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Chapter 7, Lesson 1 85

Chapter 7, Lesson 1 18. They seem to be equal (and they seem to bisect each other). Set I (pages 260-261) . The aerial photographer Georg Gerster, the 19. They seem to be equal (and the opposite author of Below from Above (Abbeville Press, sides seem to be parallel). 1986), wrote about his photograph of the freeway overpass covered by carpets: 20. They seem to be (and they "Years after having taken this photograph, I am seem to bisect each other). still amazed at this super-scale Paul Klee .... In Europe, special mats, a Swiss development, have •21. They are equiangular (or each of their been adopted. They retain the heat generated as is a right ). the concrete hardens and make it unnecessary to 22. They are equilateral. keep wetting the concrete. The new procedure is more economical, though less colorful " .

Theorem 24. 23. They seem to be parallel and equal. • 1. Two points determine a . 24. They seem to be equal. 2. The sum of the angles of a is 180°. 25. They seem to bisect each other. •3. Addition. SAT Problem. 4. Betweenness of Rays Theorem. •26. 2x + 2y = 180. (2x + 2y + 180 = 360.) 5. Substitution. 27. x + y = 90. Corollary. Set 11(pages 262-264) •6. The sum of the angles of a is Martin Gardner dedicated his book of mathemati• 360°. cal recreations, Penrose Tiles to Trapdoor Ciphers 7. Division. (W. H. Freeman and Company, 1989) to Roger Penrose with these words: 8. An angle whose measure is 90° is a right "To Roger Penrose, for his beautiful, surprising angle. discoveries in mathematics, physics, and •9. A quadrilateral each of whose angles is a cosmology; for his deep creative insights into is a . how the universe operates; and for his humility in not supposing that he is exploring only the • 10. Each angle of a rectangle is a right angle. products of human minds." 11. All right angles are equal. Two chapters of Gardner's book are about Carpets. Penrose tilings, a subject full of rich mathematical ideas. The two tiles in the exercises come from a 12. . with angles of 72° and 108°. John Norton Conway calls them "kites" and "darts." 13. Right. Amazingly, in all nonperiodic tilings of these tiles, •14. Lines that form right angles are perpen• the ratio of the number of kites to the number of dicular. darts is the golden ratio: = 1.618 ...! There 15. In a plane, two lines perpendicular to a third line are parallel. are a number of simple ways to prove that there are infinitely many ways to tile the plane with Rectangles. these two tiles. Even so, any finite region of one tiling appears infinitely many times in every other 16. They are equal and they are right angles. tiling; so no finite sample of a Penrose tiling can • 17. They seem to be parallel and equal. be used to determine which of the infinitely 86 Chapter 7, Lesson 1 nnany different tilings it is! Students interested in •40. Six sides, three diagorials, and four . such mind-boggling ideas will enjoy reading Gardner's material and doing the experiments 41. Example figure: suggested in it. The exercises on the angles of are intended to encourage students to realize that memorizing a result such as "each angle of an (n-2)i8o°„ equiangular n-gon has a measure of n is not important. What is important is for the student to feel comfortable and confident in being able to reconstruct such a result if it is 42. Eight sides, five , and six triangles. needed. •43. n-3. Penrose Tiles. 44. n-2. 28. The one labeled ABCD. 45. A quadrilateral has 4 - 3 = 1 which •29. 144°. (360°-3-72°.) forms 4-2 = 2 triangles.

30. •46. 720°. (4 X 180°.) 720° •47. 120°. (^.) 6 48. 1,080°. (6 X 180°.) 1 080° 49. 135°. (^^^.) • 31. SSS. 8 32. All are isosceles. 51. •50. (n-2)180°n . 33. They seem to be collinear. Linkage(n-2)180 Problems.° 34. They are collinear because •52. Their sum is 360°. ZACE = 72° + 108° = 180°. 53. Their sum is 180° if we think of the linkage 35. Yes. It appears that it could be folded along as a triangle. (It is 360° if we think of the the line through A, C, and E so that the two linkage as a quadrilateral, because halves would coincide. ZBCD = 180°.) 36. No. Quadrilateral ABED is equilateral but •54. They all appear to be smaller. not equiangular. 55. ZA + ZB + ZAPB = 180°, and Gentstone Pattern. ZC + ZD + ZCPD = 180°; so •37. Triangles, , and an . ZA + ZB + ZAPB = ZC + ZD + ZCPD. Because ZAPB = ZCPD, it follows that 38. Five sides, two diagonals, and three ZA + ZB = ZC + ZD by subtraction. triangles. Quadrilateral Angle Sum. 39. Example figure: 56. Addition.

57. Substitution.

58. Subtraction. Chapter 7, Lesson 2 87

59. Zl and Z2 are a linear pair; so they are 6. BC looks longer than CF, but they are supplementary and their sum is 180°. actually the same length. Likewise for Z3 and Z4. 7. The opposite angles of a are (Zl + Z2) + (Z3 + Z4) = S; so equal. 180° + 180° = S, and so S = 360°. •8. An exterior angle of a triangle is greater Set III (page 264) than either remote interior angle. This triangle-to- puzzle is the puzzle of 9. Substitution. Henry Dudeney first introduced in the Set III exercise of Chapter 1, Lesson 3. It is perhaps Theorem 26. surprising that, by carefully comparing the two 10. The opposite sides of a parallelogram are figures and knowing that the angles of an equal. and a square are 60° and 90° respectively, we can find all of the remaining • 11. The opposite sides of a parallelogram are angles from just the one given. (By using the fact parallel. that the triangle and square have equal and 12. Parallel lines form equal alternate interior applying some simple trigonometry to the angles. triangle labeled D, we can find the exact values of all of the angles in the figure. The 41° angle, 13. ASA. 4/7 14. Corresponding parts of congruent triangles for example, is rounded from Arcsin —.) are equal. • 15. A is bisected if it is divided into two equal segments. . 16.

•17. The diagonals of a parallelogram bisect each other.

Chapter 7, Lesson 2 • 18. Two points are sjmimetric with respect to a point if it is the of the line segment Set I (pages 267-268) connecting them.

1. The first card (the card with the picture of 19. Parallel lines form equal alternate interior the baseball player). angles. •2. Point. 20. Vertical angles are equal.

•3. See if the figure looks exactly the same 21. ASA. when it is turned upside down. 22. Corresponding parts of congruent triangles Optical Illusion. are equal. •4. Three. 23. Two points are symmetric with respect to a •5. Yes. The opposite sides of a parallelogram point if it is the midpoint of the line segment are equal. connecting them. 88 Chapter 7, Lesson 2

Set 11(pages 268-269) 34. Point E is the midpoint of AD. In encyclopedias and unabridged dictionaries, the •35. AD = 2DC. entry for "parallelogram" is followed by an entry for "parallelogram law." The Encyclopedia 36. They are supplementary. Parallel lines form Britannica describes it as the "rule for obtaining supplementary interior angles on the same geometrically the sum of two vectors by side of a transversal. constructing a parallelogram. A diagonal will give •37. 180. the requisite vector sum of two adjacent sides. The law is commonly used in physics to determine 38. 90. a resultant force or stress acting on a structure." The except from Newton's Mathematical Principles 39. Isosceles. of Natural Philosophy \s taken from the original 40. Right. Because x + y = 90, ZBEC = 90°. Latin edition published in 1687. It immediately follows statements of his three laws of motion. 41. BEIEC. A specific measure for ZBAD in the figure for Two Parallelograms. exercises 47 through 50 was given to make the exercises a little easier. Actually, an/measure other than 60° or 120° will work. A nice problem for better students would be to rework the exercises letting ZBAD = x°and to explain what happens when x = 60 or 120. Parallelogram Rule.

24. Corollary 43. Because x, y, and z are the measures of the angles of a triangle, x + y + z = 180. Therefore, ZABC is a straight angle; so A, B, and C are collinear. 44. Because ABDE and BCDE are parallelograms, AB II ED and BC II ED. If A, B, and C are not C D collinear, then through B there are two lines parallel to ED. This contradicts the Parallel •26. 10. (5 cm represents 50 lb; so 1 cm represents Postulate; so A, B, and C must be collinear. 10 lb.) •45. All three (AABE s ADEB = ABCD). 27. 25.(2.5x10 = 25.) 46. No. 28. About 6 cm. Hidden Triangles. 29. About 60. (6x10 = 60.) •30. About 24°. Angle Bisectors. 31.

•32. ZAEB = :c°; ZCED = y° 48. AADE, ACDF, and AEBF. 33. AB,AE,andED. 49. SAS. Chapter 7, Lesson 3 89

50. Three. In addition to AABE and ABCF, ADEF Pop-Up Parallelogram. is equilateral because its sides are the • 1. A quadrilateral is a parallelogram if its corresponding sides of the three congruent triangles. opposite sides are equal. 2. The opposite sides of a parallelogram are Set 111 (page 269) parallel.

The four symbols used on modern playing cards • 3. Parallel lines form equal corresponding were designed in 1392 by the court painter to angles. King Charles VI of France. The four suits represented four classes of French society: the •4. In a plane, if a line is perpendicular to one spades, soldiers, the clubs, farmers, the diamonds, of two parallel lines, it is also perpendicular artisans, and the hearts, the clergy. According to to the other. John Scarne in Scarne's New Complete Guide to Theorem 28. Gambling (Simon & Schuster, 1974), playing cards first arrived in the New World in 1492 with 5. The sum of the angles of a quadrilateral is Columbus and immediately found favor with the 360°. Native Americans, who made them of deerskin or sheepskin! 6. Division.

Playing Card . 7. Two angles are supplementary if their sum is 180°. 1. So that they look the same whether they are held "right side up" or "upside down." •8. Supplementary interior angles on the same side of a transversal mean that lines are 2. No playing card has line symmetry. parallel. 3. The diamonds are more symmetric because 9. A quadrilateral is a parallelogram if its the diamond symbol itself has point opposite sides are parallel. symmetry, whereas the symbols of the other suits do not. (They do have line symmetry, Theorem 29. however.) 10. Two points determine a line. 4. All of the diamond cards have point 11. Parallel lines form equal alternate interior symmetry except for the 7. All face cards angles. have point symmetry. In the club, spade, and heart suits, the 2s and 4s have point 12. Reflexive. symmetry. •13. SAS. 5. The 6s and 8s of clubs, spades, and hearts 14. Corresponding parts of congruent triangles would have point symmetry if one of the are equal. symbols in the middle row were turned upside down. (It is impossible for the odd- • 15. A quadrilateral is a parallelogram if its numbered cards in these three suits to be opposite sides are equal. redesigned so as to have point symmetry!) Theorem 30. Chapter 7, Lesson 3 16. A line segment that is bisected is divided into two equal parts. Set I (pages 272-273) 17. Vertical angles are equal. The letter p is the only letter of the English alphabet that transforms into three other letters 18. SAS. by reflections and rotations. Some students may 19. Corresponding parts of congruent triangles enjoy thinking about other examples of letters that transform into at least one other letter in are equal. this way. •20. Equal alternate interior angles mean that lines are parallel. 90 Chapter// Lesson3

•21. A quadrilateral is a parallelogram if two rigid bracing if and only if the bracing subgraph is opposite sides are both parallel and equal. a tree." Letter Transformations. Parallel Rulers. • 22. Line symmetry. The figure can be folded 31. They are connected so that AB = CD and along a vertical line so that the two halves AC = BD. A quadrilateral is a parallelogram coincide. if its opposite sides are equal. 23. Point symmetry. If the figure is turned •32. The pairs of opposite angles. upside down, it looks exactly the same. Tent Geometry. 24. Line symmetry. The figure can be folded 33. along a horizontal line so that the two halves coincide. c 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). 34. Yes. ABCD and CDEF are parallelograms; so AB = CD and CD = EF (the opposite sides of 28. Yes. Point symmetry and line symmetry a parallelogram are equal) and AB 11 CD and (four lines, one vertical, one horizontal, and CD II EF (the opposite sides of a parallelogram two at 45° angles with them). are parallel). It follows that AB = EF (substitution) and AB 11 EF (two lines parallel Economics Graph. to the same line are parallel to each other). • 29. They seem to be parallel. Therefore, ABFE is a parallelogram because two opposite sides are parallel and equal. 30. The quadrilateral formed by the four lines is a parallelogram (a quadrilateral is a 35. Yes. AD = BC, DE = CF, and AE = BF (the parallelogram if two opposite sides are opposite sides of a parallelogram are equal). equal and parallel); so the lines C and C + I Therefore, AADE = ABCF (SSS). are parallel. Rope Trick.

Set II (pages 273-275) 36. The method for constructing a parallel to a line through a given point not on it is another example of a "rusty compass" construction. Martin Gardner observes that, although the method has long been known, it is still being rediscovered and reported as new. The grid-bracing illustrations are taken from •37. A quadrilateral is a parallelogram if its the section titled "Bracing Structures" in Jay diagonals bisect each other. Kappraff's Connections—The Geometric Bridge between Art and Science (McGraw-Hill, 1991). 38. The opposite sides of a parallelogram are Kappraff cites the article "How to Brace a One- equal. Story Building" in Environment and Planning (1977) as the original source of this material, an • 39. SSS. important topic for engineers and architects. 40. Examples of pertinent theorems dealing with this Corresponding parts of congruent triangles subject in graph theory are: "In any distorted grid are equal. all the elements of a row (column) are parallel" 41. The opposite angles of a parallelogram are and "A bracing of an n by m grid is a minimum equal. Chapter?, Lesson 4 91

42. Substitution. Set III (page 275) •43. A quadrilateral is a rectangle if all of its 1. (The figures below show two possible answers angles are equal. depending on which triangles are constructed inward and outward.) Parallel Postulate. 44. Through point P, there is exactly one line parallel to line /.

45.

46. PC is parallel to line / because ABCP is a parallelogram (a quadrilateral is a parallelogram if its opposite sides are equal). Flexible Grid. •47. 12. 48. 4. 49. 12. 50. 4. 2. The quadrilateral formed seems to be a 51. They remain unchanged because their parallelogram. are determined by their sides (SSS) and their sides do not change. Chapter 7, Lesson 4 52. They remain parallel because the opposite sides of the quadrilaterals remain equal; so Set I (pages 277-279) they are always parallelograms. In his definitive book on the subject, Dissections- •53. They are the sides of the quadrilaterals Plane and Fancy (Cambridge University Press, above and below the braced square. 1997)/ Creg N. Frederickson (using radians, 2TT radians = 360°) reports: 54. They are the sides of the quadrilaterals to "We can think of a regular as a union of the right of the braced square. various rhombuses and isosceles triangles (which 55. Because all of the quadrilaterals remain are halves of rhombuses). Such a characterization parallelograms, these sides remain parallel is an internal structure of the corresponding to the horizontal and vertical sides of the polygon. If the has an even braced square. number of sides, then the internal structure needs only rhombuses A regular polygon with 2(7

sides contains q each of —-rhombuses, for each 1 q q

positive integer / that is less than —, plus —

if q is even. For example, the contains 92. Chapter 7, Lesson 4

6 each of —rhombuses and —rhombuses, plus Theorem 34. 6 3 • 17. All of the sides of a rhombus are equal. 3 squares." TT 18. In a plane, two points each equidistant from [— and ~ indicate rhombuses having angles of 6 3 the endpoints of a line segment determine the perpendicular bisector of the line 30° and 60°.] There are several ways in which the segment. rhombuses can be arranged when the dodecagon is divided in this way. Which Parts? The remark about a square being perfect • 19. Two consecutive sides. because all of its angles are "just right" was made by Albert H. Beiler at the beginning of the chapter 20. One side. titled "On The Square" in his book titled Recreations in the Theory of Numbers—The 21. One side and one angle. Queen of Mathematics Entertains (^Dover, 1966). 22. Two consecutive sides and one angle. Regular Dodecagon. "fust Right." •1. 12. 23. If all of the angles of a triangle were right •2. Three. angles, the sum of their measures would be 270°, rather than 180°. 3. A square. •24. No. It must be a rectangle. 4. Three. 25. Shown below are two of many possible examples. •5. 15. (The ambiguity arises because the right angles may be on the inside or the outside of the 6. Three. (There are six narrow, six medium. polygon.) and three square.) Theorem 31. • 7. All of the angles of a rectangle are equal. 8. A quadrilateral is a parallelogram if its opposite angles are equal. Set 11(pages 279-280) Theorem 32. Checking a Wall. 9. All of the sides of a rhombus are equal. 26. No. If the opposite sides of a quadrilateral • 10. A quadrilateral is a parallelogram if its are equal, it must be a parallelogram but opposite sides are equal. not necessarily a rectangle. Theorem 33. 27. No. We know that the diagonals of a 11. All of the angles of a rectangle are equal. rectangle are equal but to conclude that, if the diagonals of a quadrilateral are equal, •12. All rectangles are parallelograms. it is a rectangle would be to make the 13. The opposite sides of a parallelogram are converse error. equal.

14. Reflexive. 15. SAS.

16. Corresponding parts of congruent triangles are equal. 29. SSS. Chapter 7, Lesson 4 93

30. Corresponding parts of congruent triangles Rhombus Problem. are equal.

•31. A quadrilateral is a parallelogram if its opposite sides are equal. 32. The opposite angles of a parallelogram are equal. 33. Substitution.

•34. A quadrilateral all of whose angles are equal 48. Yes. Because ABCD is a rhombus, AB = AD. is a rectangle. It is also a parallelogram; so ZB = ZD. AAEB = AAFD (AAS); so AE = AF. 35. True. Triangle Problem. •36. False. 37. True. 38. False. •39. True. Square Problem.

50. Because ABDE is a parallelogram, AB = ED (the opposite sides of a parallelogram are equal). Also, because BDCE is a rectangle, ED = CB (the diagonals of a rectangle are equal). So AB = CB (substitution). Therefore, AABC is isosceles. (Additional drawings using the conditions given reveal that it isn't 41. Parallelograms. necessarily equilateral.)

42. They must be parallelograms because two Set III (page 280) opposite sides are both parallel and equal. (For example, in AMCO, AM 11 CO because Bill Leonard told the story of starting a lesson ABCD is a square and AM = CO because M with a high-school class of remedial mathematics and O are of the equal segments students with the problem of counting the AB and CD.) squares in a 4 x 4 grid. He had hoped that the students would think about it for awhile, but it •43. Its opposite sides are parallel. didn't take long for several students to start yelling out "16." Leonard then asked if they had 44. A rhombus. counted squares such as the one shaded in the 45. They seem to be collinear. second figure. The class immediately yelled out in unison, "171" 46. It appears to have point symmetry and line symmetry (with respect to the lines of the Counting Squares. diagonals of the square). 1. It contains 30 squares in all. There are 16 small squares, 9 2x2 squares, 4 3x3 squares, and 14x4 square. 2. It contains 15 squares in all. There are 7 small squares, 6 2x2 squares, 13x3 square, and 14x4 square. 94 Chapter 7, Lesson 5

Chapter 7, Lesson 5 Theorem 35. • 16. The bases of a are parallel. Set I (pages 282-284) •17. Through a point not on a line, there is Hal Morgan in his book titled Symbols of America exactly one line parallel to the line. (Viking, 1986), comments on the Chase Manhattan Bank trademark: "After World War II the 18. A quadrilateral is a parallelogram if its representational images of the nineteenth opposite sides are parallel. century began to give way to abstract designs inspired by the spare, geometric forms of the •19. The opposite sides of a parallelogram are Bauhaus. Some of these modern designs are quite equal. striking in their clarity and simplicity.... The 20. The legs of an are equal. design firm of Chermayeff & Geismar Associates created the Chase Manhattan Bank's distinctive 21. Substitution. octagon logo in i960 The abstract, interlocking design seems to have become a sort of model for •22. If two sides of a triangle are equal, the the designers of other bank and insurance logos." angles opposite them are equal. Morgan shows examples of several more-recent 23. Parallel lines form equal corresponding logos that are also point-symmetric designs of angles. interlocking solid black congruent polygons. 24. Substitution. Quadrilaterals in Perspective. • 25. Parallel lines form supplementary interior • 1. A trapezoid. angles with a transversal. •2. That it has exactly one pair of parallel sides. 26. The sum of two supplementary angles is 3. Yes. Picnic blankets are usually rectangular. 180°. (The perspective alters the blaJnket's .) 27. Substitution. •4. That all of its angles are right angles (or that 28. Subtraction. they are equal). Theorem 36. • 5. Yes. All rectangles are parallelograms. 29. The legs of an isosceles trapezoid are equal. 6. Rhombuses and parallelograms. • 30. The angles of an isosceles trapezoid 7. Squares, rectangles, rhombuses, and parallelograms. are equal. Geometric Trademark. 31. Reflexive. •8. Point. 32. SAS. 33. Corresponding parts of congruent triangles 9. It is a square because all of its sides and are equal. angles are equal.

• 10. They seem to be congruent. Set 11(pages 284-285) •11. . 12. CDandJE. • 13. No. The legs of a trapezoid must be equal for it to be isosceles. 14. No. One is a right angle and the other is acute. 15. An octagon. Chapter 7, Lesson 6 95

35. An isosceles trapezoid. Set III (page 285) •36. 75°. The Moscow Puzzles is the English language edition of Boris Kordemsky's Mathematical 37. The base angles of an isosceles trapezoid Know-how, which, according to Martin Gardner, are equal. is the best and most popular puzzle book ever published in the Soviet Union. Kordemsky taught 38. Isosceles. mathematics at a high school in Moscow for 39. 75°. many years. Gardner reported Andrew Miller's discovery •40. Parallel lines form equal corresponding of a second solution to Kordemsky's trapezoid angles. dissection problem in the February 1979 issue of Scientific American and posed it as a new puzzle •41. 105°. for the readers of that magazine. 42. 30°. Congruence Puzzle. Regular . Kordemsky's solution is shown below. 43. Examples are ABCD, ABCE, ACDE, and ABDE. Their bases are parallel (they form supplementary interior angles on the same side of a transversal) and their legs are equal (given).

44. Examples are ABCH, ABCE, and AFDE. Their opposite sides are parallel (established The solution thought of by Andrew Miller is in exercise 43). shown below. The three cuts intersect the upper 13 7 base at points —, —, and — of the way from its 45. Examples are ABCH, ABGE, and AFDE. 8 4 8 They are rhombuses because all of their left endpoint. sides are equal. In regard to ABCH, for example, AB = BC (given), AB = HQ and AH = BC (the opposite sides of a parallelo• gram are equal); so HC = AH (substitution). Trapezoid Diagonals.

•46. AC and DB bisect each other. Chapter 7, Lesson 6 •47. A quadrilateral is a parallelogram if its diagonals bisect each other. Set I (pages 288-289) 48. The opposite sides of a parallelogram are The "state capital" quadrilateral was chosen partly pa;allel. for fun but primarily to show that the original quadrilateral can have any shape. It doesn't even •49. / trapezoid has only one pair of opposite have to be convex. Not only can the quadrilateral 5 ides parallel. be concave, it can even be "self-intersecting" or skew! ^u. AC and DB bisect each other. Portuguese Theorem. 51. AC and DB do not bisect each other. 1. Midpoints. 52. Yes. One example of such a trapezoid is shown below. •2. Amidsegment. 3. Other side.

4. A midsegment of a triangle is parallel to the third side and half as long. 96 Chapter 7, Lesson 6

Two Midsegments. • 5. DE|| ACandDE= ^AC. 22. State Capitals Problem. c 6.

23. ADEC is an isosceles trapezoid because DE II AC (a midsegment of a triangle is parallel to the third side) and AD = CE. •24. ZA = ZC because the base angles of an isosceles trapezoid are equal.

7. MNPQ looks like a parallelogram. 25. AABC is isosceles because, if two angles of a triangle are equal, the sides opposite them •8. A midsegment of a triangle is parallel to the are equal. third side. (MN and QP are midsegments of (Alternatively, AB = 2AD = 2CE = CB .) ASOH and ASBH.) •9. In a plane, two lines parallel to a third line 26. PC = -AC because FG = ^DE and are parallel to each other. 4 2 DE = ^AC [so FG = ^(^AC) = ^AC]. • 10. A midsegment of a triangle is half as long as 2 2 2 4 the third side. Also, FG I! AC because FG || DE and II 11. Substitution. DE AC (in a plane, two lines parallel to a third line are parallel to each other). 12. That it is a parallelogram. • 13. MNPQ is a parallelogram because two Set 11(pages 289-291) opposite sides are both parallel and equal. Exercises 27 through 36 are intended to review 14. MP and NQ seem to bisect each other. some basics of coordinate geometry, including the distance formula, and to get students to 15. The diagonals of a parallelogram bisect each thinking about how midpoints and the directions other. of lines might be treated. Midpoint Quadrilateral. The Midsegment Theorem is, in a way, a "special case" of the fact that the line segment 16. A midsegment of a triangle is half as long as connecting the midpoints of the legs of a trapezoid the third side. is parallel to the bases and half their sum. • 17. Addition. The edition of 's E/ement5 with the fold-up figures was its first translation into •18. Substitution. English, by Henry Billingsley. After a career in business, Billingsley became sheriff of London 19. Its perimeter. and was elected its mayor in 1596. An excerpt •20. The sum of its diagonals. from The Elements of Geometrie, including an actual fold-up model, is reproduced in Edward R. 21. The perimeter of the midpoint quadrilateral Tufte's wonderful book Envisioning Information is equal to the sum of the diagonals of the (Graphics Press, 1990). original quadrilateral. In Mathematical Gems II (Mathematical Association of America, 1976), Ross Honsberger wrote: "Solid geometry pays as much attention to the tetrahedron as plane geometry does to the Chapter?, Lesson 6 97 triangle. Yet many elementary properties of the Tetrahedron. tetrahedron are not very well known." Exercises 37. 37 through 41 establish that any triangle can be folded along its midsegments to form a tetrahe• dron. Unlike tetrahedra in general, however, a tetrahedron formed in this way is always isosceles; that is, each pair of opposite edges are equal. All the faces of an isosceles tetrahedron are congruent triangles and consequently have the same ; furthermore, the sum of the face angles at each is 180°. More information on this geometric solid can be found in chapter 9 of Honsberger's book. »38. They are congruent by SSS. Midpoint Coordinates. 39. y 15 B(5,U} N(8,10) 10

40. They are equal. 0(7,5) 41. 180°. (The measures of the three angles that meet at each vertex are a, b, and c.) A(3,2) More Midpoint Quadrilaterals. 27. o 10 15 42. Example figure: II y>K^ II ij •28. (4,7). •29. (8,10). 30. (7,5). 31. Yes. The coordinates of the midpoint are "midway between" the respective coordinates •43. A rhombus. of the endpoints. (They are their averages.) 44. The line segments divide the rectangle into four right triangles and the quadrilateral. •32. MN = V(8-4)^ + (10-7)^ = V4^ + 3^ = The triangles are congruent by SAS; so their VI6+9 = V25 = 5. hypotenuses are equal. Because the hypot• enuses are the sides of the quadrilateral, it is 33. AC = V(ll-3)^ + (8-2)2 = Vs^+e^ = equilateral and must be a rhombus. V64 + 36 = VTOO = 10. 45. Example figure:

34. Yes. MN = ^AC as the Midsegment Theorem states.

•35. They should be parallel. 36. Yes. They seem to be parallel because they have the same direction; each line goes up 3 46. A rectangle. units as it goes 4 units to the right. 98 Chapter?, Lesson 6

