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CHAPTER For a complete list of the postulates and theorems in this chapter, see p. S82. Study Guide: Review 6 Organizer Vocabulary kite . . . . . . . . . . . . . . . . . . . . . . . . 427 rhombus . . . . . . . . . . . . . . . . . . . 409 base angle of a trapezoid . . . . 429 leg of a trapezoid . . . . . . . . . . . 429 side of a polygon . . . . . . . . . . . . 382 organize and review key concepts and skills presented in Chapter 6. concave . . . . . . . . . . . . . . . . . . . . 383 midsegment of a trapezoid . . 431 square . . . . . . . . . . . . . . . . . . . . . 410 convex . . . . . . . . . . . . . . . . . . . . . 383 parallelogram . . . . . . . . . . . . . . 391 trapezoid. . . . . . . . . . . . . . . . . . . 429 diagonal . . . . . . . . . . . . . . . . . . . 382 rectangle . . . . . . . . . . . . . . . . . . . 408 vertex of a polygon . . . . . . . . . . 382 isosceles trapezoid . . . . . . . . . . 429 regular polygon. . . . . . . . . . . . . 382 GI base of a trapezoid . . . . . . . . . . 429 Objective: Help students <D @<I Online Edition Multilingual Glossary Complete the sentences below with vocabulary words from the list above. 1. The common endpoint of two sides of a polygon is a(n) Resources 2. A polygon is ? . −−−− if no diagonal contains points in the exterior. ? −−−− ? is a quadrilateral with four congruent sides. −−−− 4. Each of the parallel sides of a trapezoid is called a(n) ? . −−−− 3. A(n) PuzzleView Test & Practice Generator 6-1 Properties and Attributes of Polygons (pp. 382–388) Multilingual Glossary Online ■ Tell whether the figure is a polygon. If it is a polygon, name it by the number of its sides. The figure is a closed plane figure made of segments that intersect only at their endpoints, so it is a polygon. It has six sides, so it is a hexagon. ■ Tell whether the polygon is regular or irregular. Tell whether it is concave or convex. The polygon is equilateral, but it is not equiangular. So it is not regular. No diagonal contains points in the exterior, so it is convex. Lesson Tutorial Videos CD-ROM Answers 1. vertex of a polygon 2. convex 3. rhombus 4. base of a trapezoid 5. not a polygon 6. polygon; triangle Find each measure. the sum of the interior angle measures of a convex 11-gon (n - 2)180° Polygon ∠ Sum Thm. (11 - 2)180° = 1620° Substitute 11 for n. 7. polygon; dodecagon 9. irregular; convex 10. reg.; convex ■ 11. 1800° 12. 162° the measure of each exterior angle of a regular pentagon sum of ext. = 360° 13. 90° Chapter 6 Polygon Ext. ∠ SumThm. 360° = 72° measure of one ext. ∠ = _ 5 438 Tell whether each polygon is regular or irregular. Tell whether it is concave or convex. 8. 9. 10. Find each measure. 11. the sum of the interior angle measures of a convex dodecagon 12. the measure of each interior angle of a regular 20-gon 13. the measure of each exterior angle of a regular quadrilateral 14. the measure of each interior angle of hexagon ABCDEF Çà  xànànàÇàxàChapter 6 Polygons and Quadrilaterals ge07se_c06_0438_0447.indd 438 438 Tell whether each figure is a polygon. If it is a polygon, name it by the number of its sides. 5. 6. 7. ■ 8. irregular; concave 14. m∠A = m∠D = 144°; m∠B = m∠E = 126°; m∠C = m∠F = 90° EXERCISES EXAMPLES KEYWORD: MG7 Glossary 12/3/05 10:14:46 AM Answers 6-2 Properties of Parallelograms (pp. 391–397) In PQRS, m∠RSP = 99°, PQ = 19.8, and RT = 12.3. Find PT. + , / * −− −− PT RT PT = RT PT = 12.3 ■ ■ - → diags. bisect each other Def. of segs. Substitute 12.3 for RT. JKLM is a parallelogram. Find each measure. 16. 62.4 EXERCISES EXAMPLES ■ 15. 37.5 ÝÊÊ{®Â ÓÞÊÊ ÎÝ LK ÞÊÊÇ −− −− JM LK → opp. sides JM = LK Def. of segs. 2y - 9 = y + 7 Substitute the given values. y = 16 Solve for y. LK = 16 + 7 = 23 m∠M m∠J + m∠M = 180° (x + 4) + 3x = 180 x = 44 m∠M = 3 (44) = 132° → cons. supp. Substitute the given values. Solve for x. In ABCD, m∠ABC = 79°, BC = 62.4, and BD = 75. Find each measure. 15. BE 16. AD 17. 37.5 18. 79° 19. 101° 20. 101° 17. ED 18. m∠CDA 21. 9.5 19. m∠BCD 20. m∠DAB 22. 9.5 23. 54° WXYZ is a parallelogram. Find each measure. 21. WX 22. YZ 23. m∠W 24. m∠X 25. m∠Y 26. m∠Z 8 LÊÊÈ 24. 126° £{> 9 7 È> 25. 54° xLÊÊn 26. 126° < 27. T(6, -5) 27. Three vertices of RSTV are R(-8, 1), S(2, 3), and V(-4, -7). Find the coordinates of vertex T. 28. Write a two-column proof. Given: GHLM is a parallelogram. ∠L ∠JMG Prove: GJM is isosceles. 28. 1. GHLM is a . ∠L ∠JMG (Given) 2. ∠G ∠L ( → opp. ) 3. ∠G ∠JMG (Trans. Prop. of ) −− −− 4. GJ MJ (Conv. of Isosc. Thm.) 5. GJM is isosc. (Def. of isosc. ) 29. m∠ A = m∠E = 63°; m∠G = 117°; since 117° + 63° = 180°, ∠G is supp. to ∠ A and to ∠E. So 1 ∠ of ACEG is supp. to both of its cons. . By Thm. 6-3-4, ACEG is a . −− −− 30. RS = QT = 25, so RS QT. m∠R = 76°, m∠Q = 104°, and m∠R + m∠Q = 180°, so ∠R is supp. to ∠Q. Since ∠R and ∠Q are a pair of same-side int. , −− −− and they are supp., RS QT. So 1 pair of opp. sides of QRST are and . By Thm. 6-3-1, QRST is a . 6-3 Conditions for Parallelograms (pp. 398–405) EXERCISES EXAMPLES ■ Show that MNPQ is a parallelogram for a = 6 and b = 1.6. Ó>ÊÊx ÓLÊÊn ÇL + {>ÊÊÇ * MN = 2a + 5 QP = 4a - 7 MN = 2 (6) + 5 = 17 QP = 4 (6) - 7 = 17 MQ = 7b NP = 2b + 8 MQ = 7 (1.6) = 11.2 NP = 2 (1.6) + 8 = 11.2 Since its opposite sides are congruent, MNPQ is a parallelogram. ■ Determine if the quadrilateral must be a parallelogram. Justify your answer. No. One pair of opposite angles are congruent, and one pair of consecutive sides are congruent. None of the conditions for a parallelogram are met. Show that the quadrilateral is a parallelogram for the given values of the variables. 29. m = 13, n = 27 30. x = 25, y = 7 ÎÊÊ£n®Â ÓÊʮ  , xÞÊÊ£ä ÎÝÊÊ£®Â + {ÝÊÊ{®Â ÓÞÊÊ££ / Determine if the quadrilateral must be a parallelogram. Justify your answer. 31. 32. 33. Show that the quadrilateral with vertices B(-4, 3), D(6, 5), F(7, -1), and H(-3, -3) is a parallelogram. Study Guide: Review ge07se_c06_0438_0447.indd 439 31. Yes; the diags. of the quad. bisect each other. By Thm. 6-3-5, the quad. is a . 439 32. No; a pair of alt. int. are , so 1 pair of opp. sides are . A different pair of opp. sides are . None of the conditions for a are met. −− −− 1 33. slope of BD = slope of FH = __ ; −− −− 5 slope of BH = slope of DF = -6; both pairs of opp. sides have the −− −− −− same slope, so BD FH and BH −− DF; by def., BDFH is a . 12/3/05 10:14:49 AM Study Guide: Review 439 Answers 6-4 Properties of Special Parallelograms (pp. 408–415) 34. 18 35. 39.6 In rectangle JKLM, KM = 52.8, and JM = 45.6. Find each length. 37. 19.8 38. 25.5 ■ 39. 10.5 40. 25.5 41. 21 ■ 42. 41° 43. 49° 44. 82° ■ 45. 98° 46. m∠1 = 57°; m∠2 = 66°; m∠3 = 33°; m∠4 = 114°; m∠5 = 57° 47. m∠1 = 37°; m∠2 = 53°; m∠3 = 90°; m∠4 = 37°; m∠5 = 53° −− −− , so RT SU. 48. RT = SU = 2 √10 −− Slope of RT = -3, and slope −− 1 −− −− , so RT ⊥ SU. The of SU = __ 3 −− coordinates of the mdpt. of RT −− −− and SU are (-4, -3), so RT −− and SU bisect each other. So the diags. of RSTU are ⊥ bisectors of each other. −− −− , so EG FH. 49. EG = FH = 3 √2 −− Slope of EG = -1, and slope −− −− −− of FH = 1, so EG ⊥ FH. The −− coordinates of the mdpt. of EG −− −− 7 __ 1 and FH are 2 , -__ , so EG 2 −− and FH bisect each other. So the diags. of EFGH are ⊥ bisectors of each other. ( ■ KL JKLM is a . KL = JM = 45.6 NL JL = KM = 52.8 1 JL = 26.4 NL = _ 2 ) Rect. → → opp. sides Rect. → diags. → diags. bisect each other PQRS is a rhombus. + * Find m∠QPR, given that / m∠QTR = (6y + 6)° and m∠SPR = 3y°. , m∠QTR = 90° Rhombus → diags. ⊥ 6y + 6 = 90 Substitute the given value. y = 14 Solve for y. m∠QPR = m∠SPR Rhombus → each m∠QPR = 3 (14) ° = 42° diag. bisects opp. The vertices of square ABCD are A(5, 0), B(2, 4), C(-2, 1), and D(1, -3). Show that the diagonals of square ABCD are congruent perpendicular bisectors of each other. AC = BD = 5 √ 2 Diags. are . −− 1 Product of slopes is -1, slope of AC = -_ 7 −− so diags. are ⊥. slope of BD = 7 −− mdpt. of AC −− 3, _ 1 = mdpt. of BD = _ Diags. bisect 2 2 each other. In rectangle ABCD, CD = 18, and CE = 19.8. Find each length. 34. AB 35. AC 36. BD 37. BE In rhombus WXYZ, WX = 7a + 1, WZ = 9a - 6, and VZ = 3a. Find each measure. 7 38. WZ 39. XV 40. XY 8 6 < 41. XZ In rhombus RSTV, m∠TZV = (8n + 18)°, and m∠SRV = (9n + 1)°. Find each measure. < 42. m∠TRS 43. m∠RSV 44. m∠STV 9 45. m∠TVR / , 6 Find the measures of the numbered angles in each figure. 46. rectangle MNPQ ÎΠ£ Ó { Î 47. rhombus CDGH * Ó { Î x + £ xΠx Show that the diagonals of the square with the given vertices are congruent perpendicular bisectors of each other. 48. R(-5, 0), S(-1, -2), T(-3, -6), and U(-7, -4) 49. E(2, 1), F(5, 1), G(5, -2), and H(2, -2) 6-5 Conditions for Special Parallelograms (pp. 418–425) EXERCISES EXAMPLES ■ 51. valid 52. valid 440 Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. −− −− Given: LP ⊥ KN * Conclusion: KLNP is a rhombus. The conclusion is not valid. If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. To apply this theorem, you must first know that KLNP is a parallelogram. Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. −− −− −− −− 50. Given: ER ⊥ FS, ER FS Conclusion: EFRS is a square. −− −− 51. Given: ER and FS bisect each other. −− −− ER FS Conclusion: EFRS is a rectangle. −− −− −− −− −− −− 52. Given: EF RS, FR ES, EF ES Conclusion: EFRS is a rhombus. , - Chapter 6 Polygons and Quadrilaterals ge07se_c06_0438_0447.indd 440 Chapter 6 ( ) 50. Not valid; by Thm. 6-5-2, if the diags. of a are , then the is a rect. By Thm. 6-5-4, if the diags. of a are ⊥, then the is a rhombus. If a is both a rect. and a rhombus, then the is a square. To apply this chain of reasoning, you must first know that EFRS is a . 440 EXERCISES EXAMPLES 36. 39.6 12/3/05 10:14:52 AM ■ Use the diagonals to tell whether a parallelogram with vertices P(-5, 3), Q(0, 1), R(2, -4), and S(-3, -2) is a rectangle, rhombus, or square. Give all the names that apply. PR = √ 98 = 7 √ 2 QS = √ 18 = 3 √ 2 Distance Formula Distance Formula Use the diagonals to tell whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. 53. B(-3, 0), F(-2, 7), J(5, 8), N(4, 1) Answers 54. D(-4, -3), H(5, 6), L(8, 3), P(-1, -6) 55. rect., rhombus, square 53. rhombus 54. rect. 56. 64° 55. Q(-8, -2), T(-6, 8), W(4, 6), Z(2, -4) 57. 25° Since PR ≠ QS, PQRS is not a rectangle and not a square. −− 7 = -1 slope of PR = _ Slope Formula -7 −− _ 3 Slope Formula slope of QS = = 1 3 Since the product of the slopes is -1, the diagonals are perpendicular. PQRS is a rhombus. 58. 65° 59. 123° 60. m∠R = 126°; m∠S = 54° 61. 51.6 62. 48.5 63. 3.5 64. n = 3 or n = -3 6-6 Properties of Kites and Trapezoids (pp. 427–435) + * EXAMPLES In kite PQRS, m∠SRT = 24°, and m∠TSP = 53°. Find m∠SPT. PTS is a right triangle. Kite → diags. ⊥ m∠SPT + m∠TSP = 90° Acute of rt. , are comp. m∠SPT + 53 = 90 m∠SPT = 37° ■ Substitute 53 for m∠TSP. Subtract 53 from both sides. Find m∠D. m∠C + m∠D = 180° 51 + m∠D = 180 m∠D = 129° ■ ■ 9 x£Â < 57. m∠ZWV 58. m∠VZW 59. m∠WZY Find each measure. 60. m∠R and m∠S x{ 8 {Ó Çΰx , / 9 < 63. EQ - Îä Ó°Ç * ΰ£ / + 64. Find the value of n so that PQXY is isosceles. + 8 ÈÓÊÊǮ * 9 6 7 < 6 ÈÇ 8 61. BZ if ZH = 70 and EK = 121.6 62. MN 1 (XY + WZ) AB = _ 2 1 (42 + WZ) 73.5 = _ 2 147 = 42 + WZ 105 = WZ 56. m∠XYZ - In trapezoid HJLN, * JP = 32.5, and HL = 50. Find PN. −− −− JN HL Isosc. trap. → diags. JN = HL = 50 Def. of segs. JP + PN = JN Seg. Add. Post. 32.5 + PN = 50 Substitute. PN = 17.5 Subtract 32.5 from both sides. Find WZ. 67. isosc. trap. In kite WXYZ, m∠VXY = 58°, and m∠ZWX = 50°. Find each measure. Same-Side Int. Thm. Substitute 51 for m∠C. Subtract. 66. trap. EXERCISES / ■ 65. kite nÓÊÊ££®Â 9 Trap. Midsegment Thm. Give the best name for a quadrilateral whose vertices have the given coordinates. 65. (-4, 5), (-1, 8), (5, 5), (-1, 2) Substitute. 66. (1, 4), (5, 4), (5, -4), (1, -1) Multiply both sides by 2. 67. (-6, -1), (-4, 2), (0, 2), (2, -1) 7 < Solve for WZ. Study Guide: Review ge07se_c06_0438_0447.indd 441 441 12/3/05 10:14:55 AM Study Guide: Review 441 CHAPTER 6 Organizer Tell whether each figure is a polygon. If it is a polygon, name it by the number of its sides. GI < polygon; decagon n mastery of concepts and skills in Chapter 6. D@<I not a polygon 2. 1. Objective: Assess students’ 3. The base of a fountain is in the shape of a quadrilateral, as shown. Find the measure of each interior angle of the fountain. £{  3. m∠A = 96°; m∠B = 112°; m∠C = 64°; m∠D = 88° 1260° Online Edition 4. Find the sum of the interior angle measures of a convex nonagon. 5. Find the measure of each exterior angle of a regular 15-gon. 24° £Ó ££Â 6. In EFGH, EH = 28, HZ = 9, and FH = 18; 7. JKLM is a parallelogram. m∠EHG = 145°. Find FH and m∠FEH. m∠FEH Find KL and m∠L. KL = 17; m∠L = 52° Resources Assessment Resources Chapter 6 Tests = 35° < ÞÊÊ££ {ÞÊÊÇ ÈÝÊÊ£®Â ÓÝÊʮ • Free Response (Levels A, B, C) 8. Three vertices of PQRS are P(-2, -3), R(7, 5), and S(6, 1). Find the coordinates of Q.