Patterns and Inductive Reasoning GEOMETRY LESSON 1 1

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 (For help, go the Skills Handbook, page 715. ) Here is a list of the counting numbers: 1, 2, 3, 4, 5, . . . Some are even and some are odd. 1. Make a list of the positive even numbers. 2. Make a list of the positive odd numbers. 3. Copy and extend this list to show the first 10 perfect squares. 12 = 1, 22 = 4, 32 = 9, 42 = 16, . . . 4. Which do you think describes the square of any odd number? It is odd. It is even. 1 -1

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 Solutions 1. Even numbers end in 0, 2, 4, 6, or 8: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, . . . 2. Odd numbers end in 1, 3, 5, 7, or 9: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, . . . 3. 12 = (1)(1) = 1; 22 = (2)(2) = 4; 32 = (3)(3) = 9; 42 = (4)(4) = 16; 52 = (5)(5) = 25; 62 = (6)(6) = 36; 72 = (7)(7) = 49; 82 = (8)(8) = 64; 92 = (9)(9) = 81; 102 = (10) = 100 4. The odd squares in Exercise 3 are all odd, so the square of any odd number is odd. 1 -1

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 Find a pattern for the sequence. Use the pattern to show the next two terms in the sequence. 384, 192, 96, 48, … Each term is half the preceding term. So the next two terms are 48 ÷ 2 = 24 and 24 ÷ 2 = 12. 1 -1

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 Make a conjecture about the sum of the cubes of the first 25 counting numbers. Find the first few sums. Notice that each sum is a perfect square and that the perfect squares form a pattern. 13 13 + 2 3 + 3 3 + 4 3 + 5 3 =1 =9 = 36 = 100 = 225 = 12 = 32 = 62 = 102 = 152 = 12 = (1 + 2)2 = (1 + 2 + 3 + 4)2 = (1 + 2 + 3 + 4 + 5)2 The sum of the first two cubes equals the square of the sum of the first two counting numbers. 1 -1

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 (continued) The sum of the first three cubes equals the square of the sum of the first three counting numbers. This pattern continues for the fourth and fifth rows of the table. 13 + 2 3 + 3 3 + 4 3 = 100 = 102 = (1 + 2 + 3 + 4)2 13 + 2 3 + 3 3 + 4 3 + 5 3 = 225 = 152 = (1 + 2 + 3 + 4 + 5)2 So a conjecture might be that the sum of the cubes of the first 25 counting numbers equals the square of the sum of the first 25 counting numbers, or (1 + 2 + 3 + … + 25)2. 1 -1

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 The first three odd prime numbers are 3, 5, and 7. Make and test a conjecture about the fourth odd prime number. One pattern of the sequence is that each term equals the preceding term plus 2. So a possible conjecture is that the fourth prime number is 7 + 2 = 9. However, because 3 X 3 = 9 and 9 is not a prime number, this conjecture is false. The fourth prime number is 11. 1 -1

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 The price of overnight shipping was $8. 00 in 2000, $9. 50 in 2001, and $11. 00 in 2002. Make a conjecture about the price in 2003. Write the data in a table. Find a pattern. 2000 2001 2002 $8. 00 $9. 50 $11. 00 Each year the price increased by $1. 50. A possible conjecture is that the price in 2003 will increase by $1. 50. If so, the price in 2003 would be $11. 00 + $1. 50 = $12. 50. 1 -1

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 Pages 6– 9 Exercises 1 1 , 5 6 1. 80, 160 12. 2. 33, 333; 333, 333 13. James, John 3. – 3, 4 14. Elizabeth, Louisa 4. 1 1 , 16 32 5. 3, 0 6. 1, 1 3 15. Andrew, Ulysses 16. Gemini, Cancer 21. The sum of the first 100 pos. even numbers is 100 • 101, or 10, 100. 18. 9. 720, 5040 10. 64, 128 11. 20. The sum of the first 30 pos. even numbers is 30 • 31, or 930. 17. 7. N, T 8. J, J 19. The sum of the first 6 pos. even numbers is 6 • 7, or 42. 1 1 , 36 49 1 -1

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 22. The sum of the first 100 odd numbers is 1002, or 10, 000. 28. 1 ÷ 1 = 3 and 3 is 2 3 2 2 improper. 29. 75°F 23. 555, 555 24. 123, 454, 321 25– 28. Answers may vary. Samples are given. 25. 8 + (– 5 = 3) and 3 >/ 8 26. 1 1 1 • >/ and • >/ 3 2 2 3 27. – 6 – (– 4) < – 6 and – 6 – (– 4) < – 4 31. 31, 43 32. 10, 13 33. 0. 0001, 0. 00001 30. 40 push-ups; answers may vary. Sample: Not very confident, Dino may reach a limit to the number of push-ups he can do in his allotted time for exercises. 1 -1 34. 201, 202 35. 63, 127 36. 31 63 , 32 64 37. J, S 38. CA, CO 39. B, C

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 40. Answers may vary. Sample: In Exercise 31, each number increases by increasing multiples of 2. In Exercise 33, to get the next term, divide by 10. 42. 43. 44. 41. 45. You would get a third line between and parallel to the first two lines. 46. 102 cm 1 -1 47. Answers may vary. Samples are given. a. Women may soon outrun men in running competitions. b. The conclusion was based on continuing the trend shown in past records. c. The conclusions are based on fairly recent records for women, and those rates of improvement may not continue. The conclusion about the marathon is most suspect because records date only from 1955.

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 48. a. b. about 12, 000 radio stations in 2010 c. Answers may vary. Sample: Confident; the pattern has held for several decades. 49. Answers may vary. Sample: 1, 3, 9, 27, 81, . . . 1, 3, 5, 7, 9, . . . 50. His conjecture is 52. probably false because most 53. people’s growth slows by 18 until they stop growing somewhere between 18 and 22 years. 51. a. b. H and I c. a circle 1 -1 21, 34, 55 a. Leap years are years that are divisible by 4. b. 2020, 2100, and 2400 c. Leap years are years divisible by 4, except the final year of a century which must be divisible by 400. So, 2100 will not be a leap year, but 2400 will be.

