OxfordAQA International GCSE Mathematics: Angle facts, polygons and quadrilaterals (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 references G1 to G5 in section 3.3.1 of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). All of this content is Core, and the specification says Core content can also be assessed on the Extension tier papers. Diagrams are described in words or text, so sketch them first.

If a fact is unfamiliar, start with the study guide or the revision notes. The course page is OxfordAQA IGCSE Mathematics and the printable list is the checklist.

Questions

1. The angles around a point measure 4x°, (3x + 15)°, (5x − 25)° and 70°. Find x and the smallest of the four angles. [3]

2. In the diagram, JK and LM are parallel. The line NP crosses JK at S and LM at T. Angle NSK = 128°.

        N
         \
J ------- S ------- K
           \
            \
L ---------- T ------ M
              \
               P

(a) Find angle STM. Give a reason. [2] (b) Find angle JST. Give a reason. [2] (c) Find angle KST and name the type of angles that KST and STM are. [2]

3. Two lines are parallel and are crossed by a third line. Two alternate angles are (7m − 21)° and (4m + 9)°.

(a) Find the value of m. [2] (b) Find the size of each alternate angle. [1] (c) Find the size of the interior angle that pairs with (7m − 21)°. [1]

4. In triangle PQR, PQ = PR. Angle PQR = (y + 31)° and angle QPR = (4y − 2)°. QR is extended to S.

(a) Find y. [3] (b) Find angle PRS. [2]

5. (a) A triangle has angles 38°, 38° and 104°. Give two names that describe it. [1] (b) A triangle has angles 25°, 65° and 90°. Say whether it is isosceles, equilateral or scalene, and give its other name. [1] (c) The angles of a triangle are in the ratio 2 : 3 : 5. Find the three angles. [2]

6. (a) In triangle LMN, write the angle at N between NL and NM using angle notation. [1] (b) In triangle LMN, which lower-case letter labels the side opposite angle M? [1] (c) Write, using symbols, “AB is parallel to DC and AD is perpendicular to AB”. [1] (d) How many vertices and how many edges has a hexagon? [1]

7. Seven of the interior angles of an octagon are 135°, 142°, 128°, 150°, 117°, 139° and 146°. Find the eighth angle. [3]

8. (a) Find the exterior angle and the interior angle of a regular polygon with 18 sides. [2] (b) A different regular polygon has an interior angle of 156°. Find its number of sides. [2] (c) Find the angle at the centre of the polygon in part (b). [1]

9. ABCD is a rhombus. Angle DAB = 58°. The diagonals AC and BD meet at M.

(a) Find angle ABC. [2] (b) Find angle ABD. [2] (c) Find angle AMB. [1]

10. (a) PQRS is a kite with PQ = PS and RQ = RS. Angle QPS = 84° and angle PQR = 112°. Find angle QRS. [3] (b) ABCD is a trapezium with AB parallel to DC. Angle DAB = 71°. Angle ABC is four times angle BCD. Find angle ADC and angle ABC. [3]

11. ABCDEFGHIJKL is a regular polygon with 12 sides.

(a) Show that each interior angle is 150°. [2] (b) Triangle ABC is drawn. Find angle BAC. [2] (c) Find angle LAC. [1] (d) Use exterior angles to explain why AB is parallel to GH. [2]

Answers

1. 4x + (3x + 15) + (5x − 25) + 70 = 360 (angles at a point) [1]; 12x + 60 = 360, so x = 25 [1]. The angles are 100°, 90°, 100° and 70°, so the smallest is 70° [1]. [3] Examiner insight: Check by adding your four angle sizes. A total other than 360° shows an arithmetic slip, and the check costs only a few seconds.

2. (a) Angle STM = 128° [1], because corresponding angles are equal (JK ∥ LM) [1]. [2] (b) Angle JST = 128° [1], because it is vertically opposite angle NSK [1]. [2] (c) Angle KST = 180 − 128 = 52° (angles on a straight line) [1]. KST and STM are interior angles (they add up to 180°) [1]. [2] Examiner insight: Each part asks for a value and a named fact. Writing the property in words next to each result is good habit, because a bare number may not show that you know why it is true.

3. (a) 7m − 21 = 4m + 9 (alternate angles are equal) [1]; 3m = 30, so m = 10 [1]. [2] (b) 7 × 10 − 21 = 49° [1]. [1] (c) 180 − 49 = 131° [1]. [1] Examiner insight: Substitute m back into both expressions. If they give different angles, the equation was set up wrongly, often by mixing up alternate and interior angles.

4. (a) PQ = PR, so angle PRQ = angle PQR = y + 31 (base angles of an isosceles triangle) [1]; 2(y + 31) + (4y − 2) = 180, so 6y + 60 = 180 [1]; y = 20 [1]. [3] (b) Angle PRQ = 51° [1], so angle PRS = 180 − 51 = 129° (angles on a straight line) [1]. [2] Examiner insight: The exterior angle fact gives a quick check: 129° should equal 51° + 78°. Use it to catch slips before you finalise.

5. (a) Isosceles and obtuse-angled [1]. [1] (b) All angles differ, so it is scalene [1]; it is also right-angled. [1] (c) 2 + 3 + 5 = 10 parts, so one part is 180 ÷ 10 = 18° [1]. The angles are 36°, 54° and 90° [1]. [2] Examiner insight: Give the names the specification uses. A question about a triangle may expect both a side type and an angle type, so read the wording before answering.

6. (a) ∠LNM (or ∠MNL) [1]. [1] (b) m [1]. [1] (c) AB ∥ DC and AD ⊥ AB [1]. [1] (d) 6 vertices and 6 edges [1]. [1] Examiner insight: In angle notation the middle letter is the vertex. Writing ∠LMN when you mean the angle at N changes the angle, so check the middle letter.

7. Sum of interior angles of an octagon = (8 − 2) × 180 = 1080° [1]. Seven angles total 135 + 142 + 128 + 150 + 117 + 139 + 146 = 957° [1]. Eighth angle = 1080 − 957 = 123° [1]. [3] Examiner insight: Use n − 2 = 6 triangles, not 8. Write the sum of all eight angles first so the method is easy to follow.

8. (a) Exterior angle = 360 ÷ 18 = 20° [1]. Interior angle = 180 − 20 = 160° [1]. [2] (b) Exterior angle = 180 − 156 = 24° [1]. n = 360 ÷ 24 = 15 [1]. [2] (c) Angle at the centre = 360 ÷ 15 = 24° [1]. [1] Examiner insight: Working through the exterior angle is shorter and less error-prone than forming an equation from (n − 2) × 180 ÷ n.

9. (a) AD ∥ BC, so angle ABC = 180 − 58 = 122° (interior angles) [1][1]. [2] (b) BD bisects angle ABC, so angle ABD = 122 ÷ 2 = 61° [1][1]. [2] (c) The diagonals of a rhombus cross at right angles, so angle AMB = 90° [1]. [1] Examiner insight: Say which property of the rhombus you use, such as “diagonals bisect the angles”. Triangle ABD (isosceles, since AB = AD) gives a second route to angle ABD.

10. (a) In a kite, angle PSR = angle PQR = 112° [1]. Angle QRS = 360 − 84 − 112 − 112 [1] = 52° [1]. [3] (b) Angle ADC = 180 − 71 = 109° (interior angles, AB ∥ DC) [1]. Let angle BCD = c, so angle ABC = 4c and 5c = 180 (interior angles) [1]; c = 36, so angle ABC = 144° [1]. [3] Examiner insight: In (a), say which pair of angles is equal. In (b), identify the parallel sides before you use interior angles, because the other two sides are not parallel, so those angles have no fixed total.

11. (a) Exterior angle = 360 ÷ 12 = 30° [1]. Interior angle = 180 − 30 = 150° [1]. [2] (b) Triangle ABC is isosceles because AB = BC [1]. Angle BAC = (180 − 150) ÷ 2 = 15° [1]. [2] (c) Angle LAC = 150 − 15 = 135° [1]. [1] (d) Going round the polygon, the direction of travel turns by 30° at each vertex [1]. From side AB to side GH is 6 turns, 6 × 30 = 180°, so GH runs in the opposite direction to AB, which makes them parallel [1]. [2] Examiner insight: In a “show that” question, the result is given, so every step to it needs to appear. State the exterior angle first, then subtract it from 180°.

Where marks are usually lost

• Not naming the angle fact used, such as “alternate” or “angles on a straight line”.
• Mixing up alternate angles (equal) and interior angles (total 180°).
• Using n instead of n − 2 in the interior angle sum.
• Dividing 360° by the interior angle to find the number of sides.
• Taking angles from the diagram when it may not be to scale.
• Naming an angle with the wrong middle letter.
• Giving only one name for a triangle when both a side type and an angle type apply.
• Stopping at the value of x when the question asks for an angle.

Next steps

• Review the revision notes and the study guide.
• Return to the OxfordAQA IGCSE Mathematics course page and tick off topics on the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.3.1, references G1 to G5.