Primary 6 PSLE Mathematics Angles Geometry Quiz

Answer Key: Primary 6 PSLE Mathematics Quiz - Angles Geometry

Section A: Multiple Choice (5 marks)

1. B) a and c

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

2. B) 35°

• Marks: 1
• Explanation: The angle sum of a triangle is 180°. Third angle = 180° - 65° - 80° = 35°.
• Common mistake: Students may forget the angle sum property or make an arithmetic error.

3. A) 60°

• Marks: 1
• Explanation: In a parallelogram, adjacent angles are supplementary (sum to 180°). Adjacent angle = 180° - 120° = 60°.
• Common mistake: Students may think opposite angles are supplementary (they are equal) or that adjacent angles are equal.

4. C) 360°

• Marks: 1
• Explanation: The angle sum of any quadrilateral is 360°. This can be derived by dividing the quadrilateral into two triangles (each 180°).
• Common mistake: Students may confuse this with the angle sum of a triangle (180°).

5. B) 180°

• Marks: 1
• Explanation: A straight line forms an angle of 180°. The line XY and its extension beyond Y form a straight line, so the angle between them is 180°.
• Common mistake: Students may think the angle is 0° or 90°.

Section B: Short-Answer Questions (20 marks)

6. ∠ACD = 145°

• Marks: 2 (1 mark for correct method, 1 mark for correct answer)
• Explanation: AB is a straight line, so ∠ACD + ∠DCB = 180° (angles on a straight line). ∠ACD = 180° - 35° = 145°.
• Working: 180° - 35° = 145°
• Common mistake: Students may subtract from 360° or 90° instead of 180°.

7. ∠QPR = 60°

• Marks: 2 (1 mark for correct method, 1 mark for correct answer)
• Explanation: Angle sum of triangle PQR = 180°. ∠QPR = 180° - 55° - 65° = 60°.
• Working: 180° - 55° - 65° = 60°
• Common mistake: Arithmetic error in subtraction.

8. 70°, 110°, 110°

• Marks: 2 (1 mark for identifying opposite angle, 1 mark for adjacent angles)
• Explanation: In a rhombus, opposite angles are equal, and adjacent angles are supplementary. The given angle is 70°, so the opposite angle is also 70°. The other two angles are each 180° - 70° = 110°.
• Working: Opposite angle = 70°. Adjacent angles = 180° - 70° = 110° each.
• Common mistake: Students may think all angles are equal (like a square) or only give two angles.

9. ∠BOD = 42°

• Marks: 2 (1 mark for identifying vertically opposite angles, 1 mark for correct answer)
• Explanation: When two lines intersect, vertically opposite angles are equal. ∠AOC and ∠BOD are vertically opposite, so ∠BOD = ∠AOC = 42°.
• Working: ∠BOD = ∠AOC = 42°
• Common mistake: Students may think they are supplementary (sum to 180°).

10. x = 30

• Marks: 2 (1 mark for setting up equation, 1 mark for solving correctly)
• Explanation: Angle sum of triangle = 180°. (x + 10) + (2x) + (3x - 10) = 180°. Simplify: 6x = 180°. x = 30.
• Working:
  • (x + 10) + (2x) + (3x - 10) = 180°
  • 6x = 180°
  • x = 30
• Common mistake: Students may forget to include all three angles or make an algebraic error.

11. ∠FGH = 72°

• Marks: 2 (1 mark for identifying corresponding angles, 1 mark for correct answer)
• Explanation: When a transversal crosses parallel lines, corresponding angles are equal. ∠EFG and ∠FGH are corresponding angles, so ∠FGH = ∠EFG = 72°.
• Working: ∠FGH = ∠EFG = 72° (corresponding angles, AB ∥ CD)
• Common mistake: Students may confuse corresponding angles with alternate or interior angles.

12. a = 110°

• Marks: 2 (1 mark for using angle sum of quadrilateral, 1 mark for correct answer)
• Explanation: Angle sum of quadrilateral = 360°. a = 360° - 80° - 95° - 75° = 110°.
• Working: 360° - 80° - 95° - 75° = 110°
• Common mistake: Arithmetic error or using 180° instead of 360°.

13. ∠ACB = 30°

• Marks: 2 (1 mark for identifying angle at centre theorem, 1 mark for correct answer)
• Explanation: The angle at the centre of a circle is twice the angle at the circumference subtended by the same arc. ∠AOB = 2 × ∠ACB. So ∠ACB = 60° ÷ 2 = 30°.
• Working: ∠ACB = ∠AOB ÷ 2 = 60° ÷ 2 = 30°
• Common mistake: Students may think the angle at the circumference is equal to the angle at the centre.

14. y = 10

• Marks: 2 (1 mark for setting up equation, 1 mark for solving correctly)
• Explanation: Vertically opposite angles are equal. 3y = y + 20. 2y = 20. y = 10.
• Working:
  • 3y = y + 20
  • 2y = 20
  • y = 10
• Common mistake: Students may think the angles are supplementary (sum to 180°).

15. x = 70°

• Marks: 2 (1 mark for using isosceles triangle property, 1 mark for correct answer)
• Explanation: In an isosceles triangle, base angles are equal. Angle sum = 180°. 40° + x + x = 180°. 2x = 140°. x = 70°.
• Working:
  • 40° + 2x = 180°
  • 2x = 140°
  • x = 70°
• Common mistake: Students may forget that base angles are equal or make an arithmetic error.

Section C: Problem-Solving Questions (20 marks)

16. ∠BCD = 70°, ∠CDA = 105°

• Marks: 4 (2 marks for each angle: 1 mark for correct method, 1 mark for correct answer)
• Explanation:
  • In a trapezium with AB ∥ DC, interior angles on the same side of a transversal are supplementary.
  • ∠DAB + ∠CDA = 180° (angles on same side of transversal AD). So ∠CDA = 180° - 75° = 105°.
  • ∠ABC + ∠BCD = 180° (angles on same side of transversal BC). So ∠BCD = 180° - 110° = 70°.
• Working:
  • ∠CDA = 180° - 75° = 105°
  • ∠BCD = 180° - 110° = 70°
• Common mistake: Students may think opposite angles are supplementary (they are not in a trapezium) or confuse which angles are on the same side of the transversal.

17. p = 120°

• Marks: 4 (2 marks for correct method, 2 marks for correct answer)
• Explanation: The exterior angle of a triangle is equal to the sum of the two opposite interior angles. The exterior angle p is adjacent to the 60° angle, so the two opposite interior angles are 50° and 70°. p = 50° + 70° = 120°.
• Working: p = 50° + 70° = 120°
• Alternative method: Interior angle adjacent to p = 60°. Angles on a straight line: p + 60° = 180°. p = 120°.
• Common mistake: Students may add the adjacent angle (60°) instead of the opposite angles, or think the exterior angle equals one interior angle.

18. ∠ACB = 60°

• Marks: 4 (2 marks for identifying angle at centre theorem, 2 marks for correct answer)
• Explanation: The angle at the centre of a circle is twice the angle at the circumference subtended by the same arc. ∠AOB is the angle at the centre subtended by arc AB. ∠ACB is the angle at the circumference subtended by the same arc AB. So ∠ACB = ∠AOB ÷ 2 = 120° ÷ 2 = 60°.
• Working: ∠ACB = 120° ÷ 2 = 60°
• Common mistake: Students may think ∠ACB = ∠AOB (equal) or use the wrong relationship. Note: The fact that OA = OB = OC is always true for radii and does not affect the calculation.

19. q = 120°

• Marks: 4 (2 marks for using angle sum of pentagon, 2 marks for correct answer)
• Explanation: The angle sum of a pentagon (5-sided polygon) is (5 - 2) × 180° = 3 × 180° = 540°. q = 540° - 100° - 120° - 90° - 110° = 120°.
• Working:
  • Angle sum of pentagon = (5 - 2) × 180° = 540°
  • q = 540° - 100° - 120° - 90° - 110° = 120°
• Common mistake: Students may use the wrong formula (e.g., (n-1) × 180°) or make an arithmetic error.

20. ∠GHI = 80°

• Marks: 4 (2 marks for correct method using parallel lines, 2 marks for correct answer)
• Explanation:
  • Since AB ∥ CD, ∠EFG and ∠FGH are angles on the same side of the transversal FG. They are not directly related by a simple rule.
  • We need to find ∠GHI. Consider the triangle FGH. We know ∠FGH = 42°.
  • To find ∠GFH, note that ∠EFG = 58° and AB ∥ CD. ∠GFH is the alternate angle to ∠FGH? No, let's reconsider.
  • Actually, ∠EFG and ∠FGH are interior angles on the same side of the transversal FG. Since AB ∥ CD, interior angles on the same side of a transversal are supplementary. So ∠EFG + ∠FGH should be 180°? But 58° + 42° = 100°, not 180°. This means FG is not a transversal between the parallel lines in the usual sense.
  • Let's re-examine the diagram. E is on AB above F, F is on AB, G is on CD, H is on CD. So line EG is a transversal crossing AB at F and CD at G. ∠EFG is the angle between the transversal and AB at F. ∠FGH is the angle between the transversal and CD at G.
  • Since AB ∥ CD, alternate angles are equal. ∠EFG and ∠FGD are alternate angles, so ∠FGD = 58°.
  • Now, ∠FGH = 42°. So ∠HGD = ∠FGD - ∠FGH = 58° - 42° = 16°.
  • Now, line HI is another transversal. ∠GHI is the angle between this transversal and CD at H. Since AB ∥ CD, ∠GHI and ∠GFH are corresponding angles? No, let's think differently.
  • Actually, we can use the angle sum of triangle FGH. We know ∠FGH = 42°. We need ∠GFH.
  • ∠GFH is the angle between line FG and line FH. Since F is on AB and H is on CD, and AB ∥ CD, ∠GFH and ∠FHD are corresponding angles? Not directly.
  • Let's use a different approach. Consider the triangle formed by the two transversals and the parallel lines. The angles at F and G are known. The angle at H (∠GHI) is what we need.
  • In triangle FGH, ∠FGH = 42°. ∠GFH = ∠EFG = 58° (alternate angles, AB ∥ CD). So ∠GHI = 180° - 42° - 58° = 80°.
• Working:
  • ∠GFH = ∠EFG = 58° (alternate angles, AB ∥ CD)
  • In triangle FGH: ∠GHI = 180° - ∠FGH - ∠GFH = 180° - 42° - 58° = 80°
• Common mistake: Students may struggle to identify the correct angle relationships or try to use the wrong parallel line rule.