Nets of 3D shapes: Worked examples — Maths deep dive

This page is the "put it into practice" companion to the parent note on nets. There we built each shape's net from first principles; here we use those nets on every kind of problem a test can throw at you. If you want to re-read how a formula was born, jump back to the parent. If you want to drill, stay here.

The scenario matrix

Before any worked example, let's list every kind of case this topic contains. A "case class" is a situation that behaves differently from the others — so if we solve one example from each row, nothing on a test can surprise us.

#	Case class	What makes it different	Covered by
A	Plain cube / prism	All faces flat rectangles, straight adding	Ex 1, Ex 2
B	Curved surface must be unrolled	A rectangle's width = a circumference, not a length	Ex 3 (cylinder)
C	Curved surface becomes a sector	Slant height ≠ vertical height; pie-slice geometry	Ex 4 (cone)
D	Slant height is hidden — must use Pythagoras	You are given vertical height, not slant	Ex 5 (pyramid)
E	Degenerate / zero input	A dimension is 0 → a face vanishes; limiting behaviour	Ex 6
F	Reverse problem	Given the area, find a missing length	Ex 7
G	Real-world word problem	Extra distractors, units, "how much material"	Ex 8
H	Exam twist / validity	Is this pattern even a legal net? Counting/folding logic	Ex 9

Prerequisites you'll lean on: Surface Area of 3D Shapes, Circles and Circumference, Triangles and Area, and — crucially for rows D — Pythagoras Theorem.

Recall Two number facts we reuse constantly

Circle of radius $r$ : area $::: π r^{2}$ , and distance around it (circumference) $::: 2 π r$ .

Case A — Plain prism (straight adding)

Worked example Ex 1 — Cuboid gift box (cell A)

A closed box is $l = 8$ cm long, $w = 5$ cm wide, $h = 3$ cm tall. How much cardboard (surface area) does its net contain?

Forecast: Guess — will the answer be nearer $100$ , $150$ , or $250$ cm²? Write your guess down.

Step 1. Name the three pairs of faces the net shows. A cuboid net is six rectangles: two of $l \times w$ , two of $l \times h$ , two of $w \times h$ . Why this step? Opposite faces of a cuboid are identical, so counting in pairs saves work and matches the flat net exactly.

Step 2. Compute each pair. $2 l w = 2 (8) (5) = 80, 2 l h = 2 (8) (3) = 48, 2 w h = 2 (5) (3) = 30$ Why this step? Each product is one rectangle's area; the $2$ is because that rectangle appears twice on the net.

Step 3. Add. $SA = 80 + 48 + 30 = 158 cm^{2}$ Why this step? Surface area = total area of the flat paper = sum of all pieces.

Verify: Units are cm·cm = cm² ✓. Sanity: the biggest face is $8 \times 5 = 40$ ; six mid-sized faces landing near $158$ is reasonable. Our forecast bucket was " $\approx 150$ ."

Case B — Unrolling a curved surface

Worked example Ex 3 — Full cylinder can (cell B)

A cylinder has radius $r = 4$ cm and height $h = 9$ cm. Find the total net area.

Forecast: Which is bigger — the two round lids together, or the curved wall?

Step 1. Two circular lids: $2 π r^{2} = 2 π (4)^{2} = 32 π$ cm². Why this step? The net has a top circle and a bottom circle, each of area $π r^{2}$ .

Step 2. Unroll the wall into a rectangle. Its height is $h = 9$ ; its width is the circumference $2 π r = 8 π$ . Why width = circumference? Look at the figure: the wall wraps once around the circular rim, so when flattened its width must equal the distance around that rim. This is the one idea that makes cylinders different from prisms.

Step 3. Wall area = width × height $= 8 π \times 9 = 72 π$ cm². Why this step? Area of a rectangle is base times height, and we just found both.

Step 4. Total: $32 π + 72 π = 104 π \approx 326.73$ cm². Why this step? Add all flat pieces.

Verify: Curved wall $72 π$ > lids $32 π$ , matching the tall (h=9) can — forecast confirmed. $104 π \approx 326.73$ ✓, units cm² ✓.

Case C — Curved surface becomes a sector

Before Ex 4, one word we will lean on: the radian.

Worked example Ex 4 — Ice-cream cone (cell C)

A cone has base radius $r = 5$ cm and slant height $l = 13$ cm. Find (a) the total net area and (b) the sector's opening angle.

Forecast: The unrolled curved part is a pie slice. Will it be a thin sliver or nearly a full circle?

Step 1. Base circle: $π r^{2} = π (5)^{2} = 25 π$ cm². Why this step? The flat base is an ordinary circle.

