Frustum of a cone
A frustum is a cone with its top cut off parallel to the base. Its CSA is π(R + r)l and its volume is (1/3)πh(R² + Rr + r²).
What a frustum is
Cut a cone with a plane parallel to its base. The small cone on top comes off. The part left, with two parallel circular ends, is a frustum. R is the radius of the bigger end, r of the smaller end, h the height between the ends and l the slant height. A bucket, a steel tumbler and a lampshade are frustums.
Frustum formulas
| Measure | Formula |
|---|---|
| Slant height | l = √(h² + (R − r)²) |
| Curved surface area | π(R + r)l |
| Total surface area (both ends closed) | π(R + r)l + πR² + πr² |
| Open bucket (wide top open, bottom radius r) | π(R + r)l + πr² |
| Volume | (1/3)πh(R² + Rr + r²) |
How to draw a frustum
- Draw two flattened circles (ellipses) one above the other: a big one and a small one.
- Join their left ends and their right ends with two slanting straight lines.
- Draw a dotted vertical line between the two centres and label it h.
- Label the radii R (bigger end) and r (smaller end), and one slanting side l.
- For the slant height, show the right triangle with sides h, (R − r) and l.
Example 1: a closed frustum
A frustum has end radii 7 cm and 3 cm and height 3 cm. Find its slant height, curved surface area, total surface area and volume. (Take π = 3.14.)
- Given: R = 7 cm, r = 3 cm, h = 3 cm.
- l = √(h² + (R − r)²) = √(3² + 4²) = √25 = 5 cm
- CSA = π(R + r)l = 3.14 × 10 × 5 = 157 cm²
- TSA = CSA + πR² + πr² = 157 + 3.14 × (49 + 9)
- = 157 + 182.12 = 339.12 cm²
- Volume = (1/3)πh(R² + Rr + r²) = 1/3 × 3.14 × 3 × (49 + 21 + 9)
- = 3.14 × 79 = 248.06 cm³
l = 5 cm, CSA = 157 cm², TSA = 339.12 cm², volume = 248.06 cm³
Example 2: a metal bucket
A metal bucket is a frustum 24 cm deep. The radius of its open top is 14 cm and of its closed bottom is 7 cm. Find the area of metal sheet used and the capacity of the bucket in litres. (Take π = 22/7.)
- Given: R = 14 cm, r = 7 cm, h = 24 cm.
- l = √(24² + (14 − 7)²) = √(576 + 49) = √625 = 25 cm
- Sheet = curved surface + bottom = π(R + r)l + πr²
- = 22/7 × 21 × 25 + 22/7 × 7 × 7
- = 1650 + 154 = 1804 cm²
- Volume = (1/3)πh(R² + Rr + r²) = 1/3 × 22/7 × 24 × (196 + 98 + 49)
- = 1/3 × 22/7 × 24 × 343 = 22 × 8 × 49 = 8624 cm³
- Capacity = 8624 ÷ 1000 = 8.624 litres
Metal sheet = 1804 cm²; capacity = 8.624 litres
A bucket, a steel tumbler, a flower pot and a lampshade are frustums. The frustum volume gives a bucket's capacity in litres.
1 marks: “Write the curved surface area of a frustum with end radii R and r and slant height l.” π(R + r)l.
2 marks: “Find the slant height of a frustum whose end radii are 10 cm and 4 cm and whose height is 8 cm.” l = √(h² + (R − r)²) = √(8² + 6²) = √100. Answer: 10 cm.
5 marks: “A steel bucket open at the top is a frustum of height 12 cm. The radii of its top and bottom are 15 cm and 6 cm. Find the area of steel sheet needed and the volume of water it can hold. (π = 22/7)” l = √(144 + 81) = 15 cm. Sheet = π(R + r)l + πr² = 990 + 113.14 = 1103.14 cm². Volume = (1/3)πh(R² + Rr + r²) = 30888/7 ≈ 4412.57 cm³, about 4.41 litres.
Frustum volume: R squared, R times r, r squared. Add the three, multiply by πh, then divide by 3.