What is Solid Geometry and how is it different from Plane Geometry?

Solid Geometry is the study of three-dimensional figures like cubes, spheres, and cylinders, focusing on their properties such as volume and surface area. Unlike Plane Geometry, which deals with two-dimensional shapes like squares and circles, Solid Geometry involves calculations in three dimensions, making it a crucial part of the Cambridge IGCSE Mathematics curriculum under Mensuration.

How do you calculate the volume of a cylinder in Solid Geometry?

To calculate the volume of a cylinder, use the formula V = πr²h, where V is the volume, r is the radius of the base, and h is the height of the cylinder. This formula is essential for solving problems in the Solid Geometry section of the Cambridge IGCSE Mathematics syllabus.

What is the difference between surface area and volume in Solid Geometry?

In Solid Geometry, surface area refers to the total area covered by the surfaces of a three-dimensional object, while volume measures the space inside the object. Understanding these concepts is vital for Cambridge IGCSE students, as they are frequently tested in Mensuration problems.

How can you find the surface area of a sphere?

The surface area of a sphere can be calculated using the formula A = 4πr², where A is the surface area and r is the radius of the sphere. This formula is a key component of Solid Geometry and is often included in the Cambridge IGCSE Mathematics curriculum.

Why is understanding Solid Geometry important for Cambridge IGCSE Mathematics?

Understanding Solid Geometry is crucial for Cambridge IGCSE Mathematics as it helps students solve real-world problems involving three-dimensional objects. It enhances spatial reasoning and is a fundamental part of the Mensuration topic, preparing students for advanced mathematical studies and practical applications.

Why is "Using slant height in $V = \dfrac{1}{3}\pi r^{2} h$ for a cone" a common mistake in Solid Geometry?

Why it happens: Confusing slant length with perpendicular height h. How to avoid it: Volume needs PERPENDICULAR height. Use Pythagoras to recover h from and r if needed: h = √(^2) - r^2. Source: 0580/42 — recurring

Why is "Using $\text{cm}^{2}$ instead of $\text{cm}^{3}$ for volume" a common mistake in Solid Geometry?

Why it happens: Speed. How to avoid it: Volumes are in cubed units; surface areas are in squared units.

Why is "Including both circles in surface area for an open cylinder" a common mistake in Solid Geometry?

Why it happens: Defaulting to the closed-cylinder formula. How to avoid it: Read the question. Open at the top: r^2 + 2 r h. Closed: 2 r^2 + 2 r h.

Why is "Rounding $\pi$ early" a common mistake in Solid Geometry?

Why it happens: Substituting 3.14 at step one. How to avoid it: Keep symbolic until the final line, or use the calculator's key. Then round once at the end.

MensurationCambridge IGCSE Maths 0580 Extended

Solid Geometry

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Volume and Surface Area of 3D Solids — Cambridge IGCSE 0580 Maths Extended (2026)

Cuboids, prisms, cylinders, cones, pyramids and spheres. Memorise the formulae from the formula sheet, but understand prism vs pyramid (with $\dfrac{1}{3}$ ).

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E6.5 — Calculate volumes and surface areas of cuboids, prisms, cylinders, spheres, pyramids and cones.
E6.5 — Calculate volumes and surface areas of compound shapes.

Prisms (cuboids, cylinders, triangular prisms)

$V = \text{area of cross-section} \times \text{length}$ . Surface area: net.

A prism has the same cross-section throughout its length.

Cuboid (rectangular prism): $V = lwh$ . Surface area $= 2(lw + lh + wh)$ .

Cylinder (circular prism): $V = \pi r^{2} h$ . Curved surface area $= 2\pi r h$ . Total surface area (closed cylinder) $= 2\pi r^{2} + 2\pi r h$ .

Triangular prism: $V = (\dfrac{1}{2}b h_{t}) \times \ell$ where $b, h_{t}$ are the triangle base and height, $\ell$ is the prism length.

Worked. Cylinder radius $4\,\text{cm}$ , height $10\,\text{cm}$ .

