Right circular cone

Look at a party hat, an ice-cream cone, or a heap of sand tapering to a point — every one of them is a right circular cone: a circular base that narrows smoothly to a single apex sitting directly above the centre of the base. In the Class 10 Mensuration unit, the cone sits between the cylinder and the sphere as the second of three solids whose curved surface area, total surface area, and volume you must be able to compute and apply to real problems — tents, canvas costs, and melting-and-recasting one solid into another. Only the complete cone matters here; the frustum (a cone with its tip sliced off) is outside this syllabus. The three formulas you need all fall out of one right triangle hiding inside the cone, and one relationship — the cone holds exactly one-third the volume of the cylinder that snugly contains it — ties the whole topic together.

The core insight

A right circular cone's slant height satisfies $l=\sqrt{r^2+h^2}$ , and its curved surface area, total surface area, and volume follow from unrolling that slanted surface into a flat sector, giving $\text{CSA}=\pi rl$ , $\text{TSA}=\pi r(l+r)$ , and $V=\tfrac{1}{3}\pi r^2h$ — exactly one-third of a cylinder with the same base and height.

The core idea

A right circular cone is fixed by three lengths, and any two of them determine the third.

Radius $r$ — the radius of the circular base.
Height $h$ — the perpendicular distance from the apex straight down to the centre of the base (the axis of the cone).
Slant height $l$ — the distance from the apex to any point on the rim of the base, measured along the sloping surface.

Because the axis, the base radius, and the slant height together form a right triangle inside the cone (the right angle sits at the centre of the base), Pythagoras gives the link between them directly:

l = \sqrt{r^2 + h^2}

where: $r$ is the base radius, $h$ is the vertical height, and $l$ is the slant height, all measured in the same length unit. This single relation is the gateway to every cone problem: whichever two of $r$ , $h$ , $l$ a question hands you, this equation hands back the third. The curved (lateral) surface, total surface, and volume then follow from three standard results:

\text{CSA} = \pi r l, \qquad \text{TSA} = \pi r l + \pi r^2 = \pi r(l+r), \qquad V = \frac{1}{3}\pi r^2 h

where: CSA is the area of the sloping surface alone, TSA adds the flat circular base ( $\pi r^2$ ), and $V$ is the space enclosed. Notice that CSA and TSA both need the slant height $l$ , while volume needs the vertical height $h$ — the single most common slip in this chapter is reaching for the wrong one.

Figure

A right circular cone with radius r, height h and slant height l marked as a right triangle inside it, beside its curved surface unrolled flat into a sector of radius l

The unrolling picture in the figure is why CSA has the clean form $\pi r l$ : cut the cone along a slant edge and flatten the curved surface, and it opens into a sector of a circle whose radius is $l$ and whose curved edge exactly matches the base circle's circumference, $2\pi r$ . The volume relation is worth memorising as a ratio too — a cone always holds exactly one-third the volume of the cylinder that shares its base and height, which is the fact tested in "how many conical cups fill a cylindrical tank" problems.

Derivation ACurved surface area of a cone from the unrolled sector

The goal is to show, using only the geometry of a circular sector and nothing beyond the Class 10 toolkit, why the curved surface area of a right circular cone of base radius $r$ and slant height $l$ works out to the compact formula $\text{CSA} = \pi r l$ , by relating the sector's arc length to the cone's base circumference and then finding the fraction of the full circle of radius $l$ that the sector occupies.

Step 1. Cut the cone open along one slant edge and flatten the curved (lateral) surface. It unrolls into a sector of a circle whose radius equals the cone's slant height $l$ — every point on the cone's slanted surface was at distance $l$ from the apex, and that distance is preserved when the surface is laid flat.

Step 2. Identify what the sector's curved edge must equal. Before unrolling, that edge was the rim of the cone's base — a full circle of radius $r$ , with circumference $2\pi r$ . Unrolling does not stretch the surface, so the sector's arc length after flattening is still exactly $2\pi r$ .

Step 3. Find what fraction of the full circle of radius $l$ this sector represents. A full circle of radius $l$ has circumference $2\pi l$ . The sector's arc length is $2\pi r$ , so the sector is the fraction $\dfrac{2\pi r}{2\pi l} = \dfrac{r}{l}$ of the complete circle.

Step 4. Apply that same fraction to the circle's area. The area of a full circle of radius $l$ is $\pi l^2$ , so the sector's area — which is precisely the cone's curved surface — is $\dfrac{r}{l}\times \pi l^2 = \pi r l$ . This confirms $\text{CSA} = \pi r l$ , and adding the base circle $\pi r^2$ back on (for a solid cone resting on a surface) gives $\text{TSA} = \pi r l + \pi r^2 = \pi r(l+r)$ .

Worked example

A conical tent has a base radius of $7\text{ m}$ and a height of $24\text{ m}$ . Find (a) the slant height, (b) the curved surface area of the tent, (c) the cost of canvas at ₹35 per square metre, and (d) the total surface area if the same shape were a solid metal cone instead of a hollow tent. Take $\pi = \dfrac{22}{7}$ .

Step 1. Find the slant height using $l = \sqrt{r^2+h^2}$ . With $r=7$ and $h=24$ : $l = \sqrt{7^2+24^2} = \sqrt{49+576} = \sqrt{625} = 25\text{ m}$ .

Step 2. Find the curved surface area: $\text{CSA} = \pi r l = \dfrac{22}{7}\times 7\times 25 = 22\times 25 = 550\text{ m}^2$ .

Step 3. Find the cost of canvas. A tent is open at the base, so only the curved surface needs covering: $\text{cost} = 550\times 35 = ₹19{,}250$ .

Step 4. For part (d), a solid cone (unlike a tent) has a floor, so add the base circle: $\text{TSA} = \pi r(l+r) = \dfrac{22}{7}\times 7\times(25+7) = 22\times 32 = 704\text{ m}^2$ . The tent needs $550\text{ m}^2$ of canvas costing ₹19,250, while the equivalent solid cone would have a total surface area of $704\text{ m}^2$ .

Competitive stretch

A metallic solid cone of radius $6\text{ cm}$ and height $8\text{ cm}$ is melted and recast into a solid cylinder of radius $4\text{ cm}$ . Find the cylinder's height. Melting and recasting conserves volume, so volume of cone equals volume of cylinder: $\dfrac{1}{3}\pi(6)^2(8) = \pi(4)^2 h_{\text{cyl}}$ , which gives $\dfrac{1}{3}\times 36\times 8 = 16\,h_{\text{cyl}}$ , so $96 = 16\,h_{\text{cyl}}$ , and $h_{\text{cyl}} = 6\text{ cm}$ . This "equate the volumes" move works for melting a cone into any other solid, and it can also run backwards: given a volume and a radius, rearrange $V=\tfrac{1}{3}\pi r^2h$ to get $h = \dfrac{3V}{\pi r^2}$ .

Common traps

Mixing up $h$ and $l$ — CSA and TSA need the slant height $l$ ; the volume needs the vertical height $h$ . Check which the formula is asking for before substituting.
Dropping the $\tfrac{1}{3}$ in the volume formula — a cone is not $\pi r^2 h$ ; that number belongs to a cylinder of the same base and height.
Using the diameter where the radius is required (or vice versa) when applying $l=\sqrt{r^2+h^2}$ .
Using TSA instead of CSA for open shapes like tents and caps — only closed, floored solids need the base circle added.

Ask the tutor: If your slant-height calculation or your CSA/TSA/volume choice on a cone problem isn't matching the answer key, ask the AI tutor to walk through your working step by step.

Why does the volume formula for a cone have a one-third in it while the surface area formulas do not?

The one-third comes from a genuinely different kind of reasoning: the volume result is proved by comparing a cone to a cylinder of the same base and height and showing the cone holds exactly a third as much (a fact you can verify experimentally by pouring water from a conical vessel into a cylindrical one three times). The surface area formulas, by contrast, come from unrolling the slanted surface into a flat sector and computing its area directly, as shown in the derivation above — no comparison to another solid is involved, so no extra fraction appears.

In a word problem, how do I know whether to use CSA or TSA?

Ask whether the object has a covered base in real life. A tent, a party hat, or a funnel is open underneath, so only the sloping surface needs material — use CSA. A solid cone standing on a table, a painted decoration, or anything described as "resting on a surface" has its base sitting flush against something, but if the base itself needs paint or material too (for instance a solid metal cone being fully coated), use TSA. Read the problem's wording carefully — words like "tent," "cap," or "canvas over" usually signal CSA, while "solid," "painted all over," or "total surface" signal TSA.

The problem gives me the diameter, not the radius. Does that change which formula I use?

No, the formulas are unchanged — you simply must convert first. Halve the diameter to get $r$ before substituting into $l=\sqrt{r^2+h^2}$ , $\text{CSA}=\pi rl$ , $\text{TSA}=\pi r(l+r)$ , or $V=\tfrac{1}{3}\pi r^2h$ . This is one of the most common slips in exams: students substitute the diameter directly where $r$ belongs, which throws every answer off by a factor of two (or four, inside the squared terms). Always write down $r = \dfrac{\text{diameter}}{2}$ as an explicit first line of working.

Right circular cone

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads