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Answers to Exercises CHAPTER 4 • CHAPTER 4 CHAPTER 4 • CHAPTER angles P and D can be drawn at each endpoint using the protractor. LESSON 4.1 Q 1. The angle measures change, but the sum remains 180°. 2. 73° 3. 60° 4. 110° 5. 24° 6. 3 360° 180° 900° 7. 3 180° 180° 360° 8. 69°; 47°; 116°; 93°; 86° 9. 30°; 50°; 82°; 28°; 32°; 78°; 118°; 50° 55⬚ P 10. Answers to Exercises ⬔R ⬔M 11. ⬔A ⬔G ⬔L 12. First construct E, using the method used in Exercise 10. 85⬚ 40⬚ 7 cm 17. The third angles of the triangles also have the same measures; are equal in measure 18. You know from the Triangle Sum Conjecture that mA mB mC 180°, and mD mE mF 180°. By the transitive property, mA mB mC mD mE mF. You also know that mA mD, and mB mE. You can substitute for mD and mE in the longer equation to get mA mB mC mA mB mF. Subtracting equal terms from both sides, you are left with mC mF. 19. For any triangle, the sum of the angle measures is 180°, by the Triangle Sum Conjecture. Since the triangle is equiangular, each angle has the same measure, say x. So x x x 180°, and x 60°. 20. false E R A 13. Fold ⬔M ⬔A 21. false E ⬔R ⬔G ⬔L A R 22. false 14. From the Triangle Sum Conjecture mA mS mM 180°. Because M is a right angle, mM 90°. By substitution, mA mS 90° 180°. By subtraction, mA mS 90°. So two wrongs make a right! 15. Answers will vary. See the proof on page 202. To prove the Triangle Sum Conjecture, the Linear Pair Conjecture and the Alternate Interior Angles Conjecture must be accepted as true. 16. It is easier to draw PDQ if the Triangle Sum Conjecture is used to find that the measure of can be drawn to be 7 cm, and D is 85°. Then PD 48 ANSWERS TO EXERCISES D 23. false 24. true 25. eight; 100 LESSON 4.2 F D E C A B 16. P M N K G H 17. possible answer: Fold 1 Fold 3 Fold 2 Fold 4 105⬚ 60⬚ 45⬚ 18. 19. 20. 21. 22. 23. 0 perpendicular parallel parallel neither parallelogram 40 8 16 24 32 40 24. New: (6, 3), (2, 5), (3, 0). Triangles are congruent. 25. New: (3, 3), (3, 1), (1, 5). Triangles are congruent. ANSWERS TO EXERCISES 49 Answers to Exercises 1. 79° 2. 54° 3. 107.5° 4. 44°; 35 cm 5. 76°; 3.5 cm 6. 72°; 36°; 8.6 cm 7. 78°; 93 cm 8. 75 m; 81° 9. 160 in.; 126° 10. a 124°, b 56°, c 56°, d 38°, e 38°, f 76°, g 66°, h 104°, k 76°, n 86°, p 38°; Possible explanation: The angles with measures 66° and d form a triangle with the angle with measure e and its adjacent angle. Because d, e, and the adjacent angle are all congruent, 3d 66° 180°. Solve to get d 38°. This is also the measure of one of the base angles of the isosceles triangle with vertex angle measure h. Using the Isosceles Triangle Conjecture, the other base angle measures d, so 2d h 180°, or 76° h 180°. Therefore, h 104°. 11. a 36°, b 36°, c 72°, d 108°, e 36°; none 12a. Yes. Two sides are radii of a circle. Radii must be congruent; therefore, each triangle must be isosceles. 12b. 60° 13. NCA 14. IEC 15. USING YOUR ALGEBRA SKILLS 4 1. 3. 5. 7. false not a solution not a solution y 4 true solution x7 x 8 1 10. n 2 9. x 4.2 Answers to Exercises 2. 4. 6. 8. 11. x 2 12. t 18 9 2 14a. x 4 13. n 5 14b. x 94; The two methods produce identical results. Multiplying by the lowest common denominator (which is comprised of the factors of both denominators) and then reducing common factors (which clears the denominators on either side) is the same as simply multiplying each numerator by the opposite denominator (or cross multiplying). Algebraically you could show that the two methods are equivalent as follows: c a b d c a bd b bd d abd bcd b d ad bc The method of “clearing fractions” results in the method of “cross multiplying.” 50 ANSWERS TO EXERCISES 15. You get an equation that is always false, so there is no solution to the equation. 