What is G4: Quadrilaterals and polygons?

Classify a plane shape using properties that must hold, not how the drawing looks. A parallelogram has two pairs of parallel opposite sides; its opposite sides and angles are equal, and its diagonals bisect each other.

How many marks are G4 questions worth?

In verified past-paper appearances, G4 questions on Edexcel GCSE Maths 1MA1 have been worth 2–4 marks.

Which past exam papers has G4 appeared in?

G4 has appeared in 14 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Derive and apply properties and definitions of special quadrilaterals (square, rectangle, parallelogram, trapezium, kite, rhombus), triangles and other plane figures using appropriate language

Quadrilaterals and polygons

Worked answers, methods and verified real exam appearances for G4 on Edexcel GCSE Maths 1MA1.

Explanation

Classify a plane shape using properties that must hold, not how the drawing looks. A parallelogram has two pairs of parallel opposite sides; its opposite sides and angles are equal, and its diagonals bisect each other.
A rectangle is a parallelogram with four right angles and equal diagonals. A rhombus has four equal sides and perpendicular diagonals, while a square is both a rectangle and a rhombus.
A kite has two pairs of adjacent equal sides.
An isosceles triangle has equal base angles.
Examiners expect a property to be connected to a conclusion, especially in 'explain' and proof questions.

A parallelogram and its diagonals, which bisect each other.

Worked example

The diagonals of quadrilateral $ABCD$ bisect each other and meet at right angles. State the most specific guaranteed quadrilateral and justify it.

1.Diagonals that bisect each other establish that $ABCD$ is a parallelogram.
2.A parallelogram whose diagonals are perpendicular is a rhombus.
3.No right angle or equal diagonals are given, so a square is not guaranteed.

Answer: $ABCD$ must be a rhombus, but need not be a square.

Common mistakes

Don't state that every rhombus has four right angles because the diagram resembles a square.
Don't use 'the diagonals are equal' as a property of every parallelogram instead of rectangles and squares.

Exam tip

For a classification question, give the most specific shape forced by the information and state the defining property.

Worked practice

Tier 1 · Easy

Name the quadrilateral that has exactly one pair of parallel sides.

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	Trapezium	1	A quadrilateral with exactly one pair of parallel sides is a trapezium.

Tier 2 · Standard

One interior angle of a rhombus is

68^\circ

. Work out the other three interior angles.

(2)

(Total for Question 2 is 2 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$112^\circ$ , $68^\circ$ , $112^\circ$	2	Opposite angles in a rhombus are equal and adjacent angles sum to $180^\circ$ . The opposite angle is $68^\circ$ , and each adjacent angle is $180^\circ-68^\circ=112^\circ$ .

Tier 3 · Hard

The diagonals of quadrilateral

WXYZ

bisect each other and are equal in length. Explain why

WXYZ

must be a rectangle, and why the information does not prove that it is a square.

(3)

(Total for Question 3 is 3 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Diagonals that bisect each other make the quadrilateral a parallelogram; equal diagonals then make it a rectangle. A square would additionally require four equal sides or perpendicular diagonals, which has not been given.	3	First use the converse parallelogram property: diagonals that bisect each other establish a parallelogram. In a parallelogram, equal diagonals establish a rectangle. Equal diagonals alone do not establish equal sides or perpendicular diagonals, so a non-square rectangle remains possible.

Tier 1 · Easy

Write down one property of the diagonals of a rectangle.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	They are equal in length (accept that they bisect each other).	1	A rectangle has diagonals of equal length. As a parallelogram, it also has diagonals that bisect each other, so either property is valid.

Tier 2 · Standard

A parallelogram has adjacent side lengths

7

cm and

11

cm. Work out its perimeter.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$36$ cm	2	Opposite sides of a parallelogram are equal, so the perimeter is $2\times7+2\times11=14+22=36$ cm.

Tier 3 · Hard

In kite

ABCD

AB=AD

and

CB=CD

. Angle

A

74^\circ

and angle

C

126^\circ

. Work out angles

B

and

D

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$\angle B=80^\circ$ and $\angle D=80^\circ$	3	The opposite angles between the unequal sides of a kite are equal, so let $B=D=x$ . The angles in a quadrilateral total $360^\circ$ , so $74+126+2x=360$ . Hence $2x=160$ and $x=80^\circ$ .

Tier 2 · Standard

Quadrilateral

ABCD

is a parallelogram. Also

AB=BC

and

\angle ABC=90^\circ

. Write down the most specific name of

ABCD

and explain which two properties force this conclusion.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$ABCD$ is a square; the right angle makes the parallelogram a rectangle, and the equal adjacent sides make it a rhombus.	3	A parallelogram with one right angle has four right angles, so it is a rectangle. A parallelogram with equal adjacent sides has all four sides equal, so it is a rhombus. A quadrilateral that is both a rectangle and a rhombus is a square.

Tier 3 · Hard

In parallelogram

ABCD

, diagonal

AC

bisects

\angle DAB

. Prove that

ABCD

is a rhombus. Include the geometric reason beside each equality.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$\angle DAC=\angle ACB=\angle CAB$ , so $AB=BC$ and $ABCD$ is a rhombus.	4	Since $AD\parallel BC$ , $\angle DAC=\angle ACB$ by alternate angles. Since $AC$ bisects $\angle DAB$ , $\angle DAC=\angle CAB$ . Therefore $\angle CAB=\angle ACB$ , so $AB=BC$ because equal angles in a triangle face equal sides. A parallelogram with equal adjacent sides has four equal sides, so $ABCD$ is a rhombus.

Tier 3 · Hard

In rhombus

ABCD

, diagonals

AC

and

BD

meet at

E

, and

\angle BAC=31^\circ

. Work out

\angle ABE

and all four interior angles of the rhombus. Give a reason for each stage.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$\angle ABE=59^\circ$ ; $\angle DAB=\angle BCD=62^\circ$ and $\angle ABC=\angle CDA=118^\circ$	4	The diagonals of a rhombus are perpendicular, so $\angle AEB=90^\circ$ . Angles in triangle $ABE$ total $180^\circ$ , giving $\angle ABE=180^\circ-90^\circ-31^\circ=59^\circ$ . Diagonal $AC$ bisects $\angle DAB$ , so $\angle DAB=2\times31^\circ=62^\circ$ . Opposite angles of a rhombus are equal, giving $\angle BCD=62^\circ$ , and adjacent angles in a parallelogram are supplementary, giving $\angle ABC=\angle CDA=118^\circ$ . Coordinates $A=(0,0)$ , $B=(1,0)$ , $D=(\cos62^\circ,\sin62^\circ)$ and $C=(1+\cos62^\circ,\sin62^\circ)$ verify existence, and the angle facts fix every requested value uniquely.

Q10

Tier 3 · Hard

The diagonals of convex quadrilateral

ABCD

meet at

E

at right angles. Points

A,E,C

are in order and points

B,E,D

are in order. Given

BE=DE=3

cm,

AE=4

cm and

CE=7

cm, write down the most precise name for

ABCD

and work out its perimeter. Explain why it is not a rhombus.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	Kite; perimeter $=10+2\sqrt{58}$ cm; it is not a rhombus because its adjacent side pairs have different lengths	4	The four small triangles at $E$ are right-angled. Pythagoras gives $AB=AD=\sqrt{4^2+3^2}=5$ cm and $BC=CD=\sqrt{7^2+3^2}=\sqrt{58}$ cm. Two distinct pairs of equal adjacent sides make $ABCD$ a kite. Its perimeter is $2(5)+2\sqrt{58}=10+2\sqrt{58}$ cm. Since $5\ne\sqrt{58}$ , not all four sides are equal, so it is not a rhombus. Coordinates $A=(0,4)$ , $E=(0,0)$ , $C=(0,-7)$ , $B=(-3,0)$ and $D=(3,0)$ verify the unique labelled configuration.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-06	3F	Q20	2	Allowed	Foundation	QP MS
2024-11	3F	Q7	2	Allowed	Foundation	QP MS
2021-11	1F	Q6	2	Non-calculator	Foundation	QP MS
2024-11	1H	Q4	3	Non-calculator	Higher	QP MS
2019-11	3H	Q8	4	Allowed	Higher	QP MS
2021-11	1F	Q16	4	Non-calculator	Foundation	QP MS
2019-11	3F	Q29	4	Allowed	Foundation	QP MS
2023-06	2F	Q7	3	Allowed	Foundation	QP MS
2024-06	2F	Q26	4	Allowed	Foundation	QP MS
2024-11	1F	Q21	3	Non-calculator	Foundation	QP MS
2023-11	2F	Q19	4	Allowed	Foundation	QP MS
2024-11	2H	Q9	3	Allowed	Higher	QP MS
2022-06	3F	Q6	2	Allowed	Foundation	QP MS
2023-06	3F	Q8	2	Allowed	Foundation	QP MS

Other points in G Geometry and measures

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