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UNIT 1 • SIMILARITY, CONGRUENCE, AND PROOFS Answer Key Lesson 1: Investigating Properties of Dilations Pre-Assessment, pp. 1–3 1. d 2. c 3. b 4. a 5. b Warm-Up 1.1.1, p.6 1. Reflect the garden over the line y = x: A' (3, –7), B' (7, –3), C' (3, –3), or rotate the garden about the origin 180° counterclockwise: A' (7, –3), B' (3, –7), C' (3, –3). 2. Reflections and rotations are rigid motions. They preserve angle and length measures. The result of either transformation is a congruent figure. Practice 1.1.1: Investigating Properties of Parallelism and the Center, pp. 24–28 1. The pentagon has been dilated. The corresponding sides of the pentagon are parallel and the scale factor is consistent (k = 2). Additionally, the preimage points and image points are collinear with the center of dilation. 2. The triangle has been dilated. The corresponding sides of the triangle are parallel and the scale factor is consistent (k = 1/3). Additionally, the preimage points and image points are collinear with the center of dilation. 3. The triangle has not been dilated. The scale factor is inconsistent between corresponding sides. 4. The rectangle has not been dilated. The scale factor is inconsistent between corresponding sides. 5. k = 2.5; enlargement 6. k = 1.25; enlargement 7. k = 0.4; reduction 8. k = 0.5; reduction 9. No, it is not a dilation. The scale factors are inconsistent. The scale factor of the vertical sides is 1.5. The scale factor of the horizontal sides is 4/3. 10. k = 3/5; reduction Warm-Up 1.1.2, p. 29 1. 0.85 2. 85% 3. 6/5.1 4. 118% 5. Hideki might have pushed his sister again to cause her to swing higher. Practice 1.1.2: Investigating Scale Factors, p. 42 1. 7.2 2. 10.2 3. 12.3 4. 20 5. H' (–21, –9), J' (–15, –18), K' (–18, –24) 6. P' (–3, 2), Q' (2.5, 4.5), R' (–1.5, –2) 7. M' (–3.75, 6), N' (5.25, –2.25), O' (–7.5, –3) 8. A' (8.4, 7), B' (2.8, 2.8), D' (–4.2, 5.6) 9. D" (12, 8), E" (24, 8), F" (–6, 16); k = 4 10. S' (–4.5, 3), T' (4.5, 3), U' (4.5, –3), V' (–4.5, –3); the new countertop is longer by a factor of 1.5 feet. Progress Assessment, pp. 43–47 1. c 6. c 2. d 7. d 3. a 8. a 4. b 9. a 5. b 10. b 11. Answers: a. Enlargement; the scale factor when converted to a decimal is 1.8. Since 1.8 is greater than 1, the scale factor creates an enlargement. b. G' (0, 0), H' (0, 14.4), I' (9, 14.4), and J' (9, 0) c. The perimeter of the original book’s front cover is 26 inches. The perimeter of the dilated book is 46.8 inches. The ratio 46.8/26 is equal to the scale factor of the dilation, which is 1.8 or 180%. d. The area of the original book’s front cover (the preimage) is 40 in2. The area of the dilated book cover (the image) is 129.6 in2. The ratio of the areas is 129.6/40 and is equal to the square of the scale factor of the dilation, 1.82 or 3.24. Lesson 2: Constructing Lines, Segments, and Angles Pre-Assessment, pp. 48–50 1. d 2. c 3. c 4. d 5. b Warm-Up 1.2.1, p. 54 1. The areas do not overlap. 2. Maryellen could use the leash as a compass to determine where to place the stake to minimize overlap. Practice 1.2.1: Copying Segments and Angles, p. 78 1–10. Check students’ work for accuracy. Warm-Up 1.2.2, p. 79 1. Check students’ work for accuracy. 2. 90˚ U1-743 © Walch Education CCGPS Analytic Geometry Teacher Resource Practice 1.2.2: Bisecting Segments and Angles, pp. 101–102 1–10. Check students’ work for accuracy. Warm-Up 1.2.3, p. 103 1. Check students’ work for accuracy. 2. Check students’ work for accuracy. Practice 1.2.3: Constructing Perpendicular and Parallel Lines, p. 128 1–10. Check students’ work for accuracy. Progress Assessment, pp. 129–133 1. c 6. b 2. b 7. c 3. a 8. b 4. c 9. d 5. d 10. c 11. Check students’ work for accuracy. Lesson 3: Constructing Polygons Pre-Assessment, p. 134 1. d 2. b 3. c 4. b 5. a Warm-Up 1.3.1, p. 137 1. Copy the given 60˚ angle to construct a triangle. Extend the sides of the copied angles to determine the center of the audience. Speaker 1 Speaker 2 2. No, it is not possible to construct a second non-congruent triangle using the given information. The given information includes two angle measures and a side length between those two angles. If you have two triangles and any two angles and the included side are equal, then the triangles are congruent. Practice 1.3.1: Constructing Equilateral Triangles Inscribed in Circles, pp. 159–160 1–3. Check students’ work for accuracy. Be sure each of the vertices lies on the circle. 4. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is twice the length of the given segment. 5. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment. 6–8. Check students’ work for accuracy. Be sure each of the vertices lies on the circle. 9. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is twice the length of the given segment. 10. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment. Warm-Up 1.3.2, p. 161 1. Each of the new angles measures 45˚. 2. The angle bisector intersects the square at the bisected angle and the angle opposite the bisected angle. 3. Antonia will have created 4 triangles. 4. Each triangle will have two 45˚ angles and one 90˚ angle. Practice 1.3.2: Constructing Squares Inscribed in Circles, p. 180 Center of audience 1–2. Check students’ work for accuracy. Be sure each of the vertices lies on the circle. 3–6. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to the length of the given segment. 7. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to twice the length of the given segment. 8. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment. 9–10. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment. 60˚ U1-744 CCGPS Analytic Geometry Teacher Resource © Walch Education Warm-Up 1.3.3, p. 181 1. The length of each side of the square planting bed will be 1 the length of the original plank of wood. 4 2. To determine the length of each side of the square planting bed, first find the midpoint of the plank of wood by bisecting the plank. This construction will divide the plank into two equal pieces. Next, find the midpoint of one of the halves by bisecting it. The result is a piece of the plank that 1 is the original length of the plank of wood. 4 Practice 1.3.3: Constructing Regular Hexagons Inscribed in Circles, p. 204 1. Check students’ work for accuracy. Be sure each of the vertices lies on the circle. 2–3. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to the length of the given segment. 4. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to twice the length of the given segment. 5. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment. 6. Check students’ work for accuracy. Be sure each of the vertices lies on the circle. 7–8. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to the length of the given segment. 9. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to twice the length of the given segment. 10. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment. Progress Assessment, pp. 205–209 1. a 6. b 2. b 7. c 3. a 8. a 4. c 9. c 5. d 10. d 11. a. Check students’ work for accuracy. b. F ind the perpendicular bisector of each side of the equilateral triangle created when lines are drawn to connect the vending machines. Extend the bisectors to locate intersection points with the circle. The points of intersection represent the placement of the recycling bins. Lesson 4: Exploring Congruence Pre-Assessment, pp. 210–213 1. d 2. a 3. b 4. d 5. c Warm-Up 1.4.1, p. 216 1. The coordinates: rx-axis(A (2, 5)) = A' (2, –5) rx-axis(B (3, 5)) = B' (3, –5) rx-axis(C (3, 2)) = C' (3, –2) rx-axis(D (5, 2)) = D' (5, –2) rx-axis(E (5, 1)) = E' (5, –1) rx-axis(F (2, 1)) = F' (2, –1) 2. 