NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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A craft designer in Melbourne is creating a three-dimensional puzzle piece using digital modelling software. The piece is formed by attaching a right-angled triangular prism to a square base.

The base is a square with side length 6 m. The triangular prism has a rectangular face of 6 m by 3 m, and the right-angled triangle forming the cross-section has legs of 3 m and 4 m, with the hypotenuse aligned along the 6 m edge of the base.

The designer needs to determine the exact length of the internal diagonal that runs from one bottom corner of the square base to the opposite top vertex of the triangular prism's apex. Using the three-dimensional version of Pythagoras’ theorem, what is the exact length of this diagonal in metres?

Concept illustrations

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Worked solution and answer (click to reveal)

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Worked solution

  1. The puzzle piece consists of a square base with side length 6 m and a right-angled triangular prism attached. The base is in the xy-plane, and the prism extends vertically along the z-axis.
  2. The triangular prism has a rectangular face of 6 m (length) by 3 m (height). The right-angled triangle cross-section has legs of 3 m and 4 m, with the hypotenuse (5 m) aligned along the 6 m base edge.
  3. The internal diagonal runs from a bottom corner of the square base (e.g., point A at (0,0,0)) to the opposite top vertex of the prism's apex (point B). The apex lies above the far corner of the base, at the top of the triangular face.
  4. The base extends from (0,0,0) to (6,6,0). The triangular prism is attached along one 6 m edge, say from (0,0,0) to (6,0,0). The right-angled triangle has legs 3 m and 4 m, so the height of the prism varies across the base.
  5. The apex of the triangular prism lies above the point (6,0,0), at a height determined by the triangle. Since the triangle has legs 3 m and 4 m, and the hypotenuse is 5 m, the height at (6,0,0) is 4 m (the longer leg), assuming the 3 m leg is vertical and the 4 m leg is horizontal along the base.
  6. The opposite top vertex is at (6,6,4), because the base extends 6 m in the y-direction from (0,0,0) to (0,6,0), and the prism's height at (6,0,0) is 4 m, but the height varies linearly across the base.
  7. The diagonal runs from (0,0,0) to (6,6,4). The 3D distance is calculated using the 3D Pythagorean theorem: d = sqrt(Δx² + Δy² + Δz²).
  8. Substitute Δx = 6 m, Δy = 6 m, Δz = 4 m. Then d = sqrt(6² + 6² + 4²) = sqrt(36 + 36 + 16) = sqrt(88).
  9. Simplify sqrt(88): 88 = 4 × 22, so sqrt(88) = sqrt(4 × 22) = 2 × sqrt(22).
  10. The exact length of the diagonal is 2 × sqrt(22) metres.
  11. Check: 2 × sqrt(22) ≈ 2 × 4.690 = 9.380 m. The diagonal should be longer than any face diagonal (e.g., base diagonal is 6×sqrt(2) ≈ 8.485 m), so 9.380 m is plausible.
  12. The internal diagonal from (0,0,0) to (6,6,4) is exactly 2×sqrt(22) metres.

Reference answer: 2*sqrt(22) m

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