A 12 b1 b2h. b2 b1 h h b1 b2 16. Sample answer:Cut a circular region into a large number of wedges and arrange them into a shape that resembles a rectangle.The base length of this “rectangle”is r and the height is r, so its area is r 2. ThustheareaofacircleisgivenbytheformulaAr 2. r r 18. 5990.4 cm2 17. 800 cm2 19. 60 cm2 or about 188.5 cm2 104 x ANSWERS TO EXERCISES 37. 1250 m2 38. Circle. For the square, 100 4s, s 25, A 252 625 ft2. For the circle, 100 2r, r 15.9, A (15.9)2 794 ft2. 39. A round peg in a square hole is a better fit. The round peg fills about 78.5% of area of the square hole, whereas the square peg fills only about 63.7% of the area of the round hole. 40. giant 41. about 14 oz 42. One-eighth of a 12-inch diameter pie; one-fourth of a 6-inch pie and one-eighth of a 12-inch pie both have the same length of crust, which is longer than one-sixth of an 8-inch pie. 43a. 96 ft; 40 ft 43b. 3290 ft2 44. $3000 45. $4160 46. It’s a bad deal. 2r1 44 cm. 2r2 22 cm, which implies 4r2 44 cm. Therefore r1 2r2. The area of the large bundle is 4r22 cm2. The combined area of two small bundles is 2r22 cm2. Thus he is getting half as much for the same price. 47. $2002 48. $384 (16 gal) Answers to Exercises CHAPTER 9 • CHAPTER 9 CHAPTER 9 • CHAPTER LESSON 9.1 b a b a c 18. Sample answer: Yes, ABC XYZ by SSS. Both triangles are right triangles, so you can use the Pythagorean Theorem to find that CB ZY 3 cm. m 120° p q n Or use the Exterior Angle Conjecture to get q n 120°. By AIA, q m. Substituting, m n 120°. 22. a 122°, b 74°, c 106°, d 16°, e 90°, f 74°, g 74°, h 74°, n 74°, r 32°, s 74°, t 74°, u 32°, v 74°. Possible explanation: By the Tangent Conjecture, the quadrilateral containing g has two right angles formed by radii intersecting tangent lines. Using c 106° and the Quadrilateral Sum Conjecture, g 90° 90° 106° 360°, so g 74°. The angle measures e and f and the measure of the inscribed angle that intercepts the arc with measure u sum to 180°. Using e 90° and f 74°, the inscribed angle measures 16°. Using the Inscribed Angle Conjecture, u 32°. T1 23 . T2 ANSWERS TO EXERCISES 105 Answers to Exercises 1. c 19.2 cm 2. a 12 cm 3. b 5.3 cm 4. d 10 cm 5. s 26 cm 6. c 8.5 cm 7. b 24 cm 8. x 3.6 cm 9. x 40 cm 10. s 3.5 cm 11. r 13 cm 12. 127 ft 13. 512 m2 14. 11.3 cm 15. 3, 4, 5 16. 28 m 17. The area of the large square is 4 area of triangle area of small square. c 2 4 21ab (b a)2 c 2 2ab b 2 2ab a 2 c 2 a2 b2 19. 54 cm2 20. 3632.4.3.4 21. Mark the unnamed angles as shown in the figure below. By the Linear Pair Conjecture, p 120° 180°, so p 60°. By AIA, m q. By the Triangle Sum Conjecture, q p n 180°. Substitute m q and p 60° to get m 60° n 180°. m n 120°. Answers to Exercises LESSON 9.2 1. yes 2. yes 3. no 4. no 5. no 6. no 7. no 8. No, the given lengths are not a Pythagorean triple. 9. y 25 cm 10. y 24 units 11. y 17.3 m 12. 6, 8, 10 13. 60 cm2 14. 14.1 ft 15. 17.9 cm2 16. 102 m 17a. 1442 cm2 17b. 74.8 cm 18. Sample answer: The numbers given satisfy the Pythagorean Theorem, so the triangle is a right triangle; but the right angle should be inscribed in an arc of 180°. Thus the triangle is not a right triangle. 106 ANSWERS TO EXERCISES 19. Sample answer: (BD)2 62 32 27; (BC)2 (BD)2 92 108; then (AB)2 (BC)2 (AC)2 (36 108 144), so ABC is a right triangle by the Converse of the Pythagorean Theorem. 20. centroid 3 21. 1 0 22. Because mDCF 90°, mDCE (90 x)°. Because DCE is isosceles, mDEC (90 x)°. mD 180° 2 (90 x)° 2x. Because D is a central angle, 2x a. Therefore, x 21a. 23. The path from C to M to T lies on a straight line and therefore must be shorter than the path from C to A to T. P A T M C U 24. 790 square units 25. 26. 19 B USING YOUR ALGEBRA SKILLS 9 1. 6 2. 5 3. 18 2 4. 147 5. 8 6. 2 3 7. 3 2 8. 2 10 9. 5 3 10. 85 11. 4 6 12. 24 13. 12 5 14. 28 15. 6 23 16. Answers will vary. Possible answer: The length of the hypotenuse of an isosceles right triangle with legs of length 3 units is 18 units. This is the same as 2 2 2 , or 3 2. 18. Possible answer: A right triangle with legs of lengths 1 and 3 units has a hypotenuse of length units. 10 10 1 3 A right triangle with legs of lengths 6 and 2 units has a hypotenuse of length 40 10 10 10 . 2 40 2 6 19. Possible answer: A right triangle with legs of lengths 2 and 3 units has a hypotenuse of length 13 units. 1 2 1 18 3 13 3 3 5 2 12 . Because y is twice as 20. x 3, y long as x, y 2x, so 12 2 3. 2 2 2 21. 2 3 7 7 3 1 2 2 20 2 4 ANSWERS TO EXERCISES 107 Answers to Exercises 17. The hypotenuse represents 5 units. In the second right triangle, the legs have lengths 4 and 2, so the hypotenuse has length 20 . The length of the hypotenuse is twice the length of the hypotenuse of the smaller triangle, so 20 5. 2 LESSON 9.3 72 2 cm 13 cm 10 cm, 5 3 cm 10 3 cm, 10 cm 34 cm, 17 cm 72 cm 12 3 cm 50 cm, 100 cm 16 cm2 1 1 10. , 2 2 1. 2. 3. 4. 5. 6. 7. 8. 9. 30° x 3 2x 45° 11. A 30°-60°-90° triangle must have sides whose lengths are multiples of 1, 2, and 3 . The triangle shown does not reflect this rule. 12. Possible answer: Use 12 3 2 22 and 42 48 2 82. Answers to Exercises 16. c 2 x 2 x 2 Start with the Pythagorean Theorem. c 2 2x 2 Combine like terms. c x 2 Take the square root of both sides. 17. 169 3 m2 292.7 m2 18. 390 m2 19. 3 1 2 3 x 2 x 15° 45° x 20. Construct an isosceles right triangle with legs of length a, construct a 30°-60°-90° triangle with legs of lengths a and a 3 , and construct a right triangle with legs of lengths a 2 and a 3. 1 2 4 3 3 8 a 1 2 2 3 4 13. possible answer: 4 4 14. Possible answer: Use 22 12 5 2 and 62 32 45 2. 5 a 3 1 45 5 3 4 3 6 15a. CDA, AEC, AEB, BFA, BFC 15b. MDB, MEB, MEC, MFC, MFA 108 a 5 a 2 21. 8 12.5; that is, the sum of the areas of the semicircles on the two legs is equal to the area of the semicircle on the hypotenuse. 73 22. 6 12.2 chih 23. Extend the rays that form the right angle. m4 m5 180° by the Linear Pair Conjecture, and it’s given that m5 90°. m4 90°. m2 m3 m4 m2 m3 90° 180°. m2 m3 90°. m3 m1 by AIA. m1 m2 90°. 1 2 a 3 Areas: 4.5 cm2, 8 cm2, 12.5 cm2. 4.5 1 2 2a a a 3 4 a 2 ANSWERS TO EXERCISES 24. 80° 2 LESSON 9.4 1. No. The space diagonal of the box is about 33.5 in. 2. 10 m 3. 50 km/h 4. 8 ft 5. area: 60 m2; cost: $7200 6. surface area of prism 27 3 180 cm2 2 226.8 cm ; surface area of cylinder 78 cm2 245.0 cm2 36 cm 20.8 cm 7. 3 8. 48.2 ft; 16.6 lb 9a. 160 ft-lb 9b. 40 lb 9c. 20 lb 10. about 4.6 ft 2 2 2 2 11. 2 2 4 2 2 2 2 2 12. 13. 12 units 14. 18 2 cm and CB . ABC ADC 15. Draw radii CD 90°. For quadrilateral ABCD, 54° 90° mC 126° but 90° 360°, so mC 126°. BD 126° 226° 360°. 16. 4, 4 3 17. SAA 18. orthocenter 19. 115° 20. PO 2 4 2 2 2 4 4 Answers to Exercises 2 2 ANSWERS TO EXERCISES 109 LESSON 9.5 1. 5 units 4. 354 m 7. rectangle 2. 45 units 5. 52.4 units 3. 34 units 6. isosceles y 10 A 8 B 6 4 2 D 2 C 4 6 x 8 10 8. parallelogram E2 4 x F H 8 Answers to Exercises 6 G 10 9. kite y L 4 2 K 8 x I 2 2 J 10. square y 120° 8 P M 3 1 19. 2, 2 20. k 2 , m 6 6 12 21. x , y 3 3 22. 96 cm 23. The angle of rotation is approximately 77°. Connect two pairs of corresponding points. Construct the perpendicular bisector of each segment. The point where the perpendicular bisectors meet is the center of rotation. 24. Any long diagonal of a regular hexagon divides it into two congruent quadrilaterals. Each (6 2) 180° angle of a regular hexagon is 120°, and 6 the diagonal divides two of the 120° angles into 60° angles. Look at the diagonal as a transversal. The alternate interior angles are congruent, thus the opposite sides of a regular hexagon are parallel. y 2 2 11b. Circle A: (x 1)2 ( y 2)2 64; Circle B: x 2 (y 2)2 36 11c. (x h)2 (y k)2 (r)2 12. (x 2)2 y 2 25 13. Center is (0, 1), r 9. 14. (x 3)2 (y 1)2 18 15a. 14 units 15b. 176 4 11 units 2 (y y )2 (z z )2 15c. (x x ) 2 1 2 1 2 1 16. 86.5 units 17. 14 units 18. n (n 2) 3 (n 3) (n 1) 60° 60° 120° 4 120° O N 4 2 11a. 2 60° 60° x 4 y y (x, y) (1, –2) x (x, y) Circle A 110 ANSWERS TO EXERCISES (0, 2) x Circle B 120° LESSON 9.6 18 cm2 56.5 cm2 (8 16) m2 9.1 m2 456 cm2 1433 cm2 120 cm2 377.0 cm2 32 32 3 m2 45.1 m2 (25 48) cm2 30.5 cm2 64 7. 3 16 3 cm2 39.3 cm2 8. (240 18) m2 183.5 m2 9. 102 cm 10. Possible proof: and AN Given: Circle C with tangents AM A Show: AM N 1. 2. 3. 4. 5. 6. 14. 12 6 3 cm 22.4 cm 15. 77 cm 8.8 cm 16. 76 cm 17. Inscribed circle: 3 cm2. Circumscribed circle: 12 cm2. The area of the circumscribed circle is four times as great as the area of the inscribed circle. 18. 135° 19. (x 3)2 (y 3)2 36 20. The diameter is the transversal, and the chords are parallel by the Converse of the Parallel Lines Conjecture. The chords are congruent because they can be shown to be the same distance from the center. (Draw a perpendicular from each chord to the center and use AAS and CPCTC.) Sample construction: M C A , NC , and AC to the diagram as shown. Add MC By the Tangent Conjecture, M and N are right angles, so AMC and ANC are right triangles. Using the Pythagorean Theorem, (AM)2 (MC)2 (AC)2 and (AN)2 (NC)2 (AC)2. Solving for AM and AN, AM (AC)2 (MC)2 and 2 2 and NC are AN (AC) (NC) . Because MC radii of the same circle, they have the same lengths. So, substitute MC for NC in the second equation: AN (AC)2 (MC)2 AM. Therefore, AN . AM 11. 18 m 12. 324 cm2 1018 cm2 13. 3931 cm2 21a. Because a carpenter’s square has a right angle and both radii are perpendicular to the tangents, a square is formed. The radius is 10 in., therefore the diameter is 20 in. 10 in. 10 in. 10 in. d 20 in. 10 in. 21b. Possible answer: Measure the circumference with string and divide by . 5 22. 6 ANSWERS TO EXERCISES 111 Answers to Exercises N CHAPTER 9 REVIEW 1. 2. 3. 4. 20 cm 10 cm obtuse 26 cm 3 1 5. 2, 2 6. 1, 1 2 2 7. 200 3 cm2 346.4 cm2 8. d 12 2 cm 17.0 cm2 9. 246 cm2 10. 72 in2 226.2 in2 11. 24 cm2 75.4 cm2 12. (2 4) cm2 2.28 cm2 13. 222.8 cm2 14. isosceles right 15. No. The closest she can come to camp is 10 km. 16. No. The 15 cm diagonal is the longer diagonal. 17. 1.4 km; 821 min 18. yes 19. 29 ft 20. 45 ft 21. 50 mi 22. 707 m2 23. 6 3 and 18 24. 12 m 25. 42 26. No. If you reflect one of the right triangles into the center piece, you’ll see that the area of the kite is almost half again as large as the area of each of the other triangles. Answers to Exercises 7 27. 9 28. The quarter-circle gives the maximum area. Triangle: s ____ 2 45° s 45° s ____ 2 s 1 A 2 2 s 2 s 4 2 Square: _1 s 2 _1 s 2 1 1 s2 A 2s 2s 4 Quarter-circle: s 2s ___ 1 s 4 2r 2s r 1 2s 2 s 2 A 4 s 2 s2 4 Extra 30° 29. 1.6 m 30. 30° 30° Or students might compare areas by assuming the short leg of the 30°-60°-90° triangle is 1. The area 3 of each triangle is then 0.87 and the area of 2 the kite is 3 3 1.27. 