10 9 y 8 7 6 5 4 A 3 2 1 B C F D E x -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -1 ( 5, –1 ) -2 ( 2, –1 ) ( 5, –2 ) -3 F' -4 -5 -6 -7 C' ( 3, –2 ) A' E' D' B' ( 2, –5 ) ( 3, –5 ) -8 -9 -10 U1-745 © Walch Education CCGPS Analytic Geometry Teacher Resource Practice 1.4.1: Describing Rigid Motions and Predicting the Effects, pp. 235–238 6. 1. Translation; right 4 units and up 3 units; the orientation stayed the same. 2. Reflection; the line of reflection is x = –1. The orientation changed, and the preimage and the image are mirror reflections of each other. The line of reflection is the perpendicular bisector of the segments connecting the vertices of the preimage and image. 3. Rotation; the orientation changed, but the figures are not mirror images of each other. 4. y 10 9 8 7 6 5 4 3 x -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -1 A -2 -3 -4 -5 -6 8 7 -7 -8 -9 B' 4 3 B' 2 1 B 10 9 6 5 y C' C A' -10 7. 2 1 R C' A' x -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -1 -2 C -3 -4 A B -5 -6 -7 -8 -9 B -10 5. 10 9 T C D y A 8 D' C' A' B' 7 6 5 4 3 A' B' 2 1 A B D 8. C -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -1 D' -2 C' -3 -4 N x W E S -5 -6 -7 -8 -9 -10 A' D' A B D C B' C' U1-746 CCGPS Analytic Geometry Teacher Resource © Walch Education Warm-Up 1.4.2, p. 239 9. 1. B A C A' C' TV cabinet R TV B' 10. Answers may vary. Method 1: Rotate the paddle 3 times using a 90º angle of rotation and point R. Method 2: Reflect the paddle over the line m to create paddle 2. Rotate paddle 2 90º counterclockwise about point R to create paddle 3. Reflect paddle 3 over line n to create paddle 4. m 2. Since the drawing is a scaled representation of the cabinet and TV, use the units provided on the grid and draw the diagonal. Use the Pythagorean Theorem to calculate d = 65 ≈ 8.06 , which is the length of the diagonal of the cabinet. The TV is 8 units. Therefore, the TV should fit, but it will be a tight fit. Practice 1.4.2: Defining Congruence in Terms of Rigid Motions, pp. 258–261 3 2 R 1 A n 4 1. Not congruent; a dilation occurred with a scale factor of 2/3, indicating a reduction. Dilations are non-rigid motions. 2. Congruent; a rotation occurred. Rotations are rigid motions. 3. Congruent; a translation occurred by moving the figure 9.5 units to the left and 4.5 units up. Translations are rigid motions. 4. Not congruent; a horizontal stretch has occurred with a scale factor of 1.5. Stretches are non-rigid motions. 5. Not congruent; a vertical stretch has occurred with a scale factor of 1.5. Stretches are non-rigid motions. 6. Not congruent; a horizontal compression has occurred with a scale factor of 2/3. Compressions are non-rigid motions. 7. The speaker has been rotated and translated. Since these are both rigid motions, the figures are congruent. 8. The corner cabinet has been rotated. Since rotations are rigid motions, the figures are congruent. 9. The inner square has been dilated by a scale factor of 5/3. Since dilations are non-rigid motions, the squares are not congruent. 10. Answers may vary. Sample answer: The triangles are congruent. Triangle 1 can be reflected over a vertical line passing through the top right vertex to create triangle 3. Triangle 3 can be reflected over another vertical line passing through the top right vertex to create triangle 5. Triangle 1 can be reflected over a horizontal line passing through the center of the triangle and then translated U1-747 © Walch Education CCGPS Analytic Geometry Teacher Resource 2 units to the right to create triangle 2. Triangle 2 can be reflected over a vertical line passing through the bottom right vertex to create triangle 4. Since all the transformations described are rigid motions, the triangles are congruent. Progress Assessment, pp. 262–270 1. b 6. b 2. d 7. b 3. a 8. c 4. c 9. b 5. a 10. c 11. a. A nswers may vary. Sample answer: Looking at the sugar and flour containers, the left handle can be reflected over the vertical line passing through the center of each container to create the right handle. The handles are all congruent and can be created from rigid motions. b. The rectangles of the containers have undergone a nonrigid motion of a dilation with a scale factor of 4/3. The flour container is larger than the sugar container. This means that the flour and sugar containers are not congruent. Lesson 5: Congruent Triangles Pre-Assessment, p. 271 1. d 2. c 3. c 4. a 5. d Warm-Up 1.5.1, pp. 274–275 1. The position of Piece 2 is the result of a horizontal translation of 6 units to the right and a vertical translation of 3 units down. 2. The position of Piece 3 is the reflection of Piece 2 over a vertical line. 3. A series of congruency transformations results in a congruent figure, so Piece 3 is congruent to Piece 1. Practice 1.5.1: Triangle Congruency, pp. 292–294 1. Sample answer: FHG ≅ BCD 2. Sample answer: JNP ≅ RTV 3. Sample answer: BDF ≅ PLN 4. ∠Q ≅ ∠W , ∠R ≅ ∠X , ∠S ≅ ∠Y , QR ≅ WX , RS ≅ XY , QS ≅ WY 5. ∠A ≅ ∠C , ∠F ≅ ∠G , ∠H ≅ ∠J , AF ≅ CG , FH ≅ GJ , AH ≅ CJ 6. ∠L ≅ ∠H , ∠P ≅ ∠J , ∠Q ≅ ∠K , LP ≅ HJ , PQ ≅ JK , LQ ≅ HK 7. Yes, the triangles are congruent; ADH ≅ JPK . 8. No, the triangles are not congruent. 9. Yes, the triangles are congruent; EDF ≅ IHG . 10. Yes, the triangles are congruent; RST ≅ WVU . Warm-Up 1.5.2, p. 295 1. The base of the pyramid is a square; therefore, AB ≅ BC ≅ CD ≅ DA ; the triangles are congruent, so the corresponding parts of the triangles are also congruent: AE ≅ BE ≅ CE ≅ DE . 2. The corresponding angles of the congruent triangles are congruent; therefore, ∠EBA ≅ ∠EAD ≅ ∠EDC ≅ ∠ECB and ∠EAB ≅ ∠EBC ≅ ∠ECD ≅ ∠EDA . Practice 1.5.2: Explaining ASA, SAS, and SSS, pp. 312–315 1. SSS 2. Congruency cannot be determined; the identified congruent parts form SSA, which is not a triangle congruence statement. 3. SAS 4. ASA 5. SAS 6. Congruency cannot be determined; there is not enough information about both of the triangles to determine if the triangles are congruent. 7. ABC ≅ FGE ; SAS or ASA 8. There is not enough information to determine if the sails are congruent. The information provided follows SSA, which is not a triangle congruence statement. 9. Congruency cannot be determined; there is not enough information about both of the triangles to determine if the triangles are congruent. 10. The plots of land are congruent by SAS or SSS. Progress Assessment, pp. 316–319 1. c 6. b 2. a 7. a 3. b 8. c 4. a 9. d 5. d 10. b 11. a. AB ≅ BC , ∠BAD ≅ ∠BCD b. There is not enough information to determine if the triangles are congruent. The information provided follows SSA, which is not a triangle congruence statement. c. FJ ≅ HJ , ∠FJK ≅ ∠HJE d. FKJ ≅ HKJ by SAS e. The minimum amount of information needed to prove that two triangles are congruent is two angles and the included side, or two sides and the included angle, or three sides. (Students may also recall that given two angles and a side not included, congruency may be proven by AAS.) U1-748 CCGPS Analytic Geometry Teacher Resource © Walch Education Lesson 6: Defining and Applying Similarity Pre-Assessment, pp. 320–322 1. d 2. a 3. a 4. a 5. c Warm-Up 1.6.1, p. 324 1. 2,722.5 feet 2. 1,440 feet 3. 1,282.5 feet Practice 1.6.1: Defining Similarity, pp. 342–346 1. ∠C = 133˚, ∠D = 37˚, ∠E = 10˚, EF = 4.5, DE = 5.5 2. ∠M = 26˚, ∠N = 14˚, ∠P = 140˚, NP = 9.8, PM = 5.6, KL = 4 3. Similarity transformations preserve angle measures. The triangles are not similar because the angle measures in each triangle are different. 