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Document 1806979
Chapter 5, continued Technology Activity 5.4 (pp. 326–327) 7. The orthocenter and circumcenter are sometimes outside the triangle. Both lie outside the triangle only when it is obtuse. The incenter and centroid never lie outside the triangle. The incenter is always in the interior of the three angles of the triangle so it is always inside the triangle. The centroid is always two-thirds of the distance from each vertex to the midpoint of the opposite side, a point that is always inside the triangle. 1. F C B E A D 8. C G D Yes, points D, E, and F are collinear. E 2. A B F F C B E Both the incenter and the centroid are contained by the triangle formed by the midsegments. They remain contained when the shape of the triangle is changed. D A Yes, points D, E, and F remain collinear. 3. 5.5 Guided Practice (pp. 329–330) } } } 1. ST, RS, and RT, because the side opposite the smallest } } angle is ST and the side opposite the largest angle is RT. F 2. C B 378 E A 8 ft D 10 ft 538 6 ft 4. G is not collinear with the three collinear points D, E, and F. 3. 5. 15 in. F G A 11 in. x in. 11 in. 15 in. x in. E C B D 11 1 15 > x 26 > x x 1 11 > 15 x>4 The length of the third side must be greater than 4 inches and less than 26 inches. D, E, and F remain collinear. G is collinear with D, E, and F only when the triangle is isosceles. 6. C D E A B F Yes, D, E, and F remain collinear no matter how the triangle’s vertices are moved. 140 Geometry Worked-Out Solution Key 5.5 Exercises (pp. 331–334) Skill Practice } 1. BC is opposite A. } AC is opposite B. } AB is opposite C. 2. By Theorem 5.11, you can tell that the longest side is opposite the largest angle and the shortest side is opposite the smallest angle. Copyright © by McDougal Littell, a division of Houghton Mifflin Company. G Lesson 5.5 Chapter 5, continued 3. 18. Yes, using the Triangle Inequality Theorem: 28 1 34 > 39 Blue d Re 62 > 39 Red Blue The longest side is opposite the largest angle and the shortest side is opposite the smallest angle. 4. ue Bl Bl Red Blue The longest side is opposite the obtuse angle and the two congruent legs are opposite the two congruent angles. } } } } } } 6. AB, BC, AC 7. XY, YZ, XZ C, A, B } } } 8. RT, ST, RS Z, X, Y } } } 9. KL, JL, JK S, R, T J, K, L Choice B: 21. x 1 5 > 12 M, P, N x>7 11. mD 5 1808 2 (908 1 338) 5 578 DF, FG, DG G, D, F 12. C; mR 5 1808 2 (658 1 568) 5 598 Because mR > mT, ST > RS. Therefore, ST > 8. 1238 3m 9m 13 > 9 17 > x x>1 7>x The length of the third side must be greater than 1 meter and less than 7 meters. 12 1 18 > x 30 > x The length of the third side must be greater than 6 feet and less than 30 feet. 24. x 1 10 > 23 12 ft x > 13 718 198 17 > 15 314>x x>6 418 35 ft 5 1 8 >9 5 1 12 > x 22. x 1 3 > 4 23. x 1 12 > 18 7m 9 1 8 >5 The length of the third side must be greater than 7 inches and less than 17 inches. } } } Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 13ò15 14 > 8 NP, MN, MP 10 1 23 > x 33 > x The length of the third side must be greater than 13 yards and less than 33 yards. 37 ft 718 9 ft, 4 ft, 15 ft 9 1 4 > 15 9 ft, 5 ft, 8 ft 9 1 5 >8 } } } 15. 245 > 35 20. B; 3 yd 5 9 ft 10. mm 5 1808 2 (1278 1 298) 5 248 14. 155 > 125 3 ft 5 1 yd Blue 168 35 1 120 > 125 Choice A: ue Red 13. 73 > 28 120 1 125 > 35 The longest side is opposite the 908 angle and the shortest side is the leg opposite the smallest angle. 5. 67 > 34 19. Yes, using the Triangle Inequality Theorem: 160 > 120 Red Blue 34 1 39 > 28 35 1 125 > 120 Blue d Re 28 1 39 > 34 11 in. 25. 2 ft 5 24 in. 618 24 inches, 40 inches 13 in. 488 14 in. 16. Yes, using the Triangle Inequality Theorem: 6 1 7 > 11 13 > 11 6 1 11 >7 7 1 11 >6 17 > 7 17. No, using the Triangle Inequality Theorem: 3 1 6 >9 18 > 6 x 1 24 > 40 24 1 40 > x x > 16 64 > x The length of the third side must be greater than 16 inches and less than 64 inches. 