What is the length of the longest pole that can be kept inside a room that is 12 m long, 4 m broad, and 3 m high? (Assume the pole can be placed diagonally in 3D space.)

Introduction / Context:
This question tests the concept of the space diagonal of a cuboid (rectangular box). The longest straight object that can fit inside a rectangular room is the space diagonal connecting two opposite corners in 3D. To find this length, we use the 3D Pythagoras theorem: space diagonal = sqrt(l^2 + b^2 + h^2). It is essentially applying Pythagoras twice: first to get the floor diagonal, then combining with height. Because the room dimensions are given in metres, the resulting pole length is also in metres.

Given Data / Assumptions:

• Length l = 12 m
• Breadth b = 4 m
• Height h = 3 m
• Longest pole length = space diagonal = sqrt(l^2 + b^2 + h^2)

Concept / Approach:
Compute l^2 + b^2 + h^2, then take the square root. If the sum is a perfect square, the diagonal is an integer.

Step-by-Step Solution:
Space diagonal d = sqrt(12^2 + 4^2 + 3^2) 12^2 = 144, 4^2 = 16, 3^2 = 9 Sum = 144 + 16 + 9 = 169 d = sqrt(169) = 13 m

Verification / Alternative check:
Floor diagonal = sqrt(12^2 + 4^2) = sqrt(160). Then space diagonal = sqrt(160 + 3^2) = sqrt(169) = 13. Same result confirms correctness.

Why Other Options Are Wrong:
14, 15, 16 are larger than the computed space diagonal and will not fit corner-to-corner. 12.5 can come from using only the floor diagonal and ignoring height.

Common Pitfalls:
Using only 2D diagonal of the floor, adding dimensions instead of squaring and rooting, or mixing up breadth and height values.

Final Answer:
The longest pole length is 13 m.