4. The triangles are similar. ABC can be dilated by a scale factor of 2 with the center at (0, 0) and then reflected over the line x = 6 to obtain DEF . 5. The triangles are similar. ABC can be dilated by a scale factor of 2.5 with center at (0, 0) to obtain DEF . 6. The triangles are similar. ABC can be dilated by a scale factor of 1 with the center at (0, 0), translated horizontally –4 units and vertically –4 units, and then reflected over the line y = –4 to obtain DEF . 7. The triangles are similar. ABC can be dilated by a scale factor of 1/3 with the center at (0, 0) and then rotated 90˚ clockwise about the origin to obtain DEF . 8. The triangles are similar. ABC can be dilated by a scale factor of 1 with the center at (0, 0) and then rotated 180˚ clockwise about the origin to obtain DEF . 9. Similarity transformations preserve angle measures. The triangles are not similar because the angle measures in each triangle are different. 10. The triangles are similar. ABC can be dilated by a scale factor of 1/2 with the center at (0, 0) and then translated vertically –6 units to obtain DEF . Warm-Up 1.6.2, p. 347 1. 1/12 2. 3.875 meters Practice 1.6.2: Applying Similarity Using the Angle-Angle (AA) Criterion, pp. 359–362 1. Yes, ABC ∼ ZYX because of the AA Similarity Statement. 2. Yes, ABD ∼ECD because of the AA Similarity Statement. 3. There is not enough information to determine similarity. 4. ABC ∼YZX ; x = 3 1/3 5. ABC ∼ DEA ; x = 4; BC = 3; AE = 6 6. AED ∼CBD ; x = 1; ED = 2; DB = 6 7. 6 feet 8 inches 8. 1063 1/3 feet = 1063 feet 4 inches 9. 36 feet 10. 2.56 feet Progress Assessment, pp. 363–369 1. d 6. d 2. c 7. b 3. b 8. c 4. c 9. a 5. a 10. b 11. Answers: a. 96.25 feet b. approximately 2.18 feet c. Triangles can be formed using the height of the object and the length of the shadow as two sides. The height of the object and the length of the shadow create a 90˚ angle. Connecting these two sides creates the third side of the triangle and an acute angle. This acute angle is congruent to the angle created by a nearby object and its shadow at the same time of day. Because two angles are known to be congruent, the Angle-Angle Similarity Statement affirms the triangles are similar and the side lengths are proportional. Lesson 7: Proving Similarity Pre-Assessment, pp. 370–371 1. a 2. c 3. b 4. a 5. b Warm-Up 1.7.1, p. 374 1. Two angles in each triangle are congruent, as noted by the arc marks. According to the Angle-Angle Similarity Statement, if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. 2. x = 5.5 cm; y = 8.75 cm; z = 3.5 cm Practice 1.7.1: Proving Triangle Similarity Using Side-Angle-Side (SAS) and Side-Side-Side (SSS) Similarity, pp. 388–390 1. ABC ∼ DEF by SAS 2. ABC ∼EDF by SSS 3. ABD ∼ECD by SAS 4. ABC ∼FED ; SSS 5. not similar; corresponding sides are not proportional 6. ABC ∼ DEF ; SAS 7. not similar; corresponding sides are not proportional 8. x = 9 9. x = 2 2/3 10. x = 5 Warm-Up 1.7.2, p. 391 1. x = 3 2. 7 units 3. y = 4 4. 25 units U1-749 © Walch Education CCGPS Analytic Geometry Teacher Resource Practice 1.7.2: Working with Ratio Segments, pp. 408–411 1. BD = 3.8 units 2. BE = 11 2/3 units 3. EC = 9 units 4. BD = 4 units 5. CD = 3.5 units; BD = 2.5 units 6. CD = 7.5 units; CB = 9 units 7. No; the sides are not proportional. 8. Yes; the sides are proportional. 9. No; the sides are not proportional. 10. Yes; the sides are proportional. 6. ∠ C ≅ ∠D ∠ C is a right angle. 6. Corresponding angles of congruent triangles are congruent. Warm-Up 1.7.4, p. 433 1. ACE ∼ ABD 2. 238 feet 1. 44,980 ≈ 212.1 cm 2. No, the diagonals are not congruent. 3. To be “square,” the lengths of each diagonal must be congruent. The lengths are not congruent. The door, as assembled, is not square. Practice 1.7.3: Proving the Pythagorean Theorem Using Similarity, pp. 428–432 1. x = 6 units 2. x = 10 units 3. x = 8 3 ≈ 13.9 units 4. x = 12 units; f = 28.8 units 5. b = 65 units; f = 144 units 6. x = 6 units; a = 7.5 units; b = 10 units 7. a = 4 5 ≈ 8.9 units ; f = 4 units 8. x = 4 6 ≈ 9.8 units ; a = 33 ≈ 11.5 units; b = 4 22 ≈ 18.8 units 9. a = 20 units; b = 15 units; c = 25 units 10. 1. ABC with sides of length a, b, and c, where c2 = a2 + b2 Draw DE equal in length to a on line l. Draw line m perpendicular to DE . Locate a point, F, on m so that DF is equal in length to b. Connect F and E. Call this segment x. 5. Side-Side-Side Congruence Statement 7. ABC is a right triangle 7. ∠C is a right angle. Warm-Up 1.7.3, p. 412 Statements 5. ABC ≅FED 3. 176 sections Practice 1.7.4: Solving Problems Using Similarity and Congruence, pp. 451–455 1. 6 m 2. 3 ft 3. 252 ft 4. 700 ft 5. 54 m 6. 5.7 m 7. 27.5 m 8. 33.8 ft 9. 3.2 m 10. 84 m Progress Assessment, pp. 456–461 1. c 2. b 3. d 4. a 5. c 11. Answers: 6. b 7. c 8. b 9. a 10. c a. ABC ∼ ADE because of the Side-Angle-Side (SAS) AD 14 2 = = Similarity Statement. ∠BAC ≅ ∠DAE . DB 21 3 AE 17.5 2 = = ; therefore, the side lengths are and EC 26.25 3 proportional. Reasons 1. Given b. DE BC because of the converse to the Triangle Proportionality Theorem, which states that if a line divides two sides proportionally, then the line is parallel to the third side. Lesson 8: Proving Theorems About Lines and Angles Pre-Assessment, pp. 462–464 1. d 2. c 3. b 2. FED is a right triangle. 2. Line m is perpendicular to line l. 3. a2 + b2 = x2 3. Pythagorean Theorem 4. x2 = c2, or x = c 4. Segments are congruent. 4. a 5. d Warm-Up 1.8.1, p. 468 1. 2. 3. 4. m∠1 m∠2 m∠3 m∠ 4 must be between 75º and 105º. must be between 75º and 105º. must be between 75º and 105º. must be between 75º and 105º. U1-750 CCGPS Analytic Geometry Teacher Resource © Walch Education Practice 1.8.1: Proving the Vertical Angles Theorem, pp. 494–497 1. Answers may vary. Sample answer: Adjacent angles are ∠2 and ∠3 as well as ∠3 and ∠ 4 . Nonadjacent angles are ∠1 and ∠ 4 as well as ∠5 and ∠2 . 2. Supplementary angles are ∠1 and ∠2 as well as ∠1 and ∠5 . Statement: m∠1 + m∠2 = 180; m∠1 + m∠5 = 180 . 3. ∠2 and ∠5 are vertical angles. Statement: ∠2 ≅ ∠5 . 4. ∠2 and ∠3 are complementary angles. Statement: m∠2 + m∠3 = 90 . 5. 68º 6. 98º 7. 30º 8. 37º 9. Answers may vary. Sample answer: Given that ∠1 and ∠2 form a linear pair and ∠2 and ∠3 form a linear pair, prove that ∠1 ≅ ∠3. l m 3 4 ∠1 and ∠2 are supplementary and ∠2 and ∠3 are supplementary because of the definition of a linear pair. ∠1 ≅ ∠3 because angles supplementary to the same or congruent angles are congruent. 10. Answers may vary. Sample answer: Statements 1. DB is the perpendicular bisector of AC . E is a point on DB . 2. B is the midpoint of AC . 3. AB ≅ BC 4. ∠ EBA and ∠EBC are right angles. 5. EB ≅ EB 6. EBA ≅EBC 7. EA ≅ EC 8. EA = EC 1. m∠2 = 70 2. Completed table: Angle relationship used to determine measure Angle Measure 1 110º Linear pairs are supplementary. ∠1 and ∠2 are a linear pair. 2 70º The sum of the interior angles of a triangle equals 180º. 3 110º Vertical angles are congruent. ∠1 and ∠3 are vertical angles. 4 80º Corresponding angles of similar triangles are congruent. ∠ 4 and ∠7 are corresponding angles in similar triangles. 5 100º Linear pairs are supplementary. ∠ 4 and ∠5 are a linear pair. 6 100º Vertical angles are congruent. ∠5 and ∠6 are vertical angles. 7 80º Given 8 30º Given Practice 1.8.2: Proving Theorems About Angles in Parallel Lines Cut by a Transversal, pp. 522–524 2 1 Warm-Up 1.8.2, p. 498 Reasons 1. Given 1. 