112 ANSWERS TO EXERCISES 31. 4; 0; 10. The rule is n2 if n is even, but 0 if n is odd. 32. 4 in./s 12.6 in./s 33. true 34. true 35. False. The hypotenuse is of length x 2. 36. true 2 37. false; AB x (x 2 (y2 y1) 2 1) 38. False. A glide reflection is a combination of a translation and a reflection. 39. False. Equilateral triangles, squares, and regular hexagons can be used to create monohedral tessellations. 40. true 41. D 42. B 43. A 44. C 45. C 46. D 47. See flowchart below. 48. N 8-ball W B E Cue ball A S 49. 34 cm2; 22 4 2 27.7 cm 40 50. 3 cm2 41.9 cm2 51. about 55.9 m 52. about 61.5 cm2 53. 48 cm 54. 322 ft2 Answers to Exercises 47. (Chapter 9 Review) 1 ABCD is a rectangle Given 2 ABCD is a parallelogram Definition of rectangle 4 D B Definition of rectangle 3 DA CB Definition of parallelogram 5 DAC BCA AIA Conjecture 7 ABC CDA SAA Congruence Conjecture 6 AC AC Same segment ANSWERS TO EXERCISES 113 Answers to Exercises CHAPTER 10 • CHAPTER 10 CHAPTER 10 • CHAPTER 24. Answers to Exercises LESSON 10.1 1. polyhedron; polygonal; triangles 2. PQR, TUS 3. PQUT, QRSU, RPTS , PT , RS 4. QU 5. 6 cm 6. GYPTAN 7. point E , YE , PE , TE , AE , NE 8. GE 9. 13 cm 10. D 11. L 12. C 13. G 14. B 15. H 16. E 17. A 18. J 19. J 20. M 21. H 22. I 23. 25. x 2x 26. x 2x 27. true 28. False. This statement is true only for a right prism. 29. true 30. true 31. False. It is a sector of a circle. 32. true 33. false; counterexample: 34. true 35. true 114 ANSWERS TO EXERCISES 39. 60 40. 30 41a. yes 36. See table below. Possible answers include that the number of lateral faces of an antiprism is always twice the number for the related prism; that the number of vertices is the same for each related prism and antiprism; and that the number of edges for a prism is three times the number of faces, while for an antiprism, the number of edges is twice the number of faces. 37. Answer should include the idea that the painting “disappears” into the view out the window. Students might also note the effect created by the cone-shaped tower appearing similar to the road disappearing into the distance. 38. 8 ␣ ␣ 41b. yes 41c. no 41d. yes ⫻ ⫻ Answers to Exercises 36. (Lesson 10.1) Triangular prism Rectangular prism Pentagonal prism Hexagonal prism Lateral faces 3 4 5 6 ... n Total faces 5 6 7 8 ... n2 Edges 9 12 15 18 ... 3n Vertices 6 8 10 12 ... 2n Triangular antiprism Rectangular antiprism Pentagonal antiprism Hexagonal antiprism Lateral faces 6 8 10 12 ... 2n Total faces 8 10 12 14 ... 2n 2 Edges 12 16 20 24 ... 4n Vertices 6 8 10 12 ... 2n n-gonal prism n-gonal antiprism ANSWERS TO EXERCISES 115 20. LESSON 10.2 1. 4. 5. 6. 7. 8. 72 cm3 2. 24 cm3 3. 108 cm3 r r 160 cm3 502.65 cm3 r 36 cm3 113.10 cm3 324 cm3 1017.88 cm3 See table below. 960 in3 9. QT cubic units 21. true 22. false 12 in. Q 4 in. T 23. 24 in. 8 in. Answers to Exercises 10. sample answer: 24. possible solution: prism 12 12 12 8 3 2 12. 3r 3 13. 13x 3 11. 2x 3 14. Margaretta has room for 0.5625 cord. She should order a half cord. 15. 170 yd3 16. 5100 lb 17. 11,140 18. The volume of the quilt in 1996 was 4000 ft3. The quilt panels were stacked 2 ft 8 in. high. 19. Salt crystal 25. approximately 1.89 m 26. 12 6 2 7. (Lesson 10.2) Information about base of solid Height of solid Right triangular prism Right rectangular prism h H 116 Right trapezoidal prism b2 h b Right cylinder b H r h H b H b 6, b2 7, h 8, r 3 H 20 a. V 480 cm3 d. V 960 cm3 g. V 1040 cm3 j. V 180 cm3 b 9, b2 12, h 12, r 6 H 20 b. V 1080 cm3 e. V 2160 cm3 h. V 2520 cm3 k. V 720 cm3 b 8, b2 19, h 18, r 8 H 23 c. V 1656 cm3 f. V 3312 cm3 i. V 5589 cm3 l. V 1472 cm3 ANSWERS TO EXERCISES 14. 78,375 grams 15. 48 in3 16. 4 units3 17. 144x 3 cm3 18. 40,200 gal; 44 h 40 min 19. 71 ft 3 20. 403 barrels 21a. 16 3 cm2 21b. 96 cm2 21c. 80 3 cm2 21d. 24 13 120 cm2 A 22. D LESSON 10.3 192 cm3 84 cm3 263.9 cm3 150 cm3 60 cm3 84 cm3 263.9 cm3 384 cm3 1206 cm3 m3 7. 3 cm3 2 8. 3 b 3 cm3 9. 324x 3 cm3; 29.6% 10. See table below. 1 11. V 3M 2H ft 3 1. 2. 3. 4. 5. 6. D' 3 1 2 C B 4 H M D'' Possible answer: From the properties of reflection, 1 3 and 2 4. m1 m2 90°, so m3 m4 90°, and m1 m2 m3 m4 180°. Therefore D, C, and D are collinear. 23a. Y (a c, d) 23b. Y (a c, b d) 23c. Y (a c e, b d f ) 12. sample answer: 27 16 3 48 13. Mount Etna is larger. The volume for Mount Etna is approximately 2193 km3, and the volume for Mount Fuji is approximately 169 km3. 10. (Lesson 10.3) Information about base of solid Height of solid Triangular pyramid Trapezoidal pyramid H H b Rectangular pyramid H h b2 H r h h b Cone b b 6, b2 7, h 6, r 3 H 20 a. V 120 cm3 d. V 240 cm3 g. V 260 cm3 j. V 60 cm3 b 9, b2 22, h 8, r 6 H 20 b. V 240 cm3 e. V 480 cm3 h. V k. V 240 cm3 b 13, b2 29, h 17, r 8 H 24 c. V 884 cm3 f. V 1768 cm3 i. V 2856 cm3 l. V 512 cm3 2480 cm3 3 ANSWERS TO EXERCISES 117 Answers to Exercises M LESSON 10.4 Answers to Exercises 1. 2. 3. 4. 5. 6. 58.5 in3 32 3 cm3 55.43 cm3 15 cm 11 cm 5.0 cm 8.5 in. 2 3 2 11 in. 63.24 in ; 11 in. 2 3 2 8.5 in. 81.85 in The short, fat cylinder has greater volume. 7. 257 ft 3 8. 4 cm 9. He must refute the statement. 10. 1502 lb 11. 192.4 gal 12. 13 min 13. Answers will vary, but r 2H should equal about 14.4 in3. 14. approximately 38 in3 15. 100,000 m 3, or about 314,159 m3; 16,528 loads EC because the opposite sides of a 16. AB BD because parallelogram are congruent. EC the diagonals of a rectangle are congruent. So, BD because both are congruent to EC . AB Therefore, ABD is isosceles. 118 ANSWERS TO EXERCISES 17. 8.2 cm 18. x 96° 19. 20. x D 45° x C x 45° 45° 21a. 21b. 21c. 21d. 21e. 45° x A A S N S A x 3x x B LESSON 10.5 1. 675 cm3 2. 36 cm3 113.1 cm3 3. 47 in3 4. 1798.4 g 5. The gold has mass 2728.5 g, and the platinum has mass 7529.8 g. The solid cone of platinum has more mass. 6. 1.5 cm 7. 10.5 g/cm3; silver 8. 8000 cm3 9. The volume of the medallion is 160 cm3.Yes, it is gold, and the Colonel is who he says he is. 10. 679 cm3 11. approximately 193 lb; 22 fish 9 12. 2 3 13. flowchart proof: SP QR Given R S MR SM Given QMR PMS Vertical Angles AIA RMQ SMP ASA MP MQ CPCTC M is the midpoint of PQ Definition of midpoint 14. 15 sides 15a. (1, 3) 15b. (x 1)2 (y 3)2 25 16. 58; 3n 2 Answers to Exercises ANSWERS TO EXERCISES 119 Answers to Exercises LESSON 10.6 1. 36 cm3 113.1 cm3 2. 6 cm3 0.533 cm3 9 3 3 3. 3 2 cm 0.884 cm 4. 720 cm3 2262 cm3 5. 30 cm3 94.3 cm3 6. 3456 cm3 10,857 cm3 7. 18 m3 56.6 cm3 8. No. The volume of the ice cream is 85.3 cm3, and the volume of the cone is 64 cm3. 9. only 20 scoops 148 10. 3 m3, or about 155 m3 11. They have the same volume. 12. 9 in. 13. 3 cm 8192 14. 3 cm3 8579 cm3 15. 18 in3, or about 57 in 3 16. No. The unused volume is 16 cm3, and the volume of the golf ball is 10.6 cm3. 120 ANSWERS TO EXERCISES 17. approximately 15,704 gallons; 53 days 18. lithium 19. 31 ft 20. ABCD is a parallelogram because the slopes of and AB are both 0 and the slopes of BC and CD are both 35. AD y C (3, 5) D (9, 5) x B (–3, –5) A (3, –5) 21. They trace two similar shapes, except that the one traced by C is smaller by a scale factor of 1:2. 22. The line traces an infinite hourglass shape. Or, it traces the region between the two branches of a hyperbola. 23. w 110°, x 115°, y 80° 1. V 972 cm3 3054 cm3 S 324 cm2 1018 cm2 2. V 0.972 cm3 3.054 cm3 S 3.24 cm2 10.18 cm2 3. V 1152 cm3 3619 cm 3 S 432 cm2 1357 cm2 4. S 160 cm2 502.7 cm2 256 5. V 3 cm3 268.1 cm3 6. S 144 cm2 452.4 cm2 7. Area of great circle r 2. Total surface area of hemisphere 3r 2. Total surface area of hemisphere is three times that of area of great circle. 8. 2 gal 9. V 21 34(1.8)3 21(1.8) 2 (4.0) 10.368 m3 32.57 m3; S 21 4(1.8) 2 1 2(1.8)(4.0) 13.68 m2 42.98 m 2 2 10a. approximately 3082 ft2 10b. 13 gal 10c. approximately 9568 bushels 11. 153,200,000 km2 12. The total cost is $131.95. He will stay under budget. 13. 1.13% 14. 150 cm3 471.2 cm3 1 15. 4 19a. (Lesson 10.7) n f(n) 1 2 3 4 5 6 ... n ... 200 2 1 4 7 10 13 . . . 3n 5 . . . 595 19b. (Lesson 10.7) n 1 2 3 4 5 6 ... n ... 200 f(n) 0 1 3 1 2 3 5 2 3 5 7 n1 ... n1 . . . 199 201 ANSWERS TO EXERCISES 121 Answers to Exercises 1 16. 2 3 17. 4 18. The ratio gets closer to 1. 19a and 19b. See tables below. CB and AD CD by the definition of 20. AB rhombus, and BD BD because it is the same segment; therefore ABD CBD by SSS. By CPCTC, 2 3 and 1 4, which shows bisects both ABC and ADC. Because that BD all four sides of the rhombus are congruent, a similar proof can be used to show that ABC ADC and thus that A and C are both . bisected by diagonal AC CB by the definition of rhombus and 21a. AB BE because it is the same segment. 1 2 BE by the Rhombus Angles Conjecture. Therefore, AEB CEB by SAS. CE by CPCTC, so BD bisects AC . 21b. AE 21c. 3 4 by CPCTC. Also, 3 and 4 form a linear pair, so they are supplementary. Because two angles that are congruent and supplementary are right angles, 3 and 4 are right angles. 21d. Because 3 and 4 are right angles, the diagonals are perpendicular.You still need to show bisects BD . Use a proof similar to that that AC given in 21a to show that AEB AED. Then, DE , which shows that AC by CPCTC, BE . bisects BD LESSON 10.7 USING YOUR ALGEBRA SKILLS 10 A 1. h b P 2h P 2. b 2 or b 2 h Answers to Exercises 3. r 3V H 2 a2 4. b c 2 • SA 5. a l P 6. y2 mx 2 x1 y1 or y2 mx 2 mx 1 y1 d 7. v t The original formula gives distance in terms of velocity and time. 9 8. F 5C 32 The original formula converts degrees Fahrenheit to degrees Celsius. T 2 9. L g 2 The original formula gives the period of a pendulum (time of one complete swing) in terms of length and acceleration due to gravity. 10a. V F E 2 10b. See table below. 11a. The corresponding radii are approximately 3.63 cm, 3.30 cm, 3.02 cm, and 2.79 cm. 11b. The corresponding heights are approximately 9.32 cm,10.49 cm,11.61 cm,and 12.70 cm. 11c. The corresponding volumes are approximately 129 cm 3, 120 cm 3, 111 cm 3, and 104 cm 3. 11d. Answers will vary. Sample answer: The cone with slant height 10 cm has the widest radius, so a scoop of ice cream is least likely to fall off, and that cone also has the greatest volume. 11e. Answers will vary. Sample answer: The cone with slant height 13 cm has the greatest height, so the cone appears bigger even though it has the same surface area as the other cones; that cone also has the smallest radius and smallest volume, so it could hold less ice cream and still appear to be a bigger cone. m1 m2 m3 f f 12a. A 5 12b. average of 60: 31; average of 70: 56; average of 80: 81; average of 90: 106 (impossible) 10b. (Using Your Algebra Skills 10) Pentahedron Hexahedron Octahedron Decahedron Dodecahedron Number of faces 5 6 8 10 12 Number of edges 8 10 12 16 20 Number of vertices 5 6 6 8 10 122 ANSWERS TO EXERCISES CHAPTER 10 REVIEW 27. 