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Answers to Geometry Unit 1 Practice7. BLesson 1-18. m P 5 105 , m Q 5 75 1. a. angle; T, STQ, QTS9. Four angles appear to be obtuse. They are PMS(or SMP), PNV (or MNV, VNM, VNP), RMN (or RMQ, NMR, QMR), and TNQ(or QNT).b. line segment; AC , CAc. line; line m, DF , FDd. ray; PL , PK , PHe. plane; plane TPS, plane TSP, plane PTS, planePST, plane STP, plane SPT, plane G10. a. They are both line segments and they both havean endpoint on the circle.b. A chord has both endpoints on the circle.A radius has only one endpoint on the circle.2. C3. three rays; YW , XY , XZ4. 6 angles: XTY (or YTX), YTZ (or ZTY), ZTW (or WTZ), XTZ (or ZTX), YTW(or WTY), XTW (or WTX)5. a.b.Lesson 2-111. a. 32, 37b. 62, 87SRc. 38, 47VTc.d. 7,500; 37,500e. 25, 36XWd.12. Answers may vary. Sample answers: 37, 38, 39, 40,41; 10, 20, 30, 40, 50TD13. a.Ee.CDEb. 15, 21, 28c. Sample answer: The first term is 1. For thesecond term, add 2. For the third term, add 3.For each subsequent term, add 1 more.Lesson 1-26. a. Four radii are shown. They are PA (or AP ),PB (or BP ), PC (or CP ), and PD (or DP ).d. 1, 2, 3, 4, . . . The pattern is “for the nth term,add the number n to the previous term.”b. Two diameters are shown. They are AC (or CA)and BD (or DB).14. Cc. Answers may vary. Sample answer: They aresimilar in that they are both line segments andthey both have two endpoints on the circle.They are different in that a diameter always goesthrough the center of the circle but a chord doesnot have to go through the center of the circle.15. Sample answer: Rules B and D can be used to findthe second term of the sequence but not any otherterms. Rule A does give the correct second term.Therefore, Rules A, B, and D do not correctlydescribe the sequence.d. The intersection of the diameters is the center ofthe circle. 2015 College Board. All rights reserved.A1SpringBoard Geometry, Unit 1 Practice
25. a. Sample answer: A point is a position.Lesson 2-2b. Sample answer: An angle bisector is a ray thatbisects an angle.16. Use 2p and 2q to represent two even integers.Then (2p)(2q) 5 2(2pq).We know that the expression 2pq represents aninteger because when you find the product of twoor more integers, the result is also an integer. So theexpression “2(2pq)” is an even integer because it is2 times an integer.Lesson 3-226. a. If I wake up early, then I set my alarm clock.b. If I eat breakfast at a restaurant, then it is aweekend.17. Dc. If an angle is obtuse, then its measure is between90 and 180 .18. Sample answer: 2 x 1 1 , 72x , 6 Subtraction Propertyof Inequalityx , 3 Division Property ofInequality27. a. A counterexample is x 5 25.b. Two counterexamples are a line with A betweenB and C and a line with C between A and B.5 is not less than 3 because 5 is to the right of 3 onthe number line. So x 5 5 is not in the solution setof x , 3.c. A sample counterexample is a triangle withangles of 100 , 40 , and 40 .28. D19. The student’s statement is a conjecture because it isa generalization based on a pattern of data. Thestatement is not a theorem because it has not beenproved using deductive reasoning.29. Sample answer: If30. a. 3x 1 1 5 163x 1 15 2 , then x 5 5.8b. Multiplication Property of Equality20. a. Deductive reasoning. The student’s conclusionis based on a proof, so the reasoning isdeductive rather than inductive.c. 3x 5 15d. Subtraction (or Addition) Property of Equalityb. Yes, the