Surface area & volume of 3D shapes – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator4 marks

(a) Write down the mathematical name of the solid that has one flat circular face, one curved surface and one vertex. [1]

A square-based pyramid has a square base and four triangular faces that meet at a point.

(b) For a square-based pyramid, write down the number of
(i) faces, (ii) edges, (iii) vertices. [3]

(a) coneB1
(b)(i) $5$B1
(b)(ii) $8$B1
(b)(iii) $5$B1
Answer: (a) cone (b)(i) 5 (ii) 8 (iii) 5
Examiner tip: Count edges carefully: 4 round the base plus 4 slanting up to the top makes 8, not 4 or 5.

2Grade 1–3FoundationCalculator5 marks

A gift box is a cuboid measuring $6\ \text{cm}$ by $4\ \text{cm}$ by $2.5\ \text{cm}$.

(a) Work out the volume of the box. [2]

(b) Work out the total surface area of the box. [3]

(a) $6 \times 4 \times 2.5$M1
(a) $60$ (cm³)A1
(b) area of at least two different faces, e.g. $24$, $15$ or $10$M1
(b) complete method, e.g. $2 \times 24 + 2 \times 15 + 2 \times 10$ or $2(24 + 15 + 10)$M1
(b) $98$ (cm²)A1
Answer: (a) $60\ \text{cm}^3$ (b) $98\ \text{cm}^2$
Examiner tip: A cuboid has six faces in three matching pairs; adding only the three different areas (49) loses the final marks.

3Grade 4–5FoundationCalculator5 marks

The diagram shows a solid prism. The cross-section of the prism is a right-angled triangle with sides $9\ \text{cm}$, $12\ \text{cm}$ and $15\ \text{cm}$. The length of the prism is $20\ \text{cm}$.

(a) Work out the volume of the prism. [2]

(b) Work out the total surface area of the prism. [3]

(a) $\tfrac{1}{2} \times 9 \times 12 \times 20$ oeM1
(a) $1080$ (cm³)A1
(b) area of a triangle ($54$) or area of any one rectangular face ($180$, $240$ or $300$)M1
(b) complete method, e.g. $2 \times 54 + 180 + 240 + 300$M1
(b) $828$ (cm²)A1
Answer: (a) $1080\ \text{cm}^3$ (b) $828\ \text{cm}^2$
Examiner tip: There are three different rectangles (9 by 20, 12 by 20 and 15 by 20) — the sloping 15 cm face is often forgotten.

Stuck on Surface area & volume of 3D shapes?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4Grade 4–5Calculator5 marks

Jo has two containers.

Container A is a cylinder with radius $14\ \text{cm}$ and height $32\ \text{cm}$.
Container B is a cuboid measuring $0.25\ \text{m}$ by $0.4\ \text{m}$ by $0.2\ \text{m}$.

Jo says, “Container A holds more water than container B.”

Is Jo correct? You must show all your working. [5]

$\pi \times 14^2 \times 32$ (= $19\,704$…)M1
$0.25 \times 0.4 \times 0.2$ (= $0.02$ m³) or $25 \times 40 \times 20$ (= $20\,000$ cm³)M1
correct conversion to the same units, e.g. dimensions of B written as $25$, $40$, $20$ (cm) or $0.02 \times 1\,000\,000$ or $19\,704 \div 1000$ (= $19.7$ litres) or $19\,704 \div 1\,000\,000$ (m³)M1
$19\,700$ (cm³) and $20\,000$ (cm³) oe in the same units, e.g. $19.7$ and $20$ litresA1
No, with correct supporting values (B holds more)C1
Answer: No: A holds about $19\,700\ \text{cm}^3$ (19.7 litres) and B holds $20\,000\ \text{cm}^3$ (20 litres)
Examiner tip: Change metres to centimetres BEFORE multiplying (or multiply by $100^3$ after); multiplying $0.02$ by 100 is the classic slip.

5Grade 4–5Calculator4 marks

A plant pot is a cylinder with a base but no lid.

The pot has diameter $8\ \text{cm}$ and height $11\ \text{cm}$.

The outside of the pot, including the base, is painted.

Work out the area that is painted.
Give your answer correct to 3 significant figures. [4]

$\pi \times 4^2$ (= $50.2$…) for the baseM1
$\pi \times 8 \times 11$ or $2 \times \pi \times 4 \times 11$ (= $276.4$…) for the curved surfaceM1
$\pi \times 4^2 + \pi \times 8 \times 11$ (one base only) (= $326.7$…)M1
$327$ (cm²)A1
Answer: $327\ \text{cm}^2$
Examiner tip: Use the radius (4 cm), not the diameter, in $\pi r^2$, and include only ONE circle because the pot has no lid.

6Grade 6–7HigherCalculator4 marks

A solid is made from a cone and a hemisphere.

The cone has base radius $3\ \text{cm}$ and vertical height $10\ \text{cm}$. The flat face of the hemisphere fits exactly on the base of the cone.

Work out the total volume of the solid.
Give your answer correct to 3 significant figures. [4]

$\tfrac{1}{3} \times \pi \times 3^2 \times 10$ (= $30\pi$ = $94.2$…)M1
$\tfrac{1}{2} \times \tfrac{4}{3} \times \pi \times 3^3$ (= $18\pi$ = $56.5$…)M1
$30\pi + 18\pi$ oe (dep on both previous M marks)M1
$151$ (cm³) (accept $150.7$ – $150.8$)A1
Answer: $151\ \text{cm}^3$
Examiner tip: A hemisphere is half a sphere — halve $\tfrac{4}{3}\pi r^3$; using the full sphere gives 207 and scores only 1 of the 4 marks.

7Grade 6–7HigherCalculator4 marks

A solid metal cylinder has radius $6\ \text{cm}$ and height $20\ \text{cm}$.

The cylinder is melted down. $8\%$ of the metal is lost in the process.

The rest of the metal is used to make solid spheres, each of radius $1.5\ \text{cm}$.

Work out the greatest number of spheres that can be made. [4]

$\pi \times 6^2 \times 20$ (= $720\pi$ = $2261.9$…)M1
$\tfrac{4}{3} \times \pi \times 1.5^3$ (= $4.5\pi$ = $14.13$…)M1
$\dfrac{0.92 \times 720\pi}{4.5\pi}$ oe (= $147.2$)M1
$147$ caoA1
Answer: $147$ spheres
Examiner tip: Round DOWN at the end — you cannot make 0.2 of a sphere — and remember the 8% loss means using 92% of the volume.

8Grade 6–7HigherNon-calculator4 marks

The diagram shows a solid cone.

The base radius is $5\ \text{cm}$ and the vertical height is $12\ \text{cm}$.

Work out the total surface area of the cone.
Give your answer in terms of $\pi$. [4]

$\sqrt{5^2 + 12^2}$ (= $13$) for the slant heightM1
$\pi \times 5 \times 13$ (= $65\pi$) for the curved surfaceM1
$65\pi + \pi \times 5^2$ oeM1
$90\pi$ (cm²)A1
Answer: $90\pi\ \text{cm}^2$
Examiner tip: The curved surface formula $\pi r l$ uses the SLANT height $l$, so find it with Pythagoras first — using 12 gives $60\pi$.

9Grade 8–9HigherCalculator4 marks

A cylindrical tank has internal radius $8\ \text{cm}$. It contains water to a depth of $10\ \text{cm}$.

A solid metal sphere is placed in the tank and sinks to the bottom. The sphere is completely covered by the water.

The water level rises by $2.25\ \text{cm}$.

Work out the radius of the sphere.
Give your answer correct to 3 significant figures. [4]

$\pi \times 8^2 \times 2.25$ (= $144\pi$ = $452.3$…) for the volume of water displacedM1
$\tfrac{4}{3}\pi r^3 = 144\pi$ oeM1
$r^3 = 108$ oe or $r = \sqrt[3]{\dfrac{3 \times 144}{4}}$M1
$4.76$ (cm) (accept $4.762$…)A1
Answer: $4.76\ \text{cm}$
Examiner tip: The sphere's volume equals only the extra slice of water (height 2.25 cm), not the whole new water volume.

10Grade 8–9HigherNon-calculator5 marks

A solid sphere has radius $r$ cm.

A solid cone has base radius $2r$ cm and vertical height $h$ cm.

The volume of the cone is equal to the volume of the sphere.

Show that total surface area of cone : surface area of sphere $= (2 + \sqrt{5}) : 2$ [5]

$\tfrac{1}{3}\pi (2r)^2 h = \tfrac{4}{3}\pi r^3$M1
$h = r$A1
slant height $l = \sqrt{(2r)^2 + r^2} = \sqrt{5}\,r$M1
total surface area of cone $= \pi(2r)(\sqrt{5}r) + \pi(2r)^2$ (= $(2\sqrt{5} + 4)\pi r^2$)M1
$\dfrac{(2\sqrt{5} + 4)\pi r^2}{4\pi r^2}$ simplified fully to $(2 + \sqrt{5}) : 2$ with no errors seenA1
Answer: $h = r$, $l = \sqrt{5}r$, cone $= (4 + 2\sqrt{5})\pi r^2$, sphere $= 4\pi r^2$, ratio $(2 + \sqrt{5}) : 2$
Examiner tip: In a 'show that', every step must be written — state $h = r$ and $l = \sqrt{5}r$ explicitly before forming the ratio.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session