47. The sides of the new quadrilateral are Set 111 (page 2.91) parallel to the diagonals of the rhombus The problem of the nested triangles is a preview because of the Midsegment Theorem. of the idea of a limit, a topic considered later in Consequently, they form four small the treatment of measuring the . The sum parallelograms. Each parallelogram has a of the lengths of all of the sides of the triangles right angle because the diagonals of a is the sum of the geometric series: rhombus are perpendicular. Because the opposite angles of a parallelogram are 6 + ^6 + (^)^6 + (^)^6 + • • • =2x6 equal, it follows that the four angles of the new quadrilateral are right angles; so it is a Students who tackle this problem will be rectangle. interested to know that they will explore it further in second-year algebra. 48. Another square. Infinite Series. Base Average. The perimeter of the largest triangle is 6 in. From •49. MN seems to be parallel to AB and DC, the the Midsegment Theorem, we know that the sides bases of the trapezoid. of the next triangle are half as long; so its perimeter is 3 in. Reasoning in the same way, we find that the perimeters in inches of successive triangles are I. 5, 0.75, 0.375, 0.1875, and so on. Starting with the first two triangles, we find that the successive sums of the sides are 9,10.5, II. 25,11.625,11.8125, and so on. The sums seem to be getting closer and closer to 12 inches and (as students who study infinite series will learn to 51. DC II AP; so ZC = ZNBP (parallel lines prove) the sum of all of the sides of the triangles form equal alternate interior angles). in the figure is 1 foot. CN = NB because N is the midpoint of CB. ZCND = ZBNP (vertical angles are equal). Chapter 7, Review So ADCN = APBN (ASA). [An alternative proof could use Set I (pages 292-294) ZCDN = ZBPN (equal alternate interior A question such as that in exercise 11 is intended angles) in place of either pair of angles to arouse the student's curiosity. According to above, in which case the triangles would be Jearl Walker, who posed it in his book titled The congruent by AAS.] Flying Circus of Physics (Wiley, 1977), "If the ratio 52. DN = NP (corresponding parts of congruent of the bar's density to the fluid's density is close triangles are equal); so N is the midpoint of to 1 or o, the bar floats in stable equilibrium as in DP. Therefore, MN is a midsegment of the first figure. If the ratio is some intermediate ADAPandso MN || AP. value, then the bar floats in stable equilibrium with its sides at 45° to the fluid's surface. In each 53. MN II AP and DC || AP; so MN || DC. (In a case, the orientation of stable equilibrium is plane, two lines parallel to a third line are determined by the position in which the potential parallel to each other.) energy of the system is least." This floating-bar problem is a good illustration of how scientists 54. MN = ^ AP by the Midsegment Theorem. arrive at conclusions by using inductive reasoning (performing experiments with floating bars to see Because AP = AB + BP and DC = BP what happens) and how they then "explain" the (corresponding parts of congruent triangles results by using deductive reasoning (calculating are equal), MN = |(AB + BP) = ^(AB + DC) the potential energy and its minimum value by using mathematics, including geometry'). (substitution). The frequent appearance in this chapter of theorems whose converses also are theorems can easily mislead students into forgetting that a Chapter?, Review 99

statement and its converse are not logically •13. All of the sides and angles of a square are equivalent. The intent of exercises 17 through 30 equal. is to remind students of this fact and to encourage them to draw figures to test unfamiliar conclusions 14. SAS (or SSS). (or even to remind themselves of facts that they 15. Corresponding parts of congruent triangles have proved). Knowledge of the properties of the are equal. various quadrilaterals should not be as much a matter of memorization as of looking at pictures • 16. If a line divides an angle into two equal and applying simple common sense. parts, it bisects the angle. Palm Strand Rhombus. Related Statements. 17. It is the converse of the first statement.

18. No. The first statement is true; the first figure shows that the triangles are congruent by SSS. The second figure shows that the second statement is false.

19. True.

20. True.

•21. True. •2. Both pairs of its opposite sides are parallel 22. False. (because the strips have parallel sides). Example of a quadrilateral that has two pairs of 3. The opposite angles of a parallelogram are equal angles but is not an isosceles trapezoid: equal.

•4. AAS. 5. Corresponding parts of congruent triangles are equal.

6. The opposite sides of a parallelogram are 23. False. equal. Example of a parallelogram that is not a rhombus: •7. Substitution.

8. All of its sides are equal.

Floating Bar. 24. True. •9. Four. 25. True. 10. Yes. 26. True. 11. (Student answer.) (The bar can float either •27. True. way depending on it and the liquid.) 28. False. 12. Example of a quadrilateral with perpendicular A diagonals that is not a rhombus: 100 Chapter?, Review

29. False. to form a parallelogram and a triangle; the proof is comparable to that of Theorem 35, outlined in Example of a parallelogram whose diagonals are exercises 16 through 28 of Lesson 5. The additional not equal: fact that the base opposite the larger base angles of an isosceles trapezoid is longer than the other base is the basis for the magic trick of exercise 58. Like the question about the floating bar in Set I, the question about the magic trick is included for 30. False. fun and not meant to be taken too seriously. Example of a quadrilateral whose diagonals are Dissection Puzzle. equal that is not a parallelogram: •36. D, E,andF. 37. D. 38. DandF. 39. CandD. Jumping Frog. 40. A, B, D, E, and F. •31. The sum of the angles of a quadrilateral is 360°. 41.

32. They are rhombuses and all rhombuses are G c H c I parallelograms. (Or, Both pairs of their opposite sides are equal.) •33. The opposite angles of a parallelogram are equal.

34. They are parallelograms because their diagonals bisect each other.

35. The opposite sides of a parallelogram are parallel and, in a plane, two lines parallel to a third line are parallel to each other. •42. GI ± HM and ML 1HM. In a plane, two Set 11(pages 294-295) lines perpendicular to a third line are Only the right angles in the pieces of the puzzle parallel. of exercises 36 through 50 actually have measures 43. A parallelogram. that are integers. From AABC it is evident that 44. A quadrilateral is a parallelogram if its tan A = ^ = 2; so ZA = Arctan 2 = 63.43°. opposite sides are parallel.

Although part of basic vocabulary in books of •45. 2c. the past, words such as "trapezium" and "" are now close to being obsolete. 46. 2a+ b. [Some students may recognize that Strangely, in British usage, the word "trapezium" GM can also be expressed in terms of c by is used to mean "trapezoid." using the . Even though the quadrilaterals in exercises 54 GM2 = c2 + (2c)2 = 5c2; so GM = c Vs.] through 57 are drawn to scale, it is almost impos• sible to recognize them merely by their appear• 47. A square. ance. It is tempting to conclude that, because 48. All of its sides and angles are equal. quadrilateral ABCD is a trapezoid whose base angles are equal, it is isosceles, which, of course, •49. A trapezoid. is the converse error. To prove that it is isosceles would require adding an extra line to the figure Chapter 7, Algebra Review 101

50. It has exactly one pair of parallel sides. 61. It appears to be a rhombus. APDO = ANCO (HO II IJ because, in a plane, two lines (SAS); so PO = NO. Because MN = PO and perpendicular to a third line are parallel.) MP = NO, MN = MP; so all of the sides of MNOP are equal. Words from the Past. •51. No, because a trapezoid has two parallel Algebra Review sides.

52. Yes, because, other than squares, rhombuses •1. 2.236. are parallelograms that have no right •2. 22.36. angles. •3. 223.6. 53. No, because a rectangle has four right angles. •4. 7.07.

Rectangles Not. •5. 70.7. 54. ABCD is a trapezoid because BC | j AD (they •6. Vsoo = Vioo-5 = ViooVs = loVs. form supplementary interior angles on the V50,ooo = Vioo ViooVsoo = same side of a transversal). (It is also possible •7. -500 = to show that ABCD is isosceles.) 10(10V5) = 100V5. 55. EFGH is a parallelogram because both pairs of opposite angles are equal. •8. VSO = V25^ = V25V2 =5V2.

•56. IJKL is a trapezoid because IJ || LK. •9. V5,000 = VlOO-50 = VTOOVSO =10(5V2) = 50 V2. 57. From its angles we can conclude that MNOP is neither a parallelogram nor a trapezoid; •10. V25x = ^[25^[x = 5^[x. so it is not any of the special types of •11. = -^Ix^-x = V^V^ = x^4x. quadrilaterals that we have studied. • 12. = = x-4n. 58. While you look at the card, the magician rotates the pack 180" so that the bottom •13. Vl2x^2 = V4^3^ = 4lS4x^ = Ix'S. edges of the cards are at the top. When your chosen card is put back, it is easy to slide it • 14. V147 + V3 = V493+V3 = V49V3 + V3 = up because it is now wider at the top than any of the others. 7V3 + V3 =8^3.

Another Midpoint Quadrilateral. •15. V147 + 3 = Vl50 = V25^ = V25V6 = 5^6.

•16. VTOO-VITS = Vl00-7-V25^ = V1OOV7-V25V7 = I0V7-5V7 = 5V7.

•17. V700-175 = V525 = V25-21 = V25V2T = 5V2T. DOC •18. V20+20+20 = V60 = V4I5 = ViVis = 60. MN II AC and PO II AC (a midsegment of a triangle is parallel to the third side); so 2V15.

MN II PO. MN = -AC and PO = ^AC (a • 19. V2O+V2O + V2O = 3V2O = 3V4^ = Vs = midsegment of a triangle is half as long as 3V4 6V5. the third side); so MN = PO. MNOP is a •20. V6^ + V8^ =6 + 8 = 14. parallelogram because two opposite sides are both pcirallel and equal. • 21. V6^+82 ^ V36 + 64 = VToO = 10. 102 Chapter?, Algebra Review

22. V14V2T = V14-21 = V2-7-3-7 = 7V6.

V2^ 242 •23. = VT2T =11. T

•24. (4V5)(6V2)= 24VT0.

•25. (3^7)2 = 9-7 = 63.

•26. '-^ = 3S. 4V3 •27. (5-V2) + (5+V2) = 10.

•28. (5-V2)(5 + V2) = 25-2 = 23.

•29. V3(V27 + 1)= VsT + Vs =9 + V3.

•30. V3+(V27 + 1)=V3 + V9^+ 1 = V3+ 3V3 + 1 = 4V3 + 1.

•31. (V^+Vy) + (V^+Vy) = 2V^+2^-

•32. {4~x + ^Y = {-fxy + l{4i^) + {^Y = X + 2-yjxy + y.