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 54. Answers may vary. Sample: 55. (continued) d. 100 + 99 + 98 + … + 3 + 2 + 1 1 + 2 + 3 + … + 98 + 99 + 100 101 + … + 101 56. B The sum of the first 100 numbers is 57. I 100 • 101 , or 5050. 2 The sum of the first n numbers is n(n+1). 2 55. a. 1, 3, 6, 10, 15, 21 b. They are the same. c. The diagram shows the product of n and n + 1 divided by 2 when n = 3. The result is 6. 1 -1 58. [2] a. 25, 36, 49 b. n 2 [1] one part correct

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 59. [4] a. The product of 11 and a three-digit number that begins and ends in 1 is a four-digit number that begins and ends in 1 and has middle digits that are each one greater than the middle digit of the three-digit number. (151)(11) = 1661 (161)(11) = 1771 59. (continued) [3] minor error in explanation 60 -67. [2] incorrect description in part (a) [1] correct products for (151)(11), (161)(11), and (181)(11) 68. B b. 1991 69. N c. No; (191)(11) = 2101 70. G 1 -1

Patterns and Inductive Reasoning GEOMETRY LESSON 1 -1 Find a pattern for each sequence. Use the pattern to show the next two terms or figures. Use the table and inductive reasoning. 1. 3, – 6, 18, – 72, 360 – 2160; 15, 120 2. 3. Find the sum of the first 10 counting numbers. 55 4. Find the sum of the first 1000 counting numbers. 500, 500 Show that the conjecture is false by finding one counterexample. 5. The sum of two prime numbers is an even number. Sample: 2+3=5, and 5 is not even 1 -1

Points, Lines, and Planes GEOMETRY LESSON 1 -2 (For help, go to the Skills Handbook, page 722. ) Solve each system of equations. 1. y = x + 5 y = –x + 7 2. y = 2 x – 4 y = 4 x – 10 4. Copy the diagram of the four points A, B, C, and D. Draw as many different lines as you can to connect pairs of points. 1 -2 3. y = 2 x y = –x + 15

Points, Lines, and Planes GEOMETRY LESSON 1 -2 Solutions 1. By substitution, x + 5 = –x + 7; adding x – 5 to both sides results in 2 x = 2; dividing both sides by 2 results in x = 1. Since x = 1, y = (1) + 5 = 6. (x, y) = (1, 6) 2. By substitution, 2 x – 4 = 4 x – 10; adding – 4 x + 4 to both sides results in – 2 x = – 6; dividing both sides by – 2 results in x = 3. Since x = 3, y = 2(3) – 4 = 6 – 4 = 2. (x, y) = (3, 2) 3. By substitution, 2 x = –x + 15; adding x to both sides results in 3 x = 15; dividing both sides by 3 results in x = 5. Since x = 5, y = 2(5) = 10. (x, y) = (5, 10) 4. The 6 different lines are AB, AC, AD, BC, BD, and CD. 1 -2

Points, Lines, and Planes GEOMETRY LESSON 1 -2 In the figure below, name three points that are collinear and three points that are not collinear. Points Y, Z, and W lie on a line, so they are collinear. Any other set of three points do not lie on a line, so no other set of three points is collinear. For example, X, Y, and Z and X, W, and Z form triangles and are not collinear. 1 -2

Points, Lines, and Planes GEOMETRY LESSON 1 -2 Name the plane shown in two different ways. You can name a plane using any three or more points on that plane that are not collinear. Some possible names for the plane shown are the following: plane RST plane RSU plane RTU plane STU plane RSTU 1 -2

Points, Lines, and Planes GEOMETRY LESSON 1 -2 Use the diagram below. What is the intersection of plane HGC and plane AED? As you look at the cube, the front face is on plane AEFB, the back face is on plane HGC, and the left face is on plane AED. The back and left faces of the cube intersect at HD. Planes HGC and AED intersect vertically at HD. 1 -2

Points, Lines, and Planes GEOMETRY LESSON 1 -2 Shade the plane that contains X, Y, and Z. Points X, Y, and Z are the vertices of one of the four triangular faces of the pyramid. To shade the plane, shade the interior of the triangle formed by X, Y, and Z. 1 -2

Points, Lines, and Planes GEOMETRY LESSON 1 -2 Pages 13– 16 Exercises 1. no 2. yes; line n 3. yes; line n 4. yes; line m 5. yes; line n 6. no 7. no 8. yes; line m 9. Answers may vary. Sample: AE, EC, GA 10. Answers may vary. Sample: BF, CD, DF 11. ABCD 16. BCGH 17. RS 18. VW 19. UV 20. XT 12. EFHG 21. planes QUX and QUV 13. ABHF 22. planes XTS and QTS 14. EDCG 23. planes UXT and WXT 15. EFAD 24. UVW and RVW 1 -2

Points, Lines, and Planes GEOMETRY LESSON 1 -2 25. 27. 29. 30. S 26. 28. 31. X 32. R 33. Q 34. X 1 -2

Points, Lines, and Planes GEOMETRY LESSON 1 -2 35. no 36. yes 37. no 38. coplanar 39. coplanar 40. noncoplanar 41. coplanar 42. noncoplanar 44. Answers may vary. Sample: The plane of the ceiling and the plane of a wall intersect in a line. 45. Through any three noncollinear points there is exactly one plane. The ends of the legs of the tripod represent three noncollinear points, so they rest in one plane. Therefore, the tripod won’t wobble. 43. noncoplanar 1 -2 46. Postulate 1 -1: Through any two points there is exactly one line. 47. Answer may vary. Sample: 48. 49. not possible

Points, Lines, and Planes GEOMETRY LESSON 1 -2 50. 56. 54. 51. not possible 52. no no 55. 57. yes 53. no yes 58. yes 1 -2

Points, Lines, and Planes GEOMETRY LESSON 1 -2 59. 65. never yes 60. always 61. never 68. Answers may vary. Sample: 66. a. 1 b. 1 c. 1 d. 1 e. A line and a point not on the line are always coplanar. 67. Post. 1 -3: If two planes intersect, then they intersect in exactly one line. 62. always 69. A, B, and D 63. always 70. Post. 1 -1: Through any two points there is exactly one line. 64. sometimes Post. 1 -4: Through three noncollinear points there is exactly one plane. 1 -2

Points, Lines, and Planes GEOMETRY LESSON 1 -2 71. Post. 1 -3: If two planes intersect, then they intersect in exactly one line. 72. The end of one leg might not be coplanar with the ends of the other three legs. (Post. 1 -4) 73. 74. 76. yes no 75. 77. no yes 1 -2

Points, Lines, and Planes GEOMETRY LESSON 1 -2 78. 80. no 79. Infinitely many; explanations may vary. Sample: Infinitely many planes can intersect in one line. By Post. 1 -1, points D and B determine a line and points A and D determine a line. The distress signal is on both lines and, by Post. 1 -2, there can be only one distress signal. 1 -2 81. a. Since the plane is flat, the line would have to curve so as to contain the 2 points and not lie in the plane; but lines are straight. b. One plane; Points A, B, and C are noncollinear. By Post. 1 -4, they are coplanar. Then, by part (a), AB and BC are coplanar. 82. 1

Points, Lines, and Planes GEOMETRY LESSON 1 -2 83. 1 4 90. 91. I, K 84. 1 92. 42, 56 85. A 93. 1024, 4096 86. I 94. 25, – 5 87. B 95. 34 88. H 96. 44 89. [2] a. ABD, ABC, ACD, BCD b. AD, BD, CD [1] one part correct The pattern 3, 9, 7, 1 repeats 11 times for n = 1 to 44. For n = 45, the last digit is 3. 1 -2

Points, Lines, and Planes GEOMETRY LESSON 1 -2 Use the diagram at right. 1. Name three collinear points. D, J, and H 2. Name two different planes that contain points C and G. planes BCGF and CGHD 3. Name the intersection of plane AED and plane HEG. HE 4. How many planes contain the points A, F, and H? 1 5. Show that this conjecture is false by finding one counterexample: Two planes always intersect in exactly one line. Sample: Planes AEHD and BFGC never intersect. 1 -2

Segments, Rays, Parallel Lines and Planes GEOMETRY LESSON 1 -3 (For help, go to Lesson 1 -2. ) Judging by appearances, will the lines intersect? 1. 2. 3. Name the plane represented by each surface of the box. 4. the bottom 5. the top 6. the front 7. the back 8. the left side 9. the right side 1 -3

Segments, Rays, Parallel Lines and Planes GEOMETRY LESSON 1 -3 Solutions 1. no 2. yes 3. yes 4 -9. Answers may vary. Samples given: 4. NMR 5. PQL 6. NKL 7. PQR 8. PKN 9. LQR 1 -3

Segments, Rays, Parallel Lines and Planes GEOMETRY LESSON 1 -3 Name the segments and rays in the figure. The labeled points in the figure are A, B, and C. A segment is a part of a line consisting of two endpoints and all points between them. A segment is named by its two endpoints. So the segments are BA (or AB) and BC (or CB). A ray is a part of a line consisting of one endpoint and all the points of the line on one side of that endpoint. A ray is named by its endpoint first, followed by any other point on the ray. So the rays are BA and BC. 1 -3

Segments, Rays, Parallel Lines and Planes GEOMETRY LESSON 1 -3 Use the figure below. Name all segments that are parallel to AE. Name all segments that are skew to AE. Parallel segments lie in the same plane, and the lines that contain them do not intersect. The three segments in the figure above that are parallel to AE are BF, CG, and DH. Skew lines are lines that do not lie in the same plane. The four lines in the figure that do not lie in the same plane as AE are BC, CD, FG, and GH. 1 -3

Segments, Rays, Parallel Lines and Planes GEOMETRY LESSON 1 -3 Identify a pair of parallel planes in your classroom. Planes are parallel if they do not intersect. If the walls of your classroom are vertical, opposite walls are parts of parallel planes. If the ceiling and floor of the classroom are level, they are parts of parallel planes. 1 -3

Segments, Rays, Parallel Lines and Planes GEOMETRY LESSON 1 -3 Pages 19 -23 Exercises 1. 2. 7. a. TS or TR, TW b. SR, ST 12. BC 13. BE, CF 3. 8. 4; RY, SY, TY, WY 14. DE, EF, BE 4. 9. Answers may vary. Sample: 2; YS or YR, YT or YW 15. AD, AB, AC 10. Answers may vary. Check students’ work. 17. ABC || DEF 5. RS, RT, RW, ST, SW, TW 6. RS, ST, TW, WT, TS, SR 11. DF 1 -3 16. BC, EF

Segments, Rays, Parallel Lines and Planes GEOMETRY LESSON 1 -3 18 -20 Answers may vary. Samples are given 18. BE || AD 19. CF, DE 20. DEF, BC 21. FG 22. Answers may vary. Sample: CD, AB 23. BG, DH, CL 25. true 31. False; they are ||. 26. False; they are skew. 32. False; they are ||. 27. true 33. Yes; both name the segment with endpoints X and Y. 28. False; they intersect above CG. 29. true 34. No; the two rays have different endpoints. 30. False; they intersect above pt. A. 35. Yes; both are the line through pts. X and Y. 24. AF 1 -3

Segments, Rays, Parallel Lines and Planes GEOMETRY LESSON 1 -3 41. never 36. 42. sometimes 43. always 44. sometimes 37. always 38. never 39. always 40. always 49. a. Answers may vary. Sample: northeast and southwest b. Answers may vary. Sample: northwest and southeast, east and west 50. Two lines can be parallel, skew, or intersecting in one 46. sometimes point. Sample: train tracks–parallel; vapor 47. sometimes trail of a northbound jet and an eastbound jet at 48. Answers may vary. different altitudes– Sample: (0, 0); check skew; streets that cross students’ graphs. –intersecting 45. always 1 -3

Segments, Rays, Parallel Lines and Planes GEOMETRY LESSON 1 -3 51. Answers may vary. 55. a. The lines of Sample: Skew lines intersection are cannot be contained in parallel. one plane. Therefore, they have “escaped” a b. Examples may vary. plane. Sample: The floor and ceiling are 52. ST || UV parallel. A wall intersects both. The lines of intersection 53. Answers may vary. are parallel. Sample: XY and ZW intersect at R. 56. Answers may vary. Sample: The diamond structure makes it tough, strong, hard, and durable. The graphite structure makes it soft and slippery. 57. a. one segment; EF b. 54. Planes ABC and DCBF intersect in BC. 3 segments; EF, EG, FG 1 -3

Segments, Rays, Parallel Lines and Planes GEOMETRY LESSON 1 -3 58. No; two different planes 61. QR cannot intersect in 62. Yes; no; yes; more than one line. explanations may vary. 59. yes; plane P, for 63. D example 57. c. Answers may vary. Sample: For each “new” point, the number of new segments equals the number of “old” points. d. 45 segments e. n(n – 1) 64. H 65. B 66. F 67. B 60. Answers may vary. Sample: VR, QR, SR 68. C 69. D 2 1 -3

Segments, Rays, Parallel Lines and Planes GEOMETRY LESSON 1 -3 71– 78. Answers may vary. 70. [2] a. Alike: They do Samples are given. not intersect. Different: Parallel 71. EF lines are coplanar and skew lines lie 72. A in different planes. 73. C b. No; of the 8 other lines shown, 4 intersect JM and 4 are skew to JM. [1] one likeness, one difference 79. 80. 81. 82. 1. 4, 1. 48 83. – 22, – 29 74. AEF and HEF 84. FG, GH 75. ABH 85. P, S 76. EHG 86. No; whenever you subtract a negative number, the answer is greater than the given number. Also, if you subtract 0, the answer stays the same. 77. FG 78. B 1 -3

Segments, Rays, Parallel Lines and Planes GEOMETRY LESSON 1 -3 Use the figure below for Exercises 1 -3. Use the figure below for Exercises 4 and 5. 1. Name the segments that 4. Name a pair of parallel planes. form the triangle. RS, TR, ST plane BCD || plane XWQ 2. Name the rays that have point T 5. Name a line that is skew to XW. as their endpoint. TO, TP, TR, TS AC or BD 3. Explain how you can tell that no lines in the figure are parallel or skew. The three pairs of lines intersect, so they cannot be parallel or skew. 1 -3

Measuring Segments and Angles GEOMETRY LESSON 1 -4 (For help, go to the Skills Handbook, pages 719 and 720. ) Simplify each absolute value expression. 1. |– 6| 2. |3. 5| 3. |7 – 10| 4. |– 4 – 2| 5. |– 2 – (– 4)| 6. |– 3 + 12| Solve each equation. 7. x + 2 x – 6 = 6 8. 3 x + 9 + 5 x = 81 9. w – 2 = – 4 + 7 w 1 -4