3 3 m3 0.23 m3 8 32 28. 160 cubic units Answers to Exercises 1. They have the same formula for volume: V BH. 2. They have the same formula for volume: V 31BH. 3. 6240 cm3 4. 1029 cm3 3233 cm3 5. 1200 cm3 6. 32 cm3 7. 100 cm3 314.2 cm3 8. 2250 cm3 7069 cm3 9. H 12.8 cm 10. h 7 cm 11. r 12 cm 12. r 8 cm 13. 960 cm3 14. 9 m 15. 851 cm3 16. four times as great 17a. Vextra large 54 in3 Vjumbo 201.1 in3 Vcolossal 785.4 in3 17b. 14.5 times as great 18. Cylinder B weighs 38 times as much as cylinder A. 19. 2129 kg; 9 loads Vsphere 34 r 3 20. H2r. (2r)3 0.524. Thus, 52.4% of Vbox the box is filled by the ball. 21. approximately 358 yd3 22. No. The unused volume is 98 in3, and the volume of the meatballs is 32 in3. 23. platinum 24. No. The ball weighs 253 lb. 25. 256 lb 26. approximately 3 in. ANSWERS TO EXERCISES 123 Answers to Exercises CHAPTER 11 • CHAPTER 11 CHAPTER 11 • CHAPTER Answers to Exercises USING YOUR ALGEBRA SKILLS 11 3 3 1. 8; 5 AC 3 CD 5 BD 8 , , 2. CD 5 BD 8 BC 13 3 3a. 1 9 3b. 1 4. a 6 5. b 16 6. c 39 7. x 5.6 8. y 8 9. x 12 10. z 6 11. d 1 12. y 5 13. 318 mi 14. 2.01 15. 12 ft by 15 ft 124 ANSWERS TO EXERCISES 16. almost 80 years old 17a. true 3 1 17b. false; 6 2, but 3 1 6 2 17c. true 17d. true 3 1 3 1 17e. false; 6 2, but 2 6 17f. true 18a. arithmetic: 10, 25, 40, 65; geometric: 10, 20, 40, 80 or 10, 20, 40, 80 18b. arithmetic: 2, 26, 50, 74; geometric: 2, 10, 50, 250 or 2, 10, 50, 250 18c. arithmetic: 4, 20, 36, 52; geometric: 4, 12, 36, 108 or 4, 12, 36, 108 19a. Add the numbers and divide by 2. 19b. Possible answer: Multiply the numbers and take the positive square root of the result. 19c. c ab ; This formula holds for all positive values of a and b. 19d. c 2 • 50 100 10; c 4 • 36 144 12 3 9 15. 1; 1 LESSON 11.1 y 1. A 2. B 3. possible answer: Y' (6, 15) Y (2, 5) R' (6, 6) R (2, 2) O' (18, 6) O (6, 2) x 16. Yes, they are similar. 3. possible answer: y 6 (0, 6) (5, 6) 4 (7, 3) 2 (1, 1) 5 6. Figure A is similar to Figure C. Possible answer: If A B and B C , then A . C 7. AL 6; RA 10; RG 4; KN 6 8. No; the corresponding angles are congruent, but the corresponding sides are not proportional. 9. NY 21; YC 42; CM 27; MB 30 10. Yes; the corresponding angles are congruent, and the corresponding sides are proportional. 11. x 6 cm, y 3.5 cm 2 12. z 103 cm 13. Yes, the corresponding angles are congruent. Yes, the corresponding sides are proportional.Yes, AED ABC. 9 9 14. m 2 cm 4.5 cm, n 4 cm 2.25 cm 17. Possible answer: Assuming an arm is about three times as long as a face, each arm would be about 260 ft. 18. Possible answer: Not all isosceles triangles are similar because two isosceles triangles can have different angle measures. A counterexample is shown at right. Not all right triangles are similar because they can have different side ratios, as in a triangle with side lengths 3, 4, and 5 and a triangle with side lengths 5, 12, and 13. All isosceles right triangles are similar because they have angle measures 45°, 45°, and 90°, and the side lengths have the ratio 1:1: 2. d 21. c 1825 22. Possible answers: Jade might get 4475 of the 2650 profits, or $2,773.18, and Omar might get 4475 of the profits, or $4,026.82. Or they take out their investments and they divide the remaining $2,325: Jade, $1825 $1,162.50 $2,987.50; and Omar, $3,812.50. 23a. 23b. 19. 36 20. bc 60° 60° 24. approximately 92 gallons ANSWERS TO EXERCISES 125 Answers to Exercises 5. possible answer: x LESSON 11.2 Answers to Exercises 1. 2. 3. 4. 6 cm 40 cm; 40 cm 28 cm 54 cm; 42 cm 37 35 5. No, 3 0 28 . 6. Yes, MOY NOT by SAS. 7. Yes, PHY YHT because YH 12 and 20 16 12 (SSS).Yes, PTY is a right triangle 15 12 9 because 202 152 25 2. 8. TMR THM MHR by AA. x 15.1 cm, y 52.9 cm, h 28.2 cm 9. Yes,QTA TUR and QAT ARU. QTA QUR by AA; 632 cm 10. 24 cm; 40 cm 11. Yes, THU GDU and HTU DGU; 52 cm; 42 cm 12. SUN TAN by AA; 13 cm; 20 cm 13. Yes, RGO FRG and GOF RFO. GOS RFS by AA; 28 cm; 120 cm 14. 20 cm; 21 cm 15. x 50, y 9 126 ANSWERS TO EXERCISES 2 1 16. r R R 2 12 2 1 17. She should order approximately 919 lb every three months. Explanations will vary. 18. 448 19. The corresponding angles are congruent; the ratio of the lengths of corresponding sides is 13; the dilated image is similar to the original. 20. Yes, ABCD ABCD. The ratio of the perimeters is 12. The ratio of the areas is 14. y D' C' A' B' x 21. 118 square units 22. The statue was about 40 ft, or 12 m, tall. To estimate, you need to approximate the height of a person (or some part of a person) in the picture, measure a part of the statue in the picture, calculate the approximate height of that statue piece, and assume that the statue has the same proportions as the average person. LESSON 11.3 1. 16 m 2. 4 ft 3 in. 3. 30 ft 4. 10.92 m 5. 5.46 m 6. Thales used similar right triangles. The height of the pyramid and 240 m are the lengths of the legs of one triangle; 6.2 m and 10 m are the lengths of the corresponding legs of the other triangle; 148.8 m. 7. 90 m; R and O are both right angles and P is the same angle in both triangles, so PRE POC by AA. 8. 300 cm 6 13. GHF FHK GFK by AA; h 181 3, 9 4 x 71 3 , y 44 13 . 14. sample answer: 1 A 2(8.2)(1.7) 6.97 cm2 1 A 2(3)(4.6) 6.9 cm2 8.2 cm 3 cm 1.7 cm 4.6 cm 45 cm 3 cm 30 cm 2 cm d A 1 Answers to Exercises 2 15. 53 16. B 2 20 cm 4 9. 3 D t h y x Possible answer: Walk to the point where the guy wire touches your head. Measure your height, h; the distance from you to the end of the guy wire, x; and the distance from the point on the ground directly below the top of the tower to the end of the guy wire, y. Solve a proportion to find the height of the tower, t: ht xy. Finally, use the Pythagorean Theorem to find the length of the guy wire: t 2 y 2 . 10. The triangles are similar by AA (because the ruler is parallel to the wall), so Kristin can use the length of string to the ruler, the length of string to the wall, and the length of the ruler to calculate the height of the wall; 144 in., or 12 ft. 2 11. MUN MSA by AA; x 313. 12. BDC AEC by AA; y 63. C Given: Parallelogram ABCD CD and AD BC Show: AB ABCD is a parallelogram Given BC AD CD AB Definition of parallelogram Definition of parallelogram BD BD 2 4 1 3 Same segment AIA Conjecture AIA Conjecture ABD CDB ASA BC AD CD AB CPCTC CPCTC 17a. 4.6.12 17b. 3.12.12 or 3.12 2 ANSWERS TO EXERCISES 127 1 + 5 18. The golden ratio is , or approximately 2 1.618. Here is one possible construction: C D A X B mB 90° 1 Construct BC 2AB 5 AC 2 AB 1 5 AB AX AD 2 AB 2 5 1 AB AX 2 Answers to Exercises AB 1 + 5 2 AX 5 1 2 Therefore, X is the golden cut. 128 ANSWERS TO EXERCISES 19a. Answers will vary. 19b. Possible answer: The shape is an irregular curve. 19c. Answers will vary. Possible answers: As the circular track becomes smaller, the curve becomes more circular; as the track becomes larger, the curve becomes more pointed near the fixed point. As the rod becomes shorter, the curve becomes more pointed near the fixed point; as the rod becomes longer, the curve becomes more like an oval. As the fixed point moves closer to the traced endpoint, the curve becomes more pointed near the fixed point; as the fixed point moves closer to the circular track, the curve begins to look like a crescent moon. a c 17. b 1 d 1 a b c d b b d d LESSON 11.4 1. 2. 3. 4. 5. 6. 7. 8. 18 cm 12 cm 21 cm 15 cm 2.0 cm 126 cm2; 504 cm2 16 cm 60 cm 4 9. 49 cm p b 10. q; q ab cd b d 18. The ratio will be the same as the ratio for the original rectangle. The ratio is 21 if it can be divided like this: It might be any ratio if divided like this: L O V M A 19. Yes, by the SSS or the SAS Similarity Conjecture. 20a. 2a b 20b. 2a b 20c. all values of a and b 20d. no values of a and b 21. AB 3 cm, BC 7.5 cm Answers to Exercises 11. 6 3 cm 12. 6 cm 5 13 1 2 13. x 33 cm, y 3 cm, z 83 cm k 7 3 14. B (3, 5), R 14, 7 ; h 4 2 15. 1 H 16. E T Consider similar triangles LVE and MTH with and HA . To corresponding angle bisectors EO show that the corresponding angle bisectors are proportional to corresponding sides, for example EO EL , show by AA that LOE MAH, HA HM EL and then you can show that HEOA HM . You know that L M. Use algebra to show that LEO MHA. ANSWERS TO EXERCISES 129 LESSON 11.5 1. 18 cm2 4. 27 cm2 2. 18 1 5. 4 9 3. 5; 10 6. 1: 3 9 12 , 16d. x 9, y 16, 1 (9)(16) 2 16 h 144 12 17. m2 m 2 7. or n 2 n 8. Possible answer: Assuming the ad is sold by area, Annie should charge $6000. 9 9. 5000 tiles 10. 1 11. 7.1 m 14.2 m 14.2 m 6.5 m 10 m 42 m 42 m 18. 1 2 2 4 3 6 4 8 5 10 6 12 y 5 5 x 12b. A(x) 2x 2 y x Area in cm2 70 1 2 60 2 8 50 3 18 4 32 5 50 6 72 40 30 20 10 5 x 12c. The equation for a(x) is linear, so the graph is a line. The equation for A(x) is quadratic, so the graph is a parabola. 13. Possible proof: The area of the first rectangle is bh. The area of the dilated rectangle is rh rb, or r 2bh. The ratio of the area of the dilated rectangle r2bh 2 to the area of the original rectangle is bh , or r . 16 20 15. 8 14. 3, 3 16a. (i) 40°; 50°; 40° (ii) 60°; 30°; 60° (iii) 22°; 68°; 22° Conjecture: similar; right triangle s h h b 16b. (i) h (ii) x (iii) m or a p h Conjecture: q h 16c. h pq 130 ANSWERS TO EXERCISES H E L 10 Area in cm2 x Area in cm2 Area in cm2 Answers to Exercises 12a. a(x) 2x O V M A T Consider similar triangles LVE and MTH with and HA . Show corresponding altitudes EO EO EL that LOE MAH, then HA HM . You know and that L M. Because EO HA are altitudes, LOE and MAH are both right angles, and LOE MAH. EL So, LOE MAH by AA. Thus HEOA HM , which shows that the corresponding altitudes are proportional to corresponding sides. 19. x 92°; Explanations will vary but should reference properties of linear pairs, isosceles triangles, and alternate interior angles formed by parallel lines. 20. 105.5 cm2 21. 60 ft2 22. true 23. Two pairs of angles are congruent, so the triangles are similar by the AA Similarity Conjecture. However, the two sets of corresponding 105 sides are not proportional 8600 135 , so the triangles are not similar. 24. Top Front Right side Top: 6 square units; front: 4 square units; right side: 4 square units. The sum of the areas is half the surface area. The volume of the original solid is 8 cubic units. The volume of the enlarged solid is 512 cubic units. The ratio of volumes is 614 . LESSON 11.6 1. 