Decomposing an everyday object into the basic solids

What to assume they know

• The names cuboid, cone, cylinder, sphere, hemisphere, and the fact that each is determined by very few numbers — an edge triple, a radius and height, a radius alone
• The Class IX surface-area results for those solids: the curved surface of a cylinder, the curved surface of a cone in terms of its slant height, the surface of a sphere, and the six faces of a cuboid
• That a cone's slant height is recovered from its radius and its vertical height by Pythagoras, and is not one of the numbers a problem usually gives you
• Radius as half the diameter, and the habit of checking which of the two a figure has actually labelled
• Working fluently in centimetres and metres within one problem, and converting between them before combining

What they should be able to do

• Look at a drawn or described object and name the basic solids it is assembled from, and how many of each
• Say which dimension of the composite is shared by two pieces because they meet along a face, and write the equation that says so
• Split a stated overall height or length into the heights of the separate pieces
• Recover a cone's slant height from the radius and vertical height that a problem actually supplies
• Distinguish a piece added on top of a solid from a piece hollowed out of it, and say which of the chapter's objects is which
• Read a labelled figure correctly as to whether the marked measurement is a radius or a diameter
• Set up the decomposition of an object before computing anything, and state what still has to be found before a formula can be used
• Explain why the chapter's method is a reduction to problems already solved rather than a new technique

Where it usually goes wrong

• "There must be a formula for a top / a capsule / a rocket." There is not, and looking for one is the mistake the chapter is designed to break. The object is measured by measuring its pieces.
• "The number I was given is the number the formula wants." Almost never. A problem gives the height of the whole top and the formula wants the height of the cone; it gives a diameter and the formula wants a radius. Make the translation an explicit written step before any formula appears.
• "Slant height is one of the given measurements." It is supplied in the tent question and in essentially nothing else. Everywhere else it has to be built from the radius and the vertical height, and forgetting this is where most of the chapter's arithmetic goes wrong.
• "The two pieces meeting at a join always share a radius." They share one only when they are matched deliberately, as the toy in Fig. 12.5 is. The rocket's cone and cylinder do not, and the whole point of that example is that they do not.
• "A hemisphere is a basic solid, so it will be in the opening figure." Fig. 12.1 draws four solids and the hemisphere is not among them, though the chapter uses hemispheres more than anything else and its summary on p. 170 lists five. This is worth pointing out rather than papering over.
• "Scooping and sticking are different problems." They are the same decomposition run with a different sign, and the chapter puts a hollowed bird-bath among a run of assembled objects precisely to make that point.
• "Mixed units will come out in the wash." They will not. A bird-bath given as 1.45 m and 30 cm has to be brought to one unit before the pieces can be combined, and the answer then has to be converted back to whatever the question wants.

Questions to check understanding

• Name the basic solids an illustrated object is made from, and how many of each
• Given a total height and one part's dimension, find the other part's height
• Given a diameter across the widest part of a joined solid, state the radius each piece is built on
• Find a cone's slant height from its radius and vertical height, inside a larger composite problem
• State the largest hemisphere that can sit on a given cube face, and justify the bound
• Convert between metres and centimetres inside a composite-solid problem before combining pieces
• Given a figure, say whether a labelled length is a radius or a diameter, and what goes wrong if it is read the other way

Examples worth working on the board

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

• Fig. 12.1 (§12.1, p. 161). Read from the printed page and confirmed on a close-up taken wide enough to include the caption and the panel labels: four solids in a row, labelled (i) to (iv) beneath — a cuboid drawn in three-quarter view, a cone, a cylinder, a sphere. There is no fifth panel and no hemisphere among them. The hemisphere arrives two paragraphs later, in the description of the tanker, and by the time the chapter reaches its summary on p. 170 the list of basic solids has grown to five. Worth showing the student: the chapter's own inventory expands as soon as it meets a real object.
• Fig. 12.2, the tanker (§12.1, p. 161). Read from a close-up taken wide enough to include the caption and the flatbed beneath: a cab at the left with a driver drawn in silhouette, and behind it a long tank with rounded ends resting on three saddle supports. One vertical band crosses the tank near its middle. What matters for the argument is what the artwork does not draw — there is no line where a hemisphere would meet the cylinder. The seams are invisible on the finished object, which is exactly why recognising the pieces takes an act of imagination rather than an act of reading.
• Fig. 12.3, the test tube (§12.1, p. 162). A narrow open-topped tube with a rounded bottom, drawn with liquid inside and a wisp rising from the mouth. Same decomposition as the tanker with one hemisphere instead of two. Useful because the student has held one.
• The join constrains the pieces (§12.2, p. 163, with Fig. 12.5). The chapter builds a toy by bringing the flat face of a cone against the flat face of a hemisphere, and states plainly that the cone's base radius has to be taken equal to the hemisphere's radius if the result is to have no step in it. Two solids meeting along a circular face contribute one radius between them, not two.
• Example 1, the playing top (p. 163). Inputs: the top is a cone with a hemisphere set on it; overall height 5 cm; the diameter across the widest part 3.5 cm; π taken as 22/7. Nothing else is given. Verified: the shared radius is 1.75 cm; the hemisphere contributes 1.75 cm of the overall height, so the cone's vertical height is 5 − 1.75 = 3.25 cm; the cone's slant height is √(1.75² + 3.25²) = √13.625 ≈ 3.6912, which the page rounds to 3.7 cm. Three numbers had to be manufactured before a single area formula could be applied, and every one of them came from the join.
• Example 3, the wooden rocket (p. 165). Inputs: a cone standing on a cylinder; overall height 26 cm; the conical part 6 cm tall; base diameter of the cone 5 cm; base diameter of the cylinder 3 cm; π taken as 3.14. Verified: cone radius 2.5 cm, cylinder radius 1.5 cm, cylinder height 26 − 6 = 20 cm, cone slant height √(2.5² + 6²) = √42.25 = 6.5 cm exactly. This is the counter-case to Example 1 and the reason it is in the brief: here the two radii are deliberately unequal, so the join is a step rather than a smooth continuation, and the decomposition has to record that.
• Fig. 12.8's second panel (p. 165). Beside the rocket the page prints a small plan view of the base: two concentric circles, the inner labelled 3 cm and annotated as the cylinder's base, the outer labelled 5 cm and annotated as the cone's base. Read from the printed page — the two annotations are lettering set inside the artwork.
• Example 2, the decorative block (p. 164). Inputs: a cube of edge 5 cm with a hemisphere of diameter 4.2 cm fixed on its top face; π taken as 22/7. Verified: hemisphere radius 2.1 cm. Note the shape of the join here — a circle sitting inside a square face, not a circle matched to a circle. Fig. 12.7 labels the cube 5 cm on three edges and the dome 4.2 cm across.
• Example 4, the bird-bath (p. 166). Inputs: a cylinder of height 1.45 m with a hemisphere hollowed into its upper end; the radius given as 30 cm; π taken as 22/7. Verified: the units must be reconciled before anything else — 1.45 m is 145 cm — and the hemisphere's radius is the cylinder's radius, so no new number is needed for it. Read from the printed page: Fig. 12.9 marks 30 cm with an arrow running from the axis out to the rim, so it is a radius, not a diameter, and the bath is drawn standing on three legs.
• Objects assembled from more than one piece across the exercise sets. Hand these to the explanation as recognition drills, stated as data rather than as questions to solve here: two cubes of volume 64 cm³ each set side by side (Exercise 12.1 q. 1, p. 166); a hollow hemisphere of diameter 14 cm carrying a hollow cylinder, total height 13 cm (q. 2); a cone of radius 3.5 cm on a hemisphere of the same radius, total height 15.5 cm (q. 3); a cube of side 7 cm with a dome on top (q. 4); a cube with a hemisphere of diameter equal to its edge scooped out of one face (q. 5); a capsule 14 mm long and 5 mm across, a cylinder capped by a hemisphere at each end (q. 6, Fig. 12.10); a tent, cylinder 2.1 m high and 4 m across with a conical top of slant height 2.8 m (q. 7, p. 167); a cylinder 2.4 cm high and 1.4 cm across with a cone of the same height and diameter drilled out (q. 8); a cylinder 10 cm high of radius 3.5 cm with a hemisphere scooped from each end (q. 9, Fig. 12.11). Verified as decompositions: q. 1 gives a single 8 × 4 × 4 cuboid, because each cube has edge 4 cm; q. 2 gives a cylinder of height 13 − 7 = 6 cm standing on a 7 cm hemisphere; q. 3 gives a cone of height 15.5 − 3.5 = 12 cm, whose slant height is √(12² + 3.5²) = 12.5 cm exactly; q. 4 gives the largest possible dome radius as 3.5 cm, since the dome's circle must fit inside a 7 cm square face; q. 6 gives a cylinder of length 14 − 5 = 9 mm between two hemispheres of radius 2.5 mm.
• Where the join is not a full circle. Three of the objects above have joins that do not match face to face: the block's dome on a square face, the same structure again in Exercise 12.1 q. 4 where a hemisphere is set on a 7 cm cube, and the rocket's wide cone on a narrow cylinder. The tent of q. 7 does not belong in that list — its cylindrical part and its conical top share a 4 m diameter, so the join is a full circle matched to a full circle, exactly as on the tanker and the capsule. What is distinctive about the tent is that it is a shell with no floor, and the question says so in its own closing note. (q. 5 is a hemisphere scooped out of a cube face, which is a hollowing rather than a join.) Flag them here; the arithmetic they force is the next topic's business.

Figures to have open

• Fig. 12.1 redrawn as four labelled panels with the hemisphere added in a fifth, visibly separate slot. This is the chapter's own figure (p. 161); redraw it, and keep the panel labels (i)–(iv) on the original four so the explanation can say which four the book drew.
• An exploded-then-assembled sequence for the tanker: two hemispheres and a cylinder drawn apart, then pushed together, then the finished tank with the seams fading out. The chapter's Fig. 12.4 (p. 162) is close to this; the explanation's version should end on the seamless object to make the recognition problem concrete.
• The playing top with a dimension ladder alongside: 5 cm overall, 1.75 cm of it hemisphere, 3.25 cm of it cone, and a right triangle hanging off the cone showing where 3.7 cm comes from. Standard schematic, built from Fig. 12.6's data (p. 163).
• The rocket's plan view — two concentric circles with the annulus shaded — enlarged well beyond the size it is printed at. This is the chapter's own second panel of Fig. 12.8 (p. 165) and the argument of section 9 depends on it being legible.
• A single strip showing the chapter's joins side by side: circle-to-equal-circle, circle-inside-square, wide-circle-on-narrow-circle. Standard schematic; an added arrangement.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 12 "Surface Areas and Volumes", §12.1 Introduction, pp. 161–162, with Fig. 12.1 and Fig. 12.2 (p. 161) and Fig. 12.3 (p. 162)
• §12.2 "Surface Area of a Combination of Solids", pp. 162–163, for the assembly sequence of Fig. 12.4 and the matched-radius condition illustrated by Fig. 12.5
• Example 1 and Fig. 12.6, pp. 163–164; Example 2 and Fig. 12.7, p. 164; Example 3 and Fig. 12.8, p. 165; Example 4 and Fig. 12.9, p. 166
• Exercise 12.1 questions 1–9, pp. 166–167, with Fig. 12.10 (p. 166) and Fig. 12.11 (p. 167), used here as recognition material rather than as area problems
• The chapter's summary, §12.4, p. 170, which names five basic solids where Fig. 12.1 draws four