Measuring Segments and Angles GEOMETRY LESSON 1 -4 Solutions 1. The number of units from 0 to – 6 on the number line is 6. 2. The number of units from 0 to 3. 5 on the number line is 3. 5. 3. |7 – 10| = |– 3|, and the number of units from 0 to – 3 on the number line is 3. 4. |– 4 – 2| = |– 6|, and the number of units from 0 to – 6 on the number line is 6. 5. |– 2 – (– 4)| = |– 2 + 4| = |2|, and the number of units from 0 to 2 on the number line is 2. 1 -4

Measuring Segments and Angles GEOMETRY LESSON 1 -4 Solutions (continued) 6. |– 3 + 12| = |9|, and the number of units from 0 to 9 on the number line is 9. 7. Combine like terms: 3 x – 6 = 6; add 6 to both sides: 3 x = 12; divide both sides by 3: x = 4 8. Combine like terms: 8 x + 9 = 81; subtract 9 from both sides: 8 x = 72; divide both sides by 8: x = 9 9. Add – 7 w + 2 to both sides: – 6 w = – 2; divide both sides by – 6: w = 1 3 1 -4

Measuring Segments and Angles GEOMETRY LESSON 1 -4 Find which two of the segments XY, ZY, and ZW are congruent. Use the Ruler Postulate to find the length of each segment. XY = | – 5 – (– 1)| = | – 4| = 4 ZY = | 2 – (– 1)| = |3| = 3 ZW = | 2 – 6| = |– 4| = 4 Because XY = ZW, XY ZW. 1 -4

Measuring Segments and Angles GEOMETRY LESSON 1 -4 If AB = 25, find the value of x. Then find AN and NB. Use the Segment Addition Postulate to write an equation. AN + NB = AB (2 x – 6) + (x + 7) = 25 3 x + 1 = 25 3 x = 24 x=8 AN = 2 x – 6 = 2(8) – 6 = 10 NB = x + 7 = (8) + 7 = 15 Segment Addition Postulate Substitute. Simplify the left side. Subtract 1 from each side. Divide each side by 3. Substitute 8 for x. AN = 10 and NB = 15, which checks because the sum of the segment lengths equals 25. 1 -4

Measuring Segments and Angles GEOMETRY LESSON 1 -4 M is the midpoint of RT. Find RM, MT, and RT. Use the definition of midpoint to write an equation. RM = MT Definition of midpoint 5 x + 9 = 8 x – 36 Substitute. 5 x + 45 = 8 x Add 36 to each side. 45 = 3 x Subtract 5 x from each side. 15 = x Divide each side by 3. RM = 5 x + 9 = 5(15) + 9 = 84 MT = 8 x – 36 = 8(15) – 36 = 84 Substitute 15 for x. RT = RM + M�T= 168 RM and MT are each 84, which is half of 168, the length of RT. 1 -4

Measuring Segments and Angles GEOMETRY LESSON 1 -4 Name the angle below in four ways. The name can be the number between the sides of the angle: The name can be the vertex of the angle: 3. G. Finally, the name can be a point on one side, the vertex, and a point on the other side of the angle: AGC, CGA. 1 -4

Measuring Segments and Angles GEOMETRY LESSON 1 -4 Find the measure of each angle. Classify each as acute, right, obtuse, or straight. Use a protractor to measure each angle. m 1 = 110 Because 90 < 110 < 180, m 1 is obtuse. 2 = 80 Because 0 < 80 < 90, 2 is acute. 1 -4

Measuring Segments and Angles GEOMETRY LESSON 1 -4 Suppose that m 1 = 42 and m ABC = 88. Find m 2. Use the Angle Addition Postulate to solve. m 1+m 2=m ABC Angle Addition Postulate. 42 + m 2 = 88 Substitute 42 for m m 2 = 46 Subtract 42 from each side. 1 -4 1 and 88 for m ABC.

Measuring Segments and Angles GEOMETRY LESSON 1 -4 Pages 29– 33 Exercises 1. 9; 9; yes 9. 25 15. 130 2. 9; 6; no 10. a. 13 16. XYZ, ZYX, Y 17. MCP, 1 PCM, 18. ABC, CBA 19. CBD, DBC 3. 11; 13; no 4. 7; 6; no 5. XY = ZW 6. ZX = WY b. RS = 40, ST = 24 11. a. 7 b. RS = 60, ST = 36, RT = 96 12. a. 9 b. 9; 18 7. YZ < XW 13. 33 8. 24 14. 34 1 -4 C or

Measuring Segments and Angles GEOMETRY LESSON 1 -4 20 -23. Drawings may vary. 20. 21. 24. 60; acute 33. – 2. 5, 2. 5 25. 90; right 34. – 3. 5, 3. 5 26. 135; obtuse 35. – 6, – 1, 1, 6 27. 34 36. a. 78 mi b. Answers may vary. Sample: measuring with a ruler 28. 70 22. 29. Q 37– 41. Check students’ work. 30. 6 23. 31. – 4 32. 1 1 -4

Measuring Segments and Angles GEOMETRY LESSON 1 -4 42. true; AB = 2, CD = 2 43. false; BD = 9, CD = 2 49. Answers may vary. Sample: (15, 0), (– 9, 0), (3, 12), (3, – 12) 44. false; AC = 9, BD = 9, AD = 11, and 9 + 9 =/ 11 50– 54. Check students’ work. 45. true; AC = 9, CD = 2, AD = 11, and 9 + 2 = 11 56– 58. Answers may vary. Samples are given. 55. about 42° 60. 150 61. 30 62. 100 63. 40 64. 80 65. 125 56. 3: 00, 9: 00 46. 2, 12 66. 125 57. 5: 00, 7: 00 47. 115 58. 6: 00, 12: 32 48. 65 59. 180 1 -4

Measuring Segments and Angles GEOMETRY LESSON 1 -4 67– 68. Answers may vary. Samples are given. 67. 68. 69. QVM and MNP and MQV and VPN 71. y = 15; AC = 24, DC = 12 72. ED = 10, DB = 10, EB = 20 MVN PNQ 70. a. 19. 5 b. 43; 137 c. Answers may vary. Sample: The sum of the measures should be 180. 73. a. Answers may vary. Sample: The two rays come together at a sharp point. 75. 12; m AOC = 82, m AOB = 32, m BOC = 50 76. 8; m AOB = 30, m BOC = 50, m COD = 30 77. 18; m AOB = 28, m BOC = 52, m AOD = 108 b. Answers may vary. Sample: Molly had an acute pain in her 78. 7; m AOB = 28, m BOC = 49, knee. m AOD = 111 74. 45, 75, and 165, or 135, 105, and 15 1 -4 79. 30

Measuring Segments and Angles GEOMETRY LESSON 1 -4 80. a–c. Check students’ 86. [2] a. work. 87. never 81. Angle Add. Post. 89. always 82. C 83. F 84. D 85. H 88. never b. An obtuse measures between 90 and 180 degrees; the least and greatest whole number values are 91 and 179 degrees. Part of ABC is 12°. So the least and greatest measures for DBC are 79 and 167. [1] one part correct 1 -4 90. never 91. always 92. always 93. always 94. never 95. 25, 30 96. 3125; 15, 625 97. 30, 34

Measuring Segments and Angles GEOMETRY LESSON 1 -4 Use the figure below for Exercises 1 -3. Use the figure below for Exercises 4– 6. 1. If XT = 12 and XZ = 21, then TZ = 7. 9 2. If XZ = 3 x, XT = x + 3, and TZ = 13, find XZ. 24 3. Suppose that T is the midpoint of XZ. If XT = 2 x + 11 and XZ = 5 x + 8, find the value of x. 14 4. Name 2 two different ways. DAB, BAD 5. Measure and classify 1, 2, and BAC. 90°, right; 30°, acute; 120°, obtuse 6. Which postulate relates the measures of 1, 2, and BAC? Angle Addition Postulate 1 -4

Basic Construction GEOMETRY LESSON 1 -5 (For help, go to Lesson 1 -3 and 1 -4. ) In Exercises 1 -6, sketch each figure. 1. CD 2. GH 4. line m 5. acute 3. AB ABC 6. XY || ST 7. DE = 20. Point C is the midpoint of DE. Find CE. 8. Use a protractor to draw a 60° angle. 9. Use a protractor to draw a 120° angle. 1 -5

Basic Construction GEOMETRY LESSON 1 -5 Solutions 1 -6. Answers may vary. Samples given: 1. The figure is a segment whose endpoints are C and D. 2. The figure is a ray whose endpoint is G. 3. The figure is a line passing through points A and B. 4. 5. The figure is an angle whose measure is between 0° and 90°. 6. The figure is two segments in a plane whose corresponding lines are parallel. 1 -5

Basic Construction GEOMETRY LESSON 1 -5 Solutions (continued) 7. Since C is a midpoint, CD = CE; also, CD + CE = 20; substituting results in CE + CE = 20, or 2 CE = 20, so CE = 10. 8. 9. 1 -5

Basic Construction GEOMETRY LESSON 1 -5 Construct TW congruent to KM. Step 1: Draw a ray with endpoint T. Step 2: Open the compass to the length of KM. Step 3: With the same compass setting, put the compass point on point T. Draw an arc that intersects the ray. Label the point of intersection W. TW KM 1 -5

Basic Construction GEOMETRY LESSON 1 -5 Construct Y so that Y G. Step 1: Draw a ray with endpoint Y. Step 2: With the compass point on point G, draw an arc that intersects both sides of G. Label the points of intersection E and F. Step 3: With the same compass setting, put the compass point on point Y. Draw an arc that intersects the ray. Label the point of intersection Z. 1 -5 75°

Basic Construction GEOMETRY LESSON 1 -5 (continued) Step 4: Open the compass to the length EF. Keeping the same compass setting, put the compass point on Z. Draw an arc that intersects the arc you drew in Step 3. Label the point of intersection X. Step 5: Draw YX to complete Y Y. G 1 -5

Basic Construction GEOMETRY LESSON 1 -5 1 Use a compass opening less than 2 AB. Explain why the construction of the perpendicular bisector of AB shown in the text is not possible. Start with AB. Step 1: Put the compass point on point A and draw a short arc. Make 1 sure that the opening is less than AB. 2 Step 2: With the same compass setting, put the compass point on point B and draw a short arc. Without two points of intersection, no line can be drawn, so the perpendicular bisector cannot be drawn. 1 -5

Basic Construction GEOMETRY LESSON 1 -5 m WR bisects AWB. m AWR = x and BWR = 4 x – 48. Find m AWB. Draw and label a figure to illustrate the problem m m AWR = m BWR x = 4 x – 48 Definition of angle bisector Substitute x for m AWR and 4 x – 48 for m BWR. Subtract 4 x from each side. Divide each side by – 3 x = – 48 x = 16 AWR = 16 Substitute 16 for x. BWR = 4(16) – 48 = 16 AWB = m AWR + m BWR Angle Addition Postulate AWB = 16 + 16 = 32 Substitute 16 for m AWR and for m BWR. 1 -5

Basic Construction GEOMETRY LESSON 1 -5 Construct MX, the bisector of Step 1: Put the compass point on vertex M. Draw an arc that intersects both sides of M. Label the points of intersection B and C. Step 2: Put the compass point on point B. Draw an arc in the interior of M. 1 -5 M.

Basic Construction GEOMETRY LESSON 1 -5 (continued) Step 3: Put the compass point on point C. Using the same compass setting, draw an arc in the interior of M. Make sure that the arcs intersect. Label the point where the two arcs intersect X. Step 4: Draw MX. MX is the angle bisector of M. 1 -5

Basic Construction GEOMETRY LESSON 1 -5 Pages 37 -40 Exercises 1. 9. a. 11; 30 6. b. 30 c. 60 2. 10. 5; 50 3. 7. 11. 15; 48 12. 11; 56 4. 13. 5. 8. 1 -5

Basic Construction GEOMETRY LESSON 1 -5 14. 16. Find a segment on XY so that you can construct YZ as its bisector. 15. 1 -5 17. Find a segment on SQ so that you can construct SP as its bisector. Then bisect PSQ.

Basic Construction GEOMETRY LESSON 1 -5 21. (continued) b. Infinitely many; there’s only 1 midpt. but there exist infinitely many lines 19. a-b. through the midpt. A segment has exactly one bisecting line because there can be only one line 21. Explanations may vary. to a segment at its Samples are given. midpt. a. One midpt. ; a midpt. 20. Locate points A and B divides a segment into on a line. Then c. There an infinite two segments. If construct a at A and number of lines in there were more than B as in Exercise 16. space that are to a one midpt. the Construct AD and BC segment at its midpt. segments wouldn’t be. so that AB = AD = BC. The lines are coplanar. 18. a. CBD; 41 b. 82 c. 49; 49 20. (continued) 1 -5

Basic Construction GEOMETRY LESSON 1 -5 22. 23. 24. 25. They are both correct. If you mult. each side of Lani’s eq. by 2, the result is Denyse’s eq. 27. 26. Open the compass to more than half the 28. a. measure of the segment. Swing large arcs from the endpts. to intersect above and below the segment. Draw a line through the two pts. where the arcs intersect. The pt. where They appear to the line and segment meet at one pt. intersect is the midpt. of the segment. 1 -5

Basic Construction GEOMETRY LESSON 1 -5 28. (continued) b. 30. 33. a. c. The three bisectors of a intersect in one pt. 29. 31. impossible; the short segments are not long enough to form a. 32. impossible; the short segments are not long enough to form a. 1 -5 b. They are all 60°. c. Answers may vary. Sample: Mark a pt. , A. Swing a long arc from A. From a pt. P on the arc, swing another arc the same size that intersects the arc at a second pt. , Q. Draw PAQ. To construct a 30° , bisect the 60°.

Basic Construction GEOMETRY LESSON 1 -5 34. a-c. 35, (continued) c. Point O is the center of the circle. 36. ; the line intersects. 37. D 38. F 35. a-b. 39. [2] a. Draw XY. With the compass pt. on B swing an arc that intersects BA and BC. Label the intersections P and Q, respectively. With the compass point on X, swing a arc intersecting XY. 1 -5 39. [2] (continued) Label the intersection K. Open the compass to PQ. With compass pt. on K, swing an arc to intersect the first arc. Label the intersection R. Draw XR.

Basic Construction GEOMETRY LESSON 1 -5 41. 39. [2] b. With compass 40. (continued) open to XK, put 42. c. Draw AB. Do compass point on X constructions as in 43. and swing an arc parts a and b. Open 44. intersecting XR. With the compass to the 45. compass on R and length of the shortest open to KR, swing an segment in part b. arc to intersect the first With the pt. of the arc. Label intersection compass on B, swing 46. T. Draw XT. an arc in the opp. 47. direction from A [1] one part correct intersecting AB at C. 48. 40. [4] a. Construct its AC = 1. 25 (AB). bisector. 49. b. Construct the bisector. [3] explanations are not Then construct the thorough 50. bisector of two new [2] two explanations correct segments. [1] part (a) correct 1 -5 6 10 4 3 100 20 and 180 No; they do not have the same endpt. Yes; they both represent a segment with endpts. R and S.

Basic Construction GEOMETRY LESSON 1 -5 Use the figure at right. NQ bisects DNB. For problems 1 -4, check students’ work. 1. Construct AC so that AC NB. 2. Construct the perpendicular bisector of AC. 3. Construct RST so that RST 4. Construct the bisector of RST. QNB. 5. Find x. 17 6. Find m DNB. 88 1 -5

The Coordinate Plane GEOMETRY LESSON 1 -6 (For help, go to the Skills Handbook, pages 715 and 716. ) Find the square root of each number. Round to the nearest tenth if necessary. 1. 25 2. 17 3. 123 Evaluate each expression for m = – 3 and n = 7. 4. (m – n)2 5. (n – m)2 6. m 2 + n 2 Evaluate each expression for a = 6 and b = – 8. 7. (a – b)2 8. a 2 + b 2 9. a+b 2 1 -6

The Coordinate Plane GEOMETRY LESSON 1 -6 Solutions 1. 25 = 52 = 5 3. 123 11. 0912 = 11. 1 2. 17 4. 1232 = 4. 1 4. (m – n)2 = (– 3 – 7)2 = (– 10)2 = 100 5. (n – m)2 = – 7 – (– 3))2 = (7 + 3)2 =102 = 100 6. m 2 + n 2 = (– 3)2 + (7)2 = 9 + 49 = 58 7. (a – b)2 = (6 – (– 8))2 = (6 + 8)2 =142 = 196 8. 9. a 2 + b 2 = = = (6)2 + (– 8)2 36 + 64 100 = 10 1 -6 a+b 6 + (– 8) = 2 2 – 2 = – 1

The Coordinate Plane GEOMETRY LESSON 1 -6 Find the distance between R(– 2, 6) and S(6, – 2) to the nearest tenth. Let (x 1, y 1) be the point R(– 2, 6) and (x 2, y 2) be the point S(6, – 2). d= (x 2 – x 1)2 + (y 2 – y 1)2 Use the Distance Formula. d= (6 – (– 2))2 + (– 2 – (6))2 Substitute. d= 82 + (– 8)2 Simplify. d= 64 + 64 = 128 11. 3137085 Use a calculator. To the nearest tenth, RS = 11. 3. 1 -6

The Coordinate Plane GEOMETRY LESSON 1 -6 How far is the subway ride from Oak to Symphony? Round to the nearest tenth. Oak has coordinates (– 1, – 2). Let (x 1, y 1) represent Oak. Symphony has coordinates (1, 2). Let (x 2, y 2) represent Symphony. d= (x 2 – x 1)2 + (y 2 – y 1)2 Use the Distance Formula. d= (1 – (– 1))2 + (2 – (– 2))2 Substitute. d= 22 + 42 Simplify. d= 4 + 16 20 = 20 4. 472135955 Use a calculator. To the nearest tenth, the subway ride from Oak to Symphony is 4. 5 miles. 1 -6

The Coordinate Plane GEOMETRY LESSON 1 -6 AB has endpoints (8, 9) and (– 6, – 3). Find the coordinates of its midpoint M. Use the Midpoint Formula. Let (x 1, y 1) be A(8, 9) and (x 2, y 2) be B(– 6, – 3). The midpoint has coordinates ( x 1 + x 2 2 , y 1 + y 2 2 Midpoint Formula ) The x–coordinate is 8 + (– 6) 2 = =1 2 2 Substitute 8 for x 1 and (– 6) for x 2. Simplify. The y–coordinate is 9 + (– 3) 6 = =3 2 2 Substitute 9 for y 1 and (– 3) for y 2. Simplify. The coordinates of midpoint M are (1, 3). 1 -6

The Coordinate Plane GEOMETRY LESSON 1 -6 The midpoint of DG is M(– 1, 5). One endpoint is D(1, 4). Find the coordinates of the other endpoint G. Use the Midpoint Formula. Let (x 1, y 1) be D(1, 4) and the midpoint ( x 1 + x 2 2 , y 1 + y 2 2 )be (– 1, 5). Solve for x Find the x–coordinate of G. 2 and y 2, the coordinates of G. Find the y–coordinate of G. 1 + x 2 – 1 = 2 Use the Midpoint Formula. – 2 = 1 + x 2 Multiply each side by 2. The coordinates of G are (– 3, 6). 1 -6 5= 4 + y 2 2 10 = 4 + y 2

The Coordinate Plane GEOMETRY LESSON 1 -6 Pages 46– 49 Exercises 1. 6 11. about 4. 5 mi 21. (6, 1) 2. 18 12. about 3. 2 mi 22. (– 2. 25, 2. 1) 3. 8 13. 6. 4 23. (3 7 , – 3) 4. 9 14. 15. 8 24. (10, – 20) 5. 23. 3 15. 8 25. (5, – 1) 6. 10 16. 5 26. (0, – 34) 7. 25 17. B, C, D, E, F 27. (12, – 24) 8. 12. 2 18. (4, 2) 28. (9, – 28) 9. 12. 0 19. (3, 1) 29. (5. 5, – 13. 5) 10. 9 mi 20. (3. 5, 1) 30. (8, 18) 8 1 -6

The Coordinate Plane GEOMETRY LESSON 1 -6 31. (4, – 11) 40. 2. 2; (3. 5, 1) 32. 5. 0; (4. 5, 4) 41. IV 33. 5. 8; (1. 5, 0. 5) 42. 43. 34. 7. 1; (– 1. 5, 0. 5) ST = (5 – 2)2 + (– 3 – (– 6))2 = 9 + 9 = 3 2 4. 2 TV = (6 – 5)2 + (– 6 – (– 3))2 = 1 + 9 = 10 3. 2 35. 5. 4; (– 2. 5, 3) 36. 10; (1, – 4) 37. 2. 8; (– 4, – 4) 38. 6. 7; (– 2. 5, – 2) 39. 5. 4; (3, 0. 5) The midpts. Are the same, (5, 4). The diagonals bisect each other. VW = (5 – 6)2 + (– 9 – (– 6))2 = 9 + 9 = 3 2 3. 2 SW = (5 – 2)2 + (– 9 – (– 6))2 = 9+9=3 2 4. 2 No, but ST = SW and TV = VW. 1 -6

The Coordinate Plane GEOMETRY LESSON 1 -6 44. 19. 2 units; (– 1. 5, 0) 50. 1073 mi 45. 10. 8 units; (3, – 4) 51. 2693 mi 46. 5. 4 units; (– 1, 0. 5) 52. 328 mi 47. Z; about 12 units 57. exactly one pt. , E (– 5, 2) 58. exactly one pt. , J (2, – 2) 53– 56. Answers may vary. 59. a–f. Answers may Samples are given. vary. Samples are 48. 165 units; The dist. TV given. is less than the dist. 53. (3, 6), (0, 4. 5) TU, so the airplane a. BC = AD should fly from T to V 54. E (0, 0), (8, 4) to U for the shortest b. If two opp. sides of a route. 55. (1, 0), (– 1, 4) quad. are both || and , then the other 49. 934 mi 56. (0, 10), (5, 0) two opp. sides are. 1 -6

The Coordinate Plane GEOMETRY LESSON 1 -6 59. (continued) f. If a pair of opp. sides of a quad. are both || and , then the segment joining d. If one pair of opp. the midpts. of the sides of a quad. are other two sides has both || and , then the same length as its diagonals bisect each of the first pair each other. of sides. e. EF = AB 60. A (0, 0, 0) B (6, 0, 0) C (6, – 3. 5, 0) D (0, – 3. 5, 0) E (0, 0, 9) F (6, 0, 9) G (0, – 3. 5, 9) 59. (continued) c. The midpts. are the same. 1 -6 61. 62. 6. 5 units 63. 11. 7 units 64. B 65. I

The Coordinate Plane GEOMETRY LESSON 1 -6 66. A 70. 73. 71. 74. 10 67. C 68. A 69. [2] a. (– 10, 8), (– 1, 5), (8, 2) b. Yes, R must be (– 10, 8) so that RQ = 160. 75. 10 76. 48 72. 77. TAP, 78. 150 [1] part (a) correct or plausible explanation for part (b) 1 -6 PAT

The Coordinate Plane GEOMETRY LESSON 1 -6 A has coordinates (3, 8). B has coordinates (0, – 4). C has coordinates (– 5, – 6). 1. Find the distance between A and B to the nearest tenth. 12. 4 2. Find BC to the nearest tenth. 5. 4 3. Find the midpoint M of AC to the nearest tenth. (– 1, 1) 4. B is the midpoint of AD. Find the coordinates of endpoint D. (– 3, – 16) 5. An airplane flies from Stanton to Mercury in a straight flight path. Mercury is 300 miles east and 400 miles south of Stanton. How many miles is the flight? 500 mi 6. Toni rides 2 miles north, then 5 miles west, and then 14 miles south. At the end of her ride, how far is Toni from her starting point, measured in a straight line? 13 mi 1 -6

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 (For help, go to the Skills Handbook page 719 and Lesson 1 -6. ) Simplify each absolute value. 1. |4 – 8| 2. |10 – (– 5)| 3. |– 2 – 6| Find the distance between the points to the nearest tenth. 4. A(2, 3), B(5, 9) 5. K(– 1, – 3), L(0, 0) 6. W(4, – 7), Z(10, – 2) 7. C(– 5, 2), D(– 7, 6) 8. M(– 1, – 10), P(– 12, – 3) 9. Q(– 8, – 4), R(– 3, – 10) 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 Solutions 1. | 4 – 8 | = | – 4 | = 4 2. | 10 – (– 5) | = | 10 + 5 | = | 15 | = 15 3. | – 2 – 6 | = | – 8 | = 8 4. d = d= d= (x 2 – x 1)2 + (y 2 – y 1)2 5. d = (5 – 2)2 + (9 – 3)2 3 2 + 62 d= d= d = 9 + 36 = 45 To the nearest tenth, AB = 6. 7. 6. d = d= d= (x 2 – x 1)2 + (y 2 – y 1)2 (0 – (– 1))2 + (0 – (– 3))2 1 2 + 32 d = 1 + 9 = 10 To the nearest tenth, KL = 3. 2. (x 2 – x 1)2 + (y 2 – y 1)2 7. d = (10 – 4)2 + ( – 2 –(– 7))2 6 2 + 52 d= d= d = 36 + 25 = 61 To the nearest tenth, WZ = 7. 8. (x 2 – x 1)2 + (y 2 – y 1)2 (– 7 – (– 5))2 + (6 – 2)2 (– 2)2 + 52 d = 4 + 16 = 20 To the nearest tenth, CD = 4. 5. 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 Solutions (continued) 8. d = d = (x 2 – x 1)2 + (y 2 – y 1)2 (– 12 – (– 1))2 + (– 3 – (– 10))2 (– 11)2 + 72 d = 121 + 49 = 170 To the nearest tenth, MP = 13. 0. 9. d = (x 2 – x 1)2 + (y 2 – y 1)2 d = (– 3 – (– 8))2 + (– 10 – (– 4))2 d = 52 + (– 6)2 d = 25 + 36 = 61 To the nearest tenth, QR = 7. 8. 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 Margaret’s garden is a square 12 ft on each side. Margaret wants a path 1 ft wide around the entire garden. What will the outside perimeter of the path be? Because the path is 1 ft wide, increase each side of the garden by 1 ft. s = 1 + 12 + 1 = 14 P = 4 s Formula for perimeter of a square P = 4(14) = 56 Substitute 14 for s. The perimeter is 56 ft. 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 . G has a radius of 6. 5 cm. Find the circumference of. G in terms of. Then find the circumference to the nearest tenth. C=2 r Formula for circumference of a circle. C = 2 (6. 5) Substitute 6. 5 for r. C = 13 Exact answer. C = 13 40. 840704 Use a calculator. The circumference of. G is 13 , or about 40. 8 cm. 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 Quadrilateral ABCD has vertices A(0, 0), B(9, 12), C(11, 12), and D(2, 0). Find the perimeter. Draw and label ABCD on a coordinate plane. Find the length of each side. Add the lengths to find the perimeter. AB = = (9 – 0)2 + (12 – 0)2 = 81 + 144 = 92 + 122 255 = 15 BC = |11 – 9| = |2| = 2 CD = = Ruler Postulate (2 – 11)2 + (0 – 12)2 = 81 + 144 = Use the Distance Formula. (– 9)2 + (– 12)2 Use the Distance Formula. 255 = 15 DA = |2 – 0| = |2| = 2 Ruler Postulate 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 (continued) Perimeter = AB + BC + CD + DA = 15 + 2 + 15 + 2 = 34 The perimeter of quadrilateral ABCD is 34 units. 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 To make a project, you need a rectangular piece of fabric 36 in. wide and 4 ft long. How many square feet of fabric do you need? Write both dimensions using the same unit of measurement. Find the area of the rectangle using the formula A = bh. 36 in. = 3 ft Change inches to feet using 12 in. = 1 ft. A = bh Formula for area of a rectangle. A = (4)(3) Substitute 4 for b and 3 for h. A = 12 You need 12 ft 2 of fabric. 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 Find the area of. B in terms of. In. B, r = 1. 5 yd. A= r 2 Formula for area of a circle A= (1. 5)2 Substitute 1. 5 for r. A = 2. 25 The area of. B is 2. 25 yd 2. 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 Find the area of the figure below. Draw a horizontal line to separate the figure into three nonoverlapping figures: a rectangle and two squares. 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 (continued) Find each area. Then add the areas. AR = bh AR = (15)(5) AR = 75 Formula for area of a rectangle Substitute 15 for b and 5 for h. AS = s 2 AS = (5)2 AS = 25 Formula for area of a square Substitute 5 for s. A = 75 + 25 A = 125 Add the areas. The area of the figure is 125 ft 2. 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 Pages 55– 58 Exercises 16. 1. 22 in. 9. 10 2. 36 cm 10. 3. 7 3. 56 in. 11. 2 1 ft in. m 14. 6 units 4. 78 cm 12. 56. 5 in. 5. 120 m 13. 22. 9 m 6. 48 in. 14. 1. 6 yd 7. 38 ft 15. 351. 9 cm 17. 8. 15 cm 25. 1 units 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 1 3 18. 20. 1 ft 2 or 192 in. 2 29. 9 21. 4320 in. 2 or 3 yd 2 30. 0. 25 22. 1 1 ft 2 of 162 in. 2 31. 9. 9225 23. 8000 cm 2 or 0. 8 m 2 32. 0. 01 64 8 16 units 19. 24. 5. 7 m 2 or 57, 000 25. 120, 000 26. 6000 27. 400 38 units 28. 64 ft 2 cm 2 or 12 m 2 2 or 666 yd 2 3 m 2 ft 2 1 -7 in. 2 m 2 ft 2 m 2 33. 153. 9 ft 2 34. 54. 1 m 2 35. 452. 4 cm 2 36. 452. 4 in. 2 37. 310 m 2 38. 19 yd 2

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 39. 24 cm 2 43. 3289 m 2 48. Answers may vary. Sample: For Exercise 40. 80 in. 2 44– 47. Answers may vary. 46, you use feet Samples are given. because the bulletin 41. a. 144 in. 2 board is too big for b. 1 ft 2 44. 38 in. ; 90 in. 2 inches. You do not use c. 144; a square yards because your whose sides are 12 45. 39 in. ; 93. 5 in. 2 estimated lengths in in. long and a feet were not divisible square whose sides 46. 12 ft; 8 ft 2 by 3. are 1 ft long are the same size. 47. 8 ft; 3. 75 ft 2 49. 16 cm 42. a. 30 squares b. 16; 9; 4; 1 c. They are =. Post 1 -10 50. 96 cm 2 51. 288 cm 1 -7

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 52. a. Yes; every square is a rectangle. 54. 56. 38 units 57. 54 units 2 b. Answers may vary. Sample: No, not all rectangles are squares. c. A = ( P 4 2 ) or A = P 2 16 58. 1, 620, 000 m 2 perimeter = 10 units area = 4 units 2 59. 30 m 60. (4 x – 2) units 55. 61. Area; the wall is a surface. 53. 512 tiles perimeter = 16 units area = 15 units 2 1 -7 62. Perimeter; weather stripping must fit the edges of the door.

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 63. Perimeter; the fence 66. a. base must fit the perimeter of 1 the garden. 2 3 64. Area; the floor is a surface. 24 25 65. 6. 25 units 2 26 47 48 49 height 98 96 94 : : 52 50 48 : : 6 4 2 1 -7 area 98 192 282 b. 1248 1250 1248 282 192 98 c. 25 ft by 50 ft

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 67. a. b. c. d. 9 9 3 a 68. 20 units 2 2 69. 25 a units 2 4 70. (9 m 2 – 24 mn + 16 n 2) units 2 71. Answers may vary. Sample: one 8 in. -by-8 in. square + one 5 in. by-5 in. square + two 4 in. -by-4 in. squares 72. 388. 5 yd 73. 64 74. 2336 83. 9. 2 units; (1, 6. 5) 75. 540 84. 6. 7 units; (– 2. 5, – 2) 76. 216 85. 90 77. 810 86. WI 78. (15, 13) 87. 62 units 79. 8. 5 units; (5. 5, 5) 88. 18 units 80. 5. 8 units; (1. 5, 5. 5) 89. 6 units 81. 13. 9 units; (3, 5. 5) 90. 33 units 82. 6. 4 units; (– 2, 3. 5) 1 -7 RI

Perimeter, Circumference, and Area GEOMETRY LESSON 1 -7 A rectangle is 9 ft long and 40 in. wide. 1. Find the perimeter in inches. 296 in. 2. Find the area in square feet. 30 ft 2 3. The diameter of a circle is 18 cm. Find the area in terms of. 81 cm 2 4. Find the perimeter of a triangle whose vertices are X(– 6, 2), Y(8, 2), and Z(3, 14). 42 units 5. Find the area of the figure below. All angles are right angles. 256 in. 2 1 -7

Tools of Geometry GEOMETRY CHAPTER 1 Page 64 1. Div. each preceding term by – 2; 1 , – 1 2 4 2. Add 2 to the preceding term; 10, 12 4. Answers may vary. 8. B Sample: 1, 2, 4, 8, 16, 32, . . . 9. a. 1 1, 2, 4, 7, 11, 16, . . . b. infinitely many In the first seq. double c. 1 each term. In the second d. 1 seq. , add consecutive counting numbers. 10. 29, 054. 0 ft 2 3. Rotate the U clockwise 5. A, B, C one-quarter turn. Alphabet is backwards; 6. Answers may vary. Sample: A, B, C, D 7. Answers may vary. Sample: A, B, D, E 11. never 12. sometimes 13. never 14. always 15. never 1 -A

Tools of Geometry GEOMETRY CHAPTER 1 16. 10 17. a. (11, 19) b. MC = MD = 136 24. Answers may vary. Sample: Some ways of naming an can help identify a side or vertex. 25. 18. 19. 1 units 19. 800 cm 2 or 0. 08 m 2 20. 12. 25 in. 2 21. 63. 62 cm 2 22. 7 23. 9 26. bisector 27. VW 28. 7 units 1 -A 29. AY 30. E, AY 31. 33 1 yd 2 3