Resources for Geometry a - Quadrilaterals

8/14/2019 Resources for Geometry a - Quadrilaterals

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Class i fy ing Quadr i l a te ra l sClass i fy ing Quadr i l a te ra l s

Pages 290-293 Exercises

1. , rectangle,rhombus, square

2. parallelogram

3. trapezoid

4. , rhombus

5. kite

6. trapezoid, isosc.trapezoid

7. rhombus

8. parallelogram

9. rhombus

10. rectangle

11. kite

12. isosc. trapezoid

13. rhombus

14. kite

GEOMETRY LESSON 6-1GEOMETRY LESSON 6-1


15. trapezoid

16. rectangle

17. quadrilateral

18. isos. trapezoid

19. x = 11, y = 29; 13,13, 23, 23

20. x = 4, y = 4.8; 4.5,4.5, 6.8, 6.8

21. x = 2, y = 6; 2, 7, 7, 2

22. x = 1; 4, 4, 4, 9

23. x = 3, y =5; 15, 15,15, 15

24. x = 5, y = 4; 3, 3, 3, 3

25. 40, 40, 140, 140; 11,11, 15, 32



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26. 58, 58, 122, 122; 6,6, 6, 6

27. rectangle, square,trapezoid

28. Check studentswork

29-34. Answers mayvary. Samples aregiven.

29.

30.

31. Impossible; atrapezoid with one rt.

must haveanother, since twosides are .

32.

33.



34.

35. A rhombus has 4sides, while a kitehas 2 pairs of adj.sides , but no opp.sides are . Opp.sides of a rhombus

35. (continued)are , while opp. sides of a kite are not .

36.

37. True; a square is both a rectangle and a rhombus.

38. False; a trapezoid only has one pair of sides.



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39. False; a kite doesnot have opp.sides.

40. True; all squaresare .

41. False; kites are not.

42. False; onlyrhombuses with rt.

are squares.

43. Rhombuses; all 4sides arebecause they come

43. (continued)from the same cut.

44-45. Check students

work.

46. some isos.trapezoids, sometrapezoids.

47. , rhombus,rectangle, square

48. rectangle, square

49. rhombus, square, kite,some trapezoids

50. A trapezoid has onlyone pair of sides.

51-54. Check students

sketches.

51. rectangle, , kite

52. rhombus,

53. square, rhombus,

54. rhombus, , kite

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55. Answers may vary.Sample:a.

N can be

anywhere on thecan be anywhereon the y- axisexcept (0, 0),(0, 2), and(0, 2).

b. For points N mentioned

55. b. (continued)above, KL = LM and KN = NM , butKL KN .

56-59. Explanationsmay vary. Samplesare 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

62. C

63. H

=/



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64. [2] Slope of AB is

. The slope

of BC is 1, so AB and BC are not

. Since oneis not a rightand a rectanglerequires all 4to be right ,the figure couldnot be arectangle.

[1] incorrect slopeOR failure torecognize theinformation

64. [1] (continued)provided by theslopes.

65. Yes; the sum of thelengths of any 2sides is greater thanthe third side.

66. No; 5 + 7 20

67. No; 3 + 5 8

68. 28 mm

69. 16 mm

70. 12 mm

71. 82

72. 90

73. 58

74. y = 3 x + 4

32

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s >

>


Proper t i es o f Para l l e logramsProper t i es o f Para l l e lograms


1. 127

2. 67

3. 76

4. 124

5. 100

6. 118

7.

8. 4

9. 4

10. 3; 10, 20, 20

11. 22; 18.5, 23.6, 23.6

12. 20

13. 18

14. 17

15. 12;m Q = m S = 36,m P = m R = 144

16. 6;m H = m J = 30,m I = m K = 150

17. x = 6, y = 8

18. x = 5, y = 7

19. x = 7, y = 10

20. x = 6, y = 9

21. x = 3, y = 4

22. 12; 24

34



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23. Pick 4 equallyspaced lines on thepaper. Place thepaper so that the

first button is on thefirst line and the lastbutton is on thefourth line. Draw aline between thefirst and lastbuttons. Theremaining buttonsshould be placedwhere the drawnline crosses the 2lines on the paper.

24. 3

25. 3

26. 6

27. 6

28. 9

29. 2.25

30. 2.25

31. 4.5

32. 4.5

33. 6.75

34. 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



37. a. Given

b. Def. of

c. If 2 lines are ,then alt. int. are

.

d. If 2 lines are ,then alt. int. are

.

e. Reflexive Prop. of .

37. (continued)f. ASA h. CPCTC

g. ASA i. CPCTC

38.

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39. 38, 32, 110

40. 81, 28, 71

41. 95, 37, 37

42. The lines goingacross may not besince they are notmarked as .

43. 18, 162

44. 60

45. x = 15, y = 45

46. x = 109, y = 88,z = 76

47. x = 25, y = 115

48. x = y = 6

49. x = 10, y = 4

50. x = 12, y = 4

51. x = 0, y = 5

52. x = 9, y = 6

53. The opp. are ,so they have =

53. (continued)measures.Consecutive aresuppl., so their sum

is 180.

54. a. Answers mayvary. Checkstudents work.

b. No; the corr. sidescan be but the

may not be.

55. a. Given

b. Def of as

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Proper t i es o f Para l l e logramsProper t i es o f Para l l e logramsGEOMETRY LESSON 6-2GEOMETRY LESSON 6-2

56. Answers may vary. Sample:1. LENS and NGTH are s.

(Given)

2. ELS ENS and GTH GNH (Opp. of a are .)

3. ENS GNH (Vertical are .)

4. ELS GTH (Trans. Prop. of )

57. Answers may vary. Sample: In LENS andNGTH , GT EH and EH LS by the def. of a .Therefore LS GT because if 2 lines are to thesame line then they are to each other.

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55. (continued)

c. Opp. Sides of aare .

d. Trans. Prop. of

e. If 2 lines are tothe same line,then they are

to each other.f. If 2 lines are ,

then the corr.are .

g. Trans. Prop. of

h. AAS

i. CPCTC

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58. Answers may vary. Sample:1. LENS and NGTH are . (Given)

2. GTH GNH (Opp. of a are .)

3. ENS GNH (Vertical are .)

4. LEN is supp. to ENS (Cons. in a are suppl.)

5. ENS GTH (Trans. Prop. of )

6. E is suppl. to T. (Suppl. of are suppl.)

59. Answers may vary. Sample: In RSTW and XYTZ , R T andX T because opp. of a are . Then R X by the Trans.

Prop. of .

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60. In RSTW and XYTZ , XY TW and R S TW by the def. of a

. Then XY RS because if 2lines are to the same line, thenthey are to each other.

61. AB DC and AD BC by def. of .2 3 and 1 4 because if

2 lines are , then alt. int. are .3 4 because if 2 are eachto 2 , then they are . By

Def. of bisect, AC bisects DCB .

62. a. Given: 2 sides and theincluded of ABCD areto the corr. parts of WXYZ .Let A W , AB WX and

62. a. (continued)AD WZ . Since opp. of aare , A C and W

Y . Thus C Y by theTrans. Prop. of . Similarly,opp. sides of a are , thusAB CD and WX ZY . Usingthe Trans. Prop. of , CD ZY . The same can be done toprove BC XY . Since consec.

of a are suppl., A issuppl. to D , and W issuppl. to Z . Suppls. of are , thus D Z . Thesame can be done to prove

B X . Therefore, since allcorr. and sides are ,

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62. a. (continued)ABCD WXYZ.

b. No; opp. and sides are not

necessarily in a trapezoid.

63. 10

64. 11

65. 126

66. 126

67. 160

68. 148

69. 42

70. rhombus

71. parallelogram

72. AC DB

73. 49

74. 131

75. 49

76. 131

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Proving tha t a Quadr i l a te ra l I s a Para l l e logramProving tha t a Quadr i l a te ra l I s a Para l l e logram


1. 5

2. x = 3, y = 4

3. x = 1.6, y =1

4.

5. 5

6. 13

7. Yes; both pairs of opp. sides are .

8. No; the quad. Couldbe kite.

9. Yes; both pairs of opp. are .

10. No; the quad. couldbe a trapezoid.

11. Yes; both pairs of opp. sides arebecause alt. int.are .

12. Yes; one pair of oppsides are and .

13. Yes; both pairs of opp. sides are .

14. No; opp. sides arenot .

15. No; the quad. couldbe a kite.

16. It remains abecause the shelvesand connectingpieces remain .

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17. a. bisect

b. XR

c. XYR

d. ASA

e. alt. interior

18. Opp. sides of aquad. are if andonly if the quad. is a

19. a. Distr. Prop.

b. Div. Prop. of Eq.

c. AD BC , AB DC

d. If same-side int.are suppl., thelines are .

e. Def. of

20. Yes; both pairs of opp. are .

21. No; the figure couldbe a kite.

22. Yes; a pair of opp.sides is and .

23. No; the figure could

be a trapezoid.

