Pythagoras' theorem and trigonometry - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 37 exam-style questions on Pythagoras' theorem and trigonometry for Edexcel GCSE Maths (Non-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.

37 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

The diagram shows a right-angled triangle ABC.
AB = 9 cm and BC = 12 cm.

Work out the value of \(x\).

Geometry diagram, marked "Diagram NOT accurately drawn". Triangle ABC. Angle ABC is marked as a right angle (90°). Side AB is labelled 9 cm. Side BC is labelled 12 cm. Side AC is labelled x cm.

Model answer

\(x^2 = 9^2 + 12^2 = 81 + 144 = 225\)
\(x = \sqrt{225} = 15\)

Explanation

AC is opposite the right angle, so it is the hypotenuse: add the squares of the two shorter sides and take the square root. (9, 12, 15 is the 3, 4, 5 triple multiplied by 3.)

Mark scheme

1 mark
M1 for \(9^2 + 12^2\) (= 225)
oe; correct answer with no working scores full marks
1 mark
A1 for 15
cao

Question 2

2 marks

A right-angled triangle has a hypotenuse of length 10 cm. One of the other sides has length 6 cm.
Lily works out the length of the third side. Here is her working.

\(x^2 = 10^2 + 6^2 = 136\)
\(x = \sqrt{136}\)

Explain the mistake Lily has made and work out the correct length of the third side.

Model answer

Lily has added the squares, but 10 cm is the hypotenuse (the longest side), so she should subtract.
\(x^2 = 10^2 - 6^2 = 100 - 36 = 64\)
\(x = 8\) cm

Explanation

Only the hypotenuse squared equals the sum of the other two squares. Lily's answer \(\sqrt{136}\) is more than 10, which is impossible because no side can be longer than the hypotenuse.

Mark scheme

1 mark
C1 for explaining that 10 cm is the hypotenuse, so she should subtract: \(x^2 = 10^2 - 6^2\)
oe, e.g. 'she added the squares instead of subtracting'; or 'her answer is longer than the hypotenuse'
1 mark
B1 for 8
accept 8 cm

Question 3

3 marks

The diagram shows a trapezium ABCD.
AB is parallel to DC.
Angle DAB = angle ADC = 90°.
AB = 10 cm, DC = 4 cm and AD = 8 cm.

Work out the perimeter of the trapezium.

Geometry diagram, marked "Diagram NOT accurately drawn". Trapezium ABCD. Angle DAB is marked as a right angle (90°). Angle ADC is marked as a right angle (90°). Side AB is labelled 10 cm. Side DC is labelled 4 cm. Side AD is labelled 8 cm. AB and DC are each marked with an arrowhead (parallel lines).

Model answer

Draw the perpendicular from C to AB. It has length 8 cm and meets AB 4 cm from A, leaving 10 − 4 = 6 cm.
\(BC^2 = 6^2 + 8^2 = 36 + 64 = 100\), so BC = 10 cm
Perimeter = 10 + 10 + 4 + 8 = 32 cm

Explanation

The slanted side BC is the hypotenuse of a right-angled triangle whose shorter sides are the height (8 cm) and the difference between the parallel sides (6 cm). Then add all four sides.

Mark scheme

1 mark
M1 for \(6^2 + 8^2\), using 10 − 4 = 6
oe; correct answer with no working scores full marks
1 mark
M1 for BC = 10 and adding all four sides, e.g. 10 + 10 + 4 + 8
ft their BC
1 mark
A1 for 32
cao; accept 32 cm

Question 4

4 marks

The diagram shows triangle ABC.
Angle BAC = 60°, angle ABC = 45° and BC = \(6\sqrt{2}\) cm.

Work out the length of AC.
Give your answer as a surd in its simplest form.

Geometry diagram, marked "Diagram NOT accurately drawn". Triangle ABC. Angle BAC is marked 60°. Angle CBA is marked 45°. Side BC is labelled 6√2 cm. Side AC is labelled x cm.

Model answer

Draw the perpendicular from C to meet AB at D.
In triangle BDC: CD \(= 6\sqrt{2} \times \sin 45° = 6\sqrt{2} \times \dfrac{\sqrt{2}}{2} = \dfrac{6 \times 2}{2} = 6\) cm
In triangle ADC: \(\sin 60° = \dfrac{6}{AC}\)
\(AC = \dfrac{6}{\frac{\sqrt{3}}{2}} = \dfrac{12}{\sqrt{3}} = \dfrac{12\sqrt{3}}{3} = 4\sqrt{3}\) cm

Explanation

Splitting a non-right-angled triangle with a perpendicular height creates two right-angled triangles that share the height. Find the height from the 45° triangle, then use it in the 60° triangle and rationalise the denominator.

Mark scheme

1 mark
P1 for drawing the perpendicular CD from C to AB and CD \(= 6\sqrt{2} \times \sin 45°\) (= 6)
or a correct sine rule statement, e.g. \(\dfrac{AC}{\sin 45°} = \dfrac{6\sqrt{2}}{\sin 60°}\); correct answer with no working scores full marks
1 mark
P1 for \(AC = \dfrac{6}{\sin 60°}\)
ft their CD; or \(AC = \dfrac{6\sqrt{2} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}}\)
1 mark
P1 for \(AC = \dfrac{12}{\sqrt{3}}\)
oe
1 mark
A1 for \(4\sqrt{3}\)
cao

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