What is G17: Circle and 3D mensuration?

Circle facts: C=2π r=π d and A=π r^2, with the radius half the diameter; using π r or π r^2 for a circumference scores nothing. A composite area is a sum of rectangles, triangles, trapezia and parts of circles, with a cut-out subtracted; show each area separately because each earns a process mark, and keep unrounded values until the end.

How many marks are G17 questions worth?

In verified past-paper appearances, G17 questions on Edexcel GCSE Maths 1MA1 have been worth 2–6 marks.

Which past exam papers has G17 appeared in?

G17 has appeared in 31 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

G17

Know circumference of a circle = 2πr = πd and area = πr²; calculate perimeters of 2D shapes incl. circles, areas of circles and composite shapes, surface area and volume of spheres, pyramids, cones

Circle and 3D mensuration

Worked answers, methods and verified real exam appearances for G17 on Edexcel GCSE Maths 1MA1.

Definition

Specification G17: know the formulae circumference of a circle $=2\pi r=\pi d$ and area of a circle $=\pi r^2$ ; calculate perimeters of 2D shapes, including circles; areas of circles and composite shapes; surface area and volume of spheres, pyramids, cones and composite solids.

Explanation

Circle facts: $C=2\pi r=\pi d$ and $A=\pi r^2$ , with the radius half the diameter; using $\pi r$ or $\pi r^2$ for a circumference scores nothing. A composite area is a sum of rectangles, triangles, trapezia and parts of circles, with a cut-out subtracted; show each area separately because each earns a process mark, and keep unrounded values until the end.
Perimeter counts only the outside edge, so a semicircular end contributes $\pi r$ plus nothing for the diameter joined to the rest.
Volumes: cone or pyramid $\frac13\times\text{base area}\times\text{height}$ , sphere $\frac43\pi r^3$ , with the formulae given on the paper.
Surface area of a cone is the curved part $\pi rl$ , $l$ the slant height, plus the base $\pi r^2$ ; a hemisphere's is $2\pi r^2$ plus its circular base.
Exact answers stay as multiples of $\pi$ .

A cone's perpendicular height

h

and slant height

l

have different roles.

Worked example

A solid cone has base radius

3

cm, perpendicular height

4

cm and slant height

5

cm. Curved surface area of a cone

=\pi rl

. Work out the total surface area of the cone. Give your answer in terms of

\pi

(3)

(Total for Question 1 is 3 marks)

Working

1.Curved surface area $=\pi rl=\pi\times3\times5=15\pi$ cm $^2$ .
2.Base area $=\pi r^2=\pi\times3^2=9\pi$ cm $^2$ .
3.Total surface area $=15\pi+9\pi=24\pi$ cm $^2$ .

Answer: $24\pi$ cm $^2$

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for a process to find the curved surface area, eg $\pi\times3\times5$ (= $15\pi$ )	P1
	for adding the circular base, eg “ $15\pi$ ” $+\pi\times3^2$	P1	Slant height, not perpendicular height, in $\pi rl$
	for $24\pi$	A1	cao

Common mistakes

Don't use the slant height $l$ in the cone-volume formula instead of the perpendicular height $h$ .
Don't add the circular base twice or omit it when total surface area is requested.

Mark-scheme wording to use

“for correctly finding the area of at least three sections, eg 3 of 11 × 7 (= 77), or 9 × 7 (= 63), or 1/2 × 11 × 9 (= 49.5), or 1/4 × π × 7^2 (= 38.4845..)” — Pearson 1MA1, November 2022 Paper 2H, Q4
“P1 starts process by using 250/3 π and 1/2 × 4/3 πr^3 to find radius P1 starts process using ½ curved surface area eg (4 × π × “5”^2 ) ÷ 2 P1 complete process shown eg (4 × π × “5”^2 ) ÷ 2 + ( π × “5”^2 ) A1 for 75π” — Pearson 1MA1, Specimen 1 Paper 1H, Q18
“P1 Process to find volume of cone or hemisphere P1 Process to total volume of solid P1 Process to find volume of cylinder P1 Equates 2 volumes A1 Reaches h = x/2” — Pearson 1MA1, Specimen 2 Paper 1H, Q11

What the examiner wants

The right formula at the first step. The June 2024 Paper 1H report says many with some knowledge of sectors went wrong immediately by using $\pi r^2$ or $\pi r$ instead of $2\pi r$ and gained no marks, while those who made a correct start usually finished; the Paper 3H report records volume-of-cylinder attempts that found the surface area or multiplied radius by height instead.

Exam tip

Underline 'curved' or 'total' surface area, then list the exposed faces before calculating.

Worked practice

Tier 1 · Easy

A circle has diameter

13\text{ cm}

. Write its circumference in terms of

\pi

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$13\pi\text{ cm}$	1	Use $C=\pi d$ . With $d=13\text{ cm}$ , the circumference is $13\pi\text{ cm}$ .

Tier 2 · Standard

A semicircle of diameter

10\text{ cm}

is attached to one of the shorter sides of an

18\text{ cm}

10\text{ cm}

rectangle. The shared diameter is inside the shape. Work out the area and perimeter of the composite shape in terms of

\pi

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	Area $=180+\frac{25\pi}{2}\text{ cm}^2$ Perimeter $=46+5\pi\text{ cm}$	4	The rectangle area is $18\times10=180\text{ cm}^2$ . The semicircle has radius $5\text{ cm}$ , so its area is $\frac12\pi(5^2)=\frac{25\pi}{2}\text{ cm}^2$ . For the perimeter, include the two $18\text{ cm}$ sides, the unshared $10\text{ cm}$ side and the semicircular arc $\pi r=5\pi\text{ cm}$ . This gives area $180+\frac{25\pi}{2}\text{ cm}^2$ and perimeter $46+5\pi\text{ cm}$ .

Tier 3 · Hard

A solid cone has radius

6\text{ cm}

, perpendicular height

8\text{ cm}

and slant height

10\text{ cm}

. Work out its total surface area, including the circular base, and its volume. Give both answers in terms of

\pi

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Total surface area $=96\pi\text{ cm}^2$ Volume $=96\pi\text{ cm}^3$	5	The curved surface area is $\pi rl=\pi(6)(10)=60\pi\text{ cm}^2$ and the base area is $\pi r^2=36\pi\text{ cm}^2$ , giving $96\pi\text{ cm}^2$ in total. The volume is $\frac13\pi r^2h=\frac13\pi(6^2)(8)=96\pi\text{ cm}^3$ .

Tier 1 · Easy

A circle has radius

6\text{ cm}

. Work out its area in terms of

\pi

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$36\pi\text{ cm}^2$	1	Use $A=\pi r^2$ . Therefore $A=\pi(6^2)=36\pi\text{ cm}^2$ .

Tier 2 · Standard

A square has side length

11\text{ cm}

. A circular hole of radius

4\text{ cm}

is cut from the square. Work out the area that remains in terms of

\pi

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$121-16\pi\text{ cm}^2$	2	The square has area $11^2=121\text{ cm}^2$ . The circular hole has area $\pi(4^2)=16\pi\text{ cm}^2$ . Therefore the remaining area is $121-16\pi\text{ cm}^2$ .

Tier 3 · Hard

The total surface area of a solid hemisphere, including its circular base, is

147\pi\text{ cm}^2

. Work out the volume of the hemisphere in terms of

\pi

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$\frac{686\pi}{3}\text{ cm}^3$	4	The curved surface area of a hemisphere is $2\pi r^2$ and its circular base has area $\pi r^2$ , so $3\pi r^2=147\pi$ . Hence $r^2=49$ and $r=7\text{ cm}$ . The volume is half the volume of a sphere, so $V=\frac23\pi r^3=\frac23\pi(7^3)=\frac{686\pi}{3}\text{ cm}^3$ .

Tier 2 · Standard

A circle has radius

8\text{ cm}

. A square has the same perimeter as the circle. Work out the area of the square in terms of

\pi

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$16\pi^2\text{ cm}^2$	3	The circle has perimeter $2\pi(8)=16\pi\text{ cm}$ . Each side of the square is $16\pi\div4=4\pi\text{ cm}$ . Therefore the square has area $(4\pi)^2=16\pi^2\text{ cm}^2$ .

