Volume of a right circular cone

Fill an ice-cream cone with sand and empty it into a cylindrical glass of the same base and height — you will need exactly three cones to fill the cylinder. That simple pouring experiment is the whole idea behind the volume formula for a right circular cone, one of the two summative solids (with the sphere) in the Class 9 chapter Surface Areas and Volumes. This article builds the formula $V = \tfrac13\pi r^2 h$ from that cylinder relationship, derives it properly, and works through the exam-style problems — including capacity questions that convert cubic centimetres to litres — where the single most common mistake is forgetting the factor of one-third.

The core insight

A right circular cone's volume is exactly one-third that of a cylinder sharing its base radius $r$ and perpendicular height $h$ , so $V = \tfrac13\pi r^2 h$ .

The core idea

A right circular cone has a circular base of radius $r$ , an apex directly above the centre of that base, and a perpendicular height $h$ — the straight vertical distance from apex to base centre. This is different from the slant height $\ell$ , the distance measured along the sloping surface, which is used for curved and total surface area but never for volume. The defining fact, demonstrated physically by pouring sand or water, is that three cones of a given base radius and height together fill one cylinder of the same base and height. So the cone's volume is one-third the cylinder's volume $\pi r^2 h$ :

V_{\text{cone}} = \frac{1}{3}\pi r^2 h

where: $V_{\text{cone}}$ is the volume in cubic units, $r$ is the base radius, and $h$ is the perpendicular height (same linear unit as $r$ ). Because volume multiplies three lengths together, it always comes out in cubic units — cm³, m³, and so on. Capacity problems, which ask how much liquid or grain a conical container can hold, often need the extra conversion $1000\text{ cm}^3 = 1$ litre (equivalently $1\text{ m}^3 = 1$ kilolitre). It helps to keep this formula distinct from the surface-area formulas for the same solid: curved and total surface area depend on the slant height $\ell$ , because they measure material wrapped around the outside, while volume depends only on the perpendicular height $h$ , because it measures the space enclosed inside. When a problem states both $\ell$ and $h$ , picking the wrong one is the single most frequent slip students make in this chapter. Within the Class 9 chapter Surface Areas and Volumes, the right circular cone and the sphere/hemisphere are the two solids that carry exam weight; the cube, cuboid, and cylinder reappear here only as formative review of formulas already known from earlier classes. Two extensions sit just beyond this scope and belong to Class 10 instead: combining a cone with another solid (such as a cone sitting on a hemisphere) to find a combined volume, and the frustum of a cone, which is the solid left when the pointed top of a cone is sliced off. Neither is needed to master the plain cone-volume formula developed here.

Figure

A cone beside a cylinder of the same base radius and height, with three cone-fulls of sand shown pouring into the cylinder to exactly fill it

Derivation AVolume of a right circular cone from the cylinder relationship

The goal is to establish, using only the pouring-experiment fact that three equal cones fill one matching cylinder, that a right circular cone of base radius $r$ and perpendicular height $h$ has volume $V = \tfrac13\pi r^2 h$ , and to show why this stays true for any cone regardless of how pointed or flat it is.

Step 1. Start from the volume of the cylinder that shares the cone's base radius $r$ and height $h$ . A cylinder is a stack of identical circular discs of area $\pi r^2$ , so its volume is base area times height: $V_{\text{cylinder}} = \pi r^2 h$ .

Step 2. State the experimental relationship. Filling a hollow cone of radius $r$ and height $h$ with sand and emptying it into the matching hollow cylinder, exactly three cone-fulls are needed to fill the cylinder completely, with nothing left over and no gap remaining. So $3 \times V_{\text{cone}} = V_{\text{cylinder}}$ .

Step 3. Solve for the cone's volume. Dividing both sides by 3 gives $V_{\text{cone}} = \dfrac{1}{3} V_{\text{cylinder}} = \dfrac{1}{3}\pi r^2 h$ .

Step 4. Check the relationship holds regardless of proportions. The 3-to-1 ratio does not depend on whether the cone is tall and narrow or short and wide — it is a fixed geometric fact about any cone matched to a cylinder of the same base and height, which is why the same formula $V = \tfrac13\pi r^2 h$ applies to every right circular cone. Verified: setting $r = 6$ , $h = 7$ gives cylinder volume $\pi(36)(7) = 792$ (using $\pi = 22/7$ ) and cone volume $792/3 = 264$ , matching the worked example below.