74º, because the given angles are corresponding angles and corresponding angles of a set of parallel lines intersected by a transversal are congruent. 2. 67º because alternate interior angles in a set of parallel lines intersected by a transversal are congruent. 3. 139º because same-side interior angles in a set of parallel lines intersected by a transversal are supplementary. 4. 42º because the given angle is a vertical angle with ∠7 , which is a corresponding angle with ∠5 , and in parallel lines intersected by a transversal, corresponding angles are congruent. 5. 167º because same-side exterior angles in a set of parallel lines intersected by a transversal are supplementary. 6. 115º 7. 94º 8. m ∠1 = 79, m∠2 = m∠3 = 101 , x = 19, and y = 11; same-side exterior angles are supplementary in a set of parallel lines intersected by a transversal. 2. Definition of perpendicular bisector 3. Definition of midpoint 4. Definition of perpendicular bisector 5. Reflexive Property 6. SAS 7. CPCTC 8. Definition of congruence U1-751 © Walch Education CCGPS Analytic Geometry Teacher Resource 9. l m 1 2 n 3 4 Statements Reasons 1. m n and l is the transversal. 2. ∠ 1 and ∠2 are a linear pair. ∠ 3 and ∠ 4 are a linear pair. 3. ∠ 1 and ∠2 are supplementary. ∠ 3 and ∠ 4 are supplementary. 1. Given 2. Definition of linear pair 3. If two angles form a linear pair, then they are supplementary. 4. Alternate interior angles 5. Substitution 4. ∠1 ≅ ∠ 4 ∠2 ≅ ∠3 5. ∠1 and ∠3 are supplementary. ∠ 2 and ∠ 4 are supplementary. 11. m ∠3 = 92; justifications may vary. Sample answer: First label the vertices and lines in the diagram with points. Then, draw in another line that passes through vertex C so that it is parallel to the two given parallel lines. Determine angle relationships with the angles that were given. ∠2 can now be labeled with ∠ABC. m∠ABC = 46 . ∠ABC is an alternate interior angle with ∠BCF . Therefore, they are congruent. By congruence of angles, the measures are equal. This means that m∠BCF = 46 . From the Angle Addition Postulate, m∠3 = m∠BCF + m∠FCE. Find m ∠ FCE. Using AE as the transversal to parallel lines AB and DE , m∠GAC = m∠2 = 134 and this angle is an alternate interior angle with ∠JEC . Therefore, these angles are congruent and have a measure of 134º. Now, using FC DE and AE as the transversal, ∠FCE is a same-side interior angle with ∠JEC . Same-side interior angles are supplementary. This means that m∠FCE + m∠JEC = 180. By substituting in m∠JEC = 134 and following with the Subtraction Property, m∠FCE = 46. m∠3 = m∠BCF + m∠FCE. Therefore, m∠3 = 92. G A B H 46º 134º 46º C 46º F l 10. 1 3 2 I m 4 134º E J D Lesson 9: Proving Theorems About Triangles Pre-Assessment, pp. 533–535 n 5 6 7 8 L ine l is a transversal and m n. ∠1 ≅ ∠5 and ∠2 ≅ ∠6 because they are corresponding angles. ∠5 ≅ ∠8 and ∠6 ≅ ∠7 because of vertical angles. Therefore, ∠1 ≅ ∠8 and ∠2 ≅ ∠7 by the Transitive Property. Progress Assessment, pp. 525–532 1. a 2. c 3. d 4. b 5. b 6. c 7. b 8. d 9. a 10. c 1. b 2. c 3. b 4. a 5. c Warm-Up 1.9.1, p. 539 1. 70˚ 2. 20˚ 3. Sample response: The angle created by the mirror and the reflected ray is congruent to the angle created by the mirror and the incident ray. This is true because we are told that the angle of incidence is congruent to the angle of reflection. The angle created by the mirror and the flashlight is the complement of the angle of incidence. The angle created by the mirror and the reflected ray is also the complement to the angle of reflection. By subtracting the angle of reflection from 90˚, we are able to determine that the angle measures 20˚. U1-752 CCGPS Analytic Geometry Teacher Resource © Walch Education Practice 1.9.1: Proving the Interior Angle Sum Theorem, pp. 559–561 1. m∠B = 88 2. m∠C = 37 3. m∠A = 22; m∠B = 140 4. m∠A = m∠B = m∠C = 60 5. m∠A = 62; m∠B = 37 6. m∠A = 156; m∠B = 24 7. m∠CAB = 80; m∠ABC = 46 8. m∠CAB = 55; m∠ABC = 35 9. m∠CAB = 14; m∠ABC = 155 10. A line drawn parallel to AB of ABC through C creates two exterior angles, ∠4 and ∠5. The measures of ∠4, ∠3, and ∠5 are equal to 180˚ because of the Angle Addition Postulate and the definition of a straight angle. ∠1 is congruent to ∠4 and ∠2 is congruent to ∠5 because of the Alternate Interior Angles Theorem. It follows that the measures of ∠1 and ∠4 are equal and the measures of ∠2 and ∠5 are equal because of the definition of congruent angles. By substitution, the sum of the measures of ∠1, ∠2, and ∠3 is 180˚. Warm-Up 1.9.2, p. 562 1. ABC has two congruent sides, so by definition, ABC is an isosceles triangle. 2. 47˚ 3. 94˚ Practice 1.9.2: Proving Theorems About Isosceles Triangles, pp. 582–585 1. m∠B = 80; m∠C = 50 2. m∠B = 57.5; m∠C = 65; m∠D = 57.5 3. m∠A = m∠C = 30; m∠B = 120 4. m∠A = m∠B = 24; m∠C = 132 5. x = 2 6. x = 90; y = 135 7. x = 17; y = 15 8. ABC is isosceles; ∠A ≅ ∠B . 9. ABC is not isosceles. 10. Statements 1. Draw a line AD perpendicular to BC . 2. ∠ADB and ∠ADC are right angles and are congruent. 3. ADB and ADC are right triangles. 4. AD ≅ AD 5. ∠B ≅ ∠C 6. ADB ≅ ADC 7. AB ≅ AC 8. ABC is isosceles. Reasons 1. Given a point and line, there is only one line perpendicular though the point. 2. Perpendicular lines form right angles; right angles are congruent. 3. Definition of right triangles 4. Reflexive Property 5. Given 6. AAS Congruence Statement 7. Congruent Parts of Congruent Triangles are Congruent 8. Definition of isosceles triangle Warm-Up 1.9.3, p. 586 1. approximately 3,780 feet 2. (2, 1) Practice 1.9.3: Proving the Midsegment of a Triangle, pp. 611–616 1. BC = 13.6; XZ = 7.1; m∠BZX = 20 2. AC = 15; YZ = 6.25; m∠XZY = 43 3. YZ = 18 4. BC = 7 5. (–12, –4), (3, –4), and (7, 8) 6. (–2, 3), (2, –2), and (5, 5) 7. Use the slope formula to show that EF and BC have the 1 same slope equal to − . Therefore, EF BC . Use the 2 distance formula to find EF = 17 and BC = 2 17 . 1 So, EF = BC . 2 8. Use the slope formula to show that EF and AB have the 5 same slope equal to − . Therefore, EF AB . Use the 2 distance formula to find EF = 29 and BC = 2 29 . 1 So, EF = AB . 2 9. The midpoint of AC = (–1, –2); the midpoint of BC = (7, –1); use the slope formula to show that EF 1 and AB have the same slope equal to . Therefore, 8 EF AB . Use the distance formula to find EF = 65 and 1 BC = 2 65 . So, EF = AB . 2 U1-753 © Walch Education CCGPS Analytic Geometry Teacher Resource 10. The midpoint of AC = (0, –5); the midpoint of BC = (3, 0); use the slope formula to show that EF 5 and AB have the same slope equal to . Therefore, 3 EF AB . Use the distance formula to find EF = 34 and 1 BC = 2 34 . So, EF = AB . 2 Warm-Up 1.9.4, p. 617 1 1. y = − x + 6 2 2. 178.9 yards Practice 1.9.4: Proving Centers of Triangles, pp. 656–658 1. (2, –4) is the circumcenter of ABC because the distance from this point to each of the vertices is 2 5 . 2. (8, 6) is the orthocenter of ABC because this point is a 1 26 solution to the equation of each altitude: y = − x + , 3 3 x = 8, and y = x – 2. 3. The midpoints of ABC are T (3, 4.5), U (–7.5, 0), and V (–4.5, 4.5). The equations of each of the medians of the 2 1 15 triangle are y = x + 5 , y = –x, and y = x + . (–3, 3) is 3 4 4 a solution to the equation of each median. 4. The distance from A to U is approximately 16.23 units. 2 The distance from (–3, 3) to A is 10.82. (–3, 3) is the 3 distance from A. The distance from B to V is approximately 6.36 units. The distance from (–3, 3) to B is approximately 2 4.24 units. (–3, 3) is the distance from B. The distance 3 from C to U is approximately 18.55 units. The distance 2 from (–3, 3) to C is approximately 12.37 units. (–3, 3) is 3 the distance from C. 5. ABC is an obtuse triangle. The incenter is the intersection of the angle bisectors. The incenter of a triangle is always inside the triangle. 6. ABC is an obtuse triangle. The orthocenter of an obtuse triangle is always outside of the triangle. 7. It is given that ABC has angle bisectors AN , BP , and CM , and that XR ⊥ AB , XS ⊥ BC , and XT ⊥ AC . Because any point on the angle bisector is equidistant from the sides of the angle, XR = XT, XT = XS, and XR = XS. By the Transitive Property, XR = XT = XS. 8. The circumcenter and orthocenter cannot be used to determine the location of the fire station. The triangle created when the towns are connected is obtuse. The circumcenter and orthocenter would fall outside the triangle created. 9. The incenter should be determined to achieve the largest possible pond. Once the incenter is found, a circle could be inscribed for the area of the pond. 10. The incenter should be determined. This center of the triangle is equidistant to the swings, basketball court, and gazebo. All other locations for the fountain would create different distances between each attraction. Progress Assessment, pp. 659–664 1. b 6. d 2. d 7. b 3. c 8. c 4. a 9. d 5. a 10. c 11. Answers: a. The midpoint of AB is (–1, 5.5), the midpoint of BC is (0, 3.5), and the midpoint of AC is (–2, 3). b. The length of each midsegment is one-half the length of the side it is parallel to. c. The centroid is located at (–1, 4). Lesson 10: Proving Theorems About Parallelograms Pre-Assessment, p. 665 1. a 2. b 3. a 4. d 5. c Warm-Up 1.10.1, p. 668 1. m∠1 m∠2 45º 135º 60º 120º 75º 105º 90º 90º 2. ∠1 and ∠2 are same-side interior angles. Same-side interior angles are supplementary. Therefore, use the equation as follows: m∠1 + m∠2 = 180 m∠2 = 180 − m∠1 Substitute each given measure of ∠1 into the equation in order to solve for the measure of ∠2 . Practice 1.10.1: Proving Properties of Parallelograms, pp. 693–695 1. Yes, it’s a parallelogram because opposite sides are parallel; mAB = mCD = 4 and mBC = mDA = 0 . 2. Yes, it’s a parallelogram because opposite sides are parallel; 5 mFG = mHI = 1 and mGH = mIF = − . 3 U1-754 CCGPS Analytic Geometry Teacher Resource © Walch Education 3. No, it’s not a parallelogram because opposite sides are not congruent: JK = 26 , LM = 5, KL = 10 , and MJ = 3. 4. Yes, it’s a parallelogram because opposite sides are congruent: MN = OP = 37 and NO = PM = 2 13 . 5. No, it’s not a parallelogram because the midpoints of the diagonals are not the same, indicating that the diagonals 7 do not bisect each other: M PR = 1, and M QS = ( 2,3) . 2 6. Yes, it’s a parallelogram because the midpoints of the diagonals are the same, indicating that the diagonals bisect each other: M = (6, 4) . 7. m∠A = m∠C = 104 and m∠B = m∠D = 76 ; x = 12 and y = 15 8. m∠A = m∠C = 82 and m∠B = m∠D = 98 ; x = 14 and y = 20 9. Given that quadrilateral ABCD is a parallelogram, by definition opposite sides are parallel and opposite sides are congruent. This means that AD BC , with the diagonals acting as transversals and AD ≅ BC . By alternate interior angles, ∠BDA ≅ ∠DBC and ∠DAC ≅ ∠ACB . By ASA, DPA ≅ BPC . 10. Statements Reasons 1. ABCD, EBHG , FIJG 1. Given 2. ∠D ≅ ∠B ∠B ≅ ∠G ∠G ≅ ∠I 3. ∠D ≅ ∠I 2. Opposite angles in a parallelogram are congruent. 3. Transitive Property Warm-Up 1.10.2, p. 696 1. The distance between Carrollton and Campton is about 84 miles. 2. Atlanta is at (6, 3.25) on the grid. 3. Each city is about 42 miles from Atlanta. 5 4. The slope is . 16 16 5. − 5 Practice 1.10.2: Proving Properties of Special Quadrilaterals, p. 729 1. Quadrilateral ABCD is a parallelogram and a rectangle. Justification: opposite sides are parallel, adjacent sides are perpendicular, the diagonals bisect each other, and not all four sides are congruent. 2. Quadrilateral EFGH is a kite. Justification: adjacent sides are congruent and the diagonals intersect at a right angle. 3. Quadrilateral IJKL is a parallelogram, a rectangle, a rhombus, and a square. Justification: opposite sides are parallel, consecutive sides are perpendicular, the diagonals bisect each other, and all four sides are congruent. 4. Quadrilateral MNOP is a parallelogram, a rhombus, and a square. Justification: opposite sides are parallel, adjacent sides are perpendicular, the diagonals are perpendicular, and all four sides are congruent. 5. Quadrilateral PQRS is a parallelogram. Justification: opposite sides are parallel, plus the diagonals are congruent and bisect each other. 6. Quadrilateral TUVW is an isosceles trapezoid. Justification: one pair of opposite sides is parallel and the other pair of sides is congruent. 7. Quadrilateral WXYZ is a parallelogram and a rectangle. Justification: opposite sides are parallel, adjacent sides are perpendicular, the diagonals bisect each other, and not all four sides are congruent. 8. Quadrilateral ABCD is a parallelogram, a rectangle, a rhombus, and a square. Justification: opposite sides are parallel, consecutive sides are perpendicular, the diagonals bisect each other, and all four sides are congruent. 9. Given that quadrilateral ABCD is a rhombus, the diagonals of a rhombus bisect each other and are perpendicular. Therefore, DP ≅ PB and AP ≅ PC . The diagonals of a rhombus bisect the opposite pairs of angles in the rhombus so that ∠DAP ≅ ∠PAB ≅ ∠DCP ≅ ∠BCP . The diagonals of a rhombus are perpendicular. By the definition of perpendicular lines, four right and congruent triangles are formed. Therefore, ∠DPA ≅ ∠APB ≅ ∠BPC ≅ ∠CPD . By ASA, APD ≅ APB ≅CPB ≅CPD . 10. The slopes are opposite reciprocals of each other: c −c mAC = and mBD = . The 2 2 b+ b +c b2 + c 2 − b product of the slopes is –1, showing that they are opposite reciprocals of each other. Progress Assessment, pp. 730–732 1. a 2. b 3. a 4. d 5. d 6. d 7. c 8. a 9. b 10. d 11. Quadrilateral BCDE is a parallelogram. Justifications may vary. Students should have at least two of the following justifications: Opposite sides are parallel 1 mBC = mDE = 1 and mCD = mBE = 2 ; opposite sides are congruent ED = BC = 4 2 and CD = BE = 2 5 ; and/or ( ) diagonals bisect each other because the midpoints of the diagonals are the same (M = (3, 2)). U1-755 © Walch Education CCGPS Analytic Geometry Teacher Resource Unit Assessment 14. Because they are vertical angles, m∠AHB = m∠EHD . pp. 733–741 1. b 2. b 3. b 4. b 5. d 6. a 13. a. Because AB DE , m∠ABH = m∠EDH . These two pairs of 7. a 8. c 9. d 10. b 11. d 12. c equal angles indicate that the AA Similarity Postulate can be used: HAB ∼HED . Because they are corresponding sides of similar triangles, HA = BH . Using the property DH BH DH of proportions, we conclude that = . HA HE C HE 15. A Statements B b. N o, it is not possible to construct a second noncongruent triangle because the parts of the triangle given are two sides and the included angle. The SideAngle-Side Congruence Statement ensures that two triangles are congruent given this information. Reasons 1. ZA CH 2. m∠AZO = m∠CHO 1. Given 2. Alternate Interior Angles Theorem 3. Given 3. ZC AH 4. m∠CZO = m∠AHO 4. Alternate Interior Angles Theorem 5. Reflexive Property 5. ZH = ZH 6. ASA Congruence Statement 6. CZH ≅ AHZ 7. CPCTC 7. AH = ZC 8. m∠ZCO = m∠HAO 8. Alternate Interior Angles Theorem 9. ASA Congruence Statement 9. ZCO ≅ HAO 10. CPCTC 10. ZO = OH U1-756 CCGPS Analytic Geometry Teacher Resource © Walch Education UNIT 1 • SIMILARITY, CONGRUENCE, AND PROOFS Student Book Answer Key Practice 1.1.1: Investigating Properties of Parallelism and the Center, pp. 14–18 Practice 1.3.1: Constructing Equilateral Triangles Inscribed in Circles, pp. 96–97 1. The triangle has been dilated. The corresponding sides of the triangle are parallel and the scale factor is consistent (k = 1/2). Additionally, the preimage points and image points are collinear with the center of dilation. 2. The quadrilateral has not been dilated. The scale factor is inconsistent between corresponding sides. 3. The rectangle has not been dilated. The scale factor is inconsistent between corresponding sides. 4. The rectangle has been dilated. The corresponding sides of the rectangle are parallel and the scale factor is consistent (k = 1.6). Additionally, the preimage points and image points are collinear with the center of dilation. 5. k = 1/3; reduction 6. k = 0.4; reduction 7. k = 1; congruency transformation 8. k = 1.25; enlargement; the scale factor is greater than 1; therefore, it is an enlargement. 9. No, because the scale factors of corresponding sides are inconsistent. 10. k = 1.5; enlargement; the scale factor is greater than 1; therefore, it is an enlargement. 1–3. Check students’ work for accuracy. Be sure each of the vertices lies on the circle. 4. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is twice the length of the given segment. 5. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment. 6–8. Check students’ work for accuracy. Be sure each of the vertices lies on the circle. 9. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is twice the length of the given segment. 10. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment. Practice 1.3.2: Constructing Squares Inscribed in Circles, p. 108 1. 10.125 2. 28.5 3. 4.6 4. 3/5 5. T' (–3, –1), U' (–2, –2), V' (–2/3, –1) 6. B' (–2, 0), D' (–10, –12), E' (6, –8) 7. N' (–9.6, –3.2), O' (4.8, 8), P' (6.4, –12.8) 8. E' (1.2, 2.7), F' (1.5, 0.9), G' (2.7, 3) 9. I" (3.375, 2.8125), J" (1.125, 1.125), K" (–1.6875, 2.25); k = 9/16 or 0.5625 10. 6 feet by 8 feet 1–2. Check students’ work for accuracy. Be sure each of the vertices lies on the circle. 3–6. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to the length of the given segment. 7. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to twice the length of the given segment. 8. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment. 9–10. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment. Practice 1.2.1: Copying Segments and Angles, pp. 42–43 Practice 1.3.3: Constructing Regular Hexagons Inscribed in Circles, pp. 123–124 1–10. Check students’ work for accuracy. 1. Check students’ work for accuracy. Be sure each of the vertices lies on the circle. 2–3. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to the length of the given segment. 4. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to twice the length of the given segment. Practice 1.1.2: Investigating Scale Factors, p. 24 Practice 1.2.2: Bisecting Segments and Angles, pp. 59–60 1–10. Check students’ work for accuracy. Practice 1.2.3: Constructing Perpendicular and Parallel Lines, p. 78 1–10. Check students’ work for accuracy. U1-757 © Walch Education CCGPS Analytic Geometry Teacher Resource 5. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment. 6. Check students’ work for accuracy. Be sure each of the vertices lies on the circle. 7–8. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to the length of the given segment. 9. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to twice the length of the given segment. 10. Check students’ work for accuracy. Be sure each of the vertices lies on the circle and the radius of the circle is equal to half the length of the given segment. 5. 10 9 8 4 3 2 1 C C' A' C' A' A -3 -4 -5 -6 x B C -7 -8 -9 -10 6. 10 9 8 7 6 5 4 3 2 1 A' y A C B x -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 C' B B' A R -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -1 -2 1. Rotation; the orientation changed, but the images are not mirror reflections of each other. 2. Translation; 5 units left and 3 units up; the orientation stayed the same. 3. Reflection; the line of reflection is y = –3; the orientation changed and the preimage and image are mirror reflections of each other; the line of reflection is the perpendicular bisector of the segments connecting the vertices of the preimage and image. 4. y -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -1 -2 -3 B' 7 6 5 Practice 1.4.1: Describing Rigid Motions and Predicting the Effects, pp. 141–145 10 9 8 7 6 5 4 3 2 1 y B' x 7. -4 -5 -6 -7 -8 -9 -10 U1-758 CCGPS Analytic Geometry Teacher Resource © Walch Education 10. Answers may vary. Sample answer: First, reflect the hexagon over a horizontal line just below the solar panel. Then translate the reflected hexagon about 1 unit to the right and 1 unit down. 8. B' A A B C' C R A' B F C E' F' A' E D E'' F'' A'' D' B' D'' C' C'' B'' 9. Answers may vary. Sample answer: Translate both chairs 2 units to the right. Practice 1.4.2: Defining Congruence in Terms of Rigid Motions, pp. 157–161 Fireplace Chair 1 Coffee table Couch Chair 2 1. Congruent; a translation occurred 7 units to the left and 3 units up. Translations are rigid motions. 2. Congruent; a rotation has occurred. Rotations are rigid motions. 3. Not congruent; a vertical compression has occurred with a scale factor of 1/2. Compressions are non-rigid motions. 4. Not congruent; a horizontal compression has occurred with a scale factor of 2/3. Compressions are non-rigid motions. 5. Congruent; a reflection has occurred. Reflections are rigid motions. 6. Not congruent; a vertical stretch has occurred with a scale factor of 5/3. Stretches are non-rigid motions. 7. The outer triangle has been dilated by a scale factor of 2/3. Since dilations are non-rigid motions, the triangles are not congruent. 8. The target has undergone a rotation. Since rotations are rigid motions, the targets are congruent. 9. The art is a reflection. Since reflections are rigid motions, the A+ on top is congruent to the A+ on the bottom. 10. Answers may vary. Sample answer: The windowpanes are congruent. Pane 1 can be translated to the right 2 units to create pane 2. Pane 1 can be translated 4 units to the right to create pane 3. Panes 1, 2, and 3 can be reflected over the horizontal line passing through the bottom of the panes to create panes 4, 5, and 6. Since all the transformations described are rigid motions, the panes are congruent. U1-759 © Walch Education CCGPS Analytic Geometry Teacher Resource Practice 1.5.1: Triangle Congruency, pp. 172–175 1. Sample answer: NLM ≅ HIJ 2. Sample answer: PQR ≅ TSV 3. Sample answer: USW ≅ LNQ 4. ∠H ≅ ∠M , ∠I ≅ ∠N , ∠J ≅ ∠P , HI ≅ MN , IJ ≅ NP , HJ ≅ MP 5. ∠B ≅ ∠H , ∠D ≅ ∠J , ∠E ≅ ∠L , BD ≅ HJ , DE ≅ JL , BE ≅ HL 6. ∠N ≅ ∠T , ∠P ≅ ∠V , ∠R ≅ ∠X , NP ≅ TV , PR ≅ VX , NR ≅ TX 7. Yes, the triangles are congruent; DGA ≅ LJH . 8. Yes, the triangles are congruent; MNP ≅ RTS . 9. Yes, the triangles are congruent; BCD ≅ FGE . 10. No, the triangles are not congruent. Practice 1.5.2: Explaining ASA, SAS, and SSS, pp. 185–187 1. SAS 2. ASA 3. Congruency cannot be determined; the identified congruent parts form SSA, which is not a triangle congruence statement. 4. ASA 5. Congruency cannot be determined; the identified congruent parts form SSA, which is not a triangle congruence statement. 