Cambridge IGCSE Mathematics 0580 · Extended · Syllabus 2025-2027

Solid Geometry Worked ExamplesIGCSE Maths 0580

Exam-style worked examples on volume and surface area of prisms, cylinders, cones, spheres and pyramids — including the distinction between curved and total surface area, and composite solids made from two shapes joined together. Diagram-based items add a prism with only some faces painted, a frustum-shaped bucket, a cone folded from a sector, equal volumes and equal surface areas solved in terms of π, and a cone's mass from its density. A spherical tank checks litres and percentages.

This page has 13 exam-style IGCSE Maths 0580 solid geometry worked examples — 1 Core, 9 Extended and 3 Challenge, 6 with diagrams. Each one shows the full method step by step, the final answer in the form a marker expects, and an examiner tip on where marks are commonly lost.

Example 1CorePrisms[3 marks]

Volume of a prism

A prism has a uniform cross-section of area $24\text{ cm}^2$ and a length of $15\text{ cm}$ .

(a) Calculate its volume. (b) Explain why the same formula works whatever the shape of the cross-section.

Example 2ExtendedCylinders[5 marks]

Cylinder: volume and surface area

A closed cylinder has radius $5\text{ cm}$ and height $12\text{ cm}$ .

Calculate (a) its volume, (b) its curved surface area, (c) its total surface area. Give each answer correct to $3$ significant figures.

Example 3ExtendedCones[5 marks]

Cone: volume and surface area

A cone has base radius $6\text{ cm}$ and vertical height $8\text{ cm}$ .

(a) Find the slant height. (b) Calculate the volume. (c) Calculate the curved surface area. Give (b) and (c) correct to $3$ significant figures.

Example 4ExtendedSpheres[4 marks]

Sphere

A sphere has radius $3\text{ cm}$ .

(a) Calculate its volume. (b) Calculate its surface area. Give both correct to $3$ significant figures. (c) Comment on your two answers.

Example 5ExtendedPyramids[4 marks]

Pyramid

A pyramid has a square base of side $6\text{ cm}$ and a vertical height of $10\text{ cm}$ .

(a) Calculate its volume. (b) A cuboid has the same base and height. Find the ratio of the pyramid's volume to the cuboid's.

Example 6ChallengeComposite solids[5 marks]

Composite solid

A solid is made from a cylinder of radius $5\text{ cm}$ and height $12\text{ cm}$ with a hemisphere of the same radius fixed on top.

(a) Calculate the total volume, correct to $3$ significant figures. (b) Calculate the total surface area of the solid, correct to $3$ significant figures.

Example 7ExtendedPrism in context: volume and selected faces[6 marks]

Painting a concrete ramp

A concrete ramp is a triangular prism, as shown. Its cross-section is a right-angled triangle with a horizontal base of $1.2\text{ m}$ and a vertical height of $0.5\text{ m}$ . The ramp is $2\text{ m}$ long.

(a) Calculate the volume of concrete in the ramp.

(b) Calculate the length of the sloping edge of the cross-section.

(c) Every face of the ramp is to be painted except the base, which sits on the ground. Calculate the area to be painted.

NOT TO SCALE

Example 8ExtendedFrustum as a large cone minus a small cone[8 marks]

Capacity and curved surface of a frustum

A bucket is the frustum of a cone. The top has radius $12\text{ cm}$ , the base has radius $8\text{ cm}$ and the vertical height is $10\text{ cm}$ . The frustum is what remains when the small cone, shown dashed, is removed from a large cone with vertex $V$ .

(a) Show that the vertical height of the large cone is $30\text{ cm}$ .

(b) Calculate the capacity of the bucket in litres, correct to $3$ significant figures.

(c) Calculate the area of the curved surface of the bucket, correct to $3$ significant figures.

