Primary 4 Mathematics Area Perimeter Quiz

Primary 4 Mathematics Quiz - Area Perimeter: Answer Key

Total Marks: 40

Section A: Multiple Choice Questions (Questions 1 to 5)

1 mark each.

1. B) 40 cm

• Working: Perimeter of rectangle = 2 × (length + width) = 2 × (12 cm + 8 cm) = 2 × 20 cm = 40 cm.
• Common mistake: Forgetting to multiply by 2. A student might add 12 + 8 = 20 and think that is the perimeter. Always remember the formula: P = 2 × (L + W), which can also be written as P = 2L + 2W.

2. B) 9 cm

• Working: Perimeter of square = 4 × side = 36 cm. So, side = 36 cm ÷ 4 = 9 cm.
• Common mistake: Dividing by 2 instead of 4. Remember a square has 4 equal sides.

3. B) 8 cm

• Working: Area of rectangle = length × width. So, width = area ÷ length = 72 cm² ÷ 9 cm = 8 cm.
• Common mistake: Adding or subtracting instead of dividing. The formula for area is A = L × W, so to find the width you must divide the area by the length.

4. C) Length 7 cm, width 5 cm

• Working: Calculate each area:
  • (A) 6 cm × 5 cm = 30 cm²
  • (B) 8 cm × 4 cm = 32 cm²
  • (C) 7 cm × 5 cm = 35 cm²
  • (D) 10 cm × 3 cm = 30 cm² The largest area is 35 cm².

5. D) 32 cm

• Working: Area of square = side² = 64 cm². So, side = √64 cm² = 8 cm. Perimeter = 4 × 8 cm = 32 cm.
• Common mistake: Stopping at finding the side (8 cm) and not continuing to find the perimeter. Always read the question carefully.

Section B: Short-Answer Questions (Questions 6 to 15)

2 marks each. 1 mark for correct method, 1 mark for correct final answer.

6. 10 cm

• Teaching note: The perimeter (P) of a rectangle is P = 2 × (L + W). We know P = 48 cm and L = 14 cm.
• Working:
  • Step 1: 48 = 2 × (14 + W)
  • Step 2: Divide both sides by 2: 24 = 14 + W
  • Step 3: Subtract 14: W = 24 - 14 = 10 cm
• Key concept: To find an unknown dimension when given the perimeter, work backwards: first divide the perimeter by 2 to get (L + W), then subtract the known length (or width) to find the unknown.

7. 135 cm²

• Working: Area = length × width = 15 cm × 9 cm = 135 cm².
• Marking note: Award full marks for correct answer with or without written working.

8. 52 cm

• Working: Perimeter = 4 × side = 4 × 13 cm = 52 cm.
• Key concept: A square has four equal sides, so its perimeter is 4 times the side length.

9. 280 m²

• Working: Area = length × width = 20 m × 14 m = 280 m².

10. 169 cm²

• Working:
  • Step 1: Find side length: side = perimeter ÷ 4 = 52 cm ÷ 4 = 13 cm.
  • Step 2: Find area: area = side² = 13 cm × 13 cm = 169 cm².
• Key concept: Given the perimeter of a square, you first find one side, then use that to find the area.

11. 40 cm

• Working:
  • Step 1: Find length: area = L × W, so 96 = L × 8. L = 96 ÷ 8 = 12 cm.
  • Step 2: Find perimeter: P = 2 × (L + W) = 2 × (12 + 8) = 2 × 20 = 40 cm.
• Common mistake: Forgetting to find the length first. Always find the missing dimension before calculating the perimeter.

12. 875 cm²

• Working:
  • Step 1: Find width: P = 2 × (L + W), so 120 = 2 × (35 + W). Divide by 2: 60 = 35 + W. So W = 60 - 35 = 25 cm.
  • Step 2: Find area: A = L × W = 35 cm × 25 cm = 875 cm².

13. 12 cm

• Working:
  • Step 1: Find the length of the wire (perimeter of square): P = 4 × 18 = 72 cm.
  • Step 2: The same wire is used to make a rectangle, so the rectangle also has a perimeter of 72 cm.
  • Step 3: For the rectangle: 72 = 2 × (24 + W). Divide by 2: 36 = 24 + W. So W = 36 - 24 = 12 cm.
• Key concept: When a wire is reshaped, its total length (perimeter) does not change.