16. Camella is not correct. Because the equation 0 0 is always true, the truth of the equation does not depend on the value of x. Therefore, x can be any real number. Camella’s answer, x 0, is only one of infinitely many solutions. 17. 2x x 2x If x equals the measure of the vertex angle, then the base angles each measure 2x. Applying the Triangle Sum Conjecture results in the following equation: x 2x 2x 180° 5x 180° x 36° The measure of the vertex angle is 36° and the measure of each base angle is 72°. LESSON 4.3 1. yes 2. no 4 5 9 3. no 5 6 12 Answers to Exercises 4. yes 5. a, b, c 6. c, b, a 7. b, a, c 8. a, c, b 9. a, b, c 10. v, z, y, w, x 11. 6 length 102 12. By the Triangle Inequality Conjecture, the sum of 11 cm and 25 cm should be greater than 48 cm. 13. b 55°, but 55° 130° 180°, which is impossible by the Triangle Sum Conjecture. 14. 135° 15. 72° 16. 72° 17. a b c 180° and x c 180°. Subtract c from both sides of both equations to get x 180 c and a b 180 c. Substitute a b for 180 c in the first equation to get x a b. 18. 45° 19. a 52°, b 38°, c 110°, d 35° 20. a 90°, b 68°, c 112°, d 112°, e 68°, f 56°, g 124°, h 124° 21. By the Triangle Sum Conjecture, the third angle must measure 36° in the small triangle, but it measures 32° in the large triangle. These are the same angle, so they can’t have different measures. 22. ABE 23. FNK 24. cannot be determined ANSWERS TO EXERCISES 51 13. Cannot be determined. SSA is not a congruence conjecture. 14. AIN by SSS or SAS 15. Cannot be determined. Parts do not correspond. 16. SAO by SAS 17. Cannot be determined. Parts do not correspond. 18. RAY by SAS and PR is (0, 0). 19. The midpoint of SD Therefore, DRO SPO by SAS. 20. Because the LEV is marking out two triangles that are congruent by SAS, measuring the length of the segment leading to the finish will also approximate the distance across the crater. 21. 22. LESSON 4.4 1. Answers will vary. Possible answer: If three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent (all corresponding angles are also congruent). 2. Answers will vary. Possible answer: The picture statement means that if two sides of one triangle are congruent to two sides of another triangle, and the angles between those sides are also congruent, then the two triangles are congruent. If you know this: then you also know this: Answers to Exercises 3. Answers will vary. Possible answer: 23. a 37°, b 143°, c 37°, d 58°, e 37°, f 53°, g 48°, h 84°, k 96°, m 26°, p 69°, r 111°, s 69°; Possible explanation: The angle with measure h is the vertex angle of an isosceles triangle with base angles measuring 48°, so h 2(48) 180°, and h 84°. The angle with measure s and the angle with measure p are corresponding angles formed by parallel lines, so s p 69°. 24. 3 cm third side 19 cm 25. See table below. 13 3 26a. y 6 b. y c. y 4x 2 3 27. (5, 3) 4. SAS 5. SSS 6. cannot be determined 7. SSS 8. SAS 9. SSS (and the Converse of the Isosceles Triangle Conjecture) 10. yes, ABC ADE by SAS 11. Possible answer: Boards nailed diagonally in the corners of the gate form triangles in those corners. Triangles are rigid, so the triangles in the gate’s corners will increase the stability of those corners and keep them from changing shape. 12. FLE by SSS 25. (Lesson 4.4) Side length 1 2 3 4 5 Elbows 4 4 4 4 4 T’s 0 4 8 12 16 Crosses 0 1 4 9 16 … 4n 4 52 ANSWERS TO EXERCISES n … 4 20 4 76 361 (n 1)2 LESSON 4.5 1. If two angles and the included side of one triangle are congruent to the corresponding side and angles of another triangle, then the triangles are congruent. 2. If two angles and a non-included side of one triangle are congruent to the corresponding side and angles of another triangle, then the triangles are congruent. If you know this: then you also know this: 3. Answers will vary. Possible answer: L K M 27. 28a. about 100 km southeast of San Francisco 28b. Yes. No, two towns would narrow it down to two locations. The third circle narrows it down to one. ORE GO N Boise IDAH Eureka Sacramento San Francisco Reno NEV O Elko ADA U TA H LI FO RN Las Vegas IA Los Angeles Miles 0 50 100 0 100 Kilometers 200 ARIZ ONA 400 ANSWERS TO EXERCISES 53 Answers to Exercises 22. Construction will show a similar but larger (or smaller) triangle constructed from a drawn triangle by duplicating two angles on either end of a new side that is not congruent to the corresponding side. 23. Draw a line segment. Construct a perpendicular. Bisect the right angle. Construct a triangle with two congruent sides and with a vertex that measures 135°. 24. 125 25. False. One possible counterexample is a kite. 26. None. One triangle is determined by SAS. CA 4. ASA 5. cannot be determined 6. SAA 7. cannot be determined 8. ASA 9. cannot be determined 10. FED by SSS 11. WTA by ASA or SAA 12. SAT by SAS 13. PRN by ASA or SAS; SRE by ASA 14. Cannot be determined. Parts do not correspond. 15. MRA by SAS 16. Cannot be determined.AAA does not guarantee congruence. 17. WKL by ASA 18. Yes, ABC ADE by SAA or ASA. slope CD 3 and slope 19. Slope AB BC 1 3, so AB BC , CD DA , and slope DA DA . ABC CDA by SAA. BC 20. 21. The construction is the same as the construction using ASA once you find the third angle, which is used here. (Finding the third angle is not shown.) 13. cannot be determined 14. KEI by ASA 15. UTE by SAS 16. Answers to Exercises LESSON 4.6 BD (same segment), A C 1. Yes. BD (given), and ABD CBD (given), so DBA CB by CPCTC. DBC by SAA. AB WN and C W (given), and 2. Yes. CN RNC ONW (vertical angles), CNR ON by CPCTC. WON by ASA. RN 3. Cannot be determined. The congruent parts lead to the ambiguous case SSA. IT 4. Yes. S I, G A (given), and TS (definition of midpoint), so TIA TSG by IA by CPCTC. SAA. SG FR and UO UR (given), and UF 5. Yes. FO (same segment), so FOU FRU by SSS. UF O R by CPCTC. MA and ME MR (given), and 6. Yes. MN M M (same angle), so EMA RMN by SAS. E R by CPCTC. EU and BU ET (given), and 7. Yes. BT UT UT (same segment), so TUB UTE by SSS. B E by CPCTC. 8. Cannot be determined. HLF LHA by and HF are not corresponding sides. ASA, but HA 9. Cannot be determined. AAA does not guarantee congruence. 10. Yes. The triangles are congruent by SAS. 11. Yes. The triangles are congruent by SAS, and the angles are congruent by CPCTC. and DF to form ABC and DEF. 12. Draw AC because all were drawn with AB CB DE FE DF for the same reason. the same radius. AC ABC DEF by SSS. Therefore, B E by CPCTC. 17. 18. a 112°, b 68°, c 44°, d 44°, e 136°, f 68°, g 68°, h 56°, k 68°, l 56°, m 124°; Possible explanation: f and g are measures of base angles of an isosceles triangle, so f g. The vertex angle measure is 44°, so subtract 44° from 180° and divide by 2 to get f 68°. The angle with measure m is the exterior angle of a triangle. Add the remote interior angle measures 56° and 68° to get m 124°. 19. ASA. The “long segment in the sand” is a shared side of both triangles 20. (4, 1) 21. See table below. 22. Value C is always decreasing. 23. x 3, y 10 21. (Lesson 4.6) 54 Number of sides 3 4 5 6 7 … 12 … n Number of struts needed to make polygon rigid 0 1 2 3 4 … 9 … n3 ANSWERS TO EXERCISES 3. See flowchart below. 4. See flowchart below. LESSON 4.7 1. See flowchart below. 2. See flowchart below. 1. (Lesson 4.7) 1 SE SU }? 2 ESM USO 4 E U }? 3 Given Given } } 5 ? ? MS SO }? CPCTC ASA Congruence Conjecture 1 2 }? Vertical Angles Conjecture 2. (Lesson 4.7) 1 3 CI IM Definition of midpoint Given 2 Answers to Exercises I is midpoint of CM 4 I is midpoint of BL 6 IL IB of }? Definition midpoint }? Given } } ? ? MB }? CL 7 }? CIL MIB CPCTC by SAS 5 1 2 }? Vertical Angles Conjecture 3. (Lesson 4.7) 1 3 NS is a median Given 2 Same segment 4 S is a midpoint Definition of median 6 WS SE ESN WSN ? } } ? SSS Definition of midpoint 5 7 E W ? } ? } CPCTC WN NE }? Given 4. (Lesson 4.7) 1 NS is an NS NS 2 angle bisector } 1 ? } 2 Definition of ? angle bisector Given 3 W E } ? Given 4 5 WNS ENS ? ? } } ? SAA } 6 WN NE } ? CPCTC 7 NEW is isosceles of }? Definition isosceles triangle NS NS }? Same segment ANSWERS TO EXERCISES 55 5. See flowchart below. BC , CD AD 6. Given: ABC with BA is the angle bisector of ABC. Show: BD 12. ACK by SSS 13. a 72°, b 36°, c 144°, d 36°, e 144°, f 18°, g 162°, h 144°, j 36°, k 54°, m 126° 14. The circumcenter is equidistant from all three vertices because it is on the perpendicular bisector of each side. Every point on the perpendicular bisector of a segment is equidistant from the endpoints. Similarly, the incenter is equidistant from all three sides because it is on the angle bisector of each angle, and every point on an angle bisector is equidistant from the sides of the angle. 15. ASA. The fishing pole forms the side. “Perpendicular to the ground” forms one angle. “Same angle on her line of sight” forms the other angle. 2 16. 7 17. A D 1 2 B C BA ⬵ BC CD ⬵ AD BD ⬵ BD Given Given Same segment 䉭ABD ⬵ 䉭CBD SSS ⬔1 ⬵ ⬔2 CPCTC Answers to Exercises → BD bisects ⬔ABC Definition of angle bisector 7. The angle bisector does not go to the midpoint of the opposite side in every triangle, only in an isosceles triangle. , because it is across from the smallest angle 8. NE , which is across in NAE. It is shorter than AE from the smallest angle in LAE. 9. The triangles are congruent by SSS, so the two central angles cannot have different measures. 10. PRN by ASA; SRE by ASA 11. Cannot be determined. Parts do not correspond. 18. y 4 Y X B O B' Y'4 O' x X' 5. (Lesson 4.7) 1 SA NE 3 ? Given ᎏ 2 SE NA 3 4 AIA Conjecture ESN ANS 4 ? 1 2 ᎏ ? AIA Conjecture ᎏ ? Given ᎏ 5 SN SN Same segment 56 ANSWERS TO EXERCISES 6 ? ᎏ ? ᎏ ? ASA ᎏ 7 ? SA NE ᎏ ? CPCTC ᎏ This proof shows that in a parallelogram, opposite sides are congruent. 7. LESSON 4.8 1. 2. 3. 4. 5. 6. 6 90°; 18° 45° See flowchart below. See flowchart below. 1 Isosceles 䉭ABC with AC ⬵ BC and CD bisects ⬔C 2 AD ⬵ BD AC ⬵ BC 2 Given Same segment D is midpoint of AB ⬔ACD ⬵ ⬔DCB CPCTC 䉭ADC ⬵ 䉭BDC AD ⬵ BD Def. of midpoint 7 CD is angle bisector of ⬔ACB Def. of angle bisector 8 CD is altitude of 䉭ABC Conjecture B (Exercise 5) CPCTC 4 䉭ADC ⬵ 䉭BDC SSS 6 Conjecture A (Exercise 4) 4 3 CD ⬵ CD Given 5 Given 3 1 9 CD ⬜ AB Def. of altitude 8. Yes. First show that the three exterior triangles are congruent by SAS. CD is a median Def. of median Answers to Exercises 4. (Lesson 4.8) 1 } 2 ? CD is the bisector of C Given 1 2 Definition of angle bisector 3 ABC is isosceles with AC BC 5 }? SAS Given 4 ADC BDC CD CD Same segment 5. (Lesson 4.8) 1 ABC is isosceles with AC BC, and CD is the bisector of C 2 ADC BDC 4 }? CPCTC Conjecture A Given 3 1 and 2 form a linear pair Definition of linear pair 1 2 5 6 Congruent supplementary angles are 90 1 and 2 are supplementary Linear Pair Conjecture 1 and 2 are right angles 7 AB CD } Definition of ? perpendicular } is an altitude 8 ? CD Definition of altitude ANSWERS TO EXERCISES 57 9. C A B D Drawing the vertex angle bisector as an auxiliary segment, we have two triangles. We can show them to be congruent by SAS, as we did in Exercise 4. Then, A B, by CPCTC. Therefore, base angles of an isosceles triangle are congruent. 10. The proof is similar to the one on page 245, but in reverse, and using the Converse of the Isosceles Triangle Conjecture. 11. 5 14. y 3x 16 15. 120 16. (4, 6) or (4, 0) or any point at which the x-coordinate is either 1 or 7 and the y-coordinate does not equal 3 17. Hugo and Duane can locate the site of the fireworks by creating a diagram using SSS. Fireworks 340 m/s • 3 s = 1.02 km 340 m/s • 5 s = 1.7 km Hugo 1.5 km Duane 18. Cn H2n H H H 30⬚ H Answers to Exercises H 12. a 128°, b 128°, c 52°, d 76°, e 104°, f 104°, g 76°, h 52°, j 70°, k 70°, l 40°, m 110°, n 58° 13. between 16 and 17 minutes 58 ANSWERS TO EXERCISES H H C C C C C H H C C H H H H H CHAPTER 4 REVIEW P ⬔L A y ⬔A ⬔P L 34. Construct P. Mark off the length PB on one ray. From point B, mark off the two segments that intersect the other ray of P at distance x. S2 S1 x x P z B ANSWERS TO EXERCISES 59 Answers to Exercises 1. Their rigidity gives strength. 2. The Triangle Sum Conjecture states that the sum of the measures of the angles in every triangle is 180°. Possible answers: It applies to all triangles; many other conjectures rely on it. 3. The angle bisector of the vertex angle is also the median and the altitude. 4. The distance between A and B is along the segment connecting them. The distance from A to C to B can’t be shorter than the distance from A to B. Therefore, AC CB AB. Points A, B, and C form a triangle. Therefore, the sum of the lengths of any two sides is greater than the length of the third side. 5. SSS, SAS, ASA, or SAA 6. In some cases, two different triangles can be constructed using the same two sides and non-included angle. 7. cannot be determined 8. ZAP by SAA 9. OSU by SSS 10. cannot be determined 11. APR by SAS 12. NGI by SAS 13. cannot be determined 14. DCE by SAA or ASA 15. RBO or OBR by SAS UT by 16. AMD UMT by SAS, AD CPCTC 17. cannot be determined 18. cannot be determined AL by 19. TRI ALS by SAA, TR CPCTC KV by 20. SVE NIK by SSS, EL overlapping segments property 21. cannot be determined 22. cannot be determined 23. LAZ IAR by ASA, LRI IZL by ASA, and LRD IZD by ASA 24. yes. PTS TPO by ASA or SAA 25. ANG is isosceles, so A G. However, the sum of mA mN mG 188°. The measures of the three angles of a triangle must sum to 180°. 26. ROW NOG by ASA, implying that OG . However, the two segments shown are OW not equal in measure. 27. a g e d b f c. Thus, c is the longest segment, and a and g are the shortest. 28. x 20° 29. Yes. TRE SAE by SAA, so sides are congruent by CPCTC. 30. Yes. FRM RFA by SAA. RFM FRA by CPCTC. Because base angles are congruent, FRD is isosceles. 31. x 48° 32. The legs form two triangles that are congruent by SAS. Because alternate interior angles are congruent by CPCTC, the seat must be parallel to the floor. 33. Construct P and A to be adjacent. The angle that forms a linear pair with the conjunction of P and A is L. Construct A. Mark off the length AL on one ray. Construct L. Extend the unconnected sides of the angles until they meet. Label the point of intersection P. 37. Possible method: Construct an equilateral triangle and bisect one angle to obtain 30°. Adjacent to that angle, construct a right angle and bisect it to obtain 45°. 38. d, a b, c, e, f Answers to Exercises 35. See flowchart below. 36. Given three sides, only one triangle is possible; therefore, the shelves on the right hold their shape. The shelves on the left have no triangles and move freely as a parallelogram. 35. (Chapter 4 Review) TMI RME 2 } } } ? ? Vertical angles ? ME TMI EMR TM 3 ? 5 M is midpoint ? ? ? and IR of TE ? Definition ? SAS 1 ? of midpoint ? Given 4 ? ? MR IM ? Definition of midpoint } } 60 } } } } } } ANSWERS TO EXERCISES } } } T E or R I 6 ? ? } } ? } CPCTC RE TI 7 }? }? of }? Converse AIA Conjecture