Step 2. Curved surface = a sector whose radius is the slant height $l$ , and whose arc length equals the base's circumference $2 π r = 10 π$ . Why arc = circumference? Unrolling the cone, the curved rim lies along the sector's arc; that rim was the base circle, so the arc length must equal $2 π r$ . See the red arc in the figure.

Step 3. Sector area $= π r l = π (5) (13) = 65 π$ cm². Why this formula? A sector's area is $\frac{1}{2} (radius) (arc length) = \frac{1}{2} (l) (2 π r) = π r l$ .

Step 4. Total: $25 π + 65 π = 90 π \approx 282.74$ cm². Why this step? Surface area = total flat paper = base circle + unrolled sector, so we add the two pieces we just found.

Step 5 (angle). Using the radian rule from the definition box, $arc = l θ$ , so $θ = \frac{arc}{l} = \frac{2 π r}{l} = \frac{10 π}{13} \approx 2.4166$ rad. Convert to degrees with $\times \frac{18 0 ^{\circ}}{π}$ : $θ \approx 138.4 6^{\circ}$ . Why this step? Because a radian is defined by $arc = r θ$ , dividing the known arc ( $2 π r$ ) by the sector's radius ( $l$ ) directly gives the opening angle — it measures how "open" the pie slice is.

Verify: $θ \approx 138. 5^{\circ}$ is less than $36 0^{\circ}$ — a real slice, not a full disc, as forecast. Curved area $65 π$ > base $25 π$ , sensible for a long slant. $90 π \approx 282.74$ ✓.

Case D — Slant height is hidden (Pythagoras)

Worked example Ex 5 — Square pyramid given

vertical height (cell D) A square pyramid has base side $a = 6$ cm and vertical height $H = 4$ cm. Find the net area.

Forecast: Danger! The triangular faces need the slant height $s$ , not $H$ . Will $s$ be bigger or smaller than $4$ ?

Step 1. Find the slant height $s$ with Pythagoras Theorem. Look at the right triangle in the figure: it stands the apex ( $H = 4$ ) above the midpoint of a base edge, which is $a /2 = 3$ across. $s = H^{2} + (a /2)^{2} = 4^{2} + 3^{2} = 25 = 5 cm$ Why this step? On the flat net each side triangle has height = the slanted distance up the face, which is longer than the vertical drop $H$ . That distance is the hypotenuse of the (vertical height, half-base) right triangle.

Step 2. Base area: $a^{2} = 36$ cm². Why this step? The base is a square of side $a$ .

Step 3. Four triangles: $4 \cdot \frac{1}{2} a s = 2 a s = 2 (6) (5) = 60$ cm². Why this step? Each face triangle has base $a$ and height $s$ ; there are 4 of them (Triangles and Area).

Step 4. Total: $36 + 60 = 96$ cm².

Verify: $s = 5 > 4 = H$ ✓ (slant is always the longest, as forecast). $3$ - $4$ - $5$ is a clean Pythagorean triple ✓. Total $96$ cm² ✓.

Case E — Degenerate / zero input (limiting behaviour)

Worked example Ex 6 — What happens as a dimension shrinks to 0? (cell E)

Take the cuboid formula $SA = 2 l w + 2 l h + 2 w h$ with $l = 8, w = 5$ but let the height $h \to 0$ .

Forecast: A box with zero height isn't a box — it's a flat sheet. What area should survive?

Step 1. Substitute $h = 0$ : $SA = 2 (8) (5) + 2 (8) (0) + 2 (5) (0) = 80 + 0 + 0 = 80 cm^{2}$ Why this step? Setting $h = 0$ collapses the four side walls (each had a factor of $h$ ) to zero area — they literally vanish.

Step 2. Interpret: only the top and bottom $l \times w$ rectangles remain, and they now lie exactly on top of each other. Why this step? $2 l w = 2 (40) = 80$ = two flat $8 \times 5$ sheets. The "solid" has degenerated into a doubled flat rectangle — the limiting shape of a box being squashed.

Step 3 (contrast — cone tip). For a cone $SA = π r^{2} + π r l$ , let $r \to 0$ : both terms $\to 0$ . A cone with zero base radius is just its apex point — no surface, area $0$ . Why this step? Every term of the cone's area carries a factor of $r$ ( $π r^{2}$ has $r^{2}$ , and $π r l$ has $r$ ), so both shrink to $0$ as $r \to 0$ — the base circle and the curved sector each collapse to nothing, leaving only the apex point. This confirms the formula matches the geometry at the degenerate limit.

Verify: As $h \to 0$ , side terms $\to 0$ and SA $\to 80$ ✓. As $r \to 0$ , cone SA $\to 0$ ✓. Degenerate inputs give the geometrically obvious flat/point answers — no formula blows up.

