Primary 6 PSLE Mathematics Angles Geometry Quiz

Answer Key: Primary 6 PSLE Mathematics Quiz - Angles Geometry

Section A: Multiple-Choice Questions

1. B) a and c

• Marks: 1
• Explanation: Vertically opposite angles are the angles opposite each other when two lines intersect. They are equal. In the diagram, angles a and c are opposite each other, so they are vertically opposite angles.
• Common mistake: Students may confuse adjacent angles (a and b) with vertically opposite angles.

2. B) 60°

• Marks: 1
• Explanation: The sum of angles in a triangle is 180°. Third angle = 180° - 35° - 85° = 60°.
• Common mistake: Students may forget that the sum is 180° and subtract from 360° instead.

3. B) 45°

• Marks: 1
• Explanation: A right-angled isosceles triangle has one 90° angle and two equal angles. The two equal angles sum to 90° (since 180° - 90° = 90°). Each equal angle = 90° ÷ 2 = 45°.
• Common mistake: Students may think the equal angles are 60° each (confusing with equilateral triangle).

4. B) 70°

• Marks: 1
• Explanation: AOB is a straight line, so ∠AOB = 180°. ∠AOC + ∠COD + ∠BOD = 180°. So 62° + 48° + ∠BOD = 180°. ∠BOD = 180° - 62° - 48° = 70°.
• Common mistake: Students may forget that angles on a straight line sum to 180°.

5. B) All sides are equal.

• Marks: 1
• Explanation: A rhombus is a quadrilateral with all four sides equal in length. While a square is a special type of rhombus, a general rhombus does not have equal angles or right angles. A trapezium has one pair of parallel sides, but a rhombus has two pairs of parallel sides.
• Common mistake: Students may think all angles are equal (which is only true for a square).

Section B: Short-Answer Questions

6. ∠x = 72°

• Marks: 2 (1 mark for correct method, 1 mark for correct answer)
• Explanation: AB is a straight line, so the angles on the line sum to 180°. x + 108° = 180°. x = 180° - 108° = 72°.
• Key concept: Angles on a straight line add up to 180°.

7. ∠QPR = 52°

• Marks: 2 (1 mark for correct method, 1 mark for correct answer)
• Explanation: In triangle PQR, the sum of angles is 180°. ∠QPR + ∠PQR + ∠PRQ = 180°. ∠QPR + 90° + 38° = 180°. ∠QPR = 180° - 90° - 38° = 52°.
• Key concept: Sum of angles in a triangle is 180°.

8. ∠BCD = 65°

• Marks: 2 (1 mark for correct method, 1 mark for correct answer)
• Explanation: In a parallelogram, adjacent angles are supplementary (sum to 180°). ∠ABC + ∠BCD = 180°. 115° + ∠BCD = 180°. ∠BCD = 180° - 115° = 65°.
• Key concept: Adjacent angles in a parallelogram are supplementary.
• Alternative method: Opposite angles in a parallelogram are equal, so ∠BCD = ∠DAB. But we don't know ∠DAB directly. Using supplementary angles is more direct.

9. ∠ACB = 36°

• Marks: 2 (1 mark for correct method, 1 mark for correct answer)
• Explanation: The angle at the centre (∠AOB) is twice the angle at the circumference (∠ACB) subtended by the same arc AB. So ∠ACB = ∠AOB ÷ 2 = 72° ÷ 2 = 36°.
• Key concept: Angle at centre is twice angle at circumference (for the same arc).

10. ∠RSP = 78°

• Marks: 2 (1 mark for correct method, 1 mark for correct answer)
• Explanation: In a trapezium with PQ ∥ SR, interior angles on the same side of a transversal are supplementary. Consider transversal QR: ∠PQR + ∠QRS = 78° + 102° = 180°. This confirms PQ ∥ SR. Now consider transversal RS: ∠RSP + ∠SRQ = 180°. ∠RSP + 102° = 180°. ∠RSP = 78°.
• Key concept: In a trapezium, angles between a leg and each base are supplementary if the bases are parallel.

11. ∠a = 135°

• Marks: 2 (1 mark for correct method, 1 mark for correct answer)
• Explanation: When two lines intersect, adjacent angles are supplementary (sum to 180°). So a + b = 180°. Given a = 3b, substitute: 3b + b = 180°. 4b = 180°. b = 45°. Then a = 3 × 45° = 135°.
• Key concept: Adjacent angles on intersecting lines are supplementary.

12. ∠ABC = 70°

• Marks: 2 (1 mark for correct method, 1 mark for correct answer)
• Explanation: AB = AC, so triangle ABC is isosceles with base BC. Base angles are equal: ∠ABC = ∠ACB. Sum of angles: ∠BAC + ∠ABC + ∠ACB = 180°. 40° + 2 × ∠ABC = 180°. 2 × ∠ABC = 140°. ∠ABC = 70°.
• Key concept: In an isosceles triangle, base angles are equal.

13. ∠x = 150°

• Marks: 2 (1 mark for correct method, 1 mark for correct answer)
• Explanation: The square has interior angles of 90°. The equilateral triangle has interior angles of 60°. At the junction, the full angle around the point is 360°. The angle inside the square is 90°, inside the triangle is 60°. So x = 360° - 90° - 60° = 210°. Wait, let's reconsider the diagram. The angle x is the exterior angle at the junction. The interior angles of the square and triangle at the shared vertex are 90° and 60° respectively. The angle between the square and triangle on the outside is x. So x = 360° - 90° - 60° = 210°. But this seems too large. Let's reconsider: If the square and triangle share a side, the angle between the square's top side and the triangle's side is x. The square's interior angle is 90°, the triangle's interior angle is 60°. The angle between the square's top and the triangle's side is 360° - 90° - 60° = 210°. However, if x is the angle on the outside of the figure, it might be the reflex angle. If x is the smaller angle between the square and triangle, then x = 90° + 60° = 150°.
• Corrected explanation: The square has interior angles of 90°. The equilateral triangle has interior angles of 60°. At the junction, the angle between the square's side and the triangle's side is the sum of the interior angles: 90° + 60° = 150°.
• Key concept: Understanding how shapes join and calculating angles at junctions.

14. ∠FGH = 55°

• Marks: 2 (1 mark for correct method, 1 mark for correct answer)
• Explanation: AB ∥ CD. ∠EFG = 125°. ∠EFG and ∠FGH are interior angles on the same side of the transversal FG. They are supplementary. So ∠FGH = 180° - 125° = 55°.
• Key concept: When a transversal intersects parallel lines, interior angles on the same side of the transversal are supplementary.

15. 540°

• Marks: 2 (1 mark for correct method, 1 mark for correct answer)
• Explanation: A regular pentagon has 5 sides. Sum of interior angles = (n - 2) × 180° = (5 - 2) × 180° = 3 × 180° = 540°. Alternatively, since each interior angle is 108°, sum = 5 × 108° = 540°.
• Key concept: Sum of interior angles of a polygon = (n - 2) × 180°.

Section C: Problem-Solving Questions

16. ∠DBE = 80°

• Marks: 4 (1 mark for identifying angles on straight line, 1 mark for finding ∠DBC, 1 mark for finding ∠EBA, 1 mark for correct answer)
• Explanation:
  • ABC is a straight line, so ∠ABD + ∠DBC = 180°. 130° + ∠DBC = 180°. ∠DBC = 50°.
  • Also, ∠ABE + ∠EBC = 180°. ∠ABE + 150° = 180°. ∠ABE = 30°.
  • Now, ∠ABD = ∠ABE + ∠EBD. 130° = 30° + ∠EBD. ∠EBD = 100°.
  • Alternatively, ∠DBE = 180° - ∠DBC - ∠EBA = 180° - 50° - 30° = 100°.
  • Wait, let's re-examine. ∠ABD = 130° means the angle from BA to BD is 130°. ∠CBE = 150° means the angle from BC to BE is 150°. The angle DBE is the angle between BD and BE. ∠DBC = 180° - 130° = 50°. ∠EBA = 180° - 150° = 30°. ∠DBE = 180° - 50° - 30° = 100°.
• Corrected answer: ∠DBE = 100°.
• Key concept: Angles on a straight line sum to 180°.

17. ∠PTR = 110°

• Marks: 4 (1 mark for identifying isosceles triangle, 1 mark for finding base angles, 1 mark for finding ∠PTS, 1 mark for correct answer)
• Explanation:
  • PQRS is a rectangle, so all angles are 90°. PQ = SR.
  • PT = PQ, so PT = PQ = SR.
  • In triangle PQT, PT = PQ, so it's isosceles. ∠PQT = ∠PTQ.
  • ∠QPT = 20°. So ∠PQT + ∠PTQ = 180° - 20° = 160°. Each base angle = 80°.
  • ∠PQR = 90°. So ∠TQR = 90° - 80° = 10°.
  • In rectangle, QR ∥ PS. ∠PTR is the angle between PT and TR.
  • ∠PTS = 180° - ∠PTQ = 180° - 80° = 100° (angles on a straight line).
  • In triangle PTS, ∠PST = 90° (rectangle). ∠SPT = 90° - 20° = 70°.
  • ∠PTS = 180° - 90° - 70° = 20°.
  • ∠PTR = 180° - ∠PTS = 180° - 20° = 160°.
  • Wait, let's reconsider. T is on SR. ∠PTR is the angle at T between PT and TR. ∠PTS is the angle at T between PT and TS. Since S, T, R are collinear, ∠PTS + ∠PTR = 180°. So ∠PTR = 180° - ∠PTS.
  • ∠PTS = 180° - ∠PTQ = 180° - 80° = 100°.
  • ∠PTR = 180° - 100° = 80°.
• Corrected answer: ∠PTR = 80°.
• Key concept: Properties of rectangles, isosceles triangles, and angles on a straight line.

18. ∠EDC = 15°

• Marks: 4 (1 mark for identifying equilateral triangle properties, 1 mark for finding angles in square, 1 mark for identifying isosceles triangles, 1 mark for correct answer)
• Explanation:
  • ABCD is a square, so all sides are equal and all angles are 90°.
  • Triangle ABE is equilateral, so AB = AE = BE, and all angles in triangle ABE are 60°.
  • Since AB = AD (square), and AB = AE (equilateral), we have AD = AE. So triangle ADE is isosceles with AD = AE.
  • ∠DAE = ∠DAB - ∠EAB = 90° - 60° = 30°.
  • In isosceles triangle ADE, base angles are equal: ∠ADE = ∠AED.
  • Sum of angles in triangle ADE: ∠DAE + ∠ADE + ∠AED = 180°. 30° + 2 × ∠ADE = 180°. 2 × ∠ADE = 150°. ∠ADE = 75°.
  • ∠ADC = 90° (square). So ∠EDC = ∠ADC - ∠ADE = 90° - 75° = 15°.
• Key concept: Properties of squares, equilateral triangles, and isosceles triangles.

19. ∠APB = 110°

• Marks: 4 (1 mark for finding arc measures, 1 mark for finding inscribed angles, 1 mark for using exterior angle theorem, 1 mark for correct answer)
• Explanation:
  • The angle at the centre is twice the angle at the circumference. ∠AOB = 80°, so ∠ACB = 40° (angle subtended by arc AB). ∠COD = 60°, so ∠CBD = 30° (angle subtended by arc CD).
  • In triangle PBC, ∠APB is an exterior angle. ∠APB = ∠PCB + ∠PBC = ∠ACB + ∠CBD = 40° + 30° = 70°.
  • Wait, ∠APB is the angle at P between PA and PB. In triangle PBC, the exterior angle at P is ∠APB. The two remote interior angles are ∠PCB and ∠PBC. ∠PCB = ∠ACB = 40°. ∠PBC = ∠CBD = 30°. So ∠APB = 40° + 30° = 70°.
• Corrected answer: ∠APB = 70°.
• Key concept: Angle at centre is twice angle at circumference. Exterior angle of a triangle equals sum of two remote interior angles.

20. ∠ECD = 65°

• Marks: 4 (1 mark for careful reading, 3 marks for showing working)
• Explanation: This is a trick question. The value of ∠ECD is given in the question as 65°. The question asks "Find ∠ECD" and the diagram shows ∠ECD = 65°. The correct answer is simply 65°.
• Working (for completeness):
  • AB ∥ CD. ∠BAE = 55°. ∠AEC = 120°.
  • ∠EAB = 55° (alternate angles with AB ∥ CD, but careful: alternate angles are equal when a transversal crosses parallel lines. Here, AE is a transversal, so ∠EAB = ∠AED? No, alternate angles are between the transversal and the parallel lines. ∠BAE and ∠AED are alternate angles. So ∠AED = 55°.
  • ∠AEC = 120°. So ∠DEC = 180° - 120° = 60° (angles on a straight line).
  • In triangle CDE, ∠ECD + ∠CDE + ∠DEC = 180°. ∠ECD + 55° + 60° = 180°. ∠ECD = 65°.
• Key concept: Always read the question carefully. Sometimes the answer is directly given.