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இலக்கு : 100% அன்பார்ந்த சென்னை மாவட்ட , 10ம் வகுப்பு பயிலும் மாணவச் செல்வங்களே ! PowerPoint Presentation
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இலக்கு : 100% அன்பார்ந்த சென்னை மாவட்ட , 10ம் வகுப்பு பயிலும் மாணவச் செல்வங்களே !

இலக்கு : 100% அன்பார்ந்த சென்னை மாவட்ட , 10ம் வகுப்பு பயிலும் மாணவச் செல்வங்களே !

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இலக்கு : 100% அன்பார்ந்த சென்னை மாவட்ட , 10ம் வகுப்பு பயிலும் மாணவச் செல்வங்களே !

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  1. இலக்கு: 100% அன்பார்ந்தசென்னைமாவட்ட, 10ம் வகுப்புபயிலும்மாணவச்செல்வங்களே! அரசுபொதுத்தேர்வுஎழுதத்தயாராகிக்கொண்டிருக்கும்இவ்வேளையில்மாணவர்கள்சிறந்தமதிப்பெண்கள்பெறவேண்டும்என்றநோக்கத்தில்கணிதப்பாடத்திற்கானஇந்த PowerPoint file (ppt) தயாரிக்கப்பட்டுள்ளது. மாணவர்கள்இதனைமுழுமையாகபயன்படுத்திசிறந்தமதிப்பெண்கள்பெறவாழ்த்துகின்றேன். இப்படிக்கு, த. இராஜேந்திரன்.M.A.,B.Ed., M.Phil., முதன்மைக்கல்விஅலுவலர், சென்னை.

  2. Practical geometry and 5 marks questions with answers PREPARED BY: R.RAJENDRAN. M.A., M.Sc., M.Ed., Headmaster (IC) K.C.SANKARALINGA NADAR. HR. SEC. SCHOOL, CHENNAI-21.

  3. Draw two tangents from a point which is 10cm away from the centre of a circle of radius 6cm. Also measure the lengths of the tangents. Length of the tangent segment A 8 6 10 M O P 8 B

  4. Draw a circle of diameter 8cm. Draw two tangents to the circle from a point P at a distance of 9cm from the center. CONSTRUCTION Length of the tangent segment A 4 8.1 9 P O M B

  5. Draw a circle of diameter 8cm. Draw two tangents to the circle from a point P at a distance of 9cm from the center. CONSTRUCTION Length of the tangent segment A 4 8.1 9 P O M B

  6. Construct the Cyclic Quadrilateral ABCD with AB = 8cm, BC = 7cm, AC = 6cm, AD = 4cm. Rough figure C D C 6 7 4 D 8 A B 6 7 4 8 A B

  7. Construct a cyclic quadrilateral ABCD with AB = 7cm, BC = 5cm, AC = 6cm, BD = 6.5 Rough figure D C 5 6 6.5 C D 7 A B 6 6.5 5 B 7 A

  8. Construct a cyclic quadrilateral ABCD with AB = 7.2cm, mABD = 45°, mBAD = 100° and BC = 4cm. D Rough figure D C 4 100° 45° A B 7.2 C 100° 45° A B

  9. Draw the graph of x2 – x – 8 = 0 and hence find the roots of x2 – 2x – 15 = 0. In x axis 1cm = 1 unit y axis 1cm = 2 units. x2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = –3, 5 x -4 -3 -2 -1 0 1 2 3 4 5 x2 16 9 4 1 0 1 4 9 16 25 -x 4 3 2 1 0 -1 -2 - 3 -4 -5 -8 -8 –8 –8 –8 –8 –8 –8 –8 –8 –8 y 12 4 -2 -6 -8 -8 -6 -2 4 12

  10. Solution set = {– 3, 5}

  11. A cyclist travels from a place A to a place B along the same route at a uniform speed on different days. The following table gives the speed of his travel and the corresponding time he took to cover the distance.

  12. Draw the speed-time graph and use it to find (i) the number of hours he will take if he travels at a speed of 5 km / hr (ii) the speed with which he should travel if he has to cover the distance in 40 hrs.