conclusion follows logically from thestatements that diameters bisect each other andthat RS and TV are diameters.e. x 5 5Lesson 3-331. a. Inverse: If it is not raining, then I do not stayindoors; Contrapositive: If I do not stay indoors,then it is not raining.Lesson 3-121. a. angle: defined; ray: definedb. segment: defined; points: undefinedb. Inverse: If I do not have a hammer, then I donot hammer in the morning; Contrapositive: If Ido not hammer in the morning, then I do nothave a hammer.c. triangle: defined; segments: definedd. lines: undefined; plane: undefined22. a. Given32. If people have the same ZIP code, then they live inthe same neighborhood. If people live in the sameneighborhood, then they have the same ZIP code.b. Multiplication Property of Equalityc. Addition Property of Equalityd. Division (or Multiplication) Property ofEquality23. C24. C 2015 College Board. All rights reserved.A2SpringBoard Geometry, Unit 1 Practice
Lesson 4-233. a. The statement is not true. The value x 5 0makes the hypothesis true and the conclusionfalse, so the statement is false.41. m MPQ 5 m QPN42. 35 b. Converse: If x fi 0, then 3x 5 0. A value such asx 5 5 makes the hypothesis true and theconclusion false, so the converse is false.43. A44. 11c. Inverse: If 3x fi 0, then x 5 0. A value such asx 5 5 makes the hypothesis true and theconclusion false, so the inverse is false.45. a.BCAd. Contrapositive: If x fi 0, then 3x 5 0. A valuesuch as x 5 5 makes the hypothesis true and theconclusion false, so the contrapositive is false.PD34. D35. a. contrapositiveb. inverseQLesson 4-136. BRb. 48 37. a. 5Lesson 5-1b. 346. D38. a. 9 cmb. 19.5 (or 19.5 cm)47. (1.5, 2)c. 12, 36 (or 12 cm, 36 cm)48. a.128 or 8 2b. 104 or 2 26d. 3 (or 3 cm)c. 104 or 2 2639. a. The midpoint will be positive when the distancebetween 0 and B is greater than the distancebetween 0 and A.d. isosceles49.b. The midpoint will be negative when thedistance between 0 and B is less than thedistance between 0 and A.5 1 13 1 1050. r 5 ( x 2 5)2 1 ( y 2 2)2Lesson 5-2c. The midpoint will be zero when the distancebetween 0 and B is equal to the distancebetween 0 and A.51. (4.5, 3.5)52. a. S(1, 2), T(7, 5)d. Never. Distance is always nonnegative.b.40. M is between P and T. Starting with PT 2 PM 5MT, add PM to each side to get PT 5 MT 1 PM orPT 5 PM 1 MT. That equation satisfies thesituation that M is a point between P and T.45 or 9 553. a. M(2, 1)b. AM 5 0 1 16 5 4; AB 5 4 1 25 5 29 5.4,AC 5 4 1 9 5 13 3.6 ; point A is closest topoint C.54. C 2015 College Board. All rights reserved.A3SpringBoard Geometry, Unit 1 Practice
55.d. 1 is supplementary to 3.y16e. Definition of supplementary angles(11, 16)5141263. a. m 1 5 37; m PTR 5 53b. Angle Addition Postulate(8, 12)5108c. m 2 5 16d. Subtraction Property of Equality(5, 8)64. a. m 1 1 m 2 5 m RVT56b. Definition of congruent anglesc. VS bisects RVT.4(2, 4)20d. Definition of angle bisector024681012x65. Sample answer: It is true that if A and B are bothcomplementary to T, then A B. However,the reason the student gives, the definition ofcomplementary angles, is not the correct reason forthis statement. The definition of complementaryangles states only that complementary angles aretwo angles whose sum is 90 . The definition doesnot mention a third angle. The correct justificationis the property that if two angles are complementaryto the same angle (or, to two congruent angles), thenthe two original angles are congruent to each other.Sample answer: I graphed (5, 8) and (8, 12) andcalculated that the distance between them as 5 units.Then I found two more points on the same line thatwere 5 units from (5, 8) or (8, 12). Using the MidpointFormula, I found that the coordinates of the twopoints were (2, 4) and (11, 16). So there are twopossibilities for the other endpoint, (2, 4) and (11, 16).Lesson 6-156. a. m ATB 5 m CTB or m 1 5 m 2Lesson 7-1b. m ATB 1 m BTC 5 m ATC orm 1 1 m 2 5 m ATC66. a. 55 c. 46 57. a. Definition of betweennessd. 78 67. a. corresponding anglesb. Definition of midpointb. alternate interior angles58. a. Definition of complementary angles; if twoangles are complementary, then the sum of theirmeasures is 90 .c. vertical anglesd. same-side interior anglesb. Definition of perpendicular lines; if two linesmeet to form a right angle, then the lines areperpendicular.68. A69. a. 759. Cb. 45 60. Cc. 135 d. (7x 2 4) 1 (20x 2 5) 5 180 because same-sideinterior angles are supplementary. So 27x 2 9 5180, 27x 5 189, x 5 7. Then m 1 5 7x 2 4 57(7) 2 4 5 45 and m 2 5 20x 2 5 5 20(7) 2 5 5140 2 5 5 135.Lesson 6-261. D62. a. Definition of angle bisector70. a. Givenb. Givenb. If lines are parallel, then corresponding anglesare congruent.c. Substitution 2015 College Board. All rights reserved.b. 101 A4SpringBoard Geometry, Unit 1 Practice
c. If lines are parallel, then same-side interiorangles are supplementary.83. Neither. The two lines have different slopes so theyare not parallel. The slopes are not negativereciprocals so the lines are not perpendicular.84. Perpendicular. The the slope of PQ is 5 22. That1is the opposite reciprocal of , the slope of RS, so2the lines are perpendicular.d. m 7 1 m 1 5 180Lesson 7-271. a. 50 b. 141 c. 42 d. 143 6233 15 5 . Opposite4 2 (22) 6 2sides of a rectangle are parallel, so the slope of CD1is . Adjacent sides of a rectangle are2perpendicular, so the slope of BC 5 slope of85. The slope of AB is72. 873. a. m nb. m nc. p qd. p qe. nonef. m n74. a. EPB EQDAD 5 22.b. m EPB 5 m EQDLesson 8-2c. Substitutiond. AB CD75. Bb.Lesson 7-3c.76. C77. a. 32x2632slope 5 , y-intercept 5 2632y 2 2 5 (x 2 12)3141y215 x41y5 x114y 2 9 5 24(x 1 2) or y 2 1 5 24x86. a. y 5b. 887. a.1d. 145 378. a. 1 ray; Perpendicular Postulatec. 3b.c.b. Sample answer: The two rays are opposite rays;the two rays lie on the same line.d.88. D79. a. Definition of congruent segmentsc. AP DQ89. Find the slope of the line through A and B. Use thatslope and the given point to write an equation inpoint-slope form.d. Definition of congruent segments90. The slope of PQ isb. Definition of bisector5 2 (21)655 21 and the24 2 226 24 1 2 5 1 (21) 22 4 ,, 5 midpoint of PQ is 2 2 2 25 (21, 2). The slope of the perpendicular bisectoris 1 and it goes through (21, 2), so its equation isy 2 2 5 1(x 2 (21)) or y 2 2 5 x 1 1. In slopeintercept form, that equation is y 5 x 1 3.80. Sample answer: AT 5 7, CT 5 5, BT 5 7, DT 5 9,m ATD 5 90. Since AT 5 BT and AB CD, thenCD is the perpendicular bisector of AB.Lesson 8-181. a.311b. 2 c. d. 152282. C 2015 College Board. All rights reserved.A5SpringBoard Geometry, Unit 1 Practice
LeSSon 7-1 66. a. 55 b. 101 c. 46 d. 78 67. a. corresponding angles b. alternate interior angles c.vertical angles d. same-side interior angles 68. A 69. a. 7 b. 45 c. 135 d. (7x 2 4) 1 (20x 2 5) 5 180 because same-side interior angles are supplementary. So 27x 2 9 5 180, 27x 5 189, x 5 7. Then m
