What Is A Frustum?
A frustum is the portion of a solid - a cone or a pyramid - that is left after the top is cut off by a plane parallel to the base. The word comes from Latin for "a piece broken off."
The cut has to be parallel to the base. That is what makes the two faces of a frustum - the base and the smaller top - similar shapes (two circles for a cone, two matching polygons for a pyramid). A slanted cut would give a different, harder solid.
Two everyday kinds:
Conical frustum - from cutting a cone. Two circular faces of radii $R$ (bottom) and $r$ (top). Think bucket or lampshade.
Pyramidal frustum - from cutting a pyramid. Two similar polygon faces. Think of a truncated square pyramid, like many building bases or a table lamp base.
The sloping sides connecting the two faces are the lateral surface. For a cone frustum this is a curved band; for a pyramid frustum it is a set of trapezoidal faces.
For an Example: Look at a bucket, a lampshade, a drinking glass, or a paper coffee cup - you are looking at a frustum.
None of them is a full cone, because a full cone tapers to a useless point. Chop that point off with a flat, parallel cut and you get a shape that can actually hold something and stand up. That chopped-cone solid has a name and a set of formulas worth knowing, and they are prettier than they first look.
What Is The Volume Of A Frustum?
For a conical frustum with bottom radius $R$, top radius $r$, and height $h$ (the perpendicular distance between the two faces):
$$V = \frac{1}{3}\pi h \left(R^2 + r^2 + Rr\right)$$
The symbols:
$R$ = radius of the larger (bottom) circular face.
$r$ = radius of the smaller (top) circular face.
$h$ = perpendicular height between the two faces.
$\pi \approx 3.14159$.
Where does it come from? A frustum is a big cone with a small cone removed from its tip. The big cone's volume minus the small cone's volume, after the similar-triangles algebra of the two cones is worked through, collapses to that tidy $R^2 + r^2 + Rr$ form. A useful check: if $r = R$ (no taper), the shape is a cylinder, and the formula gives $\frac{1}{3}\pi h(3R^2) = \pi R^2 h$, exactly the cylinder volume. And if $r = 0$ (cut all the way to the point), it gives $\frac{1}{3}\pi h R^2$, the full cone. The formula contains both simpler shapes.
What Is the Surface Area of a Frustum?
First the slant height $l$ of a conical frustum - the length up the sloping side, not the vertical height:
$$l = \sqrt{h^2 + (R - r)^2}$$
Then the areas:
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 is the curved band plus the two circular ends (bottom area $\pi R^2$ and top area $\pi r^2$). Whether you include one end, both, or neither depends on the object - an open bucket has one end; a sealed drum has both.
For a pyramidal frustum, each sloping face is an isosceles trapezoid, and the lateral surface area is the sum of those trapezoid areas. The total surface area adds the two polygon end-faces.
Examples of Frustum
Six worked examples, easier to harder. The problem statement is bold; the steps are not. We use $\cdot$ for every product and take $\pi \approx 3.14$ unless a fraction is cleaner.
Example 1
Find the slant height of a conical frustum with $R = 5$ cm, $r = 2$ cm, $h = 4$ cm.
$$l = \sqrt{h^2 + (R - r)^2} = \sqrt{4^2 + (5 - 2)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ cm}$$
Final answer: slant height 5 cm.
Example 2
Find the volume of a conical frustum with $R = 6$ cm, $r = 3$ cm, $h = 7$ cm. A student computes it as $\frac{1}{3}\pi h (R^2 + r^2)$, forgetting a term. What is the true volume?
The tempting move is to remember the cone-like $\frac{1}{3}\pi h$ front and pair it with $R^2 + r^2$, mirroring the two faces. That gives $\frac{1}{3}\cdot 3.14 \cdot 7 \cdot (36 + 9) = \frac{1}{3}\cdot 3.14 \cdot 7 \cdot 45 \approx 329.7$ cm³.
But the frustum volume has a cross term $Rr$ that the two-cone subtraction leaves behind:
$$V = \frac{1}{3}\pi h (R^2 + r^2 + Rr)$$ $$V = \frac{1}{3}\cdot 3.14 \cdot 7 \cdot (36 + 9 + 6\cdot 3)$$ $$V = \frac{1}{3}\cdot 3.14 \cdot 7 \cdot 63$$ $$V = \frac{1}{3}\cdot 1384.74 \approx 461.6 \text{ cm}^3$$
Final answer: about 461.6 cm³, not 329.7 cm³. Dropping the $Rr$ term is the most common frustum error - it comes from misremembering the formula as a plain average of the two circular areas. The $Rr$ is what accounts for the sloping wall between the two sizes.
Example 3
Find the curved surface area of the frustum from Example 1 ($R = 5$, $r = 2$, $l = 5$).
$$\text{CSA} = \pi l (R + r) = 3.14 \cdot 5 \cdot (5 + 2) = 3.14 \cdot 5 \cdot 7 = 109.9 \text{ cm}^2$$
Final answer: about 109.9 cm².
Example 4
Find the total surface area of the same frustum ($R = 5$, $r = 2$, $l = 5$).
Add the two circular ends to the curved surface:
$$\text{TSA} = \pi l (R + r) + \pi R^2 + \pi r^2$$ $$\text{TSA} = 109.9 + 3.14 \cdot 25 + 3.14 \cdot 4$$ $$\text{TSA} = 109.9 + 78.5 + 12.56 = 200.96 \text{ cm}^2$$
Final answer: about 200.96 cm².
Example 5
A bucket is a conical frustum, open at the top, with bottom radius 10 cm, top radius 15 cm, and height 12 cm. How much water can it hold, in litres? Use $\pi \approx 3.14$.
Here $R = 15$ (the wider open top), $r = 10$ (the narrower base), $h = 12$.
$$V = \frac{1}{3}\pi h (R^2 + r^2 + Rr)$$ $$V = \frac{1}{3}\cdot 3.14 \cdot 12 \cdot (225 + 100 + 150)$$ $$V = \frac{1}{3}\cdot 3.14 \cdot 12 \cdot 475$$ $$V = \frac{1}{3}\cdot 17898 = 5966 \text{ cm}^3$$
Since $1000 \text{ cm}^3 = 1$ litre, the bucket holds about 5.97 litres.
Final answer: roughly 5.97 litres. Which radius is $R$ and which is $r$ does not change the volume (the formula is symmetric in them), but it matters for slant height and surface area.
Example 6
A truncated square pyramid (a pyramidal frustum) has a bottom square of side 8 cm, a top square of side 4 cm, and height 9 cm. Find its volume.
The pyramid-frustum volume uses the two base areas $A_1$ and $A_2$:
$$V = \frac{1}{3}h \left(A_1 + A_2 + \sqrt{A_1 A_2}\right)$$
Bottom area $A_1 = 8^2 = 64$ cm²; top area $A_2 = 4^2 = 16$ cm².
$$V = \frac{1}{3}\cdot 9 \cdot \left(64 + 16 + \sqrt{64 \cdot 16}\right)$$ $$V = 3 \cdot (64 + 16 + \sqrt{1024})$$ $$V = 3 \cdot (64 + 16 + 32) = 3 \cdot 112 = 336 \text{ cm}^3$$
Final answer: 336 cm³. Notice the same three-term pattern - two faces plus a "geometric mean" cross term - as the cone formula.
Why the Frustum Matters
The frustum is the shape you reach for whenever a full cone or pyramid would be impractical, which turns out to be constantly.
The reason it earns its own formulas:
A full cone can't hold anything or stand up. A cone tapers to a point, so a bucket, a glass, or a flowerpot has to be a frustum - flat top, flat bottom, a body in between. The frustum is the "usable cone."
Big structures taper for strength, not to a point. Cooling towers, dam walls, and lampposts are wide at the base and narrower up top for stability, but they stop short of a spike. That is a frustum, and its volume tells engineers how much material it takes.
Bridging two sizes. Any duct, funnel, or transition piece that connects a wide opening to a narrow one is a frustum. Sheet-metal workers use the slant height and lateral-area formulas to cut the flat pattern that rolls up into one.
The idea is old and grand: the Great Pyramid of Giza and the truncated pyramids of ancient Egypt drove some of the earliest volume mathematics. The Moscow Mathematical Papyrus, from around 1850 BCE, contains a correct method for the volume of a truncated square pyramid - meaning humans could compute a frustum's volume nearly four thousand years before the modern formula was written down.
[INTERACTIVE: A 3D conical frustum the user can reshape by dragging the top radius $r$ from 0 (full cone) up to $R$ (cylinder). As $r$ changes, the volume $\frac{1}{3}\pi h(R^2 + r^2 + Rr)$ updates live, and labels flag the two limiting cases - "cone" when $r = 0$, "cylinder" when $r = R$ - so the reader sees the frustum sit between them.]
The Mistakes Students Make Most Often
Three errors cover most wrong frustum answers.
Mistake 1: Dropping the $Rr$ cross term in the volume
Where it slips in: recalling the volume formula from memory under time pressure (the memorizer who stores "$\frac{1}{3}\pi h$ times the two radii squared").
Don't do this: write $V = \frac{1}{3}\pi h (R^2 + r^2)$.
The correct way: the true formula is $V = \frac{1}{3}\pi h (R^2 + r^2 + Rr)$. The $Rr$ term is exactly what the two-cone subtraction leaves behind, and skipping it under-counts the volume every time - it is the single most common frustum mistake.
Mistake 2: Using vertical height in place of slant height (or the reverse)
Where it slips in: surface-area problems, which need slant height $l$, while volume needs vertical height $h$ (the rusher who plugs whichever height the problem gave).
Don't do this: put $h$ straight into $\pi l (R + r)$, or put $l$ into the volume formula.
The correct way: compute $l = \sqrt{h^2 + (R - r)^2}$ first, then use $l$ for surface area and $h$ for volume. They are different lengths - $l$ runs up the slope, $h$ runs straight up.
Mistake 3: Mixing up which radius is $R$ and which is $r$ in surface area
Where it slips in: open-top objects like buckets, where the wider face is the top, not the bottom (the second-guesser unsure which end is "the base").
Don't do this: assume the bottom is always the larger radius, or swap the radii mid-calculation.
The correct way: for volume it does not matter (the formula is symmetric), but for slant height use $|R - r|$ and stay consistent. Label the two radii once at the start and keep them fixed. Units stay consistent throughout: if radii are in cm, volume is in cm³ and area in cm².
Conclusion
A frustum is a cone or pyramid with its top cut off by a plane parallel to the base.
Conical-frustum volume: $\frac{1}{3}\pi h (R^2 + r^2 + Rr)$ - the $Rr$ cross term is essential.
Slant height $l = \sqrt{h^2 + (R - r)^2}$; curved surface $\pi l (R + r)$; total surface adds $\pi R^2 + \pi r^2$.
A pyramidal frustum uses the same three-term pattern with base areas and a $\sqrt{A_1 A_2}$ term.
Use vertical height for volume and slant height for surface area - they are different lengths.
To take the frustum and solid geometry further with a teacher, explore Bhanzu's geometry tutor, a high school math tutor for mensuration chapters, or math tutoring for structured practice.
Practice these problems to solidify your understanding: (1) slant height for $R = 8$, $r = 4$, $h = 3$; (2) volume for $R = 4$, $r = 2$, $h = 6$ (use $\pi \approx 3.14$); (3) truncated square pyramid, base 6 cm, top 3 cm, height 8 cm — find the volume. Answer to Question 1: 5. Answer to Question 2: about 175.8 cm³. Answer to Question 3: 168 cm³. If you get stuck on the pyramid problem, come back to Example 6. Want a live trainer to work through solids with your child? Book a free demo class.
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