Question 3

The shortest path on the box is a straight line on its net.

Step 1 — Define the Box

Let us imagine a rectangular box. Let its length be $L$. Let its width be $W$. Let its height be $H$. The spider starts at one corner. It wants to reach the farthest opposite corner.

Step 2 — Unfold the Box

The spider must walk on the box surfaces. It cannot fly through the air. To find the shortest path, we flatten the box. We unfold two adjacent faces of the box. These faces must connect the start and end corners. This unfolding creates a flat rectangle. The shortest path is a straight line across this rectangle. We use the Pythagorean theorem to find its length.

Step 3 — Calculate Path Options

There are three main ways to unfold two faces. Each way connects the opposite corners. Let us calculate the length for each way.

Path Option 1: Across the length and height faces. We unfold the 'front' face and the 'top' face. They share a common length $L$. The new rectangle has length $L$. Its width is the sum of height and width. This sum is $H+W$. Let $P_1$ be the length of this path.

$P_1 = \sqrt{L^2 + (H+W)^2}$

Path Option 2: Across the length and width faces. We unfold the 'front' face and a 'side' face. They share a common height $H$. The new rectangle has height $H$. Its length is the sum of length and width. This sum is $L+W$. Let $P_2$ be the length of this path.

$P_2 = \sqrt{(L+W)^2 + H^2}$

Path Option 3: Across the width and height faces. We unfold the 'bottom' face and a 'side' face. They share a common width $W$. The new rectangle has width $W$. Its length is the sum of length and height. This sum is $L+H$. Let $P_3$ be the length of this path.

$P_3 = \sqrt{(L+H)^2 + W^2}$

Step 4 — Find the Shortest Path Length

The shortest path is the smallest of these three values. We need to compare $P_1$, $P_2$, and $P_3$. The smallest value is the shortest path length. Let us use an example. Suppose the box has dimensions: Length $L = \mathbf{4}$ units. Width $W = \mathbf{3}$ units. Height $H = \mathbf{2}$ units.

Let us calculate $P_1$. $P_1 = \sqrt{4^2 + (2+3)^2}$ $= \sqrt{16 + 5^2}$ $= \sqrt{16 + 25}$ $= \sqrt{41}$

$\boxed{\approx \mathbf{6.40} \text{ units}}$

Let us calculate $P_2$. $P_2 = \sqrt{(4+3)^2 + 2^2}$ $= \sqrt{7^2 + 4}$ $= \sqrt{49 + 4}$ $= \sqrt{53}$

$\boxed{\approx \mathbf{7.28} \text{ units}}$

Let us calculate $P_3$. $P_3 = \sqrt{(4+2)^2 + 3^2}$ $= \sqrt{6^2 + 9}$ $= \sqrt{36 + 9}$ $= \sqrt{45}$

$\boxed{\approx \mathbf{6.71} \text{ units}}$

Comparing the three path lengths: $P_1 \approx \mathbf{6.40}$ units. $P_2 \approx \mathbf{7.28}$ units. $P_3 \approx \mathbf{6.71}$ units. The shortest path for this box is $P_1$.

Answer

(i) The shortest path is found by unfolding two adjacent faces of the box. (ii) These faces must connect the starting corner to the farthest opposite corner. (iii) The path is a straight line drawn across this unfolded flat rectangle. (iv) Its length is calculated using the Pythagorean theorem. (v) The shortest path length is the minimum of $\sqrt{L^2 + (H+W)^2}$, $\sqrt{(L+W)^2 + H^2}$, and $\sqrt{(L+H)^2 + W^2}$.