Chapter 10 Answers - BISD Moodlemoodle.bisd303.org/file.php/18/Solutions/Chapter 10 Answers.pdf · Chapter Lesson Chapter Lesson Set I ... Parallel lines form equal corresponding

Chapter �� Lesson � ��

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

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

According to Benno Artmann in Euclid—TheCreation of Mathematics (Springer� ��)� thearchitects of the Parthenon� Ictinos and Callikrates�based its dimensions on the square numbers and� Starting with a height of �� they chose � � forthe width and �� for the length� so that the widthwas the geometric mean between the height andthe length

Euclid developed the theory of proportion witha series of �� theorems in Book V of the ElementsHis definition of ratio as “a sort of relation inrespect of size between two magnitudes of thesame kind” is� as one commentator remarked�unusual for Euclid because it is so vague as to beof no practical use

The numbers connected with the TurtlesForever picture illustrate the fact that any threeconsecutive terms of a geometric sequence areproportional (as are any three evenly spacedterms) The term in the middle is consequentlythe geometric mean between the other twoterms

Parthenon Architecture.

•1. 2.25.

2. 2.25.

3. 81 × 16 = 1,296 and 36 × 36 = 1,296.

•4. The geometric mean.

•5. = .

6. 1.5 = 1.5.

7. 101 ft.

•8. h ≈ 45 ft. (Example proportion: = .)

•9. (or 0.285714 . . . ). ( .)

10. (or 0.8). ( .)

11. (or 10). ( .)

Turtles Forever.

12. = .

13. = .

14. 4 and 16, 2 and 32, 1 and 64.

Ratio.

•15. Division.

16. The second.

17. Euclid referred to “magnitudes of the samekind,” which suggests that numbers of thesame kind of units are being compared.

Four Rectangles.

18. Rectangle 1, ab; rectangle 2, ad;rectangle 3, cd; rectangle 4, bc.

•19. Rectangles 2 and 4. (ad = bc because = .)

20. Rectangles 1 and 3. ( = because ad = bc.)

21. Rectangles 1 and 3.

Enlargement.

•22. 30.

23. 36.

24. Correct. (Both ratios are equal to .)

25. Correct. ( = = = 1.5.)

26. Correct. [ = 2.25 and ( )2 = 2.25.]

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

The official design of the current United Statesflag was signed into law by President Eisenhowerin �� soon after Alaska became the ��th stateThe first chapter of Slicing Pizzas� Racing Turtles�and Further Adventures in Applied Mathematics(Princeton University Press� ��)� by RobertBanks� deals with the geometry of the flag

�� Chapter �� Lesson �

Concerning the ratio Banks writes: “Why the

relative length of the flag is precisely ��—orindeed why the relative length of the union isexactly �� (could it be �� ?)—is not known Itjust is”

As Thomas Rossing remarks in The Science ofSound (Addison Wesley� ��)� “the word scale isderived from a Latin word (scala) meaning ladderor staircase A musical scale is a succession of notesarranged in ascending or descending order” Inchapter � of his book� Rossing includes completediscussions of the scales based on just intonationand equal temperament as well as the Pythagoreanscale on which exercises � through � are basedAnother good reference on the structure ofmusical scales is Connections—The GeometricBridge between Art and Science� by Jay Kappraff(McGraw�Hill� ��) It is interesting to see fromthe relative lengths of the strings producing thenotes of the Pythagorean scale that they are allderived from the primes � and �:

In the currently used even�tempered scale� the

ratios � � � � and are ≈ �� in

contrast with the Pythagorean ratio of � ��

The “even tempered” ratios and are

≈ �� in contrast with the Pythagorean

ratio� ≈ ��

The fact that the means of a proportion canbe interchanged is proved in Book V of theElements as Proposition � : “If four magnitudesbe proportional� they will also be proportionalalternately” The addition property proved inexercises �� and �� appears in the Elements asProposition �� of the same book

Arm Spans.

27. No. The units of length being compared arenot the same; so the result is meaningless.

•28. 36 in; = 2.4.

29. 1.25 ft; = 2.4.

30. No.

United States Flag.

•31. .

32. .

33. .

34. The length of the flag.

35. 304 feet.

•36. The width of one stripe.

•37. Approximately 12.3 ft. ( ≈ 12.3.)

38. Approximately 86 ft. (7 × 12.3 ≈ 86.)

•39. Approximately 122 ft. ( = 0.76,

b = 121.6.)

40. No. ≈ ≈ 0.7 and = ≈ 0.53.

Pythagorean Tuning.

•41. = = = .

•42. = = = .

43. = = = .

44. = = = .

45. = = = .


46. = = = .

47. G is the geometric mean between F and A;A is the geometric mean between G and B.

( = = and = = .)

Geometric Proportions.

•48. The means are interchanged.

49. Multiplication (or, in a proportion, theproduct of the means is equal to the productof the extremes).

•50. Division.

51. Addition.

52. Substitution (and algebra).

Sinker or Floater.

53. Alice is a floater because her density is

≈ 61.5 pounds per cubic foot.

54. Let x = Ollie’s weight in pounds. His

density is . Because he is a

sinker, > 62.4 and x > 143.52. Ollie

weighs more than 143 pounds.

Set IIISet IIISet IIISet IIISet III (page ��)

The White Horse of Uffington was created byremoving sod from the chalky rock lying under�neath A book written as far back as the fourteenthcentury ranked it second only to Stonehengeamong the tourist attractions of Britain Thehorse is in excellent condition owing to its beingmaintained by the National Trust

The White Horse.

1. (Student answer.) (The length of the horse inthe photograph is about 95 mm and its

height is about 24 mm: = = ,

x ≈ 110.8. The horse is about 110 meters

2. (Student answer.) (The scale is about .

≈ .)



