Show that the product of the areas of the floor and two adjacent walls of a cuboid is the square of its volume.
Solution
\[\text { Suppose that the length, breadth and height of the cuboidal floor are l cm, b cm and h cm, respectively . } \]
\[\text { Then, area of the floor = l } \times b {cm}^2 \]
\[\text { Area of the wall = b } \times h {cm}^2 \]
\[\text { Area of its adjacent wall = l }\times h {cm}^2 \]
\[\text { Now, product of the areas of the floor and the two adjacent walls }= (l \times b) \times (b \times h) \times (l \times h) = l^2 \times b^2 \times h^2 = (l \times b \times h )^2 \]
\[\text { Also, volume of the cuboid = l } \times b \times h {cm}^2 \]
\[ \therefore\text { Product of the areas of the floor and the two adjacent walls } = (l \times b \times h )^2 = \text { (volume })^2\]