NOT TO SCALE

1
(a) Let the large cone have height $H$ , so the small cone has height $H - 10$ . The two cones are similar, so radius divided by height is the same for both.
$\frac{12}{H} = \frac{8}{H - 10}$
2
Cross-multiply and solve.
$12(H - 10) = 8H \implies 12H - 120 = 8H \implies 4H = 120 \implies H = 30$
3
(b) The bucket's volume is the large cone minus the small cone, whose height is $30 - 10 = 20\text{ cm}$ .
$\tfrac{1}{3}\pi(12^{2})(30) - \tfrac{1}{3}\pi(8^{2})(20) = 1440\pi - \tfrac{1280}{3}\pi$
4
Simplify and evaluate.
$= \tfrac{3040}{3}\pi = 3183.4\ldots\text{ cm}^3$
5
Convert to litres: $1\text{ litre} = 1000\text{ cm}^3$ .
$3183.4\ldots \div 1000 = 3.183\ldots\text{ litres}$
6
(c) Find the slant height of each cone with Pythagoras, using its full height.
$l_{\text{large}} = \sqrt{30^{2} + 12^{2}} = \sqrt{1044}, \qquad l_{\text{small}} = \sqrt{20^{2} + 8^{2}} = \sqrt{464}$
7
The curved surface of the bucket is the large cone's curved surface minus the small cone's, using $\pi r l$ .
$\pi(12)\sqrt{1044} - \pi(8)\sqrt{464} = 1218.0\ldots - 541.3\ldots = 676.7\ldots\text{ cm}^2$

Answer

(a) $\dfrac{12}{H} = \dfrac{8}{H - 10}$ gives $H = 30$ (b) $3.18$ litres (c) $677\text{ cm}^2$

Examiner tip

Frustum questions are 'large cone minus small cone', and the small cone's height is $30 - 10 = 20\text{ cm}$ , not $10\text{ cm}$ — the June 2025 examiner report noted that "Some mistakenly used the height of the frustum instead of the height of the small cone". The shortcut $\tfrac{1}{3}\pi(12^2 - 8^2) \times 10$ looks plausible but is wrong. Keep the slant heights as surds or unrounded decimals until the final subtraction.

Example 9ChallengeSector net to cone, then pouring into a cylinder[7 marks]

A cone folded from a sector

The sector shown, with centre $O$ , has radius $15\text{ cm}$ and angle $216^{\circ}$ . It is the net of the curved surface of a cone: the two straight edges are joined so that the arc becomes the circumference of the base.

(a) Show that the radius of the base of the cone is $9\text{ cm}$ .

(b) Find the volume of the cone, giving your answer in terms of $\pi$ .

(c) The cone is held vertex-down and filled with water. The water is then poured into an empty cylinder of radius $6\text{ cm}$ . Find the depth of the water in the cylinder.

NOT TO SCALE

Example 10ExtendedEquate two volumes and solve for a length, non-calculator[4 marks]

Melting a ball into a rod: equal volumes

A solid metal ball of radius $6\text{ cm}$ is melted down and all of the metal is recast as a solid cylindrical rod $18\text{ cm}$ long, as shown. No metal is lost.

Work out the radius, $r\text{ cm}$ , of the rod.

[The volume, $V$ , of a sphere with radius $r$ is $V = \tfrac{4}{3}\pi r^{3}$ .]

Do not use a calculator.

NOT TO SCALE

Example 11ChallengeEquate surface areas in terms of π; total height by Pythagoras[8 marks]

A spinning top and a tin with equal surface areas

A wooden spinning top is a hemisphere of radius $5\text{ cm}$ with a cone glued to its flat face. The cone has the same radius, and its sloping edge is $13\text{ cm}$ long.

The top is sold in a closed cylindrical tin of radius $5\text{ cm}$ and height $h\text{ cm}$ . Exactly the same amount of paint is needed to cover the whole outside of the top as to cover the whole outside of the tin.

(a) Work out $h$ . [5]

(b) How tall is the spinning top, from the point of the cone to the bottom of the hemisphere? [3]

[The curved surface area, $A$ , of a cone with radius $r$ and slant height $l$ is $A = \pi r l$ .] [The surface area, $A$ , of a sphere with radius $r$ is $A = 4\pi r^{2}$ .]

Do not use a calculator.

NOT TO SCALE

Example 12ExtendedCone volume then mass from density; total surface area[6 marks]

Mass of wax and surface area of a cone-shaped candle

A candle is a solid cone of wax. Its base has diameter $9\text{ cm}$ and it is $15\text{ cm}$ tall.

(a) Each $\text{cm}^3$ of wax has a mass of $0.9\text{ g}$ . Work out the mass of wax in the candle. [3]

(b) The whole surface of the candle, including its base, is sprayed with gold paint. Work out the area that is painted. [3]

[The volume, $V$ , of a cone with radius $r$ and height $h$ is $V = \tfrac{1}{3}\pi r^{2}h$ .] [The curved surface area, $A$ , of a cone with radius $r$ and slant height $l$ is $A = \pi r l$ .]

NOT TO SCALE

Example 13ExtendedShow that a volume in litres; percentage of a container filled[6 marks]

A spherical tank emptied into a trough

The inside of a rainwater tank is a sphere of radius $42\text{ cm}$ .

(a) Show that, when full, the tank holds $310$ litres, to the nearest litre. [3]

(b) A farmer empties the full tank into a drinking trough for animals. The inside of the trough is a cylinder of radius $48\text{ cm}$ and depth $70\text{ cm}$ , and it was empty to begin with.

What percentage of the trough's volume does the water now take up? [3]

[The volume, $V$ , of a sphere with radius $r$ is $V = \tfrac{4}{3}\pi r^{3}$ .]

1
(a) Volume in $\text{cm}^3$ first. Write down more figures than the answer you are showing.
$\tfrac{4}{3}\pi \times 42^{3} = 310\,339.08\ldots\text{ cm}^3$
2
Show the conversion explicitly: $1\text{ litre} = 1000\text{ cm}^3$ , so divide by $1000$ .
$310\,339.08\ldots \div 1000 = 310.339\ldots\text{ litres}$
3
Now round: $310.339\ldots$ is $310$ litres to the nearest litre, as required.
4
(b) Volume of the cylindrical trough.
$\pi \times 48^{2} \times 70 = 506\,676.06\ldots\text{ cm}^3$
5
The percentage is the water as a fraction of the trough, times $100$ . Use the unrounded water volume.
$\frac{310\,339.08\ldots}{506\,676.06\ldots} \times 100 = 61.25\%$
6
This value is exact ( $\tfrac{4}{3} \times 42^3 \div (48^2 \times 70) = 0.6125$ exactly, as the $\pi$ cancels), so give it in full rather than rounding it.

Answer

(a) $\tfrac{4}{3}\pi(42^3) = 310\,339.08\ldots\text{ cm}^3$ , $\div 1000 = 310.339\ldots \approx 310$ litres (b) $61.25\%$

Examiner tip

In a 'show that', the answer is given, so the marks are for the steps: the March 2026 examiner report found that "Few candidates showed the explicit step to convert cm3 to litres of division by 1000." Show a value more accurate than $310$ before rounding. In (b) the same report noted that "A common error seen was to invert the correct division" — the trough volume over the water gives $163\%$ , which cannot be right when the trough does not overflow. The answer is exact, and the June 2025 examiner report advised that "Where answers are exact integer or decimal values, candidates should give this value as their answer rather than rounding to 3 significant figures".

These worked examples are written by Tutopiya to cover the same syllabus content as the official Cambridge IGCSE 0580 specification (2025-2027). They are exam-style questions, not reproductions of past paper questions.

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Solid Geometry Worked ExamplesIGCSE Maths 0580

Example 1CorePrisms[3 marks]

Volume of a prism

A prism has a uniform cross-section of area $24\text{ cm}^2$ and a length of $15\text{ cm}$ .

(a) Calculate its volume. (b) Explain why the same formula works whatever the shape of the cross-section.

Example 2ExtendedCylinders[5 marks]

Cylinder: volume and surface area

A closed cylinder has radius $5\text{ cm}$ and height $12\text{ cm}$ .

Calculate (a) its volume, (b) its curved surface area, (c) its total surface area. Give each answer correct to $3$ significant figures.

Example 3ExtendedCones[5 marks]

Cone: volume and surface area

A cone has base radius $6\text{ cm}$ and vertical height $8\text{ cm}$ .

(a) Find the slant height. (b) Calculate the volume. (c) Calculate the curved surface area. Give (b) and (c) correct to $3$ significant figures.

Example 4ExtendedSpheres[4 marks]

Sphere

A sphere has radius $3\text{ cm}$ .

(a) Calculate its volume. (b) Calculate its surface area. Give both correct to $3$ significant figures. (c) Comment on your two answers.

Example 5ExtendedPyramids[4 marks]

Pyramid

A pyramid has a square base of side $6\text{ cm}$ and a vertical height of $10\text{ cm}$ .

(a) Calculate its volume. (b) A cuboid has the same base and height. Find the ratio of the pyramid's volume to the cuboid's.

Example 6ChallengeComposite solids[5 marks]

Composite solid

A solid is made from a cylinder of radius $5\text{ cm}$ and height $12\text{ cm}$ with a hemisphere of the same radius fixed on top.

(a) Calculate the total volume, correct to $3$ significant figures. (b) Calculate the total surface area of the solid, correct to $3$ significant figures.

Example 7ExtendedPrism in context: volume and selected faces[6 marks]

Painting a concrete ramp

(a) Calculate the volume of concrete in the ramp.

(b) Calculate the length of the sloping edge of the cross-section.

(c) Every face of the ramp is to be painted except the base, which sits on the ground. Calculate the area to be painted.

NOT TO SCALE

Example 8ExtendedFrustum as a large cone minus a small cone[8 marks]

Capacity and curved surface of a frustum

(a) Show that the vertical height of the large cone is $30\text{ cm}$ .

(b) Calculate the capacity of the bucket in litres, correct to $3$ significant figures.

(c) Calculate the area of the curved surface of the bucket, correct to $3$ significant figures.

NOT TO SCALE

1
(a) Let the large cone have height $H$ , so the small cone has height $H - 10$ . The two cones are similar, so radius divided by height is the same for both.
$\frac{12}{H} = \frac{8}{H - 10}$
2
Cross-multiply and solve.
$12(H - 10) = 8H \implies 12H - 120 = 8H \implies 4H = 120 \implies H = 30$
3
(b) The bucket's volume is the large cone minus the small cone, whose height is $30 - 10 = 20\text{ cm}$ .
$\tfrac{1}{3}\pi(12^{2})(30) - \tfrac{1}{3}\pi(8^{2})(20) = 1440\pi - \tfrac{1280}{3}\pi$
4
Simplify and evaluate.
$= \tfrac{3040}{3}\pi = 3183.4\ldots\text{ cm}^3$
5
Convert to litres: $1\text{ litre} = 1000\text{ cm}^3$ .
$3183.4\ldots \div 1000 = 3.183\ldots\text{ litres}$
6
(c) Find the slant height of each cone with Pythagoras, using its full height.
$l_{\text{large}} = \sqrt{30^{2} + 12^{2}} = \sqrt{1044}, \qquad l_{\text{small}} = \sqrt{20^{2} + 8^{2}} = \sqrt{464}$
7
The curved surface of the bucket is the large cone's curved surface minus the small cone's, using $\pi r l$ .
$\pi(12)\sqrt{1044} - \pi(8)\sqrt{464} = 1218.0\ldots - 541.3\ldots = 676.7\ldots\text{ cm}^2$

Answer

(a) $\dfrac{12}{H} = \dfrac{8}{H - 10}$ gives $H = 30$ (b) $3.18$ litres (c) $677\text{ cm}^2$

Examiner tip

Example 9ChallengeSector net to cone, then pouring into a cylinder[7 marks]

A cone folded from a sector

(a) Show that the radius of the base of the cone is $9\text{ cm}$ .

(b) Find the volume of the cone, giving your answer in terms of $\pi$ .

(c) The cone is held vertex-down and filled with water. The water is then poured into an empty cylinder of radius $6\text{ cm}$ . Find the depth of the water in the cylinder.

NOT TO SCALE

Example 10ExtendedEquate two volumes and solve for a length, non-calculator[4 marks]

Melting a ball into a rod: equal volumes

A solid metal ball of radius $6\text{ cm}$ is melted down and all of the metal is recast as a solid cylindrical rod $18\text{ cm}$ long, as shown. No metal is lost.

Work out the radius, $r\text{ cm}$ , of the rod.

[The volume, $V$ , of a sphere with radius $r$ is $V = \tfrac{4}{3}\pi r^{3}$ .]

Do not use a calculator.

NOT TO SCALE

Example 11ChallengeEquate surface areas in terms of π; total height by Pythagoras[8 marks]

A spinning top and a tin with equal surface areas

A wooden spinning top is a hemisphere of radius $5\text{ cm}$ with a cone glued to its flat face. The cone has the same radius, and its sloping edge is $13\text{ cm}$ long.

(a) Work out $h$ . [5]

(b) How tall is the spinning top, from the point of the cone to the bottom of the hemisphere? [3]

[The curved surface area, $A$ , of a cone with radius $r$ and slant height $l$ is $A = \pi r l$ .] [The surface area, $A$ , of a sphere with radius $r$ is $A = 4\pi r^{2}$ .]

Do not use a calculator.

NOT TO SCALE

Example 12ExtendedCone volume then mass from density; total surface area[6 marks]

Mass of wax and surface area of a cone-shaped candle

A candle is a solid cone of wax. Its base has diameter $9\text{ cm}$ and it is $15\text{ cm}$ tall.

(a) Each $\text{cm}^3$ of wax has a mass of $0.9\text{ g}$ . Work out the mass of wax in the candle. [3]

(b) The whole surface of the candle, including its base, is sprayed with gold paint. Work out the area that is painted. [3]

[The volume, $V$ , of a cone with radius $r$ and height $h$ is $V = \tfrac{1}{3}\pi r^{2}h$ .] [The curved surface area, $A$ , of a cone with radius $r$ and slant height $l$ is $A = \pi r l$ .]

NOT TO SCALE

Example 13ExtendedShow that a volume in litres; percentage of a container filled[6 marks]

A spherical tank emptied into a trough

The inside of a rainwater tank is a sphere of radius $42\text{ cm}$ .

(a) Show that, when full, the tank holds $310$ litres, to the nearest litre. [3]

(b) A farmer empties the full tank into a drinking trough for animals. The inside of the trough is a cylinder of radius $48\text{ cm}$ and depth $70\text{ cm}$ , and it was empty to begin with.

What percentage of the trough's volume does the water now take up? [3]

[The volume, $V$ , of a sphere with radius $r$ is $V = \tfrac{4}{3}\pi r^{3}$ .]

1
(a) Volume in $\text{cm}^3$ first. Write down more figures than the answer you are showing.
$\tfrac{4}{3}\pi \times 42^{3} = 310\,339.08\ldots\text{ cm}^3$
2
Show the conversion explicitly: $1\text{ litre} = 1000\text{ cm}^3$ , so divide by $1000$ .
$310\,339.08\ldots \div 1000 = 310.339\ldots\text{ litres}$
3
Now round: $310.339\ldots$ is $310$ litres to the nearest litre, as required.
4
(b) Volume of the cylindrical trough.
$\pi \times 48^{2} \times 70 = 506\,676.06\ldots\text{ cm}^3$
5
The percentage is the water as a fraction of the trough, times $100$ . Use the unrounded water volume.
$\frac{310\,339.08\ldots}{506\,676.06\ldots} \times 100 = 61.25\%$
6
This value is exact ( $\tfrac{4}{3} \times 42^3 \div (48^2 \times 70) = 0.6125$ exactly, as the $\pi$ cancels), so give it in full rather than rounding it.

Answer

(a) $\tfrac{4}{3}\pi(42^3) = 310\,339.08\ldots\text{ cm}^3$ , $\div 1000 = 310.339\ldots \approx 310$ litres (b) $61.25\%$

Examiner tip

Solid Geometry Worked ExamplesIGCSE Maths 0580

Volume of a prism

Cylinder: volume and surface area

Cone: volume and surface area

Sphere

Pyramid

Composite solid

Painting a concrete ramp

Capacity and curved surface of a frustum

A cone folded from a sector

Melting a ball into a rod: equal volumes

A spinning top and a tin with equal surface areas

Mass of wax and surface area of a cone-shaped candle

A spherical tank emptied into a trough

More IGCSE Maths 0580 worked examples

Mensuration

Number

Algebra

Functions

Coordinate Geometry

Geometry

Vectors and Transfomations

Trigonometry

Sets

Probability

Statistics

Calculator Skills

Solid Geometry Worked ExamplesIGCSE Maths 0580

Volume of a prism

Cylinder: volume and surface area

Cone: volume and surface area

Sphere

Pyramid

Composite solid

Painting a concrete ramp

Capacity and curved surface of a frustum

A cone folded from a sector

Melting a ball into a rod: equal volumes

A spinning top and a tin with equal surface areas

Mass of wax and surface area of a cone-shaped candle

A spherical tank emptied into a trough

More IGCSE Maths 0580 worked examples

Mensuration

Number

Algebra

Functions

Coordinate Geometry

Geometry

Vectors and Transfomations

Trigonometry

Sets

Probability

Statistics

Calculator Skills