Primary 6 PSLE Mathematics Area Perimeter Quiz

Primary 6 PSLE Mathematics Quiz - Area Perimeter (Answer Key)

General Note:

• $\pi$ is taken as $\frac{22}{7}$ or $3.14$ as specified in the question.
• Units must be included in the final answer.
• Method marks are awarded for correct logical steps even if the final calculation is wrong.

Section A: Multiple Choice Questions

1. Answer: (3)

• Concept: Perimeter and Area of a Square.
• Working:
  • Perimeter of square = $4 \times \text{side}$.
  • $36 = 4 \times \text{side} \Rightarrow \text{side} = 9$ cm.
  • Area = $\text{side} \times \text{side} = 9 \times 9 = 81$ cm$^2$.
• Common Mistake: Confusing perimeter with area or forgetting to divide by 4.

2. Answer: (1)

• Concept: Change in Area of a Rectangle.

• Working:

  • Original Area = $12 \times 8 = 96$ cm$^2$.
  • New Length = $12 + 3 = 15$ cm.
  • New Breadth = $8 - 2 = 6$ cm.
  • New Area = $15 \times 6 = 90$ cm$^2$.
  • Change = $96 - 90 = 6$ cm$^2$ decrease? Wait, let's re-read carefully.
  • Original: 96. New: 90. The area decreased by 6.
  • Let's check the options. Option (2) is "Decrease by 6 cm$^2$".
  • Correction in logic: The question asks for the change. $96 - 90 = 6$. It is a decrease.
  • Let's re-evaluate Option 1 vs 2.
  • Option 1: Increase by 6. Option 2: Decrease by 6.
  • My calculation shows a decrease. So Answer is (2).
  • Self-Correction during generation: I must ensure the key matches the question.
  • Original Area = 96. New Area = 90. Change is -6.
  • Therefore, the correct option is (2).
  • Note: In the quiz generation, I listed (1) as Increase and (2) as Decrease. The correct answer is (2).

  Let's double check the question text in the quiz. "What is the change in the area?" (1) Increase by 6 cm$^2$ (2) Decrease by 6 cm$^2$

  Correct Answer is (2).

3. Answer: (2)

• Concept: Perimeter of a Semicircle.
• Working:
  • Perimeter = Curved Arc + Diameter.
  • Curved Arc = $\frac{1}{2} \times \pi \times d = \frac{1}{2} \times \frac{22}{7} \times 14 = 22$ cm.
  • Diameter = 14 cm.
  • Total Perimeter = $22 + 14 = 36$ cm.
• Common Mistake: Forgetting to add the diameter (answering 22 cm).

4. Answer: (2)

• Concept: Overlapping Areas.
• Working:
  • Area of one square = $5 \times 5 = 25$ cm$^2$.
  • Total area of two separate squares = $25 + 25 = 50$ cm$^2$.
  • The resulting shape is a rectangle of $8 \times 5 = 40$ cm$^2$.
  • Area of Overlap = (Sum of individual areas) - (Area of union).
  • Overlap = $50 - 40 = 10$ cm$^2$.
• Alternative Method:
  • Length of union = 8 cm. Side of square = 5 cm.
  • Overlap length = $(5 + 5) - 8 = 2$ cm.
  • Overlap width = 5 cm.
  • Area = $2 \times 5 = 10$ cm$^2$.

5. Answer: (1)

• Concept: Ratio of Areas and Radii.
• Working:
  • Area $\propto r^2$.
  • $A_A : A_B = 4 : 9$.
  • $r_A^2 : r_B^2 = 4 : 9$.
  • $r_A : r_B = \sqrt{4} : \sqrt{9} = 2 : 3$.

6. Answer: (1)

• Concept: Area of Triangle and Parallelogram.
• Working:
  • Area of Triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 6 = 30$ cm$^2$.
  • Area of Parallelogram = $\text{base} \times \text{height}$.
  • Base = 10 cm. Area = 30 cm$^2$.
  • $10 \times h = 30 \Rightarrow h = 3$ cm.

7. Answer: (1)

• Concept: Area of Composite Shapes (Subtraction).
• Working:
  • Area of Square = $14 \times 14 = 196$ cm$^2$.
  • Four quadrants make one full circle of radius 7 cm.
  • Area of Circle = $\pi r^2 = \frac{22}{7} \times 7 \times 7 = 154$ cm$^2$.
  • Shaded Area = $196 - 154 = 42$ cm$^2$.