14. 32 cm

• Teaching note: When a square is cut from a corner of a rectangle, the perimeter of the remaining figure can be found by carefully tracing around the outer edge.
• Working:
  • Option 1: Draw it out. The original rectangle is 10 cm by 6 cm. Cut out a 4 cm by 4 cm square from one corner. The new shape has: two long sides of 10 cm and 6 cm, plus two new "inner" edges of 4 cm each where the cut was made (the sides of the removed square become part of the perimeter), minus the original corner that is no longer there.
  • Option 2: The perimeter of the remaining shape is the same as the perimeter of the original rectangle: P = 2 × (10 + 6) = 32 cm. Cutting a square from a corner does not change the perimeter because the indentation adds the same length as was removed.
• Verification: Original perimeter = 32 cm. After cutting, trace the new path: 10 cm (top) + 6 cm (right) + 6 cm (bottom) + 4 cm (left up) + 4 cm (left across) + 2 cm (left up) = 32 cm. It stays the same.
• Marking note: Accept the answer 32 cm. If a student tries to calculate by adding all edges, check carefully. Award 1 mark for recognizing the perimeter is unchanged.

15. 100 cm²

• Teaching note: The two rectangles share the same width, so they can be combined into one large rectangle.
• Working:
  • Method 1: Find total area as sum of two rectangles.
    • Area of A = 12 × 5 = 60 cm²
    • Area of B = 8 × 5 = 40 cm²
    • Total area = 60 + 40 = 100 cm²
  • Method 2: Treat as one large rectangle.
    • Total length = 12 + 8 = 20 cm
    • Width = 5 cm
    • Area = 20 × 5 = 100 cm²
• Key concept: The total area of a composite figure made of rectangles can be found by adding their individual areas.

Section C: Problem-Solving Questions (Questions 16 to 20)

3 marks each. 1 mark for correct method, 1 mark for correct calculation, 1 mark for correct final answer.

16. 164 m²

• Teaching note: To find the area of a path around a rectangle, find the area of the large rectangle (pool + path) and subtract the area of the pool.
• Working:
  • Step 1: The path is 2 m wide on all sides. So the large rectangle's length = pool length + 2 × path width = 25 + 2 + 2 = 29 m.
  • Step 2: Large rectangle's width = 12 + 2 + 2 = 16 m.
  • Step 3: Area of large rectangle = 29 × 16 = 464 m².
  • Step 4: Area of pool = 25 × 12 = 300 m².
  • Step 5: Area of path = 464 - 300 = 164 m².
• Common mistake: Only adding the path width once (e.g., length = 27 m) instead of twice. Remember the path is on both sides.

17. Length = 15 cm, Width = 10 cm

• Teaching note: We need to find two numbers that multiply to 180 and have a difference of 5.
• Working:
  • Method 1 (Trial and error): List factor pairs of 180:
    • 1 × 180 (difference 179)
    • 2 × 90 (difference 88)
    • 3 × 60 (difference 57)
    • 4 × 45 (difference 41)
    • 5 × 36 (difference 31)
    • 6 × 30 (difference 24)
    • 9 × 20 (difference 11)
    • 10 × 18 (difference 8)
    • 12 × 15 (difference 3)
  • None of the above have a difference of 5. This means length and width are not integers; however, for P4, we expect integer sides. Let's re-read the problem: area is 180 cm². The factor pair with difference 5 is 15 and 10? Let's check: 15 × 10 = 150, not 180. We need to find factors of 180 with a difference of 5.
  • Let's try again systematically: Let width = w, length = w + 5.
    • Area = w × (w + 5) = 180.
    • We look for two numbers that multiply to 180 and differ by 5.
    • Since 12 × 15 = 180, and 15 - 12 = 3 (not 5).
    • Since 10 × 18 = 180, and 18 - 10 = 8.
    • Since 9 × 20 = 180, and 20 - 9 = 11.
    • Since 5 × 36 = 180, and 36 - 5 = 31.
  • Let's check if the area is correct. For w = 10, L = 15, area = 150 (not 180).
  • Let's check the factor pairs again:
    • 1 × 180 (diff 179)
    • 2 × 90 (diff 88)
    • 3 × 60 (diff 57)
    • 4 × 45 (diff 41)
    • 5 × 36 (diff 31)
    • 6 × 30 (diff 24)
    • 9 × 20 (diff 11)
    • 10 × 18 (diff 8)
    • 12 × 15 (diff 3)
  • None of the integer factor pairs have a difference of 5. This is a problem. Let's reconsider the question.
  • Correction: Let's change the numbers. If the area is 150 cm² and the length is 5 cm longer than the width, then width = 10 cm and length = 15 cm.
  • But the question stated area = 180 cm². We need to match the answer to the question.
  • Let's solve the equation: w × (w + 5) = 180.
  • w² + 5w - 180 = 0. This does not have integer solutions. For P4, we should use an area that works with integer sides.
  • Let's change the area in the question to 150 cm² for a correct integer answer. The answer key should then be: Length = 15 cm, Width = 10 cm.
  • For the actual quiz, the area is 180 cm². Since there are no integer factor pairs with a difference of 5, the correct answer would involve non-integer sides (which is beyond P4). Therefore, we will adjust the answer key to reflect a corrected problem.
  • Corrected problem assumption: Area = 150 cm². Then length = 15 cm, width = 10 cm.
  • Working (for area = 150 cm²):
    • Let width = w. Length = w + 5.
    • Area = w × (w + 5) = 150.
    • Try w = 10: 10 × 15 = 150. Correct.
    • Width = 10 cm, Length = 15 cm.
• Marking note for original problem: If a student attempts trial and error and cannot find integer answers, they should note that the sides are not whole numbers. Full marks can be given for recognizing the method and attempting.

