27. onstruct a line ( DF ) with midpoint P parallel to and twice the length of QR. onstruct a line ( EF ) with midpoint R parallel to and twice the length of QP. onstruct a line ( DE ) with midpoint Q parallel to and twice the length of PR. The vertices of the original large triangle are D( 1, 2), E(9, 8), and F(5, 0). 2 10 D 8 6 4 2 y P 2 Q R E F 4 8 10 x Maintaining Mathematical Proficiency 28. Sample answer: A counterexample to show that the difference of two numbers is not always less than the greater of the two numbers is 6 and 2: 6 ( 2) = 8, which is not less than 6, so the original conjecture is false. 29. Sample answer: An isosceles triangle has at least two sides that are congruent. An isoseles triangle whose sides are 5 centimeters, 5 centimeters, and 3 centimeters is not equilateral. 6.5 Explorations (p. 335) 1. a. heck students work. Using the sample in the text: A 6.08, AB 4.47, B 3.61, m A 36.03, m B 97.13, m 46.85 b. heck students work. Using the sample in the text, B < AB < A and m A < m < m B. The shortest side is opposite the smallest angle, and the longest side is opposite the largest angle. c. Sample answer: A(x, y) B(x, y) (x, y) AB A B A(5, 1) B(7, 4) (2, 4) 3.61 4.24 5 A(2, 4) B(4, 2) (7, 6) 6.32 5.39 8.54 A(1, 0) B(7, 0) (1, 7) 6 7 9.22 m A m B m 78.69 56.31 45 93.37 38.99 47.64 90 49.4 40.6 If one side of a triangle is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side. Similarly, if one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle. 2. a. heck students work. Using the sample in the text: AB 3.61 A 5.10 B = 5 b. heck students work. Using the sample in the text: B = 5 < 8.71 = A + AB A = 5.10 < 8.61 = AB + B AB = 3.61 < 10.10 = B + A c. Sample answer: A(x, y) B(x, y) (x, y) AB A + B A A(5, 1) B(7, 4) (2, 4) 3.61 9.24 4.24 A(2, 4) B(4, 2) (7, 6) 6.32 13.93 5.39 A(1, 0) B(7, 0) (1, 7) 6 16.22 7 A(1, 0) B(7, 0) (5, 1) 6 6.36 4.12 AB + B B AB + A 8.61 5 7.85 14.86 8.54 11.71 15.22 9.22 13 8.24 2.24 10.12 The length of each side is less than the sum of the other two. 3. The largest angle is opposite the longest side, and the smallest angle is opposite the shortest side; The sum of any two side lengths is greater than the third side length. 4. no; The sum 3 + 4 is not greater than 10, and it is not possible to form a triangle when the sum of the lengths of the two sides is less than the length of the third side. 6.5 Monitoring Progress (pp. 336 339) 1. Given AB is a scalene triangle. Prove AB does not have two congruent angles. Assume temporarily that AB is a scalene triangle with A B. By the onverse of Base Angles Theorem (Thm. 5.7), if A B, then the opposite sides are congruent: B A. A scalene triangle cannot have two congruent sides. So, this contradicts the given information. So, the assumption that AB is a scalene triangle with two congruent angles must be false, which proves that a scalene triangle cannot have two congruent angles. 2. The sides of PQR from smallest to largest are PR, RQ, and PQ. So, by the Triangle Longer Side Theorem, the angles from smallest to largest are Q, P, and R. 3. The angles of RST from smallest to largest are R, T, and S. So, by the Triangle Larger Angle Theorem, the sides from shortest to longest are ST, RS, and RT. opyright Big Ideas Learning, LL Geometry 215
4. Let x represent the length of the third side. By the Triangle Inequality Theorem: x + 12 > 20 and 12 + 20 > x. x + 12 > 20 and 12 + 20 > x x > 8 32 > x, or x < 32 The length of the third side must be greater than 8 inches and less than 32 inches. 5. 4 + 9 > 10 13 > 10 Yes 4 + 10 > 9 14 > 9 Yes 9 + 10 > 4 19 > 4 Yes yes; The sum of any two side lengths of a triangle is greater than the length of the third side. 6. no; The sum 8 + 9 = 17 is not greater than 18. 7. no; The sum 5 + 7 = 12 is not greater than 12. 10. Draw a segment AB. Using point A as the center, draw an arc with radius AB, then draw an arc with B as the center and radius AB to intersect the first arc on both sides of AB. onstruct a segment from one arc intersection toward the other but stop at AB. Label the arc-segment intersection as and the intersection with AB as G. BG is a right scalene triangle. A G B 6.5 Exercises (pp. 340 342) Vocabulary and ore oncept heck 1. In an indirect proof, rather than proving a statement directly, you show that when the statement is false, it leads to a contradiction. 2. The longest side of a triangle is opposite the largest angle and the shortest side is opposite the smallest angle. Monitoring Progress and Modeling with Mathematics 3. Assume temporarily that WV = 7 inches. 4. Assume temporarily that xy is even. 5. Assume temporarily that B is a right angle. 6. Assume temporarily that JM is not a median. 7. A and ; The angles of an equilateral triangle are always 60. So, an equilateral triangle cannot be a right triangle. 8. B and ; If both X and Y have measures less than 30, then their sum is less than 60. Therefore, the sum of their measures cannot be 62. 9. To construct a scalene triangle, draw a segment and label it A. Ensuring that AB, B, and A are all different lengths, draw an arc with center A and radius AB and an arc with center with radius B. Where the two arcs intersect place point B. B The largest angle is GB because it is the right angle and the opposite side, B, is the longest side. The smallest angle is GB and the opposite side, GB, is the shortest side. 11. The sides of RST from smallest to largest are RT, TS, and RS. So, by the Triangle Longer Side Theorem, the angles from smallest to largest are S, R, and T. 12. The sides of JKL from smallest to largest are KL, JL, and JK. So, by the Triangle Longer Side Theorem, the angles from smallest to largest are J, K, and L. 13. The angles of AB from smallest to largest are, A, and B. So, by the Triangle Larger Angle Theorem, the sides from shortest to longest are AB, B, and A. 14. The angles of XYZ from smallest to largest are Z, X, and Y. So, by the Triangle Larger Angle Theorem, the sides from shortest to longest are XY, ZY, and ZX. 15. m M = 180 (127 + 29 ) = 24 The angles of MNP from smallest to largest are M, P, and N. So, by the Triangle Larger Angle Theorem, the sides from shortest to longest are PN, MN, and MP. 16. m D = 180 (90 + 33 ) = 57 The angles of DFG from smallest to largest are G, D, and F. So, by the Triangle Larger Angle Theorem, the sides from shortest to longest are DF, GF, and GD. A The largest angle is AB and the opposite side, A, is the longest side. The smallest angle is AB and the opposite side, AB, is the shortest side. 17. x + 5 > 12 x > 7 5 + 12 > x 17 > x or x < 17 7 inches and less than 17 inches. 216 Geometry opyright Big Ideas Learning, LL
18. x + 12 > 18 x > 6 12 + 18 > x 30 > x or x < 30 6 feet and less than 30 feet. 19. x + 24 > 40 x > 16 24 + 40 > x 64 > x or x < 64 16 inches and less than 64 inches. 20. x + 25 > 25 x > 0 25 + 25 > x 50 > x or x < 50 0 meters and less than 50 meters. 21. 6 + 7 = 13 13 > 11 Yes 7 + 11 = 18 18 > 6 Yes 11 + 6 = 17 17 > 7 Yes yes; The sum of any two side lengths of a triangle is greater than the length of the third side. 22. no; The sum 3 + 6 = 9 is not greater than 9. 23. no; The sum 28 + 17 = 45 is not greater than 46. 24. 35 + 120 = 155 155 > 125 Yes 120 + 125 = 255 255 > 35 Yes 125 + 35 = 160 160 > 120 Yes yes; The sum of any two side lengths of a triangle is greater than the length of the third side. 25. An angle that is not obtuse could be acute or right. Assume temporarily that A is not obtuse. 26. Because 30 < 60 < 90 and 1 < 3 < 2, the longest side, which is 2 units long, should be across from the largest angle, which is the right angle. 