chapter 8 answers - bisd moodlemoodle.bisd303.org/file.php/18/solutions/chapter 08 midterm...

Chapter �� Lesson � ��

Chapter �� Lesson �Chapter �� Lesson �Chapter �� Lesson �Chapter �� Lesson �Chapter �� Lesson �

Set I Set I Set I Set I Set I (pages ��–��)

In their book titled Symmetry—A UnifyingConcept (Shelter Publications� ��)� István andMagdolna Hargittai point out that it was LouisPasteur who first discovered that otherwiseidentical crystals can be mirror images of oneanother Except for glycine� all amino acids canexist in oppositehanded forms� but only the lefthanded version occurs naturally The Hargittaiswrite: “Many biologically important chemicalcompounds exist in lefthanded and righthandedforms� and the biological activity of the twoforms may be very different Humans metabolize only righthanded glucose Lefthandedglucose� although still sweet� passes through thesystem untouched” Martin Gardner’s book titledThe New Ambidextrous Universe (W H Freemanand Company� ��) is a good source of furtherinformation on the subject

At first glance� it might appear that RogerShepard’s figure (exercises �� through ��) alsoillustrates rotations because the arrows point inopposite directions The arrows pointing to theleft� however� are not congruent to the arrowspointing to the right The figure is featured onthe cover of Al Seckel’s excellent book titled TheArt of Optical Illusions (Carlton Books� ��)

Transformations in Art.

•1. A translation.

2. A rotation.

•3. A rotation.

4. No.

5. No.

Mirror Molecules.

•6. A reflection.

7. “Left-handed” and “right-handed.”

Reflections.

8.

9.

10.

11.

12.

13.

14.

15.

•16. Exercises 9, 11, and 14.

•17. Each has a vertical line of symmetry.

Down the Stairs.

18. A translation.

19. A one-to-one correspondence between twosets of points.

•20. Yes.

21. A transformation that preserves distanceand angle measure.

�� Chapter �� Lesson �

Peter Jones.

22. A rotation.

23. The image is produced from the originalfigure by rotating the figure 90° clockwise.

24. S.

Set IISet IISet IISet IISet II (pages ��–��)

As the Adobe Illustrator User Guide explains� theprogram “defines objects mathematically asvector graphics” Exactly how vector graphicscombines geometry with linear algebra and matrixtheory is explained in detail in The GeometryToolbox for Graphics and Modeling by Gerald EFarin and Dianne Hansford (A K Peters� ��)

Escalator Transformations.


26. A rotation.

27.

35.

•28. They are parallelograms because they havetwo sides that are both parallel and equal.

29. The opposite sides of a parallelogram areparallel.

30. It is a parallelogram.

•31. The opposite sides of a parallelogram areequal.

32. SSS.

33. Corresponding parts of congruent trianglesare equal.


•36. Betweenness of Rays Theorem.

•37. Substitution.

38. Subtraction.

39. SAS.


41. Distances.

Triangle Construction.

42.

•43. A dilation.

•44. They seem to be twice as long.

45. They seem to be equal.

46. No. It is not an isometry, because it doesn’tpreserve distance.

Computer Geometry.

47. A translation.

48. A rotation.

49. A dilation.


50. Set IIISet IIISet IIISet IIISet III (page ��)

Toothpick Puzzle.

1. (There are two possible ways to solve thepuzzle, one of which is a reflection of theother. The three toothpicks moved areshown as dotted lines in the figures below.)

51. A(3, 1) → D(3 + 2, 1 – 7), or D(5, –6).B(5, 2) → E(5 + 2, 2 – 7), or E(7, –5).C(2, 6) → F(2 + 2, 6 – 7), or F(4, –1).


53. A(3, 1) → G(–3, 1).B(5, 2) → H(–5, 2).C(2, 6) → I(–2, 6).

54. A reflection.

55. A(3, 1) → J(–3, –1).B(5, 2) → K(–5, –2).C(2, 6) → L(–2, –6).

•56. A rotation.

57. A(3, 1) → M(6, 2).B(5, 2) → N(10, 4).C(2, 6) → O(4, 12).

58. A dilation.

2. Yes. The reversed fish is a rotation image ofthe original fish. The center of rotation canbe the midpoint of either toothpick formingthe fish’s back.



Capital letters in many typefaces do not have thesimple symmetries suggested by exercises ��through �� For example� the horizontal bar of anH doesn’t always connect the midpoints of theside bars The upper part of an S is frequentlysmaller than the lower part

David Moser� the creator of the “China”transformation� has designed some otheramazing figures of this sort Two are included inthe book of visual illusions titled Can You BelieveYour Eyes by J R Block and Harold E Yuker(Brunner/Mazel� ��) One changes “England”from Chinese into English and the other does thesame thing with “Tokyo”! Both word transformations are accomplished� like the example in thetext� by a ��° rotation

A Suspicious Cow.

1. The water from which the cow is drinking.

2. It is upside down. The cow and barn thatwe see “above” the water are actuallyreflections.


3. •14. Its image looks the same.

15. Its image looks the same.

16. Its image looks the same.

17. In exercise 12. The figure has both a verticaland a horizontal line of symmetry.

SAT Problem.

•18. B.

19. A reflection.

20. Measure the distances from B and P to thedotted line and see if they are the same.

Can You Read Chinese?

21. A reflection.

•22. A reflection.

23. A rotation.

•24. Yes, because it is the composite of tworeflections in intersecting lines.

25. It is a translation of the word “China” fromChinese into English.

Set II Set II Set II Set II Set II (pages ��–��)

Ethologist Niko Tinbergen carried out a numberof interesting laboratory studies concerninganimal vision� including one with the “goosehawk” and an experiment with circular disksinterpreted by newly hatched blackbirds as their“mother” These examples and others are includedby cognitive scientist Donald D Hoffman inVisual Intelligence—How We Create What WeSee (Norton� ��)

Kaleidoscope Patterns.

26. B, D, and F.

•27. C and E.

•28. 120°.

29. Three. The three mirror lines. (The monkeyfaces are almost mirror symmetric; so thelines that bisect the angles formed by themirrors almost look like lines of symmetry.)

•30. No. The figure does not look exactly thesame upside down. (The monkey’s leftnostril is closer to its left eye than its rightnostril is to its right eye; so it is possible totell the difference.)

•4. 2x units.

•5. It is the segment’s perpendicular bisector.

Double Reflections.

6. b.

•7. A rotation.

8. 90°.

•9. 180°.

10.

11.

12.

13.


Scaring Chickens.

31. B.

•32. E.

33. D.

34. A.

•35. E.

36. A.


38. A translation is the composite of tworeflections in parallel lines.

39. Seeing the bird at C flying to A.

Boomerang.

40.

52. It is twice as large; ∠AOA” = 2∠XOY.

Set III Set III Set III Set III Set III (page ��)

A E W Mason� the author of The House of theArrow (��)� is best known for his novels TheFour Feathers and Fire Over England The Houseof the Arrow� which featured his detectiveInspector Hanaud investigating the murder of aFrench widow� was made into a movie threetimes Following the excerpt quoted in the Set IIIexercise� the story continues:

“It was exactly halfpast one; the long minutehand pointing to six� the shorter hour hand onthe righthand side of the figure twelve� halfwaybetween the one and the two With a simultaneousmovement they all turned again to the mirror;and the mystery was explained The shorterhourhand seen in the mirror was on the lefthandside of the figure twelve� and just where it wouldhave been if the hour had been halfpast ten andthe clock actually where its reflection was Thefigures on the dial were reversed and difficult ata first glance to read”

What Time Was It?

1.

•41. A rotation.

42. A rotation is the composite of two reflectionsin intersecting lines.

43. At the point in which the two lines intersect.

Triangle Reflections.

•44. If a point is reflected through a line, the lineis the perpendicular bisector of the segmentconnecting the point and its image.

45. A translation.

•46. Its magnitude.

47. It is twice as long; AA” = 2XY.

•48. SAS.


50. A rotation.

51. Its magnitude.

2. Examples suggest that thereflection of a clock face in avertical mirror always lookslike an actual time. This isnot true for the reflection ofa clock face in a horizontalmirror. For example, ahorizontal reflection of aclock face reading 10:30would not look like an actualtime because the hour handwould be in the wrongposition when the minutehand is pointing to the top ofthe clock.




The number “eight” seems to have a specialconnection to sports in which synchronization isimportant One of the definitions given in TheAmerican Heritage Dictionary of the EnglishLanguage for the word “eight” is “an eightoaredracing shell” Team routines in synchronizedswimming consist of eight swimmers

Prevaricator.

1. He has “liar” written all over his face!

2. A translation.

•3. A reflection.

4. A rotation.

•5. A glide reflection.

6. Yes.

7. If two figures are congruent, there is anisometry such that one figure is the imageof the other.

Synchronized Oars.

•8. ∠PAB and ∠PCD.

9. If the oars are assumed to be identical,AB = CD. A quadrilateral is a parallelogramif two opposite sides are both parallel andequal.

•10. The opposite sides of a parallelogram areequal.

11. A translation.

12. A glide reflection.

13. Synchronized swimming.

Swing Isometries.


15. A rotation.

16. Its center.

•17. The magnitude of the rotation.

•18. The lines bisect these angles.

19. They are the perpendicular bisectors ofthese line segments.

20. It is twice as large.

Quadrilateral Reflections.

21. Two points determine a line.

•22. They are the perpendicular bisectors ofthese line segments.

23. They appear to be parallel.

24. A translation.

•25. Two figures are congruent if there is anisometry such that one figure is the image ofthe other.


The symmetry of the illustration for the pianomoving problem suggests that the distances fromA and J to the corner of the room are equal If thisis the case� then it is easy to prove that ∆DGP inthe figure below is an isosceles right triangle Itfollows that ∠CDG � ∠DGH � ��° The piano�then� has been rotated � �° in all If the symmetrydoesn’t exist� ∆DGP will still be a right trianglewith complementary acute angles; so� eventhough the measures of ∠CDG and ∠DGH willchange� their sum will remain the same

Moving a Piano.

26. A rotation.

•27. D.

28. E.

•29. G.

30. D.

31. Another rotation about point G.

32. 270°.


Bulldogs.

33. Two figures are congruent if there is anisometry such that one figure is the imageof the other.

34. A translation.

•35. No, because the two dogs are not mirrorimages of each other.

36. Yes. A translation is the composite of tworeflections through parallel lines.

37. No, because, if there were three reflections,one dog would be a mirror image of theother.


•39. A translation and a reflection.

40. Three reflections.

Grid Problem.

41.

Set III Set III Set III Set III Set III (pages �� –��)

The irregular rubber stamp figure was chosen totry to impress upon the student’s mind that thenumber of reflections needed to show that twofigures are congruent has nothing to do withtheir complexity David Henderson suggests anice experiment in his book titled ExperiencingGeometry—In Euclidean� Spherical and HyperbolicSpaces (Prentice Hall� ��) He says: “Cut atriangle out of an index card and use it to drawtwo congruent triangles in different orientationson a sheet of paper Now� can you move onetriangle to the other by three (or fewer)reflections? You can use your cutout triangle forthe intermediate steps” This leads to proving that“on the plane� spheres� or hyperbolic planes� everyisometry is the composition of one� two� or threereflections” Chapter �� of Henderson’s book is agood resource for students (and teachers)wanting to learn more about isometries

Stamp Tricks.

1. Ollie stamped one of the images on the backof the tracing paper.

2. One figure is now the mirror image of theother and two reflections (of an asymmetricfigure) cannot produce a mirror image.

3. Yes. The figure below shows one of themany ways in which it can be done.

•42. B’(14, 9), C’(9, 6).

•43. 6 . (AA’ = =

= = 6 .)

44. AA’ appears to be parallel to line l.

45. A glide reflection, because it is the compos-ite of a translation and a reflection in a lineparallel to the direction of the translation.

•46. M(11, 4), N(12, 5), P(8, 1).

47. The y-coordinate is 7 less than thex-coordinate.

48. They lie on line l.




Two wonderful sources of ambigrams are ScottKim’s Inversions (Byte Books� ��) and JohnLangdon’s Wordplay (Harcourt Brace Jovanovich��) In the introduction to his book� Langdonwrote:

“Ambigrams come in a number of forms�limited only by the ambigrammist’s imagination�and usually involve some kind of symmetry Thisbook is made up of three types:

� words with rotational symmetry � words that have bilateral� or mirrorimage

symmetry � chains These are ambigrams that cannot

stand alone as single words� but depend onbeing linked to the preceding and ensuingwords”

In other words� ambigrams are based on thethree basic types of symmetry in the plane:rotation� reflection� and translation

The fact that games normally have twoopponents or two opposing teams requires thatalmost all playing fields and courts have two linesof symmetry Baseball and its related versions canuse a field with just one line of symmetry becausethe teams regularly interchange their positions inthe game

It is interesting that� although Washingtonhad the extra window painted solely to “completethe symmetry�” the two windows on either sideof the front door� with the windows above them�are not quite in the right places

Ambigrams.

