Mensuration (3D)

The cone formulas and the slant triplet

For a right circular cone of radius $r$, height $h$, slant $l$: $l = \sqrt{r^2 + h^2},\qquad V = \frac{1}{3}\pi r^2 h,\qquad \text{CSA} = \pi r l,\qquad \text{TSA} = \pi r(l + r)$ The curved surface unrolls into a sector of radius $l$ — that is why $l$ (not $h$) appears in CSA and TSA. Almost every cone question is built on a Pythagorean $(r, h, l)$ triplet, with $(7, 24, 25)$ dominating exams and $(3, 4, 5)$-multiples next. Memorise the whole 7-24-25 family at $\pi = 22/7$:

The base circle is always $\pi r^2 = 154$ here, so TSA $= 550 + 154$. A question tripling the size triples nothing — every length scales linearly, areas by $k^2$, volume by $k^3$.

Reverse routes

• From CSA: $l = \frac{\text{CSA}}{\pi r}$, then $h = \sqrt{l^2 - r^2}$.
• From volume: $h = \frac{3V}{\pi r^2}$ — the factor 3 moving to the top is the classic slip; $r = \sqrt{\frac{3V}{\pi h}}$ likewise.
• Conical heap questions: base circumference first — $r = \frac{C}{2\pi}$ (C $= 44$ m gives $r = 7$ m directly with $\pi = 22/7$).
• Tent questions: canvas $=$ CSA only (no floor). Painted/covered-including-base: add $\pi r^2$.

Recasting and inscribed relations

Melting preserves volume: a cone recast into a sphere satisfies $\frac{1}{3}\pi r^2 h = \frac{4}{3}\pi R^3$, so $R^3 = \frac{r^2 h}{4}$ — and $r^2 h$ is chosen divisible by 4 in every exam instance. A cone melted into $n$ similar cones divides every length by $\sqrt[3]{n}$ ($n = 27 \Rightarrow$ height $\div 3$). Solids on the same base with height equal to the radius: $\text{cone} : \text{hemisphere} : \text{cylinder} = \frac13\pi r^3 : \frac23\pi r^3 : \pi r^3 = 1 : 2 : 3$. The largest cone inscribed in a cylinder (same base, same height) holds exactly $\frac13$ of the cylinder.

Frustum preview

Cutting a cone parallel to its base leaves a frustum: $V = \frac{\pi h}{3}(R^2 + r^2 + Rr)$ — the bucket/tumbler shape, worked in the prism-pyramid subtopic. If instead the TOP piece is asked, it is a smaller similar cone: subtract, or use the similarity ratio directly.

Quick revision

• $l = \sqrt{r^2+h^2}$; CSA $= \pi r l$; TSA $= \pi r(l + r)$; $V = \frac13 \pi r^2 h$.
• Triplet $(7, 24, 25)$: CSA $550$, TSA $704$, $V$ $1232$, base $154$.
• From $V$: $h = \frac{3V}{\pi r^2}$ (mind the 3). From $C$: $r = \frac{C}{2\pi}$.
• Recast cone → sphere: $R^3 = \frac{r^2 h}{4}$; into $n$ similar cones: lengths $\div \sqrt[3]{n}$.
• Same base & height ($h = r$): cone : hemisphere : cylinder $= 1:2:3$.