Chapter 10.2 — Exercise 10.1 — Cube and Cuboid

Reading Both Formulas Off One Figure

A cube with side 4 cm and a cuboid with length 8 cm, breadth 6 cm, height 5 cm can both be handled with a direct substitution.

Nothing beyond careful substitution and arithmetic is needed once the right formula is picked — the actual skill this problem checks is choosing lateral versus total correctly, not the arithmetic itself. Mixing the two up is the single most common error at this stage — lateral surface area answers "how much material wraps around the sides," while total surface area answers "how much material covers the entire solid, top and bottom included," and a problem never states outright which one it wants; that has to be inferred from the situation being described.

Working Backward From Surface Area to Volume

A cube's total surface area is 1350 m²; the goal is its volume.

This problem chains two formulas that don't share a direct relationship on their own — surface area involves l², volume involves l³ — so the side length itself has to be recovered as an intermediate step before volume can be computed at all; there's no shortcut formula linking surface area straight to volume without passing through l first. This two-step structure — recover a linear dimension from an area or surface-area fact, then feed that dimension into a separate volume formula — is genuinely common across this whole chapter, and recognising it early saves a lot of confused searching for a formula that would connect the two quantities directly, because no such formula exists for any of the solids covered here.

Four Walls, No Ceiling or Floor

A room shaped like a cuboid, 12 m long, 10 m wide, 7.5 m high, needs its four wall area found (no doors or windows). This is exactly the lateral surface area, since "four walls" excludes the ceiling and floor the same way the lateral formula excludes a cuboid's top and bottom: 2h(l + b) = 2(7.5)(22) = 330 m². Recognising that "area of four walls" is genuinely just another name for lateral surface area — not some new formula that needs deriving from scratch — is the entire point of this problem; the room's specific dimensions barely matter much at all once that identification is correctly made.

Volume Given, One Dimension Missing

A cuboid has volume 1200 cm³, length 15 cm, breadth 10 cm; the height is unknown. Since lbh = 1200, 15 × 10 × h = 1200, giving h = 1200 ÷ 150 = 8 cm. Every volume problem in this chapter reduces to the same move once two of the three dimensions are already known: substitute what's given into lbh, then solve the resulting one-variable equation for whatever's missing.

Scaling Every Dimension by the Same Factor

How does a box's total surface area change if every dimension is doubled? Tripled? Scaled by n? Starting from S = 2(lh + bh + lb), doubling every dimension gives 2[(2l)(2h) + (2b)(2h) + (2l)(2b)] = 2[4lh + 4bh + 4lb] = 4 × 2(lh + bh + lb) = 4S. Tripling gives 9S by the identical process, replacing each factor of 2 with 3. Scaling by n in general gives n²S — every one of the three cross-terms picks up a factor of n from each of its two dimensions, so n × n = n² multiplies straight through the whole formula.

Surface area scales by the square of the linear factor because it's built from two-dimensional terms (a length times a length); volume, by contrast, would scale by the cube of the same factor, since it's built from three dimensions multiplied together — exactly the n² versus n³ distinction flagged in the introduction's unit-conversion table, now showing up as a genuine, provable consequence of the surface-area formula itself rather than just a rule to memorize.

A Prism Whose Base Isn't a Rectangle

A prism has a triangular base with sides 3 cm, 4 cm, 5 cm and height 10 cm; find its volume. Since the base isn't a rectangle, its area needs Heron's formula: semi-perimeter s = (3+4+5)/2 = 6, area = √(s(s−a)(s−b)(s−c)) = √(6×3×2×1) = √36 = 6 cm². Volume = base area × height = 6 × 10 = 60 cm³ exactly.

This problem is really testing whether "volume of any prism = base area × height," stated in the introduction, was actually understood as a general rule rather than a cuboid-specific one — the base here is a triangle, computed with an entirely different area formula (and, since 3-4-5 is a right triangle, could alternatively have used ½ × base × height directly), but once that base area is found, the exact same multiplication by height applies without any modification needed at all.

A Pyramid Built From Its Perimeter

A square pyramid is 3 m tall with a base perimeter of 16 m; find its volume. Since the base is a square with perimeter 4l = 16, l = 4 m, giving base area = l² = 16 m². Volume = ⅓ × base area × height = ⅓ × 16 × 3 = 16 m³. The height's factor of 3 cancels almost entirely against the pyramid's ⅓ factor, leaving just the base area itself as the answer — a coincidence specific to this problem's numbers, not a general shortcut, but a useful arithmetic check that the ⅓ factor was genuinely applied correctly rather than accidentally skipped somewhere along the way.

From Cubic Metres to Litres

An Olympic pool, 50 m long, 25 m wide, 3 m deep throughout, needs its capacity in litres. Volume = 50 × 25 × 3 = 3750 m³. Since 1 m³ = 1000 litres (from the introduction's conversion table), capacity = 3750 × 1000 = 3,750,000 litres. This closing problem leans on both halves of the chapter's opening material at once — the plain lbh volume formula, and the m³-to-litres conversion factor — showing exactly why that conversion table wasn't just a side note but a genuinely necessary tool for turning a raw cubic-metre answer into a real-world, directly usable quantity that a working pool's operator could actually act on.

Trading Flat Faces for a Curved One

Every solid in this exercise had entirely flat faces, whether rectangular or triangular. Exercise 10.2 introduces the cylinder, the first solid in this chapter built around a curved surface, bringing π into every formula for exactly that reason.