* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chapter 5
Survey
Document related concepts
Steinitz's theorem wikipedia , lookup
Multilateration wikipedia , lookup
History of geometry wikipedia , lookup
Reuleaux triangle wikipedia , lookup
Riemann–Roch theorem wikipedia , lookup
Rational trigonometry wikipedia , lookup
Noether's theorem wikipedia , lookup
Four color theorem wikipedia , lookup
Euler angles wikipedia , lookup
Brouwer fixed-point theorem wikipedia , lookup
Trigonometric functions wikipedia , lookup
History of trigonometry wikipedia , lookup
Integer triangle wikipedia , lookup
Transcript
Chapter 5 Chapter 5 Maintaining Mathematical Proficiency (p. 229) 8. −4 + 0 1 + 7 1. M —, — = M(−2, 4) 2 2 ( ) 6x + 1 = 3 —— AQ = √[ 0 − (−4) ]2 + (7 − 1)2 — — 4x + 1 = 3 − 2x 4x + 1 + 2x = 3 − 2x + 2x 6x + 1 − 1 = 3 − 1 — = √(4)2 + (6)2 = √ 16 + 36 = √52 ≈ 7.2 units ( 3 + 9 6 + (−2) 2 2 6x = 2 ) ( ) x = —13 —— GH = √(9 − 3)2 + (−2 − 6)2 — —— ( −12+ 8 −22+ 0 ) ( 72 −22 ) ( 72 ) 3. M —, — = M —, — = M —, −1 UV = = ——— √[ 8 − (−1) ] + [ 0 − (−2) ] —— — √(8 + 1)2 + (2)2 = √92 + 4 — — = √81 + 4 = √ 85 ≈ 9.2 units 4. 7x + 12 = 3x 7x − 7x + 12 = 3x − 7x 12 = − 4x −3 = x The solution is x = −3. 5. 14 − 6t = t 14 − 6t + 6t = t + 6t 14 = 7t 2=t The solution is t = 2. 6. 5p + 10 = 8p + 1 5p + 10 − 10 = 8p + 1 − 10 5p = 8p − 9 5p − 8p = 8p − 9 − 8p −3p = −9 The solution is x = —13. — = √(6)2 + (−8)2 = √ 36 + 64 = √ 100 = 10 units 2 2 6 —6 x = —6 12 4 2 2 2. M —, — = M —, — = M(6, 2) 2 9. z − 2 = 4 + 9z z − 2 + 2 = 4 + 9z + 2 z = 6 + 9z z − 9z = 6 + 9z − 9z −8z = 6 6 −8 —z = — −8 −8 6 z=— −8 3 z = −— 4 3 The solution is z = −—4. 10. yes; The length can be found using the Pythagorean Theorem. Chapter 5 Mathematical Practices (p. 230) 1. theorem; It is the Slopes of Perpendicular Lines Theorem (Thm. 3.14) studied in Section 3.5. 2. theorem; It is the Linear Pair Perpendicular Theorem (Thm. 3.10) studied in Section 3.4. 3. definition; This is the definition of perpendicular lines. 4. postulate; This is the Two Point Postulate (Post. 2.1) studied in Section 2.3. In Euclidean geometry, it is assumed, not proved, to be true. p=3 The solution is p = 3. 7. w + 13 = 11w − 7 w + 13 − w = 11w − 7 − w 13 = 10w − 7 13 + 7 = 10w − 7 + 7 20 = 10w 2=w The solution is w = 2. 5.1 Explorations (p. 231) 1. a. Check students’ work. b. Check students’ work. c. The sum of the interior angle measures of all triangles is 180°. d. Check students’ work; The sum of the measures of the interior angles of a triangle is 180°. 2. a. Check students’ work. b. Check students’ work. c. Check students’ work. d. Check students’ work; The sum is equal to the measure of the exterior angle. e. Check students’ work; The measure of an exterior angle is equal to the sum of the measures of the two nonadjacent interior angles. Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 141 Chapter 5 3. The sum of the measures of the interior angles of a triangle is 180°, and the measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles. 4. 2x° + (x − 6)° = 90° 3x − 6 = 90 3x = 96 x = 32 4. The sum of the measures of the two nonadjacent interior angles is 32° and the third angle which is adjacent to the exterior angle has a measure of 180° − 32° = 148°. These are known because of the conjectures made in Explorations 1 and 2. 2x = 2(32) = 64 x − 6 = 32 − 6 = 26 The angle measurements are 26° and 64°. 5.1 Exercises (pp. 236–238) 5.1 Monitoring Progress (pp. 232–235) Vocabulary and Core Concept Check 1. Sample answer: Obtuse isosceles triangle: 1. no; By the Corollary to the Triangle Sum Theorem (Cor. 5.1), the acute angles of a right triangle are complementary. Because their measures have to add up to 90°, neither angle could have a measure greater than 90°. Acute scalene triangle: 2. The measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles. Monitoring Progress and Modeling with Mathematics 2. AB = BC = AC = —— — 3. Two sides are congruent and one angle is a right angle. — √(0 − 3)2 + (0 − 3)2 = √9 + 9 = √18 ≈ 4.2 —— √[ 3 − (−3) ] + (3 − 2 3)2 —— √[ 0 − (−3) ] + (0 − 2 3)2 — = √(3 + 3)2 = √ — 32 + 32 — = √ 62 So, △XYZ is a right isosceles triangle. =6 — = √9 + 9 — = √18 ≈ 4.2 Because AC = AB, that indicates that △ABC is isosceles. 3−0 3 —=— =—=1 Slope of AB 3−0 3 0 3−3 —=— =—=0 Slope of BC −3 − 3 −6 3 − 0 −3 — = −— = — = −1 Slope of AC 3−0 3 — and AC — equals −1, Because the product of the slopes of AB — ⊥ AC —, therefore △ABC is a right that indicates that AB triangle. So, △ABC is a right isosceles triangle. 3. 40° + 3x° = (5x − 10)° So, △LMN is an equiangular equilateral triangle. 5. None of the sides are congruent and one angle is obtuse. So, △JKH is an obtuse scalene triangle. 6. None of the sides are congruent and all angles are acute. So, △ABC is an acute scalene triangle. —— — —— — 7. AB = √ (6 − 2)2 + (3 − 3)2 = √ 42 = 4 BC = √(2 − 6)2 + (7 − 3)2 = √(−4)2 + 42 — — = √16 + 16 = √ 32 ≈ 5.7 —— — AC = √(2 − 2)2 + (7 − 3)2 = √42 = 4 △ABC is isosceles because AB = AC. 3−3 0 —=— =—=0 Slope of AB 6−2 4 4 7−3 —=— = — = −1 Slope of BC 2 − 6 −4 50 = 2x x = 25 m∠ 1 + 3(25°) + 40° = 180° m∠ 1 + 75° + 40° = 180° m∠ 1 + 115° = 180° m∠ 1 = 65° 142 4. All sides are congruent and therefore, all angles are congruent. Geometry Worked-Out Solutions 7−3 4 —= — Slope of AC = — = undefined 2−2 0 — has a slope of 0 and AC — has an undefined slope, Because AB — ⊥ AC —. There is a right angle at ∠ A, which makes △ABC AB a right triangle. So, △ABC is a right isosceles triangle. Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 —— — — 8. AB = √ (6 − 3)2 + (9 − 3)2 = √ 32 + 62 = √ 9 + 36 11. m∠ 1 + 78° + 31° = 180° — = √45 ≈ 6.7 m∠ 1 + 109° = 180° —— —— BC = √(6 − 6)2 + [ 9 − (−3) ]2 = √ (0)2 + (9 + 3)2 = — √ 122 = 12 —— —— AC = √ (6 − 3)2 +[ 3 − (−3) ]2 = √32 + (3 + 3)2 12. m∠ 1 + 30° + 40° = 180° — — = √9 + 36 = √ 45 ≈ 6.7 m∠ 1 + 70° = 180° △ABC is isosceles because AB = AC. 9−3 6 —=— =—=2 Slope of AB 6−3 3 −3 − 9 −12 —=— = — = undefined Slope of BC 6−6 0 −3 − 3 −6 — Slope of AC = — = — = −2 6−3 3 m∠ 1 = 110° The triangle is an obtuse triangle. 13. m∠ 1 + 38° + 90° = 180° m∠ 1 + 128° = 180° m∠ 1 = 52° None of the slopes of the sides of the triangle are negative reciprocals of each other. So, the triangle is not a right triangle. △ABC is an isosceles triangle. —— — m∠ 1 = 71° The triangle is an acute triangle. — The triangle is a right triangle. 14. m∠ 1 + 60° + 60° = 180° m∠ 1 + 120° = 180° 9. AB = √ (4 − 1)2 + (8 − 9)2 = √ 32 + (−1)2 = √ 9 + 1 m∠ 1 = 60° — = √10 ≈ 3.2 The triangle is an equiangular triangle. —— BC = √(2 − 4)2 — + (5 − 8)2 —— =√ = √13 ≈ 3.6 —— (−2)2 + (−3)2 — = √4 + 9 — AC = √ (2 − 1)2 + (5 − 9)2 = √ 12 + (−4)2 — — = √1 + 16 = √ 17 ≈ 4.1 15. m∠ 2 = 75° + 64° = 139° 16. x° + 45° = (2x − 2)° x = 2x − 47 −x = −47 △ABC is a scalene triangle. 8 − 9 −1 —=— =— Slope of AB 4−1 3 5 − 8 −3 3 — Slope of BC = — = — = — 2 − 4 −2 2 5 − 9 −4 — Slope of AC = — = — = −4 2−1 1 x = 47 ⋅ 2x − 2 = 2 47 − 2 = 94 − 2 = 92 The exterior angle has a measure of 92°. 17. 24° + (2x + 18)° = (3x + 6)° 42 + 2x = 3x + 6 None of the slopes of the sides of the triangle are negative reciprocals of each other. So, the triangle is not a right triangle. △ABC is a scalene triangle. ——— — = √4 + 36 = √ 40 ≈ 6.3 ——— —— BC = √(3 − 0)2 + [ −2 − (−3) ]2 = √ (3)2 + (−2 + 3)2 — — = √9 + 1 = √ 10 ≈ 3.2 ——— —— AC = √ [ 3 − (−2) ]2 + (−2 − 3)2 = √ (3 + 2)2 + (−5)2 — −x = −36 — 10. AB = √ [ 0 − (−2) ]2 + (−3 − 3)2 = √ 22 + (−6)2 — 2x = 3x −6 — = √25 + 25 = √ 50 ≈ 7.1 △ABC is a scalene triangle. −6 −3 − 3 —=— = — = −3 Slope of AB 0 − (−2) 2 −2 − (−3) −2 + 3 1 — Slope of BC = — = — = — 3−0 3 3 −5 −5 −2 − 3 — Slope of AC = — = — = — = −1 3 − (−2) 3 + 2 5 — has a slope of −3 and BC — has a slope of —1, Because AB 3 — — AB ⊥BC . There is a right angle at ∠ B, which makes △ABC a right triangle. So, △ABC is a right scalene triangle. ⋅ x = 36 3x + 6 = 3 36 + 6 = 114 The exterior angle has a measure of 114°. 18. (x + 8)° + 4x° = (7x − 16)° 5x + 8 = 7x − 16 −2x + 8 = −16 −2x = −24 x = 12 ⋅ 7x − 16 = 7 12 − 16 = 84 − 16 = 68 The exterior angle has a measure of 68°. 19. 3x° + 2x° = 90° 5x = 90 x = 18 ⋅ ⋅ 3x = 3 18 = 54 2x = 2 18 = 36 The two acute angles measure 36° and 54°. Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 143 Chapter 5 27. The sum of the measures of the interior angles of the triangle 20. (3x + 2)° + x° = 90° is 180°, not 360°. 4x + 2 = 90 115° + 39° + m∠ 1 = 180° 4x = 88 154° + m∠ 1 = 180° x = 22 m∠ 1 = 26° ⋅ 3x + 2 = 3 22 + 2 = 68 The two acute angles measure 22° and 68°. 21. (11x − 2)° + (6x + 7)° = 90° 17x + 5 = 90 x=5 11x − 2 = 11 5 − 2 = 55 − 2 = 53 ⋅ 6x + 7 = 6 5 + 7 = 30 + 7 = 37 The two acute angles measure 37° and 53°. 22. (19x − 1)° + (13x − 5)° = 90° 32x − 6 = 90 32x = 96 m∠ 1 = 80° + 50° = 130° 30. m∠ 2 = 180° − 50° = 130° 31. m∠ 3 = m∠ 1 = 50° 32. m∠ 4 = m∠ 2 = 130° 33. m∠ 5 = 90° − 50° = 40° 34. m∠ 6 = 180° − 40° = 140° 35. m∠ 7 = 90° 36. m∠ 8 = 180° − 40° = 140° x=3 ⋅ ⋅ of the measures of the two nonadjacent interior angles. 29. m∠ 1 = 90° − 40° = 50° 17x = 85 ⋅ 28. The measure of the exterior angle should be equal to the sum 19x − 1 = 19 3 − 1 = 57 − 1 = 56 13x − 5 = 13 3 − 5 = 39 − 5 = 34 The two acute angles measure 34° and 56°. 23. x° + 5x° = 90° 6x = 90 37. None of the sides and none of the angles of the triangle are equal, therefore it is a scalene triangle. When you measure the three angles of the triangle, you find that they are all acute. Therefore the triangle is acute. So, the triangle is an acute scalene triangle. 38. The following sets of angle measures could form a triangle: x = 15 B. 96° + 74° + 10° = 180° ⋅ 5x = 5 15 = 75 The two acute angles measure 15° and 75°. 24. x° + 8x° = 90° D. 101° + 41° + 38° = 180° E. 90° + 45° + 45° = 180° F. 84° + 62° + 34° = 180° 9x = 90 39. 2 x = 10 ⋅ 6 + x = 20 12 + x = 20 ⋅ 8x = 8 10 = 80 The two acute angles measure 10° and 80°. 25. x° + [ 3(x + 8) ]° = 90° x + 3x + 24 = 90 4x = 66 x = 16.5 x=8 2x + 6 = 20 2x = 14 x=7 You could make another bend 6 inches from the first bend and leave the last side 8 inches long, or you could make another bend 7 inches from the first bend and then the last side will also be 7 inches long. 40. Sample answer: ⋅ 3(x + 8) = 3(16.5 + 8) = 3 24.5 = 73.5 1 GO TEAM! The two acute angles measure 16.5° and 73.5°. 2 3 26. x° + [2(x − 12)]° = 90° x + 2x − 24 = 90 3x = 114 x = 38 ⋅ 2(x − 12) = 2(38 − 12) = 2 26 = 52 The two acute angles measure 38° and 52°. 144 Geometry Worked-Out Solutions When a triangular pennant is on a stick, the two angles on the top edge of the pennant (∠ 1 and ∠ 2) should have measures such that their sum is equal to the measure of the angle formed by the stick and the bottom edge of the pennant (∠ 3). Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 41. Given △ABC is a right triangle. 43. It is possible to draw an obtuse isosceles triangle. A Sample answer: Prove ∠ A and ∠ B are complementary. C B STATEMENTS REASONS 1. △ABC is a right triangle. 1. Given 2. ∠ C is a right angle. 2. Given (marked in diagram) 3. m∠ C = 90° 3. Definition of a right angle 4. m∠ A + m∠ B + m∠ C = 180° 4. Triangle Sum Theorem (Thm. 5.1) 5. m∠ A + m∠ B + 90° = 180° 44. It is possible to draw a right isosceles triangle. It will have angle measurements of 45°, 45°, and 90°. A right equilateral triangle is not possible, because the hypotenuse must be longer than either leg in a right triangle. 5. Substitution Property of Equality 6. m∠ A + m∠ B = 90° 45. a. AB + AB + BC = Perimeter x + x + 2x − 4 = 32 6. Subtraction Property of Equality 7. ∠ A and ∠ B are complementary. An obtuse equilateral triangle is not possible, because when two sides form an obtuse angle the third side that connects them must be longer than the other two. 4x − 4 = 32 4x = 36 x=9 7. Definition of complementary angles AB + BC + BC = Perimeter x + 2x − 4 + 2x − 4 = 32 5x − 8 = 32 42. Given △ABC, exterior ∠ BCD 5x = 40 B Prove m∠ A + m∠ B = m∠ BCD x=8 A C STATEMENTS REASONS 1. △ABC, exterior ∠ BCD 1. Given 2. m∠ A + m∠ B + m∠ BCA = 180° 2. Triangle Sum Theorem (Thm. 5.1) 3. ∠ BCA and ∠ BCD form a linear pair. 3. Definition of linear pair 4. m∠ BCA + m∠ BCD = 180° 4. Linear Pair Postulate (Post. 2.8) 5. m∠ A + m∠ B + m∠ BCA = m∠ BCA + m∠ BCD 5. Transitive Property of Equality 6. m∠ A + m∠ B = m∠ BCD 6. Subtraction Property of Equality D b. AB + AB + BC = Perimeter x + x + 2x − 4 = 12 4x − 4 = 12 4x = 16 x=4 AB + BC + BC = Perimeter x + 2x − 4 + 2x − 4 = 12 5x − 8 = 12 5x = 20 x=4 There is one value for x, x = 4. 46. a. The triangle appears to have three congruent sides and three congruent angles. So, the triangle is equiangular, acute, equilateral, and isosceles. b. The triangle appears to have two congruent sides, no right angles, and no obtuse angle. So, the triangle is acute and isosceles. c. The triangle appears to have one obtuse angle and no congruent sides. So, the triangle is obtuse and scalene. d. The triangle appears to have a right angle and no congruent sides. So, the triangle is right and scalene. Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 145 Chapter 5 47. Exterior angle = Sum of the two nonadjacent interior angles ? A. 100° = 62° + 38° ? B. 81° = 57° + 24° 100° = 100° ✓ ? F. 149° = 101° + 48° 81° = 81° ✓ 53. Given ⃖⃗ AB ⃖⃗ CD Prove m∠ 1 + m∠ 2 + m∠ 3 = 180° B 2 149° = 149° ✓ A 1 D 3 48. no; According to the Exterior Angle Theorem (Thm. 5.2), the measure of an exterior angle of a triangle is always equal to the sum of the measures of the two nonadjacent interior angles. 49. By the Alternate Interior Angles Theorem (Thm. 3.2), x = 43. Use the Exterior Angle Theorem (Thm. 5.2) to find y. 75° = y° + 43° 32 = y So, x = 43 and y = 32. 50. By the Corresponding Angles Theorem (Thm. 3.1), x = 118. Use the Exterior Angle Theorem (Thm. 5.2) to find the value of y. 4 5 C STATEMENTS REASONS 1. ⃖⃗ AB ⃖⃗ CD 1. Given (marked in diagram) 2. ∠ ACD and ∠ 5 form a linear pair. 2. Definition of linear pair 3. m∠ ACD + m∠ 5 = 180° 3. Linear Pair Postulate (Post. 2.8) 4. m∠ 3 + m∠ 4 = m∠ ACD 4. Angle Addition Postulate (Post. 1.4) 5. m∠ 3 + m∠ 4 + m∠ 5 = 180° 5. Substitution Property of Equality 6. ∠ 1 ≅ ∠ 5 6. Corresponding Angles Theorem (Thm. 3.1) 7. ∠ 2 ≅ ∠ 4 7. Alternate Interior Angles Theorem (Thm. 3.2) 8. m∠ 1 = m∠ 5, m∠ 2 = m∠ 4 8. Definition of congruent angles 9. m∠ 3 + m∠ 2 + m∠ 1 = 180° 9. Substitution Property of Equality 118° = y° + 22° 96 = y So, x = 118 and y = 96. 51. The sum of the measures of the two acute angles of a right triangle is 90°. y° + 25° = 90° y = 65 Use the Exterior Angle Theorem (Thm. 5.2) to find the value of x. x° = 65° + 20° = 85° So, x = 85 and y = 65. 52. The sum of the measures of the two acute angles of a right Maintaining Mathematical Proficiency 54. (5x − 27)° = (3x + 1)° 2x − 27 = 1 2x = 28 x = 14 triangle is 90°. x° + 64° = 90° x = 26 By the Alternate Interior Angles Theorem (Thm. 3.2), the unmarked angle of the triangle containing y° measures 26°. Because both triangles contain a right angle and an angle measuring 26°, y° must be 64°. So, x = 26 and y = 64. E 3x + 1 = 3(14) + 1 = 42 + 1 = 43 So, m∠ KHL = 43°. 55. m∠ ABC = m∠ GHK (6x + 2)° = (5x − 27)° + (3x + 1)° 6x + 2 = 8x − 26 −2x = −28 x = 14 ⋅ 6x + 2 = 6 14 + 2 = 86 So, m∠ ABC = 86°. 56. 5y − 8 = 3y 2y = 8 y=4 ⋅ 3y = 3 4 = 12 So, GH = 12. 146 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 57. 3z + 6 = 8z − 9 −5z + 6 = −9 −5z = −15 z=3 ⋅ 3z + 6 = 3 3 + 6 = 9 + 6 = 15 So, BC = 15. — —— — — — 3. From the diagram, PS ≅ RQ , PT ≅ RT , and ST ≅ QT . Also, by the Vertical Angles Congruence Theorem (Thm. 2.6), — RQ —, and ∠ PTS ≅ ∠ RTQ. From the diagram, PS ∠ P ≅ ∠ R and ∠ S ≅ ∠ Q by the Alternate Interior Angles Theorem (Thm. 3.2). Because all corresponding parts are congruent, △PTS ≅ △RTQ. 4. ∠ NSR ≅ ∠ NDC and ∠ CND ≅ ∠ RNS, so by the Third 5.2 Explorations (p. 239) Angles Theorem (Thm. 5.4), ∠ SRN ≅ ∠ DCN. So, ∠ DCN = 75°. 1. translation, reflection, rotation; A rigid motion maps each part of a figure to a corresponding part of its image. Because rigid motions preserve length and angle measure, corresponding parts of a figure and its image are congruent. In triangles, this means that the corresponding sides and corresponding angles are congruent, which is sufficient to say that the triangles are congruent. 5. The additional information that is needed to conclude that — ≅ RS — or DN — ≅ SN —. △NDC ≅ △NSR is that CD 5.2 Exercises (pp. 243–244) Vocabulary and Core Concept Check 1. To show that two triangles are congruent, you need to show 2. a. Sample answer: Reflect △ABC in the x-axis and translate 3 units right. b. Sample answer: Rotate △ABC 180° about the origin. c. Sample answer: Rotate △ABC 270° counterclockwise about the origin and translate 3 units down. d. Sample answer: Reflect △ABC in the line y = x. 3. Look at the orientation of the original triangle and decide which rigid motion or composition of rigid motions will result in the same orientation as the second triangle. Then, if necessary, use a translation to move the first triangle so that it coincides with the second. 4. Sample answer: Reflect △ABC in the y-axis and translate 3 units right and 2 units down. 4 F −2 y C A E −2 2. “Is △JLK ≅ △STR?” is different. Because corresponding angles and sides are not congruent, △JLK is not congruent to △STR. For the other three questions, corresponding angles are congruent and corresponding sides are congruent. So, the triangles are congruent. Monitoring Progress and Modeling with Mathematics — —— —— — 3. corresponding sides: AB ≅ DE , BC ≅ EF , AC ≅ DF corresponding angles: ∠ A ≅ ∠ D, ∠ B ≅ ∠ E, ∠ C ≅ ∠ F Sample answer: △CBA ≅ △FED — —— —— — 4. corresponding sides: GH ≅ QR , HJ ≅ RS , JK ≅ ST , — ≅ QT — GK corresponding angles: ∠ G ≅ ∠ Q, ∠ H ≅ ∠ R, ∠ J ≅ ∠ S, ∠K ≅ ∠T Sample answer: GHJK ≅ QRST B D that all corresponding parts are congruent. If two triangles have the same side lengths and angle measures, then they must be the same size and shape. 4 x 5. ∠ N ≅ ∠ Y, so m∠ Y = 124°. 6. ∠ X ≅ ∠ M, so m∠ M = 33°. 5.2 Monitoring Progress (pp. 241–242) — —— —— — 1. corresponding sides: AB ≅ CD , BG ≅ DE , GH ≅ EF , — — AH ≅ CF corresponding angles: ∠ A ≅ ∠ C, ∠ B ≅ ∠ D, ∠ G ≅ ∠ E, ∠H ≅ ∠F 2. 4x + 5 = 105 4x = 100 x = 25 7. ∠ Z ≅ ∠ L and m∠ L = 180° − 124° − 33° = 23°. So, m∠ Z = 23°. — — 8. MN ≅ XY , so XY = 8. 9. 135 = 10x + 65 70 = 10x 7=x 4y − 4 = 28 4y = 32 y=8 So, x = 7 and y = 8. Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 147 Chapter 5 10. m∠ N = 180° − (142° + 24°) = 180° − 166° = 14° 16. Prove △ABG ≅ △DCF B m∠ N = m∠ U 14° = (2x − 50)° A 64 = 2x x = 32 STATEMENTS AB ≅ — DC , — AF ≅ — DG , 1. — — — BE ≅ CE ≅ — EF ≅ — EG , ∠ B ≅ ∠ C, ∠ A ≅ ∠ D NP = US 2x − y = 13 2(32) − y = 13 64 − y = 13 y = 51 So, x = 32 and y = 51. — —— —— —— —— — 11. VZ ≅ KJ , ZY ≅ JN , YX ≅ NM , XW ≅ ML , VW ≅ KL ∠ V ≅ ∠ K, ∠ Z ≅ ∠ J, ∠ Y ≅ ∠ N, ∠ X ≅ ∠ M, ∠ W ≅ ∠ L Because all corresponding parts of the polygons are congruent, VZYXW ≅ KJNML. — — — — 12. From the diagram, WX ≅ YZ and XY ≅ ZW . By the Reflexive — ≅ WY —. Also Property of Congruence (Thm. 2.1), WY from the diagram, ∠ X ≅ ∠ Z. Then, from the markings — ZW — and XW — ZY —. You can conclude in the diagram, XY that ∠ XYW ≅ ∠ ZWY and ∠ XWY ≅ ∠ ZYW by the Alternate Interior Angles Theorem (Thm. 3.2). Because all corresponding parts are congruent, △WXY ≅ △YZW. A D — DC —, AB — ≅ DC —, E 1. AB — is the midpoint of AC —. and BD 148 B 3. AF + FG = AG, DG + FG = DF, BE + EG = BG, CE + EF = CF 3. Segment Addition Postulate (Post. 1.2) 4. AF = DG, BE = CE = EF = EG 4. Definition of congruent segments 5. DG + FG = AG, BE + EG = CF 5. Substitution Property of Equality 6. DF = AG, BG = CF 6. Transitive Property of Equality 7. — DF ≅ — AG , — BG ≅ — CF 7. Definition of congruent segments 8. △ABG ≅ △DCF 8. All corresponding parts are congruent. ∠S ≅ ∠Y m∠ S = m∠ Y 18. In order to conclude that triangles are congruent, the sides must also be congruent; △MNP is not congruent to △RSP because the corresponding sides are not congruent. C REASONS 1. Given 2. ∠ AEB ≅ ∠ CED 2. Vertical Angles Congruence Theorem (Thm. 2.6) 3. ∠ BAE ≅ ∠ DCE, ∠ ABE ≅ ∠ CDE 3. Alternate Interior Angles Theorem (Thm. 3.2) 4. — AE ≅ — CE , — BE ≅ — DE 4. Definition of midpoint 5. △AEB ≅ △CED 1. Given (marked in diagram) m∠ S = 90° − 42° = 48° E Prove △AEB ≅ △CED REASONS corresponding parts are matched up correctly. (Thm. 5.4), ∠ C ≅ ∠ 1. By the Triangle Sum Theorem (Thm. 5.1), m∠ 1 = 180° − 45° − 80° = 55°. — and BD —. E is the midpoint of AC D 17. The congruence statement should be used to ensure that 14. ∠ B ≅ ∠ Q and ∠ A ≅ ∠ S, so by the Third Angles Theorem — —— — 15. Given AB DC , AB ≅ DC , G 2. Third Angles Theorem (Thm. 5.4) 13. ∠ L ≅ ∠ Z and ∠ N ≅ ∠ Y, so by the Third Angles Theorem (Thm. 5.4), ∠ 1 ≅ ∠ M. By the Triangle Sum Theorem (Thm. 5.1), m∠ 1 = 180° − 90° − 70° = 20°. F 2. ∠ BGA ≅ ∠ CFD −y = −51 STATEMENTS C E 5. All corresponding parts are congruent. Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 19. Given ∠ A ≅ ∠ D and ∠ B ≅ ∠ E 23. m∠ L + m∠ M + m∠ N = 180° Prove ∠ C ≅ ∠ F 40° + 90° + m∠ N = 180° 130° + m∠ N = 180° E B m∠ N = 50° m∠ N = m∠ R A C D F STATEMENTS REASONS 1. ∠ A ≅ ∠ D, ∠ B ≅ ∠ E 1. Given 2. m∠ A = m∠ D, m∠ B = m∠ E 2. Definition of congruent angles 3. m∠ A + m∠ B + m∠ C = 180°, m∠ D + m∠ E + m∠ F = 180° 3. Triangle Sum Theorem (Thm. 5.1) 4. m∠ D + m∠ E + m∠ F = m∠ A + m∠ B + m∠ C 4. Transitive Property of Equality 5. m∠ D + m∠ E + m∠ F = m∠ D + m∠ E + m∠ C 5. Substitution Property of Equality 6. m∠ F = m∠ C 6. Subtraction Property of Equality 7. ∠ F ≅ ∠ C 7. Definition of congruent angles 8. ∠ C ≅ ∠ F 8. Symmetric Property of Congruence (Thm. 2.2) m∠ L = m∠ P 50° = (2x + 4y)° 40° = (17x − y)° So, a system of equations is 2x + 4y = 50 . 17x − y = 40 Solve the second equation for y to get y = 17x − 40. Substitute this for y in the first equation and solve for x. 2x + 4(17x − 40) = 50 2x + 68x − 160 = 50 70x − 160 = 50 70x = 210 x=3 Substitute 3 for x in the second equation and solve for y. ⋅ y = 17 3 − 40 y = 51 − 40 = 11 So, x = 3 and y = 11. 24. m∠ S + m∠ T + m∠ U = 180° 130° + 28° + (4x + y)° = 180° 158 + 4x + y = 180 4x + y = 22 m∠ Y = m∠ T 20. Sample answer: (8x − 6y)° = 28° So, a system of equations is 4x + y = 22 . 8x − 6y = 28 The Transitive Property of Triangle Congruence (Thm. 5.3) guarantees that all the triangles are congruent. — —— —— — 21. corresponding sides: JK ≅ XY , KL ≅ YZ , JL ≅ XZ corresponding angles: ∠ J ≅ ∠ X, ∠ K ≅ ∠ Y, ∠ L ≅ ∠ Z Solve the first equation for y to get y = 22 − 4x. Substitute this for y in the second equation and solve for x. 8x − 6(22 − 4x) = 28 8x − 132 + 24x = 28 32x − 132 = 28 32x = 160 22. a. They are congruent because corresponding parts of congruent figures are congruent. b. They are congruent because they are both supplementary to congruent angles. c. ∠ GEB is also a right angle, and all right angles are congruent. — — d. yes; From parts (a)–(c), you know that BE ≅ DE , ∠ ABE ≅ ∠ CDE, ∠ GBE ≅ ∠ GDE, and — ≅ GE — by the Reflexive ∠ GEB ≅ ∠ GED. Also, GE Property of Congruence (Thm. 2.1), ∠ BGE ≅ ∠ DGE — ≅ DG — by the Third Angles Theorem (Thm. 5.4), and BG from the diagram markings. So, △BEG ≅ △DEG because all corresponding parts are congruent. Copyright © Big Ideas Learning, LLC All rights reserved. x=5 Substitute 5 for x in the first equation and solve for y. 4(5) + y = 22 20 + y = 22 y=2 So, x = 5 and y = 2. 25. A rigid motion maps each part of a figure to a corresponding part of its image. Because rigid motions preserve length and angle measure, corresponding parts of congruent figures are congruent, which means that the corresponding sides and corresponding angles are congruent. Geometry Worked-Out Solutions 149 Chapter 5 Maintaining Mathematical Proficiency 26. ∠ Z ≅ ∠ W — — 27. ∠ N ≅ ∠ T, RS ≅ PQ — — 28. ∠ J ≅ ∠ M, JK ≅ KM (K is the midpoint), and 2. Prove △BSR ≅ △DUT STATEMENTS REASONS 1. Given b. Check students’ work. 1. ∠ B and ∠ D are right angles, — BA ≅ — DA ≅ — DC ≅ — BC , R is the midpoint of — BA , U is the midpoint of — DA , T is the midpoint of — DC , and S is the midpoint of — BC . c. BC ≈ 1.95 units, m∠ B ≈ 98.79°, m∠ C ≈ 41.21° 2. ∠ B ≅ ∠ D 2. Right Angles Congruence Theorem (Thm. 2.3) 3. BA = DA = DC = BC 3. Definition of congruent segments 4. BA = BR + RA, DA = DU + UA, DC = DT + TC, BC = BS + SC 4. Segment Addition Postulate (Post. 1.2) 5. BR + RA = DU + UA = DT + TC = BS + SC 5. Transitive Property of Equality m∠ LKM = 90°. — —— — 29. ∠ D ≅ ∠ H, DF GH , DE ≅ HI , ∠ DFE ≅ ∠ HGI 5.3 Explorations (p. 245) 1. a. Check students’ work. d. Check students’ work. If two sides and the included angle of a triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. 2. Two triangles can be proved congruent if two pairs of corresponding sides and corresponding included angles are congruent. 3. Start with two triangles so that two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle. Then show that one triangle can be translated until it coincides with the other triangle by a composition of rigid motions. 5.3 Monitoring Progress (pp. 247–248) — ≅ BC — ≅ CD — ≅ AD —, and ∠ A, ∠ B, Given ABCD is a square, AB ∠ C, and ∠ D are right angles. R, S, T, and U are midpoints — ⊥ SU — and SV — ≅ VU —. of the sides of ABCD. RT S B R A C T V — ≅ VU —, RT — ⊥ SU — 1. SV VR ≅ — VR 2. — 6. Definition of midpoint 7. BR = RA, DU = UA, DT = TC, BS = SC 7. Definition of congruent segments 8. BR + BR = DU + DU = DT + DT = BS + BS 8. Substitution Property of Equality ⋅ ⋅ 9. 2 BR = 2 DU = 2 DT = 2 BS ⋅ U D 11. — BR ≅ — DU ≅ — DT ≅ — BS 12. △BSR ≅ △DUT 1. Given 2. Reflexive Property of Congruence (Thm. 2.1) 3. ∠ SVR and ∠ UVR are right angles. 3. Definition of perpendicular lines 4. ∠ SVR ≅ ∠ UVR 4. Right Angles Congruence Theorem (Thm. 2.3) 5. △SVR ≅ △UVR 5. SAS Congruence Theorem (Thm. 5.5) Geometry Worked-Out Solutions 10. Division Property of Equality 11. Definition of congruent segments 12. SAS Congruence Theorem (Thm. 5.5) REASONS — — 3. Given DA ≅ DG and D ∠ ADR ≅ ∠ GDR Prove △DRA ≅ △DRG A STATEMENTS — ≅ DG —, 1. DA ∠ ADR ≅ ∠ GDR 2. — DR ≅ — DR 3. △DRA ≅ △DRG 150 9. Distributive Property ⋅ 10. BR = DU = DT = BS 1. Prove △SVR ≅ △UVR STATEMENTS BR ≅ — RA , — DU ≅ — UA , 6. — — — — — DT ≅ TC , BS ≅ SC R G REASONS 1. Given 2. Reflexive Property of Congruence (Thm. 2.1) 3. SAS Congruence Theorem (Thm. 5.5) Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 — —— — 16. Given AB ≅ CD , AB CD 5.3 Exercises (pp. 249–250) A Prove △ABC ≅ △CDA Vocabulary and Core Concept Check D 1 1. An included angle is an angle formed by two adjacent 2 consecutive sides of a triangle. B 2. If two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then the two triangles are congruent. Monitoring Progress and Modeling with Mathematics — — 3. ∠ JKL is the included angle between JK and KL . — — — — — — — — 4. ∠ PKL is the included angle between PK and LK . 5. ∠ KLP is the included angle between LP and LK . STATEMENTS REASONS — ≅ CD —, AB — CD — 1. AB 1. Given 2. — AC ≅ — AC 2. Reflexive Property of Congruence (Thm. 2.1) 3. ∠ 1 ≅ ∠ 2 3. Alternate Interior Angles Theorem (Thm. 3.2) 4. △ABC ≅ △CDA 4. SAS Congruence Theorem (Thm. 5.5) 6. ∠ LJK is the included angle between JL and JK . 7. ∠ KLJ is the included angle between KL and JL . C D 17. Given C is the midpoint — and BD —. of AE — — 8. ∠ KPL is the included angle between KP and PL . Prove △ABC ≅ △EDC A B 10. yes; Two pairs of sides and the included angles are congruent. 11. no; One of the congruent angles is not the included angle. 12. no; The congruent angles are not the included angles. 13. yes; Two pairs of sides and the included angles are congruent. 14. no; ∠ NKM and ∠ KML are congruent by the Alternate Interior Angles Theorem (Thm. 3.2) but they are not the included angles. — 15. Given PQ bisects ∠ SPT, STATEMENTS REASONS 1. C is the midpoint of — and BD —. AE 1. Given 2. ∠ ACB ≅ ∠ ECD 2. Vertical Angles Congruence Theorem (Thm. 2.6) 3. — AC ≅ — EC , — BC ≅ — DC S Prove △SPQ ≅ △TPQ T 3. Definition of midpoint 4. △ABC ≅ △EDC P — ≅ TP — SP E C 9. no; The congruent angles are not the included angles. 4. SAS Congruence Theorem (Thm. 5.5) — —— — 18. Given PT ≅ RT , QT ≅ ST P Prove △PQT ≅ △RST Q T Q STATEMENTS 1. — SP ≅ — TP , — PQ bisects ∠ SPT. 2. — PQ ≅ — PQ S REASONS 1. Given STATEMENTS 1. 2. Reflexive Property of Congruence (Thm. 2.1) 3. ∠ SPQ ≅ ∠ TPQ 3. Definition of angle bisector 4. △SPQ ≅ △TPQ 4. SAS Congruence Theorem (Thm. 5.5) Copyright © Big Ideas Learning, LLC All rights reserved. — ≅ RT —, PT — — QT ≅ ST R REASONS 1. Given 2. ∠ PTQ ≅ ∠ RTS 2. Vertical Angles Congruence Theorem (Thm. 2.6) 3. △PQT ≅ △RST 3. SAS Congruence Theorem (Thm. 5.5) Geometry Worked-Out Solutions 151 Chapter 5 — — 19. △SRT ≅ △URT; RT ≅ RT by the Reflexive Property of Congruence (Thm. 2.1). Also, because all points on a circle — ≅ RU —. It is given are the same distance from the center, RS that ∠ SRT ≅ ∠ URT. So, △SRT and △URT are congruent by the SAS Congruence Theorem (Thm. 5.5). 26. In order to prove △ABC ≅ △DBC, you will need to know that ∠ ACB ≅ ∠ DCB. 27. Given △ABC, △BCD, and 20. △BAD ≅ △DCB; Because the sides of the square are — ≅ DC — and AD — ≅ CB —. Also, because the congruent, BA angles of the square are congruent, ∠ A ≅ ∠ C. So, △BAD and △DCB are congruent by the SAS Congruence Theorem (Thm. 5.5). 21. △STU ≅ △UVR; Because the sides of the pentagon are — ≅ UV — and TU — ≅ VR —. Also, because the angles congruent, ST of the pentagon are congruent, ∠ T ≅ ∠ V. So, △STU and △UVR are congruent by the SAS Congruence Theorem (Thm. 5.5). 22. △NMK ≅ △NLK; Because all points on a circle are the — ≅ NL — and KM — ≅ KL —. same distance from the center, NM — ⊥ MN — and KL — ⊥ NL —, ∠ M and ∠ L are right Because MK angles by the definition of perpendicular lines, which means that ∠ M ≅ ∠ L by the Right Angles Congruence Theorem (Thm. 2.3). So, △NMK and △NLK are congruent by the SAS Congruence Theorem (Thm. 5.5). — B △CDE are isosceles triangles and ∠ B ≅ ∠ D. D A C E Prove △ABC ≅ △CDE Because △ABC, △BCD, and △CDE are isosceles triangles, — ≅ BC —, BC — ≅ CD —, and CD — ≅ DE —. you know that AB So, by the Transitive Property of Congruence (Thm. 2.1), — ≅ CD — and BC — ≅ DE —. It is given that ∠ B ≅ ∠ D. So, AB △ABC ≅ △CDE by the SAS Congruence Theorem (Thm. 5.5). 28. SSS, ASA, AAS, and SAS all correspond to congruence theorems. Counterexample for SSA Counterexample for AAA — 23. Construct side DE so that it is congruent to AC . Construct ∠ D, with vertex D and side ⃗ DE so that it is congruent — so that it is congruent to AB —. Draw to ∠ A. Construct DF △DFE. By the SAS Congruence Theorem (Thm. 5.5), △ABC ≅ △DFE. 29. Prove △ABC ≅ △DEC A C B 3y + 1 D 1. B E — — REASONS — ≅ DC —, AC 1. Given (marked in diagram) — BC ≅ — EC 2. ∠ ACB ≅ ∠ DCE 2. Vertical Angles Congruence Theorem (Thm. 2.6) 3. △ABC ≅ △DEC 3. SAS Congruence Theorem (Thm. 5.5) 24. Construct side DE so that it is congruent to AC . Construct ∠ D, with vertex D and side ⃗ DE so that it is congruent — so that it is congruent to AB —. Draw to ∠ A. Construct DF △DFE. By the SAS Congruence Theorem (Thm. 5.5), △ABC ≅ △DFE. E C AC = CD B D BC = CE 4y − 6 = 2x + 6 4y = 2x + 12 y= F 1 —2 x 3y + 1 = 4x ( ) 3 —12x + 3 + 1 = 4x +3 1.5x + 9 + 1 = 4x 1.5x + 10 = 4x 10 = 2.5x 25. △XYZ and △WYZ are congruent so either the expressions for — and WZ — or the expressions for XY — and WY — should be set XZ equal to each other because they are corresponding sides. 5x − 5 = 3x + 9 2x − 5 = 9 2x = 14 x=7 152 4x E STATEMENTS A Geometry Worked-Out Solutions D 2x + 6 F C A 4y − 6 y= 1 —2 x=4 ⋅4 + 3 = 2 + 3 = 5 So, x = 4 and y = 5. — — 30. no; When you construct AB and AC , you have to construct them at an angle that is congruent to ∠ A. Otherwise, when you construct an angle congruent to ∠ C, you might not get a —. third segment that is congruent to BC Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 31. Given A reflection in line k maps point A to A′, a reflection in line m maps A′ to A″, and m∠ MPK = x°. Prove A rotation about point P maps A to A″ and the angle of rotation is 2x°. STATEMENTS m A″ M A′ P 1. Given 2. Line k is the perpendicular bisector of — AA′ , and line m is the perpendicular bisector of — A′A″ . 2. Definition of reflection 4. ∠ AKP ≅ ∠ A′KP, ∠ A′MP ≅ ∠ A″MP 5. — KP ≅ — KP 6. △AKP ≅ △A′KP, △A′MP ≅ △A″MP 7. — AP ≅ — A′P , — A′P ≅ — A″P , ∠ APK ≅ ∠ A′PK, ∠ A′PM ≅ ∠ A″PM 8. — AP ≅ — A″P 9. m∠ APK = m∠ A′PK, m∠ A′PM = m∠ A″PM 12. m∠ APA″ = 2(m∠ A′PK + m∠ A′PM) 12. Distributive Property 13. m∠ APA″ = 2(m∠ MPK) 13. Substitution Property of Equality 14. m∠ APA″ = 2(x°) = 2x° 14. Substitution Property of Equality 15. A rotation about point P maps A to A″ and the angle of rotation is 2x°. 15. Definition of rotation A 3. Definition of perpendicular bisector 4. Right Angles Congruence Theorem (Thm. 2.3) 5. Reflexive Property of Congruence (Thm. 2.1) 6. SAS Congruence Theorem (Thm. 5.5) 7. Corresponding parts of congruent triangles are congruent. Maintaining Mathematical Proficiency 32. Two sides are equivalent and one angle is a right angle. So, the triangle is a right isosceles triangle. 33. Two sides are equivalent and one angle is obtuse. So, the triangle is an obtuse isosceles triangle. 34. All sides are congruent, and therefore all angles are congruent. So, the triangle is a equiangular equilateral triangle. 35. None of the sides are congruent and one angle is obtuse. So, the triangle is an obtuse scalene triangle. 5.4 Explorations (p. 251) 1. a. Check students’ work. b. Check students’ work. c. Because all points on a circle are the same distance from — ≅ AC —. the center, AB d. ∠ B ≅ ∠ C e. Check students’ work. If two sides of a triangle are congruent, then the angles opposite them are congruent. f. If two angles of a triangle are congruent, then the sides opposite them are congruent. The converse is true. 2. In an isosceles triangle, two sides are congruent and the angles opposite them are congruent. 8. Transitive Property of Congruence (Thm. 2.1) 9. Definition of congruent angles 10. m∠ MPK = m∠ A′PK + m∠ A′PM, m∠ APA″ = m∠ APK + m∠ A′PK + m∠ A′PM + m∠ A″PM 10. Angle Addition Postulate (Post. 1.4) 11. m∠ APA″ = m∠ A′PK + m∠ A′PK + m∠ A′PM + m∠ A′PM 11. Substitution Property of Equality Copyright © Big Ideas Learning, LLC All rights reserved. REASONS REASONS 1. A reflection in line k maps point A to A′, a reflection in line m maps A′ to A″, and m∠ MPK = x°. 3. — AK ≅ — KA′ , ∠ AKP and ∠ A′KP are right angles, — A′M ≅ — MA″ , and ∠ A′MP and ∠ A″MP are right angles. k K STATEMENTS 3. Draw the angle bisector of the included angle between the congruent sides to divide the given isosceles triangle into two triangles. Use the SAS Congruence Theorem (Thm. 5.5) to show that these two triangles are congruent. Then, use properties of congruent triangles to show that the two angles opposite the shared sides are congruent. For the converse, draw the angle bisector of the angle that is not congruent to the other two. This divides the given triangle into two triangles that have two pairs of corresponding congruent angles. The third pair of angles are congruent by the Third Angles Theorem (Thm. 5.4). Also, the angle bisector is congruent to itself by the Reflexive Property of Congruence (Thm. 2.1). So, the triangles are congruent, and the sides opposite the congruent angles in the original triangle are congruent. Geometry Worked-Out Solutions 153 Chapter 5 5.4 Monitoring Progress (pp. 253–255) — — 7. △ABC is an equiangular triangle and, therefore, an equilateral triangle. So, x = 12. 1. If HG ≅ HK , then ∠ HKG ≅ ∠ G. — — 2. If ∠ KHJ ≅ ∠ KJH, then KJ ≅ KH . 3. The triangle is equiangular; therefore, by Corollary to the Converse of the Base Angles Theorem (Cor. 5.3) the length — is 5 units. of ST 4. The triangle on the right is an equilateral triangle. Each angle has a measure of 60°, therefore, x = 60. The triangle on the left is an isosceles triangle. Both base angles are 90° − 60° = 30°. So, y = 180 − 2 30 = 120. ⋅ — — 5. From Example 4, you know that PT ≅ QT and ∠ 1 ≅ ∠ 2. — ≅ QR — and ∠ QPS ≅ ∠ PQR. By the It is stated that PS definition of congruent angles, m∠ 1 = m∠ 2 and m∠ QPS = m∠ PQR. Also, by the Angle Addition Postulate (Post. 1.4), m∠ 1 + m∠ TPS = m∠ QPS and m∠ 2 + m∠ TQR = m∠ PQR. By substituting m∠ 1 for m∠ 2 and m∠ QPS for m∠ PQR, you get m∠ 1 + m∠ TQR = m∠ QPS. Then, by the Transitive Property of Equality, m∠ 1 + m∠ TPS = m∠ 1 + m∠ TQR. So, by the Subtraction Property of Equality, m∠ TPS = m∠ TQR. Because ∠ TPS ≅ ∠ TQR by the definition of congruent angles, you can conclude that △PTS ≅ △QTR by the SAS Congruence Theorem (Thm. 5.5). 5.4 Exercises (pp. 256–258) Vocabulary and Core Concept Check 8. △MLN is an equiangular triangle and, therefore, an equilateral triangle. So, x = 16. 9. △RST is an equilateral triangle and, therefore, an equiangular triangle. So, x = 60. 10. △DEF is an equilateral triangle and, therefore, each angle is 3x° = 60°, or x = 20. 11. The pennant is an isosceles triangle. So, x = 79 and ⋅ y = 180 − 2 79 = 180 − 158 = 22. 12. Sample answer: 7 cm Reuse Reduce Recycle 7 cm 13. The triangle in the center is equilateral; therefore, each angle measures 60°. So, x = 60. Because the top and bottom lines are parallel, the alternate interior angles have a measure of 60°. By the Base Angles Theorem (Thm. 5.6), y = 60. 14. The vertex angle of the isosceles triangle measures 180° − 40° = 140°. To find the base angles: 2x° + 140° = 180° 1. The vertex angle is the angle formed by the congruent sides, 2x = 40 or legs, of an isosceles triangle. 2. The base angles of an isosceles triangle are opposite the congruent sides, and they are congruent by the Base Angles Theorem (Thm. 5.6). Monitoring Progress and Modeling with Mathematics — — 3. If AE ≅ DE , then ∠ D ≅ ∠ A by the Base Angles Theorem (Thm. 5.6). x = 20 To find y: y° + x° = 90° y + 20 = 90 y = 70 So, x = 20 and y = 70. 15. The triangle on the left is equiangular and, therefore, equilateral. — — 4. If AB ≅ EB , then ∠ AEB ≅ ∠ A by the Base Angles Theorem (Thm. 5.6). — — 5. If ∠ D ≅ ∠ CED, then EC ≅ DC by the Converse of the Base Angles Theorem (Thm. 5.7). — — 6. If ∠ EBC ≅ ∠ ECB, then EC ≅ EB by the Converse of the Base Angles Theorem (Thm. 5.7). 154 7 cm Geometry Worked-Out Solutions 8y = 40 y=5 The triangle on the right is isosceles and the vertex angle measures 180° − 60° = 120°. So, by the Base Angles Theorem (Thm. 5.6), each base angle is 180° − 120° 60° —— = — = 30°. So, x = 30 and y = 5. 2 2 Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 16. By the Converse of the Base Angles Theorem (Thm. 5.7): 3x − 5 = y + 12 3x − 5 − 12 = y y = 3x − 17 The triangle on the right is equiangular and, therefore, equilateral. 3x − 5 = 5y − 4 3x − 5 = 5(3x − 17) − 4 3x − 5 = 15x − 85 − 4 3x − 5 = 15x − 89 −12x = − 84 x=7 y = 3x − 17 ⋅ y = 3 7 − 17 y = 21 − 17 = 4 So, x = 7 and y = 4. 17. Draw a segment with length 3 inches. Draw an arc with center at one endpoint and radius 3 inches. Draw an arc with center at the other endpoint and radius 3 inches. Connect the intersection of the arcs with two segments to form an equilateral triangle. 20. a. Because △ABD and △CBD are congruent and equilateral, — ≅ CB —. So, △ABC is isosceles. you know that AB b. Because △ABC is isosceles, ∠ BAE ≅ ∠ BCE by the Base Angles Theorem (Thm. 5.6). c. By the Reflexive Property of Congruence (Thm. 2.1), — ≅ BE —. Because △ABD and △CBD are congruent and BE equilateral, and also equiangular by the Corollary to the Base Angles Theorem (Cor. 5.2), you can conclude that — ≅ CB — as explained in part (a). ∠ ABE ≅ ∠ CBE. Also, AB So, by the SAS Congruence Theorem (Thm. 5.5), △ABE ≅ △CBE. d. m∠ ABE + m∠ CBE = m∠ ABC Angle Addition Postulate (Post. 1.4) m∠ ABE = 60°, m∠ CBE = 60° Definition of equiangular triangle m∠ ABC = 60° + 60° = 120° Substitution Property of Equality m∠ ABC + m∠ BAE + m∠ BCE = 180° Triangle Sum Theorem (Thm. 5.1) 120° + m∠ BAE + m∠ BCE = 180° Substitution Property of Equality m∠ BAE + m∠ BCE = 60° Subtraction Property of Equality m∠ BAE = m∠ BCE Corresponding parts of congruent triangles are congruent. m∠ BAE + m∠ BCE = 60° 3 in. 18. Draw a segment with length 1.25 inches. Draw an arc with center at one endpoint and radius 1.25 inches. Draw an arc with center at the other endpoint and radius 1.25 inches. Connect the intersection of the arcs with two segments to form an equilateral triangle. Substitution Property of Equality 2m∠ BAE = 60° Simplify. m∠ BAE = 30° Division Property of Equality 21. a. Each edge is made out of the same number of sides of the original equilateral triangle. b. The areas of the first four triangles in the pattern are 1.25 in. 19. When two angles of a triangle are congruent, the sides opposite the angles are congruent; Because ∠ A ≅ ∠ C, — ≅ BC —. So, BC = 5. AB 1 square unit, 4 square units, 9 square units, and 16 square units. c. Triangle 1 has an area of 12 = 1, Triangle 2 has an area of 22 = 4, Triangle 3 has an area of 32 = 9, and so on. So, by inductive reasoning, you can predict that Triangle n has an area of n2. Seventh triangle: 72 = 49 square units — 22. A, C; Because the base of isosceles △XYZ is YZ , the legs are — and XZ —. So, XY — ≅ XZ — and ∠ Y ≅ ∠ Z. XY Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 155 Chapter 5 29. no; The two sides that are congruent can form an obtuse 23. x + 4 = 4x + 1 angle or a right angle. 4 = 3x + 1 3 = 3x 30. no; The sum of the angles of a triangle is always 180°. So, if all three angles are congruent, then they will always be 180° — = 60°. 3 1=x Perimeter = 7 + 1 + 4 + 4 + 1 = 17 inches 24. 2x − 3 = x + 5 31. x=8 21 − x = x + 5 3t = 5t − 12 or 3t = t + 20 or 5t − 12 = t + 20 −2t = −12 2t = 20 4t = 32 t=6 t = 10 t=8 The values of t could be 6, 8, and 10. 16 = 2x 8=x 32. 21 − x = 2x − 3 x° 24 = 3x 8=x ⋅ Perimeter = (2 8 − 3) + (8 + 5) + (21 − 8) = 13 + 13 + 13 = 39 inches 25. By the Reflexive Property of Congruence (Thm. 2.1), the yellow triangle and the yellow-orange triangle share a congruent side. Because the triangles are all isosceles, by the Transitive Property of Congruence (Thm. 2.1), the yellow-orange triangle and the orange triangle share a side that is congruent to the one shared by the yellow triangle and the yellow-orange triangle. This reasoning can be continued around the wheel, so the legs of the isosceles triangles are all congruent. Because you are given that the vertex angles are all congruent, you can conclude that the yellow triangle is congruent to the purple triangle by the SAS Congruence Theorem (Thm. 5.5). 26. 180° − 30° = 150° 150 = 75 — 2 The measures of the base angles are each 75°. 27. yellowgreen yellow green orange redorange blue red bluepurple purple redpurple The three sides of the triangle are congruent. So, the triangle is an equiangular equilateral triangle. 28. Every fourth color is a triad: Yellow-green, blue-purple, red- orange; green, purple, orange; and blue-green, red-purple, yellow-orange. 156 Geometry Worked-Out Solutions The vertex angle is (180 − x)° and the base angles are 180° − (180 − x)° x ° —— = — . 2 2 () 33. If the base angles are x°, then the vertex angle is (180 − 2x)°, or [2(90 − x)]°. Because 2(90 − x) is divisible by 2, the vertex angle is even when the angles are whole numbers. 34. a. ∠ XVY, ∠ UXV; ∠ WUX ≅ ∠ XVY because they are both vertex angles of congruent isosceles triangles. Also, m∠ UXV + m∠ VXY = m∠ UXY by the Angle Addition Postulate (Post. 1.4), and m∠ UXY = m∠ WUX + m∠ UWX by the Exterior Angle Theorem (Thm. 5.2). So, by the Transitive Property of Equality, m∠ UXV + m∠ VXY = m∠ WUX + m∠ UWX. Also, m∠ UWX = m∠ VXY because they are base angles of congruent isosceles triangles. By substituting m∠ UWX for m∠ VXY, you get m∠ UXV + m∠ UWX = m∠ WUX + m∠ UWX. By the Subtraction Property of Equality, m∠ UXV = m∠ WUX, so ∠ UXV ≅ ∠ WUX. b. Because the triangles are congruent isosceles triangles yelloworange bluegreen x° The base angles are (180 − x)° and the vertex angle is 180° − 2(180 − x)° = (2x − 180)°. — ≅ VY — ≅ UW — ≅ VX —, then and from part (a), UX —≅— —. So, the distance between points U WX XY ≅ UV and V is 8 meters. 35. a. 2.1 mi; by the Exterior Angle Theorem (Thm. 5.2), m∠ L = 70° − 35° = 35°. Because m∠ SRL = 35° = m∠ RLS, by definition of congruent angles, ∠ SRL ≅ ∠ RLS. So, by the Converse of the Base Angles — ≅ SL —. So, SL = RS = 2.1 miles. Theorem (Thm. 5.7), RS b. Find the point on the shore line that has an angle of 45° from the boat. Then, measure the distance that the boat travels until the angle is 90°. That distance is the same as the distance between the boat and the shore line because the triangle formed is an isosceles right triangle. 36. no; The sum of the angle measures of a very large spherical triangle will be greater than 180°, but for smaller spherical triangles, the sum will be closer to 180°. Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 37. Given △ABC is equilateral. 41. Given △ABC is equilateral. A D B STATEMENTS REASONS 1. △ABC is equilateral. 1. Given 2. — AB ≅ — AC , — ≅ BC —, AB — — AC ≅ BC A ∠ CAD ≅ ∠ ABE ≅ ∠ BCF Prove △ABC is equiangular. C Prove △DEF is equilateral. STATEMENTS 2. Definition of equilateral triangle 3. ∠ B ≅ ∠ C, ∠ A ≅ ∠ C, ∠A ≅ ∠B 3. Base Angles Theorem (Thm. 5.6) 4. △ABC is equiangular. 4. Definition of equiangular triangle — —— — 38. a. By the markings, AE ≅ DE , AB ≅ DC , and ∠ BAE ≅ ∠ CDE. So, △ABE ≅ △DCE by SAS Congruence Theorem (Thm. 5.5). b. △AED and △BEC are isosceles triangles. c. ∠ EDA, ∠ BCA, and ∠ CBD are all congruent to ∠ EAD. A 39. Given △ABC is equiangular. Prove △ABC is equilateral. B C B E F REASONS 1. △ABC is equilateral. ∠ CAD ≅ ∠ ABE ≅ ∠ BCF 1. Given 2. △ABC is equiangular. 2. Corollary to the Base Angles Theorem (Cor. 5.2) 3. ∠ ABC ≅ ∠ BCA ≅ ∠ BAC 3. Definition of equiangular triangle 4. m∠ CAD = m∠ ABE = m∠ BCF, m∠ ABC = m∠ BCA = m∠ BAC 4. Definition of congruent angles 5. m∠ ABC = m∠ ABE + m∠ EBC, m∠ BCA = m∠ BCF + m∠ ACF, m∠ BAC = m∠ CAD + m∠ BAD 5. Angle Addition Postulate (Post. 1.4) 6. m∠ ABE + m∠ EBC = m∠ BCF + m∠ ACF = m∠ CAD + m∠ BAD 6. Substitution Property of Equality 7. m∠ ABE + m∠ EBC = m∠ ABE + m∠ ACF = m∠ ABE + m∠ BAD 7. Substitution Property of Equality STATEMENTS REASONS 1. △ABC is equiangular. 1. Given 8. m∠ EBC = m∠ ACF = m∠ BAD 8. Subtraction Property of Equality 2. ∠ B ≅ ∠ C, ∠ A ≅ ∠ C, ∠A ≅ ∠B 2. Definition of equiangular triangle 9. ∠ EBC ≅ ∠ ACF ≅ ∠ BAD 9. Definition of congruent angles 3. — AB ≅ — AC , — AB ≅ — BC , — AC ≅ — BC 3. Converse of the Base Angles Theorem (Thm. 5.7) 4. △ABC is equilateral. 4. Definition of equilateral triangle 40. no; The distance between point T(0, 6) and a point on y = x is equal to the distance between U(6, 0) and y = x (using the same point). So any point V on y = x will be the same distance from T and U. Therefore, TV = VU and △TVU is an isosceles triangle. Unless V(3, 3) is the third point, in which case, T, V, and U are collinear and perpendicular to y = x. Copyright © Big Ideas Learning, LLC All rights reserved. C 10. ∠ FEB ≅ ∠ DFC ≅ ∠ EDA 10. Third Angles Theorem (Thm. 5.4) 11. ∠ FEB and ∠ FED are supplementary, ∠ DFC and ∠ EFD are supplementary, and ∠ EDA and ∠ FDE are supplementary. 11. Linear Pair Postulate (Post. 2.8) 12. ∠ FED ≅ ∠ EFD ≅ ∠ FDE 12. Congruent Supplements Theorem (Thm. 2.4) 13. △DEF is equiangular. 13. Definition of equiangular triangle 14. △DEF is equilateral. 14. Corollary to the Converse of the Base Angles Theorem (Cor. 5.3) Geometry Worked-Out Solutions 157 Chapter 5 Maintaining Mathematical Proficiency — — 42. Reflexive Property of Congruence (Theorem 2.1): SE ≅ SE 43. Symmetric Property of Congruence (Theorem 2.1): — — — ≅ JK —. If JK ≅ RS , then RS 44. Transitive Property of Congruence (Theorem 2.1): — ≅ PQ —, and PQ — ≅ UV —, then EF — ≅ UV —. If EF 5.1–5.4 What Did You Learn? (p. 259) 5. corresponding angles: ∠ Q ≅ ∠ W, ∠ R ≅ ∠ X, ∠ S ≅ ∠ Y, ∠T ≅ ∠Z — ≅ WX —, RS — ≅ XY —, ST — ≅ YZ —, corresponding sides: QR — — QT ≅ WZ Sample answer: RSTQ ≅ XYZW 6. no; The congruent angles are not the included angle. 7. yes; △GHF ≅ △KHJ by the SAS Congruence Theorem (Thm. 5.5). 1. You are given a diagram of the triangle made from the segments that connect each person to the other two, along with the length of each segment; the people are all standing on the same stage (plane), so the points are coplanar; You are asked to classify the triangle by its sides and by measuring its angles. STATEMENTS 1. — GH ≅ — HK , — FH ≅ — HJ 3. Sample answer: a large triangle made up of 9 small triangles, a hexagon, a parallelogram 2. Vertical Angles Congruence Theorem (Thm. 2.6) 3. △GHF ≅ △KHJ 3. SAS Congruence Theorem (Thm. 5.5) 8. yes; △LMP ≅ △NMP by the SAS Congruence Theorem (Thm. 5.5). STATEMENTS 1. — LM ≅ — NM , ∠ LMP ≅ ∠ NMP 2. — MP ≅ — MP 5.1–5.4 Quiz (p. 260) 1. x° = 30° + 80° = 110°; The exterior angle measures 110°. 3. △LMP ≅ △NMP 2.(5x + 2)° + 6x° = 90° 11x = 88 x=8 y° = 180° − (5x + 2)° ⋅ = 180° − (5 8 + 2)° = 180° − 42° = 138° The exterior angle measures 138°. 3. 29° + (12x + 26)° = (15x + 34)° 55 + 12x = 15x + 34 21 = 3x 1. Given 2. ∠ GHF ≅ ∠ KHJ 2. There is a pair of congruent triangles, so all pairs of corresponding sides and angles are congruent. By the Triangle Sum Theorem (Thm. 5.1), the three angles in △LMN have measures that add up to 180°. You are given two measures, so you can find the third using this theorem. The measure of ∠ P is equal to the measure of its corresponding angle, ∠ L. The measure of ∠ R is equal to the measure of its corresponding angle, ∠ N. Once you write this system of equations, you can solve for the values of the variables. REASONS REASONS 1. Given 2. Reflexive Property of Congruence (Thm. 2.1) 3. SAS Congruence Theorem (Thm. 5.5) — — 9. If VW ≅ WX , then ∠ VXW ≅ ∠ XVW by the Base Angles Theorem (Thm. 5.6). — — 10. If XZ ≅ XY , then ∠ XYZ ≅ ∠ XZY by the Base Angles Theorem (Thm. 5.6). — — 11. If ∠ ZVX ≅ ∠ ZXV, then XZ ≅ VZ , by the Converse of the Base Angles Theorem (Thm. 5.7). — — 12. If ∠ XYZ ≅ ∠ ZXY, then XZ ≅ YZ , by the Converse of the Base Angles Theorem (Thm. 5.7). 7=x ⋅ (15x + 34)° = (15 7 + 34)° = (105 + 34)° = 139° The exterior angle measures 139°. 4. corresponding angles: ∠ C ≅ ∠ F, ∠ A ≅ ∠ D, ∠ B ≅ ∠ E —≅— — ≅ DE —, CB — ≅ FE — corresponding sides: CA FD, AB Sample answer: △CAB ≅ △FDE 158 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 5.5 Explorations (p. 261) DE = QR 13. 5y − 7 = 38 1. a. Check students’ work. 5y = 45 b. Check students’ work. — c. AB = 2 because AB has one endpoint at the origin and one y=9 endpoint on a circle with a radius of 2 units. m∠ S = m∠ F — has one endpoint at the origin and AC = 3 because AC one endpoint on a circle with a radius of 3 units. BC = 4 because it was created that way. m∠ F = 180° − (123° + 29°) = 28° 2x + 2 = 28 2x = 26 d. m∠ A = 104.43°, m∠ B = 46.61°, m∠ C = 28.96° x = 13 e. Check students’ work; If two triangles have three pairs So, x = 13 and y = 9. of congruent sides, then they will have three pairs of congruent angles. 14. 5x − 1 = 24 2. When the corresponding sides of two triangles are congruent, 5x = 25 the corresponding angles are also congruent and therefore, the triangles are congruent. x=5 6y = 120 3. Use rigid transformations to map triangles. y = 20 So, x = 5 and y = 20. 15. 5.5 Monitoring Progress (pp. 263–265) — —— — — — DG ≅ HK . So, △DFG ≅ △HJK by the SSS Congruence 1. yes; From the diagram markings, DF ≅ HJ , FG ≅ JK , and x = 4[ (90 − x) − 5 ] x = 360 − 4x − 20 Theorem (Thm. 5.8). 5x = 340 — measure. In order for two triangles to be congruent, all pairs of corresponding sides must be congruent. 90° − 68° = 22° The acute angles are 68° and 22°. — —— — — — QT ≅ RT . So, △QPT ≅ △RST by the SSS Congruence 3. yes; From the diagram markings, QP ≅ RS , PT ≅ ST , and 16. a. Triangle 1 has a right angle. So, it is a right triangle. It appears that triangle 2 has an obtuse angle. So, it is an obtuse triangle. It appears that triangle 3 has three nonequivalent acute angles. So, it is an acute triangle. Theorem (Thm. 5.8). 4. not stable; This square is not stable because there are many possible quadrilaterals with the given side lengths. 5. stable; The diagonal support in this figure forms triangles Triangle 4 has three congruent angles. So, it is an equiangular triangle. with fixed side lengths. By the SSS Congruence Theorem (Thm. 5.8), these triangles cannot change shape, so the figure is stable. b. Triangle 4 has three congruent sides. So, it is an equilateral triangle. It appears that triangle 5 has three noncongruent sides. So, it is a scalene triangle. Triangle 6 has two congruent sides. So, it is an isosceles triangle. c. yes; — 2. no; AB corresponds with CD , but they are not the same x = 68 A 6. not stable; The diagonal support in this figure forms a triangle and a quadrilateral. The triangle would be stable, but the quadrilateral is not because there are many possible quadrilaterals with the given side lengths. 7. Redraw △ABC and △DCB. C E 7 B B 8 C D F STATEMENTS REASONS 1. 1. Given — AB ≅ — DE , — BC ≅ — EF , B A C D ∠B ≅ ∠E 2. △ABC ≅ △DEF Copyright © Big Ideas Learning, LLC All rights reserved. 2. SAS Congruence Theorem (Thm. 5.5) Geometry Worked-Out Solutions 159 Chapter 5 — — 8. Given AC ≅ DB , ∠ ABC and ∠ DCB are right angles. REASONS — ≅ DB —, ∠ ABC 1. AC and ∠ DCB are right angles. 1. Given CB ≅ — BC 2. — — — by the Reflexive Property of Congruence (Thm. 2.1). So, it should say △JKL ≅ △LMJ or △JKL ≅ △JML by the SSS Congruence Theorem (Thm. 5.8). Prove △ABC ≅ △DCB STATEMENTS — — — — 10. no; You are given that JK ≅ KL ≅ LM ≅ MJ . Also, JL ≅ JL 11. yes; The diagonal supports in this figure form triangles with fixed side lengths. By the SSS Congruence Theorem (Thm. 5.8), these triangles cannot change shape, so the figure is stable. 2. Reflexive Property of Congruence (Thm. 2.1) 12. no; The support in this figure forms two quadrilaterals, which 3. △ABC and △DCB are right triangles. 3. Definition of a right triangle 4. △ABC ≅ △DCB 4. HL Congruence Theorem (Thm. 5.9) are not stable because there are many possible quadrilaterals with the given side lengths. — — — ⊥ AD —, CD — ⊥ AD — AB 13. Given AC ≅ DB , A B Prove △BAD ≅ △CDA D 5.5 Exercises (pp. 266–268) D C Vocabulary and Core Concept Check 1. The side opposite the right angle is called the hypotenuse of the right triangle. A 2. Three of the triangles are confirmed as right triangles. The second triangle from the left is the only triangle with legs that are not the legs of a right triangle. STATEMENTS REASONS — ≅ DB —, 1. AC 1. Given 2. Monitoring Progress and Modeling with Mathematics — —— —— — 3. yes; AB ≅ DB , BC ≅ BE , AC ≅ DE — 4. no; You cannot tell for sure from the diagram whether PS and — are congruent. RS — —— — 5. yes; ∠ B and ∠ E are right angles, AB ≅ FE , AC ≅ FD 6. no; the hypotenuses are not marked as congruent. — —— — — — 7. no; You are given that RS ≅ PQ , ST ≅ QT, and RT ≅ PT . So, it should say △RST ≅ △PQT by the SSS Congruence Theorem (Thm. 5.8). — — — — 8. yes; You are given that AB ≅ CD and AD ≅ CB . Also — ≅ BD — by the Reflexive Property of Congruence BD (Thm. 2.1). So, △ABD ≅ △CDB by the SSS Congruence Theorem (Thm. 5.8). — — — — — — DF ≅ DF by the Reflexive Property of Congruence — AB ⊥ — AD , — CD ⊥ — AD — AD ≅ — AD 2. Reflexive Property of Congruence (Thm. 2.1) 3. ∠ BAD and ∠ CDA are right angles. 3. Definition of perpendicular lines 4. △BAD and △CDA are right triangles. 4. Definition of a right triangle 5. △BAD ≅ △CDA 5. HL Congruence Theorem (Thm. 5.9) — 14. Given G is the midpoint of EH , — ≅ GI —, FG E F ∠ E and ∠ H are right angles. Prove △EFG ≅ △HIG G H I 9. yes; You are given that EF ≅ GF and DE ≅ DG . Also (Thm. 2.1). So, △DEF ≅ △DGF by the SSS Congruence Theorem (Thm. 5.8). 160 Geometry Worked-Out Solutions G Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 STATEMENTS REASONS 1. G is the midpoint — ≅ GI —, of — EH , FG ∠ E and ∠ H are right angles. 1. Given 2. — EG ≅ — HG — — to the length of QR . Use this length to draw an arc. Draw 17. Construct a side that is congruent to QS . Open your compass an arc with radius RS. Complete the triangle. By the SSS Congruence Theorem (Thm. 5.8), the two triangles are congruent. R 2. Definition of midpoint 3. △EFG and △HIG are right triangles. 3. Definition of a right triangle 4. △EFG ≅ △HIG 4. HL Congruence Theorem (Thm. 5.9) Q — —— — 15. Given LM ≅ JK , MJ ≅ KL — — to the length of QR . Use this length to draw an arc. Draw 18. Construct a side that is congruent to QS . Open your compass K L J M Prove △LMJ ≅ △JKL REASONS — ≅ JK —, MJ — ≅ KL — 1. LM 1. Given 3. △LMJ ≅ △JKL — — 16. Given WX ≅ VZ , Prove △VWX ≅ △WVZ V 2. — WV ≅ — WV X Y Z 1. Given △TUV ≅ △ZYX by the SSS Congruence Theorem (Thm. 5.8). 1 20. When you substitute —4 for x, KL ≠ LM. 6x = 4x + 4 2x = 4 2. Reflexive Property of Congruence (Thm. 2.1) 3. Definition of congruent segments 4. WZ = WY + YZ, VX = VY + YX 4. Segment Addition Postulate (Post. 1.2) 5. VX = WY + YZ 5. Substitution Property of Equality 6. VX = WZ 6. Transitive Property of Equality 8. △VWX ≅ △WVZ 19. The order of the points in the congruence statement should reflect the corresponding sides and angles. REASONS 3. WY = VY, YZ = YX 7. — VX ≅ — WZ S 3. SSS Congruence Theorem (Thm. 5.8) W — ≅ VZ —, WY —≅— 1. WX VY , — YZ ≅ — YX Q 2. Reflexive Property of Congruence (Thm. 2.1) — ≅ VY —, WY — — YZ ≅ YX STATEMENTS an arc with radius RS. Complete the triangle. By the SSS Congruence Theorem (Thm. 5.8), the two triangles are congruent. R STATEMENTS 2. — JL ≅ — JL S x=2 2x + 1 = 3x − 1 1=x−1 2=x 21. no; The sides of a triangle do not have to be congruent to each other, but each side of one triangle must be congruent to the corresponding side of the other triangle. 7. Definition of congruent segments 8. SSS Congruence Theorem (Thm. 5.8) Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 161 Chapter 5 — — — —— — 22. Given HF ≅ FS ≅ ST ≅ TH ; FT ≅SH ; ——— —— 26. AB = √ [ 3 − (−2) ]2 + (−3 − 1)2 = √ (3 + 2)2 + (−4)2 S ∠ H, ∠ F, ∠ S, and ∠ T are right angles. — — — = √ 52 + 16 = √25 + 16 = √41 ≈ 6.4 —— Prove △HFS ≅ △FST ≅ △STH T F —— BC = √ (7 − 3)2 + [ 5 − (−3) ]2 = √(4)2 + (5 + 3)2 — — — = √ 16 + 82 = √16 + 64 = √80 ≈ 8.9 —— —— AC = √ [ 7 − (−2) ]2 + (5 − 1)2 = √(7 + 2)2 + (4)2 — — — = √ 92 + 42 = √81 + 16 = √97 ≈ 9.8 —— H —— STATEMENTS REASONS — ≅ FS — ≅ ST — ≅ TH —; 1. HF DE = √ (8 − 3)2 + (2 − 6)2 = √ (5)2 + (−4)2 — — = √ 25 + 16 = √41 ≈ 6.4 1. Given EF = √ (10 − 8)2 + (11 − 2)2 = √ 22 + 92 — FT ≅ — SH ; ∠ H, ∠ F, ∠ S, —— — — = √ 4 + 81 = √85 ≈ 9.2 and ∠ T are right angles. —— 2. Reflexive Property of Congruence (Thm. 2.1) 3. △HFS, △FST, and △STH are right triangles. 3. Definition of a right triangle 4. △HFS ≅ △FST ≅ △STH 4. HL Congruence Theorem (Thm. 5.9) — — = √ 49 + 25 = √74 ≈ 8.6 The triangles are not congruent. —— — — = √ 36 + 25 = √61 ≈ 7.8 —— — ≅ ML —. JL b. SAS Congruence Theorem (Thm. 5.5); By definition of — ≅ MK —. Also, LK — ≅ LK —, by the Reflexive midpoint, JK Property of Congruence (Thm. 2.1), and ∠ JKL ≅ ∠ MKL by the Right Angles Congruence Theorem (Thm. 2.3). 24. a. To use the SSS Congruence Theorem (Thm. 5.8) to prove — ≅ CD —. △ABC ≅ △CDE, you need to know AB — — = √ 9 + 25 = √34 ≈ 5.8 △ABC ≅ △CDE, you need to know that ∠ B and ∠ D are right angles. — — ——— + (6 + = DE = √(5 − 5)2 √62 + 82 + (1 − =√ — (−6)2 —— — = √36 = 6 — EF = √(13 − 5)2 + (1 − 1)2 = √ 82 = 8 —— —— DF = √(13 − 5)2 + (1 − 7)2 = √(8)2 + (−6)2 — — — — = √ 9 + (−5)2 = √9 + 25 = √34 ≈ 5.8 ——— —— DF = √ (9 − 0)2 + [ −1 − (−1) ]2 = √92 + (−1 + 1)2 — AB = DE, BC = EF, and AC = DE, so △ABC ≅ △DEF. ——— —— 28. AB = √ [ −5 − (−5) ]2 + (2 − 7)2 = √ (−5 + 5)2 + (−5)2 — — —— — AC = √ [ 0 − (−5) ]2 + (2 − 7)2 = √52 + (−5)2 — — = √ 25 + 25 = √50 ≈ 7.1 — —— — —— — —— — — EF = √(4 − 0)2 + (1 − 1)2 = √42 = √16 = 4 — 7)2 —— DE = √(0 − 0)2 + (1 − 6)2 = √(−5)2 = √25 = 5 — = √36 + 64 = √ 100 = 10 —— — EF = √ (9 − 6)2 + [ −1 − (−6) ]2 = √32 + (−1 + 6)2 —— √[ 4 − (−2) ]2 + [ 6 − (−2) ]2 — — — ——— BC = √[ 0 − (−5) ]2 + (2 − 2)2 = √ 52 = √25 = 5 — BC = √(4 − 4)2 + [ 6 − (−2) ]2 = √(6 + 2)2 = √82 = 8 = √(4 + —— = √ 36 + (−5)2 = √36 + 25 = √61 ≈ 7.8 = √ 25 = 5 = √(4 + 2)2 + (−2 + 2)2 = √ 62 = 6 2)2 ——— DE = √ (6 − 0)2 + [ −6 − (−1) ]2 + √62 + (−6 + 1)2 — ——— 25. AB = √ [ 4 − (−2) ]2 + (−2 − (−2))2 2)2 — — AC = √(9 − 0)2 + (0 − 0)2 = √92 = √9 = 9 = √ 81 = 9 b. To use the HL Congruence Theorem (Thm. 5.9) to prove —— — BC = √ (9 − 6)2 + (0 − 5)2 = √ 32 + (−5)2 —— —— — 27. AB = √ (6 − 0)2 + (5 − 0)2 = √ 62 + 52 23. a. You need to know that the hypotenuses are congruent: —— — DF = √ (10 − 3)2 + (11 − 6)2 = √ (7)2 + (5)2 2. — SH ≅ — SH AC = — — = √64 + 36 = √ 100 = 10 AB = DE, BC = EF, and AC = DF, so △ABC ≅ △DEF. DF = √ (4 − 0)2 + (1 − 6)2 = √ 42 + (−5)2 — — = √ 16 + 25 = √41 ≈ 6.4 The triangles are not congruent. 29. yes; You could use the string to measure the length of each side on one triangle and compare it to the length of the corresponding side of the other triangle to determine whether SSS Congruence Theorem (Thm. 5.8) applies. 30. SAS Congruence Theorem (Thm. 5.5), HL Congruence Theorem (Thm. 5.9), SSS Congruence Theorem (Thm. 5.8) 162 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 — — 31. both; JL ≅ JL by the Reflexive Property of Congruence (Thm. 2.1), and the other two pairs of sides are marked as congruent. So, the SSS Congruence Theorem (Thm. 5.8) can be used. Also, because ∠ M and ∠ K are right angles, they are both right triangles, and the legs and hypotenuses are congruent. So, the HL Congruence Theorem (Thm. 5.9) can be used. 32. yes; They would have to be formed from circles that were at the same angle with each other. So, all corresponding parts would be congruent. 33. Sample answer: L 36. 5x = 3x + 10 5x − 2 = 4x + 3 2x = 10 x=5 x=5 ⋅ ⋅ BD = 4 ⋅ 5 + 3 = 23 CD = 3 ⋅ 5 + 10 = 25 — —— AB = 5 5 = 25 AC = 5 5 − 2 = 25 − 2 = 23 —, and BC — ≅ CB — (Reflexive For x = 5, AB ≅ CD , AC ≅ BD Property of Congruence, Thm. 2.1). Then, by the SSS Congruence Theorem (Thm. 5.8), △ABC ≅ △DCB. 5x = 4x + 3 S x=3 ⋅ ⋅ BD = 4 ⋅ 3 + 3 = 15 CD = 3 ⋅ 3 + 10 = 19 — — AB = 5 3 = 15 AC = 5 3 − 2 = 15 − 2 = 13 M N T U 34. Sample answer: L For x = 3, AC ≅ CD , so the triangles are not congruent. S 5x − 2 = 3x + 10 2x = 12 x=6 M N T U — — 35. a. BD ≅ BD by the Reflexive Property of Congruence — ≅ CB — and that ∠ ADB (Thm. 2.1). It is given that AB and ∠ CDB are right angles. So, △ABC and △CBD are right triangles and are congruent by the HL Congruence Theorem (Thm. 5.9). — — — —— — ⋅ ⋅ BD = 4 ⋅ 6 + 3 = 27 CD = 3 ⋅ 6 + 10 = 28 — — AB = 5 6 = 30 AC = 5 6 − 2 = 30 − 2 = 28 For x = 6, AB ≅ BD , so the triangles are not congruent. x = 5 is the only solution that yields △ABC ≅ △DCB. b. yes; Because AB ≅ CB ≅ CE ≅ FE , BD ≅ EG , and they are all right triangles, it can be shown that △ABD ≅ △CBD ≅ △CEG ≅ △FEG by the HL Congruence Theorem (Thm. 5.9). Maintaining Mathematical Proficiency — — — — — — — — 37. DF corresponds to AC . So, AC ≅ DF . 38. BC corresponds to EF . So, EF ≅ BC . 39. ∠ E corresponds to ∠ B. So, ∠ B ≅ ∠ E. 40. ∠ C corresponds to ∠ F. So, ∠ F ≅ ∠ C. 5.6 Explorations (p. 269) 1. a. Check students’ work. b. Check students’ work. — — — — — of AB ≅ AB . Also, BC of △ABC is congruent to BD c. The side that △ABC and △ABD share is AB . So, △ABD. The nonincluded angle that these two triangles share is ∠ BAC. d. no; The third pair of sides are not congruent. e. no; These two triangles provide a counterexample for SSA. They have two pairs of congruent sides and a pair of nonincluded congruent angles, but the triangles are not congruent. Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 163 Chapter 5 2. Possible congruence theorem Valid or not valid? SSS Valid SSA Not valid SAS Valid AAS Valid ASA Valid AAA Not valid — —— — — ≅ DC — AC 2. Given AB ⊥ AD , DE ⊥ AD , C −5 B′ A −4 D C B STATEMENTS REASONS — ⊥ AD —, DE — ⊥ AD —, 1. AB — — AC ≅ DC 1. Given 2. ∠ BAC and ∠ EDC are right angles. 2. Definition of perpendicular lines 3. ∠ BAC ≅ ∠ EDC 3. Right Angles Congruence Theorem (Thm. 2.3) 4. ∠ ACB ≅ ∠ DCE 4. Vertical Angles Congruence Theorem (Thm. 2.6) 5. △ABC ≅ △DEC 5. ASA Congruence Theorem (Thm. 5.10) y B C′ A Prove △ABC ≅ △DEC Sample answer: A counterexample for SSA is given in Exploration 1. A counterexample for AAA is shown here. 6 5 4 E 5 x A′ — — 3. Given ∠ S ≅ ∠ U, RS ≅ VU R U Prove △RST ≅ △VUT In this example, each pair of corresponding angles is congruent, but the corresponding sides are not congruent. T S 3. In order to determine that two triangles are congruent, one of the following must be true. V STATEMENTS REASONS 1. ∠ S ≅ ∠ U, — ≅ VU — RS 1. Given Two pairs of corresponding angles and the pair of included sides are congruent (ASA). 2. ∠ RTS ≅ ∠ VTU 2. Vertical Angles Congruence Theorem (Thm. 2.6) Two pairs of corresponding angles and one pair of nonincluded sides are congruent (AAS). 3. △RST ≅ △VUT 3. AAS Congruence Theorem (Thm. 5.11) All three pairs of corresponding sides are congruent (SSS). Two pairs of corresponding sides and the pair of included angles are congruent (SAS). The hypotenuses and one pair of corresponding legs of two right triangles are congruent (HL). 4. yes; Sample answer: In the diagram, △ABD ≅ △ACD by the HL Congruence Theorem (Thm. 5.9), the SSS Congruence Theorem (Thm. 5.8), and the SAS Congruence Theorem (Thm. 5.5). A 5.6 Exercises (pp. 274–276) Vocabulary and Core Concept Check 1. Both theorems are used to prove that two triangles are B D C 5.6 Monitoring Progress (pp. 271–273) congruent, and both require two pairs of corresponding angles to be congruent. In order to use the AAS Congruence Theorem (Thm. 5.11), one pair of corresponding nonincluded sides must also be congruent. In order to use the ASA Congruence Theorem (Thm. 5.10), the pair of corresponding included sides must be congruent. 1. yes; By the Alternate Interior Angles Theorem (Thm. 3.2), ∠ 1 ≅ ∠ 3 and ∠ 4 ≅ ∠ 2, and by the Reflexive Property of — ≅ WY —. So, △WXY ≅ △YZW by Congruence (Thm. 2.1), WY the ASA Congruence Theorem (Thm. 5.10). 164 Geometry Worked-Out Solutions 2. You need to know that one pair of corresponding sides are congruent. Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 Monitoring Progress and Modeling with Mathematics — — — — — — — NL ⊥ NQ , NL ⊥ MP , QM PL 17. Given M is the midpoint of NL , 3. yes; △ABC ≅ △QRS by the AAS Congruence Theorem Prove △NQM ≅ △MPL (Thm. 5.11). 4. yes; △ABC ≅ △DBC by the AAS Congruence Theorem Q P N M (Thm. 5.11). 5. no; Two sides and a nonincluded angle are not sufficient to determine congruence. 6. yes; △RSV ≅ △UTV by the ASA Congruence Theorem (Thm. 5.10). — — 7. Given GH ≅ MN , ∠ G ≅ ∠ M, ∠ F ≅ ∠ L — — 8. Given FG ≅ LM , ∠ G ≅ ∠ M, ∠ F ≅ ∠ L 9. yes; △ABC ≅ △DEF by the ASA Congruence Theorem (Thm. 5.10). 10. no; The congruence statements follow the pattern SSA, L STATEMENTS REASONS —, 1. M is the midpoint of NL — — — — NL ⊥ NQ , NL ⊥ MP , — PL — QM 1. Given 2. ∠ QNM and ∠ PML are right angles. 2. Definition of perpendicular lines 3. ∠ QNM ≅ ∠ PML 3. Right Angles Congruence Theorem (Thm. 2.3) 4. ∠ QMN ≅ ∠ PLM 4. Corresponding Angles Theorem (Thm. 3.1) which is not sufficient to conclude that the triangles are congruent. — — 11. no; AC and DE are not corresponding sides. 12. yes; △ABC ≅ △DEF by the AAS Congruence Theorem (Thm. 5.11). — 13. Construct a side that is congruent to DF . Construct an angle that is congruent to ∠ D and construct a second angle that is congruent to ∠ F. Complete the triangle. By the ASA Congruence Theorem (Thm. 5.10), △DEF ≅ △ACB. 5. — NM ≅ — ML 5. Definition of midpoint 6. △NQM ≅ △MPL — — 18. Given AJ ≅ KC , B ∠ BJK ≅ ∠ BKJ, ∠A ≅ ∠C Prove △ABK ≅ △CBJ — 14. Construct a side that is congruent to JK . Construct an angle that is congruent to ∠ J and construct a second angle that is congruent to ∠ K. Complete the triangle. By the ASA Congruence Theorem (Thm. 5.10), △JKL ≅ △BAC. 15. In the congruence statement, the vertices should be in corresponding order; △JKL ≅ △FGH by the ASA Congruence Theorem (Thm. 5.10). 16. The diagram shows ∠ Q ≅ ∠ V, ∠ R ≅ ∠ W, and — ≅ VW —. Because QR — and VW — are included sides, QR △QRS ≅ △VWX by the ASA Congruence Theorem (Thm. 5.10). STATEMENTS —≅— 1. AJ KC , ∠ BJK ≅ ∠ BKJ, ∠A ≅ ∠C A J K C REASONS 1. Given 2. AJ = KC 2. Definition of congruent segments 3. JC = JK + KC, AK = AJ + JK 3. Segment Addition Postulate (Post. 1.2) 4. AK = KC + JK 4. Substitution Property of Equality 5. AK = JK + KC 5. Commutative Property of Addition 6. AK = JC 6. Transitive Property of Equality 7. — AK ≅ — JC 8. △ABK ≅ △CBJ Copyright © Big Ideas Learning, LLC All rights reserved. 6. ASA Congruence Theorem (Thm. 5.10) 7. Definition of congruent segments 8. ASA Congruence Theorem (Thm. 5.10) Geometry Worked-Out Solutions 165 Chapter 5 — — 19. Given VW ≅ UW , Z ∠X ≅ ∠Z V Prove △XWV ≅ △ZWU STATEMENTS —, 1. — VW ≅ UW ∠X ≅ ∠Z 2. ∠ W ≅ ∠ W 3. △XWV ≅ △ZWU X Y U W REASONS 25. By finding the values of x and y in each triangle and solving for the angle measurements, △ABC ≅ △DBC by the ASA Congruence Theorem (Thm. 5.10) or the AAS Congruence Theorem (Thm. 5.11). Both triangles have a common side, — ≅ BC —. BC △ABC: (2x − 8)° + (5x + 10)° + (8x − 32)° = 180° 15x − 30 = 180 1. Given 15x = 210 2. Reflexive Property of Congruence (Thm. 2.2) ⋅ ⋅ m∠ BCA = (5 ⋅ 14 + 10)° = 80° m∠ CAB = (2 14 − 8)° = 20° m∠ ABC = (8 14 − 32)° = 80° 3. AAS Congruence Theorem (Thm. 5.11) 20. Given ∠ NKM ≅ ∠ LMK, L △DBC: N (y − 6)° + (3y + 2)° + (4y − 24)° = 180° 8y − 28 = 180 ∠L ≅ ∠N Prove △NMK ≅ △LKM STATEMENTS 1. ∠ NKM ≅ ∠ LMK, ∠L ≅ ∠N 2. — KM ≅ — MK 3. △NMK ≅ △LKM x = 14 8y = 208 K y = 26 M ⋅ ⋅ REASONS m∠ DBC = (4 26 − 24)° = 80° 1. Given m∠ BCD = (3 26 + 2)° = 80° 2. Reflexive Property of Congruence (Thm. 2.1) 3. AAS Congruence Theorem (Thm. 5.11) 21. You are given two right triangles, so the triangles have congruent right angles by the Right Angles Congruence Theorem (Thm. 2.3). Because another pair of angles and a pair of corresponding nonincluded sides (the hypotenuses) are congruent, the triangles are congruent by the AAS Congruence Theorem (Thm. 5.11). m∠ CDB = (26 − 6)° = 20° — ≅ BC —, ∠ BCA ≅ ∠ BCD ∠ ABC ≅ ∠ DBC, BC △ABC ≅ △DBC by the ASA Congruence Theorem (Thm. 5.10). — ≅ BC —, ∠ BCA ≅ ∠ BCD ∠ BAC ≅ ∠ BDC, BC △ABC ≅ △DBC by the AAS Congruence Theorem (Thm. 5.11). — ≅ BC —, ∠ ABC ≅ ∠ DBC ∠ BAC ≅ ∠ BDC, BC △ABC ≅ △DBC by the AAS Congruence Theorem (Thm. 5.11). — — 26. C; ∠ TSV ≅ ∠ VUW (given), ST ≅ UV (given), and ∠ STV ≅ ∠ UVW because they are right angles. 22. You are given two right triangles, so the triangles have congruent right angles by the Right Angles Congruence Theorem (Thm. 2.3). Because both pairs of legs are congruent, and the congruent right angles are the included angles, the triangles are congruent by the SAS Congruence Theorem (Thm. 5.5). 23. You are given two right triangles, so the triangles have congruent right angles by the Right Angles Congruence Theorem (Thm. 2.3). There is also another pair of congruent corresponding angles and a pair of congruent corresponding sides. If the pair of congruent sides is the included side, then the triangles are congruent by the ASA Congruence Theorem (Thm. 5.10). If the pair of congruent sides is a nonincluded pair, then the triangles are congruent by the AAS Congruence Theorem (Thm. 5.11). 24. D; To prove △JKL ≅ △MNL using the ASA Congruence Theorem (Thm. 5.10), you know that ∠ L ≅ ∠ L and — ≅ LJ —. So, the additional information needed is LM ∠ LKJ ≅ ∠ LNM. 166 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 27. Given ∠ B ≅ ∠ C b. Because △ABD ≅ △CBD and corresponding parts of A — ≅ AC — Prove AB congruent triangles are congruent, you can conclude — — that AD ≅ CD , which means that △ACD is isosceles by definition. c. no; For instance, because △ACD is isosceles, the girl sees her toes at the bottom of the mirror. This remains true as she moves backwards because △ACD remains isosceles. 30. △PTS ≅ △RTQ, △PTQ ≅ △RTS, △SQR ≅ △QSP, B STATEMENTS D C REASONS 1. Draw — AD , the angle bisector of ∠ ABC. 1. Construction of angle bisector 2. ∠ CAD ≅ ∠ BAD 2. Definition of angle bisector 3. ∠ B ≅ ∠ C 3. Given 4. — AD ≅ — AD 4. Reflexive Property of Congruence (Thm. 2.1) 5. △ABD ≅ △ACD 5. AAS Congruence Theorem (Thm. 5.11) 6. — AB ≅ — AC 6. Corresponding parts of congruent triangles are congruent. 28. yes; If you are using the AAS Congruence Theorem (Thm. 5.11), you have enough information to conclude that two pairs of angles are congruent and one pair of non-included sides is congruent. By the Third Angles Theorem (Thm. 5.4), the other pair of angles is congruent. Now, choose the third pair of angles, the pair of congruent sides, and the other pair of angles so that the pair of congruent sides are included sides. So, you can use the ASA Congruence Theorem (Thm. 5.10) to show that the triangles are congruent. △SRP ≅ △QPR; Because ∠ PTS ≅ ∠ RTQ by the Vertical Angles Congruence Theorem (Thm. 2.6), △PTS ≅ △RTQ by the SAS Congruence Theorem (Thm. 5.5); Because ∠ PTQ ≅ ∠ RTS by the Vertical Angles Congruence Theorem (Thm. 2.6), △PTQ ≅ △RTS by the SAS — intersects Congruence Theorem (Thm. 5.5); Transversal SQ — — parallel sides PS and QR to create congruent alternate — also interior angles ∠ SQR and ∠ QSP, and transversal SQ — — intersects parallel sides PQ and SR to create congruent — ≅ SQ — alternate interior angles ∠ QSR and ∠ SQP. Also, SQ by the Reflexive Property of Congruence (Thm. 2.1). So, △SQR ≅ △QSP by the ASA Congruence Theorem (Thm. — intersects parallel sides PS — and QR — to 5.10); Transversal PR create congruent alternate interior angles ∠ SPR and ∠ QRP, — also intersects parallel sides PQ — and SR — to and transversal PR create congruent alternate interior angles ∠ QPR and ∠ SRP. — ≅ PR — by the Reflexive Property of Congruence Also, PR (Thm. 2.1). So, △SRP ≅ △QPR by the ASA Congruence Theorem (Thm. 5.10). 31. Two triangles with congruent corresponding angles can be similar triangles without being congruent. This is based on dilating a triangle. The angles are equal and corresponding sides are proportional. B C A E — — 29. a. Given ∠ CDB ≅ ∠ ADB, DB ⊥ AC Prove △ABD ≅ △CBD D F 32. yes; Both triangles will always have three vertices. So, each STATEMENTS REASONS 1. ∠ CDB ≅ ∠ ADB, — DB ⊥ — AC 1. Given 2. ∠ ABD and ∠ CBD are right angles. 2. Definition of perpendicular lines 3. ∠ ABD ≅ ∠ CBD 3. Right Angles Congruence Theorem (Thm. 2.3) 4. — BD ≅ — BD 5. △ABD ≅ △CBD vertex of one triangle can be mapped to one vertex of the other triangle so that each vertex is only used once, and adjacent vertices of one triangle must be adjacent in the other. 4. Reflexive Property of Congruence (Thm. 2.1) 5. ASA Congruence Theorem (Thm. 5.10) Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 167 Chapter 5 33. a. The combinations that will make △TUV ≅ △XYZ by the ASA Congruence Theorem (Thm. 5.10) are: — ≅ XY —, ∠ U ≅ ∠ Y ∠ T ≅ ∠ X, TU — ≅ YZ —, ∠ V ≅ ∠ Z ∠ U ≅ ∠ Y, UV — — ∠ T ≅ ∠ X, TV ≅ XZ , ∠ V ≅ ∠ Z The combinations that will make △TUV ≅ △XYZ by the AAS Congruence Theorem (Thm. 5.11) are: 5.7 Explorations (p. 277) —). Because measure as the distance across the river (AB △ABC ≅ △DEC by the ASA Congruence Theorem (Thm. 5.10), the corresponding parts of the two triangles are also congruent. b. Given ∠ A and ∠ D — ≅ YZ —, ∠ U ≅ ∠ Y, ∠ T ≅ ∠ X UV Prove AB = DE — ≅ XZ —, ∠ T ≅ ∠ X, ∠ U ≅ ∠ Y TV — —, ∠ U ≅ ∠ Y, ∠ V ≅ ∠ Z TU ≅ XY — ≅ YZ —, ∠ V ≅ ∠ Z, ∠ T ≅ ∠ X UV — — TV ≅ XZ , ∠ V ≅ ∠ Z, ∠ U ≅ ∠ Y A C D E The combination that will make △TUV ≅ △XYZ by the SSS Congruence Theorem (Thm. 5.8) is: — ≅ XY —, UV — ≅ YZ —, TV — ≅ XZ — TU STATEMENTS 1. The combinations that will make △TUV ≅ △XYZ by the SAS Congruence Theorem (Thm. 5.5) are: — ≅ CD —, ∠ A and AC REASONS 1. Given ∠ D are right angles. — ≅ XY —, ∠ T ≅ ∠ X, TV — ≅ XZ — TU 2. ∠ A ≅ ∠ D 2. Right Angles Congruence Theorem (Thm. 2.3) — ≅ YZ —, ∠ V ≅ ∠ Z, TV — ≅ XZ — UV 3. ∠ BCA ≅ ∠ ECD 3. Vertical Angles Congruence Theorem (Thm. 2.6) ( 6C3 = 20 ). There are 13 combinations that provide 4. △BCA ≅ △ECD enough information. So, the probability of choosing at random enough information to prove that the triangles are 13 congruent is — , or 65%. 20 5. — AB ≅ — DE 4. ASA Congruence Theorem (Thm. 5.10) — ≅ YZ —, ∠ U ≅ ∠ Y, TU — ≅ XY — UV b. There are 20 ways to choose 3 items from 6 items Maintaining Mathematical Proficiency 1+5 0+4 6 4 34. Midpoint = —, — = —, — = (3, 2) 2 2 2 2 36. B are right angles. — ≅ CD — AC — ≅ XY —, ∠ T ≅ ∠ X, ∠ V ≅ ∠ Z TU 35. — 1. a. The surveyor can measure DE, which will have the same ( ) ( ) −2 + 4 3 + (−1) , Midpoint = ( ) = ( 22, 22 ) = (1, 1) 2 2 −5 + 2 −7 + (−4) , Midpoint = ( ) = ( −32, −112 ) 2 2 — — — — — — — — 37. Draw a segment. Label a point D on the segment. Draw an arc with center A, and label the intersection F points B and C. Using the same radius, draw an arc with center D. D Label the point of intersection of E the arc and the segment as E. Draw an arc with radius BC with center E. ⃗. So, ∠ FDE ≅ ∠ A. Label the intersection F. Draw DF 6. AB = DE 5. Corresponding parts of congruent triangles are congruent. 6. Definition of congruent segments c. By creating a triangle on land that is congruent to a triangle that crosses the river, you can find the distance across the river by measuring the distance of the corresponding congruent segment on land. 2. a. The officer’s height stays the same, he is standing perpendicular to the ground the whole time, and he tipped his hat the same angle in both directions. So, △DEF ≅ △DEG by the ASA Congruence Theorem (Thm. 5.10). Because corresponding parts of the two — ≅ EF —. By the definition triangles are also congruent, EG of congruent segments, EG equals EF, which is the width of the river. 38. Draw a segment. Label a point D on the segment. Draw an arc with F center B, and label the intersection points A and C. Using the same radius, draw an arc with center D. D E Label the point of intersection of the arc and the segment as E. Draw an arc with radius AC with center E. Label the intersection F. Draw ⃗ DF. So, ∠ FDE ≅ ∠ B. 168 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 — — — — — — that PT ≅ UQ to prove that △PTU ≅ △UQP by the SSS b. Given ∠ DEG is a right angle. 3. You know that TU ≅ QP and UP ≅ PU . You need to show ∠ DEF is a right angle. ∠ EDG ≅ ∠ EDF Prove EG = EF F G D Congruence Theorem (Thm. 5.8). Use the HL Congruence Theorem (Thm. 5.9) to prove that △QSP ≅ △TRU. Use the SAS Congruence Theorem (Thm. 5.5) to prove that — ≅ UQ —. Use the △PRT ≅ △USQ. Then state that PT SSS Congruence Theorem (Thm. 5.8) to prove that △PTU ≅ △UQP. — E STATEMENTS REASONS 5.7 Exercises (pp. 281–282) 1. ∠ EDG ≅ ∠ EDF, ∠ DEG and ∠ DEF are right angles. 1. Given Vocabulary and Core Concept Check 2. ∠ DEG ≅ ∠ DEF 2. Right Angles Congruence Theorem (Thm. 2.3) DE ≅ — DE 3. — 4. △FDE ≅ △GDE 5. — EG ≅ — EF 6. EG = EF — 4. Segments that can be assumed congruent are AC and AB . 1. Corresponding parts of congruent triangles are congruent. 3. Reflexive Property of Congruence (Thm. 2.1) 4. ASA Congruence Theorem (Thm. 5.10) 5. Corresponding parts of congruent triangles are congruent. 6. Definition of congruent segments c. By standing perpendicular to the ground and using the tip of your hat to gaze at two different points in such a way that the direction of your gaze makes the same angle with your body both times, you can create two congruent triangles, which ensures that you are the same distance from both points. 3. By creating a triangle that is congruent to a triangle with an unknown side length or angle measure, you can measure the created triangle and use it to find the unknown measure indirectly. 4. You do not actually measure the side length or angle measure you are trying to find. You measure the side length or angle measure of a triangle that is congruent to the one you are trying to find. 5.7 Monitoring Progress (pp. 278–280) 1. All three pairs of sides are congruent. So, by the SSS Congruence Theorem (Thm. 5.8), △ABD ≅ △CBD. Because corresponding parts of congruent triangles are congruent, ∠ A ≅ ∠ C. 2. no; As long as the rest of the steps are followed correctly, △LKM will still be congruent to △PNM. So, the corresponding parts will still be congruent, and you will still —. be able to find NP by measuring LK Copyright © Big Ideas Learning, LLC All rights reserved. 2. Sample answer: Finding the width of a body of water, a distance across a valley or canyon, or any inaccessible distance can require indirect measurement. Monitoring Progress and Modeling with Mathematics 3. All three pairs of sides are congruent. So, by the SSS Congruence Theorem (Thm. 5.8), △ABC ≅ △DBC. Because corresponding parts of congruent triangles are congruent, ∠ A ≅ ∠ D. 4. Two pairs of sides and the pair of included angles are congruent. So, by the SAS Congruence Theorem (Thm. 5.5), △QPR ≅ △TPS. Because corresponding parts of congruent triangles are congruent, ∠ Q ≅ ∠ T. 5. The hypotenuses and one pair of legs of two right triangles are congruent. So, by the HL Congruence Theorem (Thm. 5.9), △JMK ≅ △LMK. Because corresponding parts — ≅ LM —. of congruent triangles are congruent, JM 6. ∠ BAD ≅ ∠ CDA by the Alternate Interior Angles Theorem — ≅ AD — by the (Thm. 3.2). From the diagram, ∠ B ≅ ∠ C. AD Reflexive Property of Congruence (Thm. 2.1). So, by the AAS Congruence Theorem (Thm. 5.11), △ACD ≅ △DBA. Because corresponding parts of congruent triangles are — ≅ DB —. congruent, AC 7. From the diagram, ∠ JHN ≅ ∠ KGL, ∠ N ≅ ∠ L, and — ≅ KL —. So, by the AAS Congruence Theorem (Thm. 5.11), JN △JNH ≅ △KLG. Because corresponding parts of congruent — ≅ HJ —. triangles are congruent, GK 8. From the diagram, ∠ Q ≅ ∠ W ≅ ∠ RVT ≅ ∠ T and — ≅ RT —. So, by the AAS Congruence Theorem VW (Thm. 5.11), △QVW ≅ △VRT. Because corresponding — ≅ VT —. parts of congruent triangles are congruent, QW 9. Use the AAS Congruence Theorem (Thm. 5.11) to prove that △FHG ≅ △GKF. Then, state that ∠ FGK ≅ ∠ GFH. Use the Congruent Complements Theorem (Thm. 2.5) to prove that ∠ 1 ≅ ∠ 2. 10. Use the AAS Congruence Theorem (Thm. 5.11) to prove — ≅ CE — because that △ABE ≅ △DCE. Then, state that BE corresponding parts of congruent triangles are congruent. Use the Base Angles Theorem (Thm. 5.6) to prove that ∠ 1 ≅ ∠ 2. Geometry Worked-Out Solutions 169 Chapter 5 11. Use the ASA Congruence Theorem (Thm 5.10) to prove — because that △STR ≅ △QTP. Then, state that — PT ≅ RT corresponding parts of congruent triangles are congruent. Use the SAS Congruence Theorem (Thm. 5.5) to prove that △STP ≅ △QTR. So, ∠ 1 ≅ ∠ 2. — —— — 14. Given PA ≅ PB , QA ≅ QB Prove ∠ QPA and ∠ QPB are right angles. P A B Q 12. Use the ASA Congruence Theorem (Thm 5.10) to prove — ≅ CE — and that △ABE ≅ △CBE. Then, state that AE ∠ BCE ≅ ∠ BAE because corresponding parts of congruent triangles are congruent. Use the Congruent Complements Theorem (Thm. 2.5) to state that ∠ ECD ≅ ∠ EAF. So, you can use the SAS Congruence Theorem (Thm. 5.5) to prove that △ECD ≅ △EAF, so ∠ 1 ≅ ∠ 2. — — — — 13. Given AP ≅ BP and AQ ≅ BQ Prove ∠ AMP and ∠ BMP are right angles. STATEMENTS REASONS — ≅ PB —, QA — ≅ QB — 1. PA 1. Given PQ ≅ — PQ 2. — 3. △APQ ≅ △BPQ 3. SSS Congruence Theorem (Thm. 5.8) 4. ∠ QPA ≅ ∠ QPB 4. Corresponding parts of congruent triangles are congruent. 5. ∠ QPA and ∠ QPB form a linear pair. 5. Definition of a linear pair P A M B Q STATEMENTS — ≅ BP —, AQ — ≅ BQ — 1. AP 2. — PQ ≅ — PQ 1. Given 2. Reflexive Property of Congruence (Thm. 2.1) 3. △APQ ≅ △BPQ 3. SSS Congruence Theorem (Thm. 5.8) 4. ∠ APQ ≅ ∠ BPQ 4. Corresponding parts of congruent triangles are congruent. 5. — PM ≅ — PM 6. SAS Congruence Theorem (Thm. 5.5) 7. ∠ AMP ≅ ∠ BMP 7. Corresponding parts of congruent triangles are congruent. 8. ∠ AMP and ∠ BMP form a linear pair. 8. Definition of a linear pair 10. ∠ AMP and ∠ BMP are right angles. 6. — PQ ⊥ — AB 7. ∠ QPA and ∠ QPB are right angles. 6. Linear Pair Perpendicular Theorem (Thm. 3.10) 7. Definition of perpendicular lines 5. Reflexive Property of Congruence (Thm. 2.1) 6. △APM ≅ △BPM 9. — MP ⊥ — AB 170 REASONS 2. Reflexive Property of Congruence (Thm. 2.1) 9. Linear Pair Perpendicular Theorem (Thm. 3.10) 10. Definition of perpendicular lines Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 — — — — — ≅ LM — ≅ KM — ≅ NM — JM — — Prove FL ≅ HN 15. Given FG ≅ GJ ≅ HG ≅ GK , F G J H K M L STATEMENTS — ≅ GJ — ≅ HG —≅ 1. FG — — — GK , JM ≅ LM ≅ REASONS 1. Given —≅ NM — KM 16. Given ∠ PRU ≅ ∠ QVS, ∠ PUR ≅ ∠ QSV ≅ ∠ RUX ≅ ∠ VSY, — ≅ VU — RS Q P T S R U W Prove △PUX ≅ △QSY X N STATEMENTS V Y REASONS 1. ∠ PRU ≅ ∠ QVS, ∠ PUR ≅ ∠ QSV ≅ ∠ RUX ≅ ∠ VSY, — ≅ VU — RS 1. Given 2. ∠ FGJ ≅ ∠ HGK, ∠ JML ≅ ∠ KMN 2. Vertical Angles Congruence Theorem (Thm. 2.6) 2. RS = VU 2. Definition of congruent segments 3. △FGJ ≅ △HGK, △JML ≅ △KMN 3. SAS Congruence Theorem (Thm. 5.5) 3. RU = RS + SU, VS = VU + SU 3. Segment Addition Postulate (Post. 1.2) 4. ∠ F ≅ ∠ H, ∠L ≅ ∠N 4. Corresponding parts of congruent triangles are congruent. 4. VS = RS + SU 4. Substitution Property of Equality 5. FG = GJ = HG = GK 5. RU = VS 5. Definition of congruent segments 5. Transitive Property of Equality 6. HJ = HG + GJ, FK = FG + GK 6. Segment Addition Postulate (Post. 1.2) 7. FK = HG + GJ 7. Substitution Property of Equality 8. FK = HJ 8. Transitive Property of Equality 9. — FK ≅ — HJ 10. △HJN ≅ △FKL 11. — FL ≅ — HN 9. Definition of congruent segments 10. AAS Congruence Theorem (Thm. 5.11) 11. Corresponding parts of congruent triangles are congruent. Copyright © Big Ideas Learning, LLC All rights reserved. 6. — RU ≅ — VS 6. Definition of congruent segments 7. △PUR ≅ △QSV 8. ∠P ≅ ∠ Q, — PU ≅ — QS 9. m∠ PUR = m∠ QSV = m∠ RUX = m∠ VSY 7. ASA Congruence Theorem (Thm. 5.10) 8. Corresponding parts of congruent triangles are congruent. 9. Definition of congruent angles 10. m∠ PUX = m∠ PUR + m∠ RUX, m∠ QSY = m∠ QSV + m∠ VSY 10. Angle Addition Postulate (Post. 1.4) 11. m∠ QSY = m∠ PUR + m∠ RUX 11. Substitution Property of Equality 12. m∠ PUX = m∠ QSY 12. Transitive Property of Equality 13. ∠ PUX ≅ ∠ QSY 13. Definition of congruent angles 14. △PUX ≅ △QSY 14. ASA Congruence Theorem (Thm. 5.10) Geometry Worked-Out Solutions 171 Chapter 5 — — — — 17. Because AC ⊥ BC and ED ⊥ BD , ∠ ACB and ∠ EDB are —, congruent right angles. Because B is the midpoint of CD — ≅ BD —. The vertical angles ∠ ABC and ∠ EBD are BC congruent. So, △ABC ≅ △EBD by the ASA Congruence Theorem (Thm. 5.10). Then, because corresponding parts — ≅ ED —. So, you can of congruent triangles are congruent, AC —. find the distance AC across the canyon by measuring ED 18. a. Because the base of the red triangle is twice the base of 20. From the Internet, the distance form Miami to Bermuda is about 1035 miles. The distance from Bermuda to San Juan is about 956 miles. The distance form San Juan to Miami is about 1034 miles. The perimeter of the triangle created by these locations is about 1034 + 956 + 1034 = 3025 miles. To find the area of this triangle, which is close to an isosceles triangle, let the distance between Miami and Bermuda be the same as the distance between Miami and San Juan. Let h be the height of the triangle. the purple triangle, the red triangle has an area twice the area of the purple triangle. B b. Because the base of the orange triangle is twice the base of the purple triangle and the height of the orange triangle is twice the height of the purple triangle, the area of the orange triangle is four times the area of the purple triangle. — — 19. Given AD BC 1035 miles 478 miles h M J 478 miles 1035 miles —. E is the midpoint of AC S Prove △AEB ≅ △CED A B h2 = 10352 − 4782 h2 = 1,071,225 − 228,484 h2 = 842,741 E — h = √842,741 ≈ 918 The approximate height of the triangle is 918 miles. D C STATEMENTS REASONS — BC —, E is the 1. AD —. midpoint of AC 1. Given 2. — AE ≅ — CE 2. Definition of midpoint 3. ∠ AED ≅ ∠ BEC, ∠ AEB ≅ ∠ CED 3. Vertical Angles Congruence Theorem (Thm. 2.6) 4. ∠ DAE ≅ ∠ BCE 4. Alternate Interior Angles Theorem (Thm. 3.2) 5. △DAE ≅ △BCE 5. ASA Congruence Theorem (Thm. 5.10) 6. — DE ≅ — BE 7. △AEB ≅ △CED 6. Corresponding parts of congruent triangles are congruent. 7. SAS Congruence Theorem (Thm. 5.5) ⋅ ⋅ Area ≈ —12 956 918 = 438,804 mi2 The area of the triangle is about 439,000 square miles. Sample answer: Three cities with approximate distances as the three cities in the Bermuda Triangle are Erie, PA, to Orlando, FL (970 miles); Erie, PA, to Oklahoma City, OK (1056 miles); and Oklahoma City, OK, to Orlando, FL (1064 miles). Erie, PA 1056 miles Oklahoma City, OK 970 miles 1064 miles Orlando, FL 21. yes; You can show that WXYZ is a rectangle. This means that the opposite sides are congruent. Because △WZY and △YXW share a hypotenuse, the two triangles have congruent hypotenuses and corresponding legs, which allows you to use the HL Congruence Theorem (Thm. 5.9) to prove that the triangles are congruent. 22. a. The statement “If two triangles have the same perimeter, then they are congruent” is false. The converse is true: If two triangles are congruent, then their perimeters are the same. b. The statement “If two triangles are congruent, then their areas are the same” is true. 172 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 — — — ≅ GH — (given). So, △ABC ≅ △GHJ by the SAS and AB 23. AC ≅ GJ (given), ∠ A ≅ ∠ G because they are right angles, d. Reflect the triangle in the x-axis, so that △ABC is still an isosceles right triangle. So, C(3, −3). Congruence Theorem (Thm. 5.5). — ≅ QN — (given), and ∠ A ≅ ∠ N ∠ C ≅ ∠ Q (given), CA because they are right angles. So, △ABC ≅ △NPQ by the ASA Congruence Theorem (Thm. 5.10). 2 1 −2−1 Maintaining Mathematical Proficiency —— 4 3 — — — ——— — CD = √(−1 − 4)2 + [ −2 − (−2) ]2 = √(−1 − 4)2 — — = √(−5)2 = √ 25 = 5 ——— AD = √[ −1 − (−1) ]2 + (−2 − 1)2 —— — — = √(−1 + 1)2 + (−3)2 = √ (−3)2 = √9 = 3 Perimeter = AB + BC + CD + AD = 5 + 3 + 5 + 3 = 16 units ——— 25. JK = √ [ −2 − (−5) ]2 + (1 − 3)2 — =√ (3)2 — — + 4 = √ 9 + 4 = √ 13 ≈ 3.6 —— —— KL = √[ 3 − (−2) ]2 + (4 − 1)2 = √(3 + 2)2 + 32 — — = √25 + 9 = √34 ≈ 5.8 —— JL = √[ 3 − (−5) ]2 + (4 − 3)2 —— — — = √(3 + 5)2 + 12 = √ 82 + 1 = √ 65 = 8.1 Perimeter = JK + KL + JL 5.8 Explorations (p. 283) e. If C lies on the line x = 3, then the coordinates are C(3, y). Because △ABC is an isosceles triangle, y y —=— —=— , and mBC . mAC 3 −3 △ABC is a right triangle, so it must have a right angle. — and BC — are the congruent legs of △ABC, ∠ A Because AC and ∠ B are the congruent base angles by the Base Angles Theorem (Thm. 5.6). The vertex angle, ∠ C, must be the — ⊥ BC — by the definition right angle, which means that AC of perpendicular lines. By the Slopes of Perpendicular Lines Theorem (Thm. 3.14), y y — — = −1 and y = ±3. 3 −3 So, the coordinates of C must be (3, 3) or (3, −3). 3. You can position the figure in a coordinate plane and then use deductive reasoning to show that what you are trying to prove must be true based on the coordinates of the figure. —— —— — —— — 2 AC = √(3 − 0)2 + (3√3 − 0)2 = √ 32 + ( 3√3 ) ⋅ — ⋅ — — = √9 + 9 3 = √9 + 27 = √36 = 6 —— — 2 —— — 2 BC = √(6 − 3)2 + ( 0 − 3√3 ) = √32 + ( −3√3 ) — = √9 + 9 3 = √9 + 27 = √36 = 6 — △ABC with vertices A(0, 0), B(6, 0), C ( 3, 3√ 3 ) has side lengths AB = 6, AC = 6, and BC = 6. By the definition of congruent sides, △ABC is an equilateral triangle. 5.8 Monitoring Progress (pp. 284–286) 1. Another way of placing the rectangle in Example 1(a) that 2. a. Check students’ work. is convenient for finding side lengths would be to place the width (k) on the x-axis. b. Check students’ work. c. Check students’ work. y —— — —— —— BC = √ (6 − 3)2 + (0 − 3)2 = √ 32 + (−3)2 — — — = √ 9 + 9 = √18 = 3√ 2 (k, h) (0, h) h AC = √ (0 − 3)2 + (0 − 3)2 = √ (−3)2 + (−3)2 — — — = √ 9 + 9 = √18 = 3√ 2 3−0 3 —=— =—=1 m AC 3−0 3 3 3−0 —=— = — = −1 m BC 3 − 6 −3 — — = 1 and mBC — = −1, Because AC = BC = 3√ 2 and mAC — — AC ⊥ BC and △ABC is a right isosceles triangle. Copyright © Big Ideas Learning, LLC All rights reserved. — 4. AB = √ (6 − 0)2 + (0 − 0)2 = √ 62 = 6 — Check students’ work. Check students’ work. Check students’ work. — Using the Distance Formula, AC = √ 9 + y2 and — — — 2 AB = √ 9 + y . AC ≅ BC , so △ABC is an isosceles triangle. AB = ∣ 6 − 0 ∣ = ∣ 6 ∣ = 6 C — ≈ 3.6 + 5.8 + 8.1 = 17.5 units 1. a. b. c. d. 90° ⋅ —— = √(−2 + 5)2 + (−2)2 B 1 2 3 4 5 6 x −3 −4 = √52 = 5 —— A −2 — 24. AB = √ [ 4 − (−1) ]2 + (1 − 1)2 = √ (4 + 1)2 BC = √(4 − 4)2 + (−2 − 1)2 = √(−3)2 = √9 = 3 y x (0, 0) k (k, 0) Geometry Worked-Out Solutions 173 Chapter 5 2. A square with vertices (0, 0), (m, 0), and (0, m) will have a fourth vertex at (m, m). 5. Given Coordinates of △NPO are N(h, h), P(0, 2h), and O(0, 0). Coordinates of △NMO are N(h, h), M(2h, 0), and O(0, 0). y (0, m) Prove △NPO ≅ △NMO (m, m) P(0, 2h) y N(h, h) O(0, 0) (0, 0) (m, 0) x 3. Use the Distance Formula to find GO and GH in order to — ≅ GH —. State that ∠ OGJ ≅ ∠ HGJ by the show that GO — ≅ GJ — by the Reflexive definition of angle bisector and GJ Property of Congruence (Thm. 2.1). Then use the SAS Congruence Theorem (Thm. 5.5) to show that △GJO ≅ △GJH. — —— — NP = √ (0 − h)2 + (2h − h)2 = √ (−h)2 + h2 — — = √ 2h2 = h√2 PO = √ (0 − 0)2 + (0 − 2h)2 = √ (−2h)2 — = √ 4h2 = 2h —— — —— — NM = √ (2h − h)2 + (0 − h)2 = √ h2 + (−h)2 — — = √ 2h2 = h√2 MO = √ (0 − 2h)2 + (0 − 0)2 = √ (−2h)2 y 4. —— M(2h, 0) x — = √ 4h2 = 2h — ≅ MN — and PO — ≅ MO —, and by the Reflexive Property So, PN — ≅ ON —. So, you can apply of Congruence (Thm. 2.1), ON the SSS Congruence Theorem (Thm. 5.8) to conclude that △NPO ≅ △NMO. H(m, n) x O(0, 0) 5.8 Exercises (pp. 287–288) J(m, 0) Vocabulary and Core Concept Check Side lengths of △OHJ: —— 1. In a coordinate proof, you have to assign coordinates to — OH = √(m − 0)2 + (n − 0)2 = √ m2 + n2 —— — OJ = √(m − 0)2 + (0 − 0)2 = √ m2 = m —— — — HJ = √(m − m)2 + (0 − n)2 = √ (−n)2 = √n2 = n Midpoint of each side: m+0 n+0 m n —= — , — = —, — Midpoint of OH 2 2 2 2 ( ) ( ) — = m + 0, 0 + 0 = m, 0 Midpoint of OJ ( 2 2 ) (2 ) — = m + m, n + 0 = 2m, n = m, n Midpoint of HJ ( 2 2 ) ( 2 2) ( 2) — — — — — — — Slope of each side: n n−0 —=— =— Slope of OH m−0 m 0 0−0 —=— =—=0 Slope of OJ m−0 m −n 0−n Slope of — HJ = — = — = undefined m−m 0 — ⊥ HJ — at angle J. Therefore, △OHJ is a right triangle. OJ — vertices and write expressions for the side lengths and the slopes of segments in order to show how sides are related; As with other types of proofs, you still have to use deductive reasoning and justify every conclusion with theorems, proofs, and properties of mathematics. 2. When the triangle is positioned as shown, you are using zeros in your expressions, so the side lengths are often the same as one of the coordinates. Monitoring Progress and Modeling with Mathematics 3. Sample answer: Place the legs on the x- and y-axes. y 3 2 C(0, 2) 1 B(3, 0) A(0, 0) 1 2 3 x It is easy to find the lengths of horizontal and vertical segments and distances from the origin. 174 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 4. Sample answer: Place one side on the x-axis and one side on 9. the y-axis. y 3 y C(0, 7) 7 6 5 4 3 Z(3, 3) W(0, 3) 8 2 2 1 1 Y(3, 0) 1 X(0, 0) 2 3 x —— — CD = √(0 − 9)2 + (7 − 0)2 = √(−9)2 + 72 — — = √81 + 49 = √ 130 ≈ 11.4 The length of the hypotenuse is about 11.4 units. It is easy to find the lengths of horizontal and vertical segments and distances from the origin. 5. Sample answer: Place the legs on the x- and y-axes. D(9, 0) 1 2 3 4 5 6 7 8 9 x O(0, 0) 10. y y A(0, 50) 50 R(0, p) 40 30 20 10 T(p, 0) x S(0, 0) B(−30, 0) −30 It is easy to find the lengths of horizontal and vertical segments and distances from the origin. C(30, 0) −20 −10 10 20 30 x Use the Pythagorean Theorem = + b2, where the hypotenuse of the right triangle is one of the legs of the isosceles triangle. c2 6. Sample answer: Place the side with length 2m on the x-axis. y a2 AC2 = 502 + 302 AC2 = 2500 + 900 AC2 = 3400 — AC = √ 3400 ≈ 58.3 L(n, p) The length of one of the legs of the isosceles triangle is about 58.3 units. J(0, 0) K(2m, 0) x It is easy to find the lengths of horizontal segments and distances from the origin. ——— — 7. Find the lengths of OP , PM , MN , and NO to show that — ≅ PM — and MN — ≅ NO —. OP 8. Find the coordinates of G using the Midpoint Formula. Use these coordinates and the Distance Formula to show that — ≅ JG —. Show that HG — ≅ FG — by the definition of OG midpoint, and ∠ HGJ ≅ ∠ FGO by the Vertical Angles Congruence Theorem (Thm. 2.6). Then use the SAS Congruence Theorem (Thm. 5.5) to conclude that △GHJ ≅ △GFO. Copyright © Big Ideas Learning, LLC All rights reserved. 11. 6 y 5 L(0, 4) 4 3 M(5, 4) 2 1 O(0, 0) N(5, 0) 1 2 3 4 5 6 x —— — NL = √(5 − 0)2 + (0 − 4)2 = √52 + (−4)2 — — = √25 + 16 = √ 41 ≈ 6.4 The length of the diagonal is about 6.4 units. Geometry Worked-Out Solutions 175 Chapter 5 14. y 12. Y(n, n) Z(0, n) X(n, 0) x F(m, 0) x —— — XZ = √ (n − 0)2 + (0 − n)2 = √n2 + (−n)2 — — Side lengths of △DEF: — = √n2 + n2 = √2n2 = n√ 2 —— — —— — —— — DE = √(m − 0)2 + (n − n)2 = √ m2 = m — The length of the diagonal is n√ 2 units. EF = √(m − m)2 + (0 − n)2 = √n2 = n DF = √(m − 0)2 + (0 − n)2 = √m2 + n2 y B(h, h) Slope of each side of △DEF: 0 n−n —=— =—=0 Slope of DE m−0 m −n 0−n —=— = — = undefined Slope of EF m−m 0 n 0−n —=— = −— Slope of DF m m−0 C(2h, 0) x A(0, 0) Side lengths of △ABC: —— — — — AB = √(h − 0)2 + (h − 0)2 = √ h2 + h2 = √ 2h2 = h√2 —— — —— — — — BC = √(2h − h)2 + (0 − h)2 = √ h2 + h2 = √ 2h2 = h√2 AC = √ (2h − 0)2 + (0 − 0)2 = √ 4h2 = 2h Slope of each side of △ABC: h−0 h —=— =—=1 Slope of AB h−0 h h 2h − h —=— = — = −1 Slope of BC 0−h −h 0 0−0 —= — =—=0 Slope of AC 2h − 0 2h Midpoint of each side of △ABC: 0+h 0+h h h —= — , — = —, — Midpoint of AB 2 2 2 2 h + 2h h + 0 3h h — Midpoint of BC = —, — = —, — 2 2 2 2 2h + 0 0 + 0 2h — Midpoint of AC = —, — = —, 0 = (h, 0) 2 2 2 ( ( ( ) ( ) ) ( ) ) ( ) — ⊥ BC — by the Slopes of — mBC — = −1, AB Because mAB Perpendicular Lines Theorem (Thm. 3.14). So, ∠ ABC is — ≅ BC — because AB = BC. So, △ABC is a a right angle. AB right isosceles triangle. ⋅ 176 E(m, n) D(0, n) O(0, 0) 13. y Geometry Worked-Out Solutions Midpoint of each side of △DEF: 0+m n+n m 2n m —= — , — = —, — = —, n Midpoint of DE 2 2 2 2 2 m+m n+0 2m n n — Midpoint of EF = —, — = —, — = m, — 2 2 2 2 2 0+m n+0 m n —= — , — = —, — Midpoint of DF 2 2 2 2 ( ( ( ) ( ) ( ) ) ( ) ( ) ) ( ) — is horizontal. Because — = 0, DE Because mDE — — ⊥ EF — by the Slopes — = undefined, EF is vertical. So, DE mEF of Perpendicular Lines Theorem (Thm. 3.14), and ∠ DEF is a right angle by the definition of perpendicular lines. Also, none of the sides have the same length. So, △DEF is a right scalene triangle. 15. The coordinates of the unlabeled vertex is N(h, k). —— — ON = √(h − 0)2 + (k − 0)2 = √h2 + k2 —— — — MN = √(2h − h)2 + (0 − k)2 = √h2 + (−k)2 = √h2 + k2 16. The coordinates of the vertices are O(0, 0), U(k , 0), R(k, k), S(k, 2k), and T(2k, 2k). —— —— OT = √ (2k − 0)2 + (2k − 0)2 = √(2k)2 + (2k)2 — — — = √ 4k2 + 4k2 = √ 8k2 = 2k√ 2 Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 —— — 17. Find the lengths of DE , EC , and DC of △DEC. — —— DE = √(2h − h)2 + (2k − 2k)2 = √ h2 = h —— — EC = √(2h − h)2 + (2k − k)2 = √ h2 + k2 —— 19. The triangle formed by your position (Y), your cousin’s position (C), and the campsite (O) has the coordinates Y(500, 1200), C(1000, 0), and O(0, 0). y — Y(500, 1200) DC = √(h − h)2 + (2k − k)2 = √ k2 = k —, OC —, and BC — of △BOC. Find the lengths of BO —— — —— — —— — 1000 BO = √(h − 0)2 + (0 − 0)2 = √ h2 = h OC = √(h − 0)2 + (k − 0)2 = √ h2 + k2 500 BC = √(h − h)2 + (k − 0)2 = √k2 = k Because DE = BO, EC = OC, and DC = BC, by the — ≅ BO —, EC — ≅ OC —, and definition of congruent segments, DE — — DC ≅ BC . By the SSS Congruence Theorem (Thm. 5.8), △DEC ≅ △BOC. — 18. Because H is the midpoint of DA , the coordinates of 1000 x ——— OY = √(500 − 0)2 + (1200 − 0)2 —— point H are = √ 5002 + 12002 —— 0 − 2h 2k + 0 H —, — = H(−h, k). 2 2 ( ) = √ 250,000 + 1,440,000 —, the coordinates of Because G is the midpoint of EA point G are 0 + 2h 2k + 0 G —, — = G(h, k). 2 2 ( ) —— ——— (−h)]2 — + (0 − k)2 — = √ 1,690,000 = 1300 The distance between your position and the campsite is 1300 meters. ——— YC = √(1000 − 500)2 + (0 − 1200)2 —— DG = √(−2h − h)2 + (0 − k)2 = √ (−3h)2 + (−k)2 — = √9h2 + k2 EH = √[2h − C(1000, 0) 500 O(0, 0) =√ —— (3h)2 + (−k)2 = √9h2 + k2 Because DG = EH, by the definition of congruent segments, — ≅ EH —. DG —— = √5002 + (−1200)2 —— = √250,000 + 1,440,000 — = √1,690,000 = 1300 The distance between your position and your cousin’s position is 1300 meters. —— OH = √(1000 − 0)2 + (0 − 0)2 — = √10002 = 1000 The distance between the campsite and your cousin’s position is 1000 meters. — ≅ YC —, the triangle formed by your position, Because OY your cousin’s position, and the campsite is an isosceles triangle. Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 177 Chapter 5 20. 23. A y P(0, 2) S(−2, 1) −4 6 −2 2 y (−h, k) 4 x (2h, k) Q(3, −4) 2 −4 (−h, 0) R(1, −5) 2 (2h, 0) x −2 Side lengths: —— — PQ = √(3 − 0)2 + (−4 − 2)2 = √ 32 + (−6)2 — — — = √9 + 36 = √ 45 = 3√5 ——— —— QR = √(1 − 3)2 + [ −5 − (−4) ]2 = √ (−2)2 + (−5 + 4)2 — — — = √ 4 + (−1)2 = √4 + 1 = √5 ——— —— RS = √(−2 − 1)2 + [ 1 − (−5) ]2 = √ (−3)2 + (1 + 5)2 — — — = √9 + 36 = √45 = 3√5 —— 24. Sample answer: It would be easy to prove the Base Angles Theorem (Thm. 5.6) with a coordinate proof. First, position the given isosceles triangle, △ABC, on the coordinate plane so that the base is on the x-axis, and one vertex is at the origin. x=k y —— PS = √(−2 − 0)2 + (1 − 2)2 = √ (−2)2 + (−1)2 — — = √4 + 1 = √5 A(k, m) Slopes of the sides: −4 − 2 −6 —=— = — = −2 Slope of PQ 3−0 3 −5 − (−4) −5 + 4 −1 1 —=— =—=—=— Slope of QR 1−3 −2 −2 2 6 1 − (−5) 1 + 5 —=— = — = — = −2 Slope of RS −2 − 1 −3 −3 −1 1 1−2 —=— =—=— Slope of PS −2 − 0 −2 2 — ≅ SR — and SP — ≅ RQ —, which shows that opposite So, PQ — m SP — = −1, sides are congruent. Also, mPQ — — — — — m RQ — = −1. m PQ m RQ = − 1, m SR m SP = −1, and m SR ⋅ — ⋅ —— ⋅ ⋅ —, PQ — ⊥ RQ , SR ⊥ SP —, and SR — ⊥ RQ — by the Slopes So, PQ ⊥ SP of Perpendicular Lines Theorem (Thm. 3.14). So, by the definition of perpendicular lines, ∠ PSR, ∠ SRQ, ∠ RQP, and ∠ QPS are right angles. So, the quadrilateral is a rectangle. The second friend is correct. B(0, 0) D(k, 0) C(2k, 0) x — This is an isosceles triangle because BA = √ k2 + m2 and — 2 CA = √k + m2 . Draw the line x = k that intersects △ABC in point A(m, k) and the x-axis in the point (k, 0). Call this — and DC — are congruent because BD = k and point D. BD — — DC = k. AD ≅ AD by the Reflexive Property of Congruence — is vertical and BC — is horizontal, (Thm. 2.1). Because AD — ⊥ BC — by the Slopes of Perpendicular Lines Theorem AD (Thm. 3.14). So, ∠ BDA and ∠ CDA are congruent right angles. By the SAS Congruence Theorem (Thm. 5.5), △ABD ≅ △ACD. Because corresponding parts of congruent triangles are congruent, ∠ B ≅ ∠ C. 25. Sample answer: Reflect the triangle in the y-axis and 21. The endpoints of a segment with the origin as the midpoint are (x, y) and (−x, −y) because x + (−x) y + (−y) 0 0 M —, — = M —, — = M(0, 0). 2 2 2 2 ( ) ( ) 22. B (5d, −5d) → (−5d, −5d) → (0, 0) (0, −5d) → (0, −5d) → (5d, 0) (5d, 0) → (−5d, 0) → (0, 5d) — y — 26. Diagonal WU is horizontal, and diagonal TV is vertical. So, (0, v) by the Slopes of Perpendicular Lines Theorem (Thm. 3.14), — ⊥ TV —; Change the coordinates to T(0, m), U(m, 0), WU V(0, −m), and W(−m, 0). These coordinates can be used for any square, and the diagonals are still horizontal and vertical. So, the diagonals are perpendicular for any square. (w, 0) x (−w, 0) translate 5d units right and 5d units up. (0, −v) w v −w + 0 0 + (−v) Midpoint = —, — = −—, −— 2 2 2 2 ( 178 ) ( Geometry Worked-Out Solutions ) Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 27. a. 2. The theorems on page 275 correspond to other triangle y congruence theorems given. HA corresponds to AAS, LL corresponds to SAS, and AL corresponds to ASA or AAS. A(0, 2m) 3. Two sources that could be used to help solve Exercise 20 M(n, m) B(0, 0) C(2n, 0) on page 282 are a map website and a website that calculates distances. x —, the coordinates of M are Because M is the midpoint of AC — M(n, m).— Using the Distance Formula, AM = √n2 + m2 , — 2 2 2 2 BM = √n + m , and CM = √ n + m . So, the midpoint of the hypotenuse of a right triangle is the same distance from each vertex of the triangle. b. y Chapter 5 Review (pp. 290–294) 1. The triangle is an acute isosceles triangle because it has two equal sides and it appears that all angles are acute. 2. Because 46° + 86° = 132°, the measure of the exterior angle is 132°. 3. (9x + 9)° = 45° + 5x° 4x + 9 = 45 R(0, m) 4x = 36 x=9 ⋅ 9x + 9 = 9 9 + 9 S(−m, 0) O(0, 0) = 81 + 9 T(m, 0) x When any two congruent right isosceles triangles are positioned with the vertex opposite the hypotenuse on the origin and their legs on the axes as shown in the diagram, a triangle is formed and the hypotenuses of the original triangles make up two sides of the new triangle. — — SR = m√2 and TR = m√ 2 so these two sides are the same length. So, by definition, △SRT is isosceles. Maintaining Mathematical Proficiency 28. m∠ XYW = m∠ WYZ (3x − 7)° = (2x + 1)° x−7=1 = 90 The measure of the exterior angle is 90°. 4. 8x° + 7x° + 90° = 180° 15x + 90 = 180 15x = 90 x=6 ⋅ ⋅ 8x = 8 6 = 48 7x = 7 6 = 42 The measure of each acute angle is 42° and 48°. 5. (7x + 6)° + (6x − 7)° + 90° = 180° 13x + 89 = 180 x=8 13x = 91 29. m∠ XYZ = (3x − 7)° + (2x + 1)° ⋅ ⋅ x=7 m∠ XYZ = (5x − 6)° 7x + 6 = 7 7 + 6 = 49 + 6 = 55 From Exercise 28, you know that x = 8. 6x − 7 = 6 7 − 7 = 35 m∠ XYZ = (5 8 − 6)° = 34° The measure of each acute angle is 35° and 55°. ⋅ 5.5 –5.8 What Did You Learn? (p. 289) — 1. Given square ABCD with diagonal BD , prove △BAD ≅ △DCB; In this problem, the square could — represent the baseball “diamond,” and then diagonal BD would represent the distance from home plate to second base. So, you could use this problem to prove the equivalent of △HFS ≅ △STH. Then you could just redraw square — this time, so that △CBA is the ABCD with diagonal AC equivalent of △FST. It could easily be shown that the third triangle is congruent to the first two. Copyright © Big Ideas Learning, LLC All rights reserved. 6. corresponding angles: ∠ G ≅ ∠ L, ∠ H ≅ ∠ M, ∠ J ≅ ∠ N, ∠K ≅ ∠P — ≅ LM —, HJ — ≅ MN —, JK — ≅ NP —, corresponding sides: GH — ≅ LP — GK Sample answer: Another congruence statement is KJHG ≅ PNML. Geometry Worked-Out Solutions 179 Chapter 5 17. yes; 7. m∠ S = m∠ T = 74° 90° + 74° + m∠ V = 180° 164° + m∠ V = 180° m∠ V = 16° 8. no; There are two pairs of congruent sides and one pair of congruent angles, but the angles are not the included angles. STATEMENTS REASONS 1. ∠ E ≅ ∠ H, ∠ F ≅ ∠ J, — FG ≅ — JK 1. Given 2. △EFG ≅ △HJK 2. AAS Congruence Theorem (Thm. 5.11) 9. yes; STATEMENTS 1. — WX ≅ — YZ , ∠ WXZ ≅ ∠ YZX 2. — XZ ≅ — XZ 3. △WXZ ≅ △YZX REASONS 1. Given 18. no; There is only enough information to conclude that one pair of angles and one pair of sides are congruent. 19. yes; 2. Reflexive Property of Congruence (Thm. 2.1) 3. SAS Congruence Theorem (Thm. 5.5) — — 10. If QP ≅ QR , then ∠ QRP ≅ ∠ P. — — 11. If ∠ TRV ≅ ∠ TVR, then TV ≅ TR . STATEMENTS REASONS 1. ∠ PLN ≅ ∠ MLN, ∠ PNL ≅ ∠ MNL 1. Given 2. — LN ≅ — LN 2. Reflexive Property of Congruence (Thm. 2.1) 3. △LPN ≅ △LMN 3. ASA Congruence Theorem (Thm. 5.10) — — 12. If RQ ≅ RS , then ∠ RSQ ≅ ∠ RQS. — — 13. If ∠ SRV ≅ ∠ SVR, then SV ≅ SR . 14. 8x° = 180° − 60° △HJK ≅ △LMN. Because corresponding parts of congruent triangles are congruent, ∠ K ≅ ∠ N. x = 15 — — 5y + 1 = 26 22. First, state that QV ≅ QV . Then, use the SSS Congruence 5y = 25 y=5 So, x = 15 and y = 5. 15. no; There is only enough information to conclude that two pairs of sides are congruent. Theorem (Thm. 5.8) to prove that △QSV ≅ △QTV. Because corresponding parts of congruent triangles are congruent, ∠ QSV ≅ ∠ QTV. ∠ QSV ≅ ∠ 1 and ∠ QTV ≅ ∠ 2 by the Vertical Angles Congruence Theorem (Thm. 2.6). So, by the Transitive Property of Congruence (Thm. 2.2), ∠ 1 ≅ ∠ 2. —— — 23. OP = √ (h − 0)2 + (k − 0)2 = √ h 2 + k2 ——— 16. yes; — QP = √(h − h)2 + [ (k + j) − k ]2 = √j 2 = j STATEMENTS — WX ≅ — YZ , ∠ XWZ REASONS 1. Given and ∠ ZYX are right angles. XZ ≅ — XZ 2. — 180 pair of angles and one pair of sides are congruent. 21. By the SAS Congruence Theorem (Thm. 5.5), 8x = 120 1. 20. no; There is only enough information to conclude that one 2. Reflexive Property of Congruence (Thm. 2.1) 3. △WXZ and △YZX are right triangles. 3. Definition of a right triangle 4. △WXZ ≅ △YZX 4. HL Congruence Theorem (Thm. 5.9) Geometry Worked-Out Solutions —— —— —— — —— — QO = √(h − 0)2 + (k + j)2 = √h2 + (k + j)2 QR = √(h − 0)2 + (k − 0)2 = √h 2 + k2 OR = √(0 − 0)2 + (0 − j)2 = √j 2 = j — ≅ QR — and OR — ≅ QP —. Also, by the Reflexive Property So, OP — ≅ QO —. So, you can apply of Congruence (Thm. 2.1), QO the SSS Congruence Theorem (Thm. 5.8) to conclude that △OPQ ≅ △QRO. Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 24. Place the base along the x-axis and the vertex of the legs on the y-axis. — — 3. Given QR ≅ RS , ∠ P ≅ ∠ T y Q S N B(0, k) P C(p, 0) x A(−p, 0) 25. R Prove △SRP ≅ △QRT y (0, k) (2k, k) STATEMENTS REASONS — ≅ RS —, ∠ P ≅ ∠ T 1. QR 1. Given 2. ∠ R ≅ ∠ R 2. Reflexive Property of Congruence (Thm. 2.2) 3. △SRP ≅ △QRT 3. AAS Congruence Theorem (Thm. 5.11) T 4. (4x − 2)° + (3x + 8)° = 90° 7x + 6 = 90 (0, 0) (2k, 0) 7x = 84 x ⋅ ⋅ The fourth vertex of the rectangle has the vertex (2k, k). 4x − 2 = 4 12 − 2 = 48 − 2 = 46 3x + 8 = 3 12 + 8 = 36 + 8 = 44 Chapter 5 Test (p. 295) — — — — 1. Given CA ≅ CB ≅ CD ≅ CE A Prove △ABC ≅ △EDC B STATEMENTS REASONS — ≅ CB — ≅ CD — ≅ CE — 1. CA 1. Given 3. △ABC ≅ △EDC if a triangle is equilateral, then it is also equiangular. E 6. no; The Third Angles Theorem (Thm. 5.4) can be used to prove that two triangles are equiangular, but AAA is not sufficient to prove that the triangles are congruent. You need to know that at least one pair of corresponding sides are congruent. 2. Vertical Angles Congruence Theorem (Thm. 2.6) 7. First, use the HL Congruence Theorem (Thm. 5.9) to prove that △ACD ≅ △BED. Because corresponding parts of — ≅ BD —. Then, use the congruent triangles are congruent, AD Base Angles Theorem (Thm. 5.6) to prove that ∠ 1 ≅ ∠ 2. 3. SAS Congruence Theorem (Thm. 5.5) 8. Use the SSS Congruence Theorem (Thm. 5.8) to prove —— — — MJ KL 2. Given JK ML , J K Prove △MJK ≅ △KLM M STATEMENTS REASONS — ML —, 1. JK — — MJ KL 1. Given MK ≅ — KM 2. — The measures of the acute angles are 44° and 46°. 5. no; By the Corollary to the Base Angles Theorem (Cor. 5.2), C D 2. ∠ ACB ≅ ∠ ECD x = 12 L that △SVX ≅ △SZX. Use the Vertical Angles Congruence Theorem (Thm. 2.6) and the SAS Congruence Theorem (Thm. 5.5) to prove that △VXW ≅ △ZXY. Because corresponding parts of congruent triangles are congruent, ∠ SZX ≅ ∠ SVX and ∠ W ≅ ∠ Y. Use the Segment Addition — ≅ ZW —. Then, use Postulate (Post. 1.2) to show that VY the ASA Congruence Theorem (Thm. 5.10) to prove that △VYT ≅ △ZWR. Because corresponding parts of congruent triangles are congruent, ∠ 1 ≅ ∠ 2. 9. yes; HL Congruence Theorem (Thm. 5.9), ASA Congruence 2. Reflexive Property of Congruence (Thm. 2.1) 3. ∠ JKM ≅ ∠ LMK, ∠ JMK ≅ ∠ LKM 3. Alternate Interior Angles Theorem (Thm. 3.2) 4. △MJK ≅ △KLM 4. ASA Congruence Theorem (Thm. 5.10) Copyright © Big Ideas Learning, LLC All rights reserved. Theorem (Thm. 5.10), AAS Congruence Theorem (Thm. 5.11), SAS Congruence Theorem (Thm. 5.5) Geometry Worked-Out Solutions 181 Chapter 5 — — 10. Use the Distance Formula to find the lengths of PQ and ST . —— — — — 5. a. In order to prove △ABC ≅ △DEF, show that AB ≅ DE . —— — — —— — — PQ = √(21 − 3)2 + (30 − 30)2 = √ 182 = 18 AB = √(2 − 2)2 + (8 − 5)2 = √32 = √9 = 3 ST = √(21 − 3)2 + (0 − 0)2 = √ 182 = 18 — and DE — have the same measure and The segments AB are therefore congruent. Also, from the markings in — ≅ EF —. So, by the SAS the diagram, ∠ B ≅ ∠ E and BC Congruence Theorem (Thm. 5.5), △ABC ≅ △DEF. —— — — ≅ ST —. Also, the horizontal segments PQ — and ST — each So, PQ have a slope of 0, which implies that they are parallel. So, — intersects PQ — and ST — to form congruent alternate interior PS angles, ∠ P and ∠ S. By the Vertical Angles Congruence Theorem (Thm. 2.6), ∠ PRQ ≅ ∠ SRT. So, by the AAS Congruence Theorem (Thm. 5.11), △PQR ≅ △STR. 11. a. The triangle shown is an isosceles triangle because it has two congruent sides. b. By the Triangle Sum Theorem (Thm. 5.1): m∠ 1 + m∠ 2 + m∠ 3 = 180° m∠ 1 + m∠ 2 + 40° = 180° m∠ 1 + m∠ 2 = 140° By the Base Angles Theorem (Thm. 5.6) ∠ 1 ≅ ∠ 2: m∠ 2 + m∠ 2 = 140° 2m∠ 2 = 140° m∠ 2 = 70° m∠ 1 = 70° If m∠ 3 = 40°, then m∠ 1 = 70° and m∠ 2 = 70°. Chapter 5 Standards Assessment (pp. 296–297) 1. no; the Exterior Angle Theorem (Thm. 5.2) follows from the Triangle Sum Theorem (Thm. 5.1). Also, the Triangle Sum Theorem (Thm. 5.1) is used to prove the Exterior Angle Theorem (Thm. 5.2), so you cannot use the Exterior Angle Theorem (Thm. 5.2) to prove the Triangle Sum Theorem (Thm. 5.1). 2. By step 1, a line through point P intersects line m in point — ≅ QB — ≅ PC — ≅ PD — and AB — ≅ CD — Q. By steps 2 and 3, QA because congruent segments were drawn with the same — and CD — compass setting. So, in step 4, you see that if AB were drawn, then △AQB and △CPD would be congruent by the SSS Congruence Theorem (Thm. 5.8). Because corresponding parts of congruent triangles are congruent, ⃖⃗ m by the ∠ CBD ≅ ∠ AQB, which means that PD Corresponding Angles Converse (Thm. 3.5). 3. a. Sample answer: Reflect △JKL in the x-axis and then translate 4 units right. — —— —— — b. yes; corresponding sides: JK ≅ XY , KL ≅ YZ , JL ≅ XZ ; corresponding angles: ∠ J ≅ ∠ X, ∠ K ≅ ∠ Y, ∠ L ≅ ∠ Z 4. C; DE = √(8 − 5)2 + (2 − 2)2 = √32 = √9 = 3 b. Rotate △ABC 90° counterclockwise about the origin followed by a translation 13 units right. 6. yes; The coordinate rule for dilations is to multiply each coordinate of each point by the scale factor, which is 2 in this case. So, when you do this to the coordinates of point W, you get (2 0, 2 0) which is still (0, 0). ⋅ ⋅ 7. A, B, D Figure A has 180° rotational symmetry. Figure B has 72° rotational symmetry. Figure D has 90° and 180° rotational symmetry. 8. To prove quadrilateral ABCD is a rectangle, opposite sides must be equal and right angles must be formed at the vertices. Side lengths: —— — AB = √ (4 − 2)2 + (7 − 5)2 = √ 22 + 22 — — — = √ 4 + 4 = √8 = 2√ 2 —— — BC = √ (4 − 7)2 + (7 − 4)2 = √ (−3)2 + 32 — — — = √ 9 + 9 = √18 = 3√ 2 —— — CD = √ (7 − 5)2 + (4 − 2)2 = √ 22 + 22 — — — = √ 4 + 4 = √8 = 2√ 2 —— —— AD = √ (5 − 2)2 + (2 − 5)2 = √ (3)2 + (−3)2 — — — = √ 9 + 9 = √18 = 3√ 2 Slopes: 7−5 2 —=— =—=1 Slope of AB 4−2 2 3 7−4 —=— = — = −1 Slope of BC 4 − 7 −3 4−2 2 —=— =—=1 Slope of CD 7−5 2 2 − 5 −3 —=— Slope of AD = — = −1 5−2 3 — and BC — have the same measure and the same Sides AD — and DC —. So, by the Slopes of Parallel Lines slope, as do AB — BC — and AB — DC —. Because the Theorem (Thm. 3.13), AD — — — —, BC — ⊥ AB —, product of their slopes is −1, AD ⊥ AB , AD ⊥ DC — — and BC ⊥ DC . So, ABCD is a rectangle. 1 5 5−0 Slope = — = — = −— 2 −2 − 8 −10 2 x = — (10) + (−2) = 4 − 2 = 2 5 2 y = — (−5) + 5 = −2 + 5 = 3 5 ⋅ ⋅ So, Q(2, 3). 182 Geometry Worked-Out Solutions Copyright © Big Ideas Learning, LLC All rights reserved. Chapter 5 9. Prove △ABC is an equilateral triangle. STATEMENTS REASONS 1. AB = AC, BA = BC 1. By construction 2. AC = BC 2. Transitive Property of Congruence (Thm. 2.1) 3. — AB ≅ — AC , — BA ≅ — BC , — AC ≅ — BC 4. △ABC is an equilateral triangle. 3. Definition of congruent segments 4. Definition of equilateral triangle Copyright © Big Ideas Learning, LLC All rights reserved. Geometry Worked-Out Solutions 183