(-1, 1) • Multiple Choice (Levels A, B, C) 9. Show that WXYZ is a parallelogram for a = 4 and b = 3. • Performance Assessment 8 Î> 9 xL >Ê Ê ÊÎ {LÊ n 7 10. Determine if CDGH must be a parallelogram. Justify your answer. < 11. Show that a quadrilateral with vertices K(-7, -3), L(2, 0), S(5, -4), and T(-4, -7) is a parallelogram. Test & Practice Generator 12. In rectangle PLCM, LC = 19, and LM = 23. Find PT and PM. PT = 11.5; PM = 19 / * 13. In rhombus EHKN, m∠NQK = (7z + 6)°, and m∠ENQ = (5z + 1)°. Find m∠HEQ and m∠EHK. + m∠HEQ = 29°; m∠EHK = 122° Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. −− −− −−− −−− −− −−− −−− −− −−− −−− 15. Given: NP MQ, NM PQ, NQ MP 14. Given: NP PQ QM MN Conclusion: MNPQ is a square. Conclusion: MNPQ is a rectangle. valid * + not valid Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. 16. A(-5, 7), C(3, 6), E(7, -1), G(-1, 0) rhombus 18. m∠JFR = 43°, and m∠JNB = 68°. Find m∠FBN. 103° 17. P(4, 1), Q(3, 4), R(-3, 2), S(-2, -1) rect. 19. PV = 61.1, and YS = 24.7. Find MY. 36.4 , * 9 6 - 20. Find HR. 27 in. 442 Ó{Ê° −− −− −− 9. XN = ZN = 12, so XN ZN. Thus WY −− −−− bisects XZ. WN = YN = 15, so WN ge07se_c06_0438_0447.indd 442 −− −− −− YN. Thus XZ bisects WY. The diags. of WXYZ bisect each other. By Thm. 6-3-5, WXYZ is a . 442 Chapter 6 9 - Chapter 6 Polygons and Quadrilaterals Answers KEYWORD: MG7 Resources , ÓxÊÚÚ Ê£ÓÊÊÊÊ° 8 10. No; 1 pair of opp. sides of the quad. are . A pair of vert. formed by the diags. are . None of the conditions for a are met. −− 11. Possible answer: slope of KL = slope −− __ −− 1 of ST = 3 ; slope of KT = slope of −− 4 LS = -__ ; both pairs of opp. sides have 3 −− −− −− −− the same slope, so KL ST and KT LS; by def., KLST is a . 14. Possible answer: MNPQ is a rhombus by def. because its 4 sides are . To show that MNPQ is a square, you need to know that MNPQ is also a rect. 4/26/07 4:03:47 PM CHAPTER 6 Organizer FOCUS ON SAT The scores for each SAT section range from 200 to 800. Your score is calculated by subtracting a fraction for each incorrect multiple-choice answer from the total number of correct answers. No points are deducted for incorrect grid-in answers or items you left blank. Objective: Provide practice for If you have time, go back through each section of the test and check as many of your answers as possible. Try to use a different method of solving the problem than you used the first time. GI college entrance exams such as the SAT. <D @<I Online Edition You may want to time yourself as you take this practice test. It should take you about 6 minutes to complete. Resources 1. Given the quadrilateral below, what value of x would allow you to conclude that the figure is a parallelogram? ÓÝÊÊ£ ÝÊÊ£ 3. Which of the following terms best describes the figure below? College Entrance Exam Practice (A) Rhombus Questions on the SAT represent the following math strands: (B) Trapezoid ÓÝÊÊ{ ÎÝÊÊÓ (C) Quadrilateral Number and Operation, 20–25% (D) Square (A) -2 Algebra and Functions, 35–40% (E) Parallelogram Geometry and Measurement, 25–30% (B) 0 (C) 1 4. Three vertices of MNPQ are M(3, 1), N(0, 6), and P(4, 7). Which of the following could be the coordinates of vertex Q? (D) 2 (E) 3 2. In the figure below, if ABCD is a rectangle, what type of triangle must ABE be? (A) (7, 0) Items on this page focus on: (B) (–1, 1) • Geometry and Measurement (C) (7, 2) Text References: (D) (11, 3) Item (E) (9, 4) Data Analysis, Statistics, and Probability, 10–15% Lesson 1 2 3 4 5 6-3 6-4 6-6 6-2 6-1 (A) Equilateral 5. If ABCDE is a regular pentagon, what is the measure of ∠C? (B) Right (A) 45° (C) Equiangular (B) 60° (D) Isosceles (C) 90° (E) Scalene (D) 108° (E) 120° College Entrance Exam Practice 1. Students who chose A may have set the lengths of adjacent sides equal to each other. Remind students that any quadrilateral in which both pairs of opposite sides are congruent must be a parallelogram. ge07se_c06_0438_0447.indd 443 2. Students who chose B may be thinking of the properties of a rhombus, in which the diagonals are perpendicular. Remind students that the diagonals of a rectangle bisect each other, and ask students what that means about the sides of ABE. 3. Students who did not choose B should review the vocabulary from this chapter. Ask students to review their graphic organizer from page 431, which shows the relationships between the quadrilaterals studied in this chapter. 443 5. Remind students that the sum of the measures of the interior angles of a con12/3/05 AM vex10:15:01 polygon with n sides is (n - 2)180 and that a pentagon has 5 sides. 