What Is a Frustum of a Cone?
A frustum of a cone is a three-dimensional solid formed when a right circular cone is sliced by a plane parallel to its base and the top piece is removed. What remains is a frustum with these features:
Two circular bases of different radii, $R$ (bottom) and $r$ (top), lying in parallel planes.
A curved lateral surface connecting the two circles.
A perpendicular height $h$, the straight distance between the two circular faces.
A slant height $l$, measured along the slanted surface.
Because it comes from a right circular cone, the axis joining the centres of the two circles is perpendicular to both bases. The word "frustum" is Latin for "piece" or "morsel" - literally the chunk left after the tip is gone.
What Do a Bucket, a Lampshade, and a Drinking Glass Have in Common?
Slice the tip off a cone and the shape left behind has its own name and formulas.
A frustum of a cone is the portion of a cone that lies between its base and a plane cutting it parallel to the base - the cone with its top removed. It has two parallel circular faces, a larger one of radius $R$ and a smaller one of radius $r$, joined by a slanted curved surface. The volume of a frustum of a cone is $\frac{1}{3}\pi h(R^2 + Rr + r^2)$, where $h$ is the perpendicular height between the two circles.
Buckets, lampshades, drinking tumblers, and flowerpots are all frustums. The shape is popular in design for a reason: a wide base gives stability while a narrower top makes the object easier to hold or stack.
How Do You Find the Volume of a Frustum of a Cone?
This is the headline formula and the most-searched one. The volume of a frustum of a cone is:
$$V = \frac{1}{3}\pi h\left(R^2 + Rr + r^2\right)$$
Variable glossary: $R$ is the larger (bottom) radius, $r$ is the smaller (top) radius, $h$ is the perpendicular height, and $\pi \approx 3.14159$.
Where the formula comes from: a frustum is a large cone minus the small cone that was sliced off the top. If the full cone has height $H$ and the removed tip has height $H - h$, subtracting the two cone volumes and using similar-triangle ratios between $R$ and $r$ collapses neatly to the single formula above. The three terms $R^2 + Rr + r^2$ are what remain after that subtraction — you never have to reconstruct the whole cone in practice.
Notice the symmetry: if $r = R$, the formula becomes $\frac{1}{3}\pi h(3R^2) = \pi R^2 h$, the volume of a cylinder - exactly right, since a frustum with equal radii is a cylinder. If $r = 0$, it becomes $\frac{1}{3}\pi h R^2$, the volume of a full cone. At Bhanzu, our trainers show these two limiting checks so students can catch a wrong answer instantly.
How Do You Find the Slant Height and Surface Area of a Frustum?
The slant height $l$ connects the two radii through the Pythagorean relationship. The difference in radii $(R - r)$ and the height $h$ form the two legs of a right triangle whose hypotenuse is the slant height:
$$l = \sqrt{h^2 + (R - r)^2}$$
This comes straight from the Pythagoras theorem applied to the right triangle formed by dropping the slant edge onto the base.
Once you have $l$, the surface areas follow:
Curved (lateral) surface area: $\text{CSA} = \pi l (R + r)$
Total surface area: $\text{TSA} = \pi l (R + r) + \pi R^2 + \pi r^2$
The total surface area adds the two circular faces - area $\pi R^2$ at the bottom and $\pi r^2$ at the top - to the curved surface. If the top is open (like a real bucket), you drop the $\pi r^2$ term.
What Are the Properties of a Frustum of a Cone?
A few defining properties separate a frustum from the full cone it was cut from:
It keeps the base but loses the apex. A frustum contains the original cone's circular base but not its vertex - the pointed tip is the piece that was sliced away.
It has two parallel circular faces. The top and bottom are circles of different radii, $r$ at the top and $R$ at the bottom, lying in parallel planes.
The axis is perpendicular to both bases. For a right circular frustum, the line joining the two centres meets each circular face at a right angle.
Every cut parallel to the base is a circle. The radius of that cross-section grows steadily from $r$ at the top to $R$ at the bottom.
It reduces to familiar solids at its limits. When $r = R$ the frustum becomes a cylinder; when $r = 0$ it becomes the full cone again.
Three measurements fix everything. The two radii $R$ and $r$ and the perpendicular height $h$ determine the solid completely; the slant height $l$ follows from them.
Examples of the Frustum of a Cone
Example 1
Find the volume of a frustum with $R = 6$ cm, $r = 3$ cm, and height $h = 8$ cm.
$$V = \frac{1}{3}\pi h(R^2 + Rr + r^2) = \frac{1}{3}\pi (8)(36 + 18 + 9) = \frac{1}{3}\pi (8)(63) = 168\pi$$
Final answer: $168\pi \approx 527.79$ cm³.
Example 2
Find the slant height of a frustum with $R = 5$ cm, $r = 2$ cm, height $8$ cm — watch a common slip.
Wrong path. A student writes $l = \sqrt{h^2 + (R + r)^2} = \sqrt{64 + 49} = \sqrt{113}$, adding the radii.
Why it breaks. The horizontal leg of the slant-height triangle is the difference of the radii, not the sum. Using $R + r$ over-counts the horizontal offset and inflates the slant height.
Correct. $l = \sqrt{h^2 + (R - r)^2} = \sqrt{8^2 + (5 - 2)^2} = \sqrt{64 + 9} = \sqrt{73}$.
Final answer: $l = \sqrt{73} \approx 8.54$ cm. The horizontal leg is always $(R - r)$.
Example 3
Find the curved surface area of a frustum with $R = 7$ cm, $r = 4$ cm, and slant height $l = 10$ cm.
$$\text{CSA} = \pi l (R + r) = \pi (10)(7 + 4) = 110\pi$$
Final answer: $110\pi \approx 345.58$ cm².
Example 4
A bucket is a frustum with $R = 15$ cm, $r = 10$ cm, and height $20$ cm. How many litres of water does it hold?
$$V = \frac{1}{3}\pi (20)(15^2 + 15 \times 10 + 10^2) = \frac{1}{3}\pi (20)(225 + 150 + 100) = \frac{1}{3}\pi (20)(475)$$
$$V = \frac{9500}{3}\pi \approx 9948.4 \text{ cm}^3$$
Since $1000$ cm³ $= 1$ litre, the bucket holds about $9.95$ litres.
Final answer: $\approx 9.95$ litres.
Example 5
Find the total surface area of a closed frustum with $R = 6$ cm, $r = 3$ cm, and height $4$ cm.
First the slant height: $l = \sqrt{4^2 + (6 - 3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ cm.
$$\text{TSA} = \pi l(R + r) + \pi R^2 + \pi r^2 = \pi (5)(9) + \pi (36) + \pi (9) = 45\pi + 36\pi + 9\pi = 90\pi$$
Final answer: $90\pi \approx 282.74$ cm².
Example 6
A frustum has equal radii $R = r = 4$ cm and height $10$ cm. Confirm the volume matches a cylinder.
$$V = \frac{1}{3}\pi (10)(16 + 16 + 16) = \frac{1}{3}\pi (10)(48) = 160\pi$$
The cylinder check: $\pi R^2 h = \pi (16)(10) = 160\pi$. They agree.
Final answer: $160\pi$ cm³ — the frustum formula reduces to the cylinder when $R = r$.
Why Does the Frustum Shape Show Up Everywhere?
The frustum is one of the most-manufactured solids on Earth, and its geometry is exactly why.
Stability with usability. A wide base resists tipping while a narrower top is easier to grip or pour from - buckets, tumblers, and flowerpots all exploit this.
Stacking and nesting. Frustum-shaped cups nest inside one another, saving space, because each fits into the flare of the one below.
Large-scale engineering. Cooling towers and some rocket sections use frustum profiles to distribute structural load and airflow efficiently.
The engineering version makes the stakes concrete. The hyperbolic and frustum-like profiles of industrial cooling towers are chosen so that volume, airflow, and structural strength balance - a design where getting the two radii and the height right is not decoration but load-bearing necessity.
Where Do Students Trip Up on the Frustum of a Cone?
Mistake 1: Using the sum of radii in the slant height
Where it slips in: Computing $l$ with $(R + r)$ instead of $(R - r)$.
Don't do this: Writing $l = \sqrt{h^2 + (R + r)^2}$.
The correct way: The horizontal leg of the slant-height right triangle is the difference $(R - r)$ - the horizontal step in from the wide base to the narrow top. The first-instinct error is to add the radii, which is exactly the value that appears in the surface-area formula, not the slant-height one. The two get swapped constantly.
Mistake 2: Dropping the middle term $Rr$ in the volume
Where it slips in: Writing the volume as $\frac{1}{3}\pi h(R^2 + r^2)$.
Don't do this: Forgetting the cross term $Rr$.
The correct way: The volume needs all three terms: $R^2 + Rr + r^2$. The middle term is what accounts for the tapering sides - omit it and the volume comes out too small. The rusher who treats the frustum as "two cones' worth of area" loses this term.
Mistake 3: Mixing the two radii
Where it slips in: Assigning the larger radius to the top and smaller to the bottom, or swapping them mid-calculation.
Don't do this: Letting $R$ and $r$ drift between steps.
The correct way: Fix $R$ as the larger (base) radius and $r$ as the smaller (top) radius at the start and keep them consistent. For volume and CSA the order does not change the answer, but for slant height $(R - r)$ a swap flips the sign inside the square - the second-guesser who re-labels halfway gets a different number each time.
Conclusion
A frustum of a cone is a cone with its top sliced off parallel to the base, leaving two circular faces of radii $R$ and $r$.
Its volume is $\frac{1}{3}\pi h(R^2 + Rr + r^2)$ - all three terms matter.
The slant height is $l = \sqrt{h^2 + (R - r)^2}$, using the difference of the radii.
Curved surface area is $\pi l(R + r)$; total surface area adds $\pi R^2 + \pi r^2$.
Buckets, tumblers, lampshades, and cooling towers are everyday frustums.
To go further with a teacher, explore Bhanzu's geometry tutor or high school math tutor, or browse math classes online for solid-geometry practice.
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Practice These to Solidify Your Understanding
Work through these three problems in order:
Find the volume of a frustum with $R = 10$ cm, $r = 6$ cm, and height $12$ cm.
Find the slant height of a frustum with $R = 9$ cm, $r = 5$ cm, and height $3$ cm.
Find the total surface area of a closed frustum with $R = 8$ cm, $r = 4$ cm, and slant height $l = 6$ cm.
Want a live Bhanzu trainer to work through more frustum problems with you? Book a free demo class.
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