Pages Exercises 1., rectangle, rhombus, square. 8. parallelogram. 13. rhombus. 9. rhombus. 2. parallelogram. 10. rectangle. 3.

Transcription

1 Classifying Quadrilaterals GEOMETRY LESSON 6-1 Pages Exercises 1., rectangle, rhombus, square. parallelogram 3. trapezoid 4., rhombus 5. kite 6. trapezoid, isosc. trapezoid 7. rhombus 8. parallelogram 9. rhombus 10. rectangle 11. kite 1. isosc. trapezoid 13. rhombus 14. kite 6-1

2 Classifying Quadrilaterals GEOMETRY LESSON trapezoid 16. rectangle 17. quadrilateral 18. isos. trapezoid 19. x = 11, y = 9; 13, 13, 3, 3 0. x = 4, y = 4.8; 4.5, 4.5, 6.8, x =, y = 6;, 7, 7,. x = 1; 4, 4, 4, 9 3. x = 3, y =5; 15, 15, 15, x = 5, y = 4; 3, 3, 3, , 40, 140, 140; 11, 11, 15, 3 6-1

3 Classifying Quadrilaterals GEOMETRY LESSON , 58, 1, 1; 6, 6, 6, rectangle, square, trapezoid 8. Check students work Answers may vary. Samples are given Impossible; a trapezoid with one rt. must have another, since two sides are

4 Classifying Quadrilaterals GEOMETRY LESSON (continued) are, while opp. sides of a kite are not A rhombus has 4 sides, while a kite has pairs of adj. sides, but no opp. sides are. Opp. sides of a rhombus 37. True; a square is both a rectangle and a rhombus. 38. False; a trapezoid only has one pair of sides. 6-1

5 Classifying Quadrilaterals GEOMETRY LESSON False; a kite does not have opp. sides. 40. True; all squares are. s 41. False; kites are not s. 4. False; only rhombuses with rt. s are squares. 43. Rhombuses; all 4 sides are because they come 43. (continued) from the same cut Check students work. 46. some isos. trapezoids, some trapezoids. 47., rhombus, rectangle, square 48. rectangle, square 49. rhombus, square, kite, some trapezoids 50. A trapezoid has only one pair of sides Check students sketches. 51. rectangle,, kite 5. rhombus, 53. square, rhombus, 54. rhombus,, kite 6-1

6 Classifying Quadrilaterals GEOMETRY LESSON Answers may vary. Sample: a. N can be anywhere on the can be anywhere on the y-axis except (0, 0), (0, ), and (0, ). b. For points N mentioned 55. b. (continued) above, KL = LM and KN = NM, but KL =/ KN Explanations may vary. Samples are given. 56., rectangle, trapezoid 57., kite, rhombus, trapezoid, isos. Trapezoid. 58. kite,, rhombus, trapezoid, isos. trapezoid 59., rectangle, square, rhombus, kite, trapezoid 60. C 61. I 6. C 63. H 6-1

7 Classifying Quadrilaterals GEOMETRY LESSON [] Slope of AB is 3. The slope of BC is 1, so AB and BC are not. Since one is not a right and a rectangle requires all 4 s to be right s, the figure could not be a rectangle. [1] incorrect slope OR failure to recognize the information 64. [1] (continued) provided by the slopes. 65. Yes; the sum of the lengths of any sides is greater than the third side. 66. No; > No; > mm mm mm y = 3x

8 Properties of Parallelograms GEOMETRY LESSON 6- Pages Exercises ; 10, 0, ; 18.5, 3.6, ; m Q = m S = 36, m P = m R = ; m H = m J = 30, m I = m K = x = 6, y = x = 5, y = x = 7, y = x = 6, y = 9 1. x = 3, y = 4. 1; 4 6-

9 Properties of Parallelograms GEOMETRY LESSON 6-3. Pick 4 equally spaced lines on the paper. Place the paper so that the first button is on the first line and the last button is on the fourth line. Draw a line between the first and last buttons. The remaining buttons should be placed where the drawn line crosses the lines on the paper BC = AD = 14.5 in.; AB = CD = 9.5 in. 35. BC = AD = 33 cm; AB = CD = 13 cm 36. a. DC b. AD c. d. Reflexive e. ASA f. CPCTC 6-

10 Properties of Parallelograms GEOMETRY LESSON a. Given b.def. of 37. (continued) f. ASA h. CPCTC c. If lines are, then alt. int. s are. d.if lines are, then alt. int. s are. e. Reflexive Prop. of. 38. g. ASA i. CPCTC 6-

11 Properties of Parallelograms GEOMETRY LESSON , 3, , 8, , 37, The lines going across may not be since they are not marked as , x = 15, y = x = 109, y = 88, z = x = 5, y = x = y = x = 10, y = x = 1, y = x = 0, y = 5 5. x = 9, y = The opp. s are, so they have = 53. (continued) measures. Consecutive s are suppl., so their sum is a. Answers may vary. Check students work. b. No; the corr. sides can be but the s may not be. 55. a. Given b. Def of a 6-

12 Properties of Parallelograms GEOMETRY LESSON (continued) c. Opp. Sides of a are. d. Trans. Prop. of e. If lines are to the same line, then they are to each other. f. If lines are, then the corr. s are. g. Trans. Prop. of h. AAS i. CPCTC 56. Answers may vary. Sample: 1. LENS and NGTH are s. (Given). ELS ENS and GTH GNH (Opp. s of a are.) 3. ENS GNH (Vertical s are.) 4. ELS GTH (Trans. Prop. of ) 57. Answers may vary. Sample: In s LENS and NGTH, GT EH and EH LS by the def. of a. Therefore LS GT because if lines are to the same line then they are to each other. 6-

13 Properties of Parallelograms GEOMETRY LESSON Answers may vary. Sample: 1. LENS and NGTH are s. (Given). GTH GNH (Opp. s of a are.) 3. ENS GNH (Vertical s are.) 4. LEN is supp. to ENS (Cons. s in a are suppl.) 5. ENS GTH (Trans. Prop. of ) 6. E is suppl. to T. (Suppl. of sare suppl.) 59. Answers may vary. Sample: In RSTW and XYTZ, R T and X T because opp. s of a are. Then R X by the Trans. Prop. of. 6-

14 Properties of Parallelograms GEOMETRY LESSON In RSTW and XYTZ, XY TW and RS TW by the def. of a. Then XY RS because if lines are to the same line, then they are to each other. 61. AB DC and AD BC by def. of. 3 and 1 4 because if lines are, then alt. int. s are. 3 4 because if s are each to s, then they are. By Def. of bisect, AC bisects DCB. 6. a. Given: sides and the included of ABCD are to the corr. parts of WXYZ. Let A W, AB WX and 6. a. (continued) AD WZ. Since opp. s of a are, A C and W Y. Thus C Y by the Trans. Prop. of. Similarly, opp. sides of a are, thus AB CD and WX ZY. Using the Trans. Prop. of, CD ZY. The same can be done to prove BC XY. Since consec. s of a are suppl., A is suppl. to D, and W is suppl. to Z. Suppls. of s are, thus D Z. The same can be done to prove B X. Therefore, since all corr. s and sides are, 6-

15 Properties of Parallelograms GEOMETRY LESSON 6-6. a. (continued) ABCD WXYZ. b. No; opp. s and sides are not necessarily in a trapezoid rhombus 71. parallelogram 7. AC DB

16 Proving that a Quadrilateral Is a Parallelogram GEOMETRY LESSON 6-3 Pages Exercises x = 3, y = 4 3. x = 1.6, y = Yes; both pairs of opp. sides are. 8. No; the quad. Could be kite. 9. Yes; both pairs of opp. s are. 10. No; the quad. could be a trapezoid. 11. Yes; both pairs of opp. sides are because alt. int. are. 1. Yes; one pair of opp sides are and. s 13. Yes; both pairs of opp. sides are. 14. No; opp. sides are not. 15. No; the quad. could be a kite. 16. It remains a because the shelves and connecting pieces remain. 6-3

17 Proving that a Quadrilateral Is a Parallelogram GEOMETRY LESSON a. bisect b. XR c. XYR 19. a. Distr. Prop. b.div. Prop. of Eq. c. AD BC, AB DC. Yes; a pair of opp. sides is and. 3. No; the figure could be a trapezoid. d. ASA e. alt. interior d.if same-side int. are suppl., the lines are. s 4. Yes; the both pairs of opp. sides are. 18. Opp. sides of a quad. are if and only if the quad. is a e. Def. of 0. Yes; both pairs of opp. s are. 5. Yes; diag. bisect each other. 6. x = 15, y = 5 1. No; the figure could be a kite. 7. x = 3, y =