4. Students who chose A or E may have found the correct slopes but applied them incorrectly. Suggest that students graph the points in the coordinate plane to make sure they understand their relative locations. College Entrance Exam Practice 443 CHAPTER 6 Organizer Multiple Choice: Eliminate Answer Choices Objective: Provide opportunities For some multiple-choice test items, you can eliminate one or more of the answer choices without having to do many calculations. Use estimation or logic to help you decide which answer choices can be eliminated. GI to learn and practice common testtaking strategies. <D @<I Online Edition What is the value of x in the figure? This Test Tackler focuses on using logic and estimation to eliminate answer choices to multiple-choice test items. While this strategy may not always give students the specific answer, it may save students time by eliminating some of the choices. Encourage students to read a multiple-choice test item carefully. Then, before they begin to solve the problem, have them determine if any of the answer choices can be eliminated. 3° 83° 63° 153° n ÎÓ Ý ÎÓ The sum of the exterior angle measures of a convex polygon is 360°. By rounding, you can estimate the sum of the given angle measures. £Îx 100° + 30° + 140° + 30° = 300° If x = 153°, the sum of the angle measures would be far greater than 360°. So eliminate D. If x = 3°, the sum would be far less than 360°. So eliminate A. From your estimate, it seems likely that the correct choice is B, 63°. Confirm that this is correct by doing the actual calculation. 98° + 32° + 63° + 135° + 32° = 360° The correct answer is B, 63°. What is m∠B in the isosceles trapezoid? 216° 72° 108° 58° Base angles of an isosceles trapezoid are congruent. Since ∠D and ∠B are not a pair of base angles, their measures are not equal. Eliminate G, 108°. £än ∠D and ∠C are base angles, so m∠C = 108°. ∠B and ∠C are same-side interior angles formed by parallel lines. So they are supplementary angles. Therefore the measure of angle B cannot be greater than 180°. You can eliminate F. m∠B = 180° - 108° = 72° The correct answer is H, 72°. 444 Chapter 6 Polygons and Quadrilaterals ge07se_c06_0438_0447.indd 444 444 Chapter 6 12/3/05 10:15:03 AM Try to eliminate unreasonable answer choices. Some choices may be too large or too small or may have incorrect units. Item C Answers In isosecles trapezoid ABCD, AC = 18.2, and DG = 6.3. What is GB? Possible answers: Read each test item and answer the questions that follow. * + 2. Use the formula for the area of a and substitute 4 for the height 1 bh = and 2 for the base: A = __ __1 (4)(2) = 4. Eliminate C2 2 because it is much greater than 4. Item A The diagonals of rectangle MNPQ intersect at S. If MN = 4.1 meters, MS = 2.35 meters, and MQ = 2.3 meters, what is the area of MPQ to the nearest tenth? 1. Yes; D; the units are incorrect. 24.5 6.3 11.9 2.9 3. Yes; F; 180° is the sum of the int. of a , and because a hexagon has 3 more sides than a , the sum of its int. must be much greater than 180°. −− 6. Will the measure of GB be more than, less −− than, or equal to the measure of AC? What answer choices can you eliminate and why? 4. The sum of the int. of a convex polygon must be a multiple of 180°. Because 500 is not a multiple of 180, it can be eliminated. 7. Explain how to use estimation to answer this problem. 4.7 square meters 5.4 meters 9.4 square meters 12.8 meters 5. The student did not subtract 2 from the number of sides before multiplying by 180. Item D 1. Are there any answer choices you can eliminate immediately? If so, which choices and why? 2. Describe how to use estimation to eliminate at least one more answer choice. In trapezoid LMNP, XY = 25 feet. What are two −− −− possible lengths for LM and PN ? 6. Less than; A can be eliminated because 24.5 > 18.2. ÓxÊvÌ 8 7. Use compatible numbers and subtract: 18 - 6 = 12. GB is about 12, so select B as the answer. 9 * Item B What is the sum of the interior angles of a convex hexagon? 8. J; the units are incorrect. 18 feet and 32 feet 9. Yes; the midsegment of a trap. has a measure less than that of the longest base and greater than that of the shortest base. Because 10 and 15 are both less than 25, it cannot be the correct answer. 49 feet and 2 feet 10 feet and 15 feet 7 inches and 43 inches 180° 720° 500° 1080° 3. Can any of the answer choices be eliminated immediately? If so, which choices and why? 4. How can you use the fact that 500 is not a multiple of 180 to eliminate choice G? 5. A student answered this problem with J. Explain the mistake the student made. 8. Which answer choice can you eliminate immediately? Why? 10. The student used numbers com1 patible with 50 and 0: __ (50 + 0) 2 = 25. However, the student should have checked the answer, 1 because __ (49 + 2) = 25.5 ≠ 25. 2 9. A student used logic to eliminate choice H. Do you agree with the student’s decision? Explain. 10. A student used estimation and answered this problem with G. Explain the mistake the student made. Test Tackler 445 Answers to Test Items A. A B. H ge07se_c06_0438_0447.indd 445 12/3/05 10:15:05 AM C. B D. F KEYWORD: MG7 Resources Test Tackler 445 CHAPTER 6 KEYWORD: MG7 TestPrep Organizer CUMULATIVE ASSESSMENT, CHAPTERS 1–6 GI Objective: Provide review and practice for Chapters 1–6 and standardized tests. @<I <D Multiple Choice Use the figure below for Items 6 and 7. J 1. The exterior angles of a triangle have measures of (x + 10)°, (2x + 20)°, and 3x°. What is the measure of the smallest interior angle of the triangle? Online Edition 15° Resources Assessment Resources N M 55° 35° 65° 2. If a plant is a monocot, then its leaves have − KEYWORD: MG7 TestPrep you need to prove that quadrilateral JKLM is a parallelogram? − All orchids have leaves with parallel veins. − JM KL − − MN LN ∠MLK and ∠LKJ are right angles. ∠JML and ∠KLM are supplementary. The leaves of a Mexican vanilla plant have parallel veins. A Mexican vanilla plant is a monocot. L − 6. If JK ML, what additional information do parallel veins. If a plant is an orchid, then it is a monocot. A Mexican vanilla plant is an orchid. Based on this information, which conclusion is NOT valid? Chapter 6 Cumulative Test K 7. Given that JKLM is a parallelogram and that m∠KLN = 25°, m∠JMN = 65°, and m∠JML = 130°, which term best describes quadrilateral JKLM? All monocots are orchids. Rectangle Rhombus 3. If ABC PQR and RPQ XYZ, which of Square the following angles is congruent to ∠CAB? ∠QRP ∠YXZ ∠XZY ∠XYZ 4. Which line coincides with the line 2y + 3x = 4? 3y + 2x = 4 2x + 2 y=_ 3 a line through (-1, 1) and (2, 3) Trapezoid 8. For two lines and a transversal, ∠1 and ∠2 are same-side interior angles, ∠2 and ∠3 are vertical angles, and ∠3 and ∠4 are alternate exterior angles. Which classification best describes the angle pair ∠2 and ∠4? Adjacent angles a line through (0, 2) and (4, -4) Alternate interior angles Corresponding angles 5. What is the value of x in polygon ABCDEF? C D 6x˚ B (4x + 1)˚ (5x + 8)˚ A 446 − Which of the following would allow you to conclude that these triangles are congruent by AAS? ∠ACB ∠EDF 5.25x˚ ∠BAC ∠FDE F 18 36 ∠CBA ∠FED Chapter 6 Polygons and Quadrilaterals 10. H 1. A 11. C 2. J 12. H 3. D 13. B 4. J 14. 36 5. C 15. 15 6. J 16. 13.5 7. B 8. H 9. A Chapter 6 − 9. For ABC and DEF, ∠A ∠F, and AC EF. ∠ABC ∠EDF 24 GENL11S_c06_0446_0447.indd 446 446 (4.5x + 5)˚ E 12 Answers KEYWORD: MG7 Resources (5x - 8)˚ Vertical angles 7/13/09 11:10:58 PM 10. The vertices of ABCD are A(1, 4), B(4, y), C(3, -2), and D(0, -3). What is the value of y? 3 5 4 6 Short Response Short-Response Rubric 17. In ABC, AE = 9x - 11.25, and AF = x + 4. Items 17–19 2 Points = The student’s answer is an accurate and complete execution of the task or tasks. B 11. Quadrilateral RSTU is a kite. What is the length − of RV ? D 1 Point = The student’s answer contains attributes of an appropriate response but is flawed. E R F 8 in. S U V A 10 in. 0 Points = The student’s answer contains no attributes of an appropriate response. C a. Find the value of x. Show your work and T explain how you found your answer. − 4 inches 6 inches 5 inches 13 inches 12. What is the measure of each interior angle in a − b. If DF EF, show that AFD CFE. Extended-Response Rubric State any theorems or postulates used. 18. Consider quadrilateral ABCD. Item 20 y B regular dodecagon? 30° 150° 144° 162° 4 4 Points = The student correctly finds XZ and the range for AC. Explanations are complete, and work demonstrates a thorough understanding of concepts related to triangle inequality theorems. C A x 13. The coordinates of the vertices of quadrilateral -4 RSTU are R(1, 3), S(2, 7), T(10, 5), and U(9, 1). Which term best describes quadrilateral RSTU? Parallelogram Rhombus Rectangle Trapezoid -2 D 2 4 -3 3 Points = The student’s answers are correct, but explanations may contain minor flaws. Work demonstrates an understanding of major concepts related to triangle inequality theorems. a. Show that ABCD is a trapezoid. Justify your answer. Mixed numbers cannot be entered into the grid for gridded-response questions. For example, if you get an answer of 7__14 , you must grid either 29 7.25 or __ . 4 b. What are the coordinates for the endpoints of the midsegment of trapezoid ABCD? 19. Suppose that ∠M is complementary to ∠N and 2 Points = The student answers correctly, but explanations are missing or incomplete. Work demonstrates a limited understanding of triangle inequality theorems. Extended Response 1 Point = The student answers incorrectly but makes a reasonable attempt to show work. ∠N is complementary to ∠P. Explain why the measurements of these three angles cannot be the angle measurements of a triangle. Gridded Response 14. If quadrilateral MNPQ is a parallelogram, what is the value of x? M (3x + 3)˚ Q N − − a. If AB = 5, BC = 6, AC = 8, and m∠B < m∠Y, P b. If AB = 3, BC = 5, AC = 5, and m∠B > m∠Y, − find the length of XZ so that XYZ is a right triangle. Justify your reasoning and state any theorems or postulates used. determined by six coplanar points when no three are collinear? c. If AB = 8 and BC = 4, find the range of possible 16. Quadrilateral RSTU is a rectangle with − − diagonals RT and SU. If RT = 4a + 2 and SU = 6a - 25, what is the value of a? values for the length of AC. Justify your answer. Cumulative Assessment, Chapters 1–6 Answers 2 (9x - 11.25) 17a. By the Centroid Thm., __ 3 = x + 4. So 6x - 7.5 = x + 4, -11.5 = -5x, and x = 2.3. GENL11S_c06_0446_0447.indd 447 b. Possible answer: By the Centroid Thm., −− −− 2 FC = __ DC. It is given that DF EF 3 , so DF = EF. From part a, EF = 3.15, 2 and AF = 6.3. So DF = 3.15. FC = __ 3 2 __ (DF + FC) = 3 (3.15 + FC), so FC −− −− = 6.3. Thus AF = FC, and AF FC. ∠DFA ∠EFC by the Vert. Thm. Thus AFD CFE by SAS. −− −− 3 18a. Slope of AB = __ , and slope of CD = 2 −− −− −− __3 , so AB CD. Slope of BC = -2, and 2 −− −− slope of AD = 5. So BC is not to −− AD. ABCD is a trap. by def. 0 Points = The student answers incorrectly and does not attempt all parts of the problem. explain why XYZ is obtuse. Justify your reasoning and state any theorems or postulates used. 15. What is the greatest number of line segments − − and BC YZ. (2y - 5)˚ 3y ˚ − 20. Given ABC and XYZ, suppose that AB XY 1 1 1 b. -1__ , -__ , 1__ ,4 2 2 2 ( )( ) 19. ∠M ∠P by the Comps. Thm, and thus m∠M = m∠P. Also, m∠N = 90° - m∠P by the def. of comp. . If these were of a , m∠M + m∠N + m∠P = 180° by the Sum Thm. By subst., m∠P + (90° - m∠P) + m∠P = 180°. Thus, m∠P = 90°. But m∠P < 90° because ∠N is comp. to ∠P. So there is a contradiction, and therefore, these cannot be the of a . −− −− −− 20a. Since m∠B < m∠Y, AB XY, and BC −− YZ, it follows by the Hinge Thm. that XZ > AC = 8. Since 64 > 25 + 36, ABC is obtuse by the Pyth. Inequals. Thm. Since the longer side lies opp. the greater ∠, ∠B is the 447 largest ∠ in ABC. Since ABC is obtuse, ∠B must be obtuse. Since m∠B < m∠Y, m∠Y must also be 7/13/09 11:11:10 PM obtuse, so XYZ is obtuse. −− −− −− b. Since m∠B > m∠Y, AB XY, and BC −− YZ, it follows by the Hinge Thm. that XZ < AC = 5. So the hyp. of −− XYZ would be YZ, the longest side. According to the Pyth. Thm., XY 2 + XZ 2 = YZ 2 , 3 2 + XZ 2 = 5 2, XZ 2 = 25 - 9 = 16, and XZ = 4. c. 4 < AC < 12. By the Inequal. Thm., AB + AC > BC. By subst. 8 + AC > 4. This is true when AC > 0. AB + BC > AC, so 8 + 4 = 12 > AC. Also, AC + BC > AB, so AC + 4 > 8, and AC > 4. Standardized Test Prep 447 Organizer Ohio Sandusky Objective: Choose appropriate GI problem-solving strategies and use them with skills from Chapters 5 and 6 to solve realworld problems. <D @<I Handmade Tiles Online Edition During the nineteenth century, an important industry developed in east central Ohio thanks to an “earthy” discovery—clay! The region’s rich soil and easy access to river transportation helped establish Ohio as the pottery and ceramic capital of the United States. Today the majority of the earthenware clay used in handmade tiles is still mined in Ohio. Handmade Tiles Reading Strategies Have students write down the important information as they read Problem 1. Then have them summarize the problem in their own words. Choose one or more strategies to solve each problem. Using Data Ask students what else they need to know about the tile to solve Problem 1. the height of the Ask them to describe how they can find this information. Find the shorter leg of the 30°-60°-90° formed by the height of the . (2 in.) 1. In tile making, soft clay is pressed into long rectangular wooden trays. After the clay has dried, tiles are cut from the rectangular slab. A tile manufacturer wants to make parallelogram-shaped tiles with the dimensions shown. What is the maximum number of such tiles that can be cut from a 12 in. by 40 in. slab of clay? 36 tiles {Ê° Îä ÈÊ° 2. An interior designer is buying tiles that are in the shape of isosceles trapezoids. Each tile has bases that are 1 in. and 3 in. long, and the tiles can be arranged as shown to form a rectangle. How many tiles should the designer buy in order to frame a 25 in. by 49 in. window? 76 tiles 6.26 cm, 3. A tile manufacturer wants to make a tile in the shape of a rhombus where one diagonal 12.52 cm is twice the length of the other diagonal. What should the lengths of the diagonals be in order to make a tile with sides 7 cm long? Round to the nearest hundredth. 448 For Problem 2, ask students which problem-solving strategies might be helpful. Possible answer: Find a pattern. For each side, the corner tile gives 1 in. of length. After that, every 2 tiles provide another 4 in. of length. So it takes -1 1+ tiles to make a side that 2 is inches long. GENL11S_c06_0448_0449.indd 448 _ 448 Chapter 6 ÎÊ° Chapter 6 Polygons and Quadrilaterals Problem-Solving Focus KEYWORD: MG7 Resources £Ê° Discuss different strategies that could be used to solve Problem 2. Some students might make a table that shows the number of tiles and the corresponding distance. Others might start by solving a simpler problem, such as the number of tiles needed for a 5 in. by 9 in. window. 8/7/09 GENL 12:52 Problem Solving Strategies The Millennium Force Roller Coaster The Millennium Force Roller Coaster When it opened in May 2000, the Millennium Force roller coaster broke all previous records and became the tallest and fastest roller coaster in the world. One of 16 roller coasters at Cedar Point in Sandusky, Ohio, the Millennium Force takes riders on a wild journey that features 1.25 miles of track, a top speed of 93 miles per hour, and a breathtaking 310-foot drop! Reading Strategies Make sure that students understand what the words ascent and descent mean. If students are unfamiliar with the terms, ask them if they can guess their meanings from context clues in the problems. Choose one or more strategies to solve each problem. Using Data Have students identify the segments in the figure that represent the ascent and −− −− descent. AB; CD Ask students to identify a 45°-45°-90° right triangle −− in the figure. the rt. with AB as its hyp. 1. The first hill of the Millennium Force is 310 ft tall. The ascent to the top of the hill is at a 45° angle. What is the length of the ascent to the nearest tenth of a foot? 438.4 ft 2. The Millennium Force was the first coaster in which an elevator lift system was used to pull the trains to the top of the first hill. The system pulls the trains at a speed of 20 ft/s. How long does it take a train to reach the top of the hill? 22 s The figure shows the support structure for the first hill of the Millennium Force. For 3 and 4, use the figure. −− 3. The length of the first descent CD is 314.8 ft. To the nearest foot, what is the total horizontal distance AD that the train covers as it goes over the first hill? 425 ft −− 4. Engineers designed the support beam XY −− so that X is the midpoint of the ascent AB −− and Y is the midpoint of the descent CD. What is the length of the beam to the nearest foot? 242 ft 2:13 PM L11S_c06_0448_0449.indd 449 ENGLISH LANGUAGE LEARNERS ÈäÊvÌ 8 Problem-Solving Focus 9 For Problem 2, focus on the final step of the problem-solving process: (4) Look Back. In particular, ask students if their answer seems reasonable. Based on their experiences with roller coasters, does the amount of time for the ascent seem realistic? If not, encourage students to check their work for errors. ΣäÊvÌ {x ÈäÊvÌ Real-World Connections 449 7/24/09 9:37:35 PM Real-World Connections 449