In-text — Shortest Paths on a Cuboid — Exploring Some Geometric Themes

1. 14 marksGanita Prakash Cl-8 Part 2, Math Talk, pages 84-87

   An ant on the surface of a cuboid wants to reach a laddu, travelling only along the surface. How can a net be used to find the shortest path, and why does the method work? Why does the way the cuboid is unfolded matter?

   Hint. Ask what happens to the length of a path when the cuboid is unfolded flat.

   The key fact that makes the method work. Unfolding a cuboid does not stretch its surface. So a path drawn on the surface transfers to a path of exactly the same length on the net, and every path on the net transfers back to a path of the same length on the cuboid.

   That converts a hard three-dimensional question into an easy two-dimensional one. On a flat plane, the shortest route between two points is the straight line between them — so:

   Unfold the cuboid, join the ant to the laddu with a straight line, and fold back. That is the shortest path.

   How to be sure a given path is shortest. Draw the net and see what the path becomes. · If it becomes a straight segment, it is the shortest for that unfolding, because nothing beats a straight line on a plane. · If it becomes a bent path, it is not the shortest — the straight segment between the same two points is shorter.

   This is what settles the chapter's two examples: the path shown in the first case straightens out on the net and is therefore shortest, while the path in the second case stays bent and is not.

   Why the unfolding matters — two separate traps.

   Trap 1: the straight line can leave the net. In one of the chapter's examples the segment joining ant to laddu passes outside the unfolded shape. A line outside the net corresponds to no path on the cuboid at all, so that unfolding gives no answer and a different one must be used.

   Trap 2: different unfoldings give different lengths. Each way of unfolding lays the faces out differently, so the straight-line distance between the two points changes. Each unfolding gives one candidate route.

   So the full method is:

   1. List the different ways of unfolding the cuboid so that ant and laddu both lie on the flattened net.
   2. For each, measure the straight-line distance — usually with the Baudhāyana theorem, since the two points sit at the corners of a right triangle.
   3. Discard any whose segment leaves the net.
   4. The smallest surviving distance is the answer.

   Notice this is the previous chapter doing the work: once the net is drawn, every one of these is a right-triangle calculation.

   ✦ Unfolding preserves lengths, so the shortest surface path becomes a straight line on the net. Draw the net, join the points with a straight segment, and fold back. Because different unfoldings place the two points differently — and a segment can even fall outside the net — every valid unfolding must be tried and the smallest distance taken.

2. 25 marksGanita Prakash Cl-8 Part 2, Try This, page 87

   A box measures 30 cm long, 12 cm high and 12 cm deep. An ant sits 1 cm below the top of one end face and a laddu is stuck 1 cm above the bottom on the far side, each 6 cm from the edge. Two unfoldings give straight-line distances of 42 cm and 40 cm. Verify these and explain how to find the true shortest path.

   Hint. Check where each number in the right triangle comes from — the legs are built by adding face dimensions and offsets.

   The first unfolding — 42 cm. In this net the two points end up at the same height, so the straight segment joining them is horizontal and its length is read off directly. It is made of 6 + 30 + 6 = 42 cm — the 6 cm offset on the first face, the full 30 cm length of the box, and the 6 cm offset on the last face. Since the segment is horizontal, no theorem is needed; the length is just the sum.

   The second unfolding — 40 cm. Here the two points are separated both across and up the net, so they sit at opposite corners of a right triangle. Its legs come out as horizontal: 1 + 30 + 1 = 32 cm vertical: 12 + 12 = 24 cm The horizontal leg crosses the 30 cm length plus the two 1 cm offsets; the vertical leg crosses two 12 cm faces.

   Now apply the Baudhāyana-Pythagoras theorem: d² = 24² + 32² d² = 576 + 1024 = 1600 d = √1600 = 40 cm ✓

   Both of the book's figures check out.

   Which is shorter? Of these two, 40 cm is shorter — going diagonally across two faces beats going the long way round at the same level, even though the diagonal route crosses more faces.

   Is 40 cm the answer? The chapter does not say so, and deliberately: it ends by pointing out that "in each of these unfoldings, the lengths of the line segments between the ant and the laddu are different! So we have to carefully list all the possible different unfoldings to find the answer!" It is left open as a Try This.

   To finish it properly you must enumerate every unfolding that puts both points on the flattened net, compute the straight-line distance for each with the theorem, discard any whose segment leaves the net, and take the minimum. What the two worked cases establish is that the answer is at most 40 cm.

   Why this problem is worth the effort. It shows that "shortest" on a curved-or-folded surface is not obvious: the winning route crosses faces you would not expect, and the only way to be sure is to flatten every possibility and compare. That idea — flatten, then use straight lines — is how shortest paths are handled on far more complicated surfaces later on.

   ✦ The 42 cm figure is 6 + 30 + 6 along a level line; the 40 cm figure comes from a right triangle with legs 24 = 12 + 12 and 32 = 1 + 30 + 1, since 24² + 32² = 1600. Of the two, 40 cm is shorter — and the book leaves the problem open, since every possible unfolding must be checked before the true minimum is known.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 2, Reprint 2026-27 (hegp204.pdf), where this is Chapter 4 (pages 70-102) — the eleventh chapter of the Class 8 course and the longest in the book. Like the rest of Part 2 it carries NO printed answer key, so every formula and count was derived and then independently recomputed in Python before being written: the Sierpinski Carpet recurrences (R_n = 8^n, H_n = (8^n - 1)/7), the Sierpinski Triangle counts (3^n and (3^n - 1)/2), the areas (8/9)^n and (3/4)^n, the Koch side count 3 x 4^n and perimeter 3 x (4/3)^n, and the face/edge/vertex formulas for prisms and pyramids (checked against Euler's relation for every case). THIS IS A HEAVILY VISUAL CHAPTER, so figure-only items are handled in one of two ways and never guessed. (1) MEASURED FROM THE RENDERED PAGE: the cube-stack count on page 97 was settled by rendering the figure at 400 dpi and observing that each row sits one step BACK as well as one step up (every bottom cube shows its full top face), which makes it a square-layered step pyramid of 16 + 9 + 4 + 1 = 30 cubes rather than the ten visible; and the three letters in the page-96 puzzle were read off the printed pixel glyphs at 700-900 dpi as C (front), A (top) and F (side). (2) FLAGGED AND ANSWERED BY METHOD: the six candidate cube nets, the projection-matching sets, the cube-combination views, the isometric figures to copy, the rolling ball and the impossible triangle are all printed diagrams; each solution gives the full method and reasoning and says plainly that the diagram is not reproduced. ONE ITEM IS LEFT OPEN BY THE BOOK ITSELF and is reported as such: the 30 x 12 x 12 shortest-path Try This on page 87, where the book computes 42 cm and 40 cm for two unfoldings (24^2 + 32^2 = 1600 verified) and then says all unfoldings must be listed to find the answer — so the solution establishes only that the shortest path is at most 40 cm. The tetracube count in the page-100 exercise was verified by exhaustive computer enumeration: 8 arrangements up to rotation, 7 up to rotation and reflection, of which 5 are flat.. Questions are referenced from the NCERT textbook for identification.