An Ant Finds Its Shortest Route by Flattening the Box

By flattening the box in its mind. Once two faces are unfolded into one flat sheet, the shortest route between the two points is a straight line — which becomes a bent path when the box is folded back up.

A curved-looking problem turns into Pythagoras. This page covers everything in the CBSE Class 8 Mathematics chapter's second part: identifying solids from nets, counting vertices, edges and faces, testing whether a net closes, and shortest paths.

A net is a flat arrangement of faces that folds into a solid with no gaps and no overlaps. Count the shapes and their kinds, and the solid is decided.

- 6 squares — a cube
- 6 rectangles in matching pairs — a cuboid
- 2 triangles and 3 rectangles — a triangular prism
- 2 hexagons and 6 rectangles — a hexagonal prism
- 1 square and 4 triangles — a square pyramid
- 4 triangles — a triangular pyramid (tetrahedron)
- 2 circles and 1 rectangle — a cylinder, where the rectangle's length equals the circle's circumference
- 1 circle and 1 sector — a cone
- 8 equilateral triangles — an octahedron

The pattern is worth noticing. A prism has two identical end faces joined by rectangles; a pyramid has one base and triangles meeting at a single apex. So triangles all round means a pyramid or an octahedron, while rectangles in the middle means a prism.

An unfolded cardboard carton flattened for recycling is a cuboid net you can handle, and a cut-open cone from a party hat shows the sector clearly.

The count of faces alone is not enough, and that is what the next section tests. Both a square pyramid and a triangular prism have 5 faces — they are told apart by the kinds of face, since the pyramid has one square and four triangles while the prism has two triangles and three rectangles.

Count in groups rather than one at a time, and check the answer with Euler's relation $F+V-E=2$.

The standard solids:

- Cube — $F=6$, $V=8$, $E=12$. Check: $6+8-12=2$
- Cuboid — $6$, $8$, $12$
- Triangular prism — $F=5$, $V=6$, $E=9$. Check: $5+6-9=2$
- Square pyramid — $F=5$, $V=5$, $E=8$. Check: $5+5-8=2$
- Tetrahedron — $F=4$, $V=4$, $E=6$. Check: $4+4-6=2$
- Octahedron — $F=8$, $V=6$, $E=12$. Check: $8+6-12=2$
- Hexagonal prism — $F=8$, $V=12$, $E=18$. Check: $8+12-18=2$

General rules save memorising. For a prism on an $n$-sided base:

$$V=2n,E=3n,F=n+2$$

Testing on $n=3$: $V=6$, $E=9$, $F=5$ — the triangular prism, correct.

For a pyramid on an $n$-sided base:

$$V=n+1,E=2n,F=n+1$$

Testing on $n=4$: $V=5$, $E=8$, $F=5$ — the square pyramid, correct.

Counting a cuboid's edges in groups is the safe method: $4$ on the top, $4$ on the bottom and $4$ vertical, giving $12$ — far more reliable than ticking off twelve separate lines in a sketch. And Euler's relation applies to polyhedra only, so it will not check a cylinder or a cone.

Check the number of faces first, then mentally fold to see whether each face reaches a different position with no overlap.

For a cube the arrangement must have exactly 6 squares. That is necessary but not sufficient.

Works. Four squares in a row with one attached above and one below, in a cross shape. Folding the row into a ring gives the four side faces, and the two extra squares become the top and bottom. There are eleven different arrangements of six squares that fold into a cube.

Fails. Six squares in a straight line. Folding the row wraps it round, and after four squares the ring is complete — the fifth and sixth then overlap faces already in place, leaving the top and bottom open.

Fails. Six squares in a $2\times 3$ block. Folding it leaves faces overlapping and two openings.

Correcting a failed net. Take the straight line of six squares. Remove the last two from the end and attach them above and below the row instead, one each. Now four squares form the ring and the two moved squares close the top and bottom — a working net.

The same reasoning applies to other solids. A square pyramid needs one square with a triangle on each of its four sides; four triangles attached to just two sides of the square cannot close.

The test that actually decides it is folding, not counting, and it is why the eleven cube nets have to be recognised rather than derived from the number 6. Each of the six squares must land on a different face of the cube — so if two squares would arrive at the same place, the arrangement fails however many squares it has.

Unfold the cuboid so both points lie on one flat sheet, join them with a straight line, and find its length with Pythagoras. Different unfoldings give different lengths, so try each and take the smallest.

Worked example. A cuboid measures $4$ units long, $3$ wide and $2$ high. An ant travels across the surface from one corner to the opposite corner. There are three sensible unfoldings:

Unfolding 1 — across the length and width, then the height:

$$\sqrt{(4+3{)}^2+2^2}=\sqrt{49+4}=\sqrt{53}\approx 7.28$$

Unfolding 2 — across the length and height, then the width:

$$\sqrt{(4+2{)}^2+3^2}=\sqrt{36+9}=\sqrt{45}\approx 6.71$$

Unfolding 3 — across the width and height, then the length:

$$\sqrt{(3+2{)}^2+4^2}=\sqrt{25+16}=\sqrt{41}\approx 6.40$$

The shortest is $\sqrt{41}\approx 6.40$ units, from unfolding 3.

Worked example on a cube of side $2$. All three unfoldings give the same length:

$$\sqrt{(2+2{)}^2+2^2}=\sqrt{16+4}=\sqrt{20}\approx 4.47$$

A spider crossing a room from a floor corner to the opposite ceiling corner faces exactly this problem.

The pattern in the three answers is worth reading: pairing the two smallest dimensions inside the bracket gave the shortest route, because that keeps the squared sum lowest.

And note what this is not. The straight line through the inside of the cuboid is $\sqrt{4^2+3^2+2^2}=\sqrt{29}\approx 5.39$, which is shorter still — but an ant cannot tunnel, so the surface answer is the larger $\sqrt{41}$. Reading whether the path must stay on the surface decides which calculation the question wants.