Mathematics Β· AQA 8300 Β§G3
Volume and Surface Area
- Calculate volumes of prisms, cylinders, pyramids, cones and spheres
- Calculate surface areas of cuboids, cylinders, cones and spheres
- Solve problems involving composite 3D shapes
- Find dimensions given a volume or surface area (reverse problems)
- Apply 3D mensuration to real-world and algebraic contexts
π Core Concepts
2.1 Volume of Prisms
A prism is any 3D shape with a uniform (constant) cross-section along its length. The cross-section is always a 2D polygon.
DEFINITION β Prism
A solid with two identical, parallel polygonal faces (the bases) connected by rectangular lateral faces. The cross-section is constant along the length.
Volume of a Prism
$$V = A \times l$$
$A$ = area of cross-section
$l$ = length (depth) of prism
$V$ = volume
EXAM TIP
Always identify the cross-section first. It could be a triangle, trapezium, L-shape, or any polygon. Calculate its area, then multiply by the length.
COMMON MISTAKE
Multiplying the wrong dimension as the "length". The length must be perpendicular to the cross-section face.
2.2 Cylinders
A cylinder is a prism with a circular cross-section. Because the cross-section is a circle of area $\pi r^2$, the volume and curved surface area have their own formulae.
Cylinder β Volume & Surface Area
$$V = \pi r^2 h \qquad \text{SA} = 2\pi r h + 2\pi r^2$$
$r$ = radius of circular base
$h$ = height of cylinder
$2\pi r h$ = curved surface area (label unrolled = rectangle)
$2\pi r^2$ = two circular ends
MEMORY TRICK β Cylinder SA
Unroll the curved surface: it becomes a rectangle of width $2\pi r$ (circumference) and height $h$. Add two circles of area $\pi r^2$ each.
EXAM TIP
If a question asks for the curved surface area only (e.g. label on a tin), use $2\pi r h$ alone β do not add the circular ends unless instructed.
2.3 Pyramids
A pyramid has a polygonal base and triangular faces meeting at an apex. Its volume is exactly one-third of the prism with the same base and height. This factor of $\tfrac{1}{3}$ comes from calculus (integration of cross-sectional area).
Volume of a Pyramid
$$V = \tfrac{1}{3} \times A_{\text{base}} \times h$$
$A_{\text{base}}$ = area of the base polygon
$h$ = perpendicular height from base to apex
COMMON MISTAKE
Using the slant height instead of the perpendicular height $h$. Always use the vertical height for volume calculations.
EXAM TIP
For a square-based pyramid with base side $a$ and slant edge $e$: use Pythagoras to find $h = \sqrt{e^2 - (a/\sqrt{2})^2}$ before calculating volume.
2.4 Cones
A cone is a pyramid with a circular base. It has two important height measures: the perpendicular height $h$ (for volume) and the slant height $l$ (for surface area).
Cone β Volume & Surface Area
$$V = \tfrac{1}{3}\pi r^2 h \qquad \text{SA} = \pi r l + \pi r^2$$
$r$ = base radius
$h$ = perpendicular height
$l$ = slant height, $l = \sqrt{r^2 + h^2}$
$\pi r l$ = curved surface area (sector of circle)
$\pi r^2$ = base circle
IMPORTANT β Slant Height
The slant height $l$ is NOT the same as the perpendicular height $h$. Use Pythagoras: $$l = \sqrt{r^2 + h^2}$$
Volume uses $h$; surface area uses $l$.
MEMORY TRICK β Cone CSA
The curved surface of a cone unrolls into a sector of a circle with radius $l$ and arc length $2\pi r$. Area of sector = $\pi r l$.
2.5 Spheres
A sphere is perfectly symmetric in all directions. Both formulae involve $r^3$ (volume) and $r^2$ (surface area), derived via calculus, and are given on the AQA formula sheet.
Sphere β Volume & Surface Area
$$V = \tfrac{4}{3}\pi r^3 \qquad \text{SA} = 4\pi r^2$$
$r$ = radius of sphere
Note: diameter $d = 2r$ β halve $d$ first if given diameter
COMMON MISTAKE
Using the diameter instead of the radius in the sphere formulae. If given diameter $d$, always divide by 2 first: $r = d/2$.
EXAM TIP
A hemisphere has volume $\tfrac{2}{3}\pi r^3$ and total surface area $3\pi r^2$ (curved half $= 2\pi r^2$, plus flat circle $= \pi r^2$).
2.6 Surface Area of Cuboids
A cuboid has 6 rectangular faces in 3 pairs. Opposite faces are identical.
Cuboid Surface Area
$$\text{SA} = 2(lw + lh + wh)$$
$l$ = length, $w$ = width, $h$ = height
Each bracket term = area of one pair of opposite faces
2.7 Composite 3D Shapes
Many exam problems combine two or more standard shapes. The approach depends on whether you need volume (add constituent volumes) or surface area (add exposed faces, subtract hidden internal faces).
DEFINITION β Composite Shape
A 3D solid formed by combining (or subtracting) two or more standard shapes. Common examples: cone on cylinder, hemisphere on cylinder, pyramid on prism.
Identify constituent shapes
β
Calculate each part separately
β
Add volumes OR add exposed SA (subtract hidden faces)
β
State units and round as required
IMPORTANT β Surface Area of Composites
When two shapes share a face (e.g. a cone sitting on a cylinder), that circular face is internal β exclude it from the total surface area. Only count the outer surface.
2.8 Converting Volume Units
Volume is three-dimensional, so unit conversions are cubed. The most common conversion is between cmΒ³ and mΒ³.
Volume Unit Conversions
$$1\,\text{m}^3 = 1\,000\,000\,\text{cm}^3 = 10^6\,\text{cm}^3$$
$$1\,\text{cm}^3 = 10^{-6}\,\text{m}^3$$
$$1\,\text{litre} = 1000\,\text{cm}^3 \qquad 1\,\text{m}^3 = 1000\,\text{litres}$$
cm β m: divide by 100; so cmΒ³ β mΒ³: divide by $100^3 = 10^6$
mm β cm: divide by 10; so mmΒ³ β cmΒ³: divide by $10^3 = 1000$
MEMORY TRICK β Volume Conversions
"Area conversions are squared, volume conversions are cubed." If 1 m = 100 cm, then 1 mΒ² = 100Β² cmΒ² = 10 000 cmΒ², and 1 mΒ³ = 100Β³ cmΒ³ = 1 000 000 cmΒ³.
2.9 Reverse Problems β Finding Dimensions
At Grade 9, you may be given the volume or surface area and asked to find a dimension. This requires rearranging the formula algebraically.
WORKED EXAMPLE β Reverse Sphere
A sphere has volume 288Ο cmΒ³. Find its radius.
$\tfrac{4}{3}\pi r^3 = 288\pi$
$r^3 = 288 \times \tfrac{3}{4} = 216$
$r = \sqrt[3]{216} = 6\,\text{cm}$
EXAM TIP β Reverse Problems
Leave $\pi$ as a symbol throughout until the final step. This avoids rounding errors and keeps the algebra cleaner: e.g. $\tfrac{4}{3}\pi r^3 = 36\pi \Rightarrow r^3 = 27 \Rightarrow r = 3$.
2.10 Volume Ratios in Similar Shapes
When two shapes are mathematically similar with linear scale factor $k$, their volumes scale by $k^3$.
Similar Shapes β Volume Ratio
$$\frac{V_2}{V_1} = k^3 \qquad \text{where } k = \frac{l_2}{l_1}$$
$k$ = linear scale factor (ratio of corresponding lengths)
Surface areas scale by $k^2$; volumes scale by $k^3$
EXAM TIP
To find the scale factor from volumes: $k = \sqrt[3]{V_2 / V_1}$. Then use $k^2$ for the surface area ratio.
πΊοΈ Visual Notes
Volume &
Surface Area
Surface Area
Prisms & Cylinders
- $V = A \times l$ (cross-section Γ length)
- Cylinder: $V = \pi r^2 h$
- SA cylinder: $2\pi rh + 2\pi r^2$
- Cuboid SA: $2(lw + lh + wh)$
Pyramids & Cones
- $V = \tfrac{1}{3} A_{\text{base}} h$
- Cone: $V = \tfrac{1}{3}\pi r^2 h$
- Cone SA: $\pi r l + \pi r^2$
- Slant height: $l = \sqrt{r^2 + h^2}$
Spheres & Hemispheres
- $V = \tfrac{4}{3}\pi r^3$
- SA: $4\pi r^2$
- Hemisphere V: $\tfrac{2}{3}\pi r^3$
- Hemisphere SA: $3\pi r^2$
Composite Shapes
- Add individual volumes
- Subtract hidden (shared) faces from SA
- Common: cone + cylinder
- Common: hemisphere + cylinder
Unit Conversions
- $1\,\text{m}^3 = 10^6\,\text{cm}^3$
- $1\,\text{cm}^3 = 1000\,\text{mm}^3$
- $1\,\text{litre} = 1000\,\text{cm}^3$
- Cube the length conversion factor
Similar Shapes
- Linear SF: $k$
- Area SF: $k^2$
- Volume SF: $k^3$
- Find $k$ from $\sqrt[3]{V_2/V_1}$
Formula Reference Table
| Shape | Volume | Surface Area | Key variable |
|---|---|---|---|
| Cuboid | $lwh$ | $2(lw+lh+wh)$ | $l, w, h$ |
| Cylinder | $\pi r^2 h$ | $2\pi r h + 2\pi r^2$ | $r, h$ |
| Prism (general) | $A \times l$ | $2A + \text{perimeter} \times l$ | $A$ = cross-section area |
| Pyramid (general) | $\tfrac{1}{3}A_{\text{base}}h$ | Base + lateral faces | $h$ = perp. height |
| Cone | $\tfrac{1}{3}\pi r^2 h$ | $\pi r l + \pi r^2$ | $l = \sqrt{r^2+h^2}$ |
| Sphere | $\tfrac{4}{3}\pi r^3$ | $4\pi r^2$ | $r$ = radius |
| Hemisphere | $\tfrac{2}{3}\pi r^3$ | $3\pi r^2$ | Flat face + curved |
Volume vs Surface Area Decision Process
Read the question carefully
β
Is it asking for 3D space inside? β Volume
β
Is it asking for 2D outer skin? β Surface Area
β
Is it composite? β Split, calculate each part, combine
β
Check units (cmΒ³ or cmΒ²) β cube root or square root if reversing
βοΈ Worked Examples
Grade 4β5 β Cylinder Volume and Surface Area
A cylinder has radius 5 cm and height 12 cm. Calculate (a) its volume and (b) its total surface area. Give your answers to 3 significant figures.
1
Write the formula for volume
$V = \pi r^2 h$, where $r = 5$ and $h = 12$.
2
Substitute and calculate
$V = \pi \times 5^2 \times 12 = \pi \times 25 \times 12 = 300\pi \approx 942\,\text{cm}^3$
3
Surface area β curved part
Curved SA $= 2\pi r h = 2\pi \times 5 \times 12 = 120\pi$
4
Surface area β two circular ends
Two circles $= 2 \times \pi r^2 = 2 \times \pi \times 25 = 50\pi$
5
Total SA
$\text{SA} = 120\pi + 50\pi = 170\pi \approx 534\,\text{cm}^2$
(a) $V = 300\pi \approx 942\,\text{cm}^3$ (b) $\text{SA} = 170\pi \approx 534\,\text{cm}^2$
Grade 6β7 β Cone: Finding Slant Height then Surface Area
A cone has base radius 6 cm and perpendicular height 8 cm. Calculate the total surface area of the cone. Give your answer in terms of $\pi$.
1
Find the slant height using Pythagoras
$l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\,\text{cm}$
2
Curved surface area
Curved SA $= \pi r l = \pi \times 6 \times 10 = 60\pi\,\text{cm}^2$
3
Base circle area
Base $= \pi r^2 = \pi \times 36 = 36\pi\,\text{cm}^2$
4
Total surface area
$\text{SA} = 60\pi + 36\pi = 96\pi\,\text{cm}^2$
Total SA $= 96\pi\,\text{cm}^2 \approx 302\,\text{cm}^2$
Grade 9 β Composite Shape: Hemisphere on Cylinder
A solid is made by placing a hemisphere of radius 4 cm on top of a cylinder of radius 4 cm and height 10 cm. Calculate (a) the total volume and (b) the total surface area of the solid. Give your answers to 3 significant figures.
1
Identify the components
Cylinder: $r = 4$ cm, $h = 10$ cm. Hemisphere: $r = 4$ cm. (sits on top β the flat face of the hemisphere coincides with the top circle of the cylinder.)
2
Volume of cylinder
$V_{\text{cyl}} = \pi r^2 h = \pi \times 16 \times 10 = 160\pi\,\text{cm}^3$
3
Volume of hemisphere
$V_{\text{hem}} = \tfrac{2}{3}\pi r^3 = \tfrac{2}{3} \times \pi \times 64 = \tfrac{128\pi}{3}\,\text{cm}^3$
4
Total volume
$V = 160\pi + \tfrac{128\pi}{3} = \tfrac{480\pi}{3} + \tfrac{128\pi}{3} = \tfrac{608\pi}{3} \approx 637\,\text{cm}^3$
5
Surface area β identify exposed faces
The shared circular face (top of cylinder = flat of hemisphere) is internal; exclude it.Exposed faces: (i) curved surface of cylinder, (ii) bottom circle of cylinder, (iii) curved surface of hemisphere.
6
Calculate each exposed face
(i) Curved cylinder: $2\pi r h = 2\pi \times 4 \times 10 = 80\pi$(ii) Bottom circle: $\pi r^2 = 16\pi$
(iii) Curved hemisphere: $2\pi r^2 = 2\pi \times 16 = 32\pi$
7
Total surface area
$\text{SA} = 80\pi + 16\pi + 32\pi = 128\pi \approx 402\,\text{cm}^2$
(a) $V = \tfrac{608\pi}{3} \approx 637\,\text{cm}^3$ (b) $\text{SA} = 128\pi \approx 402\,\text{cm}^2$
β Exam Questions
Q11 mark
A sphere has radius 3 cm. Write down its surface area in terms of $\pi$.
Mark scheme:
$\text{SA} = 4\pi r^2 = 4\pi \times 9 = 36\pi\,\text{cm}^2$ β (1 mark for correct answer)
$\text{SA} = 4\pi r^2 = 4\pi \times 9 = 36\pi\,\text{cm}^2$ β (1 mark for correct answer)
Q22 marks
A triangular prism has a right-angled triangular cross-section with legs 5 cm and 12 cm. The prism has length 20 cm. Calculate the volume of the prism.
Mark scheme:
Area of triangle $= \tfrac{1}{2} \times 5 \times 12 = 30\,\text{cm}^2$ β (M1)
Volume $= 30 \times 20 = 600\,\text{cm}^3$ β (A1)
Area of triangle $= \tfrac{1}{2} \times 5 \times 12 = 30\,\text{cm}^2$ β (M1)
Volume $= 30 \times 20 = 600\,\text{cm}^3$ β (A1)
Q33 marks
A cone has slant height 13 cm and base radius 5 cm. Calculate the total surface area of the cone. Give your answer to 3 significant figures.
Mark scheme:
Curved SA $= \pi r l = \pi \times 5 \times 13 = 65\pi$ β (M1)
Base $= \pi r^2 = 25\pi$ β (M1)
Total $= 90\pi \approx 283\,\text{cm}^2$ β (A1)
Note: If student used Pythagoras to find $h$ instead of recognising $l$ was given β penalise by 1 mark only if SA formula correct but $l$ wrong.
Curved SA $= \pi r l = \pi \times 5 \times 13 = 65\pi$ β (M1)
Base $= \pi r^2 = 25\pi$ β (M1)
Total $= 90\pi \approx 283\,\text{cm}^2$ β (A1)
Note: If student used Pythagoras to find $h$ instead of recognising $l$ was given β penalise by 1 mark only if SA formula correct but $l$ wrong.
Q44 marks
A solid metal sphere of radius 6 cm is melted down and recast into a cylinder of radius 4 cm. Calculate the height of the cylinder. Give your answer to 3 significant figures.
