Longest rod in a 20 m × 16 m × 12 m room: Find the maximum length that can fit inside the room (the space diagonal).

Introduction / Context:
The problem asks for the space diagonal of a rectangular room. This is another direct usage of the 3D Pythagoras theorem and light numeric handling.

Given Data / Assumptions:

• L = 20 m, W = 16 m, H = 12 m
• Space diagonal d = √(L^2 + W^2 + H^2)

Concept / Approach:
Compute the sum of squares of the three edges and take the square root. Present the answer to one decimal place to match options.

Step-by-Step Solution:
d = √(20^2 + 16^2 + 12^2) = √(400 + 256 + 144)d = √800 = √(16 * 50) = 4√50 = 20√2 ≈ 28.284 mRounded to the listed precision → 28.2 m

Verification / Alternative check:
Since 20√2 ≈ 28.284, the nearest given option is 28.2 m (options do not include 28.3 m).

Why Other Options Are Wrong:
20 m and 16.4 m are underestimates; 48 m is much too large; 28.3 m would be closer numerically but is not among the provided choices.

Common Pitfalls:
Using only two dimensions; arithmetic slips when adding squares; incorrect rounding.

Final Answer:
28.2 m