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Either paper 3 marks
The diagram shows a frustum. It was made by standing a cone on its base and slicing a smaller cone off the top, parallel to the base.
The frustum has base radius \(15\) cm, top radius \(3\) cm and height \(32\) cm.
Find the height of the cone that was removed.
Show answer
\(8\ \text{cm}\)The cut is parallel to the base, so the cone that was taken off and the cone you started with are similar — the same shape at different sizes.
Similar solids have all their lengths in one ratio, and the radii are \(3\) cm and \(15\) cm. So the heights are in that same ratio.
Call the removed height \(k\) cm. The whole cone is \(k + 32\) cm tall, because the frustum is what is left below the cut.
\(\frac{k}{k + 32 } = \frac{ 3 }{ 15 }\)
Cross-multiply:
\(15k = 3(k + 32) = 3k + 96\)
\(15k - 3k = 96\)
\(12k = 96\)
\(k = 96 \div 12 = 8\)
The removed cone is \(8\) cm tall — and as a check, the whole cone is \(8 + 32 = 40\) cm, and \(8 : 40\) is the same ratio as \(3 : 15\).
\(8\) cm
- M1similar cones: \(\frac{k}{k + 32 } = \frac{ 3 }{ 15 }\)
- M1cross-multiplies and collects: \(12k = 96\)
- A1removed height \(8\) cm
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Either paper 4 marks
A cone of height \(54\) cm and base radius \(18\) cm had a smaller cone sliced off the top, parallel to the base. The solid left behind is the frustum shown.
The top of the frustum is a circle of radius \(3\) cm, and the cone that was removed is \(9\) cm tall.
Volume of a cone \(= \frac{1}{3}\pi r^2 h\)
Work out the volume of the frustum. Give your answer in terms of \(\pi\).
Show answer
\(5805\pi\ \text{cm}^3\)A frustum has no volume formula of its own. It is the whole cone take away the cone that was removed, so work out both and subtract.
The whole cone, radius \(18\) cm and height \(54\) cm:
\(V = \frac{1}{3} \times \pi \times 18^2 \times 54 = \frac{1}{3} \times \pi \times 324 \times 54 = 5832\pi\) cm³
The removed cone, radius \(3\) cm and height \(9\) cm:
\(V = \frac{1}{3} \times \pi \times 3^2 \times 9 = \frac{1}{3} \times \pi \times 9 \times 9 = 27\pi\) cm³
Subtract, and keep \(\pi\) untouched throughout:
\(5832\pi - 27\pi = 5805\pi\)
The frustum itself is \(54 - 9 = 45\) cm tall, and that height belongs to neither cone. Putting it into one of the formulae above is the commonest error in this topic.
\(5805\pi\) cm³
- M1volume of whole cone \(5832\pi\)
- M1volume of removed cone \(27\pi\)
- M1subtracts the removed cone from the whole cone
- A1volume \(5805\pi\ \text{cm}^3\)
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Either paper 5 marks
A solid cone has base radius \(15\) cm and height \(45\) cm.
A cone of height \(18\) cm is removed from the top of it, by cutting parallel to the base. The solid left behind is the frustum shown.
Volume of a cone \(= \frac{1}{3}\pi r^2 h\)
(a)Calculate the radius of the circular top of the frustum.
(b)Calculate the volume of the frustum. Give your answer in terms of \(\pi\).
Show answer
(a) \(6\ \text{cm}\)
(b) \(3159\pi\ \text{cm}^3\)The removed cone and the original cone are similar, because the cut is parallel to the base.
Their heights are \(18\) cm and \(45\) cm, so every length of the small one is \(\frac{ 18 }{ 45 }\) of the matching length of the big one — including the radius.
\(r = 15 \times \frac{ 18 }{ 45 } = 6\)
\(6\) cm
The frustum is what is left, so its volume is the original cone's take away the removed cone's.
Original cone: \(\frac{1}{3} \times \pi \times 225 \times 45 = 3375\pi\) cm³
Removed cone: \(\frac{1}{3} \times \pi \times 36 \times 18 = 216\pi\) cm³
\(3375\pi - 216\pi = 3159\pi\)
The frustum's own height, \(27\) cm, belongs to neither cone, so it is not used in either line.
\(3159\pi\) cm³
- (a) M1scale factor from the heights, \(\frac{ 18 }{ 45 }\), applied to the base radius
- (a) A1top radius \(6\) cm
- (b) M1volumes of the two cones \(3375\pi\) and \(216\pi\)
- (b) M1subtracts the removed cone from the original cone
- (b) A1volume \(3159\pi\ \text{cm}^3\)
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Either paper 4 marks
Here is a frustum, made by slicing a smaller cone off the top of a cone of slant height \(60\) cm.
The base radius is \(20\) cm and the top radius is \(5\) cm.
Curved surface area of a cone \(= \pi r l\), where \(l\) is the slant height.
Work out the curved surface area of the frustum. Give your answer in terms of \(\pi\).
Show answer
\(1125\pi\ \text{cm}^2\)The curved surface of the frustum is the whole cone's curved surface take away the removed cone's, so the removed cone's slant height is needed first.
