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Exploring the Pythagorean Theorem in Ancient Greece

Exploring the Pythagorean Theorem in Ancient Greece
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2024-09-26 21:26
TriangleCoordinate GeometryGeometric Construction
In collections Satz des Pythagoras - Einführung
This example explores the Pythagorean Theorem in ancient Greek times, helping students understand that the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. Key mathematical concepts include right triangles, the Pythagorean Theorem, and its applications. GeoGebra commands used include Segment, Angle, Point, and others. Suitable for 7-8th grade students.

GeoGebra Commands & Parameters

  1. SelectedIndex(DropDown12) → input12
  2. SelectedIndex(DropDown7) → input7
  3. SelectedIndex(DropDown3) → input3
  4. SelectedIndex(DropDown2) → input2
  5. Point(Line[A1, Vector[A1, A1 + (-4, 3)]]) → B1
  6. Angle(C1, B1, A1) → wb
  7. Segment(B1, C1) → sa
  8. Rotate(Flatten[Sequence[Sequence[B1 + (m * (eh, 0)) + (j * (0, eh)), m, 0, 4], j, 0, 4]], wb + 270°, B1) → listEndC
  9. Flatten(Sequence[Sequence[A3 + (m * (eh, 0)) + (j * (0, eh)), m, 0, 2], j, 0, 2]) → listEndA
  10. Point(Segment[listEndA(1), listEndC(9)]) → P9
  11. Point(Segment[listEndA(4), listEndC(8)]) → P8
  12. Point(Segment[listEndA(7), listEndC(7)]) → P7
  13. Point(Segment[listEndA(2), listEndC(6)]) → P6
  14. Point(Segment[listEndA(5), listEndC(5)]) → P5
  15. Point(Segment[listEndA(8), listEndC(4)]) → P4
  16. Point(Segment[listEndA(3), listEndC(3)]) → P3
  17. Point(Segment[listEndA(6), listEndC(2)]) → P2
  18. Point(Segment[listEndA(9), listEndC(1)]) → P1
  19. Flatten(Sequence[Sequence[B2 + (m * (eh, 0)) + (j * (0, eh)), m, 0, 3], j, 0, 3]) → listEndB
  20. Point(Segment[listEndB(13), listEndC(10)]) → P10
  21. Point(Segment[listEndB(14), listEndC(11)]) → P11
  22. Point(Segment[listEndB(15), listEndC(12)]) → P12
  23. Point(Segment[listEndB(16), listEndC(13)]) → P13
  24. Point(Segment[listEndB(9), listEndC(14)]) → P14
  25. Point(Segment[listEndB(10), listEndC(15)]) → P15
  26. Point(Segment[listEndB(11), listEndC(16)]) → P16
  27. Point(Segment[listEndB(12), listEndC(17)]) → P17
  28. Point(Segment[listEndB(5), listEndC(18)]) → P18
  29. Point(Segment[listEndB(6), listEndC(19)]) → P19
  30. Point(Segment[listEndB(7), listEndC(20)]) → P20
  31. Point(Segment[listEndB(8), listEndC(21)]) → P21
  32. Point(Segment[listEndB(1), listEndC(22)]) → P22
  33. Point(Segment[listEndB(2), listEndC(23)]) → P23
  34. Point(Segment[listEndB(3), listEndC(24)]) → P24
  35. Point(Segment[listEndB(4), listEndC(25)]) → P25
  36. Zip(listEndA(m), m, {9, 6, 3, 8, 5, 2, 7, 4, 1}) → listStartA
  37. Zip(C + (If[m ≤ anim2 ≤ n, anim2 - m, If[anim2 < m, 0, 1]] * Vector[C, A]), C, listEndC, A, listStartA, m, Sequence[0, 8], n, Sequence[1, 9]) → listPA2
  38. Zip(If[A ≟ B, B], A, {P1, P2, P3, P4, P5, P6, P7, P8, P9}, B, listEndC) → listPVCA
  39. Point(Line[Jo1, xAxis]) → Jo2
  40. Textfield(input8) → Eingabefeld8
  41. Textfield(input9) → Eingabefeld9
  42. Textfield(input10) → Eingabefeld10
  43. Point(Line[SR1a, xAxis]) → SR1b
  44. Segment(C1, A1) → sb
  45. Segment(B1, A1) → sc
  46. Angle(A1, C1, B1) → wc
  47. Zip(Start1 + (If[m ≤ anim1 ≤ n, anim1 - m, If[anim1 < m, 0, 1]] * Vector[Start1, P]), m, Sequence[0, 24], n, Sequence[1, 25], P, listEndA) → listPA
  48. Zip(Polygon[P, P + (eh, 0), 4], P, listPA) → listVA
  49. Zip(Start2 + (If[m ≤ anim1 ≤ n, anim1 - m, If[anim1 < m, 0, 1]] * Vector[Start2, P]), m, Sequence[9, 24], n, Sequence[10, 25], P, listEndB) → listPB
  50. Zip(Polygon[P, P + (eh, 0), 4], P, listPB) → listVB
  51. Zip(Rotate[Polygon[P, P + (eh, 0), 4], wb + 270°, P], P, Sequence[listPVCA(m), m, 1, 9]) → listVCA
  52. Zip(If[A ≟ B, B], A, {P10, P11, P12, P13, P14, P15, P16, P17, P18, P19, P20, P21, P22, P23, P24, P25}, B, Sequence[listEndC(n), n, 10, 25]) → listPVCB
  53. Zip(Rotate[Polygon[P, P + (eh, 0), 4], wb + 270°, P], P, Sequence[listPVCB(m), m, 1, 16]) → listVCB
  54. Zip(listEndB(m), m, {13, 14, 15, 16, 9, 10, 11, 12, 5, 6, 7, 8, 1, 2, 3, 4}) → listStartB
  55. Zip(C + (If[m ≤ anim2 ≤ n, anim2 - m, If[anim2 < m, 0, 1]] * Vector[C, A]), C, Sequence[listEndC(a), a, 10, 25], A, listStartB, m, Sequence[9, 24], n, Sequence[10, 25]) → listPB2
  56. Angle(B1, A1, C1) → wa
  57. Angle(C3, C2, A1) → wc2
  58. Point(Line[GR1, xAxis]) → GR2
  59. Point(Line[GR1, GR2]) → Jo3
  60. Point(Line[Jo3, xAxis]) → Jo4
  61. Point(Line[SB2, yAxis]) → SB6
  62. Mirror(SB2, Line[SB1, xAxis]) → SB7
  63. Textfield(input5) → Eingabefeld5
  64. Textfield(input6) → Eingabefeld6
  65. Textfield(input4) → Eingabefeld4
  66. Textfield(input1a) → Eingabefeld1a
  67. Zip(Polygon[P, P + (eh, 0), 4], P, listPA) → listVAgrau
  68. Zip(Polygon[P, P + (eh, 0), 4], P, listPB) → listVBgrau
  69. Textfield(input11) → Eingabefeld11
  70. Textfield(input1b) → Eingabefeld1b
  71. PointIn(Circle[SR1a, 20]) → SRPoint1
  72. Point(Line[SR1a + (3.8, 0), yAxis]) → SR2a
  73. PointIn(Circle[SR1a, 20]) → SRPoint2
  74. LaTeX(step, true, true) → Text7
  75. Point(Line[SR2a, xAxis]) → SR2b
  76. Element({PointSol3, PointSol3, PointSol3, PointSol5, PointSol5, PointSol9, PointSol9, PointSol9, PointSol9, PointSol10, PointSol10, PointSol9}, step) → PointSol

AI Drawing Prompt Reference

First, draw a right triangle with its two legs having lengths 4 and 3. Then, create a series of points P1 to P9 connected by line segments. Next, generate another set of points P10 to P25 that connect to points in listEndC to form a polygon. Finally, draw a quadrilateral polygon using points P and P shifted by (eh,0). The figure also includes text input fields for entering numbers.
The prompt above was compiled from the original drawing commands and is for reference only.