Chapter 10.3 — Exercise 10.2 — Surface Area of Combined Solids
Surface areas of combination of solids. This is Lesson 3 of 5 in Chapter 10: Mensuration.
Where Two Solids Meet, a Face Disappears
A cone glued onto a hemisphere doesn't have the flat circular face of either piece visible anymore — both flat faces sit pressed together, hidden inside the join. Exercise 10.2 measures exactly what's left once those hidden faces are accounted for. Every single-solid formula from the introduction still applies here — nothing new needs to be learned — the entire skill is deciding which parts of each formula still count once two solids are pressed together.
Curved Surfaces Only, Added Together
Total surface area of a joined solid = sum of the curved (lateral) surface areas of every visible pieceThis is the one rule every problem in this exercise follows without exception: never total surface area for a piece that's joined to another, only its curved contribution. A cuboid or a cube is the one exception worth flagging early, since it has no "curved" surface to speak of at all — when a cube shows up joined to something else, whichever of its flat faces get hidden by the join simply get subtracted from its total surface area directly, rather than swapped for a curved-surface formula the way a cone or hemisphere would be.
A Toy: Cone on a Hemisphere
Base diameter 6 cm, cone height 4 cm — radius 3 cm either way, since the cone and hemisphere share the same circular base, exactly the toy pictured in the diagram above.
slant height l = √(3²+4²) = 5 cm
CSA of cone = πrl = 3.14×3×5 = 47.1 cm²
CSA of hemisphere = 2πr² = 2×3.14×9 = 56.52 cm²
Total surface area = 47.1 + 56.52 = 103.62 cm²Notice the radius 3 cm gets reused twice over — once for the cone's own slant-height calculation, once again for the hemisphere's curved surface area — precisely because the two solids share one single common base measurement between them. Reading a combined-solid problem carefully enough to spot which single measurement is actually shared between its pieces, rather than assuming every dimension needs restating separately, is worth doing before any formula gets touched at all.
Three Pieces at Once: Cylinder Between a Cone and a Hemisphere
Radius 8 cm throughout; cylindrical section 10 cm tall, conical section 6 cm tall. Three curved surfaces from three genuinely different solids, added together in one final step.
| Piece | Formula | Result |
|---|---|---|
| Cone (l=√(8²+6²)=10) | πrl = 3.14×8×10 | 251.2 cm² |
| Cylinder | 2πrh = 2×3.14×8×10 | 502.4 cm² |
| Hemisphere | 2πr² = 2×3.14×64 | 401.92 cm² |
Total = 251.2 + 502.4 + 401.92 = 1155.52 cm²Even with three pieces instead of two, nothing about the underlying rule changes — only the cylinder's two ends are hidden (one against the cone, one against the hemisphere), so only its curved side ever enters the sum, exactly as with the two-piece toy above. The slant height calculation is worth a second glance too: it reuses 8 and 6, the same radius and cone-height already given, landing on another clean Pythagorean triple (6-8-10) rather than an ugly surd — a recurring pattern throughout this chapter's cone problems, not a coincidence specific to this one.
A Capsule: One Cylinder, Two Hemispheres
Length 14 mm, width 5 mm — the width fixes the radius (2.5 mm) for the whole capsule shape, cylinder and hemispherical ends alike. Since both hemispherical ends eat into the capsule's overall length, the cylinder's own height has to be recovered first before anything else.
cylinder height = total length − 2×radius = 14 − 5 = 9 mm
Surface area = CSA of cylinder + 2×CSA of hemisphere = 2πrh + 2×2πr² = 2πr(h+2r)
= 2×3.14×2.5×(9+5) = 5×3.14×14 = 219.8 mm²Factoring out 2πr before substituting, rather than computing the cylinder and two hemispheres as three separate products, is worth doing here purely to save arithmetic — the same shortcut works for any capsule-shaped solid, a shape genuinely common enough in these problems to be worth a name of its own. It's also worth being deliberate about the units in this particular problem: every measurement is given in millimetres rather than the centimetres most other problems in this chapter use, so the final answer is naturally in square millimetres too, not something to convert unless a problem explicitly asks for it.
Four More Combined Solids, Compressed
| Solid | Key idea | Result |
|---|---|---|
| Two 64 cm³ cubes joined end to end | Side a=4 cm each; forms an 8×4×4 cuboid, TSA=2(lb+bh+lh) | 160 cm² |
| Storage tank: cylinder (d=1.4 m, len 8 m) + 2 hemispherical ends | 2πr(h+2r), then ×₹20 painting rate | 41.36 m², costing ₹827.20 |
| Sphere, cylinder, cone sharing radius r and height 2r | Volumes ⁴⁄₃πr³ : 2πr³ : ⅔πr³ | Ratio 2 : 3 : 1 |
| Hemisphere (diameter=a) scooped from one face of a cube (side a) | 6a² − πa²/4 (flat circle removed) + πa²/2 (curved surface added) | a²(6 + π/4) |
The cube-and-hemisphere row is worth a second look, since it's the one genuine exception to "only add curved surfaces" in this whole exercise. Scooping a hemisphere out of a cube's face removes a flat circular area (πr²/4, since the diameter equals the cube's side) but simultaneously reveals a curved hemispherical surface in its place — so this problem subtracts a flat area and adds a curved one, rather than simply adding two curved areas together the way every other problem here does. It's a genuine preview of the very next problem below, which scoops hemispheres out of a cylinder in exactly the same subtract-then-add spirit.
A Cylinder With Both Ends Scooped Out
A wooden article is made by scooping a hemisphere out of each end of a solid cylinder — length 10 cm, radius 3.5 cm, with each hemisphere's radius matching the cylinder's own radius exactly.
CSA of cylinder = 2πrh = 2×(22/7)×3.5×10 = 220 cm²
CSA of each hemisphere = 2πr² = 2×(22/7)×3.5×3.5 = 77 cm²
Total surface area = 220 + 2×77 = 374 cm² (curved cylinder + two curved hemispherical dents)This problem and the capsule problem discussed earlier share the identical final formula, 2πr(h+2r), despite describing opposite physical situations — one adds two solid hemispheres onto a cylinder's ends, the other scoops two hemisphere-shaped holes out of a cylinder's ends. Either way, each flat circular end gets replaced by a curved hemispherical surface instead, which is exactly why the same formula answers both. It's a genuinely useful thing to notice: the formula itself doesn't know or care whether material was added or removed, only that a flat circular face has been swapped for a curved one — the physical direction of the change never enters the mathematics at all.
From Curved Surfaces to Whole Volumes
Every result in this exercise added curved surfaces together, carefully leaving out whatever face two solids shared. Exercise 10.3 measures the same kinds of combined solids by volume instead — a genuinely simpler rule, since volume never has a hidden face to subtract at all, no matter how the solids happen to be joined together. For the six core formulas this exercise depended on throughout, revisit the chapter introduction.