8. Answer: (3)

• Concept: Perimeter Equality and Area.
• Working:
  • Perimeter of Triangle = $3 \times 12 = 36$ cm.
  • Perimeter of Square = 36 cm.
  • Side of Square = $36 \div 4 = 9$ cm.
  • Area of Square = $9 \times 9 = 81$ cm$^2$.

9. Answer: (3)

• Concept: Circumference to Area.
• Working:
  • $C = 2 \pi r = 44$.
  • $2 \times \frac{22}{7} \times r = 44$.
  • $\frac{44}{7} r = 44 \Rightarrow r = 7$ m.
  • Area = $\pi r^2 = \frac{22}{7} \times 7 \times 7 = 154$ m$^2$.

10. Answer: (2)

• Concept: Area of Composite Figure.
• Working:
  • Area of Rectangle = $10 \times 7 = 70$ cm$^2$.
  • Diameter of Semicircle = 7 cm $\Rightarrow$ Radius = 3.5 cm.
  • Area of Semicircle = $\frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 3.5 \times 3.5$.
  • $3.5 = \frac{7}{2}$. So, $\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{11 \times 7}{4} = \frac{77}{4} = 19.25$ cm$^2$.
  • Total Area = $70 + 19.25 = 89.25$ cm$^2$.
  • Wait, let me re-calculate.
  • $\frac{1}{2} \times \frac{22}{7} \times 3.5 \times 3.5 = 11 \times 0.5 \times 3.5 = 5.5 \times 3.5 = 19.25$.
  • Total = $70 + 19.25 = 89.25$.
  • Looking at options: (1) 89.25, (2) 108.25.
  • My calculation gives 89.25. So Answer is (1).
  • Correction: In the quiz options, I put (1) as 89.25. So the answer is (1).

Section B: Short Answer Questions

11. Answer: 16 cm

• Concept: Area of Rhombus.
• Formula: Area = $\frac{1}{2} \times d_1 \times d_2$.
• Working:
  • $96 = \frac{1}{2} \times 12 \times d_2$.
  • $96 = 6 \times d_2$.
  • $d_2 = 96 \div 6 = 16$ cm.

12. Answer: 156 m$^2$

• Concept: Area of Path Around a Rectangle.
• Working:
  • Inner Dimensions: $20 \text{ m} \times 15 \text{ m}$. Area = $300$ m$^2$.
  • Path width = 2 m on all sides.
  • Outer Length = $20 + 2 + 2 = 24$ m.
  • Outer Breadth = $15 + 2 + 2 = 19$ m.
  • Outer Area = $24 \times 19 = 456$ m$^2$.
  • Area of Path = Outer Area - Inner Area = $456 - 300 = 156$ m$^2$.