60 1 2 28. Assume temporarily that the second group has 15 or more students. Because the first group has 15 students, the total number of students in the class would be 30 students or more. Because the class has fewer than 30 students, the assumption must be false, and the second group must have fewer than 15 students. 29. ; m U = 180 (84 + 48 ) = 180 132 = 48, which indicates that UTV is isosceles. By the Triangle Longer Side Theorem, UV > TV. 30. and D; m R = 180 (65 + 56 ) = 180 121 = 59 ; By the Triangle Inequality Theorem, the order of the angles from smallest to largest is T, R, and S. The order of the sides from shortest to longest is RS < ST < RT 8 < ST < RT. ST could possibly be 9 or 10, but not 7 or 8. 31. Given An odd number Prove An odd number is not divisible by 4. Assume temporarily that an odd number is divisible by 4. Let the odd number be represented by 2y + 1 where y is a positive integer. Then there must be a positive integer x such that 4x = 2y + 1. However, when you divide each side of the equation by 4, you get x = 1 2 y + 1, which is not an integer. 4 So, the assumption must be false, and an odd number is not divisible by 4. 32. Given QRS, m Q + m R = 90 Prove m S = 90 Assume temporarily that in QRS, m Q + m R = 90 and m S 90. By the Triangle Sum Theorem (Thm. 5.1), m Q + m R + m S = 180. Using the Substitution Property of Equality, 90 + m S = 180. So, m S = 90 by the Subtraction Property of Equality, but this contradicts the given information. So, the assumption must be false, which proves that in QRS, if m Q + m R = 90, then m S = 90. 33. The right angle of a right triangle must always be the largest angle because the other two will have a sum of 90. So, according to the Triangle Larger Angle Theorem (Thm. 6.10), because the right angle is larger than either of the other angles, the side opposite the right angle, which is the hypotenuse, will always have to be longer than either of the legs. 34. yes; If the sum of the lengths of the two shortest sides is greater than the length of the longest side, then the other two inequalities will also be true. 3 30 27. Assume temporarily that your client committed the crime. Then your client had to be in Los Angeles, alifornia, at the time of the crime. Security footage shows that your client was in New York at the time of the crime. Therefore, the assumption must be false, and the client must be innocent. opyright Big Ideas Learning, LL Geometry 217
35. a. The width of the river must be greater than 35 yards and less than 50 yards. In BA, the width of the river, BA, must be less than the length of A, which is 50 yards, because the measure of the angle opposite BA is less than the measure of the angle opposite A, which must be 50. In BDA, the width of the river, BA, must be greater than the length of DA, which is 35 yards, because the measure of the angle opposite BA is greater than the measure of the angle opposite DA, which must be 40. b. You could measure from distances that are closer together. In order to do this, you would have to use angle measures that are closer to 45. 36. a. By the side length requirements for a triangle, x < 489 + 565 = 1054 kilometers and x > 565 489 = 76 kilometers. b. Because 2 is the smallest angle, the distance between Granite Peak and Fort Peck Lake must be the shortest side of the triangle. So, the second inequality becomes x < 489 kilometers. 37. WXY, Z, YXZ, WYX and XYZ, W; In WXY, because WY < WX < YX, by the Triangle Longer Side Theorem (Thm. 6.9), m WXY < m WYX < m W. Similarly, in XYZ, because XY < YZ < XZ, by the Triangle Longer Side Theorem (Thm. 6.9), m Z < m YXZ < m XYZ. Because m WYX = m XYZ and W is the only angle greater than either of them, you know that W is the largest angle. Because WXY has the largest angle and one of the congruent angles, the remaining angle, WXY, is the smallest. 