$V = \pi \times 16 \times 10 = 160\pi \approx 503\,\text{cm}^{3}$ .
Curved SA: $2\pi \times 4 \times 10 = 80\pi \approx 251\,\text{cm}^{2}$ .
Total SA: $2\pi \times 16 + 80\pi = 112\pi \approx 352\,\text{cm}^{2}$ .

Surface area technique. "Unfold" the solid into its NET; sum the areas of every face. For a cylinder: two circles + a rectangle (the curved side unrolled, $2\pi r$ wide and $h$ tall).

A prism keeps the same cross-section along its length — volume is cross-section area × length.

Prism: $V =$ cross-section area $\times$ length.
Cylinder: $V = \pi r^{2} h$ , curved SA $= 2\pi r h$ .
Cuboid SA: $2(lw + lh + wh)$ .
Surface area: think NET, sum every face.

Pyramids and cones

$V = \dfrac{1}{3} \times \text{base area} \times h$ . The factor $\dfrac{1}{3}$ separates prism from pyramid.

Pyramid (square / rectangular base). $V = \dfrac{1}{3} \times \text{base area} \times h.$

The height $h$ is the perpendicular height from base to apex.

Worked. Square pyramid, base side $6\,\text{cm}$ , perpendicular height $10\,\text{cm}$ .

Base area: $36\,\text{cm}^{2}$ .
$V = \dfrac{1}{3} \times 36 \times 10 = 120\,\text{cm}^{3}$ .

Cone (circular pyramid). $V = \dfrac{1}{3}\pi r^{2} h.$ Curved surface area $= \pi r l$ , where $l$ is the slant height: $l = \sqrt{r^{2} + h^{2}}$ .

Worked. Cone, radius $5\,\text{cm}$ , perpendicular height $12\,\text{cm}$ .

$V = \dfrac{1}{3}\pi \times 25 \times 12 = 100\pi \approx 314\,\text{cm}^{3}$ .
Slant: $l = \sqrt{25 + 144} = \sqrt{169} = 13\,\text{cm}$ .
Curved SA: $\pi \times 5 \times 13 = 65\pi \approx 204\,\text{cm}^{2}$ .
Total SA (closed cone, with base disc): $65\pi + 25\pi = 90\pi \approx 283\,\text{cm}^{2}$ .

Surface area of a square pyramid. Base + 4 triangular faces. For each triangular face you need its slant height (from base midpoint up the side).

Use the perpendicular height h for volume; the slant height l for the curved surface area of a cone.

Pyramid/cone: $V = \dfrac{1}{3} \times$ base area $\times h$ .
$h$ = perpendicular height.
Cone slant: $l = \sqrt{r^{2} + h^{2}}$ .
Cone curved SA: $\pi r l$ .

Sphere and hemisphere

$V = \dfrac{4}{3}\pi r^{3}$ . Surface area $= 4\pi r^{2}$ .

Sphere. $V = \dfrac{4}{3}\pi r^{3}, \qquad SA = 4\pi r^{2}.$

Worked. Sphere radius $6\,\text{cm}$ .

$V = \dfrac{4}{3}\pi \times 216 = 288\pi \approx 905\,\text{cm}^{3}$ .
$SA = 4\pi \times 36 = 144\pi \approx 452\,\text{cm}^{2}$ .

Hemisphere (half a sphere). $V = \dfrac{2}{3}\pi r^{3}.$ Curved SA $= \dfrac{1}{2} \times 4\pi r^{2} = 2\pi r^{2}$ . Closed hemisphere also includes the flat circular face: total SA $= 2\pi r^{2} + \pi r^{2} = 3\pi r^{2}$ .

Sphere: $V = \dfrac{4}{3}\pi r^{3}$ , $SA = 4\pi r^{2}$ .
Hemisphere $V$ : half of sphere.
Closed hemisphere total SA: $3\pi r^{2}$ .

Compound solids and frustums

Add or subtract individual solid volumes. A frustum (cone with top sliced) is computed by subtracting the small cone.

Compound solid. A solid built from two or more standard pieces — e.g. a cylinder with a hemispherical cap.

Strategy.

Identify each component (cylinder + hemisphere, etc.).
Compute each volume separately.
Add (joined) or subtract (hollow / drilled).

Worked. Solid is a cylinder of radius $5\,\text{cm}$ , height $12\,\text{cm}$ , with a hemispherical cap of radius $5\,\text{cm}$ on top.

Cylinder $V$ : $\pi \times 25 \times 12 = 300\pi$ .
Hemisphere $V$ : $\dfrac{2}{3}\pi \times 125 = \dfrac{250\pi}{3}$ .
Total: $300\pi + \dfrac{250\pi}{3} = \dfrac{900\pi + 250\pi}{3} = \dfrac{1150\pi}{3} \approx 1204\,\text{cm}^{3}$ .

Frustum (truncated cone). $V_{\text{frustum}} = V_{\text{big cone}} - V_{\text{small cone}}.$ Use similar-triangle reasoning to find the small cone's height from the frustum dimensions.

Tip. When the question asks for surface area, don't double-count joined faces — they're internal and disappear in the join.

Split a compound solid into standard pieces, then add (joined) or subtract (hollowed) the volumes.

Compound solid: split into known pieces.
Add (joined) or subtract (hollow).
Frustum: big cone minus small cone.
Surface area: ignore internal joined faces.

How it’s examined

Volume / surface area appears every Paper 4 — typically a 6-8 mark compound-solid question. Cambridge gives a formula sheet with sphere, cone, pyramid formulae, but you must recognise WHICH formula to apply. Examiner reports flag forgetting the $\dfrac{1}{3}$ factor for pyramids/cones and using slant instead of perpendicular height.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E6.5); 0580/42 Oct/Nov 2024 — Q8 (compound solid); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

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Step-by-step worked examples — Solid Geometry

Step-by-step solutions to past-paper-style questions on solid geometry, written exactly the way a tutor would explain them at the board.

1Volume and surface area of a cuboid
Core• cuboid
Question
A cuboid measures $5\,\text{cm} \times 4\,\text{cm} \times 3\,\text{cm}$ . Find its volume and total surface area.
Step-by-step solution
1. Step 1
  Volume = $l \times w \times h$ .
  $V = 5 \times 4 \times 3 = 60\,\text{cm}^{3}$
2. Step 2
  Surface area = sum of three pairs of faces.
  $SA = 2(5 \times 4) + 2(5 \times 3) + 2(4 \times 3) = 40 + 30 + 24 = 94\,\text{cm}^{2}$
Answer
$V = 60\,\text{cm}^{3},\ SA = 94\,\text{cm}^{2}$
2Volume and surface area of a cylinder
Extended• Adapted from 0580/42 May/Jun 2024 Q11• cylinder
Question
Find the volume and total surface area of a closed cylinder with radius $4$ cm and height $10$ cm. Give answers in terms of $\pi$ .
Step-by-step solution
1. Step 1
  $V = \pi r^{2} h$ .
  $V = \pi \times 4^{2} \times 10 = 160\pi$
2. Step 2
  Total surface area = $2\pi r^{2} + 2\pi r h$ .
  $SA = 2\pi(16) + 2\pi(4)(10) = 32\pi + 80\pi = 112\pi$
Answer
$V = 160\pi\,\text{cm}^{3},\ SA = 112\pi\,\text{cm}^{2}$
3Volume of a cone
Extended• cone
Question
Find the volume of a cone with radius $3$ cm and perpendicular height $7$ cm. Give the answer in terms of $\pi$ .
Step-by-step solution
1. Step 1
  $V = \dfrac{1}{3}\pi r^{2} h$ .
  $V = \dfrac{1}{3} \times \pi \times 9 \times 7 = 21\pi$
Answer
$21\pi\,\text{cm}^{3}$
4Volume and surface area of a sphere
Extended• sphere
Question
Find the volume and surface area of a sphere of radius $6$ cm. Give answers in terms of $\pi$ .
Step-by-step solution
1. Step 1
  $V = \dfrac{4}{3}\pi r^{3}$ .
  $V = \dfrac{4}{3}\pi \times 216 = 288\pi$
2. Step 2
  $SA = 4\pi r^{2}$ .
  $SA = 4\pi \times 36 = 144\pi$
Answer
$V = 288\pi\,\text{cm}^{3},\ SA = 144\pi\,\text{cm}^{2}$
5Volume of a pyramid
Extended• Adapted from 0580/42 Oct/Nov 2024 Q14• pyramid
Question
Find the volume of a square-based pyramid with base side $6$ cm and perpendicular height $9$ cm.
Step-by-step solution
1. Step 1
  $V = \dfrac{1}{3} \times \text{base area} \times h$ .
  $V = \dfrac{1}{3} \times 36 \times 9 = 108$
Answer
$108\,\text{cm}^{3}$
Examiner tip
Always check the height is PERPENDICULAR — not the slant edge of the pyramid.
6Volume of a cube
Core• cube, volume
Question
A cube has edge length $4.5\,\text{cm}$ . Find its volume.
Step-by-step solution
1. Step 1
  $V = s^{3}$ .
  $V = 4.5^{3} = 91.125\,\text{cm}^{3}$
Answer
$91.125\,\text{cm}^{3}$
7Surface area of a cone
Extended• Adapted from 0580/42 Oct/Nov 2023 Q15• cone, surface area, slant
Question
A solid cone has base radius $5\,\text{cm}$ and slant height $13\,\text{cm}$ . Find the total surface area in terms of $\pi$ .
Step-by-step solution
1. Step 1
  Total surface area = curved surface + circular base.
  $SA = \pi r \ell + \pi r^{2}$
2. Step 2
  Substitute $r = 5$ , $\ell = 13$ .
  $SA = \pi(5)(13) + \pi(5)^{2} = 65\pi + 25\pi$
3. Step 3
  Add.
  $SA = 90\pi\,\text{cm}^{2}$
Answer
$90\pi\,\text{cm}^{2}$
Examiner tip
The examiner report flags candidates who use perpendicular height in $\pi r\ell$ . Curved surface needs SLANT height; if only the perpendicular height is given, use Pythagoras to find $\ell$ first.
8Volume of a hemisphere
Extended• hemisphere, volume
Question
Find the volume of a hemisphere of radius $6\,\text{cm}$ , giving your answer in terms of $\pi$ .
Step-by-step solution
1. Step 1
  Hemisphere is half a sphere.
  $V = \dfrac{1}{2}\times \dfrac{4}{3}\pi r^{3} = \dfrac{2}{3}\pi r^{3}$
2. Step 2
  Substitute $r = 6$ .
  $V = \dfrac{2}{3}\pi (216) = 144\pi\,\text{cm}^{3}$
Answer
$144\pi\,\text{cm}^{3}$
9Find a missing edge given the volume
Extended• Adapted from 0580/22 May/Jun 2023 Q8• cuboid, find edge
Question
A cuboid has volume $336\,\text{cm}^{3}$ . Its length is $8\,\text{cm}$ and its width is $6\,\text{cm}$ . Find its height.
Step-by-step solution
1. Step 1
  $V = lwh$ .
  $336 = 8 \times 6 \times h = 48h$
2. Step 2
  Divide.
  $h = \dfrac{336}{48} = 7\,\text{cm}$
Answer
$h = 7\,\text{cm}$
10Cylinder volume in litres
Extended• cylinder, volume, units
Question
A cylindrical water tank has radius $40\,\text{cm}$ and height $1.2\,\text{m}$ . Find the volume in litres, correct to 3 s.f. ( $1\,\text{litre} = 1000\,\text{cm}^{3}$ .)
Step-by-step solution
1. Step 1
  Convert height to centimetres.
  $h = 1.2\,\text{m} = 120\,\text{cm}$
2. Step 2
  Apply $V = \pi r^{2} h$ .
  $V = \pi (40)^{2}(120) = 192{,}000\pi\,\text{cm}^{3}$
3. Step 3
  Evaluate.
  $V \approx 603{,}185.8\,\text{cm}^{3}$
4. Step 4
  Convert to litres by dividing by $1000$ .
  $V \approx 603\,\text{litres}$
Answer
$V \approx 603\,\text{litres}$
Examiner tip
The examiner report flags candidates who mix metres and centimetres without converting. Standardise units BEFORE substituting.
11Volume of a compound solid (cylinder + hemisphere)
Challenge• Adapted from 0580/42 May/Jun 2024 Q13• compound solid, cylinder, hemisphere
Question
A toy consists of a cylinder of radius $4\,\text{cm}$ and height $10\,\text{cm}$ with a hemisphere of radius $4\,\text{cm}$ attached to the top. Find the total volume in terms of $\pi$ .
Step-by-step solution
1. Step 1
  Cylinder volume.
  $V_1 = \pi(4)^{2}(10) = 160\pi$
2. Step 2
  Hemisphere volume.
  $V_2 = \dfrac{2}{3}\pi(4)^{3} = \dfrac{128\pi}{3}$
3. Step 3
  Total.
  $V = 160\pi + \dfrac{128\pi}{3} = \dfrac{480\pi + 128\pi}{3} = \dfrac{608\pi}{3}\,\text{cm}^{3}$
Answer
$V = \dfrac{608\pi}{3}\,\text{cm}^{3} \approx 636.7\,\text{cm}^{3}$
Examiner tip
The examiner report flags candidates who add a full sphere volume instead of a hemisphere. Read the question — only HALF a sphere is attached.
12Surface area of a compound solid
Challenge• compound solid, surface area
Question
A solid consists of a cylinder of radius $3\,\text{cm}$ and height $8\,\text{cm}$ with a cone of radius $3\,\text{cm}$ and slant height $5\,\text{cm}$ on top (closed at the bottom). Find the total external surface area in terms of $\pi$ .
Step-by-step solution
1. Step 1
  External surface comprises: cylinder bottom (circle), cylinder curved side, cone curved side. NO top circle of cylinder (covered by cone base) and NO cone base (interior).
2. Step 2
  Cylinder bottom.
  $\pi r^{2} = 9\pi$
3. Step 3
  Cylinder curved surface.
  $2\pi r h = 2\pi(3)(8) = 48\pi$
4. Step 4
  Cone curved surface.
  $\pi r \ell = \pi(3)(5) = 15\pi$
5. Step 5
  Total.
  $SA = 9\pi + 48\pi + 15\pi = 72\pi\,\text{cm}^{2}$
Answer
$72\pi\,\text{cm}^{2}$
Examiner tip
The examiner report flags candidates who include the top of the cylinder AND the base of the cone — both are internal and must be omitted.
13Liquid-in-container word problem
Challenge• Adapted from 0580/42 Feb/Mar 2024 Q14• volume, rate, real-world
Question
A cylindrical tank of radius $25\,\text{cm}$ and height $40\,\text{cm}$ is initially empty. Water flows in at a constant rate of $\dfrac{1}{2}\,\text{litre per second}$ . Find the time, to the nearest second, taken to fill the tank.
Step-by-step solution
1. Step 1
  Volume of the tank.
  $V = \pi(25)^{2}(40) = 25{,}000\pi\,\text{cm}^{3}$
2. Step 2
  Evaluate.
  $V \approx 78{,}539.8\,\text{cm}^{3} = 78.54\,\text{litres}$
3. Step 3
  Time = volume / rate.
  $t = \dfrac{78.54}{0.5} \approx 157.08\,\text{s}$
4. Step 4
  Round to the nearest second.
  $t \approx 157\,\text{s}$
Answer
$t \approx 157\,\text{s}$
Examiner tip
The examiner report flags candidates who confuse $\text{cm}^{3}$ with litres. $1\,\text{litre} = 1000\,\text{cm}^{3}$ — always convert before dividing by the rate in litres per second.

Key Formulae — Solid Geometry

The formulae you need to memorise for solid geometry on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Cuboid
$V = lwh,\quad SA = 2(lw + lh + wh)$
$l, w, h$
length, width, height
When to use
Rectangular box volumes and surface areas.
Cylinder
$V = \pi r^{2} h,\quad SA_{\text{total}} = 2\pi r^{2} + 2\pi r h$
$r$
radius of circular face
$h$
height
When to use
Closed cylinder. For an open one, drop one $\pi r^{2}$ term.
Cone
$V = \dfrac{1}{3}\pi r^{2} h,\quad SA = \pi r^{2} + \pi r \ell$
$r$
base radius
$h$
perpendicular height
$\ell$
slant height
When to use
Cone volume needs perpendicular height; surface area needs slant.
Sphere
$V = \dfrac{4}{3}\pi r^{3},\quad SA = 4\pi r^{2}$
$r$
radius
When to use
Sphere or hemisphere (halve as needed).
Pyramid (any base)
$V = \dfrac{1}{3} \times \text{base area} \times h$
$h$
perpendicular height from apex to base
When to use
Any pyramid, including square- and triangle-based.
Prism
$V = \text{cross-sectional area} \times \text{length}$
When to use
Any prism — cross-section is constant along its length.

Key Definitions and Keywords — Solid Geometry

Definitions to memorise and the exact keywords mark schemes credit for solid geometry answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Prism
Examiner keyword
A solid with two congruent parallel ends and a constant cross-section.
Cylinder
Examiner keyword
A circular prism — two parallel circular ends with a curved surface joining them.
Slant height
Examiner keyword
The distance from the apex of a cone (or pyramid) to a point on the edge of the base, measured along the surface.
Perpendicular (vertical) height
Examiner keyword
The shortest distance from the apex to the base, perpendicular to the base.
Hemisphere
Half of a sphere. Volume is $\dfrac{2}{3}\pi r^{3}$ .

Common Mistakes and Misconceptions — Solid Geometry

The traps other students keep falling into on solid geometry questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Using slant height in $V = \dfrac{1}{3}\pi r^{2} h$ for a cone
0580/42 — recurring
Why it happens
Confusing slant length $\ell$ with perpendicular height $h$ .
How to avoid it
Volume needs PERPENDICULAR height. Use Pythagoras to recover $h$ from $\ell$ and $r$ if needed: $h = \sqrt{\ell^{2} - r^{2}}$ .
✕Using $\text{cm}^{2}$ instead of $\text{cm}^{3}$ for volume
Why it happens
Speed.
How to avoid it
Volumes are in cubed units; surface areas are in squared units.
✕Including both circles in surface area for an open cylinder
Why it happens
Defaulting to the closed-cylinder formula.
How to avoid it
Read the question. Open at the top: $\pi r^{2} + 2\pi r h$ . Closed: $2\pi r^{2} + 2\pi r h$ .
✕Rounding $\pi$ early
Why it happens
Substituting $\pi \approx 3.14$ at step one.
How to avoid it
Keep $\pi$ symbolic until the final line, or use the calculator's $\pi$ key. Then round once at the end.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

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Solid Geometry — frequently asked questions

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Solid Geometry

Short Study Notes in the form of Slides

Short Study Notes — Solid Geometry

Detailed Notes

What you’ll learn

How it’s examined

1Volume and surface area of a cuboid

2Volume and surface area of a cylinder

3Volume of a cone

4Volume and surface area of a sphere

5Volume of a pyramid

6Volume of a cube

7Surface area of a cone

8Volume of a hemisphere

9Find a missing edge given the volume

10Cylinder volume in litres

11Volume of a compound solid (cylinder + hemisphere)

12Surface area of a compound solid

13Liquid-in-container word problem

Cuboid

Cylinder

Cone

Sphere

Pyramid (any base)

Prism

Prism

Cylinder

Slant height

Perpendicular (vertical) height

Hemisphere

✕Using slant height in V=13πr2hV = \dfrac{1}{3}\pi r^{2} hV=31​πr2h for a cone

✕Using cm2\text{cm}^{2}cm2 instead of cm3\text{cm}^{3}cm3 for volume

✕Including both circles in surface area for an open cylinder

✕Rounding π\piπ early

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Solid Geometry — frequently asked questions

✕Using slant height in $V = \dfrac{1}{3}\pi r^{2} h$ for a cone

✕Using $\text{cm}^{2}$ instead of $\text{cm}^{3}$ for volume

✕Rounding $\pi$ early