18 Proving that a Quadrilateral Is a Parallelogram GEOMETRY LESSON c = 8, a = 4 9. k = 9, m = Answers may vary. Sample: 3. (4, 0) 33. (6, 6) 34. (, 4) 35. You can show a quad. is a if both opp. sides are, if both opp. s are, if opp. sides are, if diag. bisect each other, if all consecutive s are suppl., if one pair of opp. sides are both and Answers vary. Sample: 1. TRS RTW (Given). RS TW, SRT WTR (CPCTC) 3. SR WT (If alt. int. s are, then lines are.) 6-3

19 Proving that a Quadrilateral Is a Parallelogram GEOMETRY LESSON (continued) 4. RSTW is a. (If one pair of opp. sides are and, then it is a.) 37. Answers may vary. Sample: 1. AB CD, AC BD (Given) 39. C 40. F. ACDB is a. (If opp. sides are, then it is a.) 3. M is the midpoint of BC. (The diag. of a bisect each other.) 41. C 4. AM is a median. (Def. of a median) 38. G( 4, 1), H(1, 3) 4. H 6-3

20 Proving that a Quadrilateral Is a Parallelogram GEOMETRY LESSON [] Statements Reasons 1. NRJ CPT (Given). NJ CT (CPCTC) 3. NJ TC (Given) 4. JNTC is a. (If opp. sides of a quad. are both and, then the quad. is a.) [1] proof missing steps 44. [4] a. 6x = 7x 11; x = 11 b.yes; m ABC = m CDE = 66 c. Yes; BD FE and BF DE [3] one or more error in calculating x [] one explanation is incorrect [1] only part (a) answered 6-3

21 Proving that a Quadrilateral Is a Parallelogram GEOMETRY LESSON a = 8, h = 30, k = m = 9.5, x = e = 17, f = 11, c = It is given that AD BC and DAB CBA. By the Reflexive Prop. of AB AB, thus DAB CBA by SAS, so AC BD by CPCTC. 50. If two lines and a transversal form corr. s, then the two lines are ; if two lines are, then a transversal forms corr. s. 51. If the prod. of the slopes of two nonvertical lines is 1, then they are ; if two nonvertical lines are, then the prod. of their slopes os If a quad. is a, then the diag. bisect each other; if the diag. of a quad. bisect each other, then it is a. 6-3

22 Special Parallelograms GEOMETRY LESSON 6-4 Pages Exercises 1. 38, 38, 38, 38. 6, 18, , 31, , 33.5, 113, , 90, 58, , 60, 60, , 35, 55, , 90, , 55, ; LN = MP = ; LN = MP= ; LN = MP = ; LN = MP = ; LN = MP = = ; LN = MP = Impossible; if the diag. of a are, (continued) then it would have to be a rectangle and have right s. 17. Yes; diag. in a mean it can be a rectangle with opp. sides cm long. 18. Impossible; in a, consecutive s must be supp., so all s must be right s. This would make it a rectangle. 6-4

23 Special Parallelograms GEOMETRY LESSON Impossible; if the figure is a, then the opp. the bisected is also bisected, and the figure is a rhombus. But the sides are not. 0. Yes; the s are bisected so it could be a rhombus which is a. 1. Yes; the diag. are so it could be a square which is a.. The pairs of opp. sides of the frame remain, so the frame remains a. 3. After measuring the sides, she can measure the diag. If the diag. are, then the figure is a rectangle by Thm Square; a square is both a rectangle and a rhombus, so its diag. have the same 4. (continued) properties of both Symbols may vary. Samples are given: parallelogram: rhombus: R rectangle: square: S 5. R, 6., R,, 7., R,, S S S 6-4

24 Special Parallelograms GEOMETRY LESSON , R,, S , S 30., R,, 31., R,, 3., 33. R, 34. R, S S S S S 36. Diag. are, diag. are. Diag. are and. Diag. are, diag. are. 38. a. Opp. sides are and ; diag. bis. each other; opp. are ; consec. s are suppl. b. All sides are ; diag. are. c. All s are rt. s ; diag. are bis. of each other; each diag. bis two s. s 6-4

25 Special Parallelograms GEOMETRY LESSON Answers may vary. Samples are given. 39. Draw diag. 1, and construct its midpt. Draw a line through the mdpt. Construct segments of length diag. in opp. directions from mdpt. Then, bisect these segments. Connect these mdpts. with the endpts. of diag Construct a rt., and draw diag. 1 from its vertex. Construct the from the opp. end of diag. 1 to a side of the rt.. Repeat to other side. 41. Same as 39, but construct a line at the midpt. of diag Same as 41 except make the diag. =. 43. Draw diag (continued) Construct a at a pt. different than the mdpt. Construct segments on the line of length diag. in opp. directions from the pt. Then, bisect these segments. Connect these midpts. to the endpts. of diag Draw an acute with the smaller diag. as a side. Construct the line to the other 6-4

26 Special Parallelograms GEOMETRY LESSON (continued) side through the nonvertex endpt. of the smaller diag. Draw an arc with compass set to the length of the larger diag. from the non-diag. side of the, passing through the line. Draw the larger diag., and then draw the non- sides of the trapezoid. 45. Yes; since all right s are, the opp. s are 45. (continued) and it is a. Since it has all right s, it is a rectangle. 46. Yes; 4 sides are, so the opp. sides are making it a. Since it has 4 sides it is also a rhombus. 47. Yes; a quad. with 4 sides is a and a with 4 sides and 4 right s is a square x = 5, y = 3, z = x = 7.5, y = Drawings may vary. Samples are given. 51. Square, rectangle, isosceles trapezoid, kite. 6-4

27 Special Parallelograms GEOMETRY LESSON (continued) 53. For a < b: trapezoid, isosc. 1 trapezoid (a > b),, rhombus, kite. 5. Rhombus,, trapezoid, kite. For a > b: trapezoid, isos. trapezoid,, rhombus (a < b), kite, rectangle, square (if a = b) 6-4

28 Special Parallelograms GEOMETRY LESSON (continued) 54. a. Def. of a rhombus b. Diagonals of a bisect each other. c. AE AE 54. (continued) d. Reflexive Prop. of. e. ABC ADE f. CPCTC g. Add. Post. h. AEB and AED are rt. s. i. suppl. s are rt. s Thm. j. Def. of 55. Answers may vary. Sample: only one diag. is needed. 56. Given ABCD with diag. AC. Let AC bisect BAD. Because ABC DAC, AB = DA by CPCTC. But since opp. sides of a are, AB = CD and BC = DA. So AB = BC = CD = DA, and ABCD is a rhombus. The new statement is true. 6-4

29 Special Parallelograms GEOMETRY LESSON , , 59. 1, , ABC ADC (ASA) 5. AB AD (CPCTC) 6. AB DC, AD BC (Opp. sides of a are.) 7. AB BC CD AD (Trans. Prop. of ) 6. Answers may vary. Sample: The diagonals of a bisect each other so AE CE. Both AED and CED are right s because AC BD, and since DE DE by the Reflexive Prop., AED CED by SAS. By CPCTC AD CD, and since opp. sides of a are, AB BC AD. 6-4

30 Special Parallelograms GEOMETRY LESSON (continued) So ABCD is a rhombus because it has 4 sides

31 Special Parallelograms GEOMETRY LESSON D 65. I 66. [] Since diag. of a rhombus bisect each other, QS = 9 cm. Also, since all sides are, RS = 9 cm. So QRS is an equilateral and each interior is or 60º. QTS is also an equilateral, so its s are 60º. By add. (m PST + m PSR = m RST), m RST = or 10. [1] no work shown OR a response that is only partially correct 67. Yes; both pairs of opp. sides are. 68. No; there are not necess. opp. sides that are both and. 69. Yes; the diag. bisect each other RQ 74. RP 75. ST

32 Trapezoids and Kites GEOMETRY LESSON 6-5 Pages 3-35 Exercises 1. 77, 103, , 69, , 131, , 75, , 115, , 10, a. isosc. trapezoids b. 69, 69, 111, , , 45, , , 6, , 40, , 55, 90, 55, , 5, 38, 37, , 90, 90, 90, 46, 34, 56, 44, 56, , Answers may vary. Sample: 18. 1, 1, 1, No; explanations may vary. Sample: If both s are bisected, then this combined 6-5