The two cones are similar, and a slant height is a length like any other, so it is in the same ratio as the radii:
\(\frac{ 15 }{ 60 } = \frac{ 5 }{ 20 }\)
\(l = 60 \times \frac{ 5 }{ 20 } = 15\) cm
Now take one curved surface from the other. The base circle and the top circle are not part of a curved surface, so neither appears:
Whole cone: \(\pi r l = \pi \times 20 \times 60 = 1200\pi\) cm²
Removed cone: \(\pi r l = \pi \times 5 \times 15 = 75\pi\) cm²
\(1200\pi - 75\pi = 1125\pi\)
Note that \(45\) cm, the sloping edge of the frustum itself, is the difference of the two slant heights, not either of them.
\(1125\pi\) cm²
- M1slant height of removed cone from the radius ratio, \(15\) cm
- M1curved surface areas \(1200\pi\) and \(75\pi\)
- M1subtracts the removed cone from the whole cone
- A1curved surface area \(1125\pi\ \text{cm}^2\)
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Calculator 5 marks
A vase is the shape of a frustum, made by removing a cone of height \(36\) cm from the top of a cone of height \(60\) cm. It stands on its wider end, and the opening is the smaller circle on top.
The base has radius \(10\) cm, the opening has radius \(6\) cm, and the vase is \(24\) cm deep.
Volume of a cone \(= \frac{1}{3}\pi r^2 h\)
Work out how many litres of water the vase holds when full. Give your answer to 1 decimal place.
\(1\) litre \(= 1000\) cm³
Show answer
\(4.9\ \text{litres}\)Work in cubic centimetres first and convert at the very end. Rounding before the conversion is where this question is usually lost.
The vase is a whole cone with a smaller cone taken out of it, so find both volumes and subtract.
Whole cone: \(\frac{1}{3} \times \pi \times 100 \times 60 = 2000\pi\) cm³
Removed cone: \(\frac{1}{3} \times \pi \times 36 \times 36 = 432\pi\) cm³
\(2000\pi - 432\pi = 1568\pi\) cm³
Only now put a number to \(\pi\). Using the \(\pi\) key on a calculator, \(1568\pi = 4926.0\) cm³ to 1 decimal place.
There are \(1000\) cm³ in a litre, so divide by \(1000\):
\(4.9\) litres
- M1volume of whole cone \(2000\pi\) cm³
- M1volume of removed cone \(432\pi\) cm³
- M1subtracting to get \(1568\pi = 4926.0\) cm³
- M1dividing by \(1000\) to convert to litres
- A1capacity \(4.9\) litres
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Calculator 4 marks
A light shade is made from card stretched around the curved surface of a frustum. It is open at both ends.
The frustum was cut from a cone of slant height \(54\) cm. Its two circular ends have radii \(18\) cm and \(6\) cm.
Curved surface area of a cone \(= \pi r l\), where \(l\) is the slant height.
Find the area of card needed. Give your answer to 1 decimal place.
Show answer
\(2714.3\ \text{cm}^2\)The light shade is open at both ends, so only the curved surface is made of card — no circles.
That curved surface is the whole cone's take away the removed cone's, so find the removed cone's slant height first. The two cones are similar, so the slant heights are in the same ratio as the radii:
\(l = 54 \times \frac{ 6 }{ 18 } = 18\) cm
Whole cone: \(\pi r l = \pi \times 18 \times 54 = 972\pi\) cm²
Removed cone: \(\pi r l = \pi \times 6 \times 18 = 108\pi\) cm²
\(972\pi - 108\pi = 864\pi\) cm²
Keeping \(\pi\) to the end and rounding once. Using the \(\pi\) key on a calculator, \(864\pi = 2714.3\) cm² to 1 decimal place.
\(2714.3\) cm²
- M1slant height of removed cone from the radius ratio, \(18\) cm
- M1curved surface areas \(972\pi\) and \(108\pi\)
- M1subtracts to get \(864\pi\) cm²
- A1area \(2714.3\) cm²
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Either paper 3 marks
The diagram shows a frustum. It was made by standing a cone on its base and slicing a smaller cone off the top, parallel to the base.
The frustum has base radius \(12\) cm, top radius \(4\) cm and height \(20\) cm.
Calculate the height of the cone that was removed.
Show answer
\(10\ \text{cm}\)The cut is parallel to the base, so the cone that was taken off and the cone you started with are similar — the same shape at different sizes.
Similar solids have all their lengths in one ratio, and the radii are \(4\) cm and \(12\) cm. So the heights are in that same ratio.
Call the removed height \(k\) cm. The whole cone is \(k + 20\) cm tall, because the frustum is what is left below the cut.
\(\frac{k}{k + 20 } = \frac{ 4 }{ 12 }\)
Cross-multiply:
\(12k = 4(k + 20) = 4k + 80\)
\(12k - 4k = 80\)
\(8k = 80\)
\(k = 80 \div 8 = 10\)
The removed cone is \(10\) cm tall — and as a check, the whole cone is \(10 + 20 = 30\) cm, and \(10 : 30\) is the same ratio as \(4 : 12\).
