Big Ideas Math Geometry Answers Chapter 8 Similarity is the one-stop destination for all your requirements during preparation. Practicing and Solving no. of questions from chapter 8 similarity BIM Geometry Textbook Solutions is the best way to understand the concepts easily and gain more subject knowledge. Kickstart your preparation by taking the help of BigIdeas math geometry ch 8 similarity Answer key pdf and test your skills up-to-date for better scoring in the exams and become math proficient. Show
Big Ideas Math Book Geometry Answer Key Chapter 8 SimilarityGet the handy preparation material free of cost and enhance your math skills with ease. To make this possible, students have to access the given links below and download the Big Ideas Math Geometry Textbook Answers Ch 8 Similarity in Pdf. In this guide, teachers, parents, and kids can find various questions from basic to complex levels. By solving the questions of chapter 8 similarity covered in BIM Geometry Ch 8 Answer key exercises, practice tests, chapter tests, cumulative assessments, etc. helps you to improve your skills & knowledge. Similarity Maintaining Mathematical ProficiencyTell whether the ratios form a proportion. Question 1. Answer: Explanation: Question 2. Answer: Explanation: Question 3. Answer: Explanation: Question 4. Answer: Explanation: Question 5. Answer: Explanation: Question 6. Answer: Explanation: Find the scale factor of the dilation. Question 7. Answer: Explanation: Question 8. Answer: Explanation: Question 9. Answer: Explanation: Question 10. Answer: Explanation: Similarity Mathematical PracticesMonitoring Progress Question 1. Answer: Explanation: Question 2. Answer: Explanation: Question 3. Answer: Explanation: 8.1 Similar PolygonsExploration 1 Comparing Triangles after a Dilation Work with a partner: Use dynamic geometry software to draw any ∆ABC. Dilate ∆ABC to form a similar ∆A’B’C’ using an scale factor k and an center of dilation. a. Compare the corresponding angles of ∆A’B’C and ∆ABC. b. Find the ratios of the lengths of the sides of ∆A’B’C’ to the lengths of the corresponding sides of ∆ABC. What do you observe? c. Repeat parts (a) and (b) for several other triangles, scale factors, and centers of dilation. Do you obtain similar results? Exploration 2 Comparing Triangles after a Dilation Work with a partner: Use dynamic geometry Software to draw any ∆ABC. Dilate ∆ABC to form a similar ∆A’B’C’ using any scale factor k and any center of dilation. a. Compare the perimeters of ∆A’B’C and ∆ABC. What do you observe? b. Compare the areas of ∆A’B’C’ and ∆ABC. What do you observe? c. Repeat parts (a) and (b) for several other triangles, scale factors, and centers of dilation. Do you obtain similar results? Communicate Your Answer Question 3. Answer: Question 4. Answer: Explanation: Lesson 8.1 Similar PolygonsMonitoring Progress Question 1. Answer: Explanation: Question 2. ABCD ~ QRST Answer: Explanation: Question 3. ∆JKL ~ ∆EFG Answer: Explanation: Question 4. Answer: Explanation: Question 5. Area of GHJK = 84m2 Answer: Explanation: Question 6. Answer: Both the hexagons are different. On the outer side of the hexagon, all the sides are equal. In the inside hexagon among the 6-sided 3 sides are different. Question 7. Answer: Both the hexagons are similar. Because all the sides of the outer hexagon are equal and all the sides of the inner hexagon are also equal. Exercise 8.1 Similar PolygonsVocabulary and Core Concept Check Question 1. Answer: Question 2. What is the scale factor? Answer: Scale Factor = \(\frac{image-length}{actual-length}\) = \(\frac{3}{12}\)= \(\frac{4}{16}\)= \(\frac{5}{20}\) = \(\frac{1}{4}\) What is the ratio of their areas? Answer: What is the ratio of their corresponding side lengths? What is the ratio of their perimeters? Monitoring Progress and Modeling with Mathematics In Exercises 3 and 4, find the scale factor. Then list all pairs of congruent angles and write the ratios of the corresponding side lengths in a statement of proportionality. Question 3. Answer: Question 4. Answer: In Exercises 5-8, the polygons are similar. Find the value of x. Question 5. Answer: Question 6. Answer: Explanation: Question 7. Answer: Question 8. Answer: Explanation: In Exercises 9 and 10, the black triangles are similar. Identify the type of segment shown in blue and find the value of the variab1e. Question 9. Answer: Question 10. Answer: In Exercises 11 and 12, RSTU ~ ABCD. Find the ratio of their perimeters. Question 11. Answer: Question 12. Answer: \(\frac { RS + ST + TU + UR }{ AB + BC + CD + DA } \) = \(\frac { RS }{ AB } \) = \(\frac { 18 }{ 24 } \) The ratio of perimeter is \(\frac { 3 }{ 4 } \). In Exercises 13-16, two polygons are similar. The perimeter of one polygon and the ratio of the corresponding side lengths are given. Find the perimeter of the other polygon. Question 13. Question 14. Answer: Explanation: Question 15. Question 16. Answer: Explanation: Question 17. Question 18. Answer: Explanation: Perimeter of backyard = 2(45 + 20) = 2(65) = 130 ft Scale factor = \(\frac { 18 }{ 45 } \) = \(\frac { 2 }{ 5 } \) So, \(\frac { perimeter of patio }{ perimeter of backyard } \) = \(\frac { 2 }{ 5 } \) \(\frac { perimeter of patio }{ 130 } \) = \(\frac { 2 }{ 5 } \) Perimeter of patio = \(\frac { 260 }{ 5 } \) = 52 ft In Exercises 19-22, the polygons are similar. The area of one polygon is given. Find the area of the other polygon. Question 19. Answer: Question 20. Answer: Explanation: Question 21. Answer: Question 22. Answer: Explanation: Question 23. Answer: Question 24. Answer: In Exercises 25 and 26, decide whether the red and blue polygons are similar. Question 25. Answer: Question 26. Answer: Question 27. ANALYZING RELATIONSHIPS Question 28. Answer: Question 29. Question 30. Answer: Explanation: Question 31. Question 32. Answer: Question 33. Question 34. Answer: Question 35. Question 36. Answer: Explanation: MATHEMATICAL CONNECTIONS Question 37. Answer: Question 38. Answer: Explanation: ATTENDING TO PRECISION Question 39. Question 40. Answer: Explanation: Question 41. Question 42. Answer: Explanation: CRITICAL THINKING Question 43. Question 44. Answer: Question 45. Question 46. Answer: Question 47. Question 48. Answer: Question 49. Question 50. Answer: The object and image are similar. Question 51. Answer: Question 52. Answer: \(\frac { PQ + QR + RS + SP }{ KL + LM + MN + NK } \) = \(\frac { PQ }{ KL } \) = \(\frac { QR }{ LM } \) = \(\frac { RS }{ MN } \) = \(\frac { SP }{ NK } \) Question 53. Question 54.
Question 55. Answer: Question 56. Answer: The two lines slopes are equal and triangles angles are congruent and side lengths are proportional. So, triangles are similar. Maintaining Mathematical proficiency Find the value of x. Question 57. Answer: Question 58. Answer: Explanation: Question 59. Answer: Question 60. Answer: Explanation: 8.2 Proving Triangle Similarity by AAExploration 1 Comparing Triangles Work with a partner. Use dynamic geometry software. a. Construct ∆ABC and ∆DEF So that m∠A = m∠D = 106°, m∠B = m∠E = 31°, and ∆DEF is not congruent to ∆ABC. Answer: m∠C ≠ m∠F b. Find the third angle measure and the side lengths of each triangle. Copy the table below and record our results in column 1. Answer: c. Are the two triangles similar? Explain. d. Repeat parts (a) – (c) to complete columns 2 and 3 of the table for the given angle measures. e. Complete each remaining column of the table using your own choice of two pairs of equal corresponding angle measures. Can you construct two triangles in this way that are not similar? f. Make a conjecture about any two triangles with two pairs of congruent corresponding angles. Communicate Your Answer Question 2. Question 3. Lesson 8.2 Proving Triangle Similarity by AAMonitoring Progress Show that the triangles are similar. Write a similarity statement. Question 1. Answer: ∆FGH and ∆RQS are similar by the AA similarity theorem. Question 2. Answer: Question 3. Answer: You know∠SVR≅∠UVT by the vertical angles congruence theorem. The diagram shows RS ∥UT so∠S≡∠U by the Alternate Interior Angles Theorem. So,△SVR∼△UVT by the AA similarity postulate. If,SR ∦ TU it breaks the AA singularity postulate. So,△SVR≠△UVT. Question 4. Question 5. Exercise 8.2 Proving Triangle Similarity by AAVocabulary and Core Concept Check Question 1. Question 2. Monitoring Progress and Modeling with Mathematics In Exercises 3 – 6. determine whether the triangles are similar. If they are, write a similarity statement. Explain your reasoning. Question 3. Answer: Question 4. Answer: Sum of all the angles of a triangle = 180° m∠R + m∠Q + m∠S = 180 m∠R + 85 + 35 = 180 m∠R = 180 – 120 m∠R = 60 In triangle VUT, m∠V + m∠U + m∠T = 180 m∠V + 65 + 35 = 180 m∠V + 100 = 180 m∠V = 180 – 100 m∠V = 80 In triangle RQS m∠R = 60, m∠Q = 85, m∠S = 35 In VUT, m∠V = 80, m∠U = 65, m∠T = 35 Question 5. Answer: Question 6. Answer: Sum of all the angles of a triangle = 180° In triangle CDE, m∠D + m∠C + m∠E = 180 82 + 25 + m∠E = 180 107 + m∠E = 180 m∠E = 180 – 107 m∠E = 73 In triangle STU m∠S + m∠U + m∠T = 180 m∠S + 25 + 73 = 180 m∠S + 98 = 180 m∠S = 180 – 98 m∠S = 82 In Exercises 7 – 10. show that the two triangles are similar. Question 7. Answer: Question 8. Answer: From the given figure LQ ∥MP Therefore, by the corresponding angle theorem, ∠LQN=∠MPN Now, inΔLQN and ΔMNP ∠LNQ=∠MNP and ∠LQN=∠MPN Therefore, Two pairs of angles are congruent. So, ΔLQN ≈ ΔMPN by the AA similarity theorem. Question 9. Answer: Question 10. Answer: It is required to show that the two triangles △RUT and △RSV are similar. To show that the two triangles are similar, use Angle-Angle (AA) Similarity Theorem. Since RT∥SV, ∠TRU ≅ ∠SVR by the Alternate Interior Angles Theorem. List the pairs of congruent angles in △RUT and △RSV ∠RUT≅∠RSV ∠TRU≅∠SVR So, by the Angle-Angle Similarity Theorem, △RUT∼△RSV In Exercises 11 – 18, use the diagram to copy and complete the statement. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. Question 17. Question 18. Question 19. Answer: Question 20. Answer: First divide the diagram into two triangles. 4/9 = 6/x 4x = 54 x = 54/4 x = 27/2 x = 13.5 Question 21. Answer: Question 22. To estimate the height of the pole △CDE can be proved similar to △CAD as ∠CED = ∠CBA (corresponding angles) If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. Corresponding angles are congruent and corresponding sides are proportional for similar triangles. REASONING Question 23. Question 24. Question 25. Question 26. Question 27. Answer: Question 28. Answer: (x + 20)/20 = 125/25 x + 20 = 20 × 5 x + 20 = 100 x = 100 – 20 x = 80 meters Question 29. Question 30. b. AAAA Question 31. Question 32. Question 33. Answer: Maintaining Mathematical Practices Determine whether there is enough information to prove that the triangles are congruent. Explain your reasoning. Question 34. Answer: By the SAS Theorem △EFK ≅ HJK. Question 35. Answer: Question 36. Answer: By the SAS theorem ∠QSP ≅∠QSR 8.1 & 8.2 QuizList all pairs of congruent angles. Then write the ratios of the corresponding side lengths in a statement of proportionality. Question 1. The congruent angles are m∠BDG = m∠MPQ m∠DBG = m∠PMQ m∠DGB = m∠PQM The ratios of its corresponding sides is equal to BD/MP = DG/PQ = BG/MQ. Question 2. Answer: All the pairs of congruent angles ∠GFE = ∠HJK The ratios of the corresponding side lengths FE/JK = ED/KL ED/KL = DG/LH DG/LH = GF/HJ The polygons are similar. Find the value of x. Question 3. Answer: Given that the polygons are similar. From the given data RS/XY = QR/YZ 2/x = 6/6 2/x = 1 2 = x Therefore the value of x is 2. Question 4. Answer: FG/JL = HG/JK 15/x = 21/7 15/x = 3 15/3 = x x = 5 Determine whether the polygons are similar. If they are, write a similarity statement. Explain your reasoning. (Section 8.1 and Section 8.2) Question 5. Answer: QR/WR = 5/10 = 1/2 RS/XY = 3/6 = 1/2 ST/YZ = 2/4 = 1/2 TV/ZV = 3/(ZT + TV) = 3/(3+3) = 3/6 = 1/2 VQ/VW = 4/(VQ + QW) = 4/(4+4) = 4/8 = 1/2 But not all the corresponding angles are congruent. The polygons are not similar. Question 6. Answer: In ΔJKH ∠K = 180 – (37 + 90) ∠K = 180 – 127 = 53 In ΔMNL ∠L = 180 – (90 + 50) ∠L = 180 – 140 ∠L = 40 Since here two triangles dont have congruent Thus they are not similar. Question 7. Answer: ∠B = ∠E = 50 ∠A = 85 ∠C = x ∠F = 45 ∠D = y The angles of both the triangles are similar but not congruent because the angles are not the same. Show that the two triangles are similar. Question 8. Answer: The triangles that have the same shape but a different size is known as similar. ΔBCD and ΔACE are similar triangles. Question 9. Answer: The triangles that have the same shape but a different size is known as similar. ΔFGH and ΔHJK are similar triangles. Question 10. Answer: The triangles that have the same shape but a different size is known as similar. ΔEFG and ΔDFH are similar triangles. Question 11. b. If the surfaces are similar, find the ratio of their perimeters and the ratio ol their areas. If not, find the dimensions of an air hockey table that are similar to an NHL hockey rink. Question 12. Answer: x1 = 68 degrees y 1= 44 degrees ∠z1 = 180 – (68 + 44) ∠z1 = 180 – 112 = 68 degrees x2 = 68 y2 = 68 ∠z2 = 180 – (68 + 68) = 180 – 136 = 44 degrees By AAA theorem, the red and green tents are similar. Thus my friend’s tents and my tents are similar. 8.3 Proving Triangle Similarity by SSS and SASExploration 1 Deciding Whether Triangles Are Similar Work with a partner: Use dynamic geometry software. a. Construct ∆ABC and ∆DEF with the side lengths given in column 1 of the table below. Answer: b. Copy the table and complete column 1. c. Are the triangles similar? Explain your reasoning. d. Repeat parts (a) – (c) for columns 2 – 6 in the table. e. How are the corresponding side lengths related in each pair of triangles that are similar? Is this true for each pair of triangles that are not similar? f. Make a conjecture about the similarity of two triangles based on their corresponding side lengths. g. Use your conjecture to write another set of side lengths of two similar triangles. Use the side lengths to complete column 7 of the table. Exploration 2 Deciding Whether Triangles Are Similar Work with a partner: Use dynamic