According to Robert Bauval and Adrian Gilbert(The Orion Mystery� Crown� ��)� the apparentrelative positions of the three pyramids at Gizanot only match those of the three stars in Orion’sbelt but their orientation with respect to the Nilematches Orion’s apparent orientation withrespect to the Milky Way! The arrangement ofthe pyramids and their relative sizes (seeminglybased on the apparent sizes of the stars) seem toindicate that the Egyptians were trying to build areplica of heaven on Earth

A “plane table” is a drawing board and rulermounted on a tripod� used by surveyors in thefield to sight and map data More details on howit works can be found in J L Heilbron’s GeometryCivilized (Clarendon Press� ��)

According to Robert C Yates in The TrisectionProblem (Franklin Press� ��; NCTM� ��)�Alfred Kempe was “one of the cleverest amateurmathematicians of the [eighteenth] century”Kempe gave a historic lecture on linkages to agroup of science teachers in London in the summerof �� This lecture� titled “How to Draw aStraight Line�” and published in �� was reprintedby NCTM in �� The linkage invented byKempe for trisecting angles is also discussed byYates in the chapter on mechanical trisectors inThe Trisection Problem and by Martin Gardner inthe chapter titled “How to Trisect an Angle” inMathematical Carnival (Knopf� ��) As Kempehimself observed� his idea can be extended toproduce a linkage that can divide an angle intoany number of equal parts

The Pyramids and Orion.

•1. They are proportional.

•2. They are equal.

Plane Table.

•3. ∠DAC = ∠FBC.

4. ∠AEC = ∠BGC.

5. ∠GCB = ∠ECA.long.)


•6. = .

7. = .

8. = .

Kempe’s Linkage.

•9. OE = 4 and OG = 8. ( = = .)

10. They are always equal. Correspondingangles of similar triangles are equal.

11. It trisects ∠AOG.

Billboards.

12. Their corresponding angles are equalbecause billboards are rectangular and allright angles are equal. Their correspondingsides, however, are not proportional, because

≠ . As a calculator shows, ≈ 0.45,

whereas = 0.47826 . . . .

•13. 24.2. ( = , 5x = 121, x = 24.2.)

14. (or 5 or approximately 5.26).

( = , 23x = 121, x = = 5 .)

Dilation Problem.

•15. The center of the dilation.

16. AB = 5, BC = 4, AC = 6, A’B’ = 2.5, B’C’ = 2,and A’C’ = 3.

17. They are half as long.

•18. .

19. They are their midpoints.

Quadrilateral Conclusions.

•20. Corresponding sides of similar polygons areproportional.

21. Multiplication. (In a proportion, the productof the means is equal to the product of theextremes.)

22. They have equal areas.

23. They also have equal areas.αADGH = αAEFH + αEDGF andαEDCI = αEDGF + αFGCI.Because αADGH = αEDCI (exercise 22),αAEFH + αEDGF = αEDGF + αFGCI(substitution); so αAEFH = αFGCI(subtraction).

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

A good article on the instruments of the violinfamily is “The Physics of Violins” by Carleen MaleyHutchins� published in the November �� issueof Scientific American and reprinted in theScientific American Resource Library (W HFreeman and Company� ��)

The International Standard of paper sizes isnow in common use in Europe As the exercisesreveal� they are based on a sheet with an area of� square meter that can be cut in half to producea second sheet similar in shape Further sizes areproduced in the same way As the exercises alsoreveal� this pattern requires that all sizes have the

ratio The sequence beginning with size A�

continues to size A�� (a tiny � mm by �� mmsheet) Computer programs usually offer sizes Aand A� in their menus of choices for printingformats

Similar-Triangles Proof.

•24. A midsegment of a triangle is parallel tothe third side.

25. Parallel lines form equal correspondingangles.

26. Reflexive.

27. A midsegment of a triangle is half as longas the third side.

•28. Division.

29. The midpoint of a line segment divides itinto two equal segments.

•30. Betweenness of Points Theorem.

31. Substitution.


32. Division.

33. Substitution (they are all equal to ).

•34. Two triangles are similar if their correspond-ing angles are equal and their correspond-ing sides are proportional.

Violin Family.

•35. OC = 56 mm. 56 = r2 28; so r2 = = 2.0.

36. OD = 87 mm. 87 = r3 28; so r3 = ≈ 3.1.

•37. BF = 24 mm. = ≈ 1.1.

38. CG = 42 mm. = = 2.0.

39. DH = 65 mm. = ≈ 3.1.

40. Viola: 1.1 × 24 ≈ 26 inches.Cello: 2.0 × 24 = 48 inches.Bass: 3.1 × 24 ≈ 74 inches.

Paper Sizes.

41. = .

•42. Approximately 840 mm. (x2 = 1188 × 594,x ≈ 840.)

43. The geometric mean.

•44. Approximately 1 m2.(1188 mm × 840 mm = 997,920 mm2 ≈ 1 m2.)

45. Approximately 420 mm. ( ≈ = 420).

46. = .

•47. = 1, r2 = 2, r = ; so the exact ratio of

the length to width is .

Picture Frames.

48. = .

•49. = , 140 + 10w = 140 + 14w,

0 = 4w, w = 0.

50. = , ab + 2bw = ab + 2aw,

2bw = 2aw, bw = aw. If w ≠ 0, then a = b.

51. They mean that it is only possible for arectangular picture to be surrounded by aframe of constant width whose outer edgeis similar to it if the picture is square.

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

The Villa Foscari� also called “La Malcontenta�” islocated on the Brenta River near Venice It wascommissioned by the brothers Nicolo and AlviseFoscari and is still in the possession of the Foscarifamily

Similar Rectangles.

1. Ratio 1:2.

Dimensions 1 × 2:HION, JKQP, NOUT, PQWV.

Dimensions 1.5 × 3:ACIG, BDJH, CEKI, DFLJ.

Dimensions 2 × 4:GHTS, IJVU, KLXW, HKQN, NQWT.

Dimensions 4 × 8:GLXS.

2. Ratio 1:4.

Dimensions 1 × 4:HIUT, JKWV.

Dimensions 1.5 × 6:AEKG, BFLH.

Dimensions 2 × 8:GLRM, MRXS.

3. Ratio 2:3.

Dimensions 1 × 1.5:BCIH, DEKJ.

Dimensions 2 × 3:GIOM, HJPN, IKQO, JLRP,MOUS, NPVT, OQWU, PRXV.

Dimensions 4 × 6:GKWS, HLXT.


4. Ratio 3:4.

Dimensions 1.5 × 2:ABHG, BDJI, CEKI, DFLJ.

Dimensions 3 × 4:GIUS, HJVT, IKWU, JLXV.


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

Mathematics is truly a universal languageAlthough the words in the Turkish version of theSide�Splitter Theorem are incomprehensible tosomeone who does not know Turkish� the figureand symbolic statements about it are recognizableeverywhere

Descartes’s method for multiplying twonumbers geometrically appeared in Book I�“Problems the Construction of Which RequiresOnly Straight Lines and Circles�” of his Géométrie�published as Appendix I to his Discours de laméthode (� ��) As Descartes described theconstruction: “Taking one line which I shall callthe unit in order to relate it as closely as possibleto numbers� and which can in general be chosenarbitrarily� and having given two other lines� tofind a fourth line which shall be to one of thegiven lines as the other is to the unit (which isthe same as multiplication)” More on this subjectcan be found in A Source Book in Mathematics��–�� edited by D J Struik (Harvard Uni�versity Press� �� )

In his book titled Perspective in Perspective(Routledge & Kegan Paul� ��)� LawrenceWright describes the various ways in whichartists have drawn a cube� ranging from a meresquare to “spherical perspective” Although the“two�point perspective” version used for exercises�� through �� was the basis for perspectivedrawings in the Renaissance� Wright observes:“None of the three faces shown is square� theoverall shape is not even partly square� none ofthe twelve angles seen is a right angle� and of thethree sets of parallels only the verticals are shownas such”

Turkish Theorem.

1. “Paralel” and “hipotez” obviously mean“parallel” and “hypothesis.” “Distan”seems to mean “distance.” (“Hüküm”evidently means “conclusion,” but it isunrecognizable as such.)

•2. If a line parallel to one side of a triangleintersects the other two sides in differentpoints.

3. It divides the sides in the same ratio.

Errors of Omission.

4. Example answer:

5. Example answer:

6. Example answer:

Picturing Products.

•7. The area of the rectangle.

•8. As lengths: AC = 3, AD = 2, and AE = 6.

9. = .

10. c = ab.

11.

12.

Supply and Demand.

•13. = .


14. = .

15. = .

16. = .

•17. Substitution. (Both are equal to .)

Side-Splitter Practice.

•18. x = 2.4. ( = , 5x = 12, x = 2.4.)

19. x = 8. ( = , x2 = 64, x = 8.)

20. x = 7.5. ( = , 4x = 30, x = 7.5.)

21. x = 10. ( = , 36x = 360, x = 10.)

Two-Point Perspective.

•22. Reasonable because BE || CF in ∆OCF (theSide-Splitter Theorem).

23. Not reasonable, because lines AD and BCintersect in O.

24. Reasonable because of division (BC = CDand EF = FG).

25. Reasonable because DG || CF in ∆PCF (thecorollary to the Side-Splitter Theorem).

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

The Dames Point Bridge is the longest cable�stayedbridge in the United States and the only one tofeature the “harp�” or parallel�stay� arrangementon two vertical planes Completed in �� it is �miles long

Parallelogram Exercise.

•26. It is a parallelogram because DF || EB andDF = EB.

27. XE || YB in ∆AYB and FY || DX in ∆CDX. Ifa line parallel to one side of a triangleintersects the other two sides in differentpoints, it divides the sides in the same ratio.

28. AX = XY = YC. Because AE = EB, = 1;

so = 1 and AX = XY. Because DF = FC,

= 1; so = 1 and XY = YC.

Corollary to Theorem 44.

•29. If a line parallel to one side of a triangleintersects the other two sides in differentpoints, it divides the sides in the same ratio.

30. Multiplication. (In a proportion, the productof the means is equal to the product of theextremes.)

31. Addition.

32. Substitution (factoring).

33. Division (and substitution).

Parallels Path.

34. If a line parallel to one side of a triangleintersects the other two sides in differentpoints, it cuts off segments proportional tothe sides.

35. = = = = =

= .

•36. EO. (Because = .)

37. FO. (Because = .)

38. Because = , AM = GM by

multiplication.

•39. Point G is the same point as point A.

40. It would retrace itself.

Bridge Cables.

•41. EF = FG; so = 1 by division.

42. = 1. In ∆IEG, HF || IG; so, by the

Side-Splitter Theorem, = .

� Chapter �� Lesson �

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

44. ABHI and BCEH are parallelograms (bothpairs of their opposite sides are parallel).

•45. The opposite sides of a parallelogram areequal.

46. = 1. Because = 1, BC = EH and

AB = HI, = 1 by substitution.

47. Because = 1, AB = BC by multiplication.

Original Proofs.

48. Proof. (One possibility.)(1) In ∆ABC, AD bisects ∠BAC. (Given.)(2) ∠BAD = ∠DAC. (If an angle is bisected,

it is divided into two equal angles.)(3) AE = ED. (Given.)(4) ∠EDA = ∠DAC. (If two sides of a

triangle are equal, the angles oppositethem are equal.)

(5) ∠BAD = ∠EDA. (Substitution.)(6) ED || AB. (Equal alternate interior angles

mean that lines are parallel.)