24. Yes; the both pairsof opp. sides are

.

25. Yes; diag. bisecteach other.

26. x = 15, y = 25

27. x = 3, y = 11

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28. c = 8, a = 24

29. k = 9, m = 23.4

30. Answers may vary.Sample:

31.

32. (4, 0)

33. (6, 6)

34. (2, 4)

35. You can show a quad. is a if both opp. sidesare , if both opp. are , if opp. sides are ,if diag. bisect each other, if all consecutive aresuppl., if one pair of opp. sides are both and .

36. Answers vary. Sample:1. TRS RTW (Given)

2. RS TW , SRT WTR (CPCTC)

3. SR WT (If alt. int. are , then lines are .)

16

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36. (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)

2. ACDB is a . (If opp. sides are , then itis a .)

3. M is the midpoint of BC . (The diag. of abisect each other.)

4. AM is a median. (Def. of a median)

38. G (4, 1), H (1, 3)

39. C

40. F

41. C

42. H


Proving tha t a Quadr i l a te ra l I s a Para l l e logramProving tha t a Quadr i l a te ra l I s a Para l l e logramGEOMETRY LESSON 6-3GEOMETRY LESSON 6-3

43. [2] Statements Reasons

1. NRJ CPT (Given)

2. NJ CT (CPCTC)

3. NJ TC (Given)

4. JNTC is a . (If opp. sides of aquad. are bothand , thenthe quad. is a .)

[1] proof missing steps

44. [4] a. 6x = 7 x 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

[2] one explanationis incorrect

[1] only part (a)answered


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Proving tha t a Quadr i l a te ra l I s a Para l l e logramProving tha t a Quadr i l a te ra l I s a Para l l e logramGEOMETRY LESSON 6-3GEOMETRY LESSON 6-3

45. a = 8, h = 30, k = 20

46. m = 9.5, x = 15

47. e = 17, f = 11, c = 204

48. It is given that AD BC andDAB CBA . By the Reflexive

Prop. of AB AB , thus DAB CBA by SAS, so AC BD by

CPCTC.

49. If a quad. is a , then the diag.bisect each other; if the diag. of aquad. bisect each other, then it is a

.

50. If two lines and a transversal formcorr. , then the two lines are ; if two lines are , then a transversalforms corr. .

51. If the prod. of the slopes of twononvertical lines is 1, then theyare ; if two nonvertical lines are

, then the prod. of their slopes os 1.

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Spec ia l Para l l e logramsSpec ia l Para l l e logramsGEOMETRY LESSON 6-4GEOMETRY LESSON 6-4


1. 38, 38, 38, 38

2. 26, 128, 128

3. 118, 31, 31

4. 33.5, 33.5, 113,33.5

5. 32, 90, 58, 32

6. 90, 60, 60, 30

7. 55, 35, 55, 90

8. 60, 90, 30

9. 90, 55, 90

10. 4; LN = MP = 4

11. 3; LN = MP = 7

12. 1; LN = MP = 4

13. 9; LN = MP = 67

14. ; LN = MP = = 9

15. ; LN = MP = 12

16. Impossible; if thediag. of a are ,

16. (continued)then it would have tobe a rectangle andhave right .

17. Yes; diag. in amean it can be arectangle with 2 opp.sides 2 cm long.

18. Impossible; in a ,consecutive mustbe supp., so allmust be right .This would make it arectangle.

53

293

23

52

12

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19. Impossible; if thefigure is a , thenthe opp. thebisected is also

bisected, and thefigure is a rhombus.But the sides are not

.

20. Yes; the arebisected so it couldbe a rhombus whichis a .

21. Yes; the diag. areso it could be asquare which is a .

22. The pairs of opp.sides of the frameremain , so theframe remains a .

23. After measuring thesides, she canmeasure the diag. If the diag. are , thenthe figure is arectangle by Thm. 6-14.

24. Square; a square isboth a rectangle anda rhombus, so itsdiag. have the same

24. (continued)properties of both.