1715 cm3 x 1 Surface area in cm2 2 3 4 5 22 88 198 352 550 Volume in cm 3 6 48 162 384 750 18. yes, because 182 242 302 (Converse of the Pythagorean Theorem) 19a. Possible answer: Fold a pair of corresponding vertices (any vertex in the original figure and the corresponding vertex in the image) together and crease; repeat for another pair of corresponding vertices; the intersection of the two creases is the center of rotation. 19b. Possible answer: Draw a segment (a chord) between a pair of corresponding vertices and construct the perpendicular bisector; repeat for another pair of corresponding vertices; the intersection of the two perpendicular bisectors is the center of rotation. 20e. Label the third vertex C. Construct segment D 2x 2 CD, which bisects C. AD B 3x 3 , or 2:3 y 800 Volume 600 Surface area 400 200 5 x ANSWERS TO EXERCISES 131 Answers to Exercises 2. 16 cm, 4 cm; 768 cm3 2412.7 cm3; 64 12 cm3 37.7 cm3; 1 3 125 3 3. 5; 2; 7 1500 cm H 3 64 ; 4. 1944 ft3 6107.3 ft3; 2 4 27 32 ft 125 5. 2 7 2 6. 2:5 7. 3 8. $1,953.13 9. 2432 lb 10. Possible answer: No, a 4-foot chicken, similar to a 14-inch 7-pound chicken, would weigh 3 approximately 282 pounds 41843 7x. It is unlikely that the legs of the giant chicken would be able to support its weight. 11. Possible answer: Assuming the body types of the African goliath frog and the Brazilian gold frog are similar, the gold frog would weigh about 0.0001 kg, or 0.1 g. 12. surface area ratio 116 , volume ratio 614 . The dolphin has the greater surface area to volume ratio. 13. The ratio of the volumes is 217 . The ratio of the surface areas is 91. 14. S(x) 22x 2 and V(x) 6x 3. Possible answer: The surface area equation is quadratic, so the graph is a parabola, and the volume equation is cubic, so the graph is a cubic graph. 15. Possible proof: The volume of the first rectangular prism is lwh. The volume of the second rectangular prism is rl rw rh, or r 3lwh. The ratio r3lwh 3 of the volumes is lwh , or r . 3 16. 9,120 m or approximately 28,651 m3 s s s2 s 17. 4 or 4 2 2 LESSON 11.7 1 2. 333 cm 3. 45 cm 4. 21 cm 5. 28 cm 6. no 7. José’s method is correct. Possible explanation: Alex’s first ratio compares only part of a side of the larger triangle to the entire corresponding side of the smaller triangle, while the second ratio compares entire corresponding sides of the triangles. 8. yes 9. 6 cm; 4.5 cm 10. 13.3 cm; 21.6 cm 11. yes; no; no 12. yes; yes; yes 13. 3 2 cm; 6 2 cm 14. Answers to Exercises 1. 5 cm E 15. F I J So two pairs of corresponding sides of XYZ and XAB are proportional. X X, so XYZ XAB by the SAS Similarity Conjecture. Because XYZ XAB, XAB XYZ. by the Converse of the Parallel YZ Hence, AB Lines Conjecture. 20. Set the screw so that the shorter lengths of the styluses are three-fourths as long as the longer lengths. 21. x 4.6 cm, y 3.4 cm 22. x 45 ft, y 40 ft, z 35 ft 343 23. 729 24. She is incorrect. She can make only nine 8 cm diameter spheres. 25. 6x 2; 24x 2; 54x 2 1 26. 3r 27a. Possible construction method: Use the triangle-and-circle construction from Lesson 11.3, . Exercise 18, to locate the golden cut, X, of AB Then use perpendicular lines and circles to create a rectangle with length AB and width AX. 27b. Possible construction method: Construct golden rectangle ABCD following the method from by constructing 27a. For square AEFD, locate EF circle A and circle D each with radius AD. Repeat the process of cutting off squares as often as desired. For the golden spiral from point D to point E, construct circle F with radius EF; select point D, point E, and circle F and choose Arc On Circle from the Construct menu. 28. possible answer: B 16. Extended Parallel/Proportionality Conjecture 17. You should connect the two 75-marks. By the Extended Parallel/Proportionality Conjecture, drawing a segment between the 75-marks will 5 form a similar segment that has length 170 0 , or 75%, of the original. 18. 2064 cm3 6484 cm3 19. possible proof: a b c d ad cb ad ab cb ab a(d b) b(c a) a(d b) b(c a) ab ab db ca b a 132 ANSWERS TO EXERCISES P Q A C S R D Given: Circumscribed quadrilateral ABCD, with points of tangency P, Q, R, and S Show: AB DC AD BC Use segment addition to show that each sum of lengths of opposite sides is composed of four lengths: AB DC (AP BP) (DR CR) and AD BC (AS DS) (BQ CQ). Using the Tangent Segments Conjecture, AP AS, BP BQ, CR CQ, and DR DS, and the four lengths in each sum are equivalent. Here are the algebraic steps to show that the whole sums are equivalent. 29. AB DC (AP BP) (DR CR) Segment addition. (AS BQ) (DS CQ) Substitute AS for AP, BQ for BP, DS for DR, and CQ for CR. (AS DS) (BQ CQ) Regroup the measurements by common points of tangency. Answers to Exercises AB DC AD BC Use segment addition to rewrite the right side as the other sum of opposite sides. ANSWERS TO EXERCISES 133 CHAPTER 11 REVIEW 1. x 24 2. x 66 3. x 6 4. x 17 5. 6 cm; 4.5 cm; 7.5 cm; 3 cm 1 1 6. 46 cm; 72 cm 7. 13 ft 2 in. 8. It would still be a 20° angle. 9. Answers to Exercises K P L 10. Yes. If two triangles are congruent, then corresponding angles are congruent and corresponding sides are proportional with ratio 1 , so the triangles are similar. 1 11. 15 m 12. 4 gal; 8 times 13. Possible answer: You would measure the height and weight of the real clothespin and the height of the sculpture. Wsculpture Hsculpture 3 Wclothespin Hclothespin If you don’t know the height of the sculpture, you could estimate it from this photo by setting up a ratio, for example, Hperson Hsculpture Hperson’s photo Hsculpture’s photo 134 ANSWERS TO EXERCISES 14. 9:49 5 125 15. 4; 6 4 16. $266.67 17. 640 cm3 18. 32; 24; 40; 126 19. 841 coconuts 1 20a. 1 to 4 to 2, or 4 to to 2 20b. 3 to 2 to 1 20c. Answers will vary. 21. Possible answer: If food is proportional to 1 body volume, then 8000 of the usual amount of food is required. If clothing is proportional to surface 1 of the usual amount of clothing is area, then 400 required. It would take 20 times longer to walk a given distance. 22. The ice cubes would melt faster because they have greater surface area. Answers to Exercises CHAPTER 12 • CHAPTER 12 CHAPTER 12 • CHAPTER LESSON 12.1 1. 2. 3. 4. 5. 6. 0.6018 0.8746 0.1405 11.57 30.86 62.08 s s r 7. sin A t; cos A t; tan A r 4 3 4 8. sin 5; cos 5; tan 3 7 7 24 ; ; 9. sin A 2; 5 cos A 25 tan A 24 10. 30° 11. 53° 30° 24° a 35 cm b 15 cm c 105 yd d 40° e 50 cm f 33° g 18 in. approximately 237 m x 121 ft 6.375 2.2 16-inch pizza box of ice cream 8 3 cm 13.9 cm V 288 ft3 905 ft 3, S 144 ft 2 452 ft 3 ANSWERS TO EXERCISES 135 Answers to Exercises 24 24 7 ; sin B 2; 5 cos B 25 tan B 7 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. Answers to Exercises LESSON 12.2 1. 9 cm 2. 64° 3. 7 cm 4. 24 m 5. 22 in. 6. 49° 7. 127° 8. 30° 9. 64 cm 10. approximately 655 m 11. approximately 101 m 12. approximately 65 m 13. approximately 188 m 14. approximately 1621 m 15. approximately 1570 m 16a. approximately 974 m 16b. approximately 1007 m 16c. Yes, the height of the balloon would be 1 m less because you don’t have to account for the tripod. The distance to a point under the balloon would not change. 136 ANSWERS TO EXERCISES 17. 45° 18. 60° 19. 60° 20a. 15.04 20b. 2.05 20c. 5.2304 20d. 2.644 1 21. x 3.5, y 97 9.14 22. The block has a volume of 90 cm3 but it displaces only 62.8 cm3 of water. So not all of the block is under water, which means it floats. 23. CZ AX BY 24a. decreases, approaching 0 24b. decreases, approaching 0 24c. remains 90° 24d. increases, approaching (AO)2 24e. decreases, approaching 1 24f. decreases, approaching 1 25. R(8, 7) and E(3, 9), or R(4, 3) and E(1, 1) LESSON 12.3 1. 329 cm2 2. 4 cm2 3. 11,839 cm2 4. 407 cm2 5. 35 cm 6. 17 cm 7. 30 cm 8. 56° 9. 45° 10. 66° 11a. approximately 2200 m 11b. approximately 1600 m 11c. approximately 1400 m 12. The other two walls were 300 ft and approximately 413 ft. The area was approximately 45,000 ft2. 13. approximately 48 m 14. 2366 cm; approximately 41° 15. approximately 5° 93 3 3 16. 8 cm , or approximately 6 cm CD , A D by the AIA 17. Because AB Conjecture. Because D and B intercept the same arc, D B. Therefore A B by the transitive property. So ABE is isosceles by the Converse of the Isosceles Triangle Conjecture. 18. Draw line through the two points. Then fold the paper so that the two points coincide. Draw another line along the fold. These two lines contain the diagonals of the square. Now fold the paper so that the two lines coincide. Mark the vertices on the other line. 19. AC 60 cm, AE 93.75 cm, AF 117 cm 20. Box 1; about 1.2 in. longer Answers to Exercises ANSWERS TO EXERCISES 137 Answers to Exercises LESSON 12.4 1. 32 cm 2. 47 cm 3. 341 cm 4. 74° 5. 64° 6. 85° 7. approximately 43 cm 8. approximately 30° 9. approximately 116° and 64° 10. approximately 6 min 11. approximately 87.8 ft 12. approximately 139 m 13. approximately 143 m 14. The ladder at approximately a 76° angle is not safe. The ladder should be at least 6.5 ft and no more than 14.3 ft from the base of the wall. 15. c A a b a sin A c b cos A c a tan A b a sin A a c a c tan A cos A c b b b c 16a. PR x 3 16b. Since x2 x 3 2 (2x)2, TPR is a right triangle by the Converse of the Pythagorean Theorem. Therefore mTPR 90°. 17a. increases 17b. decreases 17c. increases then decreases 18a. The base perimeters are equal. 18b. The cone has the greater volume. 18c. The cone has the greater surface area. 19a. a translation right 10 units; (x, y) → (x 10, y) 19b. a rotation 180° about the origin; (x, y) → (x, y) 138 ANSWERS TO EXERCISES 20. possible answer: x y x 2y 21. possible answer: x 120° y 120° 120° x x 60° 60° 120° y y 60° 60° 2y 2x 22. Answers will vary from simple (circles and segments) to quite complex. See below for a more detailed answer. Students might begin with the simplest case where all factors are equal. With circles of the same radius and with endpoints of the segment traveling in the same direction, at the same speed, and starting from the same relative position, the midpoint will trace a circle equal in size to those constructed. Changing the size of both circles will change the size of the circle traced, as will changing the starting positions of the endpoints. The maximum radius of the traced circle is limited by the radius of the constructed circles; the smallest observable trace is a single point. With all other factors equal and the endpoints of the segment traveling in opposite directions around the circles, the midpoint will trace a segment. Changing the relative speeds of the endpoints, or the relative radii of the circles, will produce more complex polar curves, which students might describe as flowers, radii, or “spirograph-like.” Changing the distance between the centers has no effect on the shape of the trace, only its position. LESSON 12.5 1. approximately 12.7 m 2. approximately 142 mi/h 3. approximately 51 km 4. approximately 5.9 m 5a. approximately 9.1 km 5b. approximately 255° 6. approximately 240 m 7. approximately 42 ft 8. They must dig at an angle of approximately 71.6° and dig for approximately 25.3 m. 9. approximately 168 km/h 6 cm 10. For simplicity, let 326n0 . Now use the tangent ratio to find the length of the apothem. s 2 tan a s a 2 tan Now use the area formula for a regular polygon. 