33 Trapezoids and Kites GEOMETRY LESSON (continued) with KM KM by the Reflexive Prop. means KLM KNM by SAS. So by CPCTC, opp. s L and N are, so it is not an isos. trapezoid ABCD is an isos. trapezoid, AB DC. (Given). Draw AE DC. (Two points determine a line.) 3. AD EC (Def. of a trapezoid) 4. AECD is a. (Def. of a ) 5. C 1 (Corr. s are.) 6. DC AE (Opp. sides of a are.) 7. AB AE (Trans. Prop. of ) 6-5

34 Trapezoids and Kites GEOMETRY LESSON (continued) 8. AEB is an isosc.. (Def. of an isosc. ) B 1 (Base s of an isosc. are.) 10. B C (Trans. Prop. of ) 11. B and BAD are suppl., C and CDA are suppl. (Same side int. s are suppl.) 1. BAD CDA (Suppl. of s are.) 8. x = 35, y = x = 18, y = Isosc. trapezoid; all the large rt. s appear to be , 68, Yes, the s can be obtuse. 33. Yes, the s can be obtuse, as well as one other. 34. Yes; if s are rt. s, they are suppl. The other are also suppl. s 6-5

35 Trapezoids and Kits GEOMETRY LESSON No; if two consecutive s are suppl., then another pair must be also because one pair of opp. s is. Therefore, the opp. s would be, which means the figure would be a and not a kite. 36. Yes; the s must be 45 or 135 each. 37. No; if two consecutive s were 37. (continued) compl., then the kite would be concave. 38. Rhombuses and squares would be kites since opp. sides can be also. 39. D is any point on BN such that ND =/ BN and D is below N AB CD, and AD CD (Given). BD BD (Refl. Prop. of ) 3. ABD CBD (SSS) 4. A C (CPCTC) 41. Answers may vary. Sample: Draw TA and RP 6-5

36 Trapezoids and Kites GEOMETRY LESSON (continued) 1. isosc. trapezoid TRAP (Given). TA RP (Diag. of an isosc. trap. are.) 3. TR PA (Given) 4. RA RA (Refl. Prop. of ) 5. TRA PAR (SSS) 6. RTA APR (CPCTC) 4. Draw BI as described, then draw BT and BP. 1. TR PA (Given). R A (Base s of isosc. trap. are.) 3. RB AB (Def. of bisector) 4. TRB PAB (SAS) 5. BT BP (CPCTC) 6. RBT ABP (CPCTC) 6-5

37 Trapezoids and Kites GEOMETRY LESSON (continued) TBI PBI (Compl. of s are.) 8. BI BI (Refl. Prop. of ) 9. TBI PBI (SAS) 10. BIT BIP (CPCTC) 11. BIT and BIP are rt. s. ( suppl. s are rt. s.) 1. TI PI (CPCTC) 13. BI is bis. of TP. (Def. of bis.) Check students justifications. Samples are given. 43. It is one half the sum of the lengths of the bases; draw a diag. of the trap. to form s. The bases B and b of the trap. are each a base of a. Then the segment joining the midpts. of the non- sides is the sum of the midsegments of the s. 1 This sum is B + b (B + b). 44. It is one half the difference of the lengths of the bases; from Ex. 43, the length of the segment joining

38 Trapezoids and Kites GEOMETRY LESSON (continued) the midpts. of the non- sides is 1 (B + b). The middle part of this 45. B 46. I 47. C 48. C segment joins the midpts. of the diags. Each outer segment 1 measures B. So the length of the segment connecting the midpts. of the diags. is (B b) D 50. [] HRW and HBW [1] incorrect diagram OR no work shown 6-5

39 Trapezoids and Kites GEOMETRY LESSON a. 4 b. 5 c a. 3 b. 30 c SAS 6-5

40 Placing Figures in the Coordinate Plane GEOMETRY LESSON 6-6 Pages Exercises 1. W(0, h); Z(b, 0). W(a, a); Z(a, 0) 3. W( b, b); Z( b, b) 4. W(0, b); Z(a, 0) 5. W( r, 0); Z(0, t) 6. W( b, c); Z(0, c) b h 7., ; a h b 8. a, ; undefined 9. ( b, 0); undefined a 10., ; r b b 11. ; 1., c ; a. (a, 0) b.(0, b) c. (a, b) t b a d. b + a t r 13. (continued) e. b + a f. b + a g. MA = MB = MC Answers may vary. Samples are given. 14. A, C, H, F 15. B, D, H, F 16. A, B, F, E 6-6

41 Placing Figures in the Coordinate Plane GEOMETRY LESSON A, C, G, E 18. A, C, F, E 19. A, D, G, F 0. W(0, h); Z(b, 0) 1. W(a, a); Z(a, 0). W( b, b); Z( b, b) 3. W(0, b); Z(a, 0) 4. Z(0, t); W( r, 0) 5. W( b, c); Z(0, c) 6. a. Diag. of a rhombus are. b.diag. of a that is not a rhombus are not. 7. Answers may vary. Sample: r = 3, t = ; 3 3 slopes are and ; all lengths are 13; the opp. sides have the same slope, so they are. The 4 7. (continued) sides are. 8. (c a, b) 9. (a, 0) 30. ( b, 0) 31. a. b. ( b, 0), (0, b), (b, 0), (0, b) 6-6

42 Placing Figures in the Coordinate Plane GEOMETRY LESSON (continued) c. b 3. a. d. 1, 1 e. Yes, because the product of the slopes is (continued) b. 33. c. b + 4c d. b + 4c e. the lengths are =. 34. Step 1: (0, 0) Step : (a, 0) Step 3: Since m 1 + m + 90 = 180, 1 and must be compl. 3 and are the acute s of a rt.. Step 4: ( b, 0) Step 5: ( b, a) Step 6: Using the formula for slope, the 6-6

43 Placing Figures in the Coordinate Plane GEOMETRY LESSON (continued) slope for 1 = 35. B 36. F 37. C 38. C 39. A and the slope for a b =. Mult. the b a a b b a slopes, = C 41. [] (b, a); the diag. of a rectangle bisect each other. [1] no conclusion given 4. 6, 118, 118; (3, ) 44. ( 3, 4) 45. a. Reflexive b. AAS 6-6

44 Proofs Using Coordinate Geometry GEOMETRY LESSON 6-7 Pages Exercises a 1. a. W, ; c + e b Z, d b. W(a, b); Z(c + e, d) c. W(a, b); Z(c + e, d) d. c; it uses multiples of to name the coordinates of W and Z.. a. origin b. x-axis c. d. coordinates 3. a. y-axis b. Distance 4. a. rt. b. legs 4. (continued) c. multiples of d. M e. N f. Midpoint g. Distance 5. a. isos. b. x-axis c. y-axis 6-7

45 Proofs Using Coordinate Geometry GEOMETRY LESSON (continued) d. midpts. e. sides f. slopes g. the Distance Formula 6. a. (b + a) + c b. (a + b) + c 7. a. a + b b. a + b 8. a. D( a b, c), E(0, c), F(a + b, c), G(0, 0) b. (a + b) + c c. (a + b) + c d. (a + b) + c e. (a + b) + c f. g. c a + b c a + b 8. (continued) h. i. c a + b c a + b j. sides k. DEFG 9. a. (a, b) b. (a, b) c. the same point 10. Answers may vary. Sample: The 6-7

46 Proofs Using Coordinate Geometry GEOMETRY LESSON (continued) Midsegment Thm.; the segment connecting the midpts. of sides of the is to the 3rd side and half its length; you can use the Midpoint Formula and the Distance Formula to prove the statement directly. 11. a. b. midpts. 11. (continued) c. ( b, c) d. L(b, a + c), M(b, c), N( b, c), K( b, a + c) e. 0 f. vertical lines g. h Answers may vary. Samples are given. 1. yes; Dist. Formula 13. yes; same slope 14. yes; prod. of slopes = no; may not have intersection pt. 16. no; may need measures 6-7

47 Proofs Using Coordinate Geometry GEOMETRY LESSON no; may need measures 18. yes; prod. of slopes of sides of A = yes; Dist. Formula 0. yes; Dist. Formula, sides = 1. no; may need measures. yes; intersection pt. for all 3 segments 3. yes; slope of AB = slope of BC 4. yes; Dist. Formula, AB = BC = CD = AD 5. 1, 4, ,, 4, 6, , 0.4, 1.6,.8, 4, 5., 6.4, 7.6, , 1.5, 1.8,..., 9.5, , +, + 3,...., +(n 1) 30. (0, 7.5), (3, 10), (6, 1.5) 31. 1, 6, 1, 8, (3, 10), 5, 11, 7, 13 1 n 1 n n n 3. ( 1.8, 6), ( 0.6, 7), 6-7

48 Proofs Using Coordinate Geometry GEOMETRY LESSON (continued) (0.6, 8), (1.8, 9), (3, 10), (4., 11), (5.4, 1), (6.6, 13), (7.8, 14) 33. (.76, 5.), (.5, 5.4), (.8, 5.6),..., (8.5, 14.6), (8.76, 14.8) 1 n 10 n 34. 3, +, 5 +, 1 n 10 n 3 +, 5 +,....., 1 n 3 + (n 1), 5 + (n 1) 10 n 35. Assume b > a. a +, b a n a +,...., a + (n 1) b a n 36. Assume b a, d c. b a n d c n b a n a +, c +, b a n b a n d c n a +, c +,..., a + (n 1), c + (n 1) d c n 37. a. The s with bases d and b, and heights c and a, respectively, have the same area. They 6-7

49 Proofs Using Coordinate Geometry GEOMETRY LESSON a. (continued) share the small right with base d and height c, and the remaining areas are s with base c and height (b d). So ad = bc. Mult. both sides by gives ad = bc. b. The diagram shows that =, since both represent the slope of the top segment of the. So by (a), ad = bc. 38. Divide the quad. into s. Find the centroid for each and connect them. Now divide the quad. into 1 a b 1 c d 38. (continued) other s and follow the same steps. Where the two lines meet connecting the centroids of the 4 s is the centroid of the quad. 39. a. L(b, d), M(b + c, d), N(c, 0) d b + c b. AM: y = x; d b c d b c BN: y = (x c); CL: y = (x c) (b + c) 3 c. P, d 3 d. Pt. P satisfies the eqs. for AM 6-7

50 Proofs Using Coordinate Geometry GEOMETRY LESSON d. (continued) and CL. 40. a. e. AM = (b + c) + d ; AP = = (b + c) + d = (b + c) + d = AM The other distances are found similarly b c 3 (b + c) 3 3 d (continued) b. Let a pt. on line p be (x, y). Then the eq. of p is = or y = (x a). c. x = 0 d. When x = 0, y = (x a) = e. b c b c ( a) =. So p and q intersect at 0,. a c ab c b c ab c y 0 x a f. Let a pt. on line r be (x, y). y 0 x b Then the eq. of r is = b c a c 6-7

51 Proofs Using Coordinate Geometry GEOMETRY LESSON f. (continued) or y = a (x b). ab c g. = (0 b) h. 0, c ab c a c 41. a. Horiz. lines have slope 0, and vert. lines have undef. slope. Neither could be mult. to get (continued) c. Let the eq. for 1 be y = b x and for be y = a b origin be the int. point. a x and the b. Assume the lines do not intersect. Then they have the same slope, say m. Then m m = m = 1, which is impossible. So the lines must intersect. 6-7

52 Proofs Using Coordinate Geometry GEOMETRY LESSON c. (continued) Define C(a, b), A(0, 0), and a B a,. Using the Distance b Formula, AC = a + b, BA = a +, and a CB = b +. b a 4 b Then AC + BA = CB, and m A = 90 by the Conv. of the Pythagorean Thm. So A 7 + a 3 + b 44. [] a. = 3; a = 1; = 4; b = 11; (a, b) = ( 1, 11) b. (7 ( 1)) + ( 3 11) = 60 = [1] minor computational error OR no work shown 43. G 6-7

53 Proofs Using Coordinate Geometry GEOMETRY LESSON [4] a-b. Sample: 46. ( a, b) 47. a. If the sum of the s of a polygon is 360, then the polygon is a quad. c. AP = (b a) + c = RQ PQ = PQ AQ = a = RP APQ RQP by SSS [3] minor computational error [] parts (a) and (b) correct [1] one part correct b. If a polygon is a quad., then the sum of its s is a. If x =/ 51, then x =/ 10. b. If x =/ 10, then x =/ a. If a =/ 5, then a =/ 5. b. If a =/ 5, then a =/

54 Proofs Using Coordinate Geometry GEOMETRY LESSON a. If b 4, then b is not neg. b. If b is not neg., then b a. If c 0, then c is not pos. 54. LM NK, LN NL by the Refl. Prop. of, and LNK NLM by all rt. s are. So LNK NLM by SAS, and K M by CPCTC. b. If c is not pos., then c A C, AD CD and ADB CDB so by ASA ADB CDB and by CPCTC AB CB. 53. HE FG, EF GH, and HF HF by the Refl. Prop. of, so HEF FGH by SSS. Then CPCTC