To solve this problem, we need to compute the third dimension of a cuboidal box such that the body diagonal is the least possible integer.

Given that the dimensions of the box are 9 units, 12 units, and the unknown dimension $x$, we can use the formula for the body diagonal $d$ of a cuboid:

$d = \sqrt{a^2 + b^2 + c^2}$

where $a$, $b$, and $c$ are the three dimensions of the cuboid. In this case, $a = 9$, $b = 12$, and $c = x$. We want the body diagonal $d$ to be the least possible integer.

1. Plug in the known values:

$d = \sqrt{9^2 + 12^2 + x^2}$ $d = \sqrt{81 + 144 + x^2}$ $d = \sqrt{225 + x^2}$

2. We want $\sqrt{225 + x^2}$ to be an integer. Let's try different values of $x$ from the options provided:

• For $x = 8$: $d = \sqrt{225 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$

• For $x = 15$: $d = \sqrt{225 + 15^2} = \sqrt{225 + 225} = \sqrt{450}$ $\sqrt{450}$ is not an integer.

• For $x = 12$: $d = \sqrt{225 + 12^2} = \sqrt{225 + 144} = \sqrt{369}$ $\sqrt{369}$ is not an integer.

• For $x = 14$: $d = \sqrt{225 + 14^2} = \sqrt{225 + 196} = \sqrt{421}$ $\sqrt{421}$ is not an integer.

Among the given options, $x = 8$ gives us the least possible integer value for the body diagonal, which is 17.

Thus, the third dimension is $\boxed{8}$.

Would you like further details or have any questions?

Here are 5 related questions:

1. How would you calculate the volume of a cuboid with given dimensions?
2. What is the formula for the surface area of a cuboid?
3. Can the body diagonal formula be used for any three-dimensional shape?
4. How does changing one dimension of a cuboid affect its body diagonal?
5. If two dimensions are equal, what type of special cuboid is formed?

Tip: The Pythagorean theorem is extended into three dimensions for calculating the body diagonal of a cuboid.