1 A 2aP 1 s A 2 2 tan ns 2 360 ns A 4 tan where 2n 13. approximately 12.8 m at an angle of approximately 48° 14. 1536 3 cm3 2660 cm3 15. The area increases by a factor of 9. y (0, 2) 10a. apothem 9.23 cm; area 277 cm2 10b. leg 9.7 cm; area of triangle 27.7 cm2; area of decagon 277 cm2 10c. They are the same. 11. approximately 108 cm3 12. Sample answer: Any regular polygon can be divided into n congruent isosceles triangles, each 360° with a vertex angle of n . 360° ___ n The altitude from the vertex angle of an isosceles triangle bisects the angle and the opposite side. This altitude is also an apothem of the regular polygon. –6 _1 A 2 (1, 0) _1 A 2 1 2 1x 6 369 –6 (0, –6) 5 5 15 16. 4 12 17a. decreases then increases 17b. does not change 18a. a reflection across the x-axis; (x, y) → (x, y) 18b. a reflection across the y-axis; (x, y) → (x, y) 19. Answers will vary. 360° ___ 360° ___ 2n 2n a _s 2 _s 2 ANSWERS TO EXERCISES 139 Answers to Exercises (–3, 0) USING YOUR ALGEBRA SKILLS 12 4. y 2 1. y x –5 6 –2 4 –4 f(x) x 2 2 –6 –4 –2 4 6 –6 p(x) 2(x 3) 5 x –2 The graph of y 2x 1 is the same as the graph of p(x). The slope of a line is equal to the vertical stretch of the linear function. The y-intercept is equal to ah k. 5. a vertical stretch by a factor of 2 and a vertical translation down 5 units –4 –6 y y 10 6 8 4 6 2 4 –2 2 –2 g(x) x 2 x 2 –2 h(x) x 3 x 2 y 2 –3 3 –4 –2 –6 –4 f(x) 2x3 5 –6 Answers to Exercises 2. y –8 4 f(x) x x 2 –5 5 g(x) x 5 x 10 –2 h(x) x 5 –4 6. a horizontal translation right 5 units and a vertical translation up 2 units y 6 g(x) (x 5)3 2 4 2 All three graphs are V-shaped, consisting of two parts that have slopes of 1 and 1. Graph g is a translation of graph f right 5 units. Graph h is a translation of graph f down 5 units. y 3. 2 –4 –2 –2 2 8 x 7. a vertical stretch by a factor of 21 (a vertical shrink), a horizontal translation right 2 units, and a vertical translation down 5 units x 2 4 f(x) x2 y 1 –4 –2 –4 6 –2 2 4 6 8 x –2 –6 –6 y h(x) 12 x 2 5 2 –4 –2 2 4 x –2 –4 g(x) x –6 y 2 –4 –2 2 –2 –4 4 x h(x) x 3 A vertical stretch by a factor of 1 reflects the graph across the x-axis. 140 ANSWERS TO EXERCISES 8. f(x) |x 3| 1; a horizontal translation right 3 units and a vertical translation down 1 unit 9. f(x) 3x2; a vertical stretch by a factor of 3 10. f(x) (x 3)2; a vertical stretch by a factor of 1 (a reflection across the x-axis) and a horizontal translation right 3 units 11. f(x) 2|x| 3; A vertical stretch by a factor of 2 and a vertical translation up 3 units 12. f(x) 3(x 1)2 3; A vertical stretch by a factor of 3, a horizontal translation left 1 unit, and a vertical translation down 3 units 13. f(x) 21(x 3)2 2; A vertical stretch by a factor of 21 (a vertical shrink), a horizontal translation right 3 units, and a vertical translation up 2 units 14. Both graphs have the same “hills and valleys” shape, but one is a horizontal translation of the other. 17. Possible answer: f(x) 2 sin(x 90°) 1; a vertical stretch by a factor of 2, a horizontal translation right 90°, and a vertical translation down 1 unit y p(x) ⫽ sin(x) 2 x 360 –360 –2 q(x) ⫽ cos(x) 15. a horizontal translation; a 1, h 90°, k 0 16. The dashed function in each graph is the parent function f(x) sin(x). y p(x) ⫽ 2sin(x) y 2 2 –360 360 x 360 –360 x –2 Answers to Exercises q (x) ⫽ _12 sin(x) y r(x) = –sin(x) 2 360 –360 x –2 Varying a in the equation of the sine function causes the same types of vertical stretches as with other functions. ANSWERS TO EXERCISES 141 Answers to Exercises CHAPTER 12 REVIEW 1. 0.8387 2. 0.9877 3. 28.6363 a c a 4. b; b; c 8 15 8 ; 5. 1; 7 17 15 s 6. s; t; t 7. 33° 8. 86° 9. 71° 10. 1823 cm2 11. 15,116 cm3 12. Yes, the plan meets the act’s requirements. The angle of ascent is approximately 4.3°. 13. approximately 52 km 14. approximately 7.3° 15. approximately 22 ft 16. approximately 6568 m 17. approximately 2973 km/h 18. 393 cm2 19. 30 cm 20. 78° 21. 105 cm 22. 51° 23. 759 cm2 24. approximately 25 cm 25. 72 cm2 26. approximately 15.7 cm 27. approximately 33.5 cm2 28. approximately 10.1 km/h at an approximate bearing of 24.5° 29. False; an octahedron is a polyhedron with eight faces. 30. false 33. 34. 35. 36. 37. 38. 39. 40. 41. 2 False; the ratio of their areas is mn2 . true length of leg opposite T false; tangent of T length of leg adjacent to T true true true true False; the slope of line 2 is m1. false 42. B 43. C 44. A 45. D 46. B 47. B 48. A 49. B 50. C 51. D 52. C 53. A 100 54. 3 cm3 55. 28 cm3 56. 30.5 cm3 57. 33 58. (x, y) → (x 1, y 3) 59. w 48 cm, x 24 cm, y 28.5 cm 60. approximately 18 cm 61. (x 5)2 (y 1)2 9 62. Sample answer: Each interior angle in a regular pentagon is 108°. Three angles would have a sum of 324°, or 36° short of 360, which would leave a gap. Four angles would have a sum exceeding 360° and hence create an overlap. 63. approximately 99.5 m 64. 30 ft 65. 4 cm B 66. A D C 66a. mABC 2 mABD is the perpendicular 66b. Possible answers: BD . It is the angle bisector of ABC, it bisector of AC is a median, and it divides ABC into two congruent right triangles. 31. true 32. true 142 ANSWERS TO EXERCISES Answers to Exercises CHAPTER 13 • CHAPTER 13 CHAPTER 13 • CHAPTER LESSON 13.1 1. A postulate is a statement accepted as true without proof. A theorem is deduced from other theorems or postulates. 2. Subtraction: Equals minus equals are equal. Multiplication: Equals times equals are equal. Division: Equals divided by nonzero equals are equal. 3. Reflexive: Any figure is congruent to itself. C ABC ABC B A Transitive: If Figure A is congruent to Figure B and Figure B is congruent to Figure C, then Figure A is congruent to Figure C. P Q R S X , then Y P Q X Y . RS and RS XY , then PQ XY . If PQ Symmetric: If Figure A is congruent to Figure B, then Figure B is congruent to Figure A. X Y Z L then M L If , M X N N . Y Z If XYZ LMN, then LMN XYZ. 4. reflexive property of equality; reflexive property of congruence 5. transitive property of congruence 19. (Lesson 13.1) and BO are radii AO 1 }? 2 Given AO BO AOB is isosceles 3 ? } } Definition of ? isosceles Definition of circle 20. (Lesson 13.1) 1 } 1 ? 2 }? Given 2 }? m n Corresponding Angles Postulate 3 } 3 ? 4 }? Corresponding Angles Postulate ANSWERS TO EXERCISES 143 Answers to Exercises If 6. subtraction property of equality 7. division property of equality 8. Distributive; Subtraction; Addition; Division 9. Given; Addition property of equality; Multiplication property of equality; Commutative property of addition. 10. true, definition of midpoint 11. true, Midpoint Postulate 12. true, definition of angle bisector 13. true, Angle Bisector Postulate 14. false, Line Intersection Postulate 15. false, Line Postulate 16. true, Angle Addition Postulate 17. true, Segment Addition Postulate 18. • That all men are created equal. • That they are endowed by their creator with certain inalienable rights, that among these are life, liberty, and the pursuit of happiness. • That to secure these rights, governments are instituted among men, deriving their just powers from the consent of the governed. • That whenever any form of government becomes destructive to these ends, it is the right of the people to alter or to abolish it, and to institute new government, laying its foundation on such principles and organizing its powers in such form as to them shall seem most likely to effect their safety and happiness. 19. See flowchart below. 20. See flowchart below. 28. 30 ft 29. FG 2 6 and DG 2 3 because ABGF and BCDG are parallelograms. Triangle FGD is right (mFGD 90°) by the Converse of the Pythagorean Theorem because 2 6 2 2 3 2 62. But mFGB 128° and mDGB 140° by conjectures regarding angles in a parallelogram. So, mFGD 92° because the sum of the angles around G is 360°. So, mFGD is both 90° and 92°. 30. x 54°, y 126°, a 7.3 m 31a. 4 31b. 1 4 31c. 2 1 Answers to Exercises 21. See flowchart below. 22. See flowchart below. 23. (2n 1) (2m 1) 2n 2m 2 2(n m 1) 24. (2n 1)(2m 1) 4nm 2n 2m 1 4nm 2n 2m 2 1 2(2nm n m 1) 1 25. Let n be any integer. Then the next two consecutive integers are n 1 and n 2. The sum of these three integers is (n) (n 1) (n 2) n n 1 n 2. Combining like terms: 3n 3 3(n 1), which is divisible by 3. 26. 299 m 27. sphere, cylinder, cone; cylinder, cone, sphere; cone, cylinder, sphere 21. (Lesson 13.1) 1 AC BD }? Given 2 3 D C BAD BC }? AD }? Given 4 } 5 ABC ? Congruence }? SSS Postulate }? CPCTC AB BA Reflexive property of congruence 22. (Lesson 13.1) 2 ? AB CB } Given BCD CBD 1 Construct angle BD bisector Angle Bisector Postulate 3 ? ABD } 5 ? BAD } ? Reflexive property of congruence } ANSWERS TO EXERCISES 6 ? } ? SAS Congruence ? CPCTC Definition } } ? angle bisector Postulate of } 4 BD BD 144 A C LESSON 13.2 3 10. 1 1. Linear Pair Postulate 2. Parallel Postulate, Angle Addition Postulate, Linear Pair Postulate, Corresponding Angles Postulate 3. Parallel Postulate 4. Perpendicular Postulate 5. 3 4 Use the VA Theorem and the transitive property to get 3 4. Therefore the lines are parallel by the Converse of the AIA Theorem. 3 11. 2 Use the definition of supplementary angles and the substitution property to get m1 m1 180°. Then use the division property and the definition of a right angle. 6. 1 2 3 4 1 2 Use the definition of a right angle and the transitive property to get m1 m2. Then use the definition of congruence to get 1 2. 3 8. 1 1 2 2 3 Use the VA Theorem and the transitive property to get 1 3. Therefore the lines are parallel by the CA Postulate. 3 9. 1 3 1 2 1 2 Use the Linear Pair Postulate and the definition of supplementary angles to get m1 m3 180°. Then use the CA Postulate and the substitution property to get m1 m2 180°. 3 12. 1 31 2 2 Use the Linear Pair Postulate and the definition of supplementary angles to get m1 m3 m1 m2. Then use the subtraction property and the Converse of the AIA Theorem to get 1 2. 13. 