Primary 6 PSLE Mathematics Area Perimeter Quiz

Answer Key - Primary 6 PSLE Mathematics Quiz - Area Perimeter

Section A: Multiple Choice Questions (5 marks)

1. B) 8 cm (1 mark)

• Perimeter of rectangle = 2 × (length + breadth)
• 40 = 2 × (12 + breadth)
• 12 + breadth = 20
• breadth = 8 cm
• Common mistake: Students may forget to divide by 2 first.

2. C) 32 cm (1 mark)

• Area of square = side × side = 64 cm²
• side = √64 = 8 cm
• Perimeter = 4 × side = 4 × 8 = 32 cm
• Common mistake: Students may confuse area and perimeter formulas.

3. B) 40 cm² (1 mark)

• Area of triangle = ½ × base × height
• = ½ × 10 × 8 = 40 cm²
• Common mistake: Students may forget to multiply by ½.

4. B) 35.42 cm (1 mark)

• Perimeter of figure = perimeter of rectangle (excluding one breadth) + circumference of semicircle
• Rectangle perimeter (3 sides) = 10 + 6 + 10 = 26 cm
• Circumference of semicircle = ½ × π × d = ½ × 3.14 × 6 = 9.42 cm
• Total perimeter = 26 + 9.42 = 35.42 cm
• Common mistake: Students may include the full rectangle perimeter or forget that the diameter side is not part of the outer perimeter.

5. B) 8 cm (1 mark)

• Area of parallelogram = base × height
• 120 = 15 × height
• height = 120 ÷ 15 = 8 cm
• Common mistake: Students may confuse the formula with that of a rectangle.

Section B: Short-Answer Questions (20 marks)

6. Area of path = 156 m² (2 marks)

• Outer area (garden) = 25 × 18 = 450 m²
• Inner rectangle dimensions: length = 25 - 2 - 2 = 21 m, breadth = 18 - 2 - 2 = 14 m
• Inner area = 21 × 14 = 294 m²
• Area of path = 450 - 294 = 156 m²
• Marking: 1 mark for correct inner dimensions, 1 mark for correct final answer.
• Common mistake: Students may subtract the path width only once instead of twice.

7. Area of one square = 64 cm² (2 marks)

• The figure has 6 equal sides in its perimeter (the shared side is not counted).
• Each side = 48 ÷ 6 = 8 cm
• Area of one square = 8 × 8 = 64 cm²
• Marking: 1 mark for finding side length, 1 mark for correct area.
• Common mistake: Students may count all 8 sides of the two squares.

8. Radius = 10 cm (2 marks)

• Circumference = 2πr
• 62.8 = 2 × 3.14 × r
• 62.8 = 6.28 × r
• r = 62.8 ÷ 6.28 = 10 cm
• Marking: 1 mark for correct substitution, 1 mark for correct answer.
• Common mistake: Students may use the formula for area instead of circumference.

9. Area of square = 144 cm² (2 marks)

• Perimeter of rectangle = 2 × (15 + 9) = 2 × 24 = 48 cm
• This is also the perimeter of the square.
• Side of square = 48 ÷ 4 = 12 cm
• Area of square = 12 × 12 = 144 cm²
• Marking: 1 mark for finding perimeter/side, 1 mark for correct area.
• Common mistake: Students may forget that the wire length remains the same.

10. Area of triangle ABC = 24 cm² (2 marks)

• In a right-angled triangle, the two shorter sides are the base and height.
• Area = ½ × AB × BC = ½ × 6 × 8 = 24 cm²
• Marking: 1 mark for identifying base and height, 1 mark for correct answer.
• Common mistake: Students may use AC (the hypotenuse) as the height.

11. Area of remaining paper = 536 cm² (2 marks)

• Area of original paper = 30 × 20 = 600 cm²
• Area of one square = 4 × 4 = 16 cm²
• Area of four squares = 4 × 16 = 64 cm²
• Remaining area = 600 - 64 = 536 cm²
• Marking: 1 mark for finding area of squares, 1 mark for correct final answer.
• Common mistake: Students may forget to multiply by 4.

12. Area of shaded region = 42 cm² (2 marks)

• Area of square = 14 × 14 = 196 cm²
• Radius of circle = 14 ÷ 2 = 7 cm
• Area of circle = πr² = (22/7) × 7 × 7 = 154 cm²
• Shaded area = 196 - 154 = 42 cm²
• Marking: 1 mark for correct circle area, 1 mark for correct final answer.
• Common mistake: Students may use diameter instead of radius in the circle area formula.

13. Area of trapezium = 60 cm² (2 marks)

• Area = ½ × (sum of parallel sides) × height
• = ½ × (12 + 8) × 6
• = ½ × 20 × 6 = 60 cm²
• Marking: 1 mark for correct formula substitution, 1 mark for correct answer.
• Common mistake: Students may forget to multiply by ½.

14. Area of rectangle = 147 cm² (2 marks)

• Let breadth = b, then length = 3b
• Perimeter = 2 × (3b + b) = 2 × 4b = 8b
• 8b = 56, so b = 7 cm
• Length = 3 × 7 = 21 cm
• Area = 21 × 7 = 147 cm²
• Marking: 1 mark for finding dimensions, 1 mark for correct area.
• Common mistake: Students may forget to multiply by 2 in the perimeter formula.

15. Perimeter of semicircle = 36 cm (2 marks)

• Perimeter = curved part + diameter
• Curved part = ½ × 2πr = πr = (22/7) × 7 = 22 cm
• Diameter = 2 × 7 = 14 cm
• Total perimeter = 22 + 14 = 36 cm
• Marking: 1 mark for curved part, 1 mark for correct total.
• Common mistake: Students may forget to add the diameter.

Section C: Problem-Solving Questions (20 marks)

16. New height of water = 20.5 cm (4 marks)

• Base area of tank = 50 × 40 = 2000 cm²
• Volume of water = 2000 × 20 = 40,000 cm³
• Volume of cube = 10 × 10 × 10 = 1000 cm³
• Total volume (water + cube) = 40,000 + 1000 = 41,000 cm³
• New height = 41,000 ÷ 2000 = 20.5 cm
• Marking:
  • 1 mark for base area
  • 1 mark for volume of water
  • 1 mark for volume of cube
  • 1 mark for correct final answer
• Common mistake: Students may add the cube's side length to the water height instead of calculating volume displacement.

17. Area of triangle TUR = 24 cm² (4 marks)

• The rectangle PQRS has length PS = QR = PT + TS = 4 + 6 = 10 cm
• Width PQ = SR = QU + UR = 8 + 4 = 12 cm
• Area of rectangle = 10 × 12 = 120 cm²
• Area of triangle PTU = ½ × PT × QU = ½ × 4 × 8 = 16 cm²
• Area of triangle TSR = ½ × TS × SR = ½ × 6 × 12 = 36 cm²
• Area of triangle UQR = ½ × UR × QR = ½ × 4 × 10 = 20 cm²
• Area of triangle TUR = 120 - 16 - 36 - 20 = 48 cm²
• Alternative method: Area of triangle TUR = ½ × base TU × height (perpendicular distance from R to TU). Using coordinates: T(0,4), U(12,8), R(12,0). Area = ½ × |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)| = ½ × |0(8-0) + 12(0-4) + 12(4-8)| = ½ × |0 - 48 - 48| = 48 cm²
• Marking:
  • 1 mark for finding rectangle dimensions
  • 1 mark for area of rectangle
  • 1 mark for finding areas of the three right triangles
  • 1 mark for correct final answer
• Common mistake: Students may incorrectly identify the base and height of triangle TUR.

18. Total area of flower bed = 178.465 m² (4 marks)

• Area of rectangle = 20 × 7 = 140 m²
• Radius of each semicircle = 7 ÷ 2 = 3.5 m
• Area of one semicircle = ½ × πr² = ½ × 3.14 × 3.5² = ½ × 3.14 × 12.25 = 19.2325 m²
• Area of two semicircles = 2 × 19.2325 = 38.465 m²
• Total area = 140 + 38.465 = 178.465 m²
• Marking:
  • 1 mark for rectangle area
  • 1 mark for radius
  • 1 mark for area of semicircles
  • 1 mark for correct final answer
• Common mistake: Students may forget to divide the diameter by 2 to get the radius.

19. Breadth of rectangle = 8 cm (4 marks)

• Total wire length = 1 m = 100 cm
• Perimeter of square = 4 × 12 = 48 cm
• Remaining wire for rectangle = 100 - 48 = 52 cm
• Perimeter of rectangle = 2 × (length + breadth)
• 52 = 2 × (18 + breadth)
• 18 + breadth = 26
• breadth = 8 cm
• Marking:
  • 1 mark for converting metres to centimetres
  • 1 mark for perimeter of square
  • 1 mark for remaining wire length
  • 1 mark for correct final answer
• Common mistake: Students may forget to convert 1 m to 100 cm.

20. Length of chord AB = 16 cm (4 marks)

• In triangle OAM, O is the centre, M is the midpoint of AB.
• OM is perpendicular to AB, so triangle OAM is right-angled at M.
• OA = radius = 10 cm, OM = 6 cm
• Using Pythagoras' theorem: AM² = OA² - OM² = 10² - 6² = 100 - 36 = 64
• AM = √64 = 8 cm
• AB = 2 × AM = 2 × 8 = 16 cm
• Marking:
  • 1 mark for identifying the right triangle
  • 1 mark for correct application of Pythagoras' theorem
  • 1 mark for finding AM
  • 1 mark for correct final answer (AB = 2 × AM)
• Common mistake: Students may forget to multiply AM by 2 to get the full chord length.

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