Surface area of a right circular cone

Roll up a paper sector into a party hat and you have made a right circular cone — the solid traced out when a right triangle spins a full turn about one of its legs. An ice-cream cone, a traffic cone, the roof of a hut: all the same shape, described by just three lengths, the base radius $r$ , the height $h$ , and the slant height $\ell$ . This article asks a single practical question: how much material covers such a shape, and how much of that material forms the sloping side versus the flat circular base? Getting the answer right — and knowing which of the two you actually need for a given real object — is the entire skill tested in this part of the Class 9 Mensuration chapter.

The core insight

Unrolling a cone's curved surface into a flat circular sector shows that its curved area is $\pi r \ell$ , and adding the base circle $\pi r^2$ gives the total surface area.

The core idea

A right circular cone has a circular base of radius $r$ and an apex directly above the centre of that base at perpendicular height $h$ . The slant height $\ell$ is the straight-line distance from the apex to any point on the rim of the base — it is the length you would measure along the outside of the cone, not straight up through the middle. Because $h$ is perpendicular to the base, the radius, height, and slant height form a right triangle with $\ell$ as the hypotenuse. This gives the first relation every cone problem leans on:

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

where: $r$ is the base radius, $h$ is the perpendicular height (apex to base-centre), and $\ell$ is the slant height (apex to base-rim). Now picture slitting the curved surface along one slant line and pressing it flat. It opens out into a sector of a large circle whose radius equals $\ell$ . That sector's arc length exactly matches the circumference of the cone's base, $2\pi r$ . Working out the sector's area from this fact (done in full below) gives the curved surface area, and adding the one flat face — the base circle — gives the total:

\text{CSA} = \pi r \ell \qquad \qquad \text{TSA} = \pi r \ell + \pi r^2

where: CSA is the curved (lateral) surface area, TSA is the total surface area, and $\pi r^2$ is the area of the single circular base. A cone has only one flat face, unlike a cylinder's two, so TSA needs just one extra term.

Figure

A right circular cone shown beside its flattened net a sector of radius l next to a circle of radius r

Derivation ACurved surface area from the unrolled sector

The goal of this derivation is to show, using nothing beyond the ordinary formula for the area of a circle and a simple ratio of arc lengths, why slitting a cone's curved surface open into a flat sector of radius $\ell$ and then computing that sector's area gives exactly $\pi r \ell$ for the curved surface area of the original cone, with no extra assumptions about the cone's height.

Step 1. Picture the cut. Slit the cone's curved surface along one slant line from apex to rim and press it flat. Because every point on the curved surface was a distance $\ell$ from the apex, the flattened shape is a sector (a pie-slice) of a full circle of radius $\ell$ , with the apex at the sector's centre.

Step 2. Match the arc to the base circumference. The curved edge of that sector is exactly the rim of the cone's base, unrolled. Since the base is a circle of radius $r$ , its circumference is $2\pi r$ , so the sector's arc length must also equal $2\pi r$ .

Step 3. Compare the sector to the full circle of radius $\ell$ . A full circle of radius $\ell$ has circumference $2\pi \ell$ and area $\pi \ell^2$ . The sector we have is only a fraction of that full circle — specifically the fraction $\dfrac{2\pi r}{2\pi \ell} = \dfrac{r}{\ell}$ of the whole, since a sector's arc length is proportional to its share of the full circle.

Step 4. Apply the same fraction to the area. Area also scales with that same fraction of the full circle, so the sector's area is $\dfrac{r}{\ell} \times \pi \ell^2 = \pi r \ell$ . Since the sector's area is exactly the curved surface area of the cone (we only unrolled it flat; unrolling does not change area), this gives $\text{CSA} = \pi r \ell$ .

Step 5. Recover the total surface area. Re-attaching the one flat face the cone actually has — the base circle of area $\pi r^2$ — gives $\text{TSA} = \pi r \ell + \pi r^2$ , matching the formula stated above.

Worked example

Problem. A cone has base radius $7$ cm and height $24$ cm. Find its curved surface area and total surface area. (Use $\pi = \tfrac{22}{7}$ .)

Step 1. Find the slant height using $\ell = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ cm.

Step 2. Curved surface area: $\text{CSA} = \pi r \ell = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550\ \text{cm}^2$ .

Step 3. Total surface area: $\text{TSA} = \text{CSA} + \pi r^2 = 550 + \frac{22}{7} \times 49 = 550 + 154 = 704\ \text{cm}^2$ .

The CSA is $550\ \text{cm}^2$ and the TSA is $704\ \text{cm}^2$ .

Competitive stretch

A conical tent with the same slant height pattern but a base diameter of $28$ m and slant height $17.5$ m needs canvas at ₹70 per square metre. Since a tent has no canvas floor, only the CSA is relevant: radius $r = 14$ m, so $\text{CSA} = \frac{22}{7} \times 14 \times 17.5 = 22 \times 2 \times 17.5 = 770\ \text{m}^2$ , and the cost is $770 \times 70 = ₹53{,}900$ . Notice the height was never needed — only the slant height, since the tent's surface is the CSA alone.

Common traps

Mixing up $h$ and $\ell$ : the height is the vertical apex-to-centre distance, the slant height is the apex-to-rim distance along the surface; they are equal only when $r = 0$ , which never happens for a real cone.
Forgetting the base term in TSA: computing only $\pi r \ell$ and calling it the total surface area, when TSA must add $\pi r^2$ .
Adding a base that should not be added: open objects like tents, funnels, or ice-cream cones without a wafer bottom need only the CSA — read the problem before deciding.
Substituting radius and height from mismatched parts of the problem into the Pythagoras step instead of confirming both belong to the same cone.

Ask the tutor: Want to work through a few more CSA-versus-TSA word problems, or practise rearranging the slant-height formula to find a missing radius or height? Ask the AI tutor to walk you through it step by step.

Why does the curved surface unroll into a sector and not some other shape?

Every point on the cone's curved surface is exactly a distance $\ell$ apart from the apex, since $\ell$ is defined as that apex-to-rim distance and it is the same along every slant line by the symmetry of a right circular cone. When you cut and flatten a surface where every point is equidistant from one corner, you necessarily get a piece of a circle centred at that corner — a sector. The straightness of the slant lines is exactly what keeps the flattened shape a clean sector rather than something curved or irregular.

Do I always need pi times 22 over 7, or can I use 3.14?

Either value works; use whichever the question specifies, and if none is given, $\frac{22}{7}$ is the safer default in CBSE Class 9 because most textbook radii are multiples of $7$ , making the arithmetic exact rather than a long decimal. If a radius is not a multiple of $7$ , using $3.14$ often gives a tidier calculation. The formulas themselves, $\text{CSA} = \pi r \ell$ and $\text{TSA} = \pi r \ell + \pi r^2$ , do not change with your choice of $\pi$ .

How do I know whether a word problem wants CSA or TSA?

Ask whether the object physically has a solid base that is part of what you are measuring. A tent, a funnel, or a paper party hat has an open bottom, so only the curved (slanting) surface is real material — use CSA. A solid cone-shaped object, like a piece of clay or a cone meant to sit flat and be fully covered, has both the curved surface and the base as real surfaces — use TSA. When in doubt, re-read what the material is covering: canvas, paper, or paint applied "all over" usually signals TSA, while "the sloping side" or "no floor" signals CSA.

Surface area of a right circular cone

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads