chapter 7 answers - bisd moodlemoodle.bisd303.org/file.php/18/solutions/chapter 07 answers.pdf ·...

Chapter �� Lesson � ��

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

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

The aerial photographer Georg Gerster� theauthor of Below from Above (Abbeville Press��)� wrote about his photograph of thefreeway overpass covered by carpets: “Years after having taken this photograph� I amstill amazed at this superscale Paul Klee � � � � InEurope� special mats� a Swiss development� havebeen adopted� They retain the heat generated asthe concrete hardens and make it unnecessary tokeep wetting the concrete� The new procedure ismore economical� though less colorful � � � �”

Theorem 24.

•1. Two points determine a line.

2. The sum of the angles of a triangle is 180°.

•3. Addition.

4. Betweenness of Rays Theorem.

5. Substitution.

Corollary.

•6. The sum of the angles of a quadrilateral is360°.

7. Division.

8. An angle whose measure is 90° is a rightangle.

•9. A quadrilateral each of whose angles is aright angle is a rectangle.

•10. Each angle of a rectangle is a right angle.

11. All right angles are equal.

Carpets.

12. Rectangles.

13. Right.

•14. Lines that form right angles are perpen-dicular.

15. In a plane, two lines perpendicular to athird line are parallel.

Rectangles.

16. They are equal and they are right angles.

•17. They seem to be parallel and equal.

18. They seem to be equal (and they seem tobisect each other).

Rhombuses.

19. They seem to be equal (and the oppositesides seem to be parallel).

20. They seem to be perpendicular (and theyseem to bisect each other).

•21. They are equiangular (or each of theirangles is a right angle).

22. They are equilateral.

Parallelograms.

23. They seem to be parallel and equal.

24. They seem to be equal.

25. They seem to bisect each other.

SAT Problem.

•26. 2x + 2y = 180. (2x + 2y + 180 = 360.)

27. x + y = 90.

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

Martin Gardner dedicated his book of mathematical recreations� Penrose Tiles to Trapdoor Ciphers(W� H� Freeman and Company� ��) to RogerPenrose with these words:

“To Roger Penrose� for his beautiful� surprisingdiscoveries in mathematics� physics� andcosmology; for his deep creative insights intohow the universe operates; and for his humilityin not supposing that he is exploring only theproducts of human minds�”

Two chapters of Gardner’s book are aboutPenrose tilings� a subject full of rich mathematicalideas� The two tiles in the exercises come from arhombus with angles of ��° and ��°� JohnHorton Conway calls them “kites” and “darts�”Amazingly� in all nonperiodic tilings of these tiles�the ratio of the number of kites to the number of

darts is the golden ratio: �� ! There

are a number of simple ways to prove that thereare infinitely many ways to tile the plane withthese two tiles� Even so� any finite region of onetiling appears infinitely many times in every othertiling; so no finite sample of a Penrose tiling canbe used to determine which of the infinitely

�� Chapter �� Lesson �

many different tilings it is! Students interested insuch mindboggling ideas will enjoy readingGardner’s material and doing the experimentssuggested in it�

The exercises on the angles of polygons areintended to encourage students to realize thatmemorizing a result such as “each angle of an

equiangular ngon has a measure of ”

is not important� What is important is for thestudent to feel comfortable and confident inbeing able to reconstruct such a result if it isneeded�

Penrose Tiles.

28. The one labeled ABCD.

•29. 144°. (360° – 3 ⋅ 72°.)

30.

•40. Six sides, three diagonals, and four triangles.

41. Example figure:

•31. SSS.

32. All are isosceles.

33. They seem to be collinear.

34. They are collinear because∠ACE = 72° + 108° = 180°.

35. Yes. It appears that it could be folded alongthe line through A, C, and E so that the twohalves would coincide.

36. No. Quadrilateral ABED is equilateral butnot equiangular.

Gemstone Pattern.

•37. Triangles, quadrilaterals, and an octagon.

38. Five sides, two diagonals, and threetriangles.

39. Example figure:

42. Eight sides, five diagonals, and six triangles.

•43. n – 3.

44. n – 2.

45. A quadrilateral has 4 – 3 = 1 diagonal whichforms 4 – 2 = 2 triangles.

•46. 720°. (4 × 180°.)

•47. 120°. ( .)

48. 1,080°. (6 × 180°.)

49. 135°. ( .)

•50. (n – 2)180°.

51. .

Linkage Problems.

•52. Their sum is 360°.

53. Their sum is 180° if we think of the linkageas a triangle. (It is 360° if we think of thelinkage as a quadrilateral, because∠BCD = 180°.)

•54. They all appear to be smaller.

