Space Diagonal of a Cuboid $d = \sqrt{l^2 + b^2 + h^2}$

3D proof using base triangle $\Delta OAB$ and vertical slice $\Delta OBF$

Background:

Cuboid Dimensions 6.0 × 4.0 × 5.0

Length ($l$) [Along +X]: 6.0

Breadth ($b$) [Along +Z]: 4.0

Height ($h$) [Along +Y]: 5.0

Standard Educational Presets

Step-by-Step Proof

Step 1: Base Face $\Delta OAB$ Base Plane $XZ$

$OA \perp AB$ at $A$ (as base $OABC$ is a rectangle). In right-angled $\Delta OAB$, by Pythagoras Theorem:

$OB^2 = OA^2 + AB^2 = l^2 + b^2$
Base diagonal: $OB = \sqrt{l^2 + b^2}$
$OB = \sqrt{6^2 + 4^2} = \sqrt{52} \approx 7.21$

Step 2: Vertical Triangle $\Delta OBF$ Vertical Slice

Vertical edge $BF$ is perpendicular to base plane $OABC$, hence $BF \perp OB$ at vertex $B$. In right-angled $\Delta OBF$:

$OF^2 = OB^2 + BF^2$
Substituting $OB^2 = l^2 + b^2$ and $BF = h$:
$OF^2 = (l^2 + b^2) + h^2 = l^2 + b^2 + h^2$

$\therefore d = OF = \sqrt{l^2 + b^2 + h^2}$ $\approx 8.77$

Formula Verification

$\sqrt{6^2 + 4^2 + 5^2} = \sqrt{77} \approx 8.77$

Q.E.D.

Visualization Toggles

Space Diagonal ($OF$) Base Diagonal ($OB$) Shaded Base $\Delta OAB$ Shaded Vertical $\Delta OBF$ Right Angle ($\perp$) Indicators Vertex Coordinates Labels Coordinate Grid

X (Length $l$)

Y (Height $h$)

Z (Breadth $b$)

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