Question

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

Answer 1

\[\text { The areas of three adjacent faces of a cuboid are x, y and z  }. \]

\[\text { Volume of the cuboid = V }\]

\[\text { Observe that x = length } \times \text { breadth } \]

\[\text { y = breadth } \times\text {  height }, \]

\[\text { z = length } \times \text { height }\]

\[\text { Since volume of cuboid V = length } \times\text {  breadth }\times \text { height, we have: } \]

\[ V^2 = V \times V\]

\[ =\text {  (length } \times \text{breadth } \times \text { height)  }\times (\text{ length  }\times \text { breadth } \times \text { height) }\]

\[ =\text{ (length  }\times \text { breadth }) \times (\text { breadth }\times \text { height) } \times\text {  (length }\times \text { height) }\]

\[ = x \times y \times z\]

\[ = xyz\]

\[ \therefore V^2 = xyz\]