•1. Rotation (or point) symmetry.

2. See if it coincides with its rotation image.(Or, for point symmetry, see if it looks thesame upside down.)

3. Reflection (line) symmetry.

4. See if it coincides with its reflection imageor fold it to see if the two halves coincide.

Sport Symmetry.

5. Baseball.

•6. It has reflection (line) symmetry withrespect to a line through home plate andsecond base.

7. Basketball.

8. It has reflection (line) and rotation (point)symmetry. It has two lines of symmetry and2-fold rotation symmetry.

9. The same type as the basketball court.

10. (Student answer.) (It is so that each team hasthe same view of the other side.)

Mount Vernon.

11. To make his house look more symmetric.

Symmetries of Basic Figures.

•12. The point itself. (Not its “center,” because apoint does not have a center!)

13. Yes. A point is symmetric with respect toevery line that contains it (because a pointon a reflection line is its own image.)

•14. A line looks the same if it is rotated 180°.

15. Any point on the line can be chosen as itscenter of symmetry.

•16. A line has reflection symmetry because itcan be reflected (folded) onto itself.

17. Infinitely many. The line itself and everyline that is perpendicular to it.

18. Yes. A line can be translated any distancealong itself and still look the same.

•19. That rays OA and OD are opposite rays andthat rays OB and OC are opposite rays.

20. If ∠AOB is rotated 180° about point O, itcoincides with ∠COD.

21. Vertical angles are equal.

•22. In a plane, two points each equidistant fromthe endpoints of a line segment determinethe perpendicular bisector of the linesegment.

23. A and C.

24. If two sides of a triangle are equal, theangles opposite them are equal.

•25. They bisect each other.

26. CD.

27. The opposite sides of a parallelogram areequal.


28. No. BD cannot be a rotation image of ACbecause they do not have the same length.


For the piano keyboard� it is interesting to notethat all of the translation images of a given keyhave the same letter name The key that is thefirst translation image to the right of a given keyis one octave higher and has a frequency twice asgreat Without the translation pattern of theblack keys� pianists would have trouble keepingtheir place!

In his book titled Reality’s Mirror (Wiley� ��)�Bryan Bunch explains the connection of odd andeven wave functions to such topics as cold fusionand the Pauli Exclusion Principle Bunch writes:“It would be only a slight exaggeration to say thatsymmetry accounts for all the observable behaviorof the material world”

Piano Keyboard.

29. Example answer:

41. Rotation (or point).

•42. Even.

43. Neither.

44. Odd.

Cherry Orchard.

•45. Because it can be translated (in variousdirections) and look exactly the same.

46. The distance between any pair ofneighboring trees.

47.

•30. Translation.

31. No.

Water Wheel.

32. Rotation and point symmetry.

•33. 22.5°. ( .)

•34. No. ( is not an integer.)

35. Yes. ( = 10.)

•36. 16.

37. No.

Wave Functions.

•38. The y-axis.

39. Reflection (line).

40. The origin.

48. Because it can be reflected (in various lines)and look exactly the same.

49. Because it can be rotated (about variouspoints) and look exactly the same.

50. Any tree. Also, any point centered betweenthree neighboring trees. (Also, the midpointsof the segments between neighboring treesare centers of 2-fold rotation symmetry.)

51. 60°.

Set III Set III Set III Set III Set III (page ��)

Short Story.

1.

2. The figure can be read as either Scott’s firstname or his last name.

�� Chapter �� Review

Chapter �� ReviewChapter �� ReviewChapter �� ReviewChapter �� ReviewChapter �� Review

Set I Set I Set I Set I Set I (pages ��–�� )

In Visual Intelligence—How We Create What WeSee (Norton� ��)� Donald Hoffman reports thatperhaps the earliest version of the “vasefaces”illusion appeared in a picture puzzle in � ��Perceptual psychologists first began to study theillusion in �� Of the many variations that haveappeared since� surely the most clever is its use inthe vase pictured in the text

In ��–�� Escher filled a large notebookwith notes and drawings that he titled RegularDivision of the Plane with AsymmetricalCongruent Polygons This notebook is reproducedin its entirety in Visions of Symmetry—Notebooks�Periodic Drawings� and Related Work of M CEscher� by Doris Schattschneider (W H Freemanand Company� ��) She remarks about thiswork: “It is impossible to look at this notebookand not conclude that in this work� Escher was aresearch mathematician”

Remarkably� the first book to be publishedon Escher’s work was written by Caroline HMacgillavry� a professor of chemical crystallography In Symmetry Aspects of M C Escher’sPeriodic Drawings (The International Union ofCrystallography� ��)� Macgillavry wrote thatEscher’s notebook had been a “revelation” to her�commenting: “It is no wonder that xray crystallographers� confronted with the ways in whichnature solves the same problem of packingidentical objects in periodic patterns� areinterested in Escher’s work”

In lecturing on his work� Escher explicitlydiscussed his use of the four isometries of theplane in creating his mosaics For more on this�see the section titled “The geometric rules”(p ��ff) in Visions of Symmetry

Amazing Vase.

1. No. The left and right sides of the upperpart of the vase do not quite look like mirrorimages of each other.

2. The profiles of Prince Philip and QueenElizabeth can be seen in silhouette.

Double Meanings.

3. A one-to-one correspondence between twosets of points.


5. A transformation in which the image is anenlargement or reduction of the original.

•6. No.

Monkey Rug.

7. A translation.

•8. A glide reflection.

•9. A rotation.

10. A reflection.

Clover Leafs.

•11. That it can be rotated so that it looks thesame in three positions.

12. 120°.

13. No. It doesn’t look the same upside down.

14.

15.

16. 90°.

•17. Yes.

18. Good.

Musical Transformations.

19. A translation.


21. A translation.

22. A rotation.

Fish Design.

23. No. There seem to be two shapes of fish inthe mosaic. The differences in their nosesand tails are the most obvious.

Chapter �� Review ��

•24. Two figures are congruent if there is anisometry such that one figure is the image ofthe other.


26. A rotation.

27. Yes. Translations.Example answer:

32. About 24 in. (In the figure, the length of thebat is 42 mm and the magnitude of thetranslation is about 24 mm.)

33. Example figure:

Set II Set II Set II Set II Set II (pages �� –��)

Keep Your Eye on the Ball—The Science andFolklore of Baseball� by Robert G Watts and ATerry Bahill (W H Freeman and Company� ��)is a great source of examples for use in illustratingthe application of algebra� geometry� and physicsto the analysis of baseball The authors describethe figure used for exercises �� through ��: “Theswing of a baseball bat exhibits two types ofmotion: translational and rotational To movethe bat from position A to position C� one canfirst rotate the bat about the center of mass andthen translate the center of mass”

John Langdon wrote that his “past” and“future” ambigrams “are intended to representthe interminable conveyor belt of time� and theindividual letters and words are the experiences�the ‘life bites’ of our existence”

Batter’s Swing.

28. A rotation.


30. For a rotation, the two reflection linesintersect. For a translation, they are parallel.

•31. About 47°.

Construction Problem.

34.

35. The image of AB is AD and the image of BCis DC.

•36. Because ∆ADC is the reflection image of∆ABC.

•37. It has reflection (line) symmetry withrespect to line AC.

38. A rotation.

�� Chapter �� Review

From N to Z.

39.

40. A rotation.

•41. 45°.

42. 90°.

43. The image of H looks like I. The image of Wlooks somewhat like E. (And, the image of Zlooks like N.)

Past and Future.

44. Example answers:

45. Translation.

46. So that the E’s look “correct” in bothdirections.

•47. No.

48. The letters A and T in PAST.The letters T, U, and E in FUTURE.

49. Yes. The letter E.

Dice Symmetries.

50. Group 1: 1, 4, and 5.These faces have 4 lines of symmetry and4-fold rotation (as well as point) symmetry.Group 2: 2, 3, and 6.These faces have 2 lines of symmetry and2-fold rotation (as well as point) symmetry.In regard to 2 and 3, the symmetry linescontain the diagonals of the face; in regardto 6, they are parallel to the edges andmidway between them.

Midterm ReviewMidterm ReviewMidterm ReviewMidterm ReviewMidterm Review


Chill Factor.

1. A conditional statement.

2. If you don’t put your coat on.

Losers, Sleepers.

3. An Euler diagram.

4. If you snooze, you lose.

5. You lose.

Marine Logic.

