The length of the longest rod that can be placed in a room of dimensions 10 m × 10 m × 5 m is

### Concept & Diagonal of a Cuboid The longest rod that can fit inside a rectangular room must span from one bottom corner to the opposite top corner. This is exactly the main diagonal of the cuboid. $$ Diagonal = \sqrt{L^2 + B^2 + H^2} $$ ### Step-by-Step Solution 1. **Identify the Dimensions**: - Length (L) = 10 m - Breadth (B) = 10 m - Height (H) = 5 m 2. **Apply the Diagonal Formula**: - Diagonal = $\sqrt{10^2 + 10^2 + 5^2}$ - Diagonal = $\sqrt{100 + 100 + 25}$ - Diagonal = $\sqrt{225}$ 3. **Calculate Final Value**: - $\sqrt{225} = 15$ m ### Exam Strategy & Shortcut This is a standard Pythagorean quadruplet conceptually ($10, 10, 5 \rightarrow 15$). Notice that all terms share a common factor of 5. You can pull out the 5 to simplify mental math: $5 \times \sqrt{2^2 + 2^2 + 1^2} = 5 \times \sqrt{4 + 4 + 1} = 5 \times \sqrt{9} = 5 \times 3 = 15$. ### Common Pitfall Mistaking the "longest rod" for the diagonal of the floor ($10\sqrt{2}$), ignoring the height dimension completely. ### Final Answer Therefore, the correct answer is **15**.