  13. If x = 5, y = 24 Time in hr If y = 40, y = 2.4 Speed in km/hr

  14. 1. Prove That A – (B  C) = (A – B)  (A – C) (1)BC (2) A-(BC) (3) A – B A B B B A A C C C (4) A – C (5) (A – B)  (A – C) B A B A C C From the diagrams (2) and (5) A – (B  C) = (A – B)  (A – C)

  15. 2. Prove that A – (B  C) = (A – B)  (A – C) using Venn diagram (2) A – ( BC) (3) A – B (1) BC A B A B B A C C C (4) A – C (5) (A – B)(A – C) A B B A C C From the diagram (2)and(5) A – (BC) = (A – B)(A – C)

  16. 3. Prove that A  (B  C) = (A  B)  (A  C) using Venn diagram. B A A B A B C C C A  (B  C) A B B C A B A B From the figures 2 and 5 A(B  C)=(AB)(AC) C C A C (A B)(A  C)

  17. 4. Prove that A  (B  C) = (A  B)  (A  C) using Venn diagram. B A A B A B C C C A  (B  C) A B B C A B A B From the figures 2 and 5 A(BC)=(AB)(AC) C C A C (AB)  (AC)

  18. 5. Prove that (A  B)’ = A’  B’ using Venn diagram. U U U A B B A B A 2. (AB)’ 3. A’ 1. AB U U B A B A From the diagrams 2 and 5 (A  B)’ = A’  B’ 5. A’  B’ 4. B’

  19. 6. Prove that (A  B)’ = A’  B’ using Venn diagram. U U U A B B A B A 2. (A  B)’ 3. A’ 1. A  B U U B A B A From the diagrams 2 and 5 (AB)’ = A’B’ 5. A’  B’ 4. B’

  20. 7. Given, A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8}, show that (i) A (B C) = (A B) C. (ii) Verify (i) using Venn diagram. Solution (i) B C = {3, 4, 5, 6}  {5, 6, 7, 8} = {3, 4, 5, 6, 7, 8} ` A  (B C) = {1, 2, 3, 4, 5}  { 3, 4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}………… (1) A B = {1, 2, 3, 4, 5}  {3, 4, 5, 6} = {1,2,3,4,5,6} `(A B) C = {1,2,3,4,5,6}  {5,6,7,8} = {1, 2, 3, 4, 5, 6, 7, 8}……………. (2) From (1) and (2), we have A  (B C) = (A B) C.

  21. A A B Using Venn diagram, we have A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8} B 3 3 4 1 4 5 5 2 6 6 8 8 7 7 C C (1) B  C (2) A  (B  C)

  22. 8. Let A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. (i) Show that A  (B C) = (A B) C. (ii) Verify (i) using Venn diagram. Solution Given A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. B C = {a,c,e}  {a,e} = {a,e} A  (B C) = {a,b,c,d}  {a,e} = {a}…… (1) A B = {a,b,c,d.}  {a,c,e} = {a,c}. (A B) C = {a,c}  {a,e} = {a}…….. (2) From (1) and (2) A  (B C) = (A B) C.

  23. A A B B A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. a a e C C (1) BC (2) A(BC)

  24. A A B B A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. c a a C C (1) AB (2) (AB)C from (2) and (4) , it is verified that A  (B C) =(A B) C

  25. 9. Let A = {0,1,2,3,4}, B = {1, - 2, 3,4,5,6} and C = {2,4,6,7}. (i) Show that A  (B C) = (A B) (A C). (ii) Verify using Venn diagram. Solution (i) B C = {1, - 2, 3, 4, 5, 6}  {2, 4, 6, 7 } = {4, 6}; A  (B C) = {0,1, 2, 3, 4}  {4, 6} = {0,1,2,3,4,6}…….(1) A B = {0,1,2,3,4}  {1, - 2, 3,4,5,6} = {- 2, 0,1, 2, 3, 4, 5, 6}, A C = {0,1,2,3,4}  {2,4,6,7} = {0,1, 2, 3, 4, 6, 7}. (A B)  (A C) = {- 2, 0,1, 2, 3, 4, 5, 6}  {0,1, 2, 3, 4, 6, 7} = {0,1, 2, 3, 4, 6}. …….(2) From (1) and (2) ,we get A  (B C) = (A B)  (A C).

  26. A A B B A = {0, 1, 2, 3, 4}, B = {1, –2, 3, 4, 5, 6},C = {2, 4, 6, 7}. 3 1 0 4 4 6 6 2 C C (1) BC (2) A(BC)

  27. A = {0, 1, 2, 3, 4}, B = {1, –2, 3, 4, 5, 6},C = {2, 4, 6, 7}. A A -2 B B 1 3 3 5 1 0 0 4 6 4 6 2 2 7 C C (3) AB (4) A C from (2) and (4) , it is verified that A  (B C) =(A B) C

  28. A B 3 1 0 4 6 2 C (5) (AB)(AC) from (2) and (5) , it is verified that A  (B C) = (A B)  (A C)

  29. 10. Given that U = {a,b,c,d,e, f,g,h}, A = {a, b, f, g}, and B = {a, b, c}, verify De Morgan’s laws of complementation. U = {a, b, c, d, e, f, g, h} A = {a, b, f, g} B = {a, b, c}. De Morgan’s laws (AB)’ = A’  B’ (AB)’ = A’  B’ VERIFICATION A  B = {a, b, c, f, g} (A  B)’ = {d, e, h}……..(1) A’ = {c, d, e, h} B’ = {d, e, f, g, h} A’B’ = {d, e, h}……….(2) From (1) and (2) (A  B)’ = A’B’ A  B = {a, b} (A  B)’ = {c, d, e, f, g, h}....(1) A’ = {c, d, e, h} B’ = {d, e, f, g, h} A’  B’ = {c, d, e, f, g, h}….(2) From (1) and (2) (A  B)’ = A’  B’

  30. 11. Verify De Morgan’s laws for set difference using the sets given below: A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} and C = {3,9,10,12,13}.. A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} C = {3,9,10,12,13}.. De Morgan’s laws (1) A\(BC) = (A\B)(A\C) (2) A\(BC) = A\B)  (A\C) VERIFICATION B  C = {1, 2, 3, 5, 7, 9, 10, 12, 13} A\(BC) = {11, 15}……..(1) A\B = {3, 9, 11, 13, 15} A\C = {1, 5, 7, 11, 15} (A\B)(A\C) = {11, 15}……….(2) From (1) and (2) A\(BC) = (A\B)(A\C)

  31. 12. Verify De Morgan’s laws for set difference using the sets given below: A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} and C = {3,9,10,12,13}.. A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} C = {3,9,10,12,13}.. BC = { } A\(B C) = {1, 3, 5, 7, 9,11,13,15}……..(1) A\B = {3, 9, 11, 13, 15} A\C = {1, 5, 7, 11, 15} (A\B)  (A\C) = {1, 3, 5, 7, 9,11,13,15}……….(2) From (1) and (2) A\(B C) = (A\B)  (A\C)

  32. 13. Let A = {10,15, 20, 25, 30, 35, 40, 45, 50}, B = {1, 5,10,15, 20, 30}and C = {7, 8,15,20,35,45, 48}. Verify A\(B C) = (A\B)  (A\C) A = {10,15, 20, 25, 30, 35, 40, 45, 50}, B = {1, 5, 10, 15, 20, 30} C = {7, 8, 15, 20, 35, 45, 48} BC = {15, 20} A\(B C) = {10, 25, 30, 35, 40, 45, 50}……..(1) A\B = {25, 30, 35, 40, 45, 50} A\C = {10, 25, 30, 40, 50} (A\B)  (A\C) = {10, 25, 30, 35, 40, 45, 50}……….(2) From (1) and (2) A\(B C) = (A\B)  (A\C)

  33. Let A = { 6, 9, 15, 18, 21 }; B = { 1, 2, 4, 5, 6 } and f : A " B be defined by f(x) = . Represent f by (i) an arrow diagram (ii) a set of ordered pairs (iii) a table (iv) a graph . ARROW DIAGRAM A B 6 9 15 18 21 1 2 4 5 6 SET OF ORDERED PAIR f = {(6, 1), (9, 2), (15, 4), (18, 5), (21, 6)}

  34. Table Graph – 7 – 6 – 5 – 4 – 3 – 2 – 1  (21, 6)  (18, 5)  (15, 4)  (9, 2)  (6, 1) | | | | | | | | | 3 6 9 12 15 18 21 24 27

  35. 15. Let A= { 0, 1, 2, 3 } and B = { 1, 3, 5, 7, 9 } be two sets. Let f : A B be a function given by f (x) = 2x + 1. Represent this function as (i) a set of ordered pairs (ii) a table (iii) an arrow diagram and (iv) a graph. Given A= {0, 1, 2, 3}, B = {1, 3, 5, 7, 9} f(x) = 2x + 1 f(0) = 2(0) + 1 = 0 + 1 = 1 f(1) = 2(1) + 1 = 2 + 1 = 3 f(2) = 2(2) + 1 = 4 + 1 = 5 f(3) = 2(3) + 1 = 6 + 1 = 7 (i)Set of ordered pairs {(0, 1), (1, 3), (2, 5), (3, 7)} (ii) Table 0 1 2 3 x 1 3 5 7 f(x)

  36. 4 6 8 10 ARROW DIAGRAM B A 3 4 5 6 7 – 7 – 6 – 5 – 4 – 3 – 2 – 1 Graph  (10, 6)  (8, 5)  (6, 4)  (4, 3) | | | | | | | | 0 2 4 6 8 10 12 14 16

  37. 16. A function f: [1, 6)R is defined as follows f(x) = Find the value of (i) f(5), (ii) f(3), (iii) f(1), (iv) f(2) – f(4) (v) 2f(5) – 3f(1) Since 5 lies between 4 and 6 f(x) = 3x2 – 10 f(5) = 3(5)2 – 10 = 75 – 10 = 65 Since 3 lies between 2 and 4 f(x) = 2x – 1 = 2(3) – 1 = 6 – 1 = 5 Since 1 lies in the interval 1  x < 2 f(x) = 1 + x f(1) = 1 + 1 = 2 Since 2 lies in the interval 2  x < 4 f(x) = 2x – 1 = 2(2) – 1 = 4 – 1 = 3 Since 4 lies in the interval 4  x < 6 f(x) = 3x2 – 10 f(4) = 3(4)2 – 10 = 48 – 10 = 38

  38. f(2) – f(4) = 3 – 38 = – 34 2f(5) – 3 f(1) = 2 (65) – 3(2) = 130 – 6 = 124

  39. 17. A function f: [-3, 7)R is defined as follows Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv) (i) Since 5 lies between 4 and 6 f(x) = 2x – 3 f(5) = 2(5) – 3 = 10 – 3 = 7 Since 6 lies in the interval 4 < x  6 f(x) = 2x – 3 f(6) = 2(6) – 3 = 12 – 3 = 9 f(5) + f(6) = 7 + 9 = 16

  40. 18. A function f: [-3, 7)R is defined as follows f(x) = Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4)(iv) (ii) Since 1 lies in the interval -3  x < 2 f(x) = 4x2 – 1 f(1) = 4(1)2 – 1 = 4 – 1 = 3 Since –3 lies in the interval-3  x < 2 f(x) = 4x2 – 1 = 4(-3)2 – 1 = 36 – 1 = 35 f(1) – f(-3) = 3 – 35 = – 32

  41. 19. A function f: [-3, 7)R is defined as follows f(x) = Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv) (iii) Since -2 lies in the interval -3  x < 2 f(x) = 4x2 – 1 f(1) = 4(-2)2 – 1 = 16 – 1 = 15 Since 4 lies in the interval 2  x  4 f(x) = 3x – 2 = 3(4) – 2 = 12 – 2 = 10 f(-2) – f(4) = 15 – 10 = 5

  42. 20. A function f: [-3, 7)R is defined as follows f(x) = Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv) (iv) Since 3 lies in the interval 2  x  4 f(x) = 3x – 2 f(3) = 3(3) – 2 = 9 – 2 = 7 Since -1 lies in the interval -3  x < 2 f(x) = 4x2 – 1 = 4(-1)2 – 1 = 4 – 1 = 3 f(6) = 9, f(1) = 3

  43. 21. A function f: [-7, 6)R is defined as follows Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (i) Since -4 lies in the interval -5  x  2 f(x) = x + 5 f(-4) = – 4 + 5 = 1 Since 2 lies in the interval -5  x  2 f(x) = x + 5 f(2) = 2 + 5 = 7 2f(-4) + 3f(2) = 2(1) + 3(7) = 2 + 21 = 23

  44. 22. A function f: [-7, 6)R is defined as follows f(x) = Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (ii) Since -7 lies in the interval -7  x < -5 f(x) = x2 + 2x + 1 f(-7) = (–7)2 + 2(–7) + 1 = 49 – 14 + 1 = 36 Since -3 lies in the interval -5  x  2 f(x) = x + 5 f(2) = -3 + 5 = 2 f(-7) – f(–3) = 36 – 2 = 34

  45. 23. A function f: [-7, 6)R is defined as follows f(x) = Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (iii) Since -3 lies in the interval -5  x  2 f(x) = x + 5 f(-3) = -3 + 5 = 2 Since -6 lies in the interval -7  x < -5 f(x) = x2 + 2x + 1 f(-6) = (–6)2 + 2(–6) + 1 = 36 – 12 + 1 = 25 4 lies in the interval 2 < x < 6 f(x) = x – 1 f(4) = 4 – 1 = 3 Since 1 lies in the interval -5  x  2 f(x) = x + 5 f(1) = 1 + 5 = 6

  46. 24. A function f: [-7, 6)R is defined as follows f(x) = Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (iii) f(-3) = 2 f(-6) = 25 f(4) = 3 f(1) = 6

  47. LHS = RHS  (AB)T= BTAT .

  48. 26. If then show that A2 – 4A + 5I2 = O A2 – 4A + 5I2 = O

  49. 27. Find X and Y if 2X + 3Y =