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Answers to Geometry Unit 2 PracticeLesson 9-113. a. x 5 01. Bb. y 5 02. (x, y) (x 1 3, y 1 5)c. y 5 215d. x 5 22e. x 5 103. a. (23, 3)b. (26, 22)c. (3, 29)14. a. (10, 2)d. (4, 26)b. (4, 22)4. a. rigidc. (4, 2)b. nonrigidd. y 5 2c. rigid15. a. J, L, N, Pd. nonrigidb. Ke. nonrigid5. a.c. H, I, M, O41 unitsd. 1 and 3 each has four lines of symmetry;2 and 4 each has two lines of symmetry.b. A rigid transformation does not change lengths.Lesson 9-2Lesson 9-46. D16. a. right7. a. (22, 2)b. leftb. (x, y) (x 2 3, y 2 3)c. up8. a. They have the same length and they are parallel.d. rightb. R(0, 3), S(21, 9)17. D9. a. (21, 0)18. a. (22, 23)b. (x, y) (x 1 6, y 2 5)b. 90 clockwise10. a. A19. a. angle Sb. Cb. angle Ec. The image is translated 4 units to the leftand 5 units up. The only two figures that satisfythat translation are triangle A for the originaltriangle and triangle C for the image.c. QRd. In a rotation, corresponding angles have thesame measure and corresponding sides have thesame length.d. (x, y) (x 1 4, y 2 5)20. a. 180 about (22, 4)b. 90 clockwise about (22, 3)Lesson 9-3c. 90 counterclockwise about (1, 5)11. Ad. 180 about (0, 1)12. C 2015 College Board. All rights reserved.A1SpringBoard Geometry, Unit 2 Practice Answers

Lesson 10-132. a. CP or PC21. a. RO, 180 (rx 5 2)b. CBP or PBCb. ry 5 5 (T(0, 5))c. PAB or BAP22. base (4, 2), tip (4, 7)d. PBC23. a. T(25, 1)33. a. MN and RP, NT and PQ, MT and RQb. T(22, 23) (RO, 180 )b. MNT and RPQ, NTM and PQR, NMTand PRQ24. a. R(3, 4), 180 b. T(23, 2)34. D35. a. SSS25. Cb. SASLesson 10-2c. ASA26. a. line md. SASb. 90 clockwise around Cc. by directed line segment DCLesson 11-2d. across AD36. a. AASb. ASA27. Sample answer. If the diameter of circle A has thesame length as the radius of circle B, then one circlecan be transformed, using only translations androtations, so the diameter of circle A coincides withthe radius of circle B. Translations and rotations arerigid motions, so the two segments are congruent.c. SASd. SSS37. a. C F28. Bb. B E or C F Or ( B and F) or( C and E)29. a. Sample answer. T(0, 26) (rx 5 4.5)c. AC DF , BC EFb. r(0, 4.5), 180 d. AB DEc. Sample answer. ry 5 0 (rx 5 4.5)38. Dd. Sample answer. R(1.5, 3), 180 (T(0, 21))39. Sample answer. Yes, two triangles can have sidelengths 5, 5, and 9. The two triangles are congruentby SSS, and they can be described as both obtuseand isosceles.30. Sample answer. The composition involves rigidmotions so the size and shape of CHGB is notchanged. After the composition, rectangle CHGBcoincides with rectangle ACED, so they arecongruent.40. (4, 27), (4, 5)Lesson 11-341. Sample answer. You have to show that a sequenceof rigid motions maps one of the triangles to theother.1142. a. m 1 5 m ABC 5 m BCD 5 m 222b. DCBLesson 11-131. a. Qb. Xc. QRd. XZc. BCe. ZXY 2015 College Board. All rights reserved.d. ASAA2SpringBoard Geometry, Unit 2 Practice Answers

