Which faces vanish at the join, and why you cannot simply add

What to assume they know

• Decomposing an everyday object into the basic solids — recognising a composite object as an assembly of basic solids, and converting the given overall dimensions into part dimensions
• The Class IX formulas for the curved surface of a cylinder and of a cone, and for the surface of a sphere, together with the area of a circle and of a square
• That a hemisphere has two pieces of surface, a curved dome of area 2πr² and a flat circular face of area πr²
• Confidence that the area of a plane region can be added and subtracted like any other measured quantity
• Reading a figure to see which faces of a drawn object are actually on its outside

What they should be able to do

• Explain why a joined solid's total surface area falls short of what the two pieces' total surface areas would give
• Identify, for a given join, exactly which parts of which faces leave the outside of the solid
• Compute the surface area of a cylinder capped by hemispheres, where every flat face disappears
• Compute the surface area of a solid where a circular face sits inside a larger flat face, and explain why the total goes up rather than down
• Compute the surface area where two circular faces meet but differ in radius, and account for the ring that survives
• Treat a hollowed-out depression by the same reasoning as an added piece, and show that a dome added and a bowl of the same radius scooped both raise the area by πr²
• Decide, in a physical question, which faces are to be counted at all — the base of a tent, the underside of a bird-bath
• Set out a surface-area calculation as an audit of faces rather than as a remembered formula

Where it usually goes wrong

• "Total surface area of a composite = TSA of one piece + TSA of the other." This is the error the chapter interrupts itself to warn against on p. 164, and the playing top shows its size: 58.85 cm² claimed against 39.6 cm² actual, an overcount of exactly twice the contact circle.
• "So the rule is: add the two, then subtract two circles." Only when the two faces are equal circles laid on one another. For the block only one small circle goes; for the rocket the overlap is the smaller circle and a ring of the larger one survives. Memorising the subtraction is how a student gets the rocket wrong.
• "Covering part of a surface must reduce the area." Putting a dome on a cube increases it, because the dome's curved skin is twice the flat circle it covers. Ask the class to vote before revealing Example 2's 163.86 cm².
• "Hollowing something out must reduce the surface area." It increases it, and by the same πr² as adding a dome would. Solid material was removed but boundary was created.
• "The cone sits on the cylinder, so the cone's base is hidden." Only the part of it that is actually in contact. Fig. 12.8's plan view shows the ring left over and the ring has to be painted.
• "CSA and TSA are interchangeable if you are careful." They differ by exactly the flat faces, which are the objects this whole topic is about. Write which one you mean at every line.
• "Every flat face of the finished object gets counted." The tent's floor is excluded because the question says so; the bird-bath's underside is excluded because it stands on legs. Deciding what counts is part of the problem, not a detail before it.
• "The hemisphere in the bird-bath adds a flat ring at the rim." It does not — the hollow and the cylinder share a radius, so the rim has no width. If the radii differed, that ring would be real and would have to be added.

Questions to check understanding

• Find the surface area of a cylinder capped by a hemisphere at one or both ends, given an overall length
• Find the surface area of a cube carrying a hemisphere, and state the largest hemisphere the face allows
• Find the surface area of a solid with a hemispherical or conical hollow, given the dimensions of the block it was cut from
• Explain, in words, why the answer to the previous question exceeds the surface area of the uncut block
• Handle a join where the two circular faces have different radii, and identify the ring that remains exposed
• Compute an area of material and then its cost at a given rate per square metre
• Decide which faces a stated physical situation requires you to count
• Demonstrate, on a named pair of solids, that adding the two pieces' total surface areas overshoots the joined solid's own

Examples worth working on the board

Inputs only. Values marked verified are worked out here on data printed inside pp. 161–170.