26. x 1 25 > 25 x>0 25 1 25 > x 50 > x The length of the third side must be less than 50 meters. 9ò9 Geometry Worked-Out Solution Key 141 Chapter 5, continued 27. By the Exterior Angle Theorem, m1 5 mA 1 mB. 34. (6x 2 11) 1 (3x 2 1) > 2x 1 3 Therefore, m1 > mA and m1 > mB. 9x 2 12 > 2x 1 3 28. E is an exterior angle of nDEG and D is a 7x > 15 nonadjacent interior angle. Therefore, mE > mD. 15 x>} 7 29. By Theorem 5.10, QNP must be the largest angle of nNPQ because it is opposite the longest side. Because QNP and LNM are vertical, mQNP 5 mLNM 5 598. Then by the Triangle Sum Theorem, 598 1 mP 1 mQ 5 1808; so mP 1 mQ 5 1218. Then 1218 > 2mQ, mQq60.58 by Theorem 5.10. This contradicts the fact that QNP must be the largest angle of nNPQ. Therefore, the diagram must be incorrect. 30. The hypotenuse of a right triangle is the side opposite the right angle. The right angle is the largest angle of a right triangle. Therefore, by Theorem 5.11 the hypotenuse must always be longer than either leg. 31. PQ 1 QR > PR Ï58 1 2Ï13 > 5Ï 2 } 13 > x The value of x must satisfy all three inequalities, so x 15 and less than 13. must be greater than } 7 Because mWYX 5 mXYZ, mWXY 1 mW 5 mZ 1 mYXZ. Then, because mW > mYXZ, mZ > mWXY. Ï58 1 5Ï2 > 2Ï 13 } So, WXY is the smallest angle. 14.7 > 7.2 Yes, it is possible to build a triangle with the given side lengths. } PQ 5 Ï58 ø 7.6 From smallest to greatest, the angles are: WXY, Z, YXZ, WYX > XYZ, W 36. In nGJH: x13>4 } QR 5 2Ï13 ø 7.2 } PR 5 5Ï2 ø 7.1 From the least to greatest measure, the angles are Q, P, and R. 32. ST 1 TU > SU } } Ï29 1 2Ï17 > 13.9 13.6 ò13.9 No, it is not possible to build a triangle with the given side lengths. 33. (x 1 11) 1 (2x 1 10) > 5x 2 9 3x 1 21 > 5x 2 9 30 > 2x 15 > x (x 1 11) 1 (5x 2 9) > 2x 1 10 6x 1 2 > 2x 1 10 4x > 8 x>2 (2x 1 10) 1 (5x 2 9) > x 1 11 7x 1 1 > x 1 11 6x > 10 5 x > }3 The value of x must satisfy all three inequalities, so x must be greater than 2 and less than 15. 142 5x 1 2 > 6x 2 11 nXYZ: Z, YXZ, XYZ } } (3x 2 1) 1 (2x 1 3) > 6x 2 11 nWXY: WXY, WYX, W 14.3 > 7.6 PQ 1 PR > QR } 7 x > }5 triangle, the angles are as follows. > Ï 58 5Ï2 1 2Ï 13 } 5x > 7 35. By Theorem 5.10, from the smallest to largest in each } 14.8 > 7.1 PR 1 QR > PQ } 8x 2 8 > 3x 2 1 Geometry Worked-Out Solution Key 314>x x>1 7>x } The length of GH must be greater than 1 and less than 7. In nFJH: y14>5 415>y y>1 9>y } The length of FH must be greater than 1 and less than 9. In nFJG: z13>5 315>z z>2 8>z } The length of FG must be greater than 2 and less than 8. P5x1y1z P>11112 P>4 71918>P 24 > P So, the perimeter of nHGF must be between 4 and 24. Problem Solving 37. mR > mQPR and mQ > mQPR by Theorem 5.10. mR 5 mQ by the Base Angles Theorem. Copyright © by McDougal Littell, a division of Houghton Mifflin Company. } (6x 2 11) 1 (2x 1 3) > 3x 2 1 Chapter 5, continued 38. You know that AB is greater than 35 yards and less than 43. Sample answer: 50 yards. To get a closer estimate, you could sighting point C closer to A and sighting point D farther from A so the two angles are closer to 458. 39. a. By the Triangle Inequality Theorem, you know that 489 km 1 565 km > x km, or 1054 km > x km. Therefore, x cannot be 1080. 5 cm, 5 cm, 14 cm 2 cm, 9 cm, 13 cm 6 cm, 6 cm, 12 cm 44. x 1 359 > 709 359 1 709 > x x > 350 1068 > x b. No, by the Triangle Inequality Theorem, you know that 489 km 1 x km > 565 km, or x km > 76 km. Therefore, x cannot be 40. c. x 1 489 > 565 The range of possible distances you might have to walk is greater than 350 yards and less than 1068 yards. 45. 