6. A syllogism.

7. If one of its premises were false.

Finding Truth.

8. Theorems.

9. Postulates.

Only in Geometry.

10. Points that lie on the same line.

11. The side opposite the right angle in a righttriangle.

12. A triangle or trapezoid that has two equallegs.

13. A quadrilateral all of whose sides are equal.

Why Three?

14. Three noncollinear points determine aplane.

15. Point, plane.

What Follows?

16. A line.

17. 180°.

18. Bisects it.

19. Are equal.

20. Equiangular.

21. Either remote interior angle.

22. The perpendicular bisector of the linesegment.

Midterm Review ��

23. Are parallel.

24. Are parallel.

25. Also perpendicular to the other.

26. The remote interior angles.

27. Equal.

28. Are equal.

Formulas.

29. The area of a circle is π times the square ofits radius.

30. The area of a rectangle is the product of itslength and width.

31. This is the Distance Formula. The distancebetween two points is the square root of thesum of the squares of the differences of theirx-coordinates and their y-coordinates.

32. The perimeter of a triangle is the sum of thelengths of its sides.

33. The perimeter of a rectangle is the sum oftwice its length and twice its width.

Protractor Problems.

34. 84°. (123 – 39.)

35. 42°. ( 84°.)

36. 81. (39 + 42 or 123 – 42.)

37. 48°. (90° – 42°.)

38. 171. (123 + 48 = 171.)

39. Yes. OD-OC-OB because 171 > 81 > 39.

Metric Angles.

40. Two angles are supplementary iff their sumis 200 grades.

41. An angle is obtuse if it is greater than 100grades but less than 200 grades.

42. The sum of the angles of a quadrilateral is400 grades.

43. Each angle of an equilateral triangle is 66grades.

Linear Pair.

44. That the figure contains two opposite rays.

45. No. If the angles are equal, they would beright angles.

46. They are supplementary.

47. Each angle is a right angle.

Polygons.

48. or .

(c2 = 152 + 302 = 225 + 900 = 1125,

c = = .)

49. 30. (It can’t be 15, because 15 + 15 = 30.)

50. 30°, 150°, and 150°. (A rhombus is aparallelogram. The opposite angles of aparallelogram are equal and the consecutiveangles are supplementary.)

51. 30°, 150°, and 150°. (The base angles of anisosceles trapezoid are equal. The parallelbases form supplementary angles with thelegs.)

Bent Pyramid.

52. 126°. (180° – 54°.)

53. 94°. [180° – 2(43°).]

54. 54°. (∠E = ∠A.)

55. 169°. (126° + 43°.)

Six Triangles.

56. ∆AFP and ∆DEP.

57. By SAS. (ED = AP because AB = ED andAB = AP.)

58. By HL. (AF = PB because AF = CD andCD = PD.)

Italian Theorem.

59. ∠B > ∠A.

60. If two sides of a triangle are unequal, theangle opposite the greater side is greaterthan that opposite the smaller side. (If twosides of a triangle are unequal, the anglesopposite them are unequal in the sameorder.)

61. Major and minor.

Impossibly Obtuse.

62. ∠A + ∠B + ∠C > 270°.

�� Midterm Review

63. The fact that the sum of the angles of atriangle is 180°.

64. It shows that what we supposed is false.

65. Indirect.

Converses.

66. If two angles are complementary, then theyare the acute angles of a right triangle. False.

67. If the diagonals of a quadrilateral bisecteach other, then the quadrilateral is aparallelogram. True.

Construction Exercises.

68.

74. AB = DC = 12; AD = BC = or .

(AD = = =

= .

BC = = =

= .

75. (Student answer.) (PD looks longer to manypeople.)

76. PD = = =

= .

PC = = =

= .

Roof Truss.

77. ∠A = ∠L. ∠A and ∠L are correspondingparts of ∆AGF and ∆LGF, which arecongruent by SSS.

78. BC || FG. BC || DE and DE || FG becauseequal alternate interior angles mean thatlines are parallel. BC || FG because, in aplane, two lines parallel to a third line areparallel to each other.

79. Yes. CD || EF because they form equalalternate interior angles with DE.

80. No. If ∠AGF and ∠GFA were supplementary,GA would be parallel to FA (supplementaryangles on the same side of a transversalmean that lines are parallel). GA and FAintersect at A.

Angles Problem 1.

81.

69. No.

70. Point P.

71. CA, CP, and CB.


Grid Exercise.

72.

73. A parallelogram.

82. ∆ACD ≅ ∆BCE by ASA. AC = BC because∆ABC is equilateral, ∠DAC = 55° = ∠EBC,and ∠DCA = 60° = ∠ECB.

Midterm Review ��

Irregular Star.

83.

84. True. ∆EFC is isosceles because EF = FCbecause ∠C = ∠E.

85. True. FGHIJ is equiangular because each ofits angles is an angle of one of the fiveoverlapping isosceles triangles, whose baseangles are the equal angles at the corners ofthe star. If two angles of one triangle areequal to two angles of another triangle, thethird angles are equal.

86. False. (This is evident from the figure.)

87. True. There are 10 isosceles triangles in all.The base angles of the triangles that areacute (∆BFG is an example) are supplemen-tary to the equal angles of pentagon FGHIJ;so they are equal.

Angle Trisector.

88. These triangles are congruent by SSS.

89. ∠1 = ∠2 and ∠2 = ∠3 because they arecorresponding parts of the congruenttriangles; so ∠1 = ∠3.

Quadrilateral Problem.

90. x + 2x + 3x + 4x = 360, 10x = 360, x = 36.

91. 144°. [4(36°).]

92. ∠B = 72°, ∠C = 108°, AB || DC, ABCD is atrapezoid.

Angles Problem 2.

93.

94. ∆ABC is isosceles. ∠BAC = 50° = ∠ACB;so AB = BC.

A, B, C.

95. a2 + b2 = c2.

96. a + b > c, a + c > b, b + c > a.

97. No. The sum of two of its sides would equalthe third side, which would contradict theTriangle Inequality Theorem.

Midsegments.

98.

99. Applying the Midsegment Theorem to

∆ABC gives DE = AB and applying it to

∆ABF gives GH = AB; so DE = GH by

substitution.

100. Again, by the Midsegment Theorem,DE || AB and GH || AB; so DE || GH.In a plane, two lines parallel to a third lineare parallel to each other.

101. DEHG is a parallelogram because twoopposite sides, DE and GH, are both paralleland equal.

102. AG = GF and BH = HF because G and H arethe midpoints of AF and BF. GF = FE andHF = FD because the diagonals of aparallelogram bisect each other. SoAG = GF = FE and BH = HF = FD.

�� Midterm Review

On the Level.

103. The design of the swing is based on aparallelogram. The supports of the plankare equal, and the part of the plank betweenthe supports is equal to the distancebetween the supports at the top. The topremains level with the ground. The plank,the opposite side of the parallelogram, isalways parallel to the top; so it also stayslevel with the ground.

Folding Experiment.

104. Example figure (the student’s figure will dependon the point and corner chosen):

105. It appears to be the perpendicular bisectorof AB.

106. When B falls on A, CA = CB and DA = DB.In a plane, two points each equidistant fromthe endpoints of a line segment determinethe perpendicular bisector of the linesegment.

Earth Measurement.

107. 20,520 mi. (360 × 57.)

108. About 3,300 mi. (c = 2πr; so

r = = ≈ 3,300.)

Not a Square.

109. Yes. ABCD could be a rhombus if all of itssides were equal.

110. Yes. If ABCD is a rhombus, its diagonalswill be perpendicular.

111. Yes. ABCD is a parallelogram and thediagonals of a parallelogram bisect eachother.

112. No. If ABCD were a rectangle, ∠DAB wouldbe equal to ∠ABC.

113. No. If AC = DB, then ∆DAB ≅ ∆CBA by SSS(DA = CB because ABCD is a parallelogram.)If ∆DAB ≅ ∆CBA, then ∠DAB = ∠ABC but∠DAB ≠ ∠ABC.

Construction Exercise.

114. Example figure:

115. They seem to be concurrent.

SAT Problem.

116. A rotation.

117. A translation.

118. A reflection.

119. A reflection.

120. Figure E.

Quilt Patterns.

121. Rotation (point) symmetry (4-fold).

122. Rotation (point) symmetry and reflection(line) symmetry (4 lines).

123. Reflection (line) symmetry (1 line).

Dividing a Lot.

124.

125.

chapter 8 answers - bisd moodlemoodle.bisd303.org/file.php/18/solutions/chapter 08 midterm...

Documents