Nets of solids and shortest paths

Take a cardboard cube, cut along a few edges, and press it flat — you get a cross-shaped pattern of six squares. That flat pattern is a net: the unfolded skin of a solid, laid out so folding it back along the cut edges rebuilds the solid exactly, with no gaps and no overlaps. Nets let you see a 3-D shape's flat "recipe," and they solve a puzzle that looks impossible at first: if a spider must crawl along the outside surface of a box from one corner to a far corner, what is the shortest route? Straight-line, through-the-air distance is useless here, because the spider cannot fly through the box. This article covers recognising and drawing nets of common solids — cube, cuboid, prism, pyramid, cylinder, cone, tetrahedron — and shows how unfolding a solid into its net turns a surface shortest-path problem into an ordinary straight-line problem, solved with the Baudhāyana–Pythagoras theorem.

The core insight

A net is a flat, non-overlapping unfolding of a solid's faces that reconstructs the solid on folding, and unfolding two points onto the same flat net turns "shortest path along a surface" into "shortest straight line," found with the Pythagoras theorem.

The core idea

A net is what results when a hollow 3-D solid is cut open along some of its edges and pressed flat, with every face fully visible, no face overlapping another, and every cut edge free to be folded back to rejoin its matching edge. The same solid can usually be unfolded in more than one way, so a solid can have several different-looking but equally valid nets — a cube alone has 11 distinct valid arrangements of its 6 squares.

Recognising a valid net. Not every arrangement of the right number of faces works. Fold the flat pattern mentally along every internal line: each edge of the finished solid must be covered by exactly one matching pair of edges, and no face may be left uncovered, doubled up, or poking through another face. A layout of 6 squares, for instance, can easily fail to be a cube net even though it has the correct face count, because two squares end up trying to occupy the same position when folded.

Nets of common solids.

  • A cube's net is 6 identical squares; a cuboid's net is 6 rectangles in 3 matching pairs, arranged similarly.
  • A prism's net is its two polygonal end-faces (bases) plus a strip of rectangular side faces joining them.
  • A pyramid's net is one polygonal base with a triangular side face attached along each base edge, all meeting at a single apex when folded up.
  • A cylinder's net is two circles plus one rectangle — the curved surface unrolled flat, its length equal to the circle's circumference.
  • A cone's net is one circle plus a sector of a larger circle — the curved lateral surface unrolled.
  • A tetrahedron's net is 4 triangles folding together at a common apex along shared edges.
Figure
One valid net of a cube shown as a cross of six squares with matching edges marked
A cube net — 6 squares that fold into one solid top left front right back bottom Fold each square up along its shared edge with front — all six close into a cube with no gap or overlap

Shortest path on a surface. Suppose an object must crawl along a solid's surface — the classic "spider and fly" problem — from one point to another. Aerial (straight-line) distance does not apply because the path is confined to the faces. The trick is to unfold the solid into a net so that the start point, the end point, and every face the path crosses all lie together on one flat sheet. On that flat sheet the shortest surface path becomes an ordinary straight line, and its length is found with the Baudhāyana–Pythagoras theorem,

a2+b2=c2,a^2 + b^2 = c^2,

where: aa and bb are the two legs of the right triangle formed on the flattened net by the horizontal and vertical runs between the two points, and cc is the straight-line (hypotenuse) distance between them on the net — which equals the shortest crawling distance on the solid, for that particular unfolding.

Derivation AShortest surface path on an unfolded cuboid

The goal is to show, for a spider crawling along the outer surface of a rectangular box from one bottom corner to the diagonally opposite top corner, that unfolding two adjacent faces of the box into a single flat rectangle converts the surface-crawling problem into a plain right-triangle problem, and that applying the Baudhāyana–Pythagoras theorem to the flattened legs gives the true shortest length of that particular route across the box's skin.

Step 1. Choose which two faces to unfold. Pick a pair of faces that share an edge and that, laid flat together along that shared edge, contain both the start corner and the end corner. For a box, a natural choice is the front face and the top face, joined along the top edge of the front face.

Step 2. Unfold along the shared edge. Rotate the top face outward about the shared edge until it lies in the same plane as the front face, with no stretching or tearing. The two rectangles now form one larger flat rectangle, and both the start corner (bottom of the front face) and the end corner (far corner of the top face) sit on this single flat sheet.

Step 3. Identify the horizontal and vertical runs on the flattened sheet. The horizontal run is the width shared by both faces (it does not change during unfolding). The vertical run is the front face's height plus the extra depth crossed on the top face, since the two faces are now stacked end to end in a straight line on the flattened sheet.

Step 4. Apply the Baudhāyana–Pythagoras theorem to the right triangle whose legs are this horizontal run aa and vertical run bb: c2=a2+b2c^2 = a^2 + b^2, so c=a2+b2c = \sqrt{a^2 + b^2}. This cc is the straight-line distance on the flat net, and because every point of that line still lies on the box's actual surface once refolded, cc is also the length of a genuine path a spider could crawl. Different choices of which two faces to unfold in Step 1 give different flattened rectangles and generally a different value of cc; the true shortest surface path is the smallest cc among all the sensible unfoldings tried.

Worked example

Question. A cuboid box has length 4 cm, breadth 3 cm, and height 5 cm. A spider sits at one bottom corner and wants to crawl along the surface to the diagonally opposite top corner. Using the "unfold into a net" method, find the shortest surface path for the unfolding that lays the front face and the top face into one flat rectangle, and check a second unfolding to see whether it does better.

Step 1. Unfold the front face (4 cm wide, 5 cm tall) and the top face (4 cm wide, 3 cm deep) along their shared 4 cm edge, so they lie flat in one plane.

Step 2. Find the two legs on the flattened sheet. The horizontal run stays 4 cm (the shared width). The vertical run is the front face's height plus the depth crossed on the top face: 5+3=85 + 3 = 8 cm.

Step 3. Apply the theorem: c2=42+82=16+64=80c^2 = 4^2 + 8^2 = 16 + 64 = 80, so c=80≈8.94c = \sqrt{80} \approx 8.94 cm.

Step 4 (Competitive stretch). Try the side face (3 cm wide, 5 cm tall) unfolded with the top face (3 cm wide, 4 cm deep) instead, joined along their shared 3 cm edge. Now the horizontal run is 3 cm and the vertical run is 5+4=95 + 4 = 9 cm, giving c2=32+92=9+81=90c^2 = 3^2 + 9^2 = 9 + 81 = 90, so c=90≈9.49c = \sqrt{90} \approx 9.49 cm — longer than the first unfolding, so it is not the better route here.

Answer: the shortest surface path found is 80≈8.94\sqrt{80} \approx 8.94 cm, from the front-and-top unfolding; the side-and-top unfolding gives a longer 90≈9.49\sqrt{90} \approx 9.49 cm, confirming that checking more than one unfolding matters when hunting for the true shortest path.

Common traps

  • Treating any arrangement of the right number of faces as a valid net — a layout of 6 squares can fail to fold into a cube even though the face count is correct.
  • Assuming a solid has only one correct net; a cube alone has 11 valid nets, so "the" net is not unique.
  • Using straight-line, through-the-air distance for a surface shortest-path problem instead of unfolding first — the spider is confined to the faces.
  • Checking only one unfolding when solving a shortest-path problem; different unfoldings can give different lengths, and the true shortest path is the smallest of all the values checked.

Ask the tutor: Want to sketch a few different valid nets of a cube, or set up your own spider-and-fly unfolding for a box of different dimensions? Ask the AI tutor to walk through it with you step by step.

How do I know I have tried the unfolding that actually gives the shortest path, and not just a shorter one

You cannot always be certain from a single unfolding — you have to compare the sensible candidates. For a spider crawling between two opposite corners of a cuboid, there are only a few genuinely different pairs of faces that could be unfolded together, so working out the length for each of these and picking the smallest is a reliable, complete check at this level. Class 8 problems generally use box-shaped solids where two or three candidate unfoldings are enough to be confident of the shortest route.

Why can I not just use the 3-D distance formula through the inside of the box

Because the spider is stuck on the outer surface — it cannot burrow through the box or fly across empty space inside it. The straight-line distance through the box's interior is always shorter than any surface path, so using it would give a wrong, physically impossible answer for a crawling creature. Unfolding the surface into a flat net is exactly the trick that respects the "stay on the surface" rule while still letting you use a straight-line method.

Does every solid have exactly one net, like a fixed unfolding pattern

No — most solids can be unfolded along different choices of edges, giving several different-looking but equally valid nets. A cube, for example, has 11 distinct valid nets, not just the familiar cross shape. What must stay true across all of them is that folding any valid net back up along its lines reconstructs the same solid with every face in its correct place and no gaps or overlaps.

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