What is G16: Area and volume formulae?

Area formulae: triangle 1/2bh, parallelogram bh, trapezium 1/2(a+b)h, where h is the perpendicular height between the base and the parallel side, never a sloping edge. A right prism has the same cross-section all the way along, so V=area of cross-section; a cylinder is the case with a circular cross-section, V=π r^2h.

How many marks are G16 questions worth?

In verified past-paper appearances, G16 questions on Edexcel GCSE Maths 1MA1 have been worth 2–5 marks.

Which past exam papers has G16 appeared in?

G16 has appeared in 62 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

G16

Know and apply formulae to calculate: area of triangles, parallelograms, trapezia; volume of cuboids and other right prisms (including cylinders)

Area and volume formulae

Worked answers, methods and verified real exam appearances for G16 on Edexcel GCSE Maths 1MA1.

Definition

Specification G16: know and apply formulae to calculate the area of triangles, parallelograms and trapezia, and the volume of cuboids and other right prisms, including cylinders.

Explanation

Area formulae: triangle $\frac12bh$ , parallelogram $bh$ , trapezium $\frac12(a+b)h$ , where $h$ is the perpendicular height between the base and the parallel side, never a sloping edge. A right prism has the same cross-section all the way along, so $V=\text{area of cross-section}\times\text{length}$ ; a cylinder is the case with a circular cross-section, $V=\pi r^2h$ .
Identify and draw the cross-section first, find its area in square units, then multiply by the length to get cubic units.
Compound shapes are split into rectangles and triangles, or a larger shape with a piece subtracted; each area is a process mark.
Convert to consistent units before dividing one area by another, for example tiles into a wall.
Units are often a separate mark, so write cm $^2$ or cm $^3$ with the number.

A trapezium uses the perpendicular height between its parallel sides.

Worked example

The diagram shows a prism

12

cm long. The cross-section of the prism is a trapezium with parallel sides

8

cm and

5

cm and perpendicular height

6

cm. Work out the volume of the prism.

(3)

(Total for Question 1 is 3 marks)

Working

1.Area of the trapezium cross-section: $\frac12(8+5)\times6=\frac12\times13\times6=39$ cm $^2$ .
2.Volume of a prism is cross-sectional area multiplied by length: $39\times12$ .
3. $39\times12=468$ , in cubic centimetres.

Answer: $468$ cm $^3$

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for a method to find the area of the trapezium, eg $\frac12(8+5)\times6$ (= 39)	M1	oe
	(dep) for a complete method to find the volume, eg “39” $\times12$	M1
	for $468$	A1	cao

Common mistakes

Don't use the sloping side of a triangle or trapezium as $h$ without a perpendicular mark.
Don't stop after finding the cross-sectional area and fail to multiply by the prism length.

Mark-scheme wording to use

“for a complete method to find the volume eg 4 x 10 x 15 … (indep) cm3” — Pearson 1MA1, June 2020 Paper 2F, Q11
“for a process to find the area of one shape, eg 10 × 8 (= 80) … for a complete process to find the total area, eg “80” – “12” (= 68)” — Pearson 1MA1, June 2024 Paper 1H, Q3(a)
“M1 a full method to find the area of the trapezium M1 a full method to calculate both areas in consistent units M1 for the area of the trapezium ÷ area of a tile (with consistent units) M1 (dep on previous M) for complete method to find the number of packs required” — Pearson 1MA1, Specimen 1 Paper 3H, Q5(a)

What the examiner wants

The cross-section identified before any multiplying. The June 2024 Paper 3H report says many could not find the volume of a cylinder, with errors ranging from finding the surface area to multiplying the radius by the height, while the Paper 1H report notes that splitting the compound floor into two rectangles and showing each area earned the process marks even when the arithmetic slipped.

Exam tip

For a prism, write 'cross-sectional area × length' before substituting so both stages of the method are clear.

Worked practice

Tier 1 · Easy

A trapezium has parallel sides of lengths

7\text{ cm}

and

13\text{ cm}

and perpendicular height

6\text{ cm}

. Work out its area.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$60\text{ cm}^2$	2	Use $A=\frac12(a+b)h$ . Then $A=\frac12(7+13)\times6=\frac12\times20\times6=60\text{ cm}^2$ .

Tier 2 · Standard

A cylinder has radius

4\text{ cm}

and height

9\text{ cm}

. Work out its volume in terms of

\pi

(2)

(Total for Question 2 is 2 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$144\pi\text{ cm}^3$	2	The circular cross-section has area $\pi r^2=\pi(4^2)=16\pi\text{ cm}^2$ . Multiply by the height: $V=16\pi\times9=144\pi\text{ cm}^3$ .