Case F — Reverse problem (given area, find a length)

Worked example Ex 7 — Back out the height of a can (cell F)

A closed cylinder has radius $r = 3$ cm and total surface area $78 π$ cm². Find its height $h$ .

Forecast: We know the answer is "nice." Will $h$ be about $3$ , $10$ , or $30$ ?

Step 1. Write the formula and plug what we know: $2 π r^{2} + 2 π r h = 78 π \Rightarrow 2 π (9) + 2 π (3) h = 78 π$ Why this step? The net's total area is the two lids plus the wall; we set that equal to the given number.

Step 2. Simplify: $18 π + 6 π h = 78 π$ . Why this step? $2 π (9) = 18 π$ (lids), $2 π (3) = 6 π$ (wall coefficient).

Step 3. Solve: $6 π h = 60 π \Rightarrow h = \frac{60 π}{6 π} = 10 cm$ Why this step? Subtract the lid area from both sides, then divide by the wall coefficient — pure algebra to isolate $h$ .

Verify: Put $h = 10$ back: $18 π + 6 π (10) = 18 π + 60 π = 78 π$ ✓. This is exactly the soup-can from the parent note, worked backwards. Forecast " $\approx 10$ " ✓.

Case G — Real-world word problem

Worked example Ex 8 — Painting an open-top water tank (cell G)

A rectangular metal tank is $l = 2$ m, $w = 1.5$ m, $h = 1$ m. It is open at the top (no lid). Paint covers $8$ m² per litre. How many whole litres are needed to paint the outside?

Forecast: "Open top" is the trap — do we include the top face or not?

Step 1. Decide which faces exist. Open top ⇒ drop one $l \times w$ face. Full box would be $2 l w + 2 l h + 2 w h$ ; remove one $l w$ : $Area = l w + 2 l h + 2 w h$ Why this step? The net of an open box is missing the top rectangle. Never blindly reuse the closed formula — read the words.

Step 2. Substitute: $= (2) (1.5) + 2 (2) (1) + 2 (1.5) (1) = 3 + 4 + 3 = 10 m^{2}$ Why this step? Each term is one (or one pair of) real face(s) that actually needs paint.

Step 3. Litres needed: $10 \div 8 = 1.25$ litres → round up to $2$ whole litres. Why round up? You cannot buy $1.25$ litres in whole units; $1$ litre leaves bare metal, so you need $2$ .

Verify: Units m²/(m² per litre) = litres ✓. Bottom $3$ + long sides $4$ + short sides $3$ = $10$ ✓. Answer: buy 2 litres.

Case H — Exam twist: is it even a valid net?

Worked example Ex 9 — Legal net check (cell H)

A candidate pattern is six squares in a single straight strip (a 1×6 row). Two questions: (a) does it have the right count? (b) does it fold into a cube?

Forecast: Count looks right — but does counting alone prove it works?

Step 1. Count the faces: $6$ squares. Why this step? A cube must unfold into exactly 6 squares. Count passes.

Step 2. Fold-test the arrangement. Wrap the strip around: squares $1, 2, 3, 4$ form the four sides (a tube), then $5$ would be a 5th wall in the same ring — there is nowhere for it to become the top/bottom. Why it fails: to close a cube you need faces branching off the ring for the top and bottom lids. A straight strip has no branch, so squares $5$ and $6$ overlap existing walls. Look at the figure: after four folds you've made an open tube and two leftover squares.

Step 3. Conclude: not a valid cube net, despite the correct count. A correct net needs the four "wall" squares in a row plus a top and bottom hanging off the sides of that row (the classic cross/T shapes). Why this matters: the correct count of $6$ is necessary but not sufficient — connectivity/folding must also work. Only $11$ of the many 6-square arrangements are legal cube nets.

Verify: Face count $= 6$ ✓ but folding fails ✓ → invalid. This is the exam classic: "It has six squares, so it must be a cube net" is the wrong reasoning.

Recall Self-test (cover the answers)

Cuboid $8 \times 5 \times 3$ surface area? ::: $158$ cm² Cube edge $7$ surface area? ::: $294$ cm² Cylinder $r = 4, h = 9$ total area? ::: $104 π \approx 326.73$ cm² Cone $r = 5, l = 13$ : total area and sector angle? ::: $90 π \approx 282.74$ cm² and $\approx 138.4 6^{\circ}$ Pyramid $a = 6$ , vertical height $4$ : slant and total? ::: $s = 5$ ; total $96$ cm² Open-top tank $2 \times 1.5 \times 1$ paint area, and litres at 8 m²/L? ::: $10$ m²; buy $2$ litres Six squares in a straight line — valid cube net? ::: No (count ok, folding fails)

See also: Volume of 3D Shapes · Platonic Solids · Tessellations · 1.2.15 Nets of 3D shapes (Hinglish)