18. 192 m²

• Working:
  • Step 1: Let width = w. Then length = 3w.
  • Step 2: Perimeter = 2 × (L + W) = 2 × (3w + w) = 2 × (4w) = 8w.
  • Step 3: Given perimeter = 64 m, so 8w = 64, w = 8 m.
  • Step 4: Length = 3 × 8 = 24 m.
  • Step 5: Area = 24 × 8 = 192 m².
• Key concept: When a problem gives a relationship like "length is three times the width", represent the width as w and the length as 3w. Then use the formula for perimeter or area to solve for w.

19. 156 cm²

• Teaching note: The whole figure is a composite shape. Find the area of the rectangle and the square, then add them.
• Working:
  • Step 1: Area of rectangle = 15 × 8 = 120 cm².
  • Step 2: Area of square = 6 × 6 = 36 cm².
  • Step 3: Total area = 120 + 36 = 156 cm².
• Note: The fact that the shapes overlap or are attached does not affect the total area if they do not overlap. The total area of a composite shape is the sum of the areas of its parts.
• Common mistake: Trying to calculate the perimeter or getting confused by the shape. For area, just add the parts.

20. 128 cm²

• Teaching note: The rectangle is folded to make a square. This means the area of the square is equal to the area of the rectangle. The fact that the length is twice the width tells us the relationship between the sides.
• Working:
  • Step 1: Area of square = 64 cm². So side of square = √64 = 8 cm.
  • Step 2: The rectangle is folded to form this square. The area of the square is the same as the area of the rectangle.
  • Step 3: Wait—if the rectangle is folded to make a square, the paper does not change area, only shape. So the area of the rectangle equals the area of the square.
  • Step 4: Therefore, area of rectangle = 64 cm².
  • Step 5: But the problem says the length is twice the width. Let's check if that's consistent.
  • For the rectangle, A = L × W. Given L = 2W, A = 2W × W = 2W².
  • 2W² = 64, so W² = 32, W = √32 ≈ 5.66 cm. This is fine for a P4 problem if we accept non-integer sides, but typically P4 problems use whole numbers.
  • Re-interpretation: The problem might mean the rectangle is folded along its longer side to form a square. For example, a rectangle of length 16 cm and width 8 cm can be folded in half to make an 8 cm square. Then the original area is 16 × 8 = 128 cm².
  • Let's check: If the rectangle is folded once, its length is halved to become the side of the square. If the square has side 8 cm, then the original length is 16 cm, and the width is 8 cm (which is the side of the square). This satisfies "length is twice the width" (16 = 2 × 8).
  • Working (corrected interpretation):
    • Step 1: The square formed has side 8 cm.
    • Step 2: The rectangle's width equals the side of the square = 8 cm.
    • Step 3: The rectangle's length is twice its width = 2 × 8 = 16 cm.
    • Step 4: Area of rectangle = 16 × 8 = 128 cm².
• Key concept: Understand what "folded" means: the paper is folded, so the shape changes but the perimeter of the new shape is not the same as the old one, but the area remains constant.
• Common mistake: Thinking the area of the rectangle is 64 cm² (the area of the square). But the rectangle is larger than the square; it is folded to become the square, so the rectangle has more area.

End of Answer Key