Tier 3 · Hard

A right triangular prism is

10\text{ cm}

long. Its triangular cross-section has base

x\text{ cm}

and perpendicular height

(x+2)\text{ cm}

. The volume is

600\text{ cm}^3

. Find

x

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$x=10$	4	The cross-sectional area is $\frac12x(x+2)$ . Hence $10\times\frac12x(x+2)=600$ , so $x(x+2)=120$ . This gives $x^2+2x-120=0$ , which factorises to $(x+12)(x-10)=0$ . A length must be positive, so $x=10$ .

Tier 1 · Easy

A parallelogram has a base of

12\text{ cm}

and a perpendicular height of

7\text{ cm}

. Work out the area of the parallelogram.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$84\text{ cm}^2$	1	The area of a parallelogram is base multiplied by perpendicular height, so $12\times7=84\text{ cm}^2$ .

Tier 2 · Standard

A trapezium has area

76\text{ cm}^2

, perpendicular height

8\text{ cm}

and one parallel side of length

5\text{ cm}

. Work out the length of the other parallel side.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$14\text{ cm}$	3	Let the other parallel side be $x\text{ cm}$ . Then $76=\frac12(5+x)\times8$ , so $76=4(5+x)$ . Hence $19=5+x$ and $x=14\text{ cm}$ .

Tier 3 · Hard

A right prism has a rectangular cross-section measuring

12\text{ cm}

8\text{ cm}

. A rectangular hole measuring

6\text{ cm}

3\text{ cm}

runs through the full

20\text{ cm}

length of the prism, with its

6\text{ cm}

edge parallel to the

12\text{ cm}

edge of the cross-section. Work out the volume of material in the prism.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$1560\text{ cm}^3$	4	The material has cross-sectional area $12\times8-6\times3=96-18=78\text{ cm}^2$ . The volume is cross-sectional area multiplied by length, so $78\times20=1560\text{ cm}^3$ .

Tier 2 · Standard

A cylinder has radius

7\text{ cm}

and volume

588\pi\text{ cm}^3

. Work out the height of the cylinder.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$12\text{ cm}$	3	Using $V=\pi r^2h$ gives $588\pi=\pi(7^2)h=49\pi h$ . Therefore $h=588\pi\div49\pi=12\text{ cm}$ .

Tier 3 · Hard

The length of a right prism is

25\text{ cm}

. Perpendicular to this length, its constant cross-section starts as a rectangle with a horizontal side of

14\text{ cm}

and a vertical side of

10\text{ cm}

. A right-angled triangular piece is removed from the top-right corner: its

6\text{ cm}

side lies along the top edge and its

4\text{ cm}

side lies along the right edge. Work out the volume of material in the prism.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$3200\text{ cm}^3$	4	The rectangular area is $14\times10=140\text{ cm}^2$ . The removed triangular area is $\frac12\times6\times4=12\text{ cm}^2$ , so the remaining cross-sectional area is $128\text{ cm}^2$ . The volume is $128\times25=3200\text{ cm}^3$ .

Tier 3 · Hard

Right prism

P

14\text{ cm}

long and has a triangular cross-section with base

12\text{ cm}

and perpendicular height

9\text{ cm}

. Right prism

Q

13.5\text{ cm}

long and has a parallelogram cross-section with base

7\text{ cm}

and perpendicular height

8\text{ cm}

. Show that the two prisms have the same volume.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	Both volumes are $756\text{ cm}^3$ .	4	The cross-sectional area of prism $P$ is $\frac12\times12\times9=54\text{ cm}^2$ , so its volume is $54\times14=756\text{ cm}^3$ . The cross-sectional area of prism $Q$ is $7\times8=56\text{ cm}^2$ , so its volume is $56\times13.5=756\text{ cm}^3$ . Therefore the volumes are equal.

Q10

Tier 3 · Hard

The radius of a cylinder is increased by

20\%

and its height is reduced by

20\%

. Sam says that these changes leave the volume unchanged. Is Sam correct? Work out the percentage change in the volume.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	No, the volume increases by $15.2\%$ .	4	Cylinder volume is proportional to $r^2h$ . The new volume factor is $1.2^2\times0.8=1.152$ . The new volume is therefore $115.2\%$ of the original, so it increases by $15.2\%$ . Sam is not correct.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-06	1F	Q28	4	Non-calculator	Foundation	QP MS
2022-06	3F	Q16	2	Allowed	Foundation	QP MS
2023-06	2F	Q24	4	Allowed	Foundation	QP MS
2023-11	3F	Q27	3	Allowed	Foundation	QP MS
2019-11	3H	Q13	4	Allowed	Higher	QP MS
2021-11	1H	Q14	3	Non-calculator	Higher	QP MS
2022-11	2H	Q17	4	Allowed	Higher	QP MS
2024-11	2H	Q12	5	Allowed	Higher	QP MS
2021-11	1H	Q6	3	Non-calculator	Higher	QP MS
2019-06	3H	Q6	5	Allowed	Higher	QP MS
2019-06	1F	Q18	4	Non-calculator	Foundation	QP MS
2023-11	2F	Q28	4	Allowed	Foundation	QP MS
2023-06	1F	Q22	4	Non-calculator	Foundation	QP MS
2022-06	3F	Q28	3	Allowed	Foundation	QP MS
2023-06	2H	Q5	4	Allowed	Higher	QP MS
2023-06	1H	Q3	4	Non-calculator	Higher	QP MS
2022-06	3H	Q7	3	Allowed	Higher	QP MS
2021-11	3F	Q19	4	Allowed	Foundation	QP MS
2023-06	2F	Q14	4	Allowed	Foundation	QP MS
2019-06	3H	Q14	4	Allowed	Higher	QP MS
2021-11	3H	Q7	4	Allowed	Higher	QP MS
2023-11	3F	Q18	5	Allowed	Foundation	QP MS
2023-06	1H	Q8	3	Non-calculator	Higher	QP MS
2022-06	1F	Q29	3	Non-calculator	Foundation	QP MS
2019-06	2F	Q15	4	Allowed	Foundation	QP MS
2022-06	1H	Q9	3	Non-calculator	Higher	QP MS
2024-11	2H	Q7	5	Allowed	Higher	QP MS
2024-11	3H	Q13	3	Allowed	Higher	QP MS
2022-06	1H	Q7	3	Non-calculator	Higher	QP MS
2023-06	3F	Q6	2	Allowed	Foundation	QP MS
2024-11	2F	Q28	3	Allowed	Foundation	QP MS
2024-06	1F	Q22	5	Non-calculator	Foundation	QP MS
2021-11	1H	Q18	5	Non-calculator	Higher	QP MS
2019-11	1H	Q14	4	Non-calculator	Higher	QP MS
2024-06	2H	Q8	3	Allowed	Higher	QP MS
2024-11	3F	Q13	4	Allowed	Foundation	QP MS
2022-06	2F	Q24	5	Allowed	Foundation	QP MS
2022-06	1F	Q22	3	Non-calculator	Foundation	QP MS
2024-11	1F	Q14	5	Non-calculator	Foundation	QP MS
2022-11	2F	Q21	5	Allowed	Foundation	QP MS
2021-11	1F	Q25	3	Non-calculator	Foundation	QP MS
2022-11	3H	Q4	4	Allowed	Higher	QP MS
2024-06	3F	Q28	4	Allowed	Foundation	QP MS
2023-06	1F	Q27	3	Non-calculator	Foundation	QP MS
2024-06	1H	Q21	5	Non-calculator	Higher	QP MS
2021-11	3F	Q27	4	Allowed	Foundation	QP MS
2022-11	2H	Q4	5	Allowed	Higher	QP MS
2022-06	2H	Q4	5	Allowed	Higher	QP MS
2024-06	1H	Q3	5	Non-calculator	Higher	QP MS
2019-06	2F	Q23	4	Allowed	Foundation	QP MS
2023-11	3F	Q19	3	Allowed	Foundation	QP MS
2022-11	3F	Q11	2	Allowed	Foundation	QP MS
2019-06	2H	Q4	4	Allowed	Higher	QP MS
2022-06	3H	Q18	5	Allowed	Higher	QP MS
2024-11	2F	Q26	5	Allowed	Foundation	QP MS
2019-11	2F	Q29	5	Allowed	Foundation	QP MS
2019-11	2H	Q9	5	Allowed	Higher	QP MS
2022-11	3F	Q24	4	Allowed	Foundation	QP MS
2023-11	2F	Q13	2	Allowed	Foundation	QP MS
2024-06	3H	Q7	4	Allowed	Higher	QP MS
2023-11	2H	Q7	4	Allowed	Higher	QP MS
2024-06	3H	Q9	3	Allowed	Higher	QP MS

Other points in G Geometry and measures

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