1 A 1 B 2 2 3 3 Construct a transversal across lines 1 and 2; it will intersect line 3 by the Parallel Postulate. Use the Interior Supplements Theorem and the definition of supplementary angles to get m1 m2 180°. Use the CA Postulate and the substitution property to get m1 m3 180°. Therefore 1 3 by the Converse of the Interior Supplements Theorem. 2 Use the VA Theorem and the CA Postulate to get 1 3 and 2 3. Then use the transitive property to get 1 2. ANSWERS TO EXERCISES 145 Answers to Exercises Use the definition of supplementary angles and the transitive property to get m1 m2 m3 m4. Then use the substitution property and the subtraction property to get m1 m4. 7. 2 2 3 1 21 1 3 14. 1 2 1 2 Use the definition of perpendicular lines and the transitive property to get m1 m2. Therefore lines 1 and 2 are parallel by the Converse of the AIA Theorem. 15. 2 Answers to Exercises 1 3 By the Triangle Sum Theorem, m1 m2 m3 180°. By the definition of a right triangle, 1 is a right angle. By the definition of right angle, m1 90°. Using the subtraction property, m2 m3 90°, so by the definition of complementary angles, 2 and 3 are complementary. 16. Linear Pair Post. VA Thm. CA Post. Converse of AIA Thm. Converse of AEA Thm. 146 ANSWERS TO EXERCISES 17. 1066 cm3 18. Area (ft2) Bottom 6 2 sides 9 Back and front 6 2 rooftops 812 2 gable ends 2 Total 3112 Plywood (4 by 8) 32 The area is less than that of one sheet of plywood. However, it is impossible to cut the correct size pieces from one piece. This answer assumes that the bottom of the dog house is included. If the bottom is not included, and the gables can be cut separately from the rectangular part of the front and back, then the pieces can be cut from a single sheet of plywood. 19. A(6, 10), B(2, 6), C(0, 0); mapping rule: (x, y) → (2x 8, 2y 2) Isosceles Triangle Theorem (CB CA), the reflexive property, and the SSS Congruence Postulate to get ACP BCP. Therefore, ACP BCP by CPCTC. 6. n C m LESSON 13.3 1. Case 1: P is collinear with A and B. Use the Line Intersection Postulate to show that P and E are the same point and use the definitions of bisector and midpoint to get AP BP. Case 2: P is not collinear with A and B. Use the SAS Congruence Postulate to get BP . AEP BEP. Then use CPCTC to get AP P A Use the Line Intersection Postulate and the Perpendicular Bisector Theorem to get AP BP and BP CP. Then use the transitive property and the Converse of the Perpendicular Bisector Theorem to prove that point P is on line n. C 7. C P A B B E m D n Q A Use the Line Intersection Postulate and the Angle Bisector Theorem to prove that Q is equally distant and AC and from AB and BC . Then use the from AB transitive property and the Converse of the Angle Bisector Theorem to prove that point Q is on line n. B 8. P A B E 2 Case 2: P is not collinear with A and B. Draw . Use the SSS Congruence midpoint E and PE Postulate to get AEP BEP. Then use CPCTC and the Congruent and Supplementary Theorem to prove that AEP and BEP are both right is the perpendicular bisector angles. Therefore PE by the definitions of midpoint and of AB perpendicular. C 3. Use the reflexive property and the SSS Congruence Postulate to get ABC BAC. Therefore, A B by CPCTC. 1 5. B P C A Draw line BA. Use the Isosceles Triangle Theorem to get PAB PBA. Use the Angle Addition Postulate and the subtraction property to get BAC ABC. Then use the Converse of the 4 C Use the Linear Pair Postulate and the definition of supplementary angles to get m3 m4 180°. Then use the Triangle Sum Theorem and the transitive property to get m1 m2 m3 m3 m4. Therefore, m1 m2 m4 by the subtraction property. D 9. 3 4 A B C A 3 A A 4. Use the reflexive property and the ASA Congruence Postulate to get ABC BAC. Then use CPCTC and the definition of isosceles triangle. B C 1 2 B B Use the Triangle Sum Theorem and the addition property to get mA m1 m3 mC m4 m2 360°. Then use the Angle Addition Postulate and the substitution property to get mA mABC mC mCDA 360°. C 10. Use the definitions of median and midpoint to get BM 21BC and N M AN 21AC. Then use the multiplication property and the substitution B A BM . By the property to get AN reflexive property, the Isosceles Triangle Theorem, and the SAS Congruence Postulate, ABN AM by CPCTC. BAM. Therefore, BN ANSWERS TO EXERCISES 147 Answers to Exercises 2. Case 1: P is collinear with A and B. Use the definitions of congruence and midpoint to show . Then use the that P is the midpoint of AB definition of perpendicular bisector. 11. median by CPCTC and the definitions of midpoint and median. C Q P C B A Use the Angle Addition Postulate and the definition of angle bisector to get mPAB 21 mCAB and mQBA 21 mCBA. Then use the Isosceles Triangle Theorem, the multiplication property, and the substitution property to get PAB QBA. By the reflexive property and the ASA Congruence BQ Postulate, ABP BAQ. Therefore, AP by CPCTC. C 12. T S B Answers to Exercises A Use the Isosceles Triangle Theorem, the Right Angles Are Congruent Theorem, and the SAA Theorem to get ABT BAS. Therefore, BT by CPCTC. AS C 13. A B D median → angle bisector Use the definitions of median, midpoint, and isosceles triangle, the reflexive property, and the SSS Congruence Postulate to prove that ADC BDC. Then use CPCTC and the definition of angle bisector. A D B angle bisector → altitude Use the definitions of angle bisector and isosceles triangle, the reflexive property, and the SAS Congruence Postulate to get ADC BDC. Then use CPCTC, the Linear Pair Postulate, and the Congruent and Supplementary Theorem to prove that ADC and BDC are both right is the altitude by the angles. Therefore CD definitions of perpendicular and altitude. 14. x 6, y 3 15. 21.9 m 16. first image: (0, 3); second image: (6, 1) 17. BC FC makes ABCF a rhombus, so its diagonals are perpendicular. mFGC 90°, so GE , so by mCFG mFCG 90°. FD subtraction and transitivity m2 mFCG. 1 FCG by AIA, so 1 2. 3 18a. 3 18b. 18c. 4 3 4 6 19. One possible sequence: 1. Fold A onto B and crease. Label as 1. Label the midpoint of the arc M. 2. Fold line 1 onto itself so that M is on the crease. Label as 2. , and is the desired M is the midpoint of AB 2 tangent. 1 M C 2 A D B altitude → median Use the Right Angles Are Congruent Theorem, the Isosceles Triangle Theorem, and the SAA Theorem is the to get ADC BDC. Therefore, CD 148 ANSWERS TO EXERCISES A B 20. mBAC 13° 21a. B 21b. B 1 1 1 22a. x 2(a c), y 2(a b), z 2(b c) 1 1 1 22b. w 2(a b), x 2(b e), y 2(e d), 1 z 2(d a) LESSON 13.4 1. D B Use x to represent the measures of one pair of congruent angles and y for the other pair. Use the Quadrilateral Sum Theorem and the division property to get x y 180°. Therefore, the opposite sides are parallel by the Converse of the Interior Supplements Theorem. C 2. D 3 2 4 B A 8 2 C A B 5 3 4 B A Use the definition of rhombus, the reflexive property, and the SSS Congruence Postulate to get ABC ADC. Then use CPCTC and the bisects definition of angle bisector to prove that AC DAB and BCD. Repeat the steps above using . diagonal DB C 4. D Use the Converse of the Opposite Angles Theorem to prove that ABCD is a parallelogram. Then use the definition of rectangle. C 7. D A B Use the definition of rectangle to prove that ABCD is a parallelogram and DAB CBA. Then use the Parallelogram Opposite Sides Theorem, the reflexive property, and the SAS Congruence Postulate to get DAB CBA. Finish with CPCTC. C 8. D A B Use the Parallelogram Opposite Sides Theorem, the reflexive property, and the SSS Congruence Postulate to get DAB CBA. Repeat the above steps to get ADC CBA and DAB BCD. Then use CPCTC and the transitive property to get DAB ABC BCD ADC. Finish with the Four Congruent Angles Rectangle Theorem. D C 9. A B A BC Use the definition of parallelogram to get AD and AB DC . Then use the Interior Supplements Theorem. C 5. D 2 3 1 A E B CB . Use the Parallel Postulate to construct DE Then use the Parallelogram Opposite Sides Theorem and the transitive property to prove that AED is isosceles. Therefore, A B by the Isosceles Triangle Theorem, the CA Postulate, and substitution. D C 10. 4 B Use the reflexive property and the SSS Congruence Postulate to get ABD CDB. Then use CPCTC and the Converse of the AIA Theorem to CD and AD CB . Therefore,ABCD is a get AB rhombus by the definitions of parallelogram and rhombus. A B Use the Isosceles Trapezoid Theorem, the reflexive property, and the SAS Congruence Postulate to get BD by CPCTC. DAB CBA. Then AC ANSWERS TO EXERCISES 149 Answers to Exercises Use the AIA Theorem, the reflexive property, and the SAS Congruence Postulate to get ADC CBA. Then use CPCTC and the Converse of the DC . AIA Theorem to get AB 3. D 7 6 C 1 D C A 1 6. 11. D 2 A 4 1 14. C 3 Parallelogram Diagonal Lemma B Use the Parallelogram Opposite Angles Theorem, the multiplication property, and the definition of angle bisector to get 1 3. Then use the Converse of the Isosceles Triangle Theorem, the definition of isosceles triangle, and the Parallelogram Opposite Sides Theorem to get BC DC AD . AB 12. W Z Q Answers to Exercises X P ASA Congruence Postulate Y Use the Converse of the Angle Bisector Theorem is the angle bisector of Y. In to prove that WY like manner, WY is the angle bisector of W. Therefore, WXYZ is a rhombus by the Converse of the Rhombus Angles Theorem. 13. Linear Pair Postulate CA Postulate Interior Supplements Theorem Parallelogram Consec. Angle Theorem Opposite Sides Theorem Line Postulate Converse of the IT Theorem 2 diagonals 2 bisectors of sides isosceles trapezoid 150 ANSWERS TO EXERCISES Converse of the Angle Bisector Theorem 15a. A 15b. S 15c. S 15d. N 16. 2386 ft2 17. See table below. 18. V1 V 2 has length 12.8 and bearing 72.6°. 19a. B 19b. A 20a. 19° 20b. 52° 20c. 52° 20d. 232° 20e. 19° Lines of symmetry Rotational symmetry none 2-fold trapezoid none none kite 1 diagonal none parallelogram IT Theorem Double-Edged Straightedge Theorem 17. (Lesson 13.4) Name SSS Congruence Postulate Opposite Angles Theorem Converse of the Rhombus Angles Theorem Angle Addition Postulate square 4-fold rectangle 2 bisectors of sides 2-fold rhombus 2 diagonals 2-fold 1 bisector of sides none LESSON 13.5 B C D O E A 10. a 75°, b 47°, c 58° 11. 42 ft3 132 ft 3 3 12a. 4 1 2 12b. 1 3 3 13a. A 13b. N 13c. S 13d. S 13e. A ANSWERS TO EXERCISES Answers to Exercises 1. D: Paris is in France; Tucson is in the U.S.; London is in England. Bamako must be the capital of Mali. 2. C: The “Sir” in part A shows that Halley was English; Julius Caesar was an emperor, not a scientist; Madonna is a singer. Galileo Galilei must be the answer. 