Tier 3 · Hard

A circular pond has radius

7\text{ m}

. A path surrounds the pond. The outer edge of the path is a circle with circumference

18\pi\text{ m}

. Work out the area of the path in terms of

\pi

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$32\pi\text{ m}^2$	4	For the outer circle, $2\pi R=18\pi$ , so $R=9\text{ m}$ . The outer area is $81\pi\text{ m}^2$ and the pond area is $49\pi\text{ m}^2$ . Therefore the path area is $81\pi-49\pi=32\pi\text{ m}^2$ .

Tier 3 · Hard

Higher only: A solid hemisphere of radius

5\text{ cm}

is joined along its circular face to the circular base of a cone. The cone also has radius

5\text{ cm}

and has slant height

13\text{ cm}

. The joined circular faces are inside the solid. Work out the exterior surface area of the solid in terms of

\pi

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$115\pi\text{ cm}^2$	4	Only the curved surfaces are exposed. The curved surface area of the hemisphere is $2\pi r^2=2\pi(5^2)=50\pi\text{ cm}^2$ . The curved surface area of the cone is $\pi rl=\pi(5)(13)=65\pi\text{ cm}^2$ . Therefore the exterior surface area is $50\pi+65\pi=115\pi\text{ cm}^2$ .

Q10

Tier 3 · Hard

Higher only: A pyramid has a trapezium as its base. The parallel sides of the trapezium are

8\text{ cm}

and

14\text{ cm}

, and the perpendicular distance between them is

6\text{ cm}

. The volume of the pyramid is

330\text{ cm}^3

. Work out the perpendicular height of the pyramid.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$15\text{ cm}$	4	The base area is $\frac12(8+14)\times6=66\text{ cm}^2$ . If the perpendicular height is $h\text{ cm}$ , then $330=\frac13\times66\times h=22h$ . Therefore $h=15\text{ cm}$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-06	1F	Q28	4	Non-calculator	Foundation	QP MS
2023-11	3H	Q19	4	Allowed	Higher	QP MS
2021-11	2F	Q25	4	Allowed	Foundation	QP MS
2019-06	3H	Q6	5	Allowed	Higher	QP MS
2023-06	3H	Q18	5	Allowed	Higher	QP MS
2024-11	3H	Q19	4	Allowed	Higher	QP MS
2023-06	3F	Q6	2	Allowed	Foundation	QP MS
2023-06	1H	Q16	4	Non-calculator	Higher	QP MS
2022-11	1H	Q24	6	Non-calculator	Higher	QP MS
2023-11	1H	Q10	3	Non-calculator	Higher	QP MS
2022-06	1H	Q13	3	Non-calculator	Higher	QP MS
2019-11	3F	Q30	3	Allowed	Foundation	QP MS
2024-06	3H	Q21	4	Allowed	Higher	QP MS
2019-11	3H	Q23	5	Allowed	Higher	QP MS
2022-06	2H	Q17	5	Allowed	Higher	QP MS
2023-11	3F	Q29	4	Allowed	Foundation	QP MS
2022-11	2F	Q21	5	Allowed	Foundation	QP MS
2023-06	2H	Q12	4	Allowed	Higher	QP MS
2019-06	3F	Q29	5	Allowed	Foundation	QP MS
2021-11	3F	Q27	4	Allowed	Foundation	QP MS
2022-11	2H	Q4	5	Allowed	Higher	QP MS
2024-06	1F	Q12	4	Non-calculator	Foundation	QP MS
2019-06	1H	Q7	4	Non-calculator	Higher	QP MS
2024-11	2F	Q10	3	Allowed	Foundation	QP MS
2019-11	3F	Q14	2	Allowed	Foundation	QP MS
2019-11	2H	Q19	2	Allowed	Higher	QP MS
2019-06	1H	Q15	4	Non-calculator	Higher	QP MS
2019-06	3F	Q16	3	Allowed	Foundation	QP MS
2021-11	1H	Q7	4	Non-calculator	Higher	QP MS
2024-06	3H	Q9	3	Allowed	Higher	QP MS
2024-11	3H	Q11	3	Allowed	Higher	QP MS

Other points in G Geometry and measures

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