Primary 3 Mathematics Geometry Quiz

Primary 3 Mathematics Quiz - Geometry: ANSWER KEY

Total Marks: 40
Duration: 40 minutes

Section A: Multiple Choice (16 marks)

Section B: Short Answer (20 marks)

9. [3 marks total]

(a) AB and BC (or BC and CD, or CD and DA, or DA and AB) — any pair of adjacent sides [1]

(b) AB and CD (or BC and DA) — opposite sides of a rectangle are parallel [1]

(c) 4 right angles — one at each corner where two perpendicular sides meet [1]

Teaching note: In a rectangle, perpendicular means meeting at 90° (adjacent sides), and parallel means never meeting, always same distance apart (opposite sides).

10. [3 marks]

Perimeter of composite figure:

Step 1: Identify the outer edges. The square (4 cm × 4 cm) and rectangle (3 cm × 4 cm) share one side of 4 cm.

Step 2: Trace the outer boundary: top 4 cm + top right 3 cm + right side 4 cm + bottom 3 cm + bottom left 4 cm + left side 4 cm.

Or use formula: Sum of all outer sides = 4 + 3 + 4 + 3 + 4 + 4 = 22 cm

Alternative method: Perimeter of square (16 cm) + perimeter of rectangle (14 cm) − 2 × shared side (8 cm) = 16 + 14 − 8 = 22 cm. But this is complex for P3; the tracing method is preferred.

Answer: 22 cm [3]

Common mistake: Students often forget to subtract the shared side and get 16 + 14 = 30 cm. Remind them: only count the outer edges for perimeter.

11. [5 marks total]

(a) Angles P and R (both acute angles, less than 90°) [2]

(b) Angle Q (obtuse angle, greater than 90°) [1]

(c) P, R, S, Q (or using symbols: 45°, 60°, 90°, 135°) [2]

Teaching note: Acute = less than 90°; Right = exactly 90°; Obtuse = greater than 90° but less than 180°.

12. [5 marks total]

(a) Perimeter of garden = 2 × (length + width) = 2 × (12 + 8) = 2 × 20 = 40 m [2]

(b) Area of path:

• The path runs through the middle, so it is 2 m wide and 8 m long (the width of the garden)
• Area = 2 m × 8 m = 16 m² [3]

Common mistake: Students might use 12 m instead of 8 m. The path runs the width direction (vertically), so its length equals the garden's width (8 m), not the garden's length.

13. [4 marks total]

(a) Angle AOC and angle COB (or BOD and DOA, or AOD and DOC, etc.) — any two adjacent right angles [2]

(b) Angles on a straight line CD add to 180°.

• Angle AOC + Angle AOD = 180°
• 90° + Angle AOD = 180°
• Angle AOD = 180° − 90° = 90° [2]

Teaching note: When two lines are perpendicular, all four angles around the intersection are right angles (90° each).

14. [3 marks]

Perimeter of L-shape:

Method — Trace the outer edges: Starting from top left and going clockwise:

• Down: 7 cm, but the cut-out means we must be careful

Better approach: The L-shape fits in a 10 cm × 7 cm rectangle with a 3 cm × 2 cm corner removed.

Outer edges:

• Top horizontal: 10 − 3 = 7 cm, then down 2 cm (inner edge), then 3 cm to corner...

Actually, trace carefully: Starting top left of notch, go right 7 cm, down 5 cm, right 3 cm, down 2 cm, left 10 cm, up 7 cm.

Or use the trick: Perimeter of L-shape = Perimeter of bounding rectangle = 2 × (10 + 7) = 34 cm

Why? The two "steps" (3 cm and 2 cm) replace the corner, keeping total distance the same.

Answer: 34 cm [3]

Teaching note: For L-shapes made from rectangles, the perimeter equals the perimeter of the bounding rectangle. Test by tracing to verify.

15. [4 marks total]

(a) Square [1]

(b) Trapezium [1]

(c) 2 [1]

(d) Perpendicular [1]

16. [4 marks total]

(a) Area of room = 6 m × 4 m = 24 m² [1]

(b) Area of carpet = 3 m × 2 m = 6 m² [1]

(c) Area not covered = 24 − 6 = 18 m² [2]

Section C: Problem Solving (24 marks)

17. [6 marks total]

(a) Area of Rectangle PQRS = length × width = 9 cm × 6 cm = 54 cm² [2]

(b) Area of Square TUVW = side × side = 6 cm × 6 cm = 36 cm² [1]

(c) When placed side by side (sharing the 6 cm side):

• New figure dimensions: total length = 9 + 6 = 15 cm; width = 6 cm
• Perimeter = 2 × (15 + 6) = 2 × 21 = 42 cm [3]

Alternative check: Trace outer edges: 9 + 6 + 6 + 15 = 36... wait, need to be careful. The shared 6 cm side is internal.

Outer edges: 9 cm (top of rectangle) + 6 cm (top of square) = 15 cm top; 6 cm right side; 6 cm bottom of square + 9 cm bottom of rectangle = 15 cm bottom; 6 cm left side.

Perimeter = 15 + 6 + 15 + 6 = 42 cm ✓

18. [5 marks total]

(a) Figure 4 is a 5×5 square = 25 dots [1]

(b) The pattern grows by adding a border of dots. Figure n is an (n+1) × (n+1) square. Each new figure adds a row and a column, increasing by an odd number: 4, 9, 16, 25... these are square numbers: 2², 3², 4², 5²... [2]

(c) Figure 6: (6+1)² = 7² = 49 dots [2]

Or: Figure follows (n+1)², so Figure 6 = 7² = 49.

19. [6 marks total]

(a) Diagram should show:

• Inner rectangle (photo): 15 cm × 10 cm
• Outer rectangle (frame): 15+2+2 = 19 cm by 10+2+2 = 14 cm
• Frame 2 cm wide on all sides [2]

(b) Perimeter of photo = 2 × (15 + 10) = 2 × 25 = 50 cm [1]

(c) Outer dimensions of frame: 19 cm × 14 cm Perimeter = 2 × (19 + 14) = 2 × 33 = 66 cm [3]

Common mistake: Students might think frame adds 2 cm total to each dimension, not 2 cm on EACH side (so +4 cm total).

20. [8 marks total]

(a) Total length = 4 cm + 5 cm + 4 cm = 13 cm [2]

(b) Total area:

• Square: 5 cm × 5 cm = 25 cm²
• Two rectangles: 2 × (4 cm × 5 cm) = 2 × 20 = 40 cm²
• Total = 25 + 40 = 65 cm² [3]

(c) Perimeter of given figure:

• Outer edges: 4 + 5 + 4 = 13 cm (top and bottom), 5 cm (left and right sides)
• Perimeter = 2 × (13 + 5) = 2 × 18 = 36 cm

Perimeter of Jane's rectangle (13 cm × 5 cm):

• 2 × (13 + 5) = 36 cm

Jane is correct. The perimeters are equal. [3]

Teaching note: Even though the shapes look different, when rectangles are attached flush along their full height, the "lost" inner edges equal the "gained" outer edges in this symmetric arrangement. The resulting perimeter matches a single rectangle with the total length and shared height.

MARK SUMMARY