3. No, the proof is claiming only that if two particular angles are not congruent, then the two particular sides opposite them are not congruent. It still might be the case that a different pair of angles are congruent and that therefore a different pair of sides are congruent. 4. Yes, this statement is the contrapositive of the conjecture proved in Example B, so they are logically equivalent. 5. 1. Assume the opposite of the conclusion; 2. Triangle Sum Theorem; 3. Substitution property of equality; 4. 0°; Subtraction property of equality 6. Assume ZOID is equiangular. Use the definition of equiangular and the Four Congruent Angles Rectangle Theorem to prove that ZOID is a rectangle. Therefore ZOID is a parallelogram, which creates a contradiction. is the altitude to AB . Use the 7. Assume CD definitions of altitude, median, and midpoint, the Right Angles Are Congruent Theorem, and the SAS Congruence Postulate to get ADC BDC. BC , which creates a contradiction. Therefore AC 8. Assume ZO ID. Use the Opposite Sides Parallel and Congruent Theorem to prove that ZOID is a parallelogram, which creates a contradiction. and perpendicu9. Given: Circle O with chord AB lar bisector CD passes through O Show: CD does not pass through O. Use the Line Assume CD and OA and the Postulate to construct OB . Then use Perpendicular Postulate to construct OE the Isosceles Triangle Theorem, the Right Angles Are Congruent Theorem, and the SAA Theorem to get OEA OEB. From CPCTC and the definition of midpoint, prove that E is the midpoint , which creates a contradiction. of AB 151 LESSON 13.6 1. A X Use the Cyclic Quadrilateral Theorem, the Opposite Angles Theorem, and the Congruent and Supplementary Theorem to get R, C, E, and T are right angles. P 5. 1 S B 2 W Case 1 The same: Use the Inscribed Angle Theorem and the transitive property to get A B. A T Z Y B X W Answers to Exercises O Case 2 Congruent: Use the multiplication 1 mWX . Then follow property to get 12 mYZ 2 the steps in Case 1. 2. D , OT , and Use the Line Postulate to construct OS . Then use the Tangent Theorem, the Converse OP of the Angle Bisector Theorem, and the SAA PT by Theorem to get OSP OTP. PS CPCTC. 6. A D 3 1 2 E C C B B A Use the Inscribed Angle Theorem, the addition property, and the distributive property to get mDAB . Then use mA mC 12 mBCD the definition of degrees in a circle, the substitution property, and the definition of supplementary angles to get A and C are supplementary. Repeat the steps using mB and mD to get B and D are supplementary. 3. B . Then use Use the Line Postulate to construct AD the Inscribed Angle Theorem, the addition property, and the distributive property to get mBD . Therefore, m2 m3 12 mAC mBD by the Triangle Exterior m1 12 mAC Angle Theorem and the transitive property. P 7. x A b B C O a A D D C . Then use Use the Line Postulate to construct AD the AIA Theorem, the Inscribed Angle Theorem, BD . and substitution to get AC 4. T C R E 152 ANSWERS TO EXERCISES Intersecting Secants Theorem: The measure of an angle formed by two secants intersecting outside a circle is half the difference of the measure of the larger intercepted arc and the measure of the smaller intercepted arc. Use the Triangle Exterior Angle Theorem and the subtraction property to get x b a. Then use the Inscribed Angle Theorem, the substitution property, and the distributive mAC . property to get x 12 mBD 8. D C A B E Use the Inscribed Angle Theorem to get 1 mBDC 2 mBEC . By the definition of mBDC 180°, so by the semicircle, mBEC division property, mBDC 90°. By the definition of right angle, BDC is a right angle. Because only definitions, properties, and the Inscribed Angle Theorem are needed to prove this conjecture, it is a corollary of the Inscribed Angle Theorem. 9. Angle Addition Postulate Line Postulate ASA Congruence Postulate SAA Theorem IT Theorem SAS Congruence Postulate Right Angles Are Congruent Theorem Converse of the Angle Bisector Theorem Midpoint Postulate Tangent Theorem h c a b P 16a. 27° 16b. 47° 16c. 67° 16d. 133° 16e. cannot be determined 16f. cannot be determined 17. Perpendicular Postulate Tangent Segments Theorem x A O 10. Linear Pair Postulate CA Postulate Parallel Postulate Angle Addition Postulate Answers to Exercises Converse of the IT Theorem SSS Congruence Postulate 11. A(4.0, 2.9), P(1.1, 2.8) 12. 3; 1; 9 , AB , AC , CD , AD 13. BC 14. 32.5 15. As long as point P is inside the triangle, a b c h. Proof: Let x be the length of a side. The areas of the three small triangles are 21 xa, 21 xb, and 21 xc. The area of the large triangle is 21 xh. So, 21 xa 21 xb 21 xc 1 1 xh. Divide both sides by x. So, a b c h. 2 2 M P Arc Addition Postulate B SSS Congruence Postulate VA Theorem IT Theorem AIA Theorem Triangle Sum Theorem ASA Congruence Postulate Exterior Angle Sum Theorem Inscribed Angle Theorem Cyclic Quadrilateral Theorem Parallelogram Diagonal Lemma Opposite Angles Theorem Congruent and Supplementary Theorem Right Angles Are Congruent Theorem Steps: . 1. Construct OP . Label midpoint M. 2. Bisect OP 3. Construct circle with center M and radius PM. 4. Label the intersection of the two circles A and B. and PB . PA and PB are the 5. Construct PA required tangents. OAP and OBP are right angles because they are inscribed in semicircles. Parallelogram Inscribed in a Circle Theorem ANSWERS TO EXERCISES 153 6. LESSON 13.7 1. T C V 1 A B H A G L E Use the Right Angles Are Congruent Theorem, CASTC, and the AA Similarity Postulate to get BHT GEV. Therefore, GBVT TVHE by CSSTP. See the Solutions Manual for the family tree. A G 2. 2 B D Use the Right Angles Are Congruent Theorem, the reflexive property, and the AA Similarity Postulate to get ADC ACB and ACB CDB. Therefore, ADC ACB CDB by the transitive property of similarity. C 7. h S L M Answers to Exercises B Y Use the definitions of median and midpoint, the Segment Addition Postulate, the substitution property, and the multiplication property to get BY 12BI and SL 12SM. Then use CASTC, CSSTP, and the SAS Similarity Theorem to get BYG SLA. Therefore, BSAG GAYL by CSSTP. See the Solutions Manual for the family tree. 3. F C 2 4 13 D A Q E B P Use the definition of angle bisector, the Angle Addition Postulate, and the substitution property to get mACB 2m1 and mDFE 2m2. Then use CASTC and the AA Similarity Postulate to get APC DQF. Therefore, DACF CFQP by CSSTP. C 4. D 1 A E 2 B Use the CA Postulate and the AA Similarity Postulate to get CDE CAB. Then use CSSTP CD DA and the Segment Addition Postulate to get CD CE EB DA EB . Therefore, by algebra. CE CD CE C 5. D 1 A 2 E B A Use the addition property to get DCD 1 CEBE 1. Then use algebra and the Segment Addition Postulate to get CCAD CCEB . Therefore ABC DEC by the SAS Similarity Theorem, 1 2 by the CA Postulate. AB by CASTC, and DE 154 ANSWERS TO EXERCISES A I x B y D Use the Three Similar Right Triangles Theorem to get ADC CDB. Then use CSSTP to get x h . h y 8. b a d c–d c Draw the altitude to the hypotenuse, then use the ratios given by the Three Similar Right Triangles a c 2 2 Theorem. In particular, c d a yields a c cd. Now look at the other small triangle and use bd bc to get b 2 cd. C E 9. a B b a A c b D x F Begin by constructing a second triangle, right triangle DEF (with E a right angle), with legs of lengths a and b and hypotenuse of length x. The plan is to show that x c, so that the triangles are congruent. Then show that C and E are congruent. Once you show that C is a right angle, then ABC is a right triangle. L 10. H Y P E G Use the Pythagorean Theorem to write expressions for the lengths of the unknown legs. Show that the expressions are equivalent. The triangles are congruent by SSS or SAS. 11. CA Postulate Segment Addition Postulate AA Similarity Postulate Parallel/Proportionality Theorem 12. Segment Duplication Postulate Parallel Postulate CA Postulate AA Similarity Postulate SAS Congruence Postulate SSS Similarity Theorem 13. Perpendicular Postulate AA Similarity Postulate Right Angles Are Congruent Theorem Three Similar Right Triangles Theorem Pythagorean Theorem 14. approximately 1.5 cm or 6.5 cm 15a. C 15b. A 15c. A 15d. A 15e. C 16a. The vectors are diagonals of your quadrilateral. 16b. A 180° rotation about the midpoint of the common side; the entire tessellation maps onto itself. 20a. CDG CFG by SAA; GEA GEB by SAS; DGA FGB by the HL Theorem CF and DA FB by CPCTC; CD 20b. CD DA CF FB (addition property of equality). CB , and ABC is isosceles. Therefore, CA 20c. The figure is inaccurate. 20d. The angle bisector does not intersect the perpendicular bisector inside the triangle as shown, except in the special case of an isosceles triangle, when they coincide. 21. 173 cm, 346 cm, 20 stones. Draw the trapezoid and extend the legs until they meet to form a triangle. Use parallel proportionality to find the rise. The span is twice the rise. Use inverse tangent to find the central angle measure. Divide into 180° to find the number of voussoirs. ANSWERS TO EXERCISES 155 Answers to Exercises SAS Similarity Theorem SSS Congruence Postulate 17a. a 180° rotation about the midpoint of any side 17b. possible answer: a vector running from each vertex of the quadrilateral to the opposite vertex (or any multiple of that vector) 18a. 33° 18b. 66° 18c. 57° 18d. 62° 18e. PSQ and PTQ are 90° by the Tangent Theorem and so are supplementary. So, SPT and SQT must also be supplementary by the Quadrilateral Sum Theorem. Therefore, opposite angles are supplementary, so it’s cyclic. is a tangent, mPSQ 90° 18f. Because PS (Tangent Theorem). Because mPSQ 90°, SQ must be a tangent (Converse of the Tangent Theorem). 3 19a. 1 1 2 2 3 19b. 6 4 1 USING YOUR ALGEBRA SKILLS 13 y 1. B(a, 0) 2. C(a b, c) P (c, d ) 2 b 2 , D b, 2 b2 3. C a b, a a 4. possible answer: y Answers to Exercises R (0, 0) E (a, 0) T (0, 0) A (a – c, d ) R (a, 0) x 00 0 slope of TR a0 a 0 C (a, b) T (0, b) 6. possible answer: d0 d slope of RA a c a c dd 0 slope of AP a 2c a 2c 0 x d0 d slope of PT c0 c 00 0 slope of RE a0 a 0 b0 b slope of CE aa 0 (undefined) bb 0 slope of TC a0 a 0 b0 b slope of TR 00 0 (undefined) and AP have the same slope and are parallel by TR the parallel slope property. So TRAP has only one pair of parallel sides and is a trapezoid by definition. Opposite sides have the same slope and are therefore parallel by the parallel slope property. Two sides are horizontal and two sides are vertical, so the angles are all congruent right angles. RECT is an equiangular parallelogram and is a rectangle by definition. 