What is G19: Similar shapes?

Congruent figures have the same size and shape. Similar figures have equal corresponding angles and corresponding lengths in one constant ratio.

How many marks are G19 questions worth?

In verified past-paper appearances, G19 questions on Edexcel GCSE Maths 1MA1 have been worth 2–5 marks.

Which past exam papers has G19 appeared in?

G19 has appeared in 17 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

G19

Apply the concepts of congruence and similarity, including the relationships between lengths, areas and volumes in similar figures

Similar shapes

Worked answers, methods and verified real exam appearances for G19 on Edexcel GCSE Maths 1MA1.

Explanation

Congruent figures have the same size and shape. Similar figures have equal corresponding angles and corresponding lengths in one constant ratio.
Match sides by their positions opposite equal angles, then calculate a scale factor as image length divided by original length.
Multiply every original length by the same factor, or divide to reverse the enlargement.
Higher tier: if the length scale factor is $k$ , the area scale factor is $k^2$ and the volume scale factor is $k^3$ .
Examiners expect corresponding quantities to be paired consistently; a ratio written in the wrong direction gives every later value incorrectly.

Similar triangles keep equal angles while all corresponding lengths share one scale factor.

Worked example

Two similar triangles have corresponding sides $8$ cm and $12$ cm. Another side of the smaller triangle is $14$ cm. Find the corresponding larger side.

1.Scale factor from smaller to larger $=12\div8=1.5$ .
2.Multiply the corresponding smaller length: $14\times1.5=21$ .
3.Attach centimetres because a length has been scaled.

Answer: $21$ cm.

Common mistakes

Don't pair sides that do not lie opposite corresponding equal angles.
Don't use $k$ instead of $k^2$ or $k^3$ for a related area or volume (Higher tier).

Exam tip

Write the scale-factor direction in words, such as 'small to large', before multiplying or dividing.

Worked practice

Tier 1 · Easy

Triangle

ABC

has

AB=5\text{ cm}

AC=7\text{ cm}

and

\angle BAC=42^\circ

. Triangle

PQR

has

PQ=5\text{ cm}

PR=7\text{ cm}

and

\angle QPR=42^\circ

. State why the triangles are congruent.

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	They are congruent by SAS.	1	The triangles have two pairs of equal corresponding sides, $AB=PQ$ and $AC=PR$ . The equal $42^\circ$ angle is included between those sides in each triangle, so the triangles are congruent by SAS.

Tier 2 · Standard

Two similar shapes have corresponding lengths in the ratio smaller:larger

=3:5

. A side on the smaller shape is

12\text{ cm}

. Another side on the larger shape is

35\text{ cm}

. Work out the corresponding missing lengths.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$12\text{ cm}$ corresponds to $20\text{ cm}$ $35\text{ cm}$ corresponds to $21\text{ cm}$	3	The scale factor from smaller to larger is $5/3$ . Therefore $12\times\frac53=20\text{ cm}$ . To reverse the enlargement, multiply by $3/5$ , giving $35\times\frac35=21\text{ cm}$ .

Tier 3 · Hard

Triangle

ABC

has side lengths

8\text{ cm}

11\text{ cm}

and

13\text{ cm}

. Similar triangle

DEF

has longest side

19.5\text{ cm}

. Work out the other two side lengths of triangle

DEF

and its perimeter.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Other sides $=12\text{ cm}$ and $16.5\text{ cm}$ Perimeter $=48\text{ cm}$	4	The longest sides correspond, so the scale factor is $19.5/13=1.5$ . The other sides are $8\times1.5=12\text{ cm}$ and $11\times1.5=16.5\text{ cm}$ . The perimeter is $12+16.5+19.5=48\text{ cm}$ .

Tier 1 · Easy

Two shapes are similar. The length scale factor from the smaller shape to the larger shape is

4

. A side on the smaller shape is

3.5\text{ cm}

. Work out the corresponding side on the larger shape.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$14\text{ cm}$	1	Multiply the smaller length by the scale factor: $3.5\times4=14\text{ cm}$ .

Tier 2 · Standard

Two polygons are similar. Their perimeters are

24\text{ cm}

and

36\text{ cm}

. A side on the smaller polygon is

10\text{ cm}

. Work out the corresponding side on the larger polygon.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$15\text{ cm}$	2	The length scale factor is the perimeter ratio, $36\div24=1.5$ . The corresponding larger side is $10\times1.5=15\text{ cm}$ .

Tier 3 · Hard

In quadrilateral

PQRS

PQ=PS

and

RQ=RS

. The diagonal

PR

is drawn. Prove that triangle

PQR

is congruent to triangle

PSR

. Hence show that

\angle QPR=\angle SPR

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$PQ=PS$ , $RQ=RS$ and $PR$ is common, so the triangles are congruent by SSS; therefore $\angle QPR=\angle SPR$ .	3	$PQ=PS$ and $RQ=RS$ are given. Also, $PR$ is the same side in both triangles. The three corresponding sides are equal, so triangle $PQR$ is congruent to triangle $PSR$ by SSS. Corresponding angles in congruent triangles are equal, hence $\angle QPR=\angle SPR$ .

Tier 2 · Standard

Two quadrilaterals are similar. A side of length

8\text{ cm}

on the smaller quadrilateral corresponds to a side of length

12\text{ cm}

on the larger quadrilateral. The difference between their perimeters is

26\text{ cm}

. Work out both perimeters.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	Smaller perimeter $=52\text{ cm}$ Larger perimeter $=78\text{ cm}$	3	The length scale factor is $12\div8=1.5$ , so the larger perimeter is $1.5$ times the smaller perimeter. If the smaller perimeter is $p$ , then $1.5p-p=26$ , so $0.5p=26$ and $p=52$ . The larger perimeter is $1.5\times52=78\text{ cm}$ .

Tier 3 · Hard

Higher only: Solid

P

and solid

Q

are similar. For each solid, its volume is divided by its surface area. The result is

2.4\text{ cm}

for solid

P

and

4\text{ cm}

for solid

Q

. Work out the ratio of the surface area of

P

to the surface area of

Q

and the ratio of the volume of

P

to the volume of

Q

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	Surface area $P : Q=9 : 25$ Volume $P : Q=27 : 125$	4	For similar solids, volume scales with $k^3$ and surface area with $k^2$ , so volume divided by surface area scales with $k$ . Hence the length ratio is $P : Q=2.4 : 4=3 : 5$ . Squaring gives the surface-area ratio $9 : 25$ , and cubing gives the volume ratio $27 : 125$ .

Tier 3 · Hard

Convex quadrilateral

ABCD

has

AB

parallel to

CD

and

AB=CD

. The diagonal

AC

is drawn, with

B

and

D

on opposite sides of

AC

. Prove that triangles

BAC

and

DCA

are congruent. Hence prove that

ABCD

is a parallelogram.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$\angle BAC=\angle DCA$ , $AB=CD$ and $AC$ is common, so the triangles are congruent by SAS; hence $\angle BCA=\angle DAC$ , so $BC$ is parallel to $AD$ and $ABCD$ is a parallelogram.	4	Because $AB$ is parallel to $CD$ , $\angle BAC=\angle DCA$ as alternate angles. Also $AB=CD$ and $AC$ is common to both triangles. The triangles are therefore congruent by SAS. Corresponding angles give $\angle BCA=\angle DAC$ , so $BC$ is parallel to $AD$ . Both pairs of opposite sides are parallel, hence $ABCD$ is a parallelogram.

Q10

Tier 3 · Hard

Higher only: A solid model is enlarged to make a mathematically similar solid. The volume increases by

237.5\%

. Work out the percentage increase in the surface area.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$125\%$	4	An increase of $237.5\%$ gives a volume scale factor of $3.375=\frac{27}{8}=\left(\frac32\right)^3$ . The length scale factor is therefore $\frac32$ . The surface-area scale factor is $\left(\frac32\right)^2=\frac94=2.25$ , so the surface area is $225\%$ of the original. This is an increase of $125\%$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-06	2F	Q27	2	Allowed	Foundation	QP MS
2021-11	2H	Q19	3	Allowed	Higher	QP MS
2022-11	3H	Q5	4	Allowed	Higher	QP MS
2023-11	3H	Q19	4	Allowed	Higher	QP MS
2022-11	2H	Q17	4	Allowed	Higher	QP MS
2019-11	1H	Q17	4	Non-calculator	Higher	QP MS
2024-06	2H	Q17	4	Allowed	Higher	QP MS
2022-11	3F	Q25	4	Allowed	Foundation	QP MS
2023-06	3H	Q18	5	Allowed	Higher	QP MS
2023-11	1H	Q13	3	Non-calculator	Higher	QP MS
2024-11	1H	Q12	4	Non-calculator	Higher	QP MS
2024-11	3H	Q13	3	Allowed	Higher	QP MS
2019-11	1F	Q29	4	Non-calculator	Foundation	QP MS
2021-11	3H	Q10	4	Allowed	Higher	QP MS
2023-06	2H	Q23	3	Allowed	Higher	QP MS
2023-11	3H	Q15	2	Allowed	Higher	QP MS
2024-06	1H	Q19	3	Non-calculator	Higher	QP MS

Other points in G Geometry and measures

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