13. Answer: 64 cm$^2$

• Concept: Perimeter of Composite Shape.
• Working:
  • Let side of square be $s$. This is also the width of the rectangles.
  • Length of rectangle = 10 cm.
  • The figure consists of a central square and two rectangles on left/right.
  • Perimeter trace:
    • Top: $10 + s + 10 = 20 + s$.
    • Bottom: $10 + s + 10 = 20 + s$.
    • Left side: $s$ (width of rect).
    • Right side: $s$ (width of rect).
    • Wait, the rectangles are attached to the sides of the square.
    • Let's trace the boundary:
      • Top edge of left rect (10) + Top edge of square (s) + Top edge of right rect (10)? No, usually "attached to sides" means the width of the rect matches the side of the square.
      • If attached to left and right sides:
      • Top boundary: Length of left rect (10) is horizontal? No, usually length is the longer side. Let's assume the rectangles extend outwards.
      • Horizontal segments: Top of left rect (10) + Top of square (s) + Top of right rect (10)? No, if they are attached to the vertical sides of the square, the "length" 10 is the horizontal extension.
      • So, Top Edge = $10 (\text{left}) + s (\text{square top? No, covered}) + 10 (\text{right})$.
      • Actually, if attached to the side, the square's top and bottom are exposed. The rectangles' tops and bottoms are exposed.
      • Let's visualize: [Rect 1] [Square] [Rect 2].
      • Top perimeter: $10 + s + 10$.
      • Bottom perimeter: $10 + s + 10$.
      • Left vertical side: $s$ (width of rect).
      • Right vertical side: $s$ (width of rect).
      • Total Perimeter = $2(20 + s) + 2s = 40 + 2s + 2s = 40 + 4s$.
      • Given Perimeter = 64.
      • $40 + 4s = 64$.
      • $4s = 24 \Rightarrow s = 6$ cm.
      • Area of Square = $s^2 = 6 \times 6 = 36$ cm$^2$.
  • Re-evaluating the diagram description: "Central square with two identical rectangles attached to opposite sides (left and right)."
  • If the rectangles are attached to the vertical sides of the square, the vertical sides of the square are internal.
  • The perimeter consists of:
    • Top of Left Rect (10)
    • Top of Square (s) -- Wait, is the rect width equal to square side? Yes.
    • So the top edge is continuous? No, they are attached side-by-side.
    • Top edge = Length of Rect 1 + Side of Square + Length of Rect 2?
    • If the rectangle's width is attached to the square's side, then the rectangle's length extends out.
    • Top boundary: Top of Rect 1 (10) + Top of Square (s) + Top of Rect 2 (10). Total = $20 + s$.
    • Bottom boundary: Same = $20 + s$.
    • Left vertical edge: Width of Rect 1 ($s$).
    • Right vertical edge: Width of Rect 2 ($s$).
    • Total P = $2(20 + s) + 2s = 40 + 4s$.
    • $64 = 40 + 4s \Rightarrow 24 = 4s \Rightarrow s = 6$.
    • Area = 36.
  • Let's check alternative interpretation: What if the "Length 10" is the vertical dimension? Unlikely for "attached to side".
  • Let's check if the answer 64 in my draft was a typo.
  • If $s=8$, $P = 40 + 32 = 72$.
  • If $s=6$, $P = 40 + 24 = 64$. Correct.
  • So Area = 36 cm$^2$.
  • Correction: The answer is 36 cm$^2$.

14. Answer: 14 cm

• Concept: Circumference to Radius.
• Working:
  • $C = 2 \pi r = 88$.
  • $2 \times \frac{22}{7} \times r = 88$.
  • $\frac{44}{7} r = 88$.
  • $r = 88 \times \frac{7}{44} = 2 \times 7 = 14$ cm.

15. Answer: 15 cm

• Concept: Area of Triangle.
• Working:
  • Area = $\frac{1}{2} \times \text{base} \times \text{height}$.
  • $120 = \frac{1}{2} \times 16 \times h$.
  • $120 = 8 \times h$.
  • $h = 120 \div 8 = 15$ cm.

Section C: Long Answer Questions

16. Answer: 112 cm$^2$

• Concept: Area of Overlapping Quadrants (Leaf Shape).
• Working:
  • Area of Square = $14 \times 14 = 196$ cm$^2$.
  • Area of one Quadrant = $\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{4} \times 22 \times 28 = 154$ cm$^2$.
  • Area of two Quadrants = $154 \times 2 = 308$ cm$^2$.
  • The two quadrants cover the square, but the overlapping region is counted twice.
  • Area of Overlap = (Area of Quad 1 + Area of Quad 2) - Area of Square.
  • Overlap = $308 - 196 = 112$ cm$^2$.

17. Answer: (a) 756 cm$^3$, (b) 492 cm$^2$

• Concept: Volume and Surface Area of Open Box.
• Working:
  • Original Cardboard: $30 \text{ cm} \times 20 \text{ cm}$.
  • Cutouts: 3 cm squares from corners.
  • Box Dimensions:
    • Length = $30 - 3 - 3 = 24$ cm.
    • Breadth = $20 - 3 - 3 = 14$ cm.
    • Height = 3 cm.
  • (a) Volume = $L \times B \times H = 24 \times 14 \times 3$.
    • $24 \times 14 = 336$.
    • $336 \times 3 = 1008$ cm$^3$.
    • Wait, recalc: $24 \times 10 = 240$, $24 \times 4 = 96$. $240+96=336$. $336 \times 3 = 1008$.
    • My previous mental check was wrong. Answer is 1008.
  • (b) Surface Area of Inside:
    • This is the area of the cardboard remaining after cuts.
    • Original Area = $30 \times 20 = 600$ cm$^2$.
    • Area of 4 cutouts = $4 \times (3 \times 3) = 36$ cm$^2$.
    • Remaining Area = $600 - 36 = 564$ cm$^2$.
    • Alternative Check: Base + 4 Walls.
    • Base = $24 \times 14 = 336$.
    • 2 Long Walls = $2 \times (24 \times 3) = 144$.
    • 2 Short Walls = $2 \times (14 \times 3) = 84$.
    • Total = $336 + 144 + 84 = 564$ cm$^2$.
  • Corrected Answers:
    • (a) 1008 cm$^3$
    • (b) 564 cm$^2$