55. ∠A + ∠B + ∠APB = 180°, and∠C + ∠D + ∠CPD = 180°; so∠A + ∠B + ∠APB = ∠C + ∠D + ∠CPD.Because ∠APB = ∠CPD, it follows that∠A + ∠B = ∠C + ∠D by subtraction.

Quadrilateral Angle Sum.

56. Addition.

57. Substitution.

58. Subtraction.

Chapter �� Lesson � ��

59. ∠1 and ∠2 are a linear pair; so they aresupplementary and their sum is 180°.Likewise for ∠3 and ∠4.(∠1 + ∠2) + (∠3 + ∠4) = S; so180° + 180° = S, and so S = 360°.

Set IIISet IIISet IIISet IIISet III (page ��)

This triangletosquare puzzle is the puzzle ofHenry Dudeney first introduced in the Set IIIexercise of Chapter �� Lesson �� It is perhapssurprising that� by carefully comparing the twofigures and knowing that the angles of anequilateral triangle and a square are ��° and �°respectively� we can find all of the remainingangles from just the one given� (By using the factthat the triangle and square have equal areas andapplying some simple trigonometry to thetriangle labeled D� we can find the exact valuesof all of the angles in the figure� The ��° angle�

for example� is rounded from Arcsin �)

Triangle into Square.

6. BC looks longer than CF, but they areactually the same length.

7. The opposite angles of a parallelogram areequal.

•8. An exterior angle of a triangle is greaterthan either remote interior angle.

9. Substitution.

Theorem 26.

10. The opposite sides of a parallelogram areequal.

•11. The opposite sides of a parallelogram areparallel.

12. Parallel lines form equal alternate interiorangles.

13. ASA.

14. Corresponding parts of congruent trianglesare equal.

•15. A line segment is bisected if it is dividedinto two equal segments.

Point Symmetry.

16.



1. The first card (the card with the picture ofthe baseball player).

•2. Point.

•3. See if the figure looks exactly the samewhen it is turned upside down.

Optical Illusion.

•4. Three.

•5. Yes. The opposite sides of a parallelogramare equal.

•17. The diagonals of a parallelogram bisect eachother.

•18. Two points are symmetric with respect to apoint if it is the midpoint of the line segmentconnecting them.


20. Vertical angles are equal.

21. ASA.


23. Two points are symmetric with respect to apoint if it is the midpoint of the line segmentconnecting them.

�� Chapter �� Lesson �

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

In encyclopedias and unabridged dictionaries� theentry for “parallelogram” is followed by an entryfor “parallelogram law�” The EncyclopediaBritannica describes it as the “rule for obtaininggeometrically the sum of two vectors byconstructing a parallelogram� A diagonal will givethe requisite vector sum of two adjacent sides�The law is commonly used in physics to determinea resultant force or stress acting on a structure�”The except from Newton’s Mathematical Principlesof Natural Philosophy is taken from the originalLatin edition published in �� It immediatelyfollows statements of his three laws of motion�

A specific measure for ∠BAD in the figure forexercises �� through �� was given to make theexercises a little easier� Actually� any measureother than ��° or ��° will work� A nice problemfor better students would be to rework theexercises letting ∠BAD x° and to explain whathappens when x �� or ��

Parallelogram Rule.

24. Corollary.

25.

34. Point E is the midpoint of AD.

•35. AD = 2DC.

36. They are supplementary. Parallel lines formsupplementary interior angles on the sameside of a transversal.

•37. 180.

38. 90.

39. Isosceles.

40. Right. Because x + y = 90, ∠BEC = 90°.

41. BE ⊥ EC.

Two Parallelograms.

42.

•26. 10. (5 cm represents 50 lb; so 1 cm represents10 lb.)

27. 25. (2.5 × 10 = 25.)

28. About 6 cm.

29. About 60. (6 × 10 = 60.)

•30. About 24°.

Angle Bisectors.

31.

•32. ∠AEB = x°; ∠CED = y°.

33. AB, AE, and ED.

43. Because x, y, and z are the measures of theangles of a triangle, x + y + z = 180.Therefore, ∠ABC is a straight angle; so A, B,and C are collinear.

44. Because ABDE and BCDE are parallelograms,AB || ED and BC || ED. If A, B, and C are notcollinear, then through B there are two linesparallel to ED. This contradicts the ParallelPostulate; so A, B, and C must be collinear.

•45. All three (∆ABE ≅ ∆DEB ≅ ∆BCD).

46. No.

Hidden Triangles.

47.

48. ∆ADE, ∆CDF, and ∆EBF.

49. SAS.

Chapter �� Lesson � �

50. Three. In addition to ∆ABE and ∆BCF, ∆DEFis equilateral because its sides are thecorresponding sides of the three congruenttriangles.

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

The four symbols used on modern playing cardswere designed in �� by the court painter toKing Charles VI of France� The four suitsrepresented four classes of French society: thespades� soldiers� the clubs� farmers� the diamonds�artisans� and the hearts� the clergy� According toJohn Scarne in Scarne’s New Complete Guide toGambling (Simon & Schuster� ��)� playing cardsfirst arrived in the New World in �� withColumbus and immediately found favor with theNative Americans� who made them of deerskin orsheepskin!

Playing Card Symmetries.

1. So that they look the same whether they areheld “right side up” or “upside down.”

2. No playing card has line symmetry.

3. The diamonds are more symmetric becausethe diamond symbol itself has pointsymmetry, whereas the symbols of the othersuits do not. (They do have line symmetry,however.)

4. All of the diamond cards have pointsymmetry except for the 7. All face cardshave point symmetry. In the club, spade,and heart suits, the 2s and 4s have pointsymmetry.

5. The 6s and 8s of clubs, spades, and heartswould have point symmetry if one of thesymbols in the middle row were turnedupside down. (It is impossible for the odd-numbered cards in these three suits to beredesigned so as to have point symmetry!)



The letter p is the only letter of the Englishalphabet that transforms into three other lettersby reflections and rotations� Some students mayenjoy thinking about other examples of lettersthat transform into at least one other letter inthis way�

Pop-Up Parallelogram.

•1. A quadrilateral is a parallelogram if itsopposite sides are equal.

2. The opposite sides of a parallelogram areparallel.

•3. Parallel lines form equal correspondingangles.

•4. In a plane, if a line is perpendicular to oneof two parallel lines, it is also perpendicularto the other.

Theorem 28.

5. The sum of the angles of a quadrilateral is360°.

6. Division.

7. Two angles are supplementary if their sumis 180°.

•8. Supplementary interior angles on the sameside of a transversal mean that lines areparallel.

9. A quadrilateral is a parallelogram if itsopposite sides are parallel.

Theorem 29.

10. Two points determine a line.


12. Reflexive.

•13. SAS.



Theorem 30.

16. A line segment that is bisected is dividedinto two equal parts.

17. Vertical angles are equal.

18. SAS.


•20. Equal alternate interior angles mean thatlines are parallel.

� Chapter �� Lesson �

•21. A quadrilateral is a parallelogram if twoopposite sides are both parallel and equal.

Letter Transformations.

•22. Line symmetry. The figure can be foldedalong a vertical line so that the two halvescoincide.

23. Point symmetry. If the figure is turnedupside down, it looks exactly the same.

24. Line symmetry. The figure can be foldedalong a horizontal line so that the twohalves coincide.

Quadrilateral Symmetries.

25. Yes. Point symmetry.

•26. No. Line symmetry (one vertical line).

27. Yes. Point symmetry and line symmetry(two lines, one vertical and one horizontal).

28. Yes. Point symmetry and line symmetry(four lines, one vertical, one horizontal, andtwo at 45° angles with them).

Economics Graph.

•29. They seem to be parallel.

30. The quadrilateral formed by the four lines isa parallelogram (a quadrilateral is aparallelogram if two opposite sides areequal and parallel); so the lines C and C + Iare parallel.


The method for constructing a parallel to a linethrough a given point not on it is another exampleof a “rusty compass” construction� MartinGardner observes that� although the method haslong been known� it is still being rediscovered andreported as new�

The gridbracing illustrations are taken fromthe section titled “Bracing Structures” in JayKappraff’s Connections—The Geometric Bridgebetween Art and Science (McGrawHill� ��)�Kappraff cites the article “How to Brace a OneStory Building” in Environment and Planning(��) as the original source of this material� animportant topic for engineers and architects�Examples of pertinent theorems dealing with thissubject in graph theory are: “In any distorted gridall the elements of a row (column) are parallel”and “A bracing of an n by m grid is a minimum

rigid bracing if and only if the bracing subgraph isa tree�”

Parallel Rulers.

31. They are connected so that AB = CD andAC = BD. A quadrilateral is a parallelogramif its opposite sides are equal.

•32. The pairs of opposite angles.

Tent Geometry.

33.

34. Yes. ABCD and CDEF are parallelograms; soAB = CD and CD = EF (the opposite sides ofa parallelogram are equal) and AB || CD andCD || EF (the opposite sides of a parallelogramare parallel). It follows that AB = EF(substitution) and AB || EF (two lines parallelto the same line are parallel to each other).Therefore, ABFE is a parallelogram becausetwo opposite sides are parallel and equal.

35. Yes. AD = BC, DE = CF, and AE = BF (theopposite sides of a parallelogram are equal).Therefore, ∆ADE ≅ ∆BCF (SSS).

Rope Trick.

36.

•37. A quadrilateral is a parallelogram if itsdiagonals bisect each other.


•39. SSS.




42. Substitution.

•43. A quadrilateral is a rectangle if all of itsangles are equal.

Parallel Postulate.

44. Through point P, there is exactly one lineparallel to line l.

45.


1. (The figures below show two possible answersdepending on which triangles are constructedinward and outward.)

46. PC is parallel to line l because ABCP is aparallelogram (a quadrilateral is aparallelogram if its opposite sides areequal).

Flexible Grid.

•47. 12.

48. 4.

49. 12.

50. 4.

51. They remain unchanged because theirshapes are determined by their sides (SSS)and their sides do not change.

52. They remain parallel because the oppositesides of the quadrilaterals remain equal; sothey are always parallelograms.

•53. They are the sides of the quadrilateralsabove and below the braced square.

54. They are the sides of the quadrilaterals tothe right of the braced square.

55. Because all of the quadrilaterals remainparallelograms, these sides remain parallelto the horizontal and vertical sides of thebraced square.

2. The quadrilateral formed seems to be aparallelogram.



In his definitive book on the subject� Dissections—Plane and Fancy (Cambridge University Press��)� Greg N� Frederickson (using radians��π radians ��°) reports: “We can think of a regular polygon as a union ofvarious rhombuses and isosceles triangles (whichare halves of rhombuses)� Such a characterizationis an internal structure of the correspondingpolygon� If the regular polygon has an evennumber of sides� then the internal structure needsonly rhombuses � � � � A regular polygon with �q

sides contains q each of rhombuses� for each

positive integer i that is less than � plus squares

if q is even� For example� the dodecagon contains

π


� each of rhombuses and rhombuses� plus

� squares�”

[ and indicate rhombuses having angles of

��° and ��°�] There are several ways in which therhombuses can be arranged when the dodecagonis divided in this way�

The remark about a square being perfectbecause all of its angles are “just right” was madeby Albert H� Beiler at the beginning of the chaptertitled “On The Square” in his book titledRecreations in the Theory of Numbers—TheQueen of Mathematics Entertains (Dover� ��)�

Regular Dodecagon.

•1. 12.

•2. Three.

3. A square.

4. Three.

•5. 15.

6. Three. (There are six narrow, six medium,and three square.)

Theorem 31.

•7. All of the angles of a rectangle are equal.

8. A quadrilateral is a parallelogram if itsopposite angles are equal.

Theorem 32.

9. All of the sides of a rhombus are equal.


Theorem 33.

11. All of the angles of a rectangle are equal.

•12. All rectangles are parallelograms.


14. Reflexive.

15. SAS.


Theorem 34.

•17. All of the sides of a rhombus are equal.

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

Which Parts?

•19. Two consecutive sides.

20. One side.

21. One side and one angle.

22. Two consecutive sides and one angle.

“Just Right.”

23. If all of the angles of a triangle were rightangles, the sum of their measures would be270°, rather than 180°.

•24. No. It must be a rectangle.

25. Shown below are two of many possible examples.(The ambiguity arises because the right anglesmay be on the inside or the outside of thepolygon.)


Checking a Wall.

26. No. If the opposite sides of a quadrilateralare equal, it must be a parallelogram butnot necessarily a rectangle.

27. No. We know that the diagonals of arectangle are equal but to conclude that, ifthe diagonals of a quadrilateral are equal,it is a rectangle would be to make theconverse error.

28.

29. SSS.

π π

π π





33. Substitution.

•34. A quadrilateral all of whose angles are equalis a rectangle.

35. True.

•36. False.

37. True.

38. False.

•39. True.

Square Problem.

40.

Rhombus Problem.

47.

41. Parallelograms.

42. They must be parallelograms because twoopposite sides are both parallel and equal.(For example, in AMCO, AM || CO becauseABCD is a square and AM = CO because Mand O are midpoints of the equal segmentsAB and CD.)

•43. Its opposite sides are parallel.

44. A rhombus.

45. They seem to be collinear.

46. It appears to have point symmetry and linesymmetry (with respect to the lines of thediagonals of the square).

48. Yes. Because ABCD is a rhombus, AB = AD.It is also a parallelogram; so ∠B = ∠D.∆AEB ≅ ∆AFD (AAS); so AE = AF.

Triangle Problem.

49.

50. Because ABDE is a parallelogram, AB = ED(the opposite sides of a parallelogram areequal). Also, because BDCE is a rectangle,ED = CB (the diagonals of a rectangle areequal). So AB = CB (substitution). Therefore,∆ABC is isosceles. (Additional drawingsusing the conditions given reveal that it isn’tnecessarily equilateral.)


Bill Leonard told the story of starting a lessonwith a highschool class of remedial mathematicsstudents with the problem of counting thesquares in a � x � grid� He had hoped that thestudents would think about it for awhile� but itdidn’t take long for several students to startyelling out “��” Leonard then asked if they hadcounted squares such as the one shaded in thesecond figure� The class immediately yelled out inunison� “��!”

Counting Squares.

1. It contains 30 squares in all. There are 16small squares, 9 2 × 2 squares,4 3 × 3 squares, and 1 4 × 4 square.

2. It contains 15 squares in all. There are 7small squares, 6 2 × 2 squares,1 3 × 3 square, and 1 4 × 4 square.




Hal Morgan in his book titled Symbols of America(Viking� ��)� comments on the Chase ManhattanBank trademark: “After World War II therepresentational images of the nineteenthcentury began to give way to abstract designsinspired by the spare� geometric forms of theBauhaus� Some of these modern designs are quitestriking in their clarity and simplicity� � � � Thedesign firm of Chermayeff & Geismar Associatescreated the Chase Manhattan Bank’s distinctiveoctagon logo in �� The abstract� interlockingdesign seems to have become a sort of model forthe designers of other bank and insurance logos�”Morgan shows examples of several morerecentlogos that are also pointsymmetric designs ofinterlocking solid black congruent polygons�

Quadrilaterals in Perspective.

•1. A trapezoid.

•2. That it has exactly one pair of parallel sides.

3. Yes. Picnic blankets are usually rectangular.(The perspective alters the blanket’s shape.)

•4. That all of its angles are right angles (or thatthey are equal).

•5. Yes. All rectangles are parallelograms.

6. Rhombuses and parallelograms.

7. Squares, rectangles, rhombuses, andparallelograms.

Geometric Trademark.

•8. Point.

9. It is a square because all of its sides andangles are equal.

•10. They seem to be congruent.

•11. Trapezoids.

12. CD and JE.

•13. No. The legs of a trapezoid must be equalfor it to be isosceles.

14. No. One is a right angle and the other isacute.

15. An octagon.

Theorem 35.

•16. The bases of a trapezoid are parallel.

•17. Through a point not on a line, there isexactly one line parallel to the line.


•19. The opposite sides of a parallelogram areequal.

20. The legs of an isosceles trapezoid are equal.

21. Substitution.

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

23. Parallel lines form equal correspondingangles.

24. Substitution.

•25. Parallel lines form supplementary interiorangles with a transversal.

26. The sum of two supplementary angles is180°.

27. Substitution.

28. Subtraction.

Theorem 36.

29. The legs of an isosceles trapezoid are equal.

•30. The base angles of an isosceles trapezoidare equal.

31. Reflexive.

32. SAS.



Stepladder.

34.


35. An isosceles trapezoid.

•36. 75°.

37. The base angles of an isosceles trapezoidare equal.

38. Isosceles.

39. 75°.

•40. Parallel lines form equal correspondingangles.

•41. 105°.

42. 30°.

Regular Pentagon.

43. Examples are ABCD, ABCE, ACDE, andABDE. Their bases are parallel (they formsupplementary interior angles on the sameside of a transversal) and their legs are equal(given).

44. Examples are ABCH, ABGE, and AFDE.Their opposite sides are parallel (establishedin exercise 43).

45. Examples are ABCH, ABGE, and AFDE.They are rhombuses because all of theirsides are equal. In regard to ABCH, forexample, AB = BC (given), AB = HC, andAH = BC (the opposite sides of a parallelo-gram are equal); so HC = AH (substitution).

Trapezoid Diagonals.

•46. AC and DB bisect each other.

•47. A quadrilateral is a parallelogram if itsdiagonals bisect each other.

48. The opposite sides of a parallelogram areparallel.

•49. A trapezoid has only one pair of oppositesides parallel.

50. AC and DB bisect each other.

51. AC and DB do not bisect each other.

52. Yes. One example of such a trapezoid isshown below.


The Moscow Puzzles is the English languageedition of Boris Kordemsky’s MathematicalKnowhow� which� according to Martin Gardner�is the best and most popular puzzle book everpublished in the Soviet Union� Kordemsky taughtmathematics at a high school in Moscow formany years�

Gardner reported Andrew Miller’s discoveryof a second solution to Kordemsky’s trapezoiddissection problem in the February �� issue ofScientific American and posed it as a new puzzlefor the readers of that magazine�

Congruence Puzzle.

Kordemsky’s solution is shown below.

The solution thought of by Andrew Miller isshown below. The three cuts intersect the upper

base at points , , and of the way from its

left endpoint.



The “state capital” quadrilateral was chosen partlyfor fun but primarily to show that the originalquadrilateral can have any shape� It doesn’t evenhave to be convex� Not only can the quadrilateralbe concave� it can even be “selfintersecting” orskew!

Portuguese Theorem.

1. Midpoints.

•2. A midsegment.

3. Other side.

4. A midsegment of a triangle is parallel to thethird side and half as long.


•5. DE || AC and DE = AC.

State Capitals Problem.

6.

Two Midsegments.

22.

7. MNPQ looks like a parallelogram.

•8. A midsegment of a triangle is parallel to thethird side. (MN and QP are midsegments of∆SOH and ∆SBH.)

•9. In a plane, two lines parallel to a third lineare parallel to each other.

•10. A midsegment of a triangle is half as long asthe third side.

11. Substitution.

12. That it is a parallelogram.

•13. MNPQ is a parallelogram because twoopposite sides are both parallel and equal.

14. MP and NQ seem to bisect each other.

15. The diagonals of a parallelogram bisect eachother.

Midpoint Quadrilateral.

16. A midsegment of a triangle is half as long asthe third side.

•17. Addition.

•18. Substitution.

19. Its perimeter.

•20. The sum of its diagonals.

21. The perimeter of the midpoint quadrilateralis equal to the sum of the diagonals of theoriginal quadrilateral.

23. ADEC is an isosceles trapezoid becauseDE || AC (a midsegment of a triangle isparallel to the third side) and AD = CE.

•24. ∠A = ∠C because the base angles of anisosceles trapezoid are equal.

25. ∆ABC is isosceles because, if two angles of atriangle are equal, the sides opposite themare equal.(Alternatively, AB = 2AD = 2CE = CB .)

26. FG = AC because FG = DE and

DE = AC [so FG = ( AC) = AC].

Also, FG || AC because FG || DE andDE || AC (in a plane, two lines parallel to athird line are parallel to each other).


Exercises �� through �� are intended to reviewsome basics of coordinate geometry� includingthe distance formula� and to get students tothinking about how midpoints and the directionsof lines might be treated�

The Midsegment Theorem is� in a way� a“special case” of the fact that the line segmentconnecting the midpoints of the legs of a trapezoidis parallel to the bases and half their sum�

The edition of Euclid’s Elements with thefoldup figures was its first translation intoEnglish� by Henry Billingsley� After a career inbusiness� Billingsley became sheriff of Londonand was elected its mayor in �� An excerptfrom The Elements of Geometrie� including anactual foldup model� is reproduced in Edward R�Tufte’s wonderful book Envisioning Information(Graphics Press� ��)�

In Mathematical Gems II (MathematicalAssociation of America� ��)� Ross Honsbergerwrote: “Solid geometry pays as much attention tothe tetrahedron as plane geometry does to the


triangle� Yet many elementary properties of thetetrahedron are not very well known�” Exercises�� through �� establish that any triangle can befolded along its midsegments to form a tetrahedron� Unlike tetrahedra in general� however� atetrahedron formed in this way is always isosceles;that is� each pair of opposite edges are equal� Allthe faces of an isosceles tetrahedron are congruenttriangles and consequently have the same area;furthermore� the sum of the face angles at eachvertex is ��°� More information on this geometricsolid can be found in chapter of Honsberger’sbook�

Midpoint Coordinates.

Tetrahedron.

37.

27.

•28. (4, 7).

•29. (8, 10).

30. (7, 5).

31. Yes. The coordinates of the midpoint are“midway between” the respective coordinatesof the endpoints. (They are their averages.)

•32. MN = = =

= = 5.

33. AC = = =

= = 10.

34. Yes. MN = AC as the Midsegment

Theorem states.

•35. They should be parallel.

36. Yes. They seem to be parallel because theyhave the same direction; each line goes up 3units as it goes 4 units to the right.

•38. They are congruent by SSS.

39.

40. They are equal.

41. 180°. (The measures of the three angles thatmeet at each vertex are a, b, and c.)

More Midpoint Quadrilaterals.

42. Example figure:

•43. A rhombus.

44. The line segments divide the rectangle intofour right triangles and the quadrilateral.The triangles are congruent by SAS; so theirhypotenuses are equal. Because the hypot-enuses are the sides of the quadrilateral, it isequilateral and must be a rhombus.

45. Example figure:

46. A rectangle.


47. The sides of the new quadrilateral areparallel to the diagonals of the rhombusbecause of the Midsegment Theorem.Consequently, they form four smallparallelograms. Each parallelogram has aright angle because the diagonals of arhombus are perpendicular. Because theopposite angles of a parallelogram areequal, it follows that the four angles of thenew quadrilateral are right angles; so it is arectangle.

48. Another square.

Base Average.

•49. MN seems to be parallel to AB and DC, thebases of the trapezoid.

50.


The problem of the nested triangles is a previewof the idea of a limit� a topic considered later inthe treatment of measuring the circle� The sumof the lengths of all of the sides of the trianglesis the sum of the geometric series:

� � � � ( )�� ( )�� ⋅ ⋅ ⋅ � × �

Students who tackle this problem will beinterested to know that they will explore itfurther in secondyear algebra�

Infinite Series.

The perimeter of the largest triangle is 6 in. Fromthe Midsegment Theorem, we know that the sidesof the next triangle are half as long; so its perimeteris 3 in. Reasoning in the same way, we find thatthe perimeters in inches of successive triangles are1.5, 0.75, 0.375, 0.1875, and so on.

Starting with the first two triangles, we findthat the successive sums of the sides are 9, 10.5,11.25, 11.625, 11.8125, and so on. The sums seemto be getting closer and closer to 12 inches and(as students who study infinite series will learn toprove) the sum of all of the sides of the trianglesin the figure is 1 foot.

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


A question such as that in exercise �� is intendedto arouse the student’s curiosity� According toJearl Walker� who posed it in his book titled TheFlying Circus of Physics (Wiley� ��)� “If the ratioof the bar’s density to the fluid’s density is closeto � or �� the bar floats in stable equilibrium as inthe first figure� If the ratio is some intermediatevalue� then the bar floats in stable equilibriumwith its sides at ��° to the fluid’s surface� In eachcase� the orientation of stable equilibrium isdetermined by the position in which the potentialenergy of the system is least�” This floatingbarproblem is a good illustration of how scientistsarrive at conclusions by using inductive reasoning(performing experiments with floating bars to seewhat happens) and how they then “explain” theresults by using deductive reasoning (calculatingthe potential energy and its minimum value byusing mathematics� including geometry)�

The frequent appearance in this chapter oftheorems whose converses also are theorems caneasily mislead students into forgetting that a

51. DC || AP; so ∠C = ∠NBP (parallel linesform equal alternate interior angles).CN = NB because N is the midpoint of CB.∠CND = ∠BNP (vertical angles are equal).So ∆DCN ≅ ∆PBN (ASA).

[An alternative proof could use∠CDN = ∠BPN (equal alternate interiorangles) in place of either pair of anglesabove, in which case the triangles would becongruent by AAS.]

52. DN = NP (corresponding parts of congruenttriangles are equal); so N is the midpoint ofDP. Therefore, MN is a midsegment of∆DAP and so MN || AP.

53. MN || AP and DC || AP; so MN || DC. (In aplane, two lines parallel to a third line areparallel to each other.)

54. MN = AP by the Midsegment Theorem.

Because AP = AB + BP and DC = BP(corresponding parts of congruent triangles

are equal), MN = (AB + BP) = (AB + DC)

(substitution).

Chapter �� Review

statement and its converse are not logicallyequivalent� The intent of exercises �� through ��is to remind students of this fact and to encouragethem to draw figures to test unfamiliar conclusions(or even to remind themselves of facts that theyhave proved)� Knowledge of the properties of thevarious quadrilaterals should not be as much amatter of memorization as of looking at picturesand applying simple common sense�

Palm Strand Rhombus.

1.

•13. All of the sides and angles of a square areequal.

14. SAS (or SSS).


•16. If a line divides an angle into two equalparts, it bisects the angle.

Related Statements.

17. It is the converse of the first statement.

18. No. The first statement is true; the firstfigure shows that the triangles are congruentby SSS. The second figure shows that thesecond statement is false.

19. True.

20. True.

•21. True.

22. False.Example of a quadrilateral that has two pairs ofequal angles but is not an isosceles trapezoid:

•2. Both pairs of its opposite sides are parallel(because the strips have parallel sides).


•4. AAS.



•7. Substitution.

8. All of its sides are equal.

Floating Bar.

•9. Four.

10. Yes.

11. (Student answer.) (The bar can float eitherway depending on it and the liquid.)

12.

23. False.Example of a parallelogram that is not a rhombus:

24. True.

25. True.

26. True.

•27. True.

28. False.Example of a quadrilateral with perpendiculardiagonals that is not a rhombus:

�� Chapter �� Review

29. False.Example of a parallelogram whose diagonals arenot equal:

to form a parallelogram and a triangle; the proofis comparable to that of Theorem �� outlined inexercises �� through �� of Lesson �� The additionalfact that the base opposite the larger base anglesof an isosceles trapezoid is longer than the otherbase is the basis for the magic trick of exercise ��Like the question about the floating bar in Set I�the question about the magic trick is included forfun and not meant to be taken too seriously�

Dissection Puzzle.

•36. D, E, and F.

37. D.

38. D and F.

39. C and D.

40. A, B, D, E, and F.

41.

30. False.Example of a quadrilateral whose diagonals areequal that is not a parallelogram:

Jumping Frog.

•31. The sum of the angles of a quadrilateral is360°.

32. They are rhombuses and all rhombuses areparallelograms. (Or, Both pairs of theiropposite sides are equal.)

•33. The opposite angles of a parallelogram areequal.

34. They are parallelograms because theirdiagonals bisect each other.

35. The opposite sides of a parallelogram areparallel and, in a plane, two lines parallel toa third line are parallel to each other.


Only the right angles in the pieces of the puzzleof exercises �� through �� actually have measuresthat are integers� From ∆ABG it is evident that

tan A �; so ∠A Arctan � ≈ ��°�

Although part of basic vocabulary in books ofthe past� words such as “trapezium” and“rhomboid” are now close to being obsolete�Strangely� in British usage� the word “trapezium”is used to mean “trapezoid�”

Even though the quadrilaterals in exercises ��through �� are drawn to scale� it is almost impossible to recognize them merely by their appearance� It is tempting to conclude that� becausequadrilateral ABCD is a trapezoid whose baseangles are equal� it is isosceles� which� of course�is the converse error� To prove that it is isosceleswould require adding an extra line to the figure

•42. GI ⊥ HM and ML ⊥ HM. In a plane, twolines perpendicular to a third line areparallel.

43. A parallelogram.


•45. 2c.

46. 2a + b. [Some students may recognize thatGM can also be expressed in terms of c byusing the Pythagorean Theorem.GM2 = c2 + (2c)2 = 5c2; so GM = c .]

47. A square.

48. All of its sides and angles are equal.

•49. A trapezoid.

Chapter �� Algebra Review ��

50. It has exactly one pair of parallel sides.(HO || IJ because, in a plane, two linesperpendicular to a third line are parallel.)

Words from the Past.

•51. No, because a trapezoid has two parallelsides.

52. Yes, because, other than squares, rhombusesare parallelograms that have no rightangles.

53. No, because a rectangle has four rightangles.

Rectangles Not.

54. ABCD is a trapezoid because BC || AD (theyform supplementary interior angles on thesame side of a transversal). (It is also possibleto show that ABCD is isosceles.)

55. EFGH is a parallelogram because both pairsof opposite angles are equal.

•56. IJKL is a trapezoid because IJ || LK.

57. From its angles we can conclude that MNOPis neither a parallelogram nor a trapezoid;so it is not any of the special types ofquadrilaterals that we have studied.

58. While you look at the card, the magicianrotates the pack 180° so that the bottomedges of the cards are at the top. When yourchosen card is put back, it is easy to slide itup because it is now wider at the top thanany of the others.

Another Midpoint Quadrilateral.

59.

60. MN || AC and PO || AC (a midsegment of atriangle is parallel to the third side); so

MN || PO. MN = AC and PO = AC (a

midsegment of a triangle is half as long asthe third side); so MN = PO. MNOP is aparallelogram because two opposite sidesare both parallel and equal.

61. It appears to be a rhombus. ∆PDO ≅ ∆NCO(SAS); so PO = NO. Because MN = PO andMP = NO, MN = MP; so all of the sides ofMNOP are equal.

Algebra ReviewAlgebra ReviewAlgebra ReviewAlgebra ReviewAlgebra Review (page ��)

•1. 2.236.

•2. 22.36.

•3. 223.6.

•4. 7.07.

•5. 70.7.

•6. = = = 10 .

•7. = = =

10(10 ) = 100 .

•8. = = = 5 .

•9. = = = 10(5 ) =

50 .

•10. = = 5 .

•11. = = = x2 .

•12. = = .

•13. = = = 2x6 .

•14. + = + = + =

7 + = 8 .

•15. = = = = .

•16. – = – =

– = – = .

•17. = = = =

.

•18. = = = =

.

•19. + + = = =

= .

•20. + = 6 + 8 = 14.

•21. = = = 10.

π ππ

�� Chapter �� Algebra Review

•22. = = = .

•23. = = = 11.

•24. ( )( ) = .

•25. ( )2 = 9 ⋅ 7 = 63.

•26. = .

•27. (5 – ) + (5 + ) = 10.

•28. (5 – )(5 + ) = 25 – 2 = 23.

•29. ( + 1) = + = 9 + .

•30. + ( + 1) = + + 1 =

+ + 1 = + 1.

•31. ( + ) + ( + ) = + .

•32. ( + )2 = ( )2 + 2( ) + ( )2 =

x + + y.

chapter 7 answers - bisd moodlemoodle.bisd303.org/file.php/18/solutions/chapter 07 answers.pdf ·...

Documents