Chapter 10.2 — Exercise 10.1 — Surface Areas and Volumes

Two Joker's Caps, Same Method

A joker's cap is just a cone, and finding how much sheet material it takes always starts with the slant height, computed via the same right-triangle relationship introduced at the start of this chapter.

The second row asks a genuinely different question from the first — not "how much sheet for a fixed number of caps" but "how many caps from a fixed sheet." Dividing 1000 by the per-cap area gives 21.21, and since a fraction of a cap can't actually be cut and used, the answer rounds down to 21, not to the nearer whole number. Whenever a division like this counts discrete physical objects, rounding down is the only answer that respects reality — claiming 22 caps would need more sheet material than is actually available to work with.

Both rows are also worth comparing for their radius-height pairs: 7-24-25 and 3-4-5 are both genuine Pythagorean triples, which is exactly why both slant heights came out as clean whole numbers rather than surds. Textbook cone problems lean on recognisable triples like these constantly, precisely so the resulting arithmetic stays clean and manageable entirely by hand.

A Cylinder's Total Surface, Scaled Up

100 paper cylinders are needed, each with radius 7 cm and height 35 cm. Since the cylinders need to be fully enclosed (not just wrapped around the side), this problem calls for total surface area, not just curved.

It's worth being deliberate about which surface area formula this problem actually needs. A cylinder open at both ends (a pipe, say) would only need the curved surface area, 2πrh — but a cylinder meant to physically contain something, like a packing tube for shuttlecocks, needs both circular ends covered too, which is exactly why total surface area is the correct choice here.

A Cone's Volume, Directly

For a cone with radius 6 cm and height 7 cm, the volume formula substitutes in directly, with no slant height needed at all — volume only ever depends on radius and height, never on the slanted side. Compare this with the two cap problems just above, both of which needed the slant height first: whether a cone problem needs l at all depends entirely on whether it's asking about surface area or about volume.

Why a Cone Is Always a Third of Its Matching Cylinder

Two proof-style problems in this exercise establish general ratios rather than specific numbers — genuinely useful facts to carry forward rather than re-derive every time.

• Equal curved surfaces, same base: if a cylinder's curved surface area equals a cone's curved surface area, and both share the same base radius, then 2πrh = πrl simplifies directly to h/l = 1/2 — the cylinder's height is always exactly half the cone's slant height in this situation.
• Equal radius, equal height: a cylinder and cone sharing the same radius and the same height always have volumes in ratio πr²h : ⅓πr²h, which cancels down to 3 : 1. The cone's ⅓ factor in its own volume formula is exactly where that ratio comes from — nothing about the specific radius or height chosen ever changes it.

Both results are proved with letters instead of numbers, which is exactly why they hold universally: any cylinder and any cone satisfying the stated condition land on the identical ratio, no matter their actual size. Neither proof needs a single substituted number anywhere — every r and every h cancels out completely, leaving a pure ratio behind, which is precisely what makes both results reusable facts rather than one-off calculations tied to specific measurements.

Fifty Rods, One Volume Formula

A cylindrical iron rod has height 11 cm and base diameter 7 cm (so radius 3.5 cm). Finding the total volume of 50 such rods is just the single-rod volume, scaled up at the very last step.

Converting the given diameter to a radius before substituting anything is the one step worth double-checking here — using 7 cm directly in place of the radius, instead of halving it to 3.5 cm first, would overstate the volume by a factor of four, since radius appears squared in the formula. This diameter-to-radius conversion shows up constantly throughout this entire chapter, in every exercise, whenever a problem states a diameter rather than a radius directly.

A Heap of Rice, Two Questions at Once

A conical heap of rice has diameter 12 m and height 8 m, and this problem explicitly instructs using π=3.14 rather than 22/7 — worth following exactly, since the two approximations give slightly different final numbers.

This single problem is really two separate questions bundled together: the volume answers "how much rice is there," while the lateral surface area answers an entirely different question, "how much canvas covers the heap from outside." Both use the same radius and height, but they're two genuinely different formulas answering two genuinely different real-world needs.

It's worth noticing, too, why this problem needs the slant height at all when the earlier volume calculation never touched it. Volume only ever cares about how much space a solid encloses — radius and height alone fully describe that. Surface area, by contrast, cares about the actual physical surface a covering would have to follow, and for a cone that surface runs along the slanted side, not straight up the vertical height — which is exactly why l, not h, is the length that belongs in the lateral surface area formula.

From Single Solids to Combined Ones

Every solid measured in this exercise stood alone, never touching another shape. Exercise 10.2 starts joining solids together — a cone stuck onto a hemisphere, a cylinder capped at both ends — and asks for the surface area of the result, using exactly the curved-surface-area-only approach previewed in the chapter introduction — a genuinely different challenge from anything in this exercise, since no single solid here ever had a hidden, joined-away face to account for. For the six core formulas and the slant-height relationship this entire exercise depended on, revisit the chapter introduction.