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Primary 5 Mathematics Area Perimeter Quiz
Free P5 Maths Area Perimeter quiz, Exam version, with questions, answers, and syllabus-aligned practice for Singapore students.
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Primary 5 Mathematics Quiz - Area Perimeter: Answer Key
Total Marks: 50
Section A: Multiple Choice Questions (Questions 1 to 5)
Each question carries 2 marks.
1. Answer: C) 96 cm²
- Working: Area of rectangle = length × breadth = 12 cm × 8 cm = 96 cm².
- Explanation: Area is measured in square units (cm²). We multiply the length and breadth.
- Marking: 1 mark for correct method, 1 mark for correct answer and units.
2. Answer: C) 9 cm
- Working: Perimeter of square = 4 × side. So, side = Perimeter ÷ 4 = 36 cm ÷ 4 = 9 cm.
- Explanation: A square has 4 equal sides. The perimeter is the total distance around the shape.
- Marking: 1 mark for correct method, 1 mark for correct answer.
3. Answer: B) 56 cm²
- Working: Area of rectangle = 10 cm × 4 cm = 40 cm². Area of square = 4 cm × 4 cm = 16 cm². Total area = 40 cm² + 16 cm² = 56 cm².
- Explanation: The total area of a composite figure is the sum of the areas of its parts.
- Marking: 1 mark for correct area of rectangle, 1 mark for correct total area.
4. Answer: B) 60 cm²
- Working: Area of triangle = ½ × base × height = ½ × 15 cm × 8 cm = 60 cm².
- Explanation: The area of a triangle is half the area of a rectangle with the same base and height.
- Marking: 1 mark for correct formula, 1 mark for correct answer.
5. Answer: A) 10 cm
- Working: Perimeter of rectangle = 2 × (length + breadth). So, 48 cm = 2 × (14 cm + breadth). 14 cm + breadth = 24 cm. Breadth = 24 cm - 14 cm = 10 cm.
- Explanation: We work backwards from the perimeter formula to find the missing dimension.
- Marking: 1 mark for correct method, 1 mark for correct answer.
Section B: Short Answer Questions (Questions 6 to 15)
Each question carries 3 marks.
6. Answer: 300 m²
- Working: Area = length × breadth = 25 m × 12 m = 300 m².
- Explanation: The area of a rectangle is found by multiplying its length and breadth.
- Marking: 1 mark for correct formula, 1 mark for correct substitution, 1 mark for correct answer and units.
7. Answer: 28 cm
- Working: Length of wire = Perimeter of square = 4 × side = 4 × 7 cm = 28 cm.
- Explanation: The wire is bent into a square, so the length of the wire is equal to the perimeter of the square.
- Marking: 1 mark for correct method, 1 mark for correct answer, 1 mark for correct units.
8. Answer: 15 cm²
- Working: Area of triangle = ½ × base × height = ½ × 6 cm × 5 cm = 15 cm².
- Explanation: The base and height must be perpendicular to each other. Here, they are.
- Marking: 1 mark for correct formula, 1 mark for correct substitution, 1 mark for correct answer.
9. Answer: 9 cm
- Working: Area of rectangle = length × breadth. So, length = Area ÷ breadth = 72 cm² ÷ 8 cm = 9 cm.
- Explanation: We rearrange the area formula to find the missing length.
- Marking: 1 mark for correct method, 1 mark for correct answer, 1 mark for correct units.
10. Answer: 185 cm²
- Working: Area of square = 15 cm × 15 cm = 225 cm². Area of rectangle = 8 cm × 5 cm = 40 cm². Remaining area = 225 cm² - 40 cm² = 185 cm².
- Explanation: The area of the remaining paper is the area of the square minus the area of the rectangle cut out.
- Marking: 1 mark for area of square, 1 mark for area of rectangle, 1 mark for correct remaining area.
11. Answer: 54 cm²
- Working: Area of Rectangle A = 9 cm × 4 cm = 36 cm². Area of Rectangle B = 6 cm × 3 cm = 18 cm². Total area = 36 cm² + 18 cm² = 54 cm².
- Explanation: The total area is the sum of the areas of the two rectangles.
- Marking: 1 mark for area of A, 1 mark for area of B, 1 mark for correct total.
12. Answer: 304 m²
- Working: Area of field = 50 m × 30 m = 1500 m². The inner rectangle (field minus path) has length = 50 m - 2 m - 2 m = 46 m and width = 30 m - 2 m - 2 m = 26 m. Area of inner rectangle = 46 m × 26 m = 1196 m². Area of path = 1500 m² - 1196 m² = 304 m².
- Explanation: The path runs inside the field, so we subtract the area of the inner rectangle from the area of the field.
- Marking: 1 mark for area of field, 1 mark for correct inner dimensions, 1 mark for correct path area.
13. Answer: The triangle has a larger area by 30 cm².
- Working: Area of rectangle = 10 cm × 6 cm = 60 cm². Area of triangle = ½ × 10 cm × 12 cm = 60 cm². Difference = 60 cm² - 60 cm² = 0 cm². (Correction: Both have the same area.)
- Explanation: The triangle's area is ½ × base × height = ½ × 10 × 12 = 60 cm². The rectangle's area is 10 × 6 = 60 cm². They are equal.
- Marking: 1 mark for area of rectangle, 1 mark for area of triangle, 1 mark for correct comparison.
14. Answer: 250 m
- Working: Perimeter of rectangle = 2 × (length + breadth) = 2 × (80 m + 45 m) = 2 × 125 m = 250 m.
- Explanation: The length of the fence is the perimeter of the rectangular land.
- Marking: 1 mark for correct formula, 1 mark for correct substitution, 1 mark for correct answer.
15. Answer: 88 cm²
- Working: Area of square = 8 cm × 8 cm = 64 cm². Area of triangle = ½ × 8 cm × 6 cm = 24 cm². Total area = 64 cm² + 24 cm² = 88 cm².
- Explanation: The total area is the sum of the areas of the square and the triangle.
- Marking: 1 mark for area of square, 1 mark for area of triangle, 1 mark for correct total.
Section C: Problem-Solving Questions (Questions 16 to 20)
Each question carries 4 marks.
16. Answer: 20 litres
- Working: Volume of water = length × width × height of water = 40 cm × 25 cm × 20 cm = 20,000 cm³. Since 1 litre = 1000 cm³, volume in litres = 20,000 cm³ ÷ 1000 = 20 litres.
- Explanation: The volume of water is the volume of the rectangular prism formed by the water. We then convert cubic centimetres to litres.
- Marking: 1 mark for correct volume in cm³, 1 mark for correct conversion factor, 1 mark for correct answer in litres, 1 mark for correct units.
17. Answer: 20 cm
- Working: Total length of wire = 1 m = 100 cm. Perimeter of rectangle = 2 × (length + breadth) = 100 cm. So, 2 × (30 cm + breadth) = 100 cm. 30 cm + breadth = 50 cm. Breadth = 50 cm - 30 cm = 20 cm.
- Explanation: The wire length is the perimeter. We convert metres to centimetres first, then solve for the breadth.
- Marking: 1 mark for correct conversion, 1 mark for correct perimeter formula, 1 mark for correct method, 1 mark for correct answer.
18. Answer: 176 cm²
- Working: Area of rectangle = 18 cm × 12 cm = 216 cm². Area of triangle = ½ × 10 cm × 8 cm = 40 cm². Area of shaded part = 216 cm² - 40 cm² = 176 cm².
- Explanation: The shaded area is the area of the rectangle minus the area of the triangle cut out.
- Marking: 1 mark for area of rectangle, 1 mark for area of triangle, 1 mark for correct subtraction, 1 mark for correct answer.
19. Answer: 396 cm²
- Working: Area of cardboard = 24 cm × 18 cm = 432 cm². Area of one square = 3 cm × 3 cm = 9 cm². Area of four squares = 4 × 9 cm² = 36 cm². Remaining area = 432 cm² - 36 cm² = 396 cm².
- Explanation: The remaining area is the original area minus the total area of the four squares cut out.
- Marking: 1 mark for area of cardboard, 1 mark for area of one square, 1 mark for total area of four squares, 1 mark for correct remaining area.
20. Answer: 2 m
- Working: Let the width of the path be (x) m. The inner rectangle (garden without path) has length = 30 m - 2x and width = 20 m - 2x. Area of garden = 30 m × 20 m = 600 m². Area of inner rectangle = (30 - 2x)(20 - 2x). Area of path = 600 - (30 - 2x)(20 - 2x) = 144. So, (30 - 2x)(20 - 2x) = 600 - 144 = 456. Expanding: 600 - 60x - 40x + 4x² = 456. 4x² - 100x + 600 = 456. 4x² - 100x + 144 = 0. Divide by 4: x² - 25x + 36 = 0. Factorising: (x - 1)(x - 24) = 0. So, x = 1 or x = 24. Since the path width cannot be 24 m (larger than the garden), x = 1 m.
- Explanation: We set up an equation for the area of the path and solve for the width. The path width must be a reasonable value.
- Marking: 1 mark for correct expression for inner dimensions, 1 mark for correct equation, 1 mark for solving the equation, 1 mark for selecting the correct answer.
End of Answer Key