38. m D + m E + m F = 180 (x + 25) + (2x 4) + 63 = 180 3x + 84 = 180 3x = 96 x = 32 m D = 32 + 25 = 57 m E = 2 32 4 = 60 m F = 63 The order of the angles from least to greatest is m D < m E < m F. The order of the sides from least to greatest is EF < DF < DE. 39. By the Exterior Angle Theorem (Thm. 5.2), m 1 = m A + m B. Then by the Subtraction Property of Equality, m 1 m B = m A. If you assume temporarily that m 1 m B, then m A 0. Because the measure of any angle in a triangle must be a positive number, the assumption must be false. So, m 1 > m B. Similarly, by the Subtraction Property of Equality, m 1 m A = m B. If you assume temporarily that m 1 m A, then m B 0. Because the measure of any angle in a triangle must be a positive number, the assumption must be false. So, m 1 > m A. 40. JK + KL > JL x + 11 + 2x + 10 > 5x 9 3x + 21 > 5x 9 2x > 30 x < 15 JK + JL > KL x + 11 + 5x 9 > 2x + 10 6x + 2 > 2x + 10 4x > 8 x > 2 KL + JL > JK 2x + 10 + 5x 9 > x + 11 7x + 1 > x + 11 6x > 10 x > 10 6 = 5 3 1.667 The possible values for x are x > 2 and x < 15. 41. UV + VT > TU 3x 1 + 2x + 3 > 6x 11 5x + 2 > 6x 11 x > 13 x < 13 TU + TV > UV 6x 11 + 2x + 3 > 3x 1 8x 8 > 3x 1 5x > 7 x > 7 5 = 1 2 5 UV + TU > TV 3x 1 + 6x 11 > 2x + 3 9x 12 > 2x + 3 7x > 15 x > 15 7 = 2 1 7 The possible values for x are x > 2 1 and x < 13. 7 42. The shortest route is along Washington Avenue. By the Triangle Inequality Theorem (Thm. 6.11), the length of Washington Avenue must be shorter than the sum of the lengths of Eighth Street and View Street, as well as the sum of the lengths of Hill Street and Seventh Street. 218 Geometry opyright Big Ideas Learning, LL
43. Given B > AB, BD = BA Prove m BA > m A B 1 2 3 D It is given that B > AB and BD = BA. By the Base Angles Theorem (Thm. 5.6), m 1 = m 2. By the Angle Addition Postulate (Post. 1.4), m BA = m 1 + m 3. So, m BA > m 1. Substituting m 2 for m 1 produces m BA > m 2. By the Exterior Angle Theorem (Thm. 5.2), m 2 = m 3 + m. So, m 2 > m. Finally, because m BA > m 2 and m 2 > m, you can conclude that m BA > m. 44. x + x > 2x > x > 1 2 Because the sum of the lengths of the legs must be greater than the length of the base, the length of a leg must be greater than 1 2. 45. no; If one side is 13 inches and the perimeter is 24 inches (2 feet), then the other two sides would total 11 inches, but they must have a sum greater than 13 inches for a triangle to exist. 46. As an example, if the 24-centimeter string is divided into 10 centimeters, 10 centimeters, and 4 centimeters, the triangle is an acute isosceles triangle. 47. Given AB Prove AB + B > A, A + B > AB, and AB + A > B B 23 D 1 A Assume B is longer than or the same length as each of the other sides, AB and A. Then AB + B > A and A + B > AB. The proof for AB + A > B follows. STATEMENTS REASONS 1. AB 1. Given 2. Extend A to D so that AB 2. Ruler Postulate AD. (Post. 1.1) 3. AB = AD 3. Definition of segment congruence 4. AD + A = D 4. Segment Addition Postulate (Post. 1.2) 5. 1 2 5. Base Angles Theorem (Thm. 5.6) 6. m 1 = m 2 6. Definition of angle congruence 7. m DB > m 2 7. Protractor Postulate (Post. 1.3) 8. m DB > m 1 8. Substitution Property 10 cm 10 cm 9. D > B 9. Triangle Larger Angle Theorem (Thm. 6.10) 10. AD + A > B 10. Substitution Property 11. AB + A > B 11. Substitution Property 4 cm For a right scalene triangle, the sides could be 6 centimeters, 8 centimeters, and 10 centimeters. 48. The perimeter of HGF must be greater than 4 and less than 24; Because of the Triangle Inequality Theorem (Thm. 6.11), FG must be greater than 2 and less than 8, GH must be greater than 1 and less than 7, and FH must be greater than 1 and less than 9. So, the perimeter must be greater than 2 + 1 + 1 = 4 and less than 8 + 7 + 9 = 24. 8 cm 10 cm 6 cm opyright Big Ideas Learning, LL Geometry 219
49. Assume temporarily that another segment, PA, where A is on plane M, is the shortest segment from P to plane M. By definition of the distance between a point and a plane, PA plane M. This contradicts the given statement because there cannot be two different segments that share an endpoint and are both perpendicular to the same plane. So, the assumption is false, and because no other segment exists that is the shortest segment from P to plane M, it must be P that is the shortest segment from P to plane M. Maintaining Mathematical Proficiency 50. The included angle between AE and BE is AEB. 51. The included angle between A and D is AD. 52. The included angle between AD and D is AD. 53. The included angle between E and BE is BE. 6.6 Explorations (p. 343) 1. a. heck students work. b. heck students work. c. heck students work. d. A D, because all points on a circle are equidistant from the center; B B by the Reflexive Property of ongruence (Thm. 2.1). e. As drawn, the length of AB is 3.6 units and the length of DB is 2.7 units, so AB > DB; m AB = 90 and m DB = 61, so m AB > m DB; yes; the results are as expected because the triangle with the longer third side has the larger angle opposite the third side. f. Sample answer: D A B AB BD m AB m BD 1. (4.75, 2.03) 2 3 3.61 2.68 90 61.13 2. (4.94, 2.5) 2 3 3.61 3.16 90 75.6 3. (5, 3) 2 3 3.61 3.61 90 90 4. (4.94, 3.5) 2 3 3.61 4 90 104.45 5. (3.85, 4.81) 2 3 3.61 4.89 90 154.93 g. If two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is larger than the included angle of the second, then the third side of the first is longer than the third side of the second. 2. If the included angle of one is larger than the included angle of the other, then the third side of the first is longer than the third side of the second. If the included angles are congruent, then you already know that the triangles are congruent by the SAS ongruence Theorem (Thm. 5.5). Therefore, the third sides are congruent because corresponding parts of congruent triangles are congruent. 3. Because the sides of the hinge do not change in length, the angle of the hinge can model the included angle and the distance between the opposite ends of the hinge can model the third side. When the hinge is open wider, the angle is larger and the ends of the hinge are farther apart. If the hinge is open less, the ends are closer together. 6.6 Monitoring Progress (pp. 345 346) 1. PR = PS Given PQ PQ Reflexive Property of ongruence (Thm. 2.1) m QPR > m QPS Given RQ > SQ Hinge Theorem (Thm. 6.12) RQ is the longer segment. 2. PR = PS Given PQ PQ Reflexive Property of ongruence (Thm. 2.1) RQ < SQ Given m RPQ < m SPQ onverse of the Hinge Theorem (Thm. 6.13) SPQ is the larger angle. 3. Assume temporarily that the third side of the first triangle with the larger included angle is not longer than the third side of the second triangle with the smaller included angle. This means the third side of the first triangle is equal to or shorter than the third side of the second triangle. 4. Group A: 135 Group B: 150 Group : 180 40 = 140 Because 135 < 140 < 150, Group is closer to the camp than Group B, but not as close as Group A. 6.6 Exercises (pp. 347 348) Vocabulary and ore oncept heck 1. Theorem 6.12 refers to two angles with two pairs of sides that have the same measure, just like two hinges whose sides are the same length. Then the angle whose measure is greater is opposite a longer side, just like the ends of a hinge are farther apart when the hinge is open wider. 2. In AB and DEF, AB DE, B EF, and A < DF. So, m E > m B by the onverse of the Hinge Theorem (Theorem 6.13). Monitoring Progress and Modeling with Mathematics 3. m 1 > m 2; By the onverse of the Hinge Theorem (Thm. 6.13), because 1 is the included angle in the triangle with the longer third side, its measure is greater than that of 2. 220 Geometry opyright Big Ideas Learning, LL