OCR GCSE Mathematics: Angle facts, polygons, triangle and quadrilateral properties, and symmetry (J560) – 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 follow OCR GCSE (9-1) Mathematics, specification Version 2.1 (August 2026), sections 8.03 and 8.04. Questions 1 to 10 and 12 are for both tiers. Question 11 is Higher tier only. Diagrams are described in words: sketch each one before you start, and write a reason beside every angle you find.

Revise first with the study guide or the revision notes. See the course hub and the checklist.

Questions

1. Four angles meet at a point. They are 5a°, (2a + 10)°, (3a − 20)° and 100°. Find the value of a and the size of the smallest angle. [3]

2. Lines PQ and RS are parallel. A line crosses PQ at X and RS at Y. Angle QXY is (3t + 14)° and angle XYR is (5t − 30)°. These two angles are alternate.

(a) Find the value of t, and the size of angle QXY. [3] (b) Angle XYS lies on the line RS, next to angle XYR. Find angle XYS, giving a reason. [2]

3. A regular polygon has an exterior angle of 15°.

(a) Work out the number of sides. [2] (b) Work out the sum of its interior angles. [2]

4. A pentagon has angles 2x°, (x + 30)°, 115°, (3x − 25)° and 90°. Find x and the largest angle. [4]

5. (a) A right-angled triangle has its two acute angles equal to (2z + 5)° and 3z°. Find z. [2] (b) Triangle ABC has AB = AC. Angle ABC is (3p + 8)° and angle BAC is (4p + 14)°. Find p and angle BAC. [3]

6. (a) ABCD is a rhombus with perimeter 36 cm. The diagonals meet at M. Angle BAC is 29°. Find the length AB, angle ABC and angle AMB. [4] (b) In trapezium PQRS, PQ is parallel to SR. Angle SPQ is (3x + 20)° and angle PSR is (2x + 10)°. Find x, then find angle PQR when angle QRS is 120°. [3]

7. (a) For each shape give the number of lines of symmetry and the order of rotation symmetry: a rectangle, a kite, a regular octagon. [3] (b) Name a quadrilateral with rotation symmetry of order 2 but no line of symmetry. [1] (c) A regular polygon has interior angle 140°. State its number of lines of symmetry. [2]

8. A regular hexagon, a square and a regular polygon with n sides fit together exactly around a point, one corner of each meeting there with no gaps. Find n. [4]

9. ABCD is a parallelogram, with AB parallel to DC and AD parallel to BC. Prove that angle ABC equals angle ADC. [4]

10. Kite ABCD has AB = AD and CB = CD. Prove that angle ABC = angle ADC by considering the diagonal BD. [5]

11. (Higher tier only) A, B and C are points on a circle with centre O. O lies inside the acute angle ABC. Prove that angle AOC = 2 × angle ABC. [5]

12. ABCDE is a regular pentagon. An equilateral triangle CDF is drawn outside the pentagon on side CD.

(a) Work out the interior angle of the pentagon. [2] (b) Work out angle BCF. [2] (c) Work out angle CBF, giving a reason. [2]

Answers

1. 5a + (2a + 10) + (3a − 20) + 100 = 360 (angles at a point) [1]; 10a + 90 = 360, so a = 27 [1]. The angles are 135°, 64°, 61°, 100°, so the smallest is 61° [1]. [3] Examiner insight: name the fact you use to form the equation, and check your answer by adding all the angles back to 360°.

2. (a) Alternate angles are equal: 3t + 14 = 5t − 30 [1]; 44 = 2t, so t = 22 [1]; angle QXY = 3 × 22 + 14 = 80° [1]. (b) Angle XYS = 180 − 80 [1] = 100° (angles on a line) [1]. Examiner insight: a reason is part of the answer on a “give a reason” part, so quote the named fact rather than only the number.

3. (a) n = 360 ÷ 15 [1] = 24 [1]. (b) (24 − 2) × 180 [1] = 3960° [1]. Examiner insight: show the (n − 2) × 180 substitution. A bare number is harder to credit if it is wrong.

4. Sum of interior angles = (5 − 2) × 180 = 540° [1]; 2x + x + 30 + 115 + 3x − 25 + 90 = 540, so 6x + 210 = 540 [1]; x = 55 [1]. Angles: 110°, 85°, 115°, 140°, 90°. Largest = 140° [1]. [4] Examiner insight: the question asks for the largest angle, not x. Reread the question and answer what it asks.

5. (a) (2z + 5) + 3z = 90 (the acute angles of a right-angled triangle sum to 90°) [1]; 5z = 85, so z = 17 [1]. (b) Base angles equal and angles sum to 180°: 2(3p + 8) + 4p + 14 = 180 [1]; 10p + 30 = 180, p = 15 [1]; angle BAC = 4 × 15 + 14 = 74° [1]. Examiner insight: in an isosceles triangle, say which two angles are equal and why, before you write the equation.

6. (a) All sides of a rhombus are equal, so AB = 36 ÷ 4 = 9 cm [1]. Triangle ABC is isosceles (AB = BC), so angle BCA = 29° [1]; angle ABC = 180 − 29 − 29 = 122° [1]. The diagonals of a rhombus meet at right angles, so angle AMB = 90° [1]. (b) Co-interior angles between parallel lines: 3x + 20 + 2x + 10 = 180 [1]; x = 30, so angle SPQ = 110° [1]. Angle PQR + angle QRS = 180° (co-interior), so angle PQR = 180 − 120 = 60° [1]. Examiner insight: quote the property by name (diagonals of a rhombus are perpendicular) rather than relying on how the diagram looks.

7. (a) Rectangle: 2 lines, order 2 [1]. Kite: 1 line, order 1 [1]. Regular octagon: 8 lines, order 8 [1]. (b) Parallelogram [1]. (c) Exterior angle = 180 − 140 = 40° [1], so n = 360 ÷ 40 = 9 and there are 9 lines of symmetry [1]. Examiner insight: give both numbers for each shape. Order 1 is the correct way to say a shape has no rotation symmetry.

8. The interior angle of a regular hexagon is 120° [1] and of a square is 90° [1]. The third angle is 360 − 120 − 90 = 150° [1]. Exterior angle = 30°, so n = 360 ÷ 30 = 12 [1]. [4] Examiner insight: the angle at the point is an interior angle of the new polygon. Convert it to an exterior angle before dividing.

9. Let angle ABC = b. AD ∥ BC, so angle DAB + angle ABC = 180° (co-interior angles) [1]; so angle DAB = 180 − b [1]. AB ∥ DC, so angle DAB + angle ADC = 180° (co-interior angles) [1]; so angle ADC = 180 − (180 − b) = b. Hence angle ABC = angle ADC [1]. [4] Examiner insight: a proof needs a reason on every line and a closing statement of what has been shown. Do not use numbers from the diagram.

10. In triangle ABD, AB = AD, so angle ABD = angle ADB (base angles of an isosceles triangle) [1]. In triangle CBD, CB = CD, so angle CBD = angle CDB (base angles of an isosceles triangle) [1]. Angle ABC = angle ABD + angle CBD [1] and angle ADC = angle ADB + angle CDB [1]. Adding equal angles gives angle ABC = angle ADC [1]. [5] Examiner insight: write which sides are equal and name the triangle each time. A proof needs both the statement and its reason to be visible.

11. (Higher tier only) OA = OB = OC (radii) [1]. Let angle ABO = u and angle CBO = v, so angle ABC = u + v. Triangle OAB is isosceles, so angle OAB = u (base angles) [1]. Extend BO to meet the circle at D. Angle AOD = u + u = 2u (exterior angle of triangle OAB) [1]. Similarly angle OCB = v, so angle COD = 2v [1]. Angle AOC = 2u + 2v = 2 × angle ABC [1]. [5] Examiner insight: introduce your own letters for the base angles and carry them through. Each line needs its reason, and the last line must state the result you were asked to prove.

12. (a) (5 − 2) × 180 ÷ 5 [1] = 108° [1]. (b) Angle BCF = 108 + 60 [1] = 168° (angles at C, with angle DCF = 60° in an equilateral triangle) [1]. (c) CF = CD = CB (sides of the equilateral triangle and the regular pentagon), so triangle BCF is isosceles [1]; angle CBF = (180 − 168) ÷ 2 = 6° [1]. Examiner insight: look for equal lengths first. Spotting CB = CF turns a hard-looking angle into a base-angle calculation.

Where marks are usually lost

• Giving an angle with no reason, or a vague reason such as “parallel lines”.
• Using the wrong parallel-line fact (alternate and corresponding mixed up).
• Using 360° for a straight line or 180° for angles at a point.
• Finding an exterior angle and stopping before the interior angle is asked for.
• Answering for x when the question asks for an angle.
• Treating a diagram as accurate when it is not.
• Writing “no symmetry” when the shape has rotation symmetry of order 2, as a parallelogram does.
• Ending a proof without stating what has been proved, or using measured values.

Next steps

• Review the revision notes and the study guide.
• Go to the course hub and tick topics on the checklist.
• Try the free 10-minute diagnostics, or book a free trial class.

Official syllabus

OCR, GCSE (9-1) Mathematics J560, specification Version 2.1 (August 2026), for first assessment in 2017: sections 8.03 and 8.04.