Chapter 10.4 — Exercise 10.3 — Cone

Three Measurements, One Right Triangle Hidden Inside

This right triangle — radius along the base, height straight up the middle, slant height as the hypotenuse running down the outside — is the single fact that makes every "given two, find the third" cone problem solvable, since l, r, and h are never all handed over at once; one of them almost always has to be recovered from the other two first. It's worth noticing that this right triangle is a purely internal, invisible feature of the solid — it never appears as an actual edge or face of the cone itself, only as a cross-section running straight down through the cone's center from apex to base — yet every single problem in this exercise that involves the slant height in any way leans on it as the connecting fact between what's given and what's asked for.

Base Area Given Directly, No Radius Needed

A cone's base area is 38.5 cm² and its volume is 77 cm³; find its exact height. Since volume = ⅓ × base area × height, ⅓ × 38.5 × h = 77, giving h = 77 × 3 ÷ 38.5 = 6 cm. This problem is deliberately easier than most in this exercise: because the base area is already given rather than the radius, there's no need to touch πr² at all — the volume formula's "base area × height" structure, established in the introduction, can be used exactly as it stands for any base shape, circular or otherwise. Remembering that the ⅓ factor belongs specifically to tapering solids — pyramids and cones alike, as the introduction pointed out — rather than to circular bases specifically is what keeps this shortcut from being confused with the plain "base area × height" formula that applies instead to a cylinder or any other straight-sided prism.

A closely related problem swaps base area for radius directly: volume 462 m³, base radius 7 m, find the height. Substituting into ⅓πr²h = 462 gives ⅓ × (22/7) × 49 × h = 462, so h = 9 m — the identical formula, just requiring πr² to be computed first since only the radius, not the base area itself, was given directly this time around.

Curved Surface Area to Total Surface Area

A cone's curved surface area is 308 cm² and its slant height is 14 cm; find both the radius and the total surface area. Since πrl = 308, (22/7)(r)(14) = 308, giving r = 7 cm. Total surface area = πr(l + r) = (22/7)(7)(21) = 462 cm² exactly. Once the radius is recovered from the lateral formula, it plugs directly into the total formula alongside the slant height that was already given — no further unknowns remain. This is the same "recover one quantity, then substitute it forward into a second formula" pattern that ran through the cuboid and cylinder exercises before this one — the lateral surface formula, containing only r and l, is solved first because it's the simpler of the two available equations, and only once r is pinned down does the more demanding total surface formula, which needs both r and l together, actually become usable.

A Cost Problem That Hides a Quadratic

Painting a cone's total surface area at 25 paise per cm² costs ₹176 total, with slant height 25 cm; find its exact volume. Since ₹176 at ₹0.25/cm² covers 176 ÷ 0.25 = 704 cm² of total surface area, πr(l + r) = 704 becomes (22/7)r(25 + r) = 704, which expands into a genuine quadratic.

This problem strings together nearly every technique this exercise has introduced: a cost-to-area conversion, the total-surface-area formula rearranged into a genuine quadratic (rather than a plain linear equation, unlike every previous problem here), factoring to discard a physically impossible negative root, the l²=r²+h² right-triangle relationship to recover the height, and finally the volume formula itself — five separate steps chained into one single answer. Discarding r = −32 is worth pausing on rather than treating it as an automatic formality: the quadratic itself has no idea that r represents a physical radius, so it hands back both mathematically valid roots without preference, and it's only the real-world context — a radius can never be a negative length — that rules one of the two answers out. This is a genuinely common pattern whenever a physical, real-world quantity gets modeled algebraically: the equation solves cleanly, but interpreting which of its solutions actually makes sense requires reasoning that lives outside the algebra itself.

Folding a Flat Sector Into a Cone

A flat circular sector of radius 15 cm and sector angle 216° is rolled up carefully into a cone; find its base radius and its height. The key insight is that the sector's curved edge becomes the cone's base circle exactly, so the sector's arc length must equal the cone's base circumference: arc length = (216/360) × 2π(15) = cone's 2πr. This simplifies directly to πrl = (216/360)πR², where R = 15 is the sector's own radius and l = 15 is automatically the cone's slant height too (since the sector's straight edges become the cone's slant sides once rolled). Solving this out gives r = 9 cm, and then l² = r² + h² gives 225 = 81 + h², so h² = 144 and h = 12 cm exactly.

This is a genuinely different kind of problem from everything else in this exercise: instead of measuring an already-formed cone directly, it starts from a flat 2D sector and asks what cone results once that sector is physically rolled into shape — the sector's own radius becomes the cone's slant height, and only a fraction of the full circle's circumference (set by the 216° sector angle) becomes the cone's much smaller base circumference. This construction is genuinely how paper cones — the kind used for party hats or ice-cream cups — are actually manufactured in practice: a flat sector is cut from a larger circle, then the two straight edges are brought together and joined, and the sector angle chosen determines exactly how tall and narrow, or how short and wide, the resulting cone ends up being.

Losing the Apex, Gaining Full Symmetry

A cone tapers from a circular base to a single point. Exercise 10.4 closes this chapter with the sphere and hemisphere, solids with no base, no apex, and no straight edge anywhere at all — just one radius, wrapped uniformly around a single center point.