Question

The maximum length of a pencil that can be kept in a rectangular box of dimensions 8 cm x 6 cm x 2 cm, is

• A. \(2\sqrt{13}\) cm
• B. \(2\sqrt{14}\) cm
• C. \(2\sqrt{26}\) cm
• D. \(10\sqrt{2}\) cm

Hint:

The longest object that fits in a box lies along its space diagonal, found using the 3D Pythagoras formula.

Answer 1

The Correct Option is C

Instead of jumping straight to the three-dimensional formula, we can build the space diagonal in two stages using plain Pythagoras twice, which also shows why that formula works in the first place.

1. Step A, find the diagonal of the base: Take the base of the box as an 8 cm by 6 cm rectangle. Its diagonal is $\sqrt{8^2+6^2} = \sqrt{64+36} = \sqrt{100} = 10$ cm.
2. Step B, use this base diagonal with the height: The space diagonal of the box, the height (2 cm) and the base diagonal (10 cm) form a right triangle, since the height stands perpendicular to the base. So the space diagonal is $\sqrt{10^2+2^2} = \sqrt{100+4} = \sqrt{104}$.
3. Simplify: $\sqrt{104} = \sqrt{4 \times 26} = 2\sqrt{26}$ cm.

Both routes agree because $\sqrt{l^2+b^2+h^2}$ is really just Pythagoras applied twice, once across the base and once up through the height.

Let's summarize:

• The longest pencil that fits is $2\sqrt{26}$ cm, matching option C.

Building the answer in two steps like this is a good check whenever the single formula feels error prone.