Longest pole inside a cuboidal room\nWhat is the maximum possible length of a pole that can be placed inside a room of dimensions 12 m (length), 4 m (breadth), and 3 m (height)?

Introduction / Context:
The longest rod/pole that fits inside a cuboidal room equals the space diagonal of the cuboid. This is a classic 3D mensuration result derived from the Pythagorean theorem applied twice.

Given Data / Assumptions:

• Length = 12 m, Breadth = 4 m, Height = 3 m
• Pole length sought = space diagonal of cuboid

Concept / Approach:
Space diagonal d of a cuboid with dimensions l, b, h is given by d = sqrt(l^2 + b^2 + h^2).

Step-by-Step Solution:

Verification / Alternative check:
The diagonal of the floor is sqrt(12^2 + 4^2) = sqrt(160). Extending to 3D with height 3 gives sqrt(160 + 9) = sqrt(169) = 13, confirming the result.

Why Other Options Are Wrong:
14 m, 15 m, and 16 m exceed the exact space diagonal and cannot fit entirely within the room.

Common Pitfalls:
Confusing the space diagonal with a face diagonal (e.g., sqrt(12^2 + 4^2)) leads to an underestimate. Always include all three dimensions.

Final Answer:
13 m