Preparing for the Regents Examination Geometry, AK

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6. AP ES (Distances between parallel lines are 7. equal.) −− AP −− ES 8. PAQ SER 9. (Leg–Acute Angle) −− QP −− RS (CPCTC) 10-8 Kites (page 239) 1 A kite is a quadrilateral with only two pairs of adjacent congruent sides. A rhombus has four congruent sides. 2 True 3 True 4 False 5 True 6 True 7 False 8 False 9 True 10 a −− AC b 1 8; 2 7; 3 6; 4 5 11 a 90 b 45 c 17 d m4 25 and mKLM 50 Note: Since there are many variations of proofs, the following is simply one set of acceptable statements to complete each proof. Depending on the textbook used, the wording and format of reasons may differ, so they have not been supplied for the method of congruence applied in each problem. (These solutions are intended to be used as a guide—other possible solutions may vary.) 12 1. −− QS is the perpendicular bisector of −− PR . 2. −− PT −− TR 3. QTP QTR 4. −− QT −− QT 5. QPT QRT (SAS SAS) 6. STP STR 7. −− TS −− TS 8. STP STR (SAS SAS) 9. −− QP −−− QR (CPCTC) 10. −− SP −− SR (CPCTC) 11. −− QT −− ST 12. PQRS is not ( Definition of a rhombus. a rhombus) 13. PQRS is a kite. (Definition of a kite) 10-9 Areas of Polygons (pages 243–245) 1 4 2 16 3 9, 12, 15 4 Length is 4 ft and width is 5 ft. 5 a h 10 x b A x(10 x) 6 a x 2 b 9 x 2 c x 2 4x 4 d x 2 4x 4 e 4 x 2 4x 1 7 a 25 _ 2 b 18 49 c _ 2 d 1 8 11 e 9 9 a 12 b 45 c 20 d 18 √ 3 e 45 _ 2 10 √ 36 144 6 √ 5 11 a BD 10 12 72 √ 3 b 120 13 s 6 14 a 13 b 120 c 120 _ 13 15 a 21x b 10 x 2 c 4x 12 d 18x 12 e 2 x 2 16 a x 4 b 4(4) 4 12 17 8 18 27 √ 3 19 30 √ 2 20 10 and 20 6x 21 a 96 22 72 23 100 24 a 6 b 260 c 60 b SMP and AML are similar. Let x be the perpendicular drawn from M to −− AP . Then x _ 20 6 x _ , and x 5. 4 25 a _ x 10 12 x _ 20 b x 4 c Area of RCS 20. Area of QCT 80. 4(8 11) 26 a A _ 2 38 b 9 ( 5, _ 2 ) 27 a Slope of −− JE 3 _ , slope of 4 −− EN 4 _ . The 3 slopes of the two legs are negative reciprocals, therefore perpendicular, forming a right angle. b A 1 _ (JE)(EN) 25 2 10-9 Areas of Polygons 65
28 Enclose PAT in a large rectangle and subtract the excess areas. a Area of PAT 120 60 24 6 12 18 b AT 10 c 3.6 29 a SA SM √ 52 ; sides are congruent. Slope of −− SA 3 _ , slope of 2 −−− SM 2 _ ; 3 slopes are negative reciprocals, therefore perpendicular and forming a right angle. b A 1 _ (SA)(SM) 26 2 c √ 26 30 Subdivide pentagon SIMON into two triangles and a trapezoid and take the sum of all the areas. Area 72 31 Divide pentagon JANET into two triangles and take the sum of the two areas. Area 64 32 Enclose quadrilateral RYAN in a large rectangle and subtract the excess areas. Area 72 13 _ 6 12 12 35.5 2 33 −− NI and −− CK have zero slope, the two sides are parallel. −−− NK −− IC m √ 10 . Therefore, NICK is an isosceles trapezoid. 34 Slope of −− JO slope of −−− HN 1 _ ; the sides are 3 parallel. Slope of −− NJ 3. The slopes are negative reciprocals, therefore perpendicular, forming a right angle. Slope of −−− OH is not equal to the slope of −− NJ , therefore, only one pair of sides are parallel. JOHN is a right trapezoid. 35 a 38 b 28 c 28 d (5, 4) Chapter Review (pages 245–248) 1 (4) The diagonals of a parallelogram bisect each other. 2 (4) mB mC 360 3 (1) rhombus 4 (1) a rhombus 5 (3) A rhombus is a square. 6 (4) parallelogram 7 (3) 3 8 (4) −− QS and −− PR bisect each other. 9 (1) the diagonals are congruent 10 a square b rhombus c parallelogram d rectangle 66 Chapter 10: Quadrilaterals 11 3x 4 0 x 50 x 5 12 5 13 5 x 90 x 18 14 (3x 20) (7x 4 0) 180 x 20 15 2 16 mDAE 90 75 15 17 (4, 3) 18 3x 1 5 7x 55 x 10 19 a mACD mCAB 30 b rectangle 180 75 20 mAEK _ 52.5 2 21 5 ft 22 (1, 1) 23 3 √ 2 24 mA 45 25 h 8 26 Rhombus. Since the triangles are congruent, the opposite angles that are originally vertex angles are congruent. By the addition postulate, the opposite angles formed by joining the triangles at their bases are congruent. Thus the quadrilateral is a parallelogram. All sides are congruent. The parallelogram is a rhombus. 27 PS QR 2x 3 x 2 x 5 PS QR SR PQ 7 28 a (2x 8) 3(x 3 4) 180 x 14 mABC mCDA 144 mDAB mBCD 36 b AE EC 4 y 6y 36 y 18 BE ED 3x 1 x 13 x 7 AC 144, BD 40 29 mABP mBAP 90 (5x 10) (2x 4) 90 x 12 mDCB 140, mAPB 90 30 a mR 150 b mRAD 150 c mGAD 135 d mD 30
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ANSWER KEY Preparing for the REGENT
Page 3 and 4:
Contents Chapter 1: Essentials of G
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1-4 Angles (pages 9-10) 1 (4) It is
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6 True 7 True 8 True 9 True 10 Answ
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3 1. e ∨ ~f 2. ~f → g 3. ~e 4.
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CHAPTER 3 3-1 Inductive Reasoning (
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6. BAC DAE 6. Transitive property.
Page 15 and 16:
CHAPTER 4-1 Setting Up a Valid Proo
Page 17 and 18: 6 Given: BA bisects CBE 1 2 2 4
Page 19 and 20: 1 1. −−−− ABCD 2. 1 2 3. _
Page 21 and 22: 18 1. I is the midpoint of −− E
Page 23 and 24: 5-3 Isosceles and Equilateral Trian
Page 25 and 26: 12. −−− HG −−− DC 13.
Page 27 and 28: 5 1. −− FG is the perpendicular
Page 29 and 30: Chapter Review (pages 84-85) Note:
Page 31 and 32: 20 Use constructing congruent angle
Page 33 and 34: 6-3 Line Reflections and Symmetry (
Page 35 and 36: 11 A(0, 8), B(2, 2), C(6, 4) (8) 10
Page 37 and 38: 5 (3) (x, y) → (x, 2y) 6 (1) tran
Page 39 and 40: 11. mABC mADC 12. 2mABD 2mADB 13.
Page 41 and 42: 6. mDAB mCAD mDCB mACD 7. mCAB
Page 43 and 44: g e 17 a _ f d b Undefined c a _
Page 45 and 46: 5 BIG is isosceles because it has t
Page 47 and 48: ___ 27 a M KA (5, 1), M ___ AT (4
Page 49 and 50: 8 106 9 mA 75, mC 67 10 57 11 60
Page 51 and 52: 9-5 The Sum of the Measures of the
Page 53 and 54: 9-7 The Converse of the Isosceles T
Page 55 and 56: 12 1. ___ CE ___ BA , ___ BD ___
Page 57 and 58: 12 a mx 45, my 45 b mx 98, my 8
Page 59 and 60: Quadrilaterals 10-2 The Parallelogr
Page 61 and 62: 7. MAD RCB 8. MAD RCB (SAS SAS)
Page 63 and 64: 5. RSQ TSV (Vertical angles) 6. QR
Page 65 and 66: 3 4x 2 3x 3 x 5 RS 18 4 Perime
Page 67: Note: Since there are many variatio
Page 71 and 72: Geometry of Three Dimensions 11-1 P
Page 73 and 74: 11-6 Volume of a Prism (pages 269-2
Page 75 and 76: 14 47.5 in. 2 15 h 4 in. 16 25 cm
Page 77 and 78: 14 a 3 : 2 b QR 10, RS 20, ST 12
Page 79 and 80: 4. A A 5. ADE ABC b (AA) AC _ AB
Page 81 and 82: 12 7 √ 2 13 4 14 5 √ 3 15 x
Page 83 and 84: 13-2 Arcs and Chords (pages 350-351
Page 85 and 86: 27 1. Common external tangents, −
Page 87 and 88: 8. mHCT 1 _ m 2 TH mCHT 1 _ CT 2
Page 89 and 90: 16 a (2 √ 2 , 2 √ 2 ), (2 √
Page 91 and 92: 15 a Use constructing a congruent a
Page 93 and 94: 5 Two horizontal lines y 11 and y
Page 95 and 96: Each review has a total of 58 possi
Page 97 and 98: Part III For each question, use the
Page 99 and 100: Chapters 1-3 (pages 438-441) Part I
Page 101 and 102: Part IV For each question, use the
Page 103 and 104: 16 Score Explanation 4 a 91, b 35
Page 105 and 106: Chapters 1-5 (pages 446-449) Part I
Page 109 and 110: 20 Score Explanation 6 The followin
Page 111 and 112: 17 Score Explanation 4 The followin
Page 113 and 114: Part II For each question, use the
Page 117 and 118: Part III For each question, use the
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Chapters 1-9 (pages 461-463) Part I
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19 Score Explanation 6 The followin
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12 Score Explanation 2 31, and an a
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16 Score Explanation 4 DN √ 106
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20 Score Explanation 2 a 3, and an
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17 Score Explanation 4 6 √ 5 , a
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16 Score Explanation 4 mADE 93, mB
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20 Score Explanation Chapters 1-14
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Part IV For each question, use the
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Practice Regents Examination One (p
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36 Score Explanation 4 36, and an a
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32 Score Explanation 2 108, and an
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Part II For each question, use the
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Part IV For each question, use the
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