OxfordAQA International GCSE Mathematics: Pythagoras' theorem and right-angled trigonometry (9260) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover statement G19 in section 3.3.2 of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1: Pythagoras’ theorem and the sine, cosine and tangent ratios in right-angled triangles. Questions 1 to 9 are Core content in 2D figures. Questions 10 and 11 are marked (Extension only) because they use 3D figures. All content can be assessed on the Extension papers. Unless told otherwise, round lengths to 3 significant figures and give angles correct to 1 decimal place. A scientific calculator is allowed on every paper.

The study guide and revision notes cover the method. The course hub and printable checklist show the wider course, and the free 10-minute diagnostics are a quick check on where you stand.

Questions

1. A right-angled triangle has short sides of 15.4 cm and 9.8 cm. Calculate the length of the hypotenuse. [2]

2. The hypotenuse of a right-angled triangle is 26.5 cm. One short side is 17.2 cm. Calculate the length of the other short side. [3]

3. A rhombus has diagonals of length 16 cm and 30 cm. The diagonals cross at right angles and cut each other in half. Show that each side of the rhombus is 17 cm. [3]

4. In triangle PQR, angle Q = 90°, angle R = 47° and PR = 19.3 cm. Calculate the length of PQ. [3]

5. In triangle ABC, angle B = 90°, angle A = 64° and AB = 8.6 cm. Calculate the length of AC. [3]

6. The cross-section of a roof is an isosceles triangle. Its base is 8.4 m wide and its peak is 1.7 m above the base. Calculate the angle the roof makes with the horizontal. [3]

7. A playground slide is a straight slope 3.8 m long. The top is 2.1 m above the level ground.

(a) Calculate the angle between the slide and the ground. [2] (b) Calculate the horizontal distance from the foot of the slide to the point directly below the top. [3]

8. An isosceles triangle has two equal sides of 11.6 cm and a base of 8.4 cm.

(a) Calculate the perpendicular height of the triangle. [3] (b) Calculate the size of one of the base angles. [2]

9. In the diagram, triangle ABC has a right angle at B, with AB = 8.5 cm and BC = 6 cm. Triangle ACD has a right angle at C, with CD = 4.8 cm. The points B and D are on opposite sides of AC.

(a) Calculate AC. [2] (b) Calculate AD. [3] (c) Calculate angle CAD. [2]

10. (Extension only) A cuboid ABCDEFGH has a horizontal rectangular base ABCD, with AB = 11 cm and BC = 6 cm. The vertical edge CG is 4 cm.

(a) Calculate the length of the diagonal AG. [3] (b) Calculate the angle between AG and the base ABCD. [3]

11. (Extension only) A pyramid has a horizontal rectangular base 12 cm by 8 cm. Its apex V is vertically above the centre of the base. Each sloping edge is 11 cm long.

(a) Calculate the distance from the centre of the base to one corner. [2] (b) Calculate the vertical height of the pyramid. [3] (c) Calculate the angle between a sloping edge and the base. [2]

Answers

1. c² = 15.4² + 9.8² [1] = 237.16 + 96.04 = 333.2, so c = √333.2 = 18.3 cm [1] [2] Examiner insight: Writing the sum of squares before taking the root shows your method clearly, so a slip in the final arithmetic is easy to spot and fix.

2. Hypotenuse is the longest side, so subtract: b² = 26.5² − 17.2² [1] = 702.25 − 295.84 = 406.41 [1], so b = √406.41 = 20.2 cm [1] [3] Examiner insight: Adding the squares here gives a short side longer than the hypotenuse, so check that your answer is smaller than 26.5.

3. Half the diagonals are 8 cm and 15 cm, and they form a right angle at the centre [1]. Side² = 8² + 15² = 64 + 225 [1] = 289, so side = √289 = 17 cm [1] [3] Examiner insight: A “show that” question needs the final value reached from your working, and half the diagonals must be used, not the full lengths.

4. PR is the hypotenuse and PQ is opposite angle R, so use sine [1]. sin 47° = PQ/19.3, so PQ = 19.3 × sin 47° [1] = 14.1 cm [1] [3] Examiner insight: Naming the sides you are using (opposite and hypotenuse) shows the choice of ratio was deliberate.

5. AB is adjacent to angle A and AC is the hypotenuse, so use cosine [1]. cos 64° = 8.6/AC, so AC = 8.6 / cos 64° [1] = 19.6 cm [1] [3] Examiner insight: Multiplying by cos 64° instead of dividing gives a hypotenuse shorter than a short side, which is a quick warning sign.

6. Split the roof triangle in half: each half is right-angled with base 4.2 m and height 1.7 m [1]. tan θ = 1.7/4.2 [1], so θ = tan⁻¹(0.4048) = 22.0° [1] [3] Examiner insight: The base of the right-angled triangle is half of 8.4 m, and using the full width is a frequent error.

7. (a) The height is opposite the angle and the slide is the hypotenuse, so sin θ = 2.1/3.8 [1]; θ = sin⁻¹(0.5526) = 33.5° [1] [2] (b) d² = 3.8² − 2.1² [1] = 14.44 − 4.41 = 10.03 [1], so d = √10.03 = 3.17 m [1] [3] Examiner insight: In part (b) you can also use cos 33.5° × 3.8, but using the exact given lengths avoids carrying a rounded angle forward.

8. (a) Splitting the triangle gives a right-angled triangle with hypotenuse 11.6 cm and base 4.2 cm [1]. h² = 11.6² − 4.2² = 134.56 − 17.64 [1] = 116.92, so h = √116.92 = 10.8 cm [1] [3] (b) cos θ = 4.2/11.6 [1], so θ = cos⁻¹(0.3621) = 68.8° [1] [2] Examiner insight: Using 4.2 and 11.6 directly in part (b) is cleaner than using the rounded height from part (a).

9. (a) AC² = 8.5² + 6² [1] = 72.25 + 36 = 108.25, so AC = √108.25 = 10.4 cm [1] [2] (b) Triangle ACD is right-angled at C, so AD² = AC² + CD² [1]. Using AC² = 108.25: AD² = 108.25 + 4.8² = 108.25 + 23.04 [1] = 131.29, so AD = √131.29 = 11.5 cm [1] [3] (c) In triangle ACD, CD is opposite angle CAD and AC is adjacent: tan(CAD) = 4.8/10.404 [1]; angle CAD = 24.8° [1] [2] Examiner insight: Keeping AC² as 108.25 rather than a rounded AC protects the accuracy of AD, and your answer to (c) needs the unrounded or 4 s.f. value of AC.

10. (Extension only) (a) Base diagonal AC² = 11² + 6² [1] = 157. Then AG² = AC² + CG² = 157 + 4² [1] = 173, so AG = √173 = 13.2 cm [1] [3] (b) The angle between AG and the base is angle GAC, in the right-angled triangle ACG with CG vertical [1]. tan(GAC) = 4/√157 [1], so angle GAC = 17.7° [1] [3] Examiner insight: A sketch of the flat triangle ACG, with CG, AC and AG labelled, is the clearest evidence that you found the right triangle in the 3D figure.

11. (Extension only) (a) Base diagonal² = 12² + 8² [1] = 208, so the centre-to-corner distance is √208 ÷ 2 = 7.21 cm [1] [2] (b) The height is perpendicular to the base, so the edge, the height and the centre-to-corner distance form a right-angled triangle with hypotenuse 11 cm [1]. h² = 11² − (centre-to-corner)² [1], where (centre-to-corner)² = 208/4 = 52, so h² = 121 − 52 = 69 and h = √69 = 8.31 cm [1] [3] (c) cos θ = 7.211/11 [1], so θ = 49.0° [1] [2] Examiner insight: The centre-to-corner distance is half the base diagonal, not the whole diagonal, and using the whole diagonal is the usual way to lose marks here.

Where marks are usually lost

• Adding the squares when you should subtract, so a short side comes out longer than the hypotenuse.
• Stopping at the value of c² or b² without taking the square root.
• Taking opposite and adjacent from the wrong angle, which gives the wrong ratio.
• Writing the wrong fraction when finding an angle, such as adjacent/opposite for a tangent.
• Rounding in the middle of a two-step problem and passing a rounded value on.
• Using the full base instead of half the base after splitting an isosceles triangle.
• Giving an angle without checking it is plausible: acute angles in a right-angled triangle must each be less than 90°.
• In 3D, using a sloping edge as the vertical height or the base diagonal as a side.
• Leaving out units, or giving a length to more or fewer significant figures than the question asks.

Next steps

• Go back to the revision notes for the one-page methods.
• Reread the study guide for any question you could not start.
• See the OxfordAQA IGCSE Mathematics hub and the printable checklist.
• Try the free 10-minute diagnostics to find your next weak spot.
• Book a free trial class if you would like help from a teacher.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.3.2 Mensuration and calculation, statement G19.