18. Answer: (a) 420 m, (b) 1260 m

• Concept: Perimeter of Stadium Shape.
• Working:
  • (a) Perimeter of one lap:
    • Two straight sections: $100 + 100 = 200$ m.
    • Two semicircles form one circle with diameter 70 m.
    • Circumference = $\pi d = \frac{22}{7} \times 70 = 220$ m.
    • Total Lap Distance = $200 + 220 = 420$ m.
  • (b) 3 Laps:
    • $3 \times 420 = 1260$ m.

19. Answer: (a) 88 cm, (b) 154 cm$^2$

• Concept: Composite Semicircles.
• Working:
  • Large Diameter = 28 cm $\Rightarrow$ Large Radius $R = 14$ cm.
  • Two small semicircles fit along the diameter, so each small diameter = $28 \div 2 = 14$ cm.
  • Small Radius $r = 7$ cm.
  • (a) Perimeter of the figure:
    • Curved part of large semicircle: $\frac{1}{2} \times \pi \times 28 = 14\pi$.
    • Curved parts of two small semicircles: $2 \times (\frac{1}{2} \times \pi \times 14) = 14\pi$.
    • Note: The straight diameter is internal/not part of the perimeter if the small semicircles are "inside" or if the figure is the boundary. The question says "figure is made up of...". Usually, this implies the outer boundary.
    • However, the diagram description says "small semicircles are placed along the diameter... rest is shaded". The perimeter of the entire figure usually includes all boundary lines.
    • If the small semicircles are cutouts or just drawn inside, the "Perimeter of the entire figure" typically refers to the outer boundary plus any internal boundaries if they are distinct regions. But standard PSLE "Perimeter of the figure" for this shape (Arbelos-like) usually means the sum of all arcs.
    • Arc Large = $\frac{1}{2} \times \frac{22}{7} \times 28 = 44$ cm.
    • Arc Small 1 = $\frac{1}{2} \times \frac{22}{7} \times 14 = 22$ cm.
    • Arc Small 2 = 22 cm.
    • Total Perimeter = $44 + 22 + 22 = 88$ cm. (The straight line segments are covered by the small semicircles' diameters if they are solid, or if it's just the arcs). Given the visual of "Yin Yang" or similar, the perimeter is the sum of the three arcs.
  • (b) Shaded Area:
    • Area of Large Semicircle = $\frac{1}{2} \pi R^2 = \frac{1}{2} \times \frac{22}{7} \times 14 \times 14 = 308$ cm$^2$.
    • Area of two Small Semicircles = Area of one small circle = $\pi r^2 = \frac{22}{7} \times 7 \times 7 = 154$ cm$^2$.
    • Shaded Area = Large Semicircle - Two Small Semicircles.
    • Shaded Area = $308 - 154 = 154$ cm$^2$.

20. Answer: (a) 2246 m$^2$, (b) 244 m

• Concept: Area and Perimeter of Composite Field.
• Working:
  • (a) Area of Grass:
    • Area of Rectangle = $60 \times 40 = 2400$ m$^2$.
    • Area of Fountain (Circle) = $\pi r^2 = \frac{22}{7} \times 7 \times 7 = 154$ m$^2$.
    • Grass Area = $2400 - 154 = 2246$ m$^2$.
  • (b) Total Fence Length:
    • Perimeter of Rectangle = $2 \times (60 + 40) = 200$ m.
    • Circumference of Fountain = $2 \pi r = 2 \times \frac{22}{7} \times 7 = 44$ m.
    • Total Fence = $200 + 44 = 244$ m.