geometry software. Construct any ∆ABC. b. Is ∆DEF similar to ∆ABC? Explain your reasoning. c. Repeat parts (a) and (b) several times by changing ∆ABC and k. Describe your results. Communicate Your Answer Question 3. Lesson 8.3 Proving Triangle Similarity by SSS and SASMonitoring progress Use the diagram. Question 1. Answer: Explanation: Question 2. Answer: Explanation: Explain how to show that the indicated triangles are similar. Question 3. Answer: Question 4. Answer: Explanation: Exercise 8.3 Proving Triangle Similarity by SSS and SASVocabulary and Core Concept Check Question 1. Answer: Question 2. Answer: Among the four triangles the second triangle is different. The second triangle is not a right-angled triangle. Except for the second triangle all triangles are right angled triangles. Monitoring progress and Modeling with Mathematics In Exercises 3 and 4, determine whether ∆JKL or ∆RST is similar to ∆ABC. Question 3. Answer: Question 4. Answer: Compare the corresponding sides of ∆ABC and ∆JKL 14/17.5 = 0.8 shortest sides 16/20 = 0.8 longest sides 20/25 = 0.8 remaining sides All the ratios of corresponding length sides is equal, it is called the scale factor and it is equal to 0.8 ∆ABC ∼ ∆JKL by the SSS similarity theorem 14/10.5 ≈ 1.33 shortest sides 16/12 ≈ 1.33 longest sides 20/16 = 1.25 remaining sides The ratio of corresponding length sides are not equal. So, ∆ABC is not similar to ∆JKL In Exercises 5 and 6, find the value of x that makes ∆DEF ~ ∆XYZ. Question 5. Answer: Question 6. Answer: 3(x -1)/7.5 = 8/4 (3x – 3) × 4 = 8 × 7.5 12x – 12 = 60 12x = 60 + 12 12x = 72 x = 72/12 x = 6 In Exercises 7 and 8, verify that ∆ABC ~ ∆DEF Find the scale factor of ∆ABC to ∆DEF Question 7. Question 8. In Exercises 9 and 10. determine whether the two triangles are similar. If they are similar, write a similarity statement and find the scale factor of triangle B to triangle A. Question 9. Answer: Question 10. Answer: Use SAS similarity theorem to check the similarity of triangles ∠T = ∠L ST/KL = 8/10 = 4/5 RT/JL = 18/24 = 3/4 Here the lengths of the sides that include ∠T and ∠L are not proportional Since the triangles are not similar, thus the scale factor cannot be found. Hence the scale factor does not similar in this case. In Exercises 11 and 12, sketch the triangles using the given description. Then determine whether the two triangles can be similar. Question 11. Question 12. Answer: In Exercises 13 – 16. show that the triangles are similar and write a similarity statement. Explain your reasoning. Question 13. Answer: Question 14. Answer: ACB ∼ DCE CD/AB = 18/27 CD/AB = 2/3 EC/BC = 14/21 EC/BC = 2/3 ∠ACB = ∠CED Question 15. Answer: Question 16. Answer: In ∆SRQ SR = 12 RQ = 16 QS = 24 SR/VU = 12/9 = 4/3 QS/TV = 24/18 = 4/3 RQ/UT = 16/12 = 4/3 All the ratios of the corresponding sides of ∆SRQ and ∆VUT are equal. Therefore, by SSS similarity theorem ∆SRQ ∼ ∆VUT. The ratio of the corresponding shortest sides, longest sides and remaining sides are equal which satisfies the condition of SSS similarity theorem. Thus by SSS similarity condition, the given triangles are similar. In Exercises 17 and 18, use ∆XYZ. Question 17. Question 18. Question 19. Answer: Question 20. Answer: ATTENDING TO PRECISION Question 21. Question 22. Question 23. Question 24. Answer: Explanation: Question 25. Question 26. Answer: Explanation: Question 27. Question 28. Answer: Out of three triangles, violet and blue triangles are similar. Explanation: Question 29. (A) ∠1 ≅∠2 (B) \(\overline{Q R}\) || \(\overline{N P}\) (C)∠1 ≅ ∠4 (D) ∆MNP ~ ∆MRQ Answer: Question 30. Answer: Question 31. a. What additional information do you need to show that ∆BED ~ ∆ACE using the SSS Similarity Theorem (Theorem 8.4)? b. What additional information do, you need to show that ∆BCD ~ ∆ACE using the SAS Similarity Theorem (Theorem 8.5)? Answer: Question 32. Answer: Question 33. Answer: Question 34. b. Write the ratio of the lengths of the second pair of corresponding legs. c. Are these triangles similar? Does this suggest a Hypotenuse-Leg Similarity Theorem for right triangles? Explain. Question 35. Question 36. Answer: The corresponding angle theorem states that ∆OPQ is similar to ∆OMN. Question 37. Question 38. b. SASAS c. SSSS d. SASSS Question 39. Question 40. Answer: Here we have to check the similarity statement for △ABC, △DEF. The scale factor k = \(\frac { AB }{ DE } \) = \(\frac { 8 }{ 6 } \) = \(\frac { 4 }{ 3 } \) Question 41. By the Right Angles Congruence Theorem (Thin. 2.3), ___________, So. ∆ABC ~ ∆DEF by ___________ . Because corresponding angles of similar triangles are congruent, ∠BAC ≅∠EDF. By ___________, l || n. Answer: Question 42. Answer: StatementsReasons1. mlmn = – 11. Given2. ml = \(\frac{D E}{A D}\), mn = \(\frac{A B}{B C}\)2. Definition of slope3. \(\frac{D E}{A D} \cdot-\frac{A B}{B C}\) = – 13. Correspomsding sides are opposite4. \(\frac{D E}{A D}=\frac{B C}{A B}\)4. Multiply each side of statement 3 by –\(\frac{B C}{A B}\).5. \(\frac{D E}{B C}\) = \(\frac { AB }{ AD } \)5. Rewrite proportion.6. Two right-angled triangles are said to be congruent to each other if the hypotenuse and one side of the right triangle are equal to the hypotenuse and the corresponding side of the other right-angled triangle.6. Right Angles Congruence Theorem (Thm. 2.3)7. ∆ABC ~ ∆ADE7. According to the side angle side theorem.8. ∠BAC ≅ ∠DAE8. Corresponding angles of similar figures are congruent.9. ∠BCA ≅ ∠CAD9. Alternate Interior Angles Theorem (Thm. 3.2)10. m∠BAC = m∠DAE, m∠BCA = m∠CAD10. Congruent angles11. m∠BAC + m∠BCA + 90° = 180°11. △ABC is a right-angled triangle12. m∠CAD + m∠DAE = 90°12. Subtraction Property of Equality13. m∠CAD + m∠DAE = 90°13. Substitution Property of Equality14. m∠CAE = m∠DAE + m∠CAD14. Angle Addition Postulate (Post. 1.4)15. m∠CAE = 90°15. Right Angle16. If two lines meet each other a an angle of 90°, then they are called the perpendicular lines.16. Definition of perpendicular linesMaintaining Mathematical proficiency Find the coordinates of point P along the directed line segment AB so that AP to PB is the given ratio. Question 43. Question 44. Question 45. 8.4 Proportionality TheoremsExploration 1 Discovering a Proportionality Relationship Work with a partner. Use dynamic geometry software to draw any ∆ABC. Answer: b. Compare the ratios of AD to BD and AE to CE. c. Move \(\overline{D E}\) to other locations Parallel to \(\overline{B C}\) with endpoints on \(\overline{A B}\) and \(\overline{A C}\), and repeat part (b). d. Change ∆ABC and repeat parts (a) – (c) several times. Write a conjecture that summarizes your results. Exploration 2 Discovering a Proportionality Relationship Work with a partner. Use dynamic geometry software to draw any AABC. a. Bisect ∆B and plot point D at the intersection of the angle bisector and \(\overline{A C}\). b. Compare the ratios of AD to DC and BA to BC. c. Change ∆ABC and repeat parts (a) and (b) several times. Write a conjecture that summarizes your results. Communicate Your Answer Question 3. Answer:
Question 4. Answer: The line segment DE is parallel to the side AC then it divides the other two sides DA and EC proportionally. DA/DB = EC/EB respectively Lesson 8.4 Proportionality TheoremsMonitoring Progress Question 1. Answer: YZ = \(\frac { 315 }{ 11 } \) Explanation: Question 2. Answer: \(\frac { PQ }{ PN } \) = \(\frac { 50 }{ 90 } \) = \(\frac { 5 }{ 9 } \) \(\frac { SR }{ SN } \) = \(\frac { 40 }{ 72 } \) = \(\frac { 5 }{ 9 } \) \(\frac { PQ }{ PN } \) = \(\frac { SR }{ SN } \) so PS is parallel to QR Find the length of the given line segment. Question 3. Answer: Explanation: Question 4. Answer: Explanation: Find the value of the variable. Question 5. Answer: Explanation: Question 6. Answer: Explanation: Exercise 8.4 Proportionality TheoremsVocabulary and Core Concept Check Question 1. Question 2. Answer: Monitoring Progress and Modeling with Mathematics In Exercises 3 and 4, find the length of \(\overline{A B}\) . Question 3. Answer: Question 4. Answer: In Exercises 5 – 8, determine whether \(\overline{K M}\) || \(\overline{J N}\). Question 5. Answer: Question 6. Answer: If \(\frac { JK }{ KL } \) = \(\frac { NM }{ ML } \), then KM || JN \(\frac { JK }{ KL } \) = latex]\frac { 22.5 }{ 25 } [/latex] = latex]\frac { 9 }{ 10 } [/latex] \(\frac { NM }{ ML } \) = \(\frac { 18 }{ 20 } \) = latex]\frac { 9 }{ 10 } [/latex] \(\frac { JK }{ KL } \) = \(\frac { NM }{ ML } \) Hence KM || JN Question 7. Answer: Question 8. Answer: If \(\frac { JK }{ KL } \) = \(\frac { NM }{ ML } \), then KM || JN \(\frac { JK }{ KL } \) = latex]\frac {35 }{ 16 } [/latex] \(\frac { NM }{ ML } \) = \(\frac { 34 }{ 15 } \) \(\frac { JK }{ KL } \) ≠ \(\frac { NM }{ ML } \) So, KM is not parallel to JN CONSTRUCTION Question 10. Answer: Question 11. Question 12. In Exercises 13 – 16, use the diagram to complete the proportion. Question 13. Answer: Question 14. Answer: \(\frac { CG }{ EG } \) = \(\frac { BF }{ DF } \) Question 15. Answer: Question 16. Answer: \(\frac { BF }{ BD } \) = \(\frac { CG }{ CE } \) In Exercises 17 and 18, find the length of the indicated line segment. Question 17. Answer: Question 18. Answer: \(\frac { SU }{ NS } \) = \(\frac { RT }{ PR } \) \(\frac { SU }{ 10 } \) = \(\frac { 12 }{ 8 } \) SU = \(\frac { 12 }{ 8 } \) • 10 SU = 10 In Exercises 19 – 22, find the value of the variable. Question 19. Answer: Question 20. Answer: \(\frac { z }{ 1.5 } \) = \(\frac { 3 }{ 4.5 } \) z = \(\frac { 3 }{ 4.5 } \) • 1.5 z = 1 Question 21. Answer: Question 22. Answer: \(\frac { q }{ 16 – q } \) = \(\frac { 36 }{ 28 } \) 28q = 36 (16 – q) 28q = 576 – 36q 28q + 36q = 576 64q = 576 q = 9 Question 23. Answer: Question 24. Answer: \(\frac { BD }{ CD } \) = \(\frac { AB }{ AC } \) BD = CD So, 1 = \(\frac { AB }{ AC } \) AC = AB MATHEMATICAL CONNECTIONS Question 25. Answer: Question 26. Answer: \(\frac { PR }{ RT } \) = \(\frac { QS }{ ST } \) \(\frac { 12 }{ 2x – 2 } \) = \(\frac { 21 }{ 3x – 1 } \) 12(3x – 1) = 21(2x – 2) 36x – 12 = 42x – 42 42x – 36x = 42 – 12 6x = 30 x = 5 Question 27. Given \(\overline{Q S}\) || \(\overline{T U}\) Prove \(\frac{Q T}{T R}=\frac{S U}{U R}\) Answer: Question 28. Given \(\frac{Z Y}{Y W}=\frac{Z X}{X V}\) Prove \(\overline{Y X}\) || \(\overline{W V}\) Answer: ZY/YW = ZX/XV — (1) 1 + ZY/YW = 1 + ZX/XV (YW + ZY)/YW = (XV + ZX)/XV ZW/YW = ZV/XV (∠Z is common for ΔZXY and ΔZVW) ZW/YW = ZV/XV ΔZXY and ΔZVW are proportional ΔZXY ≈ ΔZVW ∠ZYW = ∠ZWV YX is paralle to WV. Question 29. a. Find the lake frontage (to the nearest tenth) of each lot shown. b. In general, the more lake frontage a lot has, the higher its selling price. Which lot(s) should be listed for the highest price? c. Suppose that low prices are in the same ratio as lake frontages. If the least expensive lot is $250,000, what are the prices of the other lots? Explain your reasoning. Answer: Question 30. Answer: \(\frac { 5 }{ 2 } \) = \(\frac { x }{ 1.5 } \) x = \(\frac { 5 }{ 2 } \) • 1.5 x = 3.75 \(\frac { 3 }{ 7 } \) = \(\frac { y }{ 5.25 } \) y = \(\frac { 3 }{ 7 } \) • 5.25 y = 2.25 Question 31. Question 32. Answer: From the diagram, we can see that K₁ || K₂ || K₃ Those three parallel lines interest two traversals t₁, t₂ So, \(\frac{C B}{B A}=\frac{D E}{E F}\) Question 33. Question 34. Answer: As per the image, player 1 is closer to the receiver. So, player 1 will reach the receiver first. Question 35. Answer: Question 36. Question 37. Question 38. Answer: My self starts walking from point A with a speed of 3 miles per hour and reaches point C. My friend starts walking from point B with x speed and reaches point C. \(\frac { AD }{ DC } \) = \(\frac { BE }{ CE } \) I have to travel from A to C. So, I need to know distance between AC. Therefore, my friend is correct. Question 39. Answer: Question 40. (Hint: Draw segments parallel to \(\overline{B Y}\) through A and C, as shown. Apply the Triangle Proportionality Theorem (Theorem 8.6) to ∆ACM. Show that ∆APN ~ ∆MPC, ∆CXM ~ ∆BXP, and ∆BZP ~ ∆AZN.) Answer: Drawing line segments AN and CM such that AN ∣∣ YB ∣∣ CM and using the Triangle Proportionality Theorem Maintaining Mathematical Proficiency Use the triangle. Question 41. Question 42. Solve the equation. Question 43. Question 44. Question 45. Similarity Review8.1 Similar PolygonsFind the scale factor. Then list all pairs of congruent angles and write the ratios of the corresponding side lengths in a statement of proportionality. Question 1. Answer: \(\frac { BC }{ CD } \) = \(\frac { 8 }{ 12 } \) = \(\frac { 2 }{ 3 } \) \(\frac { EH }{ GH } \) = \(\frac { 6 }{ 9 } \) = \(\frac { 2 }{ 3 } \) So, scale factor = \(\frac { 2 }{ 3 } \) Question 2. Answer: longer sides: \(\frac { 10 }{ 25 } \) = \(\frac { 2 }{ 5 } \) shorter sides: \(\frac { 6 }{ 15 } \) = \(\frac { 2 }{ 5 } \) remaining sides: \(\frac { 8 }{ 20 } \) = \(\frac { 2 }{ 5 } \) So, scale factor = \(\frac { 2 }{ 5 } \) Question 3. Question 4. Explanation: 8.2 Proving Triangle Similarity by AAShow that the triangles are similar. Write a similarity statement. Question 5. Answer: m∠RQS = m∠UTS = 30°. △QRS and △STU are similar as per the AA similarity theorem. Question 6. Answer: m∠CAB = 60°, m∠DEF = 30° △ABC and △DEF are not similar as per the AA similarity theorem. Question 7. 8.3 Proving Triangle Similarity by SSS and SASUse the SSS Similarity Theorem (Theorem 8.4) or the SAS Similarity Theorem (Theorem 8.5) to show that the triangles are similar. Question 8. Answer: \(\frac { DE }{ CD } \) = \(\frac { 7 }{ 3.5 } \) = 2 \(\frac { AB }{ BC } \) = \(\frac { 8 }{ 4 } \) = 2 \(\frac { DE }{ CD } \) = \(\frac { AB }{ BC } \) So, BD is parallel to AE. Question 9. Answer: \(\frac { QU }{ TU } \) = \(\frac { 9 }{ 4.5 } \) = 2 \(\frac { QR }{ SR } \) = \(\frac { 14 }{ 7 } \) = 2 \(\frac { QU }{ TU } \) = \(\frac { QR }{ SR } \) So, ST is parallel to RU. Question 10. Answer: 8.4 Proportionality TheoremsDetermine whether \(\overline{A B}\) || \(\overline{C D}\) Question 11. Answer: Question 12. Answer: Question 13. Answer: Find the length of \(\overline{A B}\). Question 14. Answer: Question 15. Answer: Similarity TestDetermine whether the triangles are similar. If they are, write a similarity statement. Explain your reasoning. |