3D Raumgeometrie-Engined = √(x² + y² + z²)
3D Raumdiagonale & Quader-Rechner
Calculate the longest straight rod, beam, or internal vector fitting inside a 3D rectangular prism with exact simplified radicals and cuboid surface metrics. Working purely in a flat 2D plane instead? Use our flagship Pythagoras equation solver.
Interactive 3D Wireframe Box2-Stage KaTeX DerivationVolume & Surface Area
Result updated: 3D space diagonal is 13, approximately 13. Base diagonal is 5.
3D Raumdiagonale & Quader-Rechner
d_base = √(x² + y²) · d_space = √(x² + y² + z²)
Enter strictly positive cuboid side dimensions.
3D Raumdiagonale (d_space)
13
≈ 13
Bodendiagonale (d_base)
5
≈ 5
Volumen
144 units³
Oberfläche
192 units²
Zweistufige 3D-Herleitung (KaTeX)
Stage 1: Solve the 2D Base Face Diagonal (XY Plane)
d_base = √(3² + 4²)
d_base = 5
d_{\text{base}} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5
Stage 2: Solve the 3D Internal Space Diagonal (Vertical Right Triangle)
d_space = √((√25)² + 12²) = √(25 + 144)
d_space = 13
d_{\text{space}} = \sqrt{d_{\text{base}}^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13
Warum 3D-Diagonalen Pythagoras Zweimal Nutzen
First, apply 2D Pythagoras across the floor rectangle (x by y) to find the base face diagonal d_base² = x² + y². Because the vertical height edge z stands strictly perpendicular (90°) to every line on the floor, d_base and z form a second vertical right triangle whose hypotenuse is the 3D space diagonal.
d_space² = d_base² + z² = (x² + y²) + z² ⟹ d_space = √(x² + y² + z²)