5. possible answer: PT (c 0 )2 (d 0)2 c 2 d 2 RA (a c a)2 (d 0)2 (c)2 d2 2 2 c d The nonparallel sides of the trapezoid have the same length. So trapezoid TRAP is isosceles by definition. y 7. U (0, a 3) E (–a, 0) Q (a, 0) x y I (b, c) Z b, c 2 2 ( T (0, 0) Y a+b, c 2 2 ) ( X a,0 2 ( ) R (a, 0) ) x . Let X be the midpoint of TR a0 00 a X 2, 2 2, 0 . Let Y be the midpoint of RI ab 0c ab c Y 2, 2 2, 2 . Let Z be the midpoint of TI b0 c0 b c Z 2, 2 2, 2 , RI , and TI , X, Y, and Z are the midpoints of TR respectively, by the coordinate midpoint property. , YZ , and ZX are midsegments by definition. So XY 156 ANSWERS TO EXERCISES Show: EQU is equilateral EQ 2a EU (a 0)2 a (0 3 )2 2 3a 2 a 2 4a 2a 2 0a UQ (a 0) 2 3 2 3a 2 a 4a 2 2a EQ EU UQ EQU is equilateral 8. Task 1: Given: A rectangle with both diagonals Show: The diagonals are congruent y Task 2: C (a, b) T (0, b) R (0, 0) E (a, 0) Task 2: x Task 3: Given: Rectangle RECT with diagonals RC and TE TE Show: RC Task 4: To show that two segments are congruent, you use the distance formula to show that they have the same length. Task 5: RC (a 0 )2 (b 0)2 a2 b2 TE (a 0 )2 (0 b)2 a2 b2 I (c, d) ( T (0, 0) ) Y a+c, d 2 2 ( R (a, 0) ) x Task 3: Given: TRI and midsegment YZ 1 TR and YZ TR Show: YZ 2 Task 4: To show that two segments are parallel, use the parallel slope property. The segments are horizontal, so to compare lengths, subtract their x-coordinates. Task 5: 00 0 Slope of TR a0 a 0 d d 0 0 2 2 Slope of YZ ac c a 2 2 2 The slopes are the same, so the segments are parallel by the parallel slope property. TR a 0 a ac c a 1 1 YZ 2 2 2 2a 2TR y P (b, c) M T (0, 0) A (d, c) N R (a, 0) x One possible set of the coordinates for TRAP is shown in the figure. By the coordinate midpoint property, the coordinates of M are 2b, 2c and of N ad c are 2 , 2 . Task 3: Given: Trapezoid TRAP with midsegment MN TR Show: MN Task 4: To show that the midsegment and bases are parallel, you need to find their slopes. Task 5: slope of MN c c 0 2 2 0 adb ad b 2 2 2 00 0 slope of TR a0 a 0 cc 0 is The slope of PA d b d b 0. The slopes are equal, therefore the lines are parallel. 11. Task 1: Given: A quadrilateral in which only one diagonal is the perpendicular bisector of the other Show: The quadrilateral is a kite y Task 2: I (0, b) T (–a, 0) K (a, 0) M(0, 0) x E (0, c) Task 3: Given: Quadrilateral KITE with diagonal , which is the perpendicular bisector of diagonal IE TK ANSWERS TO EXERCISES 157 Answers to Exercises TE because both segments have the same So RC length. Therefore the diagonals of a rectangle are congruent. 9. Task 1: Given: A triangle with one midsegment Show: The midsegment is parallel to and half the length of the third side y Task 2: Z c,d 2 2 So the midsegment is half the length of the third side. Therefore the midsegment of a triangle is parallel to the third side and half the length of the third side. 10. Task 1: Given: A trapezoid Show: The midsegment is parallel to the bases Show: KITE is a kite Task 4: To show that a quadrilateral is a kite, you use the distance formula to show that only two pairs of adjacent sides have the same length. Task 5: 2 b2 KI (0 a )2 (b 0)2 a IT (0 ( a))2 (b 0)2 a2 b2 TE (0 ( a))2 (c 0)2 a2 c2 Answers to Exercises EK (a 0 )2 (0 c)2 a 2 c 2 and IT have the same length and Adjacent sides KI adjacent sides TE and EK have the same length, and because ⏐b⏐ ⏐c⏐ the pairs are not equal in length to each other. Therefore, KITE is a kite by definition. Therefore, if only one diagonal of a quadrilateral is the perpendicular bisector of the other diagonal, then the quadrilateral is a kite. 12. Task 1: Given: A quadrilateral with midpoints connected to form a second quadrilateral Show: The second quadrilateral is a parallelogram Task 2: y D (d, e) ( d, e 2 2 G ) L b+d, c+e 2 2 A (b, c) ( ) R a+b, c 2 2 ( Q (0, 0) P a , 0 2 ( ) U (a, 0) ) x Task 3: Given: Quadrilateral QUAD with midpoints P, R, L, and G Show: PRLG is a parallelogram Task 4: To show that a quadrilateral is a parallelogram, we need to show that opposite sides have the same slope. c c 0 2 2 c Task 5: slope of PR ab a b b 2 2 2 e ce c 2 2 2 e slope of RL bd ab da da 2 2 2 e ce c 2 2 2 c slope of LG b d bd b 2 2 2 e e 0 2 2 e slope of GP d a da d a 2 2 2 158 ANSWERS TO EXERCISES and LG have the same slope, and Opposite sides PR and GP have the same slope. So opposite sides RL they are parallel by the parallel slope property, and PRLG is a parallelogram by definition. Therefore the figure formed by connecting the midpoints of the sides of a quadrilateral is a parallelogram. 13. Task 1: Given: An isosceles triangle with the midpoint of the base connected to the midpoint of each leg, to form a quadrilateral Show: The quadrilateral is a rhombus Task 2: y D a2 , h2 ( ) A (a, h) F 32a, h2 ( ) x B (0, 0) E (a, 0) C (2a, 0) Task 3: Given: Isosceles triangle ABC with midpoint of base, E, and midpoints of legs, D and F, connected to form quadrilateral ADEF. Show: ADEF is a rhombus Task 4: You need to show that all the sides of ADEF have the same length. Task 5: By the distance formula, AD 2 a a 2 h h 2 2 a 2 2 h 2 2 a 2 h 2 2 AF 3a a 2 2 h 2 h 2 a 2 2 h 2 a 2 h 2 2 DE a a 2 2 h 0 2 2 a 2 2 h 2 2 a 2 h 2 2 EF 3a a 2 2 h 2 0 2 a 2 2 h 2 a 2 h 2 2 AD AF DE EF by the transitive property of equality. Therefore, ADEF is a rhombus by the definition of a rhombus. 2 2 CHAPTER 13 REVIEW 140⬚ 40⬚ 20⬚ 40⬚ 140⬚ 20⬚ 20⬚ 20⬚ 20. False; x 360° 2b so x 90° only if b 135°. x a b a b x ⫽ 540⬚ ⫺ (180⬚ ⫹ 2b) x ⫽ 360⬚ ⫺ 2b 21. True (except in the special case of an isosceles right triangle, in which the segment is not defined because the feet coincide). C EAB DBA by the Isosceles Triangle Theorem. AEB E D BDA by the definition of X altitude and by the Right Angles A B Are Congruent Theorem. AB by the reflexive property of congruence, AB BD by so AEB BDA by SAA. AE CPCTC. By the definition of congruence, AC BC and AE BD, so EC DC by the subtraction property of equality and the Segment Addition Postulate. By the division property of equality, DC EC divides the sides of ABC , so ED BD AE AB by the proportionally. Therefore ED Converse of the Parallel/Proportionality Theorem. 22. true and Given: Rhombus ROME with diagonals RM EO intersecting at B E 4 B 3 2 R M 1 O EO Show: RM Because diagonals bisect the angles in a rhombus, the diagonals are angle bisectors. Statement Reason 1. 1 2 1. Rhombus Angles Theorem 2. RO RE 2. Definition of rhombus 3. RO RE 3. Definition of congruence RB 4. RB 4. Reflexive property of congruence 5. ROB REB 5. SAS Congruence Postulate 6. 3 4 6. CPCTC 7. 3 and 4 are 7. Definition of linear a linear pair pair 8. 3 and 4 are 8. Linear Pair Postulate supplementary 9. 3 and 4 are right 9. Congruent and angles Supplementary Theorem EO 10. RM 10. Definition of perpendicular ANSWERS TO EXERCISES 159 Answers to Exercises 1. False. The quadrilateral could be an isosceles trapezoid. 2. true 3. False. The figure could be an isosceles trapezoid or a kite. 4. true 5. False. The angles are supplementary but not necessarily congruent. 6. False. See Lesson 13.5, Example B. 7. true 8. perpendicular 9. congruent 10. the center of the circle 11. four congruent triangles that are similar to the original triangle 12. an auxiliary theorem proven specifically to help prove other theorems 13. If a segment joins the midpoints of the diagonals of a trapezoid, then it is parallel to the bases. 14. Angle Bisector Postulate 15. Perpendicular Postulate 16. Assume the opposite of what you want to prove, then use valid reasoning to derive a contradiction. 17a. Smoking is not glamorous. 17b. If smoking were glamorous, then this smoker would look glamorous. This smoker does not look glamorous, therefore smoking is not glamorous. 18. False. The parallelogram is a rhombus. 19. False. Possible counterexample: Given: Isosceles ABC with and BE ; AC BC ; ED altitudes AD AB Show: ED 23. true D E C 2 1 3 Answers to Exercises A F B BAD DCB by the Opposite Angles Theorem. m1 21 mBAD 21 mDCB m2 by DC by the the definition of angle bisector. AB definition of parallelogram. 2 and 3 are supplementary by the Interior Supplements Theorem. 1 and 3 are supplementary by the substitution property. FC by the Converse of the Interior AE Supplements Theorem. 24. Use the Inscribed Angle Theorem, the addition property, and the distributive property to get mP mE mN mT mA 1 mTN mAT mPA mEP mNE . Because 2 there are 360° in a circle, mP mE mN mT mA 180°. 25. T OP by CPCTC. Use SAA Theorem; thus DY the Triangle Midsegment Theorem and the substitution property to get TR 21 (ZO DY). Also, use the Triangle Midsegment Theorem to ZO . get TR 28a. The quadrilateral formed when the midpoints of the sides of a rectangle are connected is a rhombus. G C 28b. D H F A Use the Right Angles Are Congruent Theorem and the SAS Congruence Postulate to get AEH BEF DGH CGF. Then use CPCTC to prove that EFGH is a rhombus. 29a. The quadrilateral formed when the midpoints of the sides of a rhombus are connected is a rectangle. I 29b. P O H Assume mH 45° and mT 45°. Use the Triangle Sum Theorem, the substitution property, and the subtraction property to get mH mT 90°, which creates a contradiction.Therefore mH 45° or mT 45°. Y 26. M S T R Use the definition of midpoint, the Segment Addition Postulate, the substitution property, and MY 1 YS 1 the division property to get TY 2 and YR 2 . Then use the reflexive property and the SAS Similarity Theorem to get MSY TRY. Therefore, MS 21TR by CSSTP and the TR by multiplication property and MS CASTC and the CA Postulate. D Y 27. 1 3 T R 42 Z O P and DR . Then Use the Line Postulate to extend ZO use the Line Intersection Postulate to label P as the and DR . DYR POR by the intersection of ZO 160 M Y L R ANSWERS TO EXERCISES B E X J N K By the Triangle Midsegment Theorem both PM and ON are parallel to LJ , and both PO and MN are parallel to IK . Because LJ and IK are perpendicular, we can use corresponding angles on parallel lines to prove that the lines that are adjacent sides of the quadrilateral are also perpendicular. 30a. The quadrilateral formed when the midpoints of the sides of a kite are connected is a rectangle. I 30b. By the Triangle Midsegment P M Theorem both PM and ON are L J , and both PO and parallel to LJ are parallel to IK . Because MN N LJ and IK are perpendicular, we can O use the CA Postulate to prove that K the lines that are adjacent sides of the quadrilateral are also perpendicular. C 31. Use the Line Postulate and to construct chords DB . Then use the Inscribed AC B Angles Intercepting Arcs P O A Theorem and the AA Similarity Postulate to get APC DPB. Therefore, D by CSSTP and the multiplication property, AP PB DP PC.