25-34. Symbols may

vary. Samples aregiven:parallelogram:rhombus:rectangle:square:

25. ,

26. , , ,

27. , , ,

s

R

S

R

S

R S

R S


28. , , ,

29. ,

30. , , ,

31. , , ,

32. ,

33. ,

34. ,

35.

Diag. are , diag.are .

36.

Diag. are and .

37.

Diag. are , diag.are .

38. a. Opp. sides areand ; diag. bis.each other; opp.are ; consec.are suppl.

b. All sides are ;diag. are .

c. All are rt. ;diag. are bis. of each other; eachdiag. bis two .

R S

S

R S

R S

S

R S

R S

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s s

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39-44. Answers mayvary. Samples aregiven.

39. Draw diag. 1, andconstruct its midpt.Draw a line throughthe mdpt. Constructsegments of lengthdiag. 2 in opp.directions from mdpt.Then, bisect thesesegments. Connectthese mdpts. with the

endpts. of diag. 1 .

40. Construct a rt. ,and draw diag. 1from 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, butconstruct a line atthe midpt. of diag. 1.

42. Same as 41 exceptmake the diag. =.

43. Draw diag. 1.

43. (continued)Construct a at a pt.different than themdpt. Construct

segments on theline of length diag. 2in opp. directionsfrom the pt. Then,bisect thesesegments. Connectthese midpts. to theendpts. of diag. 1.

44. Draw an acute withthe smaller diag. asa side. Construct theline to the other


44. (continued)side through the non-vertex endpt. of thesmaller diag. Drawan arc with compassset to the length of the larger diag. fromthe non-diag. side of the , passingthrough the line.Draw the larger diag.,and then draw thenon- sides of thetrapezoid.

45. Yes; since all rightare , the opp. are

45. (continued)and it is a .

Since it has all right, it is a rectangle.

46. Yes; 4 sides are ,so the opp. sides are

making it a .Since it has 4sides it is also arhombus.

47. Yes; a quad. with 4sides is a and awith 4 sides and 4right is a square.

48. 30

49. x = 5, y = 32, z = 7.5

50. x = 7.5, y = 3

51-53. Drawings mayvary. Samples aregiven.

51. Square, rectangle,isosceles trapezoid,kite.

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51. (continued)

52. Rhombus, , trapezoid, kite.

53. For a < b : trapezoid, isosc.trapezoid ( a > b ), , rhombus ,kite.

For a > b : trapezoid, isos. trapezoid,, rhombus ( a < 2 b ), kite,

rectangle, square (if a = 2 b )

12


55. Answers may vary.Sample: only onediag. is needed.

56. Given ABCD withdiag. AC. Let AC bisect BAD .Because ABC

DAC , AB = DA byCPCTC. Butsince opp. sides of a

are , AB = CD and BC = DA.So AB = BC = CD =DA, and ABCD isa rhombus. The newstatement is true.

53. (continued)

54. a. Def. of a rhombus

b. Diagonals of abisect eachother.

c. AE AE

54. (continued)d. Reflexive Prop. of

.

e. ABC ADE

f. CPCTC

g. Add. Post.

h. AEB andAED are rt. .

i. suppl. arert. Thm.

j. Def. of

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62. Answers may vary.Sample: Thediagonals of abisect each other so

AE CE . BothAED and CED are right becauseAC BD , and sinceDE DE by theReflexive Prop.,

AED CED bySAS. By CPCTC AD

CD , and sinceopp. sides of aare , AB BC AD .

57. 16, 16

58. 2, 2

59. 1, 1

60. 1, 1

61. 4. 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 )

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62. (continued)So ABCD is arhombus because ithas 4 sides.

63.


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Trapezo ids and Ki t esTrapezo ids and Ki t esGEOMETRY LESSON 6-5GEOMETRY LESSON 6-5

19. (continued)with KM KM bythe Reflexive Prop.means KLM

KNM by SAS. Soby CPCTC, opp.L and N are , so itis not an isos.trapezoid.

20. 12

21. 15

22. 15

23. 3

24. 4

25. 1

26. 1. ABCD is an isos. trapezoid, AB DC . (Given)

2. 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. are .)

6. DC AE (Opp. sides of a are .)

7. AB AE (Trans. Prop. of )

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26. (continued)8. AEB is an isosc. . (Def. of an isosc. )

9. B 1 (Base of an isosc. are .)

10. B C (Trans. Prop. of )

11. B and BAD are suppl., C and CDAare suppl. (Same side int. are suppl.)

12. BAD CDA (Suppl. of are .)

27. 28

28. x = 35, y = 30

29. x = 18, y = 108

30. Isosc. trapezoid; allthe large rt.appear to be .

31. 112, 68, 68

32. Yes, the canbe obtuse.

33. Yes, the canbe obtuse, as wellas one other .

34. Yes; if 2 arert. , they aresuppl. The other 2are also suppl.

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Trapezo ids and Ki t sTrapezo ids and Ki t sGEOMETRY LESSON 6-5GEOMETRY LESSON 6-5

s

35. No; if twoconsecutive aresuppl., then another pair must be also

because one pair of opp. is .Therefore, the opp.

would be ,which means thefigure would be aand not a kite.

36. Yes; the mustbe 45 or 135 each.

37. No; if twoconsecutive were

40. 1. AB CD , andAD CD (Given)

2. BD BD (Refl. Prop. of )

3. ABD CBD (SSS)

4. A C (CPCTC)

41. Answers may vary.Sample: Draw TAand RP

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37. (continued)compl., then the kitewould be concave.

38. Rhombuses andsquares would bekites since opp.sides can be also.

39. D is any point onBN such that ND BN and D is belowN .

=/


41. (continued)1. isosc. trapezoid TRAP (Given)

2. 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)

42. Draw BI as described, then drawBT and BP .

1. TR PA (Given)

2. R A (Base of isosc.trap. are .)

3. RB AB (Def. of bisector)

4. TRB PAB (SAS)

5. BT BP (CPCTC)

6. RBT ABP (CPCTC)

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42. 7. (continued)TBI PBI

(Compl. of are .)

8. BI BI (Refl. Prop. of )

9. TBI PBI (SAS)

10. BIT BIP (CPCTC)

11. BIT and BIP are rt. .( suppl. are rt. .)

12. TI PI (CPCTC)

13. BI is bis. of TP .(Def. of bis.)

43-44. Check studentsjustifications. Samples aregiven.

43. It is one half the sum of the lengthsof the bases; draw adiag. of the trap. to form 2 . Thebases B and b of the trap. areeach a base of a . Then thesegment joining the midpts. of thenon- sides is the sum of themidsegments of the .

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

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44. (continued)the midpts. of the non- sides is

(B + b ). The middle part of thissegment joins the midpts. of thediags. Each outer segmentmeasures B . So the length of the segment connecting themidpts. of the diags. is ( B b ).

45. B

46. I

47. C

48. C

49. D

50. [2]

HRW and HBW

[1] incorrect diagram OR no workshown

51. 126

52. 27

53. 27

12

12

1

2


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54. a . 4

b. 5

c. 5

55. a. 3

b. 30

c. 30

56. SAS

Plac ing F igures in the Coord ina te P lanePlac ing F igures in the Coord ina te P laneGEOMETRY LESSON 6-6GEOMETRY LESSON 6-6


1. W (0, h ); Z (b , 0)

2. 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 )

7. , ;

8. a , ; undefined

9. (b , 0); undefined

10. , ;

11. ;

12. , c ; 0

13. a. (2a , 0)

b. (0, 2 b )

c. (a , b )

d. b 2 + a 2

13. (continued)e. b 2 + a 2

f. b 2 + a 2

g. MA = MB = MC

1419. Answers mayvary. Samples aregiven.

14. A, C , H , F

15. B , D , H , F

16. A, B , F , E

b 2

h 2

h b

a 2

a 2

b 2

b a

r 2

t 2

t r

b 2


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17. A, C , G , E

18 . A, C , F , E

19. A, D , G , F

20. W (0, 2 h ); Z (2b , 0)

21. W (2a , 2 a ); Z (2a , 0)

22. W (2 b , 2 b );Z (2 b , 2 b )

23. W (0, b ); Z (2a , 0)

24. Z (0, 2 t ); W(2 r , 0)

25. W (2 b , 2c); Z (0, 2 c )

26. a. Diag. of a rhombusare .

b. Diag. of a thatis not a rhombusare not .

27. Answers may vary.Sample: r = 3, t = 2;

slopes are and ;

all lengths are 13;the opp. sides havethe same slope, sothey are . The 4

27. (continued)sides are .

28. (c a , b )

29. (a , 0)

30. (b , 0)

31. a.

b. (b , 0), (0, b ),(b , 0), (0, b )

23

23


31. (continued)c. b 2

d. 1, 1

e. Yes, because theproduct of theslopes is 1.

32. a.

32. (continued)b.

c. b 2 + 4 c 2

d. b 2 + 4 c 2

e. the lengths are =.

33.

34. Step 1: (0, 0)

Step 2: ( a , 0)

Step 3: Since m 1 +m 2 + 90 = 180, 1and 2 mustbe compl. 3 and

2 are the acuteof a rt. .

Step 4: ( b , 0)

Step 5: ( b , a)

Step 6: Using theformula for slope, the

s


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34. (continued)slope for 1 =

and the slope for 2= . Mult. the

slopes, = 1.

35. B

36. F

37. C

38. C

39. A

40. C

41. [2] (b , a ); the diag. of a rectangle bisect

each other.

[1] no conclusiongiven

42. 62, 118, 118; 2.5

43. (3, 2)

44. (3, 4)

45. a. Reflexive

b. AAS

b a

a b b

a a b

Proofs Us ing Coord ina te Geomet ryProofs Us ing Coord ina te Geomet ryGEOMETRY LESSON 6-7GEOMETRY LESSON 6-7


1. a. W , ;

Z ,

b. W (a , b );Z(c + e , d )

c. W (2a , 2 b );Z (2c + 2 e , 2 d )

d. c; it usesmultiples of 2 toname thecoordinates of W and Z .

2. a. origin

b. x -axis

c. 2

d. coordinates

3. a. y -axis

b. Distance

4. a. rt.

b. legs

4. (continued)c. multiples of 2

d. M

e. N

f. Midpoint

g. Distance

5. a. isos.

b. x -axis

c. y -axis

a 2

b 2

c + e 2

d 2


23/27


5. (continued)d. midpts.

e. sides

f. slopes

g. the DistanceFormula

6. a. (b + a )2 + c 2

b. (a + b )2 + c 2

7. a. a 2 + b 2

b. 2 a 2 + b 2

8. a. D (a b , c ),E (0, 2 c ),F (a + b , c ),G (0, 0)

b. (a + b )2 + c 2

c. (a + b )2 + c 2

d. (a + b )2 + c 2

e. (a + b )2 + c 2

f.

g.

8. (continued)h.

i.

j. sides

k. DEFG

9. a. (a , b )

b. (a , b )

c. the same point

10. Answers may vary.Sample: The

c a + b

c

a + b

c a + b

c a + b


10. (continued)Midsegment Thm.;the segmentconnecting themidpts. of 2 sides of the is tothe 3rd side and half its length; you canuse the MidpointFormula and theDistance Formula toprove the statementdirectly.

11. a.

b. midpts.

11. (continued)c. (2 b , 2 c )

d. L(b , a + c ),M (b , c ), N (b , c ),K (b , a + c )

e. 0

f. vertical lines

g.

h.

1224. Answers mayvary. Samples aregiven.

12. yes; Dist. Formula

13. yes; same slope

14. yes; prod. of slopes= 1

15. no; may not haveintersection pt.

16. no; may needmeasures


24/27



18. yes; prod. of slopes

of sides of A = 1

19. yes; Dist. Formula

20. yes; Dist. Formula,2 sides =


22. yes; intersection pt.for all 3 segments

23. yes; slope of AB =slope of BC

24. yes; Dist. Formula,

AB = BC = CD = AD

25. 1, 4, 7

26. 0, 2, 4, 6, 8

27. 0.8, 0.4, 1.6, 2.8, 4,5.2, 6.4, 7.6, 8.8

28. 1.76, 1.52, 1.28, . . . , 9.52, 9.76

29. 2 + , 2 + 2 ,

2 + 3 , . . . . ,

2 +( n 1)

30. (0, 7.5), (3, 10),(6, 12.5)

31. 1, 6 , 1, 8 ,

(3, 10), 5, 11 ,

7, 13

32. (1.8, 6), (0.6, 7),

12n

12n

12n

12n

23

13

23

13


32. (continued)(0.6, 8), (1.8, 9), (3, 10), (4.2, 11),(5.4, 12), (6.6, 13), (7.8, 14)

33. (2.76, 5.2), (2.52, 5.4),(2.28, 5.6), . . . , (8.52, 14.6),(8.76, 14.8)

34. 3, + , 5 + ,

3 + 2 , 5 + 2 , . . . . . ,

3 + ( n 1) , 5 + (n 1)

35. Assume b > a. a + ,

a + 2 , . . . . ,

a + (n 1)

36. Assume b a , d c .

a + , c + ,

a + 2 , c + 2 , . . . ,

a + (n 1) , c + (n 1)

37. a. The with bases d and b , andheights c and a , respectively,have the same area. They

12

n

10

n 12n

10n

12n

10n

b a n

b a n

b a n

b a n

d c n

b a n

d c n

b a n

d c n

s


25/27


37. a. (continued)share the small right with based and height c , and the remainingareas are with base c and

height ( b d ). So ad = bc.Mult. both sides by 2 givesad = bc .

b. The diagram shows that = ,

since both represent theslope of the top segment of the

. So by (a), ad = bc .

38. Divide the quad. into 2 . Find the

centroid for each and connectthem. Now divide the quad. into 2

38. (continued)other and follow the samesteps. Where the two lines meetconnecting the centroids of the 4

is the centroid of the quad.

39. a. L(b , d ), M (b + c , d ), N (c , 0)

b. AM : y = x ;

BN : y = (x c );

CL: y = (x 2c )

c. P ,

d. Pt. P satisfies the eqs. for AM

s

12 12

a b

c d

s

s

s

d b + c

2d 2b c

2(b + c )3

2d 3

d b 2c


39. d. (continued)and CL.

e. AM = (b + c )2 + d 2 ;

AP = =

(b + c )2 + d 2 =

(b + c )2 + d 2 = AM

The other 2 distances are foundsimilarly

40. a.

40. (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 ) =

(a ) = . So p and q

intersect at 0, .

e.

f. Let a pt. on line r be ( x , y ).

Then the eq. of r is =

2(b + c ) 23

2d 23

2 2

3

23

23

b c

y 0x a

b c b

c

ab c

b c

a c

ab c

y 0x b

a c

b c


26/27


40. f. (continued)or y = (x b ).

g. = (0 b )

h. 0,

41. a. Horiz. lines have slope 0, andvert. lines have undef. slope.Neither could be mult. to get 1.

b. Assume the lines do notintersect. Then they have thesame slope, say m . Then m m =m 2 = 1, which is impossible. Sothe lines must intersect.

41. (continued)c. Let the eq. for 1 be y = x

and for 2 be y = x and the

origin be the int. point.

a c

ab c

a c

ab c

b a

a b


41. c. (continued)Define C (a , b ), A(0, 0), and

B a , . Using the Distance

Formula, AC = a 2 + b 2,

BA = a 2 + , and

CB = b + .

Then AC 2 + BA2 = CB 2, andm A = 90 by the Conv. of thePythagorean Thm. So 1 2 .

42. A

43. G

44. [2] a. = 3; a = 1;

= 4; b = 11;

(a , b ) = (1, 11)

b. (7 (1)) 2 + (3 11) 2

= 260 = 2 65

16.12

[1] minor computational error ORno work shown

a 2 b

a 4b 2

a 2b

7 + a 2

3 + b 2


27/27


45. [4] a-b. Sample:

c. AP = (b a )2 + c 2 = RQ PQ = PQ AQ = a = RP

APQ RQP by SSS[3] minor computational error

[2] parts (a) and (b) correct

[1] one part correct

46. (a , b )

47. a. If the sum of the of a polygonis 360, then the polygon is a

quad.

b. If a polygon is a quad., then thesum of its is 360.

48. a. If x 51, then 2 x 102.

b. If 2x 102, then x 51.

49. a. If a 5, then a 2 25.

b. If a 2 25, then a 5.

s

s

=/ =/

=/ =/

=/ =/

=/ =/


50. a. If b 4, then b is not neg.

b. If b is not neg., then b 4.

51. a . If c 0, then c is not pos.

b. If c is not pos., then c 0.

52. 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. ThenCPCTC 1 2.

54. LM NK , LN NL by the Refl.Prop. of , and LNK NLM by all rt. are . So LNK

NLM by SAS, and K M byCPCTC.

s

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