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Question
The areas of three adjacent faces of a cuboid are x, y and z. If the volume is V, prove that V2 = xyz.
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Solution
\[\text { The areas of three adjacent faces of a cuboid are x, y and z }. \]
\[\text { Volume of the cuboid = V }\]
\[\text { Observe that x = length } \times \text { breadth } \]
\[\text { y = breadth } \times\text { height }, \]
\[\text { z = length } \times \text { height }\]
\[\text { Since volume of cuboid V = length } \times\text { breadth }\times \text { height, we have: } \]
\[ V^2 = V \times V\]
\[ =\text { (length } \times \text{breadth } \times \text { height) }\times (\text{ length }\times \text { breadth } \times \text { height) }\]
\[ =\text{ (length }\times \text { breadth }) \times (\text { breadth }\times \text { height) } \times\text { (length }\times \text { height) }\]
\[ = x \times y \times z\]
\[ = xyz\]
\[ \therefore V^2 = xyz\]
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