Worked example

Problem: Find the volume of a cone with base radius $6$ cm and height $7$ cm. Use $\pi = \tfrac{22}{7}$ .

Step 1. Write the formula and identify the values: $V = \tfrac13\pi r^2 h$ , with $r = 6$ cm and $h = 7$ cm.

Step 2. Substitute: $V = \dfrac13 \times \dfrac{22}{7} \times 6^2 \times 7$ .

Step 3. Simplify. The $7$ in the height cancels the $7$ in the denominator of $\pi$ , leaving $V = \dfrac13 \times 22 \times 36 = \dfrac{792}{3} = 264$ . So $V = 264\text{ cm}^3$ .

Competitive stretch

Problem: A conical pit of top radius $3.5$ m is $12$ m deep. Find its capacity in kilolitres, using $\pi = \tfrac{22}{7}$ and $1\text{ m}^3 = 1$ kilolitre.

Step 1. Identify $r = 3.5$ m and $h = 12$ m — the pit's depth is its perpendicular height.

Step 2. Substitute into the formula: $V = \dfrac13 \times \dfrac{22}{7} \times (3.5)^2 \times 12$ , and compute $(3.5)^2 = 12.25$ .

Step 3. Multiply through: $\dfrac{22}{7} \times 12.25 = 22 \times 1.75 = 38.5$ , so $V = \dfrac13 \times 38.5 \times 12 = 38.5 \times 4 = 154$ .

$V = 154\text{ m}^3 = 154$ kilolitres.

Common traps

Forgetting the factor of $\tfrac13$ : computing $\pi r^2 h$ (the cylinder's volume) and stopping there is the single most common cone-volume error.
Using slant height $\ell$ in place of perpendicular height $h$ . If a problem gives $\ell$ and $r$ instead of $h$ , first find $h = \sqrt{\ell^2 - r^2}$ before substituting into the volume formula.
Squaring the diameter instead of the radius, or forgetting to halve a given diameter before substituting into $r^2$ .
Mixing units within one calculation, such as radius in centimetres and height in metres — convert everything to a single unit before substituting.
Forgetting that "depth" in a well or pit problem, and "height" of a heap or tent-shaped pile, both mean the perpendicular height $h$ that the formula needs — not a slant measurement, even if the word used is not literally "height".

Ask the tutor: Want to practise finding an unknown radius or height when the volume is given, or work through more slant-height-to-height conversions with the AI tutor?

Why is the cone's volume exactly one-third of the cylinder's, and not some other fraction?

It comes from a physical fact you can verify by pouring sand or water: a hollow cone and a hollow cylinder sharing the same base radius and height, when the cone is filled and emptied into the cylinder exactly three times, fill the cylinder with nothing left over. This 3-to-1 ratio is constant for every right circular cone matched to a cylinder of the same base and height, however tall or flat the shape is, which is why the formula always carries the factor $\tfrac13$ .

The question gives me the slant height but not the height, what do I do?

Use the right triangle formed by the radius, the perpendicular height, and the slant height, related by $\ell^2 = r^2 + h^2$ (Pythagoras applied to the cone's cross-section). Rearranging gives $h = \sqrt{\ell^2 - r^2}$ . Compute $h$ first, then substitute it into $V = \tfrac13\pi r^2 h$ . Never substitute the slant height directly for $h$ in the volume formula — that is the most common mark-losing slip in this topic.

How do I convert a volume answer in cubic centimetres or cubic metres into litres for a capacity question?

Use the standard conversions $1000\text{ cm}^3 = 1$ litre and $1\text{ m}^3 = 1$ kilolitre. So a volume of $264\text{ cm}^3$ is $264/1000 = 0.264$ litres, and a volume of $154\text{ m}^3$ is $154$ kilolitres directly. Always check the unit the question asks for (litres versus kilolitres versus cm³/m³) before finishing, since capacity questions frequently expect the converted unit, not the raw cubic-unit answer.

Volume of a right circular cone

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads