Mensuration • Topic 2 of 3

Frustum of a Cone

What is a frustum of a cone? Take a right circular cone and slice it with a plane parallel to the base. Throw away the small cone at the top. What is left — the lower portion with two flat circular faces of different sizes — is a frustum of a cone (from the Latin for "a piece broken off"). A bucket, a drinking glass, a Turkish lamp shade, a flower pot and a frying-pan lid are all everyday frustums.

The four measurements.

  • Larger radius $R$ — radius of the bigger circular face.
  • Smaller radius $r$ — radius of the smaller circular face.
  • Vertical height $h$ — straight perpendicular distance between the two faces.
  • Slant height $l$ — the distance measured along the slanted side wall.

Where the slant-height formula comes from. Drop a perpendicular from the smaller circle to the larger circle. The horizontal gap between the two edges is $R-r$, the vertical gap is $h$. These form a right triangle whose hypotenuse is the slant side, so by Pythagoras:

$l=\sqrt{h^2+(R-r)^2}$.

The frustum formulas. Each can be derived by subtracting the small cone from the big cone, but for Class 10 you may quote them directly.

  • Slant height: $l=\sqrt{h^2+(R-r)^2}$.
  • Curved (lateral) surface area: $\text{CSA}=\pi(R+r)\,l$.
  • Total surface area: $\text{TSA}=\pi(R+r)\,l+\pi R^2+\pi r^2$ — the curved wall plus both circular faces.
  • Volume: $V=\tfrac13\pi h\,(R^2+Rr+r^2)$.

Sketch of the volume derivation. If the full cone has height $H$ and base radius $R$, and the removed top cone has height $H-h$ and radius $r$, similar triangles give $\dfrac{R}{H}=\dfrac{r}{H-h}$. Subtracting the two cone volumes $\tfrac13\pi R^2 H-\tfrac13\pi r^2(H-h)$ and simplifying collapses neatly to $\tfrac13\pi h(R^2+Rr+r^2)$. You do not need to reproduce this in an exam, but knowing it exists explains why the $Rr$ cross-term appears.

Which surfaces to use, by object.

  • Open bucket / glass / mug: metal or material needed $=$ CSA $+$ area of the smaller base only ($\pi r^2$), because the wide top is open.
  • Lamp shade / open frustum tube: only the CSA $\pi(R+r)l$ (both ends open).
  • Capacity of a bucket: use the volume formula; convert to litres with $1000\text{ cm}^3=1$ litre.
  • Closed frustum (rare): the full TSA with both circles.

Common mistakes. (i) Writing $(R+r)$ inside the square root — it must be $(R-r)$ for slant height but $(R+r)$ for CSA. (ii) Using diameters instead of radii. (iii) Adding both circular faces for an open bucket when only the base is closed. (iv) Forgetting to convert cubic cm to litres for capacity questions. (v) Mixing the vertical height $h$ and slant height $l$: volume uses $h$, surface area uses $l$.

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A frustum formed by cutting a small cone off the top of a full cone with a plane parallel to the base A frustum = full cone with the top cone sliced off small cone removed cut ∥ base the cutting plane is parallel to the base Anatomy of a frustum: small base radius r on top, large base radius R below, height h and slant height l Anatomy of a frustum h r R l l = √(h^2 + (R−r)^2)
Solved Examples
1
Worked Example
A bucket is a frustum of a cone. The radii of its top and bottom are 20 cm and 8 cm, and its vertical height is 16 cm. Find its slant height.
Solution
  1. Step 1: $R=20$ cm, $r=8$ cm, $h=16$ cm.
  2. Step 2: $l=\sqrt{h^2+(R-r)^2}=\sqrt{16^2+(20-8)^2}$.
  3. Step 3: $=\sqrt{256+144}=\sqrt{400}$.
  4. Step 4: $l=20$ cm.

Answer: $20$ cm.

2
Worked Example
A drinking glass shaped like a frustum has height 14 cm and the diameters of its two ends are 4 cm and 2 cm. Find its capacity. (Take $\pi=22/7$.)
Solution
  1. Step 1: $R=4/2=2$ cm, $r=2/2=1$ cm, $h=14$ cm.
  2. Step 2: $V=\tfrac13\pi h(R^2+Rr+r^2)=\tfrac13\cdot\tfrac{22}{7}\cdot 14\cdot(4+2+1)$.
  3. Step 3: $=\tfrac13\cdot\tfrac{22}{7}\cdot 14\cdot 7=\tfrac13\cdot 22\cdot 14=\tfrac{308}{3}$.
  4. Step 4: $V=\tfrac{308}{3}\approx 102.67$ cubic cm.

Answer: $\tfrac{308}{3}\approx 102.67$ cubic cm.

3
Worked Example
A lamp shade is a frustum with small radius 6 cm, large radius 14 cm and slant height 10 cm. Find the area of material needed to cover its curved surface. (Take $\pi=22/7$.)
Solution
  1. Step 1: Both ends are open, so only the CSA is needed.
  2. Step 2: $\text{CSA}=\pi(R+r)l=\tfrac{22}{7}\cdot(14+6)\cdot 10$.
  3. Step 3: $=\tfrac{22}{7}\cdot 20\cdot 10=\tfrac{4400}{7}$.
  4. Step 4: $\approx 628.57$ sq cm.

Answer: $\tfrac{4400}{7}\approx 628.57$ sq cm.

4
Worked Example
The radii of the ends of a frustum are 14 cm and 6 cm and its height is 6 cm. Find its curved surface area. (Take $\pi=22/7$.)
Solution
  1. Step 1: $R=14$, $r=6$, $h=6$. First find slant height.
  2. Step 2: $l=\sqrt{h^2+(R-r)^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$ cm.
  3. Step 3: $\text{CSA}=\pi(R+r)l=\tfrac{22}{7}\cdot 20\cdot 10=\tfrac{4400}{7}$.
  4. Step 4: $\approx 628.57$ sq cm.

Answer: $\tfrac{4400}{7}\approx 628.57$ sq cm.

5
Worked Example
A metal bucket open at the top is a frustum with radii 20 cm and 12 cm and height 15 cm. Find the area of the metal sheet used (including the closed bottom). (Take $\pi=3.14$.)
Solution
  1. Step 1: $R=20$, $r=12$, $h=15$. Slant height $l=\sqrt{15^2+(20-12)^2}=\sqrt{225+64}=\sqrt{289}=17$ cm.
  2. Step 2: Metal $=$ CSA $+$ area of smaller base (the top is open).
  3. Step 3: CSA $=\pi(R+r)l=3.14\cdot 32\cdot 17=3.14\cdot 544=1708.16$ sq cm.
  4. Step 4: Bottom $=\pi r^2=3.14\cdot 144=452.16$ sq cm.
  5. Step 5: Total metal $=1708.16+452.16=2160.32$ sq cm.

Answer: $2160.32$ sq cm.

6
Worked Example
A bucket is a frustum with radii 20 cm and 12 cm and height 15 cm. Find its capacity in litres. (Take $\pi=3.14$.)
Solution
  1. Step 1: $R=20$, $r=12$, $h=15$.
  2. Step 2: $V=\tfrac13\pi h(R^2+Rr+r^2)=\tfrac13\cdot 3.14\cdot 15\cdot(400+240+144)$.
  3. Step 3: $R^2+Rr+r^2=400+240+144=784$.
  4. Step 4: $V=\tfrac13\cdot 3.14\cdot 15\cdot 784=3.14\cdot 5\cdot 784=3.14\cdot 3920=12308.8$ cubic cm.
  5. Step 5: Capacity $=12308.8\div 1000=12.3088\approx 12.31$ litres.

Answer: $\approx 12308.8$ cubic cm $\approx 12.31$ litres.

7
Worked Example
The slant height of a frustum is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find its curved surface area.
Solution
  1. Step 1: Perimeter $=2\pi R=18\Rightarrow \pi R=9$; perimeter $=2\pi r=6\Rightarrow \pi r=3$.
  2. Step 2: $\text{CSA}=\pi(R+r)l=(\pi R+\pi r)\,l$.
  3. Step 3: $=(9+3)\cdot 4=12\cdot 4=48$ sq cm.

Answer: $48$ sq cm.

8
Worked Example
A frustum has $R=10$ cm, $r=4$ cm and $h=9$ cm. Find its volume. (Take $\pi=3.14$.)
Solution
  1. Step 1: $R^2+Rr+r^2=100+40+16=156$.
  2. Step 2: $V=\tfrac13\pi h(R^2+Rr+r^2)=\tfrac13\cdot 3.14\cdot 9\cdot 156$.
  3. Step 3: $=3.14\cdot 3\cdot 156=3.14\cdot 468=1469.52$ cubic cm.

Answer: $1469.52$ cubic cm.

9
Worked Example
A frustum has $R=33$ cm, $r=27$ cm and slant height 10 cm. Find its total surface area. (Take $\pi=22/7$.)
Solution
  1. Step 1: $\text{TSA}=\pi(R+r)l+\pi R^2+\pi r^2$.
  2. Step 2: CSA $=\tfrac{22}{7}\cdot(33+27)\cdot 10=\tfrac{22}{7}\cdot 600=\tfrac{13200}{7}\approx 1885.71$ sq cm.
  3. Step 3: $\pi R^2=\tfrac{22}{7}\cdot 1089=\tfrac{23958}{7}\approx 3422.57$ sq cm.
  4. Step 4: $\pi r^2=\tfrac{22}{7}\cdot 729=\tfrac{16038}{7}\approx 2291.14$ sq cm.
  5. Step 5: TSA $\approx 1885.71+3422.57+2291.14=7599.42$ sq cm.

Answer: $\approx 7599.42$ sq cm.

10
Worked Example
A full cone of height 20 cm and base radius 10 cm is cut by a plane parallel to the base, halfway up. Find the volume of the frustum so formed. (Take $\pi=3.14$.)
Solution
  1. Step 1: The cut is halfway up, so by similar triangles the small cone has height 10 cm and radius 5 cm.
  2. Step 2: The frustum has $R=10$ cm (base), $r=5$ cm (cut face), height $h=10$ cm.
  3. Step 3: $R^2+Rr+r^2=100+50+25=175$.
  4. Step 4: $V=\tfrac13\pi h(175)=\tfrac13\cdot 3.14\cdot 10\cdot 175=3.14\cdot\tfrac{1750}{3}\approx 1831.67$ cubic cm.

Answer: $\approx 1831.67$ cubic cm.

11
Worked Example
An open frustum-shaped container (bucket) has $R=15$ cm, $r=5$ cm and height 24 cm. Find (a) its slant height and (b) the cost of metal sheet at $₹$10 per 100 sq cm, including the closed base. (Take $\pi=3.14$.)
Solution
  1. Step 1: $l=\sqrt{h^2+(R-r)^2}=\sqrt{24^2+10^2}=\sqrt{576+100}=\sqrt{676}=26$ cm.
  2. Step 2: CSA $=\pi(R+r)l=3.14\cdot 20\cdot 26=3.14\cdot 520=1632.8$ sq cm.
  3. Step 3: Base (smaller end closed) $=\pi r^2=3.14\cdot 25=78.5$ sq cm.
  4. Step 4: Total sheet $=1632.8+78.5=1711.3$ sq cm.
  5. Step 5: Cost $=\dfrac{1711.3}{100}\times 10=$ ₹$171.13$.

Answer: (a) $26$ cm; (b) area $1711.3$ sq cm, cost $₹ 171.13$.

12
Worked Example
An oil funnel is made of a cylindrical portion of radius 6 cm and height 10 cm attached to a frustum of a cone. The frustum has radii 6 cm (top, where it meets the cylinder) and 3 cm (bottom) and slant height 5 cm. Find the area of tin sheet needed to make the funnel (both ends of the funnel are open). (Take $\pi=22/7$.)
Solution
  1. Step 1: Both open ends mean no circular caps; sheet $=$ CSA of cylinder $+$ CSA of frustum.
  2. Step 2: CSA cylinder $=2\pi r h=2\cdot\tfrac{22}{7}\cdot 6\cdot 10=\tfrac{2640}{7}\approx 377.14$ sq cm.
  3. Step 3: CSA frustum $=\pi(R+r)l=\tfrac{22}{7}\cdot(6+3)\cdot 5=\tfrac{22}{7}\cdot 45=\tfrac{990}{7}\approx 141.43$ sq cm.
  4. Step 4: Total tin $=\tfrac{2640}{7}+\tfrac{990}{7}=\tfrac{3630}{7}\approx 518.57$ sq cm.

Answer: $\tfrac{3630}{7}\approx 518.57$ sq cm.

Key Points

  • A frustum is formed by cutting a cone parallel to its base and removing the top vertex portion.
  • It features two separate circular base faces with different radii, labeled $R$ (large) and $r$ (small).
  • The slant height is calculated as $l = \sqrt{h^2 + (R - r)^2}$.
  • The curved wall area depends on both radii: $\text{CSA} = \pi(R + r)l$.
  • The volume formula modifies the standard cone formula to include both bases: $V = \frac{1}{3}\pi h(R^2 + r^2 + Rr)$.
  • For an open bucket the material is CSA plus the smaller base only ($\pi r^2$), not both circles.
  • Volume uses the vertical height $h$; surface area uses the slant height $l$ — never swap them.
  • For an open bucket the material is CSA plus the smaller base only ($\pi r^2$), not both circles.
  • Volume uses the vertical height $h$; surface area uses the slant height $l$ — never swap them.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.A frustum is formed by cutting a cone with a plane:
Explanation: Parallel cut removes the top cone.
Q2.The volume of a frustum is:
Explanation: Standard frustum volume.
Q3.The slant height of a frustum is:
Explanation: From the right triangle.
Q4.The CSA of a frustum is:
Explanation: $\pi(R+r)l$.
Practice more MCQs → 📝 Assignment · 35 marks