(7) = . (If a line parallel to one side

of a triangle intersects the other twosides in different points, it divides themin the same ratio.)

49. Proof.(1) In ∆ABC, AB = AC and DE || BC. (Given.)

(2) = . (If a line parallel to one side

of a triangle intersects the other twosides in different points, it cuts offsegments proportional to the sides.)

(3) = . (Substitution.)

(4) AD = AE. (Multiplication.)

Set III Set III Set III Set III Set III (page ��) According to an article on the reproducing panto�graph by Jack W Jacobsen (wwwcarouselscom)�“the reproducing pantograph first saw use as anearly copying machine� making exact duplicatesof written documents Artists soon adopted its

use to duplicate drawings It is known that da Vinciused one to make duplicates of his drawings andpossibly to duplicate those drawings onto canvas It was not long before sculptors and carversadapted the pantograph’s use for tracing drawingsonto blocks of marble or wood They would thenuse the reproduced lines as guidelines for carvingTrue advancement in the pantograph design cameabout late in the ��th century with the advent oftypeset printing A pantograph was used to cutout the typeset letters ”

Pantograph.

1.

2. ABCD is always a parallelogram becauseboth pairs of its opposite sides are equal.

3. ∠2, ∠4, and ∠6.

4. ∠1 = ∠3 = ∠5 = ∠7 = ( )° or (90 – )°.

5. 180°.

6. 180°. (∠4 = ∠2 and ∠5 = ∠1; so∠3 + ∠4 + ∠5 = ∠3 + ∠2 + ∠1 by substitution).

7. It is always a straight angle.

8. They are always collinear.

9. always stays the same because

= = and PA, AB, BC, and CE

are fixed lengths.

10. = = = 1.75.

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


It is a remarkable thing that� if two similartriangles are in any position but have the sameorientation� the “average” triangle formed bytaking the midpoints of the line segments

Chapter �� Lesson ��

connecting their corresponding vertices is similarto them David Wells points out in The PenguinDictionary of Curious and Interesting Geometry(Penguin Books� ��) that “the same is true ofpolygons in general It is also true if� instead oftaking the midpoints of the lines� they are justdivided in the same ratio”

The apple on the stamp honoring Sir IsaacNewton symbolizes the story that an apple fallingfrom a tree led Newton to the idea of universalgravitation by getting him to thinking that� if theearth pulls on an apple� it also pulls on moredistant objects such as the moon (The moondoesn’t fall as does the apple� because it fallsaround the earth Its sideways motion makes itmiss) The geometric figure on the stamp is fromNewton’s Principia Mathematica As WilliamDunham remarks in The Mathematical Universe(Wiley� ��)� “This work presented Newtonianmechanics in a precise� careful� and mathematicalfashion In it he introduced the laws of motionand the principle of universal gravitation anddeduced� mathematically� everything from tidalflows to planetary orbits Principia Mathematicais regarded by many as the greatest scientificbook ever written”

The figure for exercises � through �� is moreremarkable than it might seem As recently as�� C Stanley Ogilvy observed in Tomorrow’sMath (Oxford University Press) that� although away to dissect a right triangle into five trianglessimilar to itself was known� a way to do this fornonright triangles was not The division of the��°��°��° triangle on which exercises � through�� are based was discovered soon afterward andhas been proved to be the only such solutionZalman Usiskin and S G Wayment reported thisresult in an article published in MathematicsMagazine titled “Partitioning a Triangle intoFive Triangles Similar to It” All of this is reportedby Martin Gardner in Wheels� Life and OtherMathematical Amusements (W H Freeman andCompany� ��)

The chessboard puzzle was first introduced inthe Set III exercises of Chapter �� Lesson

Two entire chapters of J V Field’s TheInvention of Infinity (Oxford University Press��) deal with Piero della Francesca’s mathematicsand his treatise on perspective� a work stronglyinfluenced by Euclid

Thales’ Method.

•1. Parallel lines form equal correspondingangles.

2. All right angles are equal.

•3. AA.

4. Corresponding sides of similar triangles areproportional.

Triangle Average.

5. They appear to be their midpoints.

•6. Yes. Two triangles similar to a third triangleare similar to each other.

Newton’s Figure.

7. ∆PRI ~ ∆PDF and ∆PQI ~ ∆PES. (Also,∆PDF ~ ∆SEF, from which it follows that∆PRI ~ ∆SEF.)

8. AA. (All these triangles are right triangles.The triangles in the first two pairs shareacute angles at P. The triangles in the thirdpair have equal vertical angles at F. Thefourth pair follows from the fact that twotriangles similar to a third triangle aresimilar to each other.)

Nine Triangles.

9. ∆ABC, ∆ADC, ∆BCE, ∆CDF, ∆CEF, ∆DEF.

•10. 30°, 30°, and 120°. (It is easiest to see this bystarting with the three equal angles at F.)

11. ∆CDE is equiangular (and equilateral).∆ACE and ∆BCD are right triangles withacute angles of 30° and 60°.

Chessboard Puzzle.

12. The square has an area of 64 units and theother arrangement seems to have an area of65 units.

13. AA. [∠CAE = ∠DAF (reflexive), and∠AEC = ∠AFD (all right angles are equal.]

•14. Corresponding sides of similar trianglesare proportional.

15. EC = 3 and FD = 5.

16. No. ≠ .

•17. They are not collinear.

18. Indirect.

� Chapter �� Lesson

Piero’s Theorem.

•19. That they are parallel.

20. That lines contain the same point.

•21. ∆AHK ~ ∆ABD and ∆AKL ~ ∆ADE.

22. = and = .

•23. = .

24. = = .


Students who have solved “work problems” inalgebra may be interested in knowing that thefigure illustrating two electrical resistances inparallel also illustrates problems with two workers

working simultaneously The equation R �

is equivalent to the equation � � � a

form more familiar in the work�problem context:R� and R� become the times needed for the twoworkers to complete the job individually and Rthe time for them to complete it working together

Exercises �� through � are based on aninteresting fact that is not well known Althoughthe angles of a �� triangle are not related in anyinteresting way� the largest angle of a �� triangle is exactly twice as large as its smallestangle

Mascheroni’s work Geometria del Compasso�in which he proved that every straightedge andcompass construction can be done with anadjustable compass alone� was published in ��For more on Mascheroni constructions� seechapter �� of Martin Gardner’s MathematicalCircus (Knopf� ��)

Electrician’s Formula.

25. AA.

26. Corresponding sides of similar trianglesare proportional.

•27. Addition.

•28. Multiplication.

29. Substitution (factoring).

30. Division.

Sides and Angles.

•31. ∆ABD ~ ∆CBA.

32. = = .

33. = = .

•34. x = 8, y = 10, and z = 10. (z = 18 – x.)

35. ∠2 = ∠C. (y = 10 = z.)

36. ∠BAC = 2∠C. (∠BAC = ∠1 + ∠2, ∠1 = ∠C,∠2 = ∠C; so ∠BAC = ∠C + ∠C.)

Mascheroni Construction.

37.

38. Isosceles.

•39. AA (∠A = ∠EFA and ∠FHA = ∠A).


41. It is equal to . FH = FA = AB = AE; so

= .

42. = = (exercises 40 and 41); so

= and AH = FA. Because FA = AB,

AH = AB by substitution.

Deja Vu.

43. AA (∠A = ∠A and ∠C = ∠ADE).

44. = .

•45. = .


46. ab – as = bs; so ab = as + bs; so s(a + b) = ab;

so s = .

47. It represents a single resistance to whichresistances a and b are equivalent.

Original Proofs.

48. Proof.(1) ∆ACD with BE || CD. (Given.)(2) ∠ABE = ∠C and ∠AEB = ∠D. (Parallel

lines form equal corresponding angles.)(3) ∆ABE ~ ∆ACD. (AA.)

49. Proof.(1) Trapezoid ABCD with bases AB and DC

and diagonals AC and BD meeting at E.(Given.)

(2) AB || DC. (The bases of a trapezoid areparallel.)

(3) ∠BAC = ∠ACD and ∠ABD = ∠BDC.(Parallel lines form equal alternateinterior angles.)

(4) ∆AEB ~ ∆CED. (AA.)

(5) = . (Corresponding sides of

similar triangles are proportional.)(6) AE × ED = BE × EC. (Multiplication.)

50. Proof.(1) ∆ABC with midsegments MN, MO, and

NO. (Given.)(2) MN || AB, NO || BC, and MO || AC.

(A midsegment of a triangle is parallelto the third side.)

(3) ANMO and BMNO are parallelograms.(A quadrilateral is a parallelogram ifboth pairs of opposite sides are parallel.)

(4) ∠OMN = ∠A and ∠ONM = ∠B.(The opposite angles of a parallelogramare equal.)

(5) ∆MNO ~ ∆ABC. (AA.)

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

Dividing Line.

1. 4.5 square units each.

2. The area of the purple region is

(4 × 2) + 1 = 5 square units; so the area of

the yellow region is 9 – 5 = 4 square units.

3. AA (∠DBQ = ∠CBA and ∠QDB = ∠C).


5. QD = . Because = , = .

6. The area of the purple region is

(4 × 1.75) + 1 = 4.5 square units.

7. Yes.



Exercises �� through �� are a good basis forfurther exploration After confirming the answersto exercises �� and �� it might be fun toexperiment with other linear transformations ofthe coordinates; that is� (a� b) → (ca � e� db � f )�to see what choices of c� d� e� and f producedilations and what sorts of transformations areproduced when c ≠ d

Fish Story.

•1. CG and CF.

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

3. ∆CDE ~ ∆CAB by AA (∠CDE = ∠A and∠CED = ∠B because parallel lines formequal corresponding angles.)

•4. Corresponding altitudes.

5. = .

6. Corresponding altitudes of similar triangleshave the same ratio as that of thecorresponding sides.

•7. CG, the distance of the fish from the camera.

8. The fish is 1 foot long. ( = , DE = 1.)

9. The center.

•10. 7.5 cm.

11. 3 cm.

•12. 2.5. (r1 = = .)


13. It indicates that the larger fish is 2.5 times aslong as the smaller fish.

14. AB = 5 cm. CD = 2 cm. = = 2.5.

•15. 0.4. (r2 = = .)

16. It indicates that the smaller fish is 0.4 timesas long as the larger fish.

17. = = 0.4.

•18. r2 is the reciprocal of r1.

Drawing Conclusions.

19. No. ∆AGB appears to be larger than ∆DGE.Also, we do not know that any sides areequal.

20. Yes. The triangles are similar by AA. (Theright angles at G and the alternate interiorangles formed by AB and ED.)

•21. Yes. Corresponding sides of similar trianglesare proportional.

22. Yes. Because = , AG × GE = GB × DG.

•23. No. These areas cannot be equal if ∆AGB islarger.

24. Yes. Adding α∆DGE to each side ofαAGEF = αBGDC gives αADEF = αBEDC.

Grid Exercise.

25.

•27. ∆A’B’C’ appears to be an enlargement of∆ABC. (Also, the corresponding sides of thetwo triangles appear to be parallel.)

28. A dilation.

29. (3, 4).

30. 2.


The story of the Washington Monument is toldby former engineering professor Robert Banks inSlicing Pizzas� Racing Turtles� and FurtherAdventures in Applied Mathematics (PrincetonUniversity Press� ��) A summary of the storyfollows: The monument� �� feet � inches inheight� is made of marble and granite Thecornerstone for its construction was laid on July� �� when James Polk was president Asoriginally planned� the tower was not taperedand would have had a star rather than a pyramidat its top After � years� it had risen to a height of�� feet� but then the money ran out andconstruction didn’t start up again until �� Itwas completed in �� As Banks points out� if themonument were not capped with the pyramid butextended until the tapered sides of the maincolumn met at a point� it would be more thantwice as tall� taller than the Empire State Building

The rest of the exercises in Set II consider theSAS and SSS similarity theorems� which are niceto know but have been omitted from the mainsequence of theorems In most applications ofsimilar triangles� it is angles that we know to beequal rather than sides being proportional It isunfortunate that the proofs of these theoremsare so cumbersome with so many algebraic stepsAs is often the case with proofs of this nature�after the stage has been set with the extra lineforming the third triangle� it is probably easier toconstruct the proofs rather than to read them

Washington Monument.

•31. 555.4 ft.

32. ∆AGF ~ ∆EHF, ∆BGF ~ ∆CHF, and∆ABF ~ ∆ECF.

33. = , = ,

34.4x = 55x – 27,522, 20.6x = 27,522,

x ≈ 1,336 ft.•26. B(5, 3) → B’(2 · 5 – 3, 2 · 3 – 4), or B’(7, 2).

C(10, 6) → C’(2 · 10 – 3, 2 · 6 – 4), or C’(17, 8).


•34. Corresponding altitudes of similar triangleshave the same ratio as that of thecorresponding sides.

Congruence and Similarity.

•35. ASA.

•36. AA.

37. AAS.

38. AA.

39. Yes. If = = 1, then a = b and c = d.

The triangles are congruent by SAS.

SAS Similarity Theorem.

•40. Ruler Postulate.

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

42. A line parallel to one side of a triangle cutsoff segments on the other two sidesproportional to the sides.

43. Substitution.

44. Multiplication.

45. SAS.

46. Corresponding parts of congruent trianglesare equal.

•47. Parallel lines form equal correspondingangles.

48. Substitution.

•49. AA.

50. Yes. If = = = 1, then a = b, c = d, and

e = f. The triangles are congruent by SSS.

SSS Similarity Theorem.

51. Ruler Postulate.

52. Two points determine a line.

•53. Reflexive.

54. If an angle of one triangle is equal to anangle of another triangle and the sidesincluding these angles are proportional,then the triangles are similar.

•55. Corresponding sides of similar trianglesare proportional.

56. Substitution.

57. Multiplication.

•58. SSS Congruence Theorem.

59. Corresponding parts of congruent trianglesare equal.

60. AA Similarity Theorem.

61. Two triangles similar to a third triangleare similar to each other.


Camera Experiment.

1. It gets larger.

2. If ∆AHB ~ ∆CHD, = . So x CD = yAB,

and CD = AB.

3. It gets bigger.

4. It gets smaller.

5. It gets bigger.

6. It gets bigger.

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

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

Judo� which means “gentle way” in Japanese�didn’t become an Olympic sport until �� Thethree regions of its competition area� althoughsquare in shape� are covered with “tatami�”rectangular mats � meter wide and � meters long(Tatami are the same mats used by Japanesearchitects in creating floor plans) A consequenceof the use of these mats is that� when the judocompetition area was first designed� its innerspace� the “contest area�” had to have sides thatwere an even number of meters Interestedstudents might enjoy exploring this requirementand explaining why

Lewis Carroll included whimsical references tomaps in two of his books In The Hunting of theSnark (�� )� he wrote: “He had brought a largemap representing the sea� Without the leastvestige of land: And the crew were much pleased

�� Chapter �� Lesson

when they found it to be A map they could allunderstand” Martin Gardner remarks in LewisCarroll’s The Hunting of the Snark—CentennialEdition (William Kaufman� Inc� ��): “Incontrast� a map in Carroll’s Sylvie and BrunoConcluded� Chapter �� has everything on itThe German Professor explains how his country’scartographers experimented with larger andlarger maps until they finally made one with ascale of a mile to the mile”

Triangle Ratios.

•1. .

•2. .

3. .

4. .

5. .

6. .

7. . ( .)

8. . ( .)

9. The ratio of the areas of two similarpolygons is equal to the square of the ratioof the corresponding sides.

Judo Mat.

•10. side of 1 = 8 m.

•11. ρ1 = 32 m.

•12. α1 = 64 m2.

13. side of 2 = 10 m.

14. ρ2 = 40 m.

15. α2 = 100 m2.

16. side of 16 m.

17. ρ3 = 64 m.

18. α3 = 256 m2.

•19. . ( = .)

20. . ( = .)

•21. . ( = .)

22. . ( = .)

23. . ( = .)

24. . ( = .)

Map Scaling.

•25. They must be proportional.

26. They must be equal.

27. They must be proportional to the square ofthe ratio of the corresponding distances.

•28. 440. (The scale is , and so 1 inch

represents 5,280 inches; = 440 ft.)

29. 193,600. [(440 ft)2 = 193,600 ft2.]

•30. . ( .)

31. or 1.


The question asked in the SAT problem ofexercises �� through � is the area of ∆ADE� forwhich four numerical choices were listed�followed by the choice� “It cannot be determinedfrom the information given” This choice� thoughincorrect� is somewhat tempting in that there isno way to know the lengths of ∆ADE’s sides

Cranberries are grown commercially in levelbog areas that can be drained When the berriesare ready to harvest� the bog is flooded and theyare shaken loose from the vines by tractors Theythen float to the surface where they are corralledas shown in the photograph before being trans�ported


43. or .

Cranberry Circles.

•44. Because all circles have the same shape.

45. 45 ft and 60 ft.

•46. 90π ≈ 283 ft and 120π ≈ 377 ft.

47. 2,025π ≈ 6,362 ft2 and 3,600π ≈ 11,310 ft2.

•48. = 0.75. ( .)

49. = 0.75. ( .)

50. = 0.75. ( .)

51. = 0.5625. [ or ( )2.]

Pentagon Measurements.

52. 108°.

•53. 921 ft. (The scale is ; so 1 inch

represents 11,052 in; = 921 ft.)

54. 4,605 ft. (5 × 921.)

55. Approximately 29 acres.

[ = ( )2, x = 183,220,056 in2,

= 1,272,361.5 ft2,

≈ 29 acres.]


This exercise is based on Proposition �� of Book VIof the Elements Euclid stated it as follows:“In right�angled triangles the figure on the sidesubtending the right angle is equal to the similarand similarly described figures on the sidescontaining the right angle”

The pattern used by Barbara Dean in designingher quilt is called the “Virginia Reel” Copies of itare turned through ��°� ��°� and ��° to formthe four�fold rotation�symmetric design shownabove Jinny Beyer presents this design and othersin Chapter �� “Creating Geometric Tessellations�”of her beautiful book titled DesigningTessellations (Contemporary Books� ��)

SAT Problem.

•32. Yes. ∆ADE ~ ∆ABC by AA.

33. Yes. Corresponding sides of similar trianglesare proportional.

•34. No.

35. No.

36. Yes. The ratio of the areas of two similarpolygons is equal to the square of the ratioof the corresponding sides; so

= ( )2.

α∆ADE = α∆ABC = (54) = 6.

Quilt Pattern.

•37. Six.

38. .

39. .

•40. .

41. or 2.

42. = or .

ππ

ππ

�� Chapter �� Lesson

Pythagoras on the Sides.

1. The ratio of the areas of two similar polygonsis equal to the square of the ratio of thecorresponding sides.

2. Adding = ( )2 and = ( )2

gives + = ( )2 + ( )2, or

= . Because a, b, and

c are the lengths of the sides of the right

triangle, a2 + b2 = c2; so = 1.

Multiplying, αPy.a + αPy.b = αPy.c.

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


J L Heilbron describes Galileo’s figure for objectsin free fall in Geometry Civilized (ClarendonPress� ��): “After many false starts� he [Galileo]discovered that the fundamental characteristic offree fall is constant acceleration� which meansthat the velocity with which the body moves atany time is proportional to the elapsed time Galileo derived his rule relating distance and timeusing geometry The angle EOF is drawn sothat AC � gOA [where g is a constant numberthat measures the pull of gravity] In a momentof great insight� he identified the areas of thetriangles AOC� BOD� with the distances coveredby the falling body in the times tA� tB” Heilbrongoes on to explain the rest of Galileo’s argumentBy associating distances covered with areas oftriangles� Galileo reasoned that� becauseOA � tA and AC � g OA � gtA�

dA � α∆AOC � OA � AC � tA (gtA) � gtA�

Because A is an arbitary constant� it follows that

d � gt �

Body Ratios.

•1. . ( .)

2. .

3. . ( .)

4. . ( .)

Perspective Law.

5. If a line parallel to one side of a triangleintersects the other two sides in differentpoints, it cuts off segments proportional tothe sides.

6. AA. (∠A = ∠A and the correspondingangles formed by BC and DE with AD andAE are equal.)

7. AC = 2 cm, AE = 5 cm, DE = 4 cm.

8. = ; so = , 5BC = 8, BC = 1.6.

•9. 2.5. ( = = 2.5.)

•10. It is 6.25 times as great. (2.52 = 6.25.)

Proportion Practice.

11. 8. ( = , 15x = 120, x = 8.)

•12. 6.4. ( = , 5x = 32, x = 6.4.)

13. 5.6. ( = , 10x = 56, x = 5.6.)

•14. 12.8. ( = , 5x = 64, x = 12.8.)

Similar Triangles.

15. Nine.

16.

Reptiles.

•17. Concave pentagons.

•18. ABHIG ≅ EKJGI.

Chapter �� Review ��

19. AFEDC ~ GABHI.

•20. GFEI is a parallelogram. Both pairs of itsopposite sides are equal becauseGFEKJ ≅ EIGJK.

21. αAFEDC = 4αABHIG.

22. ρAFEDC = 2ρABHIG.

23. , or 2.

Free Fall.

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

•25. It increases. (For example, BD > AC.)

26. AA. (∠O = ∠O and ∠OAC and ∠OBD areright angles.)

27. Corresponding sides of similar trianglesare proportional.

28. The ratio of the areas of two similartriangles is equal to the square of the ratioof their corresponding sides.

Type Transformations.

29. Dilations.

•30. 2. ( .)

31. or 0.75. ( .)

•32. They are multiplied by 3. ( .)

33. They do not change.

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

The crossed�ladders problem on which exercises� through �� are based is a famous puzzle inrecreational mathematics The puzzle concernstwo crossed ladders of unequal length that leanagainst two buildings as shown in the figure:given the lengths of the ladders and the heightof their crossing point� the problem is to find thedistance between the two buildings The puzzle isfamous because the solution� based on similartriangles� leads to a difficult quartic equationDifferent numbers appear in various versions ofthe puzzle and the solution is generally not aninteger The simplest set of numbers for which all

of the distances in the figure are integers wasreported in the American Mathematical Monthlyin �� and these numbers are used in the versionof the puzzle presented in the exercises More onthe crossed�ladders problem can be found inChapter � of Martin Gardner’s MathematicalCircus (Knopf� ��)

Trick Card.

•34. = , x2 = 50, x = = ≈ 7.1.

The width of the original card isapproximately 7.1 cm.

35. ≈ 71 cm2.

•36. = = ≈ 0.71.

37. ( )2 = = .

38. Yes. The second card is formed by foldingthe first card in half.

Catapult.

39. AB = 108. (∆CFD ~ ∆CAB by AA; so

= . = , 60AB = 6,480,

AB = 108.)

Leg Splitter.

40.

•41. In a plane, two lines parallel to a third lineare parallel to each other. (EF || BC andAD || BC.)

42. If a line parallel to one side of a triangleintersects the other two sides in differentpoints, it divides the sides in the same ratio.

43. Substitution.

44. They prove that the line divides the legs ofthe trapezoid in the same ratio.

�� Chapter �� Review

Crossed Ladders.

45.•53. 15. (∆EFD ~ ∆BDA; so = , 8EF = 120,

EF = 15.)

54. 18. (∆EFC ~ ∆BDC; so = ,

20FC = 90 + 15FC, 5FC = 90, FC = 18.)

•55. . (∆ABD ~ ∆DEF; so = = .)

•56. 80. [α∆ABD = (8 · 20).]

57. 45. [α∆DEF = (6 · 15).]

•58. . ( .)

59. . (In ∆ BCD, EF || BD; so = = .)

60. 60. [α∆BED = (20 · 6).]

61. 180. [α∆DEC = (24 · 15).]

62. . ( .)

•46. 30 ft. (∆PED ~ ∆CBD; so = .)

47. 105 ft. (∆ADB ~ ∆PEB; so = .)

•48. 70 ft. (CD2 = 562 + 422; so CD = = 70.)

49. 119 ft. (AB2 = 1052 + 562; so AB = =119.)

50. PA = 85, PB = 34, PC = 20, PD = 50.

(PD2 = 302 + 402; so PD = = 50.

PB2 = 302 + 162; so PB = = 34.

In ∆ADB, PE || AD; so = ,

16PA = 1,360, PA = 85.

In ∆CBD, PE || CB; so = , 40PC = 800,

PC = 20.)

Seven Triangles.

51. ∆ABD ~ ∆DEF, ∆ABC ~ ∆DEC,∆BDC ~ ∆EFC.

52.

Algebra Review �Algebra Review �Algebra Review �Algebra Review �Algebra Review � (page ��)

•1. (x + 7)(x – 4) = 0,x + 7 = 0 or x – 4 = 0,x = –7 or x = 4,4 and –7.

•2. (5x + 1)(x – 1) = 0,5x + 1 = 0 or x – 1 = 0,

x = – or x = 1,

1 and – .

•3. x2 – 8x + 5 = 0,

x = = =

= = 4 ± ,

4 + and 4 – .

Chapter �� Algebra Review ��

•4. x2 – 49 = 0,(x + 7)(x – 7) = 0,x + 7 = 0 or x – 7 = 0,x = –7 or x = 7.7 and –7.

•5. x2 + 4x – 20 = 0,

x = =

= = = –2 ± .

–2 + and –2 – .

•6. (3x – 1)2 = 0,3x – 1 = 0,

x = .

.

•7. 3x2 + 5x – 12 = 0,(3x – 4)(x + 3) = 0,3x – 4 = 0 or x + 3 = 0,

x = or x = –3.

and –3.

•8. x2 – 7x + 12 = 2,x2 – 7x + 10 = 0,(x – 2)(x – 5) = 0,x – 2 = 0 or x – 5 = 0,x = 2 or x = 5.2 and 5.

•9. x2 + 10x + 25 = x + 17,x2 + 9x + 8 = 0,(x + 1)(x + 8) = 0,x + 1 = 0 or x + 8 = 0,x = –1 or x = –8.–1 and –8.

Algebra Review �Algebra Review �Algebra Review �Algebra Review �Algebra Review � (page � )

•1. r = .

•2. p1 = .

•3. v2 = .

•4. c = .

•5. c2 – b2 = a2,

a = .

•6. 2A = bh,

h = .

•7. V2 = 2gh,

h = .

•8. = r2,

r = or r = .

•9. IR = E,

R = .

•10. 3V = πr2 h,

h = .

•10. 5x2 – 5 = 2x,5x2 – 2x – 5 = 0,

x = =

= = =

.

and .

π

π

π π

πr

�� Chapter �� Algebra Review

•11. S(1 – r) = a,a = S(1 – r) or a = S – Sr.

•12. S(1 – r) = a,S – Sr = a,S – a = Sr,

r =

•21. r1r2 = Rr2 + Rr1, orR(r1 + r2) = r1r2,

R = or S(1 – r) = a,

1 – r = ,

r = 1 – .

•13. 8M = wL2,

w = .

•14. W2 = w1w2,

w1 = .

•15. r = .

•16. R = rA – 150r, orrA = R + 150r,

A =

= A – 150,

A = + 150.

•17. Ed2 = I,

d2 = ,

d = .

•18. = ,

= ,

l = .

•19. In – Ix = Ad2,

A = .

•20. In – Ix = Ad2,

= d2,

d = .

R = ,

R = .

•22. r1r2 = Rr2 + Rr1, orr1r2 – Rr1 = Rr2,r1(r2 – R) = Rr2,

r1 =

= – ,

r1 = ,

r1 = .

•23. nur + vr = vu(n – 1),nur + vr = vun – vu,nur – vun = –vr – vu,n(ur – vu) = –(vr + vu),

n = – or n = – or

n = or n = .

•24. F – 32 = C, or

C = (F – 32)

F – 32 = C,

9C = 5(F – 32),

C = (F – 32).

•25. = ,

= ,

f02(c – v) = fs

2(c + v),

f02c – f0

2v = fs2c + fs

2v,

f02c – fs

2c = f02v + fs

2v,

c(f02 – fs

2) = v(f02 + fs

2),

v = c .

π

π

π

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Chapter 10 Answers - BISD Moodlemoodle.bisd303.org/file.php/18/Solutions/Chapter 10 Answers.pdf · Chapter Lesson Chapter Lesson Set I ... Parallel lines form equal corresponding