53. The preceding statements must list two pairs ofcongruent sides and a pair of congruent angles thatare between the sides.43. B44. a. QR RQb. HL54. Dc. QPR55. CPCTCd. PQ SR, P S, PRQ SQR45. 11Lesson 12-256. The boxes correspond to Statements; the linesbelow the boxes correspond to Reasons.Lesson 11-446. A57. Sample answer. The arrows show the logical flowbetween statements.47. a. Yes. If a triangle has a right angle, then it is aright triangle.258. A2b. BC 2 (3 2 15) 1 (9 2 4) 5 144 1 25 5 13;59. C60. Sample answer. Yes. It is given that FJ HG andFG HJ. Also, FJ FH by the Reflexive Property.So JFH GHF by SSS. Since the two trianglesare congruent, 1 2 by CPCTC.YZ 5 (216 2 (24))22 1 (21 2(26))2 5144 1 25 5 13c. AB 5 (3 2 3)2 1 (9 2 4)2 5 5;XY 5 (24 2 (24)2 1 (21 2 (26))2 5 5Lesson 13-1d. Yes, the triangles are congruent by the HLcongruence criterion.61. a. 100 b. 40 48. Bc. 120 49. a. It bisects the vertex angle.d. 60 b. It is a right angle.62. a. 40 50. a. SRPb. 25 b. PS is perpendicular to QR and PS bisects QR .c. 115 d. 65 Lesson 12-163. a.51. a. 1 4, MN PTb. Vertical angles are congruent.Am50 Bc. 2 3nd. MT TM50 30 30 Ce. MNT TPM52. a. m QPS 5 m QPR 1 m RPS, m RPT m TPS 1 m RPSThe two angles formed at B are 50 and 30 because they are corresponding angles forparallel lines. So the measure of ABC is 80 .b. Subtraction Property of Equalityc. Reflexive Propertyd. m QPR m TPSe. ASA 2015 College Board. All rights reserved.A3SpringBoard Geometry, Unit 2 Practice Answers

b.70. a. The two base angles are congruent and theirsum is equal to the exterior angle at the vertex,so (2x 1 5) 1 (2x 1 5) 5 5x 2 3. So 4x 1 10 55x 2 3 and 13 5 x.A50 m30 50 80 Bb. 2x 1 5 5 31. The three angle measures are 31 ,31 , and 180 2 62 5 118 .n30 CLesson 14-1Angle ABC is an exterior angle of a trianglewhose remote interior angles measure 30 and50 . So the measure of angle ABC is 80 .71. a.yC64. a. ADC 2, ACD 1, CAD 3Ab. DCP 2, CDP 1, P 3Bxc. BCA, ACD, PCDd. Sample answer. The three angles with vertex Cform a straight angle, so m BCA 1 m ACD 1m PCD 5 180 . Those same angles representthe sum of the interior angles of a triangle, sothis diagram is another way to prove theTriangle Sum Theorem.65. DLesson 13-266. a. Definition of right triangleb. outsideb. TP 5 TPc. Yes, ABC is an obtuse triangle, and thealtitudes of an obtuse triangle intersect outsidethe triangle.c. HLd. M Nd. (11, 29)e. CPCTC72. (3, 1)67. 63 73. a. 4 68. a. If two angles of a triangle are congruent, thenthe sides opposite those angles are congruent.b. 36 c. 50 b. Sample answer. It is given that ZW bisects angleXYZ, so 1 2 by the definition of anglebisector. It is also given that m X 5 m Y, so X Y by the definition of angle congruence.Since WZ WZ by the Reflexive Property, WXZ WYZ by AAS. XZ and YZ arecorresponding sides in those triangles, soXZ YZ by CPCTC.d. 54 e. 126 74. D75. Sample answer. Select one side of the triangle.Using the two vertices of that side, find an equationof the line that contains that side. Then use thenegative reciprocal of the slope of that line, alongwith the third vertex, to find an equation for thealtitude to that side.69. B 2015 College Board. All rights reserved.A4SpringBoard Geometry, Unit 2 Practice Answers

Lesson 14-2Lesson 14-376. a.81. a.yPQy109876(0, 5)54321(7, 0)x(0, 0)1 2 3 4 5 6 7 8 9 10xRb. insideb. onc. No. The medians of any triangle meet insidethe triangle.c. Yes, VWX is a right triangle, and theperpendicular bisectors of the sides of a righttriangle meet on the triangle.d. (2, 0)77. (3, 2)d. (3.5, 2.5)78. a. 1.5e.b. 13.5c. 6d. 4.579. B80. Sample answer. Find the midpoints of the sides.Then use the midpoints to draw two (or three)medians. The centroid is at the intersection of themedians.y109876(0, 5)54321(7, 0)(0, 0)x1 2 3 4 5 6 7 8 9 10The angle bisectors intersect at approximately(1.8, 1.8).82. a. 8b. 21c. 4d. 4283. D84. Sample answer. Find the angle bisectors for two(or three) of the angles of the triangle. The incenteris at the intersection of the angle bisectors.85. Sample answer. Find the perpendicular bisectorsfor two (or three) of the sides of the triangle. Thecircumcenter is at the intersection of theperpendicular bisectors of the sides. 2015 College Board. All rights reserved.A5SpringBoard Geometry, Unit 2 Practice Answers

93. a. corresponding anglesLesson 15-186. a. 26 in.b. AY DYb. 13 in.c. DC 1 CYc. 13 in.d. AB 5 DCd. 65 e. def. of congruent segments87. a. kite94. Cb. TPS and TQS95. a. 55 b. 70 c. Sample answer. TS is the perp. bisector of PQ, soPR RQ and PRT QRT by the def. of perp.bisector. Also, TR TR by the ReflexiveProperty. So PTR QTR by SAS.c. 55 d. 85 e. 140 d. Sample answer. By a proof similar to the one inPart c, we can show that PRS QRS by SAS.Then PT BT and PS QS by CPCTC in thetwo pairs of congruent triangles. So PTQS is akite by the def. of kite.Lesson 15-396. a. 13b. 100 c. 688. Sample answer. It is given that ABC is isosceleswith base BC, so AB 5 AC by the def. of isos.triangle. Also, it is given that BMC is equilateral soMB 5 MC by the def. of equilateral triangle. Thatmeans each of points A and M is equidistant frompoints B and C, so AB is the perp. bisector of BC.d. 100 97. TP 5 RQ 5 18, TR 5 PQ 5 1098. a. ABCD is a parallelogram because both pairs ofopposite sides are congruent.89. Ab. They are the diagonals of the parallelogram.90. a. 50 b. 110 c. The diagonals of a parallelogram bisect eachother.c. 30 d. Find point E so that CE 5 AB and BE 5 AC.99. Bd. 60 100. a. parallelogramLesson 15-2b. X, Y, Z, and W are midpoints.91. a. 100 c. WZ AC , XY ACd. WXYZ is a parallelogram.b. 25 c. 30 d. 125 Lesson 15-4e. 55 101. a. (1, 9.5)b. RE 5 TC 5 13, RT 5 EC 5 2692. a. 23c. The length of each segment is about 14.5 units.512d. The slope of RE is 2 ; the slope of EC is ;125the product of those two fractions is 21.b. 31c. 2 2d. 118 e. 45 2015 College Board. All rights reserved.A6SpringBoard Geometry, Unit 2 Practice Answers

102. DLesson 16-1103. a. 6106. Sample answer. The midpoint of AC is 11 2 525 , 5 (1.5, 0) and the midpoint of BD22 b. 8c. 16 2 1 5 3 1 (23) ,is 5 (1.5, 0). The midpoints 22d. 53 e. 106 are the same so the diagonals of ABCD bisecteach other. We can conclude that ABCD is aparallelogram.104. a. 15b. 2, 5, 6107. a. 11c. TBE, UBR, UBE1d. area of TRUE 5 (TU )( ER )2e. Sample answer. The area of the rhombus is11(TU )( ER ) 5 (24)(18) 5 216. A formula22for the area of any parallelogram (or anyrhombus) is A 5 b h, where b is a base and his the height for that base. Using that formulawith A 5 216 and b 5 15 from Part a, then216 5 15h and h 5 14.4.b. m A 5 125 , m B 5 55 , m C 5 130 ,m D 5 50 c. No. There are no interior same-side angles thatare supplementary, so no lines are parallel.d. No. The figure does not have any pairs ofparallel sides, so it is not a parallelogram.108. a. Both pairs of opposite sides are congruent.b. Yes. If both pairs of opposite sides of aquadrilateral are congruent, the quadrilateralis a parallelogram.105. a. 6 2 or 8.49 units. Sample answer. Thediagonal of a square is the length of a sidetimes the square root of 2.109. n 5 14, m 5 20b. 45 . Sample answer. The diagonal of a squarebisects its two angles, and half of 90 is 45 .110. Dc. 22.5 . Sample answer. The diagonal of arhombus bisects its two angles, and half of 45 is 22.5 .Lesson 16-2111. Sample answer. The midpoint of diagonal MP is 4 13 8 10 , 5 (3.5, 4) and the midpoint of22 diagonal NQ is 7 1 0 , 2 1 6 5 (3.5, 4). Since 22 the diagonals bisect each other, MNPQ is aparallelogram. Also, the length of MPd. 18 square units. Sample answer. The area oftriangle ABD is half the area of the square, andthe area of the square is 36 square units.e. 18 square units. Sample answer. Each diagonalof a rhombus divides it into two congruenttriangles. Using BCD and Part d, we canconclude that half of the area of the rhombusis 18 square units. Then, noting that BCE isalso half of the rhombus, we can conclude thatthe area of BCE is 18 square units.is (4 2 3)2 1 (8 2 0)2 5 12 1 82 5 65, andthe length of NQ is (7 2 0)2 1 (2 2 6)2 57 2 1 (24)2 5 65. Using the result that thediagonals of parallelogram MNPQ are congruent,we can conclude that MNPQ is a rectangle.112. (5, 1) 2015 College Board. All rights reserved.A7SpringBoard Geometry, Unit 2 Practice Answers