Mark scheme:
Volume of sphere $= \tfrac{4}{3}\pi \times 6^3 = \tfrac{4}{3}\pi \times 216 = 288\pi$ β (M1)
Volume of cylinder $= \pi \times 4^2 \times h = 16\pi h$ β (M1)
Setting equal: $16\pi h = 288\pi$ β (M1 β equating volumes)
$h = \dfrac{288}{16} = 18\,\text{cm}$ β (A1)
Volume of sphere $= \tfrac{4}{3}\pi \times 6^3 = \tfrac{4}{3}\pi \times 216 = 288\pi$ β (M1)
Volume of cylinder $= \pi \times 4^2 \times h = 16\pi h$ β (M1)
Setting equal: $16\pi h = 288\pi$ β (M1 β equating volumes)
$h = \dfrac{288}{16} = 18\,\text{cm}$ β (A1)
Q53 marks
A water tank is in the shape of a cylinder with internal diameter 80 cm and internal height 120 cm. How many complete litres of water does it hold when full? (1 litre = 1000 cmΒ³)
Mark scheme:
Radius $= 40\,\text{cm}$ β (B1 β correct radius from diameter)
$V = \pi \times 40^2 \times 120 = 192\,000\pi \approx 603\,186\,\text{cm}^3$ β (M1)
$\div 1000 = 603.186\ldots \Rightarrow 603$ complete litres β (A1 β must be complete litres, i.e. floor/truncate)
Radius $= 40\,\text{cm}$ β (B1 β correct radius from diameter)
$V = \pi \times 40^2 \times 120 = 192\,000\pi \approx 603\,186\,\text{cm}^3$ β (M1)
$\div 1000 = 603.186\ldots \Rightarrow 603$ complete litres β (A1 β must be complete litres, i.e. floor/truncate)
Q66 marks
A pencil is modelled as a cylinder of radius 0.4 cm and length 17 cm, topped by a cone of the same radius and height 1.5 cm. A rubber (eraser) in the shape of a hemisphere of radius 0.4 cm is attached to the bottom of the cylinder. Calculate the total surface area of the pencil (excluding the flat circular end where the rubber meets the cylinder and the flat circular end of the cone base). Give your answer to 3 significant figures.
Mark scheme:
Slant height of cone: $l = \sqrt{0.4^2 + 1.5^2} = \sqrt{0.16 + 2.25} = \sqrt{2.41} \approx 1.552\,\text{cm}$ β (M1)
Curved cone SA: $\pi \times 0.4 \times 1.552 = 0.6208\pi$ β (M1)
Curved cylinder SA: $2\pi \times 0.4 \times 17 = 13.6\pi$ β (M1)
Curved hemisphere SA: $2\pi \times 0.4^2 = 0.32\pi$ β (M1)
(The shared circular faces between cone/cylinder and cylinder/hemisphere are excluded as stated.)
Total SA $= (0.6208 + 13.6 + 0.32)\pi = 14.5408\pi \approx 45.7\,\text{cm}^2$ ββ (A2 β award A1 for $14.5\pi$ to $14.6\pi$, A2 for correct final answer)
Slant height of cone: $l = \sqrt{0.4^2 + 1.5^2} = \sqrt{0.16 + 2.25} = \sqrt{2.41} \approx 1.552\,\text{cm}$ β (M1)
Curved cone SA: $\pi \times 0.4 \times 1.552 = 0.6208\pi$ β (M1)
Curved cylinder SA: $2\pi \times 0.4 \times 17 = 13.6\pi$ β (M1)
Curved hemisphere SA: $2\pi \times 0.4^2 = 0.32\pi$ β (M1)
(The shared circular faces between cone/cylinder and cylinder/hemisphere are excluded as stated.)
Total SA $= (0.6208 + 13.6 + 0.32)\pi = 14.5408\pi \approx 45.7\,\text{cm}^2$ ββ (A2 β award A1 for $14.5\pi$ to $14.6\pi$, A2 for correct final answer)
β Grade 9 Model Answers
Full annotated solution to Q4 (sphere melted into cylinder) β a classic Grade 9 problem requiring students to set two volume expressions equal and solve.
Grade 9 Model β Sphere Recast as Cylinder
A solid metal sphere of radius 6 cm is melted and recast into a solid cylinder of radius 4 cm. Find the height of the cylinder.
1
State the key principle β conservation of volume
When a solid is melted and recast, the volume is conserved (assuming no material is lost). Therefore:$$V_{\text{sphere}} = V_{\text{cylinder}}$$
2
Calculate volume of sphere
$$V_{\text{sphere}} = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3} \times \pi \times 6^3 = \tfrac{4}{3} \times 216\pi = 288\pi\,\text{cm}^3$$
Mark earned: correct formula applied with $r=6$.
3
Write expression for cylinder volume with unknown $h$
$$V_{\text{cylinder}} = \pi r^2 h = \pi \times 4^2 \times h = 16\pi h$$
Mark earned: correct formula with $r=4$, leaving $h$ as unknown.
4
Set equal and solve for $h$
$$16\pi h = 288\pi$$
Divide both sides by $\pi$: $\quad 16h = 288$
$$h = \frac{288}{16} = 18\,\text{cm}$$
Mark earned: correct algebraic manipulation, cancellation of $\pi$.
5
Why each part earns marks
- M1: Correct sphere volume formula used with $r = 6$.
- M1: Correct cylinder volume formula used with $r = 4$.
- M1: Equating the two volume expressions (demonstrating conservation of volume).
- A1: Correct answer $h = 18$ cm with appropriate units.
Height of cylinder $= 18$ cm
EXAM TIP β What makes a Grade 9 response
- Explicitly states the conservation of volume principle.
- Cancels $\pi$ symbolically rather than approximating to 3.14 early.
- Shows all intermediate steps clearly β method marks are available even if arithmetic slips.
- States the answer with correct units (cm, not cmΒ³, because the answer is a length).
π Revision Sheet
Key Definitions
| Term | Meaning |
|---|---|
| Volume | 3D space enclosed by a shape (units: cmΒ³, mΒ³) |
| Surface area | Total area of all outer faces (units: cmΒ², mΒ²) |
| Prism | Solid with constant polygonal cross-section |
| Slant height $l$ | Distance along slope of cone/pyramid face |
| Perp. height $h$ | Vertical distance from base to apex |
| Composite | Shape made from two or more standard shapes |
| Similar shapes | Same shape, different size; scale factor $k$ |
Essential Formulae
Cylinder: $V = \pi r^2 h$, $\;\text{SA} = 2\pi rh + 2\pi r^2$
Cone: $V = \tfrac{1}{3}\pi r^2 h$, $\;\text{SA} = \pi rl + \pi r^2$, $\;l = \sqrt{r^2+h^2}$
Sphere: $V = \tfrac{4}{3}\pi r^3$, $\;\text{SA} = 4\pi r^2$
Prism: $V = A \times l$
Pyramid: $V = \tfrac{1}{3}A_\text{base}h$
Cuboid SA: $2(lw + lh + wh)$
Hemisphere: $V = \tfrac{2}{3}\pi r^3$, $\;\text{SA} = 3\pi r^2$
Similar vol. ratio: $V_2/V_1 = k^3$
Memory Hooks
- "Prism = Cross Γ Length" β any prism, any base shape
- "Pyramids and Cones have β " β one third of corresponding prism
- "Four Three Pi r cubed" β $\tfrac{4}{3}\pi r^3$ for sphere volume
- "Four Pi r squared" β $4\pi r^2$ for sphere surface area
- "$l$ not $h$ for cone SA" β slant height in $\pi rl$, perpendicular height in $V$
- "Volume conversions: cube it" β 1 m = 100 cm, so 1 mΒ³ = 100Β³ cmΒ³ = 10βΆ cmΒ³
- "Unroll the cylinder" β CSA = rectangle: $2\pi r \times h$
Exam Tips
- Always halve the diameter to get the radius before substituting.
- Work in exact form ($\pi$) until the final step to avoid rounding errors.
- For composite shapes: draw and label the components separately.
- In surface area: exclude shared internal faces between joined components.
- For reverse problems: rearrange algebraically, then substitute numbers.
- Check units β volume in cmΒ³ or mΒ³, surface area in cmΒ² or mΒ².
- For similar shapes: linear SF $k$ β area SF $k^2$ β volume SF $k^3$.
- "Melted and recast" always means: set volumes equal.
π Flashcards
Click a card to reveal the answer. Use these to test your recall of key formulae and concepts.
β Common Mistakes
MISTAKE 1 β Using Diameter Instead of Radius
What students do wrong: Substituting $d$ (e.g. 10 cm) directly into $\pi r^2 h$ to get $\pi \times 10^2 \times h$.Why marks are lost: The answer is 4 times too large; all subsequent working is incorrect.
How to avoid it: Always check: "Is this the radius or the diameter?" If it says diameter, write $r = d/2$ before proceeding.
MISTAKE 2 β Using Perpendicular Height for Cone SA
What students do wrong: Writing $\text{SA} = \pi r h + \pi r^2$, using $h$ instead of the slant height $l$.Why marks are lost: The curved surface area formula specifically requires the slant height β $h$ gives a meaningless result.
How to avoid it: Always compute $l = \sqrt{r^2 + h^2}$ first. Write it down as a separate line. Then substitute $l$ in $\pi r l$.
MISTAKE 3 β Including Internal (Shared) Faces in Composite SA
What students do wrong: Adding the full surface areas of both shapes without subtracting the shared circular interface.Why marks are lost: The surface area is over-counted by $2 \times \pi r^2$ (one circular face counted twice).
How to avoid it: Draw the composite shape, then mentally "paint" the outside. Ask: "Is this face visible from outside?" Internal joins are not visible β exclude them.
MISTAKE 4 β Forgetting the Factor of β
for Cones and Pyramids
What students do wrong: Writing $V = \pi r^2 h$ for a cone or $V = A_{\text{base}} h$ for a pyramid β omitting the $\tfrac{1}{3}$.Why marks are lost: The answer is 3 times too large; the M1 formula mark is lost immediately.
How to avoid it: Recite the rule: "Pointy shapes have $\tfrac{1}{3}$." Cones and pyramids both taper to a point β they hold a third as much as the corresponding prism or cylinder.
MISTAKE 5 β Incorrect Volume Unit Conversion
What students do wrong: Dividing cmΒ³ by 100 (rather than 10βΆ) to convert to mΒ³, or dividing by 1000 when converting cmΒ³ to litres (correct) but applying the wrong factor.Why marks are lost: Final answer is wrong by a factor of 10 000; unit conversion marks are lost.
How to avoid it: Always cube the length conversion. 1 m = 100 cm, so 1 mΒ³ = 100Β³ cmΒ³ = 1 000 000 cmΒ³. For litres: 1 litre = 1000 cmΒ³ exactly.
MISTAKE 6 β Rounding $\pi$ Too Early
What students do wrong: Substituting $\pi \approx 3.14$ at the start of multi-step problems, accumulating rounding errors.Why marks are lost: Final answer differs from the expected value; accuracy mark (A1) is lost even though method is correct.
How to avoid it: Keep $\pi$ as a symbol until the final calculation. Write $= 128\pi$ and only evaluate $128\pi \approx 402\,\text{cm}^2$ as the last step using the calculator's $\pi$ button.
β Final Checklist
Click each item when you are confident. Your progress is saved automatically.
- I can state the formula $V = A \times l$ for any prism and identify the cross-section.
- I can calculate the volume and total surface area of a cylinder.
- I can apply the $\tfrac{1}{3}$ rule to pyramids and cones without a formula sheet.
- I can find the slant height $l = \sqrt{r^2 + h^2}$ of a cone and use it in the SA formula.
- I can calculate the volume and surface area of a sphere and hemisphere.
- I can calculate the surface area of a cuboid using $2(lw + lh + wh)$.
- I can decompose a composite shape and add volumes correctly.
- I know to exclude shared internal faces when finding surface area of composite shapes.
- I can convert between cmΒ³ and mΒ³ by dividing or multiplying by $10^6$.
- I can convert between cmΒ³ and litres using 1 litre = 1000 cmΒ³.
- I can solve reverse problems: given volume or SA, find an unknown dimension.
- I can apply the volume scale factor $k^3$ for similar shapes.
- I keep $\pi$ as a symbol throughout calculations and only evaluate at the final step.
- I always halve the diameter to get the radius before substituting into any formula.
- I can solve "melted and recast" problems by equating volumes and solving for the unknown dimension.