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1 CHAPTER 10 Chapter Opener Chapter Readiness Quiz (p. 536) 1. B; G; B; 5 2 z z 2 81 z 2 56 z 56 Lesson Checkpoint (p. 538) ; This is reasonable because 27 is between the perfect squares 25 and 36. So, 27 should be between 25 and 36, or 5 and 6. The answer 5.2 is between 5 and ; This is reasonable because 46 is between the perfect squares 36 and 49. So, 46 should be between 36 and 49, or 6 and 7. The answer 6.8 is between 6 and ; This is reasonable because 8 is between the perfect squares 4 and 9. So, 8 should be between 4 and 9, or 2 and 3. The answer 2.8 is between 2 and ; This is reasonable because 97 is between the perfect squares 81 and 100. So, 97 should be between 81 and 100, or 9 and 10. The answer 9.8 is between 9 and Guided Practice (p. 539) 1. The radicand in the epression 25 is D; A; B; C; a 2 b 2 c c c 2 20 c 2 20 c 4 5 c 4 5 c 25 c 7. c Practice and Applications (pp ) ; This is reasonable because 13 is between the perfect squares 9 and 16. So, 13 should be between 9 and 16, or 3 and 4. The answer 3.6 is between 3 and ; This is reasonable because 6 is between the perfect squares 4 and 9. So, 6 should be between 4 and 9, or 2 and 3. The answer 2.4 is between 2 and ; This is reasonable because 91 is between the perfect squares 81 and 100. So, 91 should be between 81 and 100, or 9 and 10. The answer 9.5 is between 9 and ; This is reasonable because 34 is between the perfect squares 25 and 36. So, 34 should be between 25 and 36, or 5 and 6. The answer 5.8 is between 5 and ; This is reasonable because 106 is between the perfect squares 100 and 121. So, 106 should be between 100 and 121, or 10 and 11. The answer 10.3 is between 10 and 11. Geometry, Concepts and Skills 163

2 ; This is reasonable because 148 is between the perfect squares 144 and 169. So, 148 should be between 144 and 169, or 12 and 13. The answer 12.2 is between 12 and ; This is reasonable because 62 is between the perfect squares 49 and 64. So, 62 should be between 49 and 64, or 7 and 8. The answer 7.9 is between 7 and ; This is reasonable because 186 is between the perfect squares 169 and 196. So, 186 should be between 169 and 196, or 13 and 14. The answer 13.6 is between 13 and a 2 b 2 c c c 2 7 c 2 7 c 21. a 2 b 2 c c c 2 49 c 2 49 c 7 c 22. a 2 b 2 c c c c c Yes, the epression can be simplified further No, the epression cannot be simplified further. 46. A lw 47. A lw A lw 49. A lw Geometry, Concepts and Skills

3 50. A lw 51. A lw A 1 4 s (30) (900) ft (1 m) 2 (100 cm) 2 10,000 cm; A 1 4 s2 3 10, s2 3 40,000 s ,094 s 2 23,094 s 152 cm s 10.1 Standardized Test Practice (p. 541) 54. C; H; , D; a 2 b 2 c c c c c 4 26 c 4 26 c 226 c 10.2 c 10.1 Mied Review (p. 541) ma1 90 ma ma ma1 68ma ma ma1 68ma Algebra Skills (p. 541) 63. ( 5) (2 1) (3 4) ( 2) (1 ) ( 6) Lesson Geo-Activity (p. 542) 2. 45, 45, Answers may vary. 4. Answers may vary, but will be times the answer in Step The answer should be the same as in Step Checkpoint (pp ) The triangle is an isosceles right triangle. By the Base Angles Theorem, its acute angles are congruent. By the Triangle Sum Theorem, So, 290, and 45. So, the triangle is a triangle. By the Triangle Theorem, The triangle is an isosceles right triangle. By the Converse of the Base Angles Theorem its sides opposite the congruent angles are congruent. By the Triangle Sum Theorem, So, 290, and 45. So, the triangle is a triangle. By the Triangle Theorem, Geometry, Concepts and Skills 165

4 10.2 Guided Practice (p. 545) 1. An isosceles right triangle has 2 congruent sides. 2. An isosceles right triangle has 2 congruent angles. The measures of the angles are 45, 45, and Practice and Applications (pp ) (1.4) cm 19. No; you cannot determine the measures of the other two angles. 20. Yes; the triangle has congruent acute angles, so 290 and 45. So, the triangle is a triangle. 21. No; the triangle is isosceles, but there is no information about the angle measures. 22. The triangle is an isosceles right triangle. By the Base Angles Theorem, its acute angles are congruent. By the Triangle Sum Theorem, So, 290, and 45. Since the measure of each acute angle is 45, the triangle is a triangle. By the Triangle Theorem, The right triangle has congruent acute angles. By the Triangle Sum Theorem, So, 290, and 45. Since the measure of each acute angle is 45, the triangle is a triangle. By the Triangle Theorem, By the Triangle Sum Theorem, So, 90180, and 90. Therefore the triangle is a triangle. By the Triangle Theorem, The right triangle has congruent acute angles. By the Triangle Sum Theorem, So, 290, and 45. Since the measure of each acute angle is 45, the triangle is a triangle. By the Triangle Theorem, 2 8 y2 8 y y. 26. The triangle is an isosceles right triangle. By the Base Angles Theorem, its acute angles are congruent. By the Triangle Sum Theorem, So, 290, and 45. Since the measure of each acute angle is 45, the triangle is a triangle. By the Triangle Theorem, Geometry, Concepts and Skills

5 27. By the Triangle Sum Theorem, So, , and 45. Therefore the triangle is a triangle. By the Triangle Theorem, T ADB, T ACD, and T BCD; since D is on the perpendicular bisecr of **** AB, **** AD DB **** by the Perpendicular Bisecr Theorem. mdb 90, so T ADB is an isosceles right triangle. By the Base Angles Theorem, ab. By the Triangle Sum Theorem, Since m mab 45, T ADB is a triangle. mcd 90, so by the Triangle Sum Theorem, mdc 180mCD m So, T ACD is a triangle. mabcd 90, so by the Triangle Sum Theorem, mabdc 180maBCD mab So, T BCD is a triangle. 29. Lengths may vary, but AC, CB, and CD are all equal. Since T ACD and T BCD are triangles, they are isosceles triangles by the Converse of the Base Angles Theorem. Therefore, AC CD and CD BC. 30. Lengths may vary, but AD DB and each is equal times the length in Eercise If the has length 52, the s should have length 5 2, or 5. If the s have length 5, then the 2 has length 5 2, or r r 2; s s 3; t t 4 2; u u 5; v v 6; w w The right triangle with s of length 1 and of length r 2 is a triangle Standardized Test Practice (p. 547) 34. a ; m 3(3 15)45; mab 3(3 15)45; mac 6(6 15)90 b. b 12; c 122 c. Check using the Pythagorean Theorem Mied Review (p. 547) 35. a 2 b 2 c a 2 b 2 c The triangle is right. The triangle is acute. 37. a 2 b 2 c The triangle is right Algebra Skills (p. 547) Geometry, Concepts and Skills 167

6 Lesson Activity (p. 548) AC AC AD CD A D 4. Lengths will vary, but AC 2 AD and CD 3 AD 1.73 AD. AC 5. The ratio is 2:1 and A D the ratio C D is approimately 1.73:1. AD 10.3 Checkpoint (pp ) 1. 2 shorter longer shorter longer shorter longer shorter shorter ; longer shorter 3 y Guided Practice (p. 552) 1. Two special right triangles are triangle and triangle. 2. true 3. false 4. false 5. true 6. true 7. true 8. 2 shorter ; longer shorter 3 y 53 CD A D shorter 8 2 h 4 h shorter 4 2 a 2 a; longer shorter 3 b Practice and Applications (pp ) shorter shorter longer shorter shorter 3 11 shorter 2 shorter shorter shorter longer shorter shorter 3 12 shorter 2 shorter longer shorter longer shorter longer shorter longer shorter Geometry, Concepts and Skills

7 21. longer shorter longer shorter longer shorter longer shorter longer shorter longer shorter longer shorter longer shorter shorter 60 2 h 30 ft h shorter ; longer shorter 3 y shorter ; longer shorter 3 y shorter ; longer shorter 3 y shorter ; 2 longer shorter 3 y shorter ; longer shorter 3 y shorter ; 2 longer shorter 3 y shorter 18 2 h 9 in. h Your shoulders should be lifted a height of 9 inches. 37. Divide the triangle in two triangles. The length of the shorter of each triangle is 15 feet. The length of the longer of each triangle is 153 feet, by the Triangle Theorem. Area 1 2 bh ft Divide the triangle in two triangles. The length of the shorter of each triangle is 9 inches. The length of the longer of each triangle is 93 inches, by the Triangle Theorem. Area 1 2 bh in.2 2 Geometry, Concepts and Skills 169

8 39. Divide the triangle in two triangles. The length of the shorter of each triangle is 3.5 centimeters. The length of the longer of each triangle is 3.53 centimeters, by the Triangle Theorem. Area 1 2 bh cm The area of the triangle is given by Area 1 2 bh. If an equilateral triangle with sides of length s is divided in two triangles, then the length of the s base is s and the height is 3. Then 2 Area 1 2 bh 1 2 s s s shorter cm 30 cm; longer shorter cm 42. Divide one of the si equilateral triangles in two triangles. The length of the shorter of each triangle is 1 cm. The length of the longer of 2 each triangle is longer cm 10.3 Standardized Test Practice (pp ) 43. C 44. F; 2 shorter 12 2 shorter 6 shorter longer shorter 3 63 perimeter cm 10.3 Mied Review (p. 555) wins l osses l osses wins wins numbe r of games losses numbe r of games Yes; by the AA Similarity Postulate, T ABC T FHG. 50. Yes; by the SAS Similarity Theorem, T PQR T STU Algebra Skills (p. 555) (4) (4) (4) ( 3)( 3) (4 3)(4 3) (1)(7) (4) 2 (4) (4) 2 2(4) Quiz 1 (p. 555) longer shorter shorter ; longer shorter 3 Lesson 10.4 y Activity (p. 556) 1. Triangle longer shorter 2. Each ratio of the lengths is s horter longer A 5 cm 4.2 cm 0.84 B 10 cm 8.4 cm 0.84 C 15 cm 12.6 cm 0.84 D 20 cm 16.8 cm Answers will vary, but the ratios in the fourth column will always be equal; the ratio of lengths depends on the measures of its angles, not the size of the right triangle Checkpoint (pp ) opposite as 1. tan S 6 le g adjacent as ; 4 opposite ar tan R 8 le g adjacent ar Geometry, Concepts and Skills

9 opposite as 2. tan S 1 0 le g adjacent as ; 12 opposite ar tan R 2 4 le g adjacent ar tan tan tan tan 44 8 and tan tan 37 4 and tan tan 59 5 and tan tan 34 o pposite 10. tan 55 o pposite adjacent adjacent tan 34 7 tan tan 347 tan tan 34 tan tan 60 o pposite adjacent tan tan Guided Practice (p. 560) 1. The acute angles in T DEF are ad and ae. 2. The opposite ad is EF ****, and the adjacent ad is DF ****. 3. tan A o pposi adjace te nt tan A o pposi adjace te nt tan A o pposite adjacent tan tan tan tan Practice and Applications (pp ) 10. tan A o pposite adjacent tan A o pposi adjace te nt tan A o pposite 2 adjacent tan P o pposite ; adjacent 2 4 tan R o pposi adjace te nt tan P o pposi adjace te nt ; 5 tan R o pposi adjace te nt tan P o pposi adjace te nt ; 3 tan R o pposi adjace te nt tan tan tan tan tan tan tan tan The measure of the other acute angle is tan 56 9 tan tan , tan tan The measure of the other acute angle is tan tan tan , tan tan The measure of the other acute angle is tan tan tan , tan tan Geometry, Concepts and Skills 171

10 27. tan 41 o pposite 28. tan 29 o pposite adjacent adjacent tan tan 29 tan 4152 tan tan 41 tan tan 53 o pposite adjacent tan tan tan The tangent ratio can only be used for any acute angle of a right triangle. This triangle is not a right triangle. 31. tan 13 o pposite adjacent h tan (tan 13) h 13.4 h The height h of the slide is about 13.4 meters. 32. tan 70 o pposite adjacent tan tan tan 49 o pposite adjacent tan tan tan tan 34 o pposite adjacent tan (tan 34) tan 40 o pposite adjacent tan (tan 40) tan 35 o pposite adjacent tan (tan 35) tan 50 o pposite adjacent tan 50 oppo site 10 10(tan 50) opposite 11.9 opposite tan 45 o pposite adjacent tan (tan 45) ta n tan 42 o pposite adjacent d tan (tan 42) d 36 d The distance d is about 36 meters Standardized Test Practice (p. 562) 39. C; tan tan tan F; tan 50 6 y 6 tan 50y 7.2 y 10.4 Mied Review (p. 562) 41. V πr 2 h 42. V lwh π(6) 2 (9) π ft m V 1 3 πr2 h 1 3 π (7) in Geometry, Concepts and Skills

11 10.4 Algebra Skills (p. 562) ( 3) Lesson Checkpoint (pp ) 1. sin A opposite cos A a djacent sin A oppos ite ; cos A a djacent sin A opposite ; cos A a djacent sin A oppos ite ; 1 cos A a djacent sin A oppos ite ; cos A a djace hypot nt e sin A oppos ite ; 8 cos A a djace hypot nt e sin cos sin cos cos sin cos sin sin A opposite sin 34 a 7 7(sin 34) a 3.9 a; cos A a djacent cos 34 b 7 7(cos 34) b 5.8 b 16. sin A opposite a sin (sin 65) a 10.9 a; cos A a djacent b cos (cos 65) b 5.1 b 17. sin 43 opposite sin 43 a 5 5(sin 43) a 3.4 a; cos 43 a djacent cos 43 b 5 5(cos 43) b 3.7 b 10.5 Guided Practice (p. 566) 1. B; cos D a djacen t ad DE D F 2. C; sin D o pposite ad EF D F 3. A; tan D o pposite EF adjacent D E 4. The value of sin 37 is constant, so sin D sin A. The sine of an acute angle of a right triangle depends on the measure of the angle, not on the size of the triangle. 5. sin A oppos ite cos A a djacent a A Geometry, Concepts and Skills 173

12 7. tan A o pposi adjace te nt sin B oppos ite cos B a djace hypot nt ab e tan B o pposi adjace te nt Practice and Applications (pp ) 11. sin A oppos ite ; cos A a djace hypot nt e sin A oppos ite ; 0 2 cos A a djacent sin A oppos ite ; cos A a djace hypot nt e sin P oppos ite ap ; 7 cos P a djace hypot nt ap e sin P oppos ite ap ; 1 cos P a djace hypot nt ap e sin P oppos ite ap ; cos P a djacent ap sin cos sin cos sin cos sin cos sin 46 o pposite sin (sin 46) 5.8 ; cos 46 a djacent y cos sin 54 o pposite sin 54 1 y4 14(sin 54) y 11.3 y; cos 54 a djacent cos (cos 54) sin 29 o pposite sin (sin 29) 5.3 ; cos 29 a djacent cos 29 1 y1 11(cos 29) y 9.6 y 28. sin 24 o pposite sin 24 1 y6 16(sin 24) y 6.5 y; cos 24 a djacent cos (cos 24) sin 37 o pposite sin (sin 37) 9.0 ; cos 37 a djacent cos 37 1 y5 15(cos 37) y 12.0 y 8(cos 46) y 5.6 y 174 Geometry, Concepts and Skills

13 30. sin 68 o pposite sin (sin 68) 24.1 ; cos 68 a djacent cos 68 2 y6 26(cos 68) y 9.7 y 31. sin 70 o pposite 15 h sin 70 1 h (sin 70) h 14 h The ladder reaches about 14 feet up the wall ft 1 2 in. 96 in. 1 ft sin 22 o pposite y sin (sin 22) y 36 in. y; cos 22 a djacent cos (cos 22) 89 in. 33. Answers may vary. 34. (sin A) 2 (cos A) When you drag point C, (sin A) 2 (cos A) sin 42 3 r4 34(sin 42) r 22.8 r, cos 48 3 r4 34(cos 48) r 22.8 r Both students get the correct answer Standardized Test Practice (p. 568) 38. C; sin 25 8 (sin 25) 8 8 sin H; the sine of an angle cannot be greater than Mied Review (p. 568) 40. longer shorter shorter y ; longer shorter shorter ; longer shorter 3 y tan tan tan tan tan tan Sample answer: A c b C a B 10.5 Algebra Skills (p. 568) , 8, 1.8, 0.8, 0, 0.08, , 2164, 2416, 2461, 2614, , 0.6, 0.56, 0.5, 0.47 In T ABC shown, sin A a c, cos A b c, and tan A a b. sin A Then c os A a c b c a c c ac b a tan A. bc b Lesson Checkpoint (pp ) 1. m tan m tan m tan Geometry, Concepts and Skills 175

14 4. Since tan A , 9 m tan Since tan A , 20 m tan Since tan A , 10 m tan m sin m cos m sin m cos m sin m cos ; tan A m tan ; mab 90m y y 2 36 y 2 20 y ; cos E mae cos ; mad 90maE z z z 2 24 z ; sin H mah sin ; mag 90maH Guided Practice (p. 572) 1. Solving a right triangle means finding the measures of both acute angles and the lengths of all three sides. 2. true 3. false m tan m sin m cos max ; 2 shortest ; longest shortest 3 y d d d 2 52 d ; cos D mad cos ; mae 90maD Practice and Applications (pp ) 12. m tan m tan m tan m tan m tan m tan Geometry, Concepts and Skills

15 18. Use the Pythagorean Theorem; (QS) (QS) QS 73 QS 19. Sample answer: Use the inverse tangent function; tan Q maq tan Sample answer: Use the fact that aq and as are complements; mas 90maQ ; 6.3 ; tan A tan m tan ; tan A m tan m sin m cos m sin m cos m cos m sin (2.5) 2 (horizontal distance) A (horizontal distance) (horizontal distance) horizontal distance m tan horizontal distance 19.8 ft sin(ramp angle) sin , so the measure of the ramp angle is about 7.2. This ramp meets the standards. 31. Sample answer: 28 ft (Any ramp greater than or equal 27.6 ft will meet the standards) 32. Answers may vary. altitude 33. tan(glide angle) distan ce runway tan(glide angle) tan(glide angle) tan , so the measure of the glide angle is about sin A m sin 1 (0.5) cos A m cos cos A m cos (LM) (LM) (LM) 2 33 LM ; cosak mak cos ; mal 90maK (NQ) (NQ) (NQ) NQ ; 4 sin N man sin ; map 90maN Geometry, Concepts and Skills 177

16 39. (ST) (ST) (ST) ST ; 6 sin S mas sin ; mar 90maS tan 1 BC ; the diagram gives the lengths AB and BC, A B but does not give the length AC Standardized Test Practice (p. 574) 41. B 42. G 10.6 Mied Review (p. 575) 43. C 2πr 2π(8) 16π 50 cm; A πr 2 π(8) 2 64π 201 cm C 2πr 2π(15) 30π 94 in.; A πr 2 π(15) 2 225π 707 in C πd π(34) 107 yd; A πr 2 π(17) 2 289π 908 yd V 4 3 πr3 47. V 4 3 πr3 4 3 π(10)3 4 3 π(14) ft 3 11,494 cm πr3 48. V π(5) π in Algebra Skills (p. 575) $8.42 $2.95 $ $ $1.80 Quiz 2 (p. 575) 1. tan 42 6 tan tan tan (tan 63) 3.9 y 3. sin sin (sin 40) y 19(sin 21) 7.7 y; 6.8 ; tan tan 21 y tan y (tan 21) y tan 40 tan y tan (tan 29) sin (sin 35) 4.6 ; tan y y tan y tan 35 y tan sin cos man ; m tan tan 40m 13.4 m; 16 2 (13.4) 2 q q q 20.9 q 178 Geometry, Concepts and Skills

17 11. q q q 2 15 q ; cos Q maq cos ; map 90maQ k k k k ; 3 sin L mal sin ; mak 90maL Chapter 10 Summary and Review (pp ) 1. A radical is an epression written with a radical symbol. 2. The number or epression inside the radical symbol is the radicand. 3. A trigonometric ratio is a ratio of the lengths of two sides of a right triangle. 4. A right triangle with side lengths 9, 93, and 18 is a triangle. 5. To solve a right triangle means determine that lengths of all three sides of the triangle and the measures of both acute angles. 6. A right triangle with side lengths 4, 4, and 42 is a triangle. 7. If af is an acute angle of a right triangle, then opposite af tangent of af. le g adjacent af 8. If af is an acute angle of a right triangle, then cosine of af a djacent af. 9. If af is an acute angle of a right triangle, then sine of af opposite af A lw A lw A lw shorter ; longer shorter 3 y shorter ; longer shorter 3 y Geometry, Concepts and Skills 179

18 37. longer shorter ; 2 shorter y 2 94 y longer shorter ; 2 shorter y y opposite 39. tan A 1 6 le g adjacent ; 15 opposite ab tan B 3 0 le g adjacent ab opposite tan A ; le g adjacent opposite ab tan B le g adjacent ab 20 opposite 41. tan A ; le g adjacent 8 2 opposite ab tan B le g adjacent ab tan tan tan tan sin A oppos ite ; 5 cos A a djace hypot nt e sin A oppos ite ; cos A a djace hypot nt e sin A oppos ite ; 0 2 cos A a djace hypot nt e sin sin cos cos sin (sin 31) 5.2 ; cos 31 1 y0 10(cos 31) y 8.6 y 54. m tan m sin m cos m tan c 2 45 c 2 45 c 6.7 c; tan A m tan ; mab 90m q q q 26.9 q; tan R mar tan ; map 90maR maz 90maX ; sin sin 55 3(0.8192) 2.5 ; cos 55 3 z 3 cos 55 z 3(0.5736) z 1.7 z 180 Geometry, Concepts and Skills

19 Chapter 10 Test (p. 580) D 5. A 6. B 7. C tan y tan (tan 38) 8(tan 70) y 2.3 ; 22.0 y; cos 38 3 y cos 70 8 y cos 383 cos y cos 38 cos 70 y y 11. sin sin (sin 35) y 15(sin 50) 5.7 y; 11.5 ; cos cos 50 y 15 10(cos 35) 15 cos 50y y 13. tan cos tan sin cos tan sin tan m tan m tan m sin m sin m cos m cos mak ; 2 shortest 9 2 KL 9 KL KL; longest shortest 3 JL JL maf ; 12 tan 25 D E DE(tan 25) DE tan 25 DE 25.7; 12 sin 25 D F DF(sin 25) 12, 12 DF sin 25 DF QR QR 2 36 QR 2 20 QR 20 QR 4.5; cos P map cos ; mar 90maP tan X max tan Chapter 10 Standardized Test (p. 581) 1. A; J; B; tan J o pposite 15 adjacent J; sin C; longer shorter 3 PS 483 Geometry, Concepts and Skills 181

20 6 6. G; sin A m sin B; sin 32 P R PR sin PR sin J; tan 72 3 h0 30(tan72) h 92.3 h 9. D; tan (tan 67) 18.8, cos 67 8 y y cos y cos 67 y 20.5 Chapter 10 Algebra Review (p. 583) 1. y 1 4 5y 7 Solve for y in Equation 1. y 1 y 1 Substitute for y in Equation y 7 4 5(1 ) Substitute for in the revised Equation 1. y 1 1 (2) Check that (2, 3) is a solution. y 1 (2) y 7 4(2) 5(3) The solution is (2, 3) 182 Geometry, Concepts and Skills 2. 2y 9 3 y 1 Solve for y in Equation 2. 3 y y Substitute for y in Equation 1. 2(3 1) Substitute for in revised Equation 2. y 3 1 3(1) 1 4 Check that (1, 4) is a solution. 1 2(4) (1) The solution is (1, 4) y 3 7 2y 1 Solve for y in Equation 1. 3 y 3 y 3 3 Substitute for y in Equation 2 7 2(3 3) Substitute for in revised Equation 1. y (5) Check that (5, 18) is a solution. 3(5) (5) 2(18) The solution is (5, 18).

21 4. y 4 y 16 Solve for y in Equation 2. y 16 y 16 Substitute for y in Equation 1. (16 ) Substitute for in revised Equation 2. y Check that (6, 10) is a solution. y y The solution is (6, 10). 5. y 1 2 y 4 Solve for y in Equation 1. y 1 y 1 Substitute for y in Equation 2. 2 y 4 2 (1 ) Substitute for in revised Equation 1. y Check that (1, 2) is a solution. y y 4 2(1) The solution is (1, 2) y 2 4 3y 6 Solve for y in Equation 1. 6 y y Substitute for y in Equation y 6 4 3(6 2) Substitute for in revised Equation 1. y 6 2 6(0) 2 2 Check that (0, 2) is a solution. 6 y 2 6(0) (2) y 6 4(0) 3(2) The solution is (0, 2) y 5 4y 3 Solve for in Equation 2. 4y 3 3 4y Substitute for in Equation y 5 2(3 4y) 3y 5 6 8y 3y 5 11y 11 y 1 Substitute for y in revised Equation y 3 4(1) 1 Check that (1, 1) is a solution. 2 3y 5 2(1) 3(1) y 3 (1) 4(1) The solution is (1, 1). Geometry, Concepts and Skills 183

22 8. 3 2y 5 3y 9 Solve for in Equation 2. 3y 9 3y 9 Substitute for in Equation 1. 3(3y 9) 2y 5 9y 27 2y 5 11y 22 y 2 Substitute for y in revised Equation 2. 3y 9 3(2) Check that (3, 2) is a solution. 3 2y 5 3(3) 2(2) y 9 (3) 3(2) The solution is (3, 2) y 7 2 4y 22 Solve for in Equation y y 11 2y Substitute for in Equation y 7 5(11 2y) 2y y 2y 7 12y 48 y 4 Substitute for y in revised Equation y 11 2(4) Check that (3, 4) is a solution. 5 2y 7 5(3) 2(4) y 22 2(3) 4(4) The solution is (3, 4). 184 Geometry, Concepts and Skills

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untitled 數 Deductive Geometry 聯 Contents 錄 : Deductive Geometry Summary of Geometry Proving Skills Method for solving Geometry Questions Construction of Geometry figures Deductive Geometry 1 (A) Summary of Geometry