6. SSS 7. DEF ≅ TVS ; SAS 8. There is not enough information to determine if the pieces of wood are congruent. The information provided follows AAA, which is not a triangle congruence statement. 9. The pieces of fabric are congruent; SAS 10. The pieces of fabric are congruent; SAS or SSS Practice 1.6.1: Defining Similarity, pp. 200–204 1. ∠C = 44˚, ∠D = ∠E = 68˚, CB = 4.5, DF = 2.1 2. ∠L = ∠M = ∠P = 45˚, MP = 9, NP = 6 3. The triangles are similar. ABC can be dilated by a scale factor of 2 with the center at (0, 0) to obtain DEF . 4. The triangles are similar. ABC can be dilated by a scale factor of 1 with center at (0, 0) and then translated –3 units vertically and 2 units horizontally to obtain DEF . 5. Similarity transformations preserve angle measures. The triangles are not similar because the angle measures in each triangle are different. 6. The triangles are similar. ABC can be dilated by a scale factor of 2.5 with the center at (0, 0) and then translated 3 units vertically and 5 units horizontally to obtain DEF . 7. The triangles are similar. ABC can be dilated by a scale factor of 2/7 with the center at (0, 0) and then rotated 180˚ clockwise about the origin to obtain DEF . 8. The triangles are similar. ABC can be dilated by a scale factor of 3/2 with the center at (0, 0) and then reflected over the line x = 0 to obtain DEF . 9. Similarity transformations preserve angle measures. The triangles are not similar because the angle measures in each triangle are different. 10. The triangles are similar. ABC can be dilated by a scale factor of 5 with the center at (0, 0) and then rotated 90˚ counterclockwise about the origin to obtain DEF . Practice 1.6.2: Applying Similarity Using the Angle-Angle (AA) Criterion, pp. 211–215 1. There is not enough information to determine similarity. 2. Yes, ABC ∼ XYZ because of the AA Similarity Statement. 3. Yes, ABC ∼YXZ because of the AA Similarity Statement. 4. ABC ∼YXZ ; x = 3 1/3 5. ABC ∼ ZYX ; x = 2 6. BCD ∼BDA ; x = 10 2/3 7. 3 feet 8. 9 feet 9. 2.16 meters 10. 60 feet Practice 1.7.1: Proving Triangle Similarity Using Side-Angle-Side (SAS) and Side-Side-Side (SSS) Similarity, pp. 226–229 1. ABC ∼FDE by SSS 2. ACE ∼BCD by SAS 3. ABC ∼ DEF by SSS 4. not similar; corresponding sides are not proportional 5. ABC ∼EFD ; SAS 6. not similar; corresponding sides are not proportional 7. ABC ∼ DEF ; SSS 8. x = 7 9. x = 5 10. x = 6 Practice 1.7.2: Working with Ratio Segments, pp. 239–242 1. CD = 12 3/8 units 2. BC = 3 1/3 units 3. DE = 8.4 units 4. CD = 6 units 5. BC = 10 units; CD = 12 units 6. CB = 21 units; CD = 35 units 7. Yes; the sides are proportional. 8. No; the sides are not proportional. 9. No; the sides are not proportional. 10. Yes; the sides are proportional. Practice 1.7.3: Proving the Pythagorean Theorem Using Similarity, pp. 252–256 1. x = 4 2 ≈ 5.7 units 2. x = 3 26 2 ≈ 7.6 units U1-760 CCGPS Analytic Geometry Teacher Resource © Walch Education Subtraction Property m∠ 4 = 90 . By definition, ∠ 4 is a right angle. Therefore, ∠2 , ∠3 , and ∠ 4 are right angles. 10. x = 13 3. x = 3 ≈ 1.73 units 4. x = 5 6 2 ≈ 19.4 units 5. a = 2 6 ≈ 4.9 units; e = 1/5; f = 4 4/5 6. a = 600; b = 175; x = 168 7. c = 4 2 ≈ 5.7 units; e = f = 2 2 ≈ 2.8 units 8. a = 6 ≈ 2.4 units; b = 3 ≈ 1.7 units; x = 2 ≈ 1.4 units 9. e = 3.6; f = 6.4 10. A segment, DE , drawn on line l is equal in length to a. Locate a point F on a line m equal in length to b, which is perpendicular to DE . Connect points F and E. Call this segment x. FED is a right triangle, so a2 + b2 = x2. It is also true that a2 + b2 = c2, so x2 = c2, or x = c. ABC ≅FED by the Side-Side-Side Congruence Statement. For this reason, ∠C ≅ ∠D and ∠C must be a right angle, making ABC a right triangle. Practice 1.7.4: Solving Problems Using Similarity and Congruence, pp. 266–271 1. 33 ft 2. 8.75 ft 3. 58 1/3 ft 4. 649.6 m 5. 23 m 6. 3.2 ft 7. 33.8 ft 8. 22.3 m 9. 7.8 m 10. 30 m Practice 1.8.1: Proving the Vertical Angles Theorem, pp. 292–295 1. Answers may vary. Sample answer: Adjacent angles are ∠2 and ∠3 as well as ∠3 and ∠ 4 . Nonadjacent angles are ∠1 and ∠ 4 as well as ∠5 and ∠2 . 2. Supplementary angles are ∠7 , ∠1 , and ∠2 . Statement: m∠7 + m∠1 + m∠2 = 180 . 3. ∠1 and ∠ 4 are vertical angles. Statement: ∠1 ≅ ∠ 4 . 4. ∠ 4 and ∠5 are complementary angles. Statement: m∠ 4 + m∠5 = 90 . 5. 131º 6. 79º 7. 102º 8. 48º 9. Since AB ⊥ CD as this was given, ∠1 and ∠2 , ∠2 and ∠3, and ∠1 and ∠ 4 all form linear pairs. This means that those pairs of angles are also supplementary by the Supplement Theorem. Therefore, m∠1 + m∠2 = 180 , m∠2 + m∠3 = 180 , and m∠1 + m∠ 4 = 180 . Given that ∠1 is a right angle, by the definition of right angles, m∠1 = 90 . Use substitution so that 90 + m∠2 = 180 , and by the Subtraction Property, m∠2 = 90 . By definition, ∠2 is a right angle. Use substitution so that 90 + m∠3 = 180 , and by the Subtraction Property m∠3 = 90 . By definition, ∠3 is a right angle. Use substitution so that 90 + m∠ 4 = 180 , and by the Practice 1.8.2: Proving Theorems About Angles in Parallel Lines Cut by a Transversal, pp. 312–316 1. 88º, because alternate interior angles in a set of parallel lines intersected by a transversal are congruent. 2. 90º, because same-side interior angles in a set of lines intersected by a transversal are supplementary. 3. 86º, because alternate interior angles in a set of lines intersected by a transversal are congruent. 4. 57º, because same-side exterior angles in a set of parallel lines intersected by a transversal are supplementary. 5. 144º, because corresponding angles in a set of lines intersected by a transversal are congruent. 6. 111º 7. 95º 8. m∠1 = 100, m∠2 = m∠3 = 80 , x = 11, and y = 12 9. Statements 1. m n and l is the transversal. 2. ∠2 ≅ ∠6 3. ∠ 6 and ∠8 are a linear pair. 4. ∠ 6 and ∠8 are supplementary. 5. m∠6 + m∠8 = 180 6. m∠2 + m∠8 = 180 7. ∠2 and ∠8 are supplementary. Reasons 1. Given 2. Corresponding Angles Postulate 3. Definition of a linear pair 4. If two angles form a linear pair, then they are supplementary. 5. Supplement Theorem 6. Substitution 7. Supplement Theorem 10. Since l is a transversal and l ⊥ m, ∠1 is a right angle. m∠1 = 90 because of the definition of a right angle. Since lines m and n are parallel, ∠1 ≅ ∠2 because of the Corresponding Angles Postulate. m∠1 = m∠2 because of the definition of congruent angles. m∠2 = 90 by substitution. ∠2 is a right angle. Therefore, l ⊥ m because of the definition of perpendicular lines. Practice 1.9.1: Proving the Interior Angle Sum Theorem, pp. 333–336 1. m∠B = 20 2. m∠B = 56 3. m∠B = 81; m∠C = 21 4. m∠A = 85; m∠B = 52; m∠C = 43 5. m∠A = 92; m∠B = 29 6. m∠A = 128; m∠B = 52 7. m∠CAB = 26; m∠ABC = 24 8. m∠CAB = 25; m∠ABC = 65 9. m∠CAB = 10; m∠ABC = 125 U1-761 © Walch Education CCGPS Analytic Geometry Teacher Resource 10. ABC is a triangle. By the Triangle Sum Theorem, the sum of the measures of ∠1, ∠2, and ∠3 is equal to 180˚. If two angles form a linear pair, they are supplementary, so ∠3 and ∠4 are also supplementary. The sum of the measures of ∠3 and ∠4 is equal to 180˚ by the definition of supplementary angles. The sum of the measures of ∠3 and ∠4 is equal to the sum of the measures of ∠1, ∠2, and ∠3 by substitution. The measure of ∠4 is equal to the sum of the measures of ∠1 and ∠2 by the Subtraction Property. Practice 1.9.2: Proving Theorems About Isosceles Triangles, pp. 350–353 1. m∠A = 40; m∠C = 70 2. m∠ADB = 108; m∠DCB = 72; m∠DBC = 36 3. m∠A = 120; m∠B = m∠C = 30 4. m∠A = m∠C = 45; m∠B = 90 5. x = 3 6. x = 78; y = 51 7. x = 13 8. ABC is not isosceles. 9. ABC is isosceles; ∠B ≅ ∠C . 10. Given ABC is equiangular, prove ABC is equilateral. Since ABC is equiangular, ∠A ≅ ∠B and ∠B ≅ ∠C . By the converse of the Isosceles Triangle Theorem, AB ≅ BC and AC ≅ BC , so by the Transitive Property, AB ≅ BC ≅ AC . Therefore, ABC is equilateral. Practice 1.9.3: Proving the Midsegment of a Triangle, pp. 372–378 1. BC = 12; XZ = 7.5; m∠BZX = 55 2. BC = 21; YZ = 5.625; m∠AXY = 72 3. XY = 5 4. AB = 42 5. (–9, –2), (5, 8), (5, –2) 6. (2, 6), (4, 3), (8, 5) 7. Use the slope formula to show that EF and BC have 3 the same slope equal to − . Therefore, EF BC . 4 Use the distance formula to find EF = 5 and BC = 10. 1 So, EF = BC . 2 8. Use the slope formula to show that EF and BC have the same slope equal to –1. Therefore, EF BC . Use the distance formula to find EF = 2 and BC = 2 2 . 1 So, EF = BC . 2 9. The midpoint of AC = (1, 1); the midpoint of BC = (4, 2); use the slope formula to show that EF 1 and AB have the same slope equal to . Therefore, 3 EF AB . Use the distance formula to find EF = 10 and 1 AB = 2 10 . So, EF = AB . 2 10. The midpoint of AC = (1, 1); the midpoint of BC = (–1, 6); use the slope formula to show that EF 5 and AB have the same slope equal to − . Therefore, 2 EF AB . Use the distance formula to find EF = 29 and 1 AB = 2 29 . So, EF = AB . 2 Practice 1.9.4: Proving Centers of Triangles, pp. 406–408 1. (–1, –3) is the circumcenter of ABC because the distance from this point to each of the vertices is 10 . 2. (–2, 1) is the orthocenter of ABC because this point is a solution to the equation of each altitude: x = –2 and y = 1. ABC is a right triangle, so the orthocenter is the vertex B. 3. The midpoints of ABC are T (7.5, –0.5), U (4, –3), and V (3.5, 0.5). The equations of each of the medians of the 1 triangle are y = 2x – 11, y = –x + 4, and y = x − 2 . (5, –1) 5 is a solution to the equation of each median. 4. The distance from A to U is approximately 6.71 units. The 2 distance from (5, –1) to A is 4.47 units. (5, –1) is the 3 distance from A. The distance from B to V is approximately 6.36 units. The distance from (5, –1) to B is approximately 2 4.24 units. (5, –1) is the distance from B. The distance 3 from C to T is approximately 7.65 units. The distance from 2 (5, –1) to C is approximately 5.10 units. (5, –1) is the 3 distance from C. 5. ABC is an obtuse triangle. The incenter is the intersection of the angle bisectors. The incenter of a triangle is always inside the triangle. 6. ABC is a right triangle. The orthocenter of a right triangle is always on a vertex of the triangle. 7. It is given that ABC has perpendicular bisectors p, q, and r of AB , BC , and AC . X is on the perpendicular bisector of AB , so it is equidistant from A and B. AX = BX by the definition of equidistant. The perpendicular bisector of BC also has the point X, so BX = CX. AX = CX by the Transitive Property of Equality; therefore, AX = BX = CX. 8. The circumcenter and orthocenter cannot be used to determine the location of the trauma center. The triangle created when the towns are connected is obtuse. The circumcenter and orthocenter would fall outside the triangle created. 9. The incenter should be determined. This center of the triangle is equidistant to the each location within the park. All other locations for the first aid station would create different distances between each location. 10. The incenter should be determined to achieve the maximum amount of space for the dog. Once the incenter is found, the dog’s leash can be staked in the ground. U1-762 CCGPS Analytic Geometry Teacher Resource © Walch Education Practice 1.10.1: Proving Properties of Parallelograms, pp. 428–430 1. No, it’s not a parallelogram because opposite sides are not 2 1 parallel: mTU = 5 , mVW = 2 , mUV = − , mWT = . 3 3 2. Yes, it’s a parallelogram because opposite sides are parallel: 2 mWX = mYZ = 5 and mXY = mZW = . 5 3. Yes, it’s a parallelogram because opposite sides are congruent: GH = IJ = 10 and HI = JG = 26 . 4. No, it’s not a parallelogram because opposite sides are not congruent: AB = 13 , CD = 2 , BC = 29 , and DA = 3 2 . 5. Yes, it’s a parallelogram because the midpoints of the diagonals are the same, indicating the diagonals bisect each 3 other: M = 3, − . 2 6. No, the midpoints are not the same, indicating the diagonals do not bisect each other: M MO = (1,1) and 3 3 M NP = , . 2 2 7. m∠A = m∠C = 165 and m∠B = m∠D = 15 ; x = 15 and y = 21 8. m∠A = m∠C = 92 and m∠B = m∠D = 88; x = 18 and y = 24 9. Given that AB is parallel to DE , we can use the Alternate Interior Angles Theorem to show that m∠BAH = m∠DEH . Given that AD is parallel to BC , the same theorem indicates that m∠ABF = m∠EAD . From these pairs of angles, it can be seen that ABF ∼EDA . Using the full meaning of the triangles being similar, we can conclude that corresponding sides are in a constant ratio; or, in other AB BF = words, . ED DA b 10. Opposite sides are parallel: mAB = mCD = and a mBC = mDA = 0 . The diagonals bisect each other because a+c b they have the same midpoint: M = , . 2 2 Practice 1.10.2: Proving Properties of Special Quadrilaterals, pp. 453–454 1. Quadrilateral ABCD is a parallelogram, a rectangle, a rhombus, and a square. Justification: opposite sides are parallel, consecutive sides are perpendicular, the diagonals bisect each other, and all four sides are congruent. 2. Quadrilateral EFGH is a parallelogram and a rectangle. Justification: opposite sides are parallel, adjacent sides are perpendicular, the diagonals bisect each other, and not all four sides are congruent. 3. Quadrilateral JKLM is an isosceles trapezoid. Justification: one pair of opposite sides is parallel and the other pair of sides is congruent. 4. Quadrilateral NOPQ is a parallelogram. Justification: opposite sides are parallel, plus the diagonals are congruent and bisect each other. 5. Quadrilateral STUV is a parallelogram and a rhombus. Justification: opposite sides are parallel, adjacent sides are not perpendicular, the diagonals are perpendicular, and all four sides are congruent. 6. Quadrilateral WXYZ is a parallelogram, a rhombus, and a square. Justification: opposite sides are parallel, adjacent sides are perpendicular, the diagonals are perpendicular, and all four sides are congruent. 7. Quadrilateral ABCD is a kite. Justification: adjacent sides are congruent and the diagonals intersect at a right angle. 8. Quadrilateral FGHJ is a parallelogram and a rectangle. Justification: opposite sides are parallel, adjacent sides are perpendicular, the diagonals bisect each other, and not all four sides are congruent. 9. Statements Reasons 1. Quadrilateral ABCD is a square. 2. DP ≅ PB AP ≅ PC 1. Given 2. The diagonals are congruent and bisect each other. 3. The diagonals 3. AC ⊥ DB of a square are perpendicular. 4. Definition of 4. ∠APD ≅ ∠APB ≅ ∠BPC ≅ ∠CPD, perpendicular lines and they are right angles. 5. ∠APD ≅ ∠APB ≅ ∠BPC ≅ ∠CPD 5. All right angles are congruent. 6. SAS Congruence 6. APD ≅ APB ≅CPB ≅CPD Statement 10. Given APD ≅ APB ≅CPB ≅CPD , Corresponding Parts of Congruent Triangles are Congruent. This means that AD ≅ AB ≅ CB ≅ CD . Also by CPCTC, ∠PAD ≅ ∠PCB. This means that BC AD since these are alternate interior angles. Then, by CPCTC again, ∠BAP ≅ ∠PCD. This means that AB CD because these angles are alternate interior angles. Since opposite sides are parallel and the sides are all congruent, by definition, quadrilateral ABCD is a rhombus. U1-763 © Walch Education CCGPS Analytic Geometry Teacher Resource