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Visual Proof of Pyramid Volume: One Third Base Area Times Height

Visual Proof of Pyramid Volume: One Third Base Area Times Height
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2016-04-25 00:41
PolygonVectorSolid GeometryTransformationGeometric Construction
① This GeoGebra example demonstrates the classic formula that the volume of a pyramid equals one third of its base area multiplied by its height, using a 3D geometric construction. ② Core mathematical concepts include pyramid volume, decomposition of a cuboid into pyramids, congruence, vector translation, and dilation. ③ Construction highlights: the Pyramid command creates the main pyramid, Translate and Dilate generate auxiliary pyramids and scaled copies, Intersect finds key points, and Polygon/If build faces and conditional objects. Multiple faces (e.g., faceABD, faceBCD) and edges (edgeAD, edgeBD) are used to compare volumes and illustrate the one-third relationship. ④ Suitable for high school or early college students learning solid geometry, this example serves as an interactive visual proof for teachers and learners to explore the volume formula.

GeoGebra Commands & Parameters

  1. If(s ≤ 0, (-s), 0) → s2
  2. If(s ≥ 0, s, 0) → s1
  3. Translate(C, Vector[B, D]) → C'
  4. Intersect(OrthogonalPlane[B, Segment[A, C]], Line[A, C]) → G
  5. Translate(A, Vector[B, D]) → A'
  6. Intersect(OrthogonalPlane[B, Segment[A, C]], Line[A', C']) → H
  7. Intersect(OrthogonalPlane[C, Segment[A, B]], Line[A', D]) → F
  8. Intersect(OrthogonalPlane[C, Segment[A, B]], Line[A, B]) → E
  9. Translate(Rotate[Translate[Rotate[C, (α * s1), Segment[A, B]], Vector[(s1 * Vector[(0, 0, 0), (0, z(D), 0)])]], ((-β) * s2), Segment[A, C]], Vector[(s2 / 2 * Vector[(0, 0, 0), (z(D), 0, 0)])]) → L
  10. Translate(Rotate[Translate[Rotate[Polygon[A, D, A'], (α * s1), Segment[A, B]], Vector[(s1 * Vector[(0, 0, 0), (0, z(D), 0)])]], ((-β) * s2), Segment[A, C]], Vector[(s2 / 2 * Vector[(0, 0, 0), (z(D), 0, 0)])]) → poly2
  11. Pyramid(Polygon[A, B, C], D) → a, faceABD, faceBCD, faceACD_1, edgeAD, edgeBD, edgeCD_1
  12. Translate(Rotate[Translate[Rotate[D, (α * s1), Segment[A, B]], Vector[(s1 * Vector[(0, 0, 0), (0, z(D), 0)])]], ((-β) * s2), Segment[A, C]], Vector[(s2 / 2 * Vector[(0, 0, 0), (z(D), 0, 0)])]) → K
  13. Translate(Rotate[Translate[Rotate[Polygon[A, C, A'], (α * s1), Segment[A, B]], Vector[(s1 * Vector[(0, 0, 0), (0, z(D), 0)])]], ((-β) * s2), Segment[A, C]], Vector[(s2 / 2 * Vector[(0, 0, 0), (z(D), 0, 0)])]) → poly3
  14. Pyramid(Polygon[A', D, C'], C) → b, faceA'CD_1, faceCC'D, faceA'CC', edgeA'C_1, edgeCD_2, edgeCC'
  15. Translate(Translate[Rotate[a, (α * s1), Segment[A, B]], Vector[(s1 * Vector[(0, 0, 0), (0, z(D), 0)])]], Vector[(o2 * Vector[A, B])]) → a'
  16. Pyramid(Polygon[A, D, A'], C) → c, faceACD, faceA'CD, faceAA'C, edgeAC, edgeCD, edgeA'C
  17. Translate(Rotate[c, (α * s1), Segment[A, B]], Vector[(s1 * Vector[(0, 0, 0), (0, z(D), 0)])]) → c'
  18. Translate(Rotate[Translate[b, Vector[(o1 * Vector[A, C])]], (s1 * α), Segment[A, B]], Vector[(s1 * Vector[(0, 0, 0), (0, z(D), 0)])]) → b'
  19. Translate(Rotate[a', ((-β) * s2), Segment[A, C]], Vector[(s2 / 2 * Vector[(0, 0, 0), (z(D), 0, 0)])]) → a''
  20. Translate(Rotate[b', ((-β) * s2), Segment[A, C]], Vector[(s2 / 2 * Vector[(0, 0, 0), (z(D), 0, 0)])]) → b''
  21. Translate(Rotate[c', ((-β) * s2), Segment[A, C]], Vector[(s2 / 2 * Vector[(0, 0, 0), (z(D), 0, 0)])]) → c''
  22. Translate(Rotate[Translate[Translate[Rotate[Polygon[A, B, D], (α * s1), Segment[A, B]], Vector[(s1 * Vector[(0, 0, 0), (0, z(D), 0)])]], Vector[(o2 * Vector[A, B])]], ((-β) * s2), Segment[A, C]], Vector[(s2 / 2 * Vector[(0, 0, 0), (z(D), 0, 0)])]) → poly1
  23. Translate(Rotate[Translate[Translate[Rotate[C, (α * s1), Segment[A, B]], Vector[(s1 * Vector[(0, 0, 0), (0, z(D), 0)])]], Vector[(o2 * Vector[A, B])]], ((-β) * s2), Segment[A, C]], Vector[(s2 / 2 * Vector[(0, 0, 0), (z(D), 0, 0)])]) → I
  24. Dilate(poly2, 1 - h, L) → poly2'
  25. Dilate(poly1, 1 - h, I) → poly1'
  26. Translate(Rotate[Translate[Rotate[Translate[Polygon[A', C, C'], Vector[(o1 * Vector[A, C])]], (s1 * α), Segment[A, B]], Vector[(s1 * Vector[(0, 0, 0), (0, z(D), 0)])]], ((-β) * s2), Segment[A, C]], Vector[(s2 / 2 * Vector[(0, 0, 0), (z(D), 0, 0)])]) → poly4
  27. Translate(Rotate[Translate[Rotate[Translate[D, Vector[(o1 * Vector[A, C])]], (s1 * α), Segment[A, B]], Vector[(s1 * Vector[(0, 0, 0), (0, z(D), 0)])]], ((-β) * s2), Segment[A, C]], Vector[(s2 / 2 * Vector[(0, 0, 0), (z(D), 0, 0)])]) → J
  28. Dilate(poly3, 1 - h, K) → poly3'
  29. Dilate(poly4, 1 - h, J) → poly4'
  30. Polygon((-6, -6), (6, -6), (6, 6), (-6, 6)) → poly5, d, e, f, g

AI Drawing Prompt Reference

In 3D space, start with a rectangular cuboid whose base is a rectangle and top is parallel. From one base vertex, connect to the opposite top vertex to form three edges, creating a triangular pyramid. Then, by translating and rotating this pyramid, generate three more congruent pyramids that together fill the entire cuboid. Color the four pyramids differently and add sliders to control rotation angles and scaling, visually demonstrating that each pyramid has equal volume, proving the pyramid volume is one-third of base area times height.
The prompt above was compiled from the original drawing commands and is for reference only.