AQA GCSE Mathematics 8300: Geometry and measures – 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 Topic 4, Geometry and measures (references G1–G25), of the AQA GCSE Mathematics (8300) specification, Version 1.0, for exams in May/June 2017 onwards and every May/June and November series after that. Parts marked Higher tier only use content from the specification’s “Higher content only” column; everything else is on both tiers. Each question is labelled “(non-calculator)” for Paper 1 style or “(calculator allowed)” for Paper 2 and 3 style. Diagrams are described in words, so sketch each one first. Give answers to 3 significant figures and angles to 1 decimal place unless told otherwise.

Learn the content in the study guide and the revision notes. The course hub is AQA GCSE Mathematics, and the printable checklist lists every statement.

Questions

1. (non-calculator) In triangle ABC, AB = AC and angle BAC = 44°. A straight line DAE is drawn through A parallel to BC, with D on the same side of A as B.

(a) Work out the size of angle ABC. Give reasons for your answer. [2] (b) Work out the size of angle DAB. Give a reason for your answer. [2]

2. (non-calculator) Each interior angle of a regular polygon is 4 times the size of each exterior angle. Work out the number of sides of the polygon. [3]

3. (non-calculator)

(a) The bearing of Q from P is 118°. Work out the bearing of P from Q. [1] (b) A garden has a tree T and a gate G. A bench must be closer to G than to T, and less than 4 m from T. Describe the two constructions you would draw on a scale plan to show where the bench can go. [2]

4. (non-calculator)

(a) Point A(2, 5) is translated by the column vector (−3, 4). Write down the coordinates of the image. [1] (b) Higher tier only. Triangle T has vertices (1, 1), (3, 1) and (1, 2). Enlarge T by scale factor −2, centre (0, 0). Write down the vertices of the image. [2]

5. (calculator allowed) A shape is made from a rectangle 12 cm by 8 cm with a semicircle on one of its 8 cm sides, outside the rectangle.

(a) Work out the area of the shape. [2] (b) Work out the perimeter of the shape. [2]

6. (calculator allowed) A cylinder has radius 4 cm and height 15 cm.

(a) Work out its volume. Give your answer in terms of π and to 3 significant figures. [2] (b) A mathematically similar cylinder has height 22.5 cm. Work out its radius. [1] (c) Higher tier only. Use a volume scale factor to work out the volume of the larger cylinder. [2]

7. (non-calculator) Triangle ABC has a right angle at B. AC = 12 cm and angle BAC = 60°.

(a) Work out the length of AB. [2] (b) Work out the length of BC. Give your answer as a surd in its simplest form. [2]

8. (calculator allowed)

(a) A ladder 7.2 m long leans against a vertical wall. Its foot is on horizontal ground, 2.1 m from the wall. (i) Work out how far up the wall the ladder reaches. [2] (ii) Work out the angle the ladder makes with the ground. [2] (b) Higher tier only. A cuboid ABCDEFGH has a base ABCD with AB = 12 cm and BC = 9 cm, and height CG = 8 cm. Work out the angle between the diagonal AG and the base ABCD. [3]

9. (non-calculator, Higher tier only) Points A, B, C and D lie in that order on a circle with centre O. D is on the minor arc AC. The angle AOC facing the minor arc is 128°.

(a) Work out angle ABC. Give a reason. [2] (b) Work out angle ADC. Give a reason. [2]

10. (calculator allowed) A solid metal cone has base radius 10 cm and perpendicular height 24 cm. The top is cut off by a plane parallel to the base, halfway up, leaving a frustum. Volume of a cone = ⅓πr²h.

(a) Work out the volume of the whole cone in terms of π. [2] (b) Show that the volume of the frustum is 700π cm³. [3] (c) The metal has density 7.8 g/cm³. Work out the mass of the frustum in kilograms. [2]

11. (calculator allowed, Higher tier only) In triangle PQR, PQ = 8.5 cm, PR = 11.2 cm and angle QPR = 57°.

(a) Work out the length of QR. [3] (b) Work out the area of the triangle. [2] (c) Work out the size of angle PQR. [2]

12. (non-calculator, Higher tier only) In triangle OAB, OA = a and OB = b. P lies on AB with AP : PB = 1 : 2.

(a) Write AB in terms of a and b. [1] (b) Show that OP = ⅔a + ⅓b. [2] (c) Point X is such that OX = 4a + 2b. Prove that O, P and X lie on a straight line. [2]

Answers

1. (a) (180° − 44°) ÷ 2 = 68° [1]; base angles of an isosceles triangle are equal and angles in a triangle add up to 180° [1] (b) 68° [1]; alternate angles are equal (DAE is parallel to BC) [1] Examiner insight: Each reason mark needs the property written in words; “Z angles” is not accepted for alternate angles.

2. e + 4e = 180° [1]; e = 36° [1]; 360° ÷ 36° = 10 sides [1] Examiner insight: Using 4e = 360° uses the exterior angle sum wrongly and scores no method mark; the link is interior + exterior = 180°.

3. (a) 118° + 180° = 298° [1] (b) The perpendicular bisector of TG, keeping the side nearer G [1]; a circle of radius 4 m (to scale) centred on T, keeping the inside, so the region is where both overlap [1] Examiner insight: Each mark needs the correct construction and the correct side; “draw a line between T and G” is not a perpendicular bisector.

4. (a) (−1, 9) [1] (b) Multiply each coordinate by −2 [1]; (−2, −2), (−6, −2), (−2, −4) [1] Examiner insight: Using scale factor +2 gives an image on the wrong side of the centre; this scores no marks because the negative sign is the whole point of the question.

5. (a) 12 × 8 + ½ × π × 4² [1] = 96 + 8π = 121 cm² [1] (b) 12 + 12 + 8 + ½ × π × 8 [1] = 44.6 cm [1] Examiner insight: The side under the semicircle is not part of the perimeter; including it gives 52.6 cm and loses the accuracy mark.

6. (a) π × 4² × 15 [1] = 240π = 754 cm³ [1] (b) k = 22.5 ÷ 15 = 1.5, radius = 6 cm [1] (c) k³ = 1.5³ = 3.375 [1]; 240π × 3.375 = 810π = 2540 cm³ [1] Examiner insight: In (c) multiplying by 1.5 instead of 1.5³ scores 0; the method mark is for cubing the length scale factor.

7. (a) AB = 12 × cos 60° [1] = 12 × ½ = 6 cm [1] (b) BC² = 12² − 6² = 108 [1]; BC = √108 = 6√3 cm [1] Examiner insight: On a non-calculator paper, “simplest form” means the final mark needs 6√3; √108 alone earns only the method mark.

8. (a)(i) √(7.2² − 2.1²) [1] = 6.89 m [1] (a)(ii) cos θ = 2.1 ÷ 7.2 [1]; θ = 73.0° [1] (b) AC = √(12² + 9²) = 15 cm [1]; tan θ = 8 ÷ 15 [1]; θ = 28.1° [1] Examiner insight: In (b) the first mark is for the base diagonal; using AB or BC as the adjacent side finds a different angle and loses both remaining marks.

9. (a) 128° ÷ 2 = 64° [1]; the angle at the centre is twice the angle at the circumference [1] (b) 180° − 64° = 116° [1]; opposite angles of a cyclic quadrilateral add up to 180° [1] Examiner insight: A correct angle with no reason earns the accuracy mark only; the reason must name the theorem, not just say “circle theorem”.

10. (a) ⅓ × π × 10² × 24 [1] = 800π cm³ [1] (b) The removed cone is similar with scale factor ½, so radius 5 cm and height 12 cm [1]; its volume = ⅓ × π × 5² × 12 = 100π [1]; 800π − 100π = 700π cm³ [1] (c) 700π × 7.8 = 17 153 g [1]; = 17.2 kg [1] Examiner insight: A “show that” needs every step; stating 700π after 800π with no small-cone volume scores at most the first mark.

11. (a) QR² = 8.5² + 11.2² − 2 × 8.5 × 11.2 × cos 57° [1]; = 93.99… [1]; QR = 9.69 cm [1] (b) ½ × 8.5 × 11.2 × sin 57° [1] = 39.9 cm² [1] (c) sin Q = 11.2 × sin 57° ÷ 9.694… [1]; Q = 75.7° [1] Examiner insight: Keep the unrounded QR in (c); using 9.7 gives about 75.5°, which may fall outside the accepted range for the accuracy mark.

12. (a) AB = b − a [1] (b) OP = OA + ⅓AB [1]; = a + ⅓(b − a) = ⅔a + ⅓b [1] (c) OX = 4a + 2b = 6(⅔a + ⅓b) = 6OP, so OX is parallel to OP [1]; OX and OP share the point O, so O, P and X are collinear [1] Examiner insight: The final mark needs both facts: parallel and a common point; “parallel” alone does not prove collinearity.

Where marks are usually lost

• Angle answers given without reasons, or with colloquial reasons, when reasons are asked for.
• Taking the scale factor as positive in a negative enlargement.
• Including an internal edge in a perimeter.
• Using the length scale factor for a volume or area in similar solids.
• Leaving a surd unsimplified on the non-calculator paper.
• Rounding the base diagonal or an unknown side too early in a multi-step calculation.
• Naming “circle theorem” instead of stating the theorem.
• Missing steps in a “show that”, which must lead to the exact given result.
• Stopping at “parallel” in a collinearity proof.
• Giving mass in grams when kilograms are asked for.

Next steps

• Recap with the revision notes.
• Re-learn anything you got wrong in the study guide.
• See the whole course on the AQA GCSE Mathematics hub.
• Tick off statements on the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA GCSE Mathematics (8300) specification, Version 1.0 (12 September 2014), for teaching from September 2015 and exams in May/June 2017 onwards, published by AQA. Section 3.4, Geometry and measures, G1–G25.