Unroll a solid's skin onto the table and the area is one you already know.
A solid's surface is a thing you can touch — the cardboard of a box, the label on a soup can, the paper cone of an ice-cream wrapper. Cut that surface along a few edges and lay it flat on the table, and you get a net: the whole skin spread into one flat figure. Nothing was added and nothing was lost, so the surface area of the solid is just the area of its net — and area in the plane is something you have known since Stage 16. The whole job of this lesson is to unroll well: a box becomes six rectangles, a can becomes a rectangle, a cone becomes a sector. Once the net is flat, the formulas almost write themselves.
Take an ordinary cardboard box and run a knife along enough of its edges to open it out flat. The six faces fall into a connected paper shape — a net. Press it back together along the cut edges and you rebuild the box exactly, because cutting and folding move paper around but never stretch it. So the area of the flat net and the area of the box's surface are the same number.
That single observation is the engine of the whole lesson. Surface area = the total area of every face of the solid = the area of any net of it. For a flat-faced solid (a polyhedron) a net is literally its faces hinged at shared edges; for a curved solid the "net" is what you get by unrolling the curved part onto the plane, which we will do carefully for the cylinder and the cone.
Cutting and folding preserve area. So surface area = the area of the net. Every formula in this lesson is just "unroll, then add up flat areas."
One caution that will matter for a curved surface: a solid can have many different nets (a cube has eleven!), but they all have the same total area, because they are all the same skin laid out differently. We only need one good net for each solid.
The interesting unrollings are the lateral (side) surfaces. Let us watch three of them come off.
A prism. The side faces of a right prism are rectangles, all the same height h (the prism's height). Standing side by side around the base, they unroll into one long strip — a single rectangle whose height is h and whose width is the whole way around the base, the base perimeter. So the lateral area of a right prism is (base perimeter) × h.
A cylinder. A cylinder is a "prism with a circle for a base." Peel the label off a can and it lies flat as a rectangle. Its height is the can's height h; its width is once around the rim — the base circumference 2πr. Hence the lateral area of a cylinder is
lateral area = 2πr · h.
A cone. Here is the one that surprises people. Cut a paper cone up one slant line and flatten it. It does not become a triangle — it becomes a sector (a pizza slice). Why? Every point of the rim is the same straight-line distance from the apex, namely the slant height l; flattened, all those rim points sit at distance l from one centre — that is an arc of a circle of radius l. And the arc's length is exactly the rim it came from, the base circumference 2πr. A full circle of radius l has circumference 2πl and spans 360°, so the fraction we keep, and therefore the sector's angle, is
θ = 360° · 2πr2πl = 360° · rl (this is GX.coneNetAngle(r, l)).
The slant l is not the height h! The apex sits straight up over the centre, so l, h, and the base radius r form a right triangle and l = √(r² + h²) (GX.coneSlant(r, h)), straight from the Pythagorean theorem.
The cone's net is a sector, not a triangle, and its radius is the slant l, not the height h. If you ever flatten a cone into a triangle, you have torn the paper.
For any solid, gather the net's pieces into two groups and add:
S = (lateral area) + (area of the base or bases).
Right prism. Lateral area = base perimeter × height; then add the two congruent bases. For a box a × b × c that is 2(ab + bc + ca) — count the three pairs of opposite faces.
Regular pyramid. Its side faces are congruent isosceles triangles. Each has base = one base-edge and height = the slant height l of the pyramid (the apex-to-base-edge distance, measured up the middle of a face — not a lateral edge, and not the vertical height). The triangles together have total base = the whole base perimeter, so
lateral area of a regular pyramid = ½ · (base perimeter) · l,
and the full surface adds the one base. Notice the same flavour as the prism: "half the perimeter times a slant" instead of "perimeter times a height," because triangles carry a factor of ½.
A square pyramid has base edge 6 and face slant height l = 5. Base perimeter = 24, so lateral area = ½ · 24 · 5 = 60. The base is 6 · 6 = 36. Total surface = 60 + 36 = 96 square units.
Now we cash in the unrollings of §28.3.2. Add the flat lateral area to the circular base(s).
Cylinder. Two circular caps (each πr²) plus the rolled-out rectangle (2πr · h):
Scyl = 2πr² + 2πrh (GX.saCyl(r, h)).
Cone. One circular base (πr²) plus the sector, whose area is ½ · (arc) · (radius) = ½ · 2πr · l = πrl:
Scone = πr² + πrl, lateral area πrl (GX.saCone(r, h), with l = √(r²+h²)).
Truncated cone (frustum). Slice the tip off a cone parallel to the base and the lateral surface becomes a ring of a sector — the big sector minus the small one. Working it out, the lateral area is π(r₁ + r₂) l, where l is the frustum's slant height and r₁, r₂ are the two base radii; add both circles for the full surface.
A cone has r = 3, h = 4. Then l = √(3² + 4²) = 5. Lateral area = πrl = π·3·5 = 15π; base = πr² = 9π; total = 15π + 9π = 24π ≈ 75.4. Its net is a sector of radius 5 with angle 360°·(3/5) = 216°.
A sphere has no flat net at all — you cannot peel an orange and lay the peel flat without tearing it (mapmakers fight this same battle every day). So the sphere's surface area must be argued, not unrolled. The clean result, which we will take as a fact here and prove with slices in 28.4 and calculus later, is beautifully simple:
Ssphere = 4πr² (GX.saSphere(r)).
Read it as "exactly four great circles." A great circle (the equator, say) is a disk of area πr²; the whole sphere's skin equals four of them. That is a memorable check: the sphere is not πr² (one circle) and not 2πr² (two) — it is 4πr².
Three classic slips, all in red because they are traps: the cone's net is a sector of radius l (not a triangle, not radius h); a cylinder's lateral area is 2πr·h while a cone's is πr·l (don't swap the h and the l); and the sphere is 4πr², never πr² or 2πr².
One move runs the whole lesson: flatten the skin, then add up flat areas. The lateral surface of a prism or cylinder unrolls into a rectangle; a cone's into a sector of radius the slant l; the sphere alone has no net and is reasoned to 4πr². Here are the formulas worth memorizing.
| Solid | Lateral area | Total surface S | Net of the side |
|---|---|---|---|
| Right prism | (base perimeter) · h | lateral + 2·(base) | row of rectangles |
| Cylinder | 2πr·h | 2πr² + 2πrh | rectangle 2πr × h |
| Regular pyramid | ½·(base perimeter)·l | lateral + base | triangles round the base |
| Cone | πr·l | πr² + πrl | sector, radius l, angle 360°·r/l |
| Truncated cone | π(r₁+r₂)·l | lateral + πr₁² + πr₂² | ring of a sector |
| Sphere | — | 4πr² | no flat net |
Two safety rails: the cone uses the slant l = √(r²+h²), not the height; and a cone's net angle comes from arc = base circumference, giving θ = 360°·r/l.
S = 2πr² + 2πrh = 2π·16 + 2π·4·10 = 32π + 80π = 112π cm² ≈ 351.9 cm². (Two caps of 16π each, plus a rolled-out rectangle 2πr × h = 8π × 10 = 80π.)
(a) l = √(6² + 8²) = √100 = 10. (b) Lateral area = πrl = π·6·10 = 60π ≈ 188.5. (c) The arc equals the base circumference 2πr = 12π, and a full circle of radius 10 has circumference 20π, so θ = 360°·(12π/20π) = 360°·(6/10) = 216°. (Equivalently θ = 360°·r/l = 360°·6/10.)
All six faces are congruent a × a squares, so unfolding gives six equal squares hinged along shared edges, and S = 6a². A cube has 11 distinct nets — all different layouts of the same six squares, hence all of area 6a². (Different folding, same skin, same area.)
The face slant goes from the apex straight up over the centre, down to the midpoint of a base edge, which is half the base edge = 5 from the centre. So l = √(height² + 5²) = √(12² + 5²) = √169 = 13. Lateral area = ½·(base perimeter)·l = ½·40·13 = 260; base = 10·10 = 100; total = 260 + 100 = 360.
Sphere: S = 4πr². Cylinder lateral area = 2πr·h = 2πr·(2r) = 4πr². They are equal! (This is Archimedes' famous result — the sphere's skin exactly matches the side of its snug cylinder. The cylinder's two caps, 2πr² more, are the extra.)
Two mistakes. First, a cone's lateral surface is a sector, not a triangle — its rim points are all distance l from the apex, so they bow out into an arc. Second, the relevant length is the slant l, not the height h: l = √(3²+4²) = 5. Correct lateral area = πrl = π·3·5 = 15π ≈ 47.1, far from 6.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops surface area through nets, matching CCSS 6.G.A.4 (represent three-dimensional figures using nets and use them to find surface area) and the high-school modeling standards G-MG.A.1–3 (use geometric shapes and their measures to model objects). The cone and sphere formulas connect to G-GMD.A.1 (give informal arguments for the formulas — here, the cone net's arc equals the base circumference, and the sphere reasoned as four great circles). It leans on the Pythagorean theorem for the slant height and on arc length and sectors from Stage 18.
A good kitchen-table activity: cut open a cereal box, a paper towel tube, and a paper party hat, and lay each flat to see the rectangle and the sector appear in your hands. The single big idea — surface area is the area of the flattened skin — is worth saying out loud more than once.