The areas of three adjacent faces of a cuboid are x, y and z.

Question:

The areas of three adjacent faces of a cuboid are xy and z. If the volume is V, prove that V2 = xyz.

Solution:

The areas of three adjacent faces of a cuboid are $x, y$ and $z$.

Volume of the cuboid $=\mathrm{V}$

Observe that $x=$ length $\times$ breadth

$y=$ breadth $\times$ height

$z=$ length $\times$ height

Since volume of cuboid $V=$ length $\times$ breadth $\times$ height, we have :

$V^{2}=V \times V$

$=($ length $\times$ breadth $\times$ height $) \times($ length $\times$ breadth $\times$ height $)$

$=($ length $\times$ breadth $) \times($ breadth $\times$ height $) \times($ length $\times$ height $)$

$=x \times y \times z$

$=x y z$

$\therefore V^{2}=x y z$

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