What is R12: Similarity and scale factors?

Similar shapes have equal corresponding angles and proportional corresponding lengths. If the linear scale factor from one shape to another is k, the length ratio is k, the area ratio is k^2, and the volume ratio is k^3.

How many marks are R12 questions worth?

In verified past-paper appearances, R12 questions on Edexcel GCSE Maths 1MA1 have been worth 3–4 marks.

Which past exam papers has R12 appeared in?

R12 has appeared in 10 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

R12

Compare lengths, areas and volumes using ratio notation; make links to similarity (including trigonometric ratios) and scale factors

Similarity and scale factors

Worked answers, methods and verified real exam appearances for R12 on Edexcel GCSE Maths 1MA1.

Explanation

Similar shapes have equal corresponding angles and proportional corresponding lengths. If the linear scale factor from one shape to another is $k$ , the length ratio is $k$ , the area ratio is $k^2$ , and the volume ratio is $k^3$ .
Match corresponding measurements and keep the comparison order consistent.
To recover a length factor from an area ratio take a square root; from a volume ratio take a cube root.
Trigonometric ratios remain constant in similar right-angled triangles.
Examiners expect you to identify whether the measurements are lengths, areas or volumes before applying the scale factor.

Corresponding lengths in similar shapes scale by

k

, so their areas scale by

k^2

Worked example

Two similar solids have smaller-to-larger volume ratio $125:216$ . The smaller surface area is $275\,\text{cm}^2$ . Find the larger surface area.

1. $125:216=5^3:6^3$ , so the length ratio is $5:6$ .
2.The surface-area ratio is $5^2:6^2=25:36$ .
3.Larger area $=275\times\dfrac{36}{25}=396\,\text{cm}^2$ .

Answer: $396\,\text{cm}^2$ .

Common mistakes

Don't use the linear scale factor directly for an area or volume.
Don't pair non-corresponding sides when forming the scale factor.
Don't take a square root when recovering a length factor from a volume ratio.

Exam tip

Annotate the scale factor as $k$ , $k^2$ or $k^3$ before calculating.

Worked practice

Tier 1 · Easy

Two similar shapes have corresponding sides of

6\,\text{cm}

and

15\,\text{cm}

. Write the smaller-to-larger length ratio in simplest form.

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$2:5$	1	$6:15$ simplifies by dividing both parts by $3$ , giving $2:5$ .

Tier 2 · Standard

The corresponding length ratio of two similar tiles is

3:7

. The smaller tile has area

54\,\text{cm}^2

. Find the area of the larger tile.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$294\,\text{cm}^2$	3	The area ratio is $3^2:7^2=9:49$ . Therefore the larger area is $54\times\frac{49}{9}=294\,\text{cm}^2$ .

Tier 3 · Hard

Two similar solids have smaller-to-larger volume ratio

125:216

. The smaller solid has surface area

275\,\text{cm}^2

. Work out the larger surface area.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$396\,\text{cm}^2$	4	Since $125:216=5^3:6^3$ , the length ratio is $5:6$ . The surface-area ratio is therefore $25:36$ . The larger area is $275\times\frac{36}{25}=396\,\text{cm}^2$ .

Tier 1 · Easy

Two similar posters have lengths in the ratio

2 : 5

. Write down the ratio of their areas.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$4 : 25$	1	For similar shapes, areas are in the ratio of the squares of corresponding lengths: $2^2 : 5^2 = 4 : 25$ .

Tier 2 · Standard

Two similar right-angled triangles have the same acute angle. In the smaller triangle, the side opposite this angle is

4\,\text{cm}

and the hypotenuse is

5\,\text{cm}

. Work out the opposite side in the larger triangle when its hypotenuse is

17.5\,\text{cm}

(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	For the same acute angle, the ratio $\dfrac{\text{opposite}}{\text{hypotenuse}}$ is constant. Therefore the larger opposite side is $17.5\times\dfrac{4}{5}=14\,\text{cm}$ .

Tier 3 · Hard

A rectangular photograph measures

8\,\text{cm}

12\,\text{cm}

and is enlarged so that its perimeter is

75\,\text{cm}

. Work out the area of the enlarged photograph.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$337.5\,\text{cm}^2$	4	The original perimeter is $2(8+12)=40\,\text{cm}$ , so the length scale factor is $75\div40=1.875$ . The enlarged dimensions are $8\times1.875=15\,\text{cm}$ and $12\times1.875=22.5\,\text{cm}$ . Its area is $15\times22.5=337.5\,\text{cm}^2$ .

Tier 2 · Standard

Two similar shapes have areas

96\,\text{cm}^2

and

150\,\text{cm}^2

. A side of the smaller shape is

12\,\text{cm}

. Work out the length of the corresponding side of the larger shape.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$15\,\text{cm}$	3	The area ratio is $96 : 150=16 : 25$ . Therefore the corresponding length ratio is $4 : 5$ . The larger side is $12\times\dfrac{5}{4}=15\,\text{cm}$ .

Tier 3 · Hard

Two similar rectangles have corresponding lengths in the ratio

3 : 5

. The two corresponding longer sides differ by

14\,\text{cm}

. The smaller rectangle has area

189\,\text{cm}^2

. Work out the area of the larger rectangle.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$525\,\text{cm}^2$	4	The difference of $5-3=2$ ratio parts is $14\,\text{cm}$ , so the corresponding longer sides are $21\,\text{cm}$ and $35\,\text{cm}$ . The area scale factor is $\left(\dfrac{5}{3}\right)^2=\dfrac{25}{9}$ . The larger area is $189\times\dfrac{25}{9}=525\,\text{cm}^2$ .

Tier 3 · Hard

Two similar solids have smaller-to-larger surface-area ratio

81 : 144

. The larger solid has volume

512\,\text{cm}^3

. Work out the volume of the smaller solid.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$216\,\text{cm}^3$	4	The surface-area ratio $81:144$ simplifies to $9:16=3^2:4^2$ , so the length ratio is $3:4$ . The volume ratio is $3^3:4^3=27:64$ . The smaller volume is $512\times\dfrac{27}{64}=216\,\text{cm}^3$ .

Q10

Tier 3 · Hard

Two similar containers have volumes in the ratio

64 : 125

. Their corresponding heights differ by

9\,\text{cm}

. Work out the sum of their heights.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$81\,\text{cm}$	4	Since $64:125=4^3:5^3$ , the corresponding height ratio is $4:5$ . The difference of one ratio part is $9\,\text{cm}$ , so the heights are $36\,\text{cm}$ and $45\,\text{cm}$ . Their sum is $81\,\text{cm}$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2021-11	2H	Q19	3	Allowed	Higher	QP MS
2019-11	1H	Q17	4	Non-calculator	Higher	QP MS
2024-06	2H	Q17	4	Allowed	Higher	QP MS
2019-06	2H	Q9	4	Allowed	Higher	QP MS
2019-06	1F	Q20	3	Non-calculator	Foundation	QP MS
2023-11	1H	Q13	3	Non-calculator	Higher	QP MS
2024-11	1H	Q12	4	Non-calculator	Higher	QP MS
2023-06	2H	Q23	3	Allowed	Higher	QP MS
2023-11	3H	Q11	3	Allowed	Higher	QP MS
2024-06	1H	Q19	3	Non-calculator	Higher	QP MS

Other points in R Ratio, proportion and rates of change

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