\(10\) cm
- M1similar cones: \(\frac{k}{k + 20 } = \frac{ 4 }{ 12 }\)
- M1cross-multiplies and collects: \(8k = 80\)
- A1removed height \(10\) cm
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Either paper 4 marks
A cone of height \(36\) cm and base radius \(6\) cm had a smaller cone sliced off the top, parallel to the base. The solid left behind is the frustum shown.
The top of the frustum is a circle of radius \(4\) cm, and the cone that was removed is \(24\) cm tall.
Volume of a cone \(= \frac{1}{3}\pi r^2 h\)
Calculate the volume of the frustum. Give your answer in terms of \(\pi\).
Show answer
\(304\pi\ \text{cm}^3\)A frustum has no volume formula of its own. It is the whole cone take away the cone that was removed, so work out both and subtract.
The whole cone, radius \(6\) cm and height \(36\) cm:
\(V = \frac{1}{3} \times \pi \times 6^2 \times 36 = \frac{1}{3} \times \pi \times 36 \times 36 = 432\pi\) cm³
The removed cone, radius \(4\) cm and height \(24\) cm:
\(V = \frac{1}{3} \times \pi \times 4^2 \times 24 = \frac{1}{3} \times \pi \times 16 \times 24 = 128\pi\) cm³
Subtract, and keep \(\pi\) untouched throughout:
\(432\pi - 128\pi = 304\pi\)
The frustum itself is \(36 - 24 = 12\) cm tall, and that height belongs to neither cone. Putting it into one of the formulae above is the commonest error in this topic.
\(304\pi\) cm³
- M1volume of whole cone \(432\pi\)
- M1volume of removed cone \(128\pi\)
- M1subtracts the removed cone from the whole cone
- A1volume \(304\pi\ \text{cm}^3\)
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Either paper 5 marks
A solid cone has base radius \(18\) cm and height \(54\) cm.
A cone of height \(24\) cm is removed from the top of it, by cutting parallel to the base. The solid left behind is the frustum shown.
Volume of a cone \(= \frac{1}{3}\pi r^2 h\)
(a)Work out the radius of the circular top of the frustum.
(b)Work out the volume of the frustum. Give your answer in terms of \(\pi\).
Show answer
(a) \(8\ \text{cm}\)
(b) \(5320\pi\ \text{cm}^3\)The removed cone and the original cone are similar, because the cut is parallel to the base.
Their heights are \(24\) cm and \(54\) cm, so every length of the small one is \(\frac{ 24 }{ 54 }\) of the matching length of the big one — including the radius.
\(r = 18 \times \frac{ 24 }{ 54 } = 8\)
\(8\) cm
The frustum is what is left, so its volume is the original cone's take away the removed cone's.
Original cone: \(\frac{1}{3} \times \pi \times 324 \times 54 = 5832\pi\) cm³
Removed cone: \(\frac{1}{3} \times \pi \times 64 \times 24 = 512\pi\) cm³
\(5832\pi - 512\pi = 5320\pi\)
The frustum's own height, \(30\) cm, belongs to neither cone, so it is not used in either line.
\(5320\pi\) cm³
- (a) M1scale factor from the heights, \(\frac{ 24 }{ 54 }\), applied to the base radius
- (a) A1top radius \(8\) cm
- (b) M1volumes of the two cones \(5832\pi\) and \(512\pi\)
- (b) M1subtracts the removed cone from the original cone
- (b) A1volume \(5320\pi\ \text{cm}^3\)
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Either paper 4 marks
Here is a frustum, made by slicing a smaller cone off the top of a cone of slant height \(60\) cm.
The base radius is \(20\) cm and the top radius is \(8\) cm.
Curved surface area of a cone \(= \pi r l\), where \(l\) is the slant height.
Calculate the curved surface area of the frustum. Give your answer in terms of \(\pi\).
Show answer
\(1008\pi\ \text{cm}^2\)The curved surface of the frustum is the whole cone's curved surface take away the removed cone's, so the removed cone's slant height is needed first.
The two cones are similar, and a slant height is a length like any other, so it is in the same ratio as the radii:
\(\frac{ 24 }{ 60 } = \frac{ 8 }{ 20 }\)
\(l = 60 \times \frac{ 8 }{ 20 } = 24\) cm
Now take one curved surface from the other. The base circle and the top circle are not part of a curved surface, so neither appears:
Whole cone: \(\pi r l = \pi \times 20 \times 60 = 1200\pi\) cm²
Removed cone: \(\pi r l = \pi \times 8 \times 24 = 192\pi\) cm²
\(1200\pi - 192\pi = 1008\pi\)
Note that \(36\) cm, the sloping edge of the frustum itself, is the difference of the two slant heights, not either of them.
\(1008\pi\) cm²
- M1slant height of removed cone from the radius ratio, \(24\) cm
- M1curved surface areas \(1200\pi\) and \(192\pi\)
- M1subtracts the removed cone from the whole cone
- A1curved surface area \(1008\pi\ \text{cm}^2\)