2 mi 489 1 565 > x x > 76 d. Because 2 is the smallest angle, the distance must be less than 489 kilometers. 3 3 1 2 }4 > x Special cases for collinear locations: 2 mi. House 3 4 Grocery Store mi. Grocery Store mi. Library The distance from your house to the grocery store is Try the Pythagorean Theorem: 72 1 72 0 102 1 greater than or equal to 1 }4 miles and less than or equal 98 Þ 100 3 The longest side is too long to satisfy the Pythagorean Theorem, so the angle is obtuse. So, the triangle is obtuse. 6.5cm 9 cm 6.5cm 8 cm 8 cm 1cm 8 cm 6 cm Obtuse isosceles Acute isosceles 42. Sample answer: 4cm 9cm 8 cm 7cm Obtuse triangle to 2 }4 miles. 46. The congruent sides are shorter than the base in an isosceles triangle when the measure of the angle opposite the base is greater than 608. The congruent sides are longer than the base when the measure of the angle opposite the base is less than 608. 47. Given: n ABC B Prove: (1) AB 1 BC > AC 9cm 10 cm 3 4 Library 7 cm 11 cm mi 3 4 House 10 cm 3 4 }12>x x > 1 }4 41. Sample answer: Acute isosceles 2 mi 2 mi. greater than the longest side length, then the side lengths form a triangle. 9 cm mi x 1 }4 > 2 40. Yes. If the sum of the shortest two side lengths is Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 3 4 x mi 1054 > x The range of possible distances from Granite Peak to Fort Peck Lake is between 76 kilometers and 1054 kilometers. 7 cm x mi Acute triangle 2 3 (2) AC 1 BC > AB (3) AB 1 AC > BC 1 D A C } One side, say BC, is longer than or at least as long as each of the other sides. Then (1) and (2) are true. 8 cm 6 cm 2 6 1 82 0 102 100 5 100 This is a right triangle. Geometry Worked-Out Solution Key 143 Chapter 5, continued Proof of (3): 1. n ABC 1. Given Statements } 1. AX, BY, and CZ are medians of n ABC. } 2. Extend AC to D so } } that AB > AD. 2. Ruler Postulate 2. AM 5 }3 AX, 3. AD 1 AC 5 DC 3. Segment Addition Postulate Reasons Reasons }} 1. Given 2 2. Concurrency of Medians of a Triangle Theorem 2 CM 5 }3 CZ, 2 BM 5 }3 BY 4. 1 > 2 4. Base Angles Theorem 5. mDBC > m2 5. Protractor Postulate 6. mDBC > m1 6. Substitution 7. DC > BC 7. 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. 8. AD 1 AC > BC 3. AM 1 BM > AB, AM 1 CM > AC, BM 1 CM > BC 2 2 3 3 2 2 } BY 1 } CZ > BC 3 3 4 AB 1 AC 1 BC A 6. AX 1 BY 1 CZ > 6. Multiplication Property of Inequality 7. AX 1 BY 1 CZ > 7. Rewrite right side as a sum. 1 } (AB 1 AC 1 BC) 1 2 1 } (AB 1 AC 1 BC) 4 of n ABC. 1 Statements 5. If a > b and c > d, then a 1 c > b 1 d. 5. }3 (AX 1 BY 1 CZ) > 3 } (AB 1 AC 1 BC) 4 48. a. Given: AD is a median B D C 8. If a > b and c is positive, then a > b 2 c. 8. AX 1 BY 1 CZ > 1 }(AB 1 AC 1 BC) 2 Reasons } 4. Substitution } AX 1 } CZ > AC, 9. Substitution Prove: AD < }2 (AB 1 BC 1 AC) 2 2 4. }3 AX 1 }3 BY > AB, 8. Substitution 9. AB 1 AC > BC 3. Triangle Inequality Theorem 1. AD is a median of n ABC. 1. Given 2. BD 1 DC 5 BC 2. Segment Addition Postulate 3. AB 1 BD > AD AC 1 DC > AD 3. Triangle Inequality Theorem 4. AB 1 AC 1 BD 1 DC > AD 1 AD 4. If a > b and c > d, then a 1 c > b 1 d. 5. AB 1 AC 1 BC > AD 1 AD 5. Substitution Property 6. AB 1 AC 1 BC > 2AD 6. Simplify. 2. By the Converse of the Hinge Theorem, SPQ is larger 7. Division Property of Inequality 3. Mixed Review for TAKS 1 7. }2 (AB 1 AC 1 BC) > AD } } 49. A; P(star tile and then blue tile) 5 P(star tile) + P(blue tile) 3 than RPQ. Group C 1.2 mi 40 } 1 140 N A Group A 2 mi M Z X Geometry Worked-Out Solution Key 135 2 mi B 1.2 mi Group B 144 1 5.6 Guided Practice (pp. 336–338) } } 1. By the Hinge Theorem, RQ is longer than SQ. Prove: AX 1 BY 1 CZ > }2 (AB 1 BC 1 AC) C 3 Lesson 5.6 b. Given: AX, BY, and CZ are medians of n ABC. Y 1 5} 5 }6 + }6 5 } 36 12 150 2 mi 1.2 mi Copyright © by McDougal Littell, a division of Houghton Mifflin Company. Statements