Geometry and MeasuresAQA GCSE Mathematics (8300)

Mensuration and Calculation

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

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Master the topic

Worked examples4 Key formulae4 Definitions6 Common mistakes4

Step-by-step worked examples

01.Surface area and volume of a cone
Higher
Question
A cone has base radius 5 cm and perpendicular height 12 cm. Find its total surface area and volume. Give exact answers in terms of $\pi$ .
Solution
1. 1
  Find the slant height using Pythagoras.
  $l = \sqrt{r^2 + h^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm}$
2. 2
  Curved surface area of cone.
  $A_{\text{curved}} = \pi r l = \pi(5)(13) = 65\pi \text{ cm}^2$
3. 3
  Add the base area for total surface area.
  $A_{\text{total}} = 65\pi + \pi r^2 = 65\pi + 25\pi = 90\pi \text{ cm}^2$
4. 4
  Volume of cone.
  $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(25)(12) = 100\pi \text{ cm}^3$
Answer
Total surface area $= 90\pi$ cm²; Volume $= 100\pi$ cm³.
02.3D Pythagoras — space diagonal of a cuboid
Higher
Question
A cuboid has length 6 cm, width 4 cm and height 3 cm. Find the length of the space diagonal.
Solution
1. 1
  Find the diagonal of the base rectangle.
  $d_{\text{base}} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52}$
2. 2
  The space diagonal is the hypotenuse of the right-angled triangle with legs $d_{\text{base}}$ and height 3 cm.
  $d = \sqrt{(\sqrt{52})^2 + 3^2} = \sqrt{52 + 9} = \sqrt{61}$
3. 3
  Calculate.
  $d = \sqrt{61} \approx 7.81 \text{ cm}$
Answer
$d = \sqrt{61} \approx 7.81$ cm.
03.Cosine rule — finding a side
Challenge
Question
In triangle $ABC$ , $b = 7$ cm, $c = 10$ cm and $\angle A = 48°$ . Find side $a$ .
Solution
1. 1
  Write the cosine rule formula.
  $a^2 = b^2 + c^2 - 2bc\cos A$
2. 2
  Substitute the known values.
  $a^2 = 7^2 + 10^2 - 2(7)(10)\cos 48° = 49 + 100 - 140\cos 48°$
3. 3
  Evaluate $\cos 48° \approx 0.6691$ .
  $a^2 = 149 - 140(0.6691) = 149 - 93.67 = 55.33$
4. 4
  Take the square root.
  $a = \sqrt{55.33} \approx 7.44 \text{ cm}$
Answer
$a \approx 7.44$ cm.
04.Exact trig values — finding an exact length
Higher
Question
A right-angled triangle has a hypotenuse of $8$ cm and one angle of $30°$ . Find the exact length of the side opposite the $30°$ angle.
Solution
1. 1
  Identify the required ratio: opposite and hypotenuse → use $\sin$ .
  $\sin 30° = \frac{\text{opp}}{\text{hyp}} = \frac{x}{8}$
2. 2
  Substitute the exact value $\sin 30° = \frac{1}{2}$ .
  $\frac{1}{2} = \frac{x}{8}$
3. 3
  Solve.
  $x = 4 \text{ cm}$
Answer
The opposite side $= 4$ cm (exact).

Key formulae

Arc Length and Sector Area
$l = \frac{\theta}{360} \times 2\pi r, \qquad A = \frac{\theta}{360} \times \pi r^2$
- $\theta$
  — angle at the centre in degrees
- $r$
  — radius of the circle
- $l$
  — arc length
- $A$
  — sector area
When to use
Any problem involving part of a circle (arc, sector, perimeter of sector). Rearrange to find $\theta$ or $r$ if either is unknown.
Volume of Sphere and Cone
$V_{\text{sphere}} = \frac{4}{3}\pi r^3, \qquad V_{\text{cone}} = \frac{1}{3}\pi r^2 h$
- $r$
  — radius
- $h$
  — perpendicular height (cone)
When to use
Volume of spheres, cones and hemispheres ( $V = \frac{2}{3}\pi r^3$ ). Use the same $\frac{1}{3}$ factor for any pyramid: $V = \frac{1}{3} A_{\text{base}} h$ .
Cosine Rule
$a^2 = b^2 + c^2 - 2bc\cos A \qquad \Leftrightarrow \qquad \cos A = \frac{b^2 + c^2 - a^2}{2bc}$
- $a, b, c$
  — side lengths (side $a$ is opposite angle $A$)
- $A$
  — angle opposite side $a$
When to use
SAS (finding the third side) or SSS (finding any angle). Use the left form to find a side; the right form to find an angle.
Sine Rule and Triangle Area
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}, \qquad \text{Area} = \frac{1}{2}ab\sin C$
- $a, b, c$
  — side lengths opposite angles $A, B, C$ respectively
- $C$
  — included angle between sides $a$ and $b$ in the area formula
When to use
Sine rule: AAS or SSA configurations. Area formula: any triangle where two sides and the included angle are known.

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Sector: A region bounded by two radii and an arc of a circle. It looks like a 'pizza slice'. The angle at the centre determines what fraction of the full circle the sector represents.
Slant height: The distance from the apex of a cone to any point on the base circle, measured along the surface. It is longer than the perpendicular height and is calculated using Pythagoras: $l = \sqrt{r^2 + h^2}$ .
Hypotenuse: The longest side of a right-angled triangle, always opposite the right angle. It is the side used in Pythagoras' theorem as $c$ in $a^2 + b^2 = c^2$ .
Included angle: The angle formed between two given sides of a triangle. In SAS, the included angle is the one sandwiched between the two known sides. It is the angle used in the cosine rule and the area formula $\frac{1}{2}ab\sin C$ .
Exact value: A value expressed without rounding, typically as a fraction, surd or in terms of $\pi$ . For trigonometry, exact values are the sine, cosine and tangent of 0°, 30°, 45°, 60° and 90° expressed as exact fractions or surds.
Ambiguous case (SSA): A situation in the sine rule where two sides and a non-included angle are given. Two different triangles can satisfy these conditions (giving two possible angles), so there may be two valid solutions. The supplement of the calculated angle must also be checked.

Common mistakes and misconceptions

Mistake
Using the slant height instead of the perpendicular height in the cone volume formula
Why it happens
The slant height is visible on the diagram and often explicitly given, so students substitute it directly into $\frac{1}{3}\pi r^2 h$ .
How to avoid it
Label both heights on the diagram. If only the slant height is given, find the perpendicular height first: $h = \sqrt{l^2 - r^2}$ . The volume formula always needs the perpendicular height.
Mistake
Forgetting to include both radii in the perimeter of a sector
Why it happens
Students calculate the arc length correctly but present it as the full perimeter, forgetting the two straight edges (radii).
How to avoid it
Write the formula $P = 2r + l$ before calculating. The perimeter of a sector always has three parts: two radii and one arc.
Mistake
Using the wrong angle when applying the area formula $\frac{1}{2}ab\sin C$
Why it happens
Students use any angle in the triangle rather than the angle that is between the two given sides.
How to avoid it
Identify the two sides $a$ and $b$ first, then find the angle $C$ that is directly between them (the included angle). Label the triangle clearly before applying the formula.
Mistake
Confusing exact trig values: swapping $\sin 30°$ and $\cos 30°$
Why it happens
Both equal $\frac{1}{2}$ or $\frac{\sqrt{3}}{2}$ and students confuse which is which, especially under time pressure.
How to avoid it
Remember: $\sin 30° = \frac{1}{2}$ (smaller angle, smaller sine); $\cos 30° = \frac{\sqrt{3}}{2}$ (larger value). Sketch the 30-60-90 triangle (sides 1, $\sqrt{3}$ , 2) and read off the ratios directly.

Geometry and MeasuresAQA GCSE Mathematics (8300)

Mensuration and Calculation

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

Previous Next

Master the topic

Worked examples4 Key formulae4 Definitions6 Common mistakes4

Step-by-step worked examples

01.Surface area and volume of a cone
Higher
Question
A cone has base radius 5 cm and perpendicular height 12 cm. Find its total surface area and volume. Give exact answers in terms of $\pi$ .
Solution
1. 1
  Find the slant height using Pythagoras.
  $l = \sqrt{r^2 + h^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm}$
2. 2
  Curved surface area of cone.
  $A_{\text{curved}} = \pi r l = \pi(5)(13) = 65\pi \text{ cm}^2$
3. 3
  Add the base area for total surface area.
  $A_{\text{total}} = 65\pi + \pi r^2 = 65\pi + 25\pi = 90\pi \text{ cm}^2$
4. 4
  Volume of cone.
  $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(25)(12) = 100\pi \text{ cm}^3$
Answer
Total surface area $= 90\pi$ cm²; Volume $= 100\pi$ cm³.
02.3D Pythagoras — space diagonal of a cuboid
Higher
Question
A cuboid has length 6 cm, width 4 cm and height 3 cm. Find the length of the space diagonal.
Solution
1. 1
  Find the diagonal of the base rectangle.
  $d_{\text{base}} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52}$
2. 2
  The space diagonal is the hypotenuse of the right-angled triangle with legs $d_{\text{base}}$ and height 3 cm.
  $d = \sqrt{(\sqrt{52})^2 + 3^2} = \sqrt{52 + 9} = \sqrt{61}$
3. 3
  Calculate.
  $d = \sqrt{61} \approx 7.81 \text{ cm}$
Answer
$d = \sqrt{61} \approx 7.81$ cm.
03.Cosine rule — finding a side
Challenge
Question
In triangle $ABC$ , $b = 7$ cm, $c = 10$ cm and $\angle A = 48°$ . Find side $a$ .
Solution
1. 1
  Write the cosine rule formula.
  $a^2 = b^2 + c^2 - 2bc\cos A$
2. 2
  Substitute the known values.
  $a^2 = 7^2 + 10^2 - 2(7)(10)\cos 48° = 49 + 100 - 140\cos 48°$
3. 3
  Evaluate $\cos 48° \approx 0.6691$ .
  $a^2 = 149 - 140(0.6691) = 149 - 93.67 = 55.33$
4. 4
  Take the square root.
  $a = \sqrt{55.33} \approx 7.44 \text{ cm}$
Answer
$a \approx 7.44$ cm.
04.Exact trig values — finding an exact length
Higher
Question
A right-angled triangle has a hypotenuse of $8$ cm and one angle of $30°$ . Find the exact length of the side opposite the $30°$ angle.
Solution
1. 1
  Identify the required ratio: opposite and hypotenuse → use $\sin$ .
  $\sin 30° = \frac{\text{opp}}{\text{hyp}} = \frac{x}{8}$
2. 2
  Substitute the exact value $\sin 30° = \frac{1}{2}$ .
  $\frac{1}{2} = \frac{x}{8}$
3. 3
  Solve.
  $x = 4 \text{ cm}$
Answer
The opposite side $= 4$ cm (exact).

Key formulae

Arc Length and Sector Area
$l = \frac{\theta}{360} \times 2\pi r, \qquad A = \frac{\theta}{360} \times \pi r^2$
- $\theta$
  — angle at the centre in degrees
- $r$
  — radius of the circle
- $l$
  — arc length
- $A$
  — sector area
When to use
Any problem involving part of a circle (arc, sector, perimeter of sector). Rearrange to find $\theta$ or $r$ if either is unknown.
Volume of Sphere and Cone
$V_{\text{sphere}} = \frac{4}{3}\pi r^3, \qquad V_{\text{cone}} = \frac{1}{3}\pi r^2 h$
- $r$
  — radius
- $h$
  — perpendicular height (cone)
When to use
Volume of spheres, cones and hemispheres ( $V = \frac{2}{3}\pi r^3$ ). Use the same $\frac{1}{3}$ factor for any pyramid: $V = \frac{1}{3} A_{\text{base}} h$ .
Cosine Rule
$a^2 = b^2 + c^2 - 2bc\cos A \qquad \Leftrightarrow \qquad \cos A = \frac{b^2 + c^2 - a^2}{2bc}$
- $a, b, c$
  — side lengths (side $a$ is opposite angle $A$)
- $A$
  — angle opposite side $a$
When to use
SAS (finding the third side) or SSS (finding any angle). Use the left form to find a side; the right form to find an angle.
Sine Rule and Triangle Area
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}, \qquad \text{Area} = \frac{1}{2}ab\sin C$
- $a, b, c$
  — side lengths opposite angles $A, B, C$ respectively
- $C$
  — included angle between sides $a$ and $b$ in the area formula
When to use
Sine rule: AAS or SSA configurations. Area formula: any triangle where two sides and the included angle are known.

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Sector: A region bounded by two radii and an arc of a circle. It looks like a 'pizza slice'. The angle at the centre determines what fraction of the full circle the sector represents.
Slant height: The distance from the apex of a cone to any point on the base circle, measured along the surface. It is longer than the perpendicular height and is calculated using Pythagoras: $l = \sqrt{r^2 + h^2}$ .
Hypotenuse: The longest side of a right-angled triangle, always opposite the right angle. It is the side used in Pythagoras' theorem as $c$ in $a^2 + b^2 = c^2$ .
Included angle: The angle formed between two given sides of a triangle. In SAS, the included angle is the one sandwiched between the two known sides. It is the angle used in the cosine rule and the area formula $\frac{1}{2}ab\sin C$ .
Exact value: A value expressed without rounding, typically as a fraction, surd or in terms of $\pi$ . For trigonometry, exact values are the sine, cosine and tangent of 0°, 30°, 45°, 60° and 90° expressed as exact fractions or surds.
Ambiguous case (SSA): A situation in the sine rule where two sides and a non-included angle are given. Two different triangles can satisfy these conditions (giving two possible angles), so there may be two valid solutions. The supplement of the calculated angle must also be checked.

Common mistakes and misconceptions

Mistake
Using the slant height instead of the perpendicular height in the cone volume formula
Why it happens
The slant height is visible on the diagram and often explicitly given, so students substitute it directly into $\frac{1}{3}\pi r^2 h$ .
How to avoid it
Label both heights on the diagram. If only the slant height is given, find the perpendicular height first: $h = \sqrt{l^2 - r^2}$ . The volume formula always needs the perpendicular height.
Mistake
Forgetting to include both radii in the perimeter of a sector
Why it happens
Students calculate the arc length correctly but present it as the full perimeter, forgetting the two straight edges (radii).
How to avoid it
Write the formula $P = 2r + l$ before calculating. The perimeter of a sector always has three parts: two radii and one arc.
Mistake
Using the wrong angle when applying the area formula $\frac{1}{2}ab\sin C$
Why it happens
Students use any angle in the triangle rather than the angle that is between the two given sides.
How to avoid it
Identify the two sides $a$ and $b$ first, then find the angle $C$ that is directly between them (the included angle). Label the triangle clearly before applying the formula.
Mistake
Confusing exact trig values: swapping $\sin 30°$ and $\cos 30°$
Why it happens
Both equal $\frac{1}{2}$ or $\frac{\sqrt{3}}{2}$ and students confuse which is which, especially under time pressure.
How to avoid it
Remember: $\sin 30° = \frac{1}{2}$ (smaller angle, smaller sine); $\cos 30° = \frac{\sqrt{3}}{2}$ (larger value). Sketch the 30-60-90 triangle (sides 1, $\sqrt{3}$ , 2) and read off the ratios directly.

Mensuration and Calculation

Master the topic

Step-by-step worked examples

01.Surface area and volume of a cone

02.3D Pythagoras — space diagonal of a cuboid

03.Cosine rule — finding a side

04.Exact trig values — finding an exact length

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions

Mensuration and Calculation

Master the topic

Step-by-step worked examples

01.Surface area and volume of a cone

02.3D Pythagoras — space diagonal of a cuboid

03.Cosine rule — finding a side

04.Exact trig values — finding an exact length

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions