**Question:**

Show that the product of the areas of the floor and two adjacent walls of a cuboid is the square of its volume.

**Solution:**

Suppose that the length, breadth and height of the cuboidal floor are $l \mathrm{~cm}, b \mathrm{~cm}$ and $h \mathrm{~cm}$, respectively.

Then, area of the floor $=l \times b \mathrm{~cm}^{2}$

Area of the wall $=b \times h \mathrm{~cm}^{2}$

Area of its adjacent wall $=l \times h \mathrm{~cm}^{2}$

Now, product of the areas of the floor and the two adjacent walls $=(l \times b) \times(b \times h) \times(l \times h)=l^{2} \times b^{2} \times h^{2}=(l \times b \times h)^{2}$

Also, volume of the cuboid $=l \times b \times h \mathrm{~cm}^{2}$

$\therefore$ Product of the areas of the floor and the two adjacent walls $=(l \times b \times h)^{2}=(\text { volume })^{2}$