Geometric notation, polygons and symmetry - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 20 exam-style questions on Geometric notation, polygons and symmetry for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

20 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

A quadrilateral has exactly one line of symmetry.
Its diagonals cross at right angles.
It has no parallel sides.

Write down the mathematical name of the quadrilateral and give a reason why it cannot be a rhombus.

Model answer

It is a kite.
It cannot be a rhombus because a rhombus has two lines of symmetry (and two pairs of parallel sides), but this shape has only one line of symmetry and no parallel sides.

Explanation

A kite has two pairs of equal adjacent sides, one line of symmetry (along one diagonal) and diagonals that cross at right angles. A rhombus also has perpendicular diagonals, but it has two lines of symmetry, rotational symmetry of order 2 and opposite sides parallel.

Mark scheme

1 mark
B1 for kite
accept "arrowhead" or "delta" only if they also say it is a kite
1 mark
C1 for a correct reason, e.g. "a rhombus has two lines of symmetry" or "a rhombus has two pairs of parallel sides"
oe

Question 2

3 marks

ABCDEF is a hexagon.
The hexagon has rotational symmetry of order 2 about the point M(3, 2).

A is (1, 1), B is (2, 4) and C is (5, 4).

Work out the coordinates of D, E and F.

Model answer

Rotational symmetry of order 2 about M means a half turn about M maps the hexagon onto itself, so each vertex maps to the opposite vertex: A to D, B to E, C to F.
A half turn about M(3, 2) maps \((x, y)\) to \((6 - x, 4 - y)\).
D = (6 − 1, 4 − 1) = (5, 3)
E = (6 − 2, 4 − 4) = (4, 0)
F = (6 − 5, 4 − 4) = (1, 0)

Explanation

Under a half turn about M, each point moves to the point the same distance from M on the opposite side, so M is the midpoint of AD, BE and CF. For example, from A(1, 1) to M is 2 right and 1 up, so D is a further 2 right and 1 up from M. Drawing a quick sketch from the description helps you check the order of the vertices.

Mark scheme

1 mark
B1 for D(5, 3)
correct answer with no working scores full marks
1 mark
B1 for E(4, 0)
1 mark
B1 for F(1, 0)
award B1 B1 B0 for all three correct points given in the wrong order

Question 3

3 marks

ABCD is a rhombus with sides of length 7.5 cm.
Angle DAB = 64°.

Work out the length of the diagonal BD.
Give your answer correct to 3 significant figures.

Model answer

The diagonals of a rhombus bisect each other at right angles and bisect the angles of the rhombus.
Let the diagonals meet at X. In triangle ABX, AB = 7.5 cm, angle BAX = 64° ÷ 2 = 32° and angle AXB = 90°.
\(BX = 7.5 \sin 32^\circ = 3.9743...\)
\(BD = 2 \times 3.9743... = 7.9487...\)
BD = 7.95 cm (3 s.f.)

Explanation

Use the symmetry properties of a rhombus: its diagonals cross at right angles, cut each other in half and bisect the corner angles. BD is opposite angle A, so in the right-angled triangle use half the angle, 32°, and then double the half-diagonal. The cosine rule in triangle ABD gives the same answer.

Mark scheme

1 mark
M1 for using the properties of a rhombus, e.g. a right-angled triangle with hypotenuse 7.5 cm and angle 32°
or the cosine rule: \(BD^2 = 7.5^2 + 7.5^2 - 2 \times 7.5 \times 7.5 \times \cos 64^\circ\)
1 mark
M1 for \(7.5 \sin 32^\circ\) (= 3.974...) or \(BD^2 = 63.18...\)
1 mark
A1 for 7.95 cm
awrt 7.95 (7.948...); correct answer with no working scores full marks

Question 4

4 marks

ABCDEFGH is a regular octagon with sides of length 6 cm.

Work out the length of the diagonal AD.
Give your answer correct to 3 significant figures.

Model answer

Interior angle of a regular octagon \(= \dfrac{6 \times 180^\circ}{8} = 135^\circ\)
By symmetry ABCD is an isosceles trapezium with AD parallel to BC, so angle BAD = 180° − 135° = 45°.
Drop perpendiculars from B and C to AD. Each end piece of AD is \(6\cos 45^\circ = 4.2426...\)
\(AD = 4.2426... + 6 + 4.2426... = 14.485...\)
AD = 14.5 cm (3 s.f.)

Explanation

Sketch the octagon with its vertices in order, so AD joins A to the vertex three places round. Using the interior angle of 135°, the quadrilateral ABCD is an isosceles trapezium, and two right-angled triangles at its ends each add \(6\cos 45^\circ\) to the length of BC. The exact answer is \(6(1 + \sqrt{2})\) cm.

Mark scheme

1 mark
P1 for the interior angle 135° (e.g. angle ABC = 135°)
correct answer with no working scores full marks
1 mark
P1 for recognising that ABCD is an isosceles trapezium with AD parallel to BC, so angle BAD = 45°
or another complete strategy, e.g. the circumradius \(3 \div \sin 22.5^\circ\) and angle AOD = 135°
1 mark
P1 for \(6 + 2 \times 6\cos 45^\circ\) oe
or \(AD^2 = 2r^2 - 2r^2\cos 135^\circ\) with r = 7.83...
1 mark
A1 for 14.5 cm
awrt 14.5 (14.485...)

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