What is G16: Area and volume formulae?

Use A=1/2bh for a triangle, A=bh for a parallelogram and A=1/2(a+b)h for a trapezium, where h is perpendicular to the base or parallel sides.

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.

Explanation

Use $A=\frac12bh$ for a triangle, $A=bh$ for a parallelogram and $A=\frac12(a+b)h$ for a trapezium, where $h$ is perpendicular to the base or parallel sides.
A right prism has a constant cross-section, so $V=\text{cross-sectional area}\times\text{length}$ .
A cylinder is a right prism with circular cross-section, giving $V=\pi r^2h$ .
Mark the cross-section first, calculate its area with square units, then multiply by the prism length to obtain cubic units.
Examiners expect correct substitution and units; a sloping side is not a perpendicular height unless a right angle establishes it.

A trapezium uses the perpendicular height between its parallel sides.

Worked example

A triangular prism is $10$ cm long. Its triangular cross-section has base $6$ cm and perpendicular height $4$ cm. Find the volume.

1.Cross-sectional area $=\frac12\times6\times4=12\text{ cm}^2$ .
2.Multiply by prism length: $V=12\times10$ .
3.Use cubic units for volume.

Answer: $120\text{ cm}^3$ .

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.

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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