• Fig. 12.4, the assembly strip (§12.2, p. 162). Read from a close-up taken wide enough to include the caption and both arrows. Three stages left to right. First: a hemisphere with its flat face drawn as a hatched ellipse facing right, a bare cylinder, and a second hemisphere with its flat face hatched and facing left. An arrow. Second: the three pushed together, the two hatched ellipses still drawn, now squeezed at the two joins. A second arrow. Third: the finished tank, with the joins shown as plain curves and no hatching anywhere. This is the entire argument of the topic printed as a picture — the hatching marks precisely the faces that are about to stop existing on the outside.
• The general statement, in the explanation's words. The finished solid's outside is the two pieces' outsides minus the region where they are in contact, counted once for each piece. So the correction to subtract is twice the contact region, never twice the larger face and never a fixed πr². Everything below is an instance.
• Case one, the perfect match. For the tanker (§12.2, p. 162) and for the capsule of Exercise 12.1 q. 6 (p. 166, Fig. 12.10), the hemispheres' flat faces and the cylinder's ends are equal circles laid exactly on one another, so all four discs leave the outside and the answer is built only from curved parts. Capsule inputs: overall length 14 mm, diameter 5 mm, π as 22/7. Verified: radius 2.5 mm, cylinder length 14 − 5 = 9 mm, and the surface works out to 2πr(h + 2r) = 2 × (22/7) × 2.5 × 14 = 220 mm². Note the shape of that expression: h + 2r is the whole 14 mm length, so the capsule's outside measures exactly what the curved wall of a plain 14 mm cylinder of the same radius would — a small result the explanation can pose as a puzzle before proving it.
• The chapter's own warning (p. 164, immediately after Example 1). Having computed the playing top's area as about 39.6 cm², the page stops to say in as many words that this is not what you get by adding the total surface areas of a cone and a hemisphere. Verified, so the explanation can show the gap: the top has r = 1.75 cm, cone height 3.25 cm and slant height taken as 3.7 cm, giving 2πr² + πrl = 19.25 + 20.35 = 39.6 cm². Adding the two solids' total areas instead gives (2πr² + πr²) + (πrl + πr²) = 39.6 + 2πr² = 39.6 + 19.25 = 58.85 cm², too big by 19.25 cm² — which is exactly twice the 9.625 cm² of the contact circle. The overcount has a name and a size, and it is the thing the student can see.
• Case two, a circle inside a square — Example 2 (p. 164, Fig. 12.7). Inputs: cube of edge 5 cm, hemisphere of diameter 4.2 cm on its top face, π as 22/7. Here the dome's flat face lies wholly inside the cube's top face, so only a 4.2 cm circle leaves the outside while the rest of the top face stays. Verified: cube surface 6 × 25 = 150 cm²; remove the circle, πr² = (22/7)(2.1)² = 13.86 cm²; add the dome, 2πr² = 27.72 cm²; total 150 − 13.86 + 27.72 = 163.86 cm². The two corrections collapse to a single +πr², so the block's surface is larger than the bare cube's by 13.86 cm². Students expect covering something up to reduce the area; make them predict first, then show them.
• The same structure in Exercise 12.1 q. 4 (p. 166). Inputs: cube of side 7 cm with a hemisphere on top, and the question first asks for the largest dome the face can carry. Verified: the dome's circle must sit inside a 7 cm square, so the greatest diameter is 7 cm and the radius 3.5 cm; the surface is then 6(49) + π(3.5)² = 294 + 38.5 = 332.5 cm². The same +πr² again.
• Case three, an imperfect match — Example 3 (p. 165, Fig. 12.8). Inputs: cone of base diameter 5 cm and height 6 cm standing on a cylinder of base diameter 3 cm, overall height 26 cm, π as 3.14; the cone is to be painted one colour and the cylinder another. Verified: the cone's base circle, radius 2.5 cm, is wider than the cylinder it rests on, radius 1.5 cm, so a flat ring of the cone's base is still exposed and still needs paint. Its area is π(2.5² − 1.5²) = 4π cm². The orange region is πrl + that ring = π(16.25 + 4) = 3.14 × 20.25 = 63.585 cm² with l = 6.5 cm. The yellow region is the cylinder's curved surface plus the one circular end that is on the bottom of the whole rocket, π(1.5)(2 × 20 + 1.5) = 4.71 × 41.5 = 195.465 cm². Two different faces of the same solid are handled two different ways in one example, which is why this is the example to build the section on.
• Hollowed solids, from the same principle. Exercise 12.1 q. 8 (p. 167): cylinder of height 2.4 cm and diameter 1.4 cm with a cone of matching height and diameter drilled out, answer wanted to the nearest cm². Verified: r = 0.7, cone slant height √(0.7² + 2.4²) = √6.25 = 2.5 exactly; the outside is the cylinder's curved surface, its intact bottom disc, and the cone's curved surface where the flat top used to be, giving πr(2h + r + l) = 2.2 × 8 = 17.6 cm², so 18 cm² to the nearest square centimetre. Exercise 12.1 q. 9 (p. 167, Fig. 12.11): a cylinder 10 cm tall, radius 3.5 cm, with a hemisphere scooped from each end. Verified: one curved cylinder and two domes give 2πr(h + 2r) = 2 × 11 × 17 = 374 cm².
• Example 4, the bird-bath (p. 166, Fig. 12.9). Inputs: cylinder of height 1.45 m, radius 30 cm, with a hemisphere hollowed into the upper end, π as 22/7. Verified: in centimetres, curved wall plus curved bowl come to 2πr(h + r) = 2 × (22/7) × 30 × 175 = 33 000 cm², which is 3.3 m². Two judgement calls sit inside that. First, the depression has the same radius as the cylinder, so the rim is a circle of no width and there is no flat annulus on top to add. Second, the cylinder's flat underside is not counted; read from the printed page, the figure draws the bath raised on three legs, which is presumably the reason, but the page does not argue for it.
• The equality worth the whole topic. Take any flat face and a radius r that fits on it. Stick a hemisphere on: lose πr², gain 2πr², net +πr². Scoop an identical hemisphere in: lose πr² of flat face, gain 2πr² of curved bowl, net +πr². Verified as algebra, and confirmed numerically against the chapter's own pair: Exercise 12.1 q. 4, a 7 cm cube with a dome added, gives 294 + πr²; q. 5, a cube of edge l with a hemisphere of diameter l hollowed out of one face, gives 6l² + π(l/2)² = (l²/4)(24 + π). Set l = 7 in the second and it returns 294 + 38.5 = 332.5 cm², the identical answer. The chapter prints the two questions one after the other and never remarks on it.
• Which faces the question wants at all. Exercise 12.1 q. 7 (p. 167): a tent of cylindrical wall 2.1 m high and 4 m across with a conical top of slant height 2.8 m; canvas costs ₹500 per m² and the floor is explicitly not covered. Verified: πr(2h + l) = (22/7)(2)(4.2 + 2.8) = 44 m², so the canvas costs ₹22 000. The floor exclusion is not geometry, it is the physical question, and it is stated in the problem rather than derived.
• Two further inputs for the same case-one drill. Exercise 12.1 q. 1 (p. 166): two cubes of volume 64 cm³ each set end to end. Verified: each has edge 4 cm, so the result is an 8 × 4 × 4 cuboid with surface 2(32 + 16 + 32) = 160 cm². Exercise 12.1 q. 3 (p. 166): a cone of radius 3.5 cm on a hemisphere of matching radius, overall height 15.5 cm. Verified: cone height 12 cm, slant height 12.5 cm exactly, and the surface is πr(l + 2r) = 11 × 19.5 = 214.5 cm².
• Exercise 12.1 q. 2 (p. 166), the one that tests the word inner: a hollow hemisphere of diameter 14 cm with a hollow cylinder standing on it, overall height 13 cm, inner surface wanted. Verified: r = 7 cm, cylinder height 13 − 7 = 6 cm, inner surface 2πr(r + h) = 2 × (22/7) × 7 × 13 = 572 cm². A vessel has an inside as well as an outside, and the question chooses.

Figures to have open

• Fig. 12.4 redrawn as a movement rather than a strip: the hatched flat faces should visibly vanish at the instant of contact. This is the chapter's own figure (p. 162) and it is the single most important picture in the topic.
• The playing top with its contact circle drawn as a separate disc lifted out of the solid, so the doubled overcount can be pointed at. Standard schematic, built on Fig. 12.6's data (p. 163).
• The cube-and-dome with the top face treated as a running ledger: full square, then square with a hole, then square with a hole and a dome. Built on Fig. 12.7 (p. 164).
• The rocket's plan view — two concentric circles, the annulus shaded — enlarged well beyond its printed size. This is the chapter's own second panel of Fig. 12.8 (p. 165) and section 7 cannot be taught without it.
• A side-by-side of a dome added to a face and a bowl scooped into the same face, both annotated −πr² then +2πr². Standard schematic; not in the book, since the chapter never puts the two together.
• The bird-bath (Fig. 12.9, p. 166) drawn with its legs, so the excluded underside is visibly off the ground.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 12 "Surface Areas and Volumes", §12.2 "Surface Area of a Combination of Solids", pp. 162–166, with Fig. 12.4 (p. 162), Fig. 12.5 (p. 163), Fig. 12.6 (p. 163), Fig. 12.7 (p. 164), Fig. 12.8 (p. 165) and Fig. 12.9 (p. 166)
• The remark closing Example 1, p. 164, which denies that the top's area can be had by adding the two solids' total surface areas
• Exercise 12.1 questions 1–9, pp. 166–167, with Fig. 12.10 (p. 166) and Fig. 12.11 (p. 167)
• §12.3's opening paragraph, p. 167, which states in retrospect that part of the surface was lost in joining — the chapter's clearest sentence on this topic sits in the next section, and the explanation may want to quote its substance here
• The chapter's summary, §12.4, p. 170, first point