Primary 6 PSLE Mathematics Geometry Quiz

Primary 6 PSLE Mathematics Quiz - Geometry (Answer Key)

General Note:

• $\pi$ is taken as $\frac{22}{7}$ unless specified as $3.14$.
• Steps are shown for clarity. Students may use alternative valid methods (e.g., model drawing vs algebra).

Section A: Angles and Triangles

1. Answer: $45^\circ$

• Concept: Angles on a straight line add up to $180^\circ$.
• Working: $\angle x + 135^\circ = 180^\circ$ $\angle x = 180^\circ - 135^\circ$ $\angle x = 45^\circ$
• Marking: 1 mark for subtraction setup, 1 mark for correct answer.

2. Answer: $70^\circ$

• Concept: Base angles of an isosceles triangle are equal. Sum of angles in a triangle is $180^\circ$.
• Working: Since $PQ = PR$, $\angle PQR = \angle PRQ$. Sum of angles $= 180^\circ$. $\angle PQR + \angle PRQ + 40^\circ = 180^\circ$ $2 \times \angle PQR = 180^\circ - 40^\circ = 140^\circ$ $\angle PQR = 140^\circ \div 2 = 70^\circ$
• Marking: 1 mark for finding sum of base angles ($140^\circ$), 1 mark for dividing by 2.

3. Answer: $70^\circ$

• Concept: Consecutive interior angles between parallel lines add up to $180^\circ$.
• Working: $AB \parallel DC$, so $\angle DAB + \angle ADC = 180^\circ$. $110^\circ + \angle ADC = 180^\circ$ $\angle ADC = 180^\circ - 110^\circ = 70^\circ$
• Marking: 1 mark for property identification/subtraction, 1 mark for answer.

4. Answer: $35^\circ$

• Concept: Sum of angles in a triangle is $180^\circ$. Complementary angles.
• Working: In $\triangle XYZ$, $\angle Y = 90^\circ$ and $\angle X = 35^\circ$. $\angle Z = 180^\circ - 90^\circ - 35^\circ = 55^\circ$ In $\triangle WYZ$ (right-angled at $W$): $\angle WYZ + \angle Z + 90^\circ = 180^\circ$ $\angle WYZ + 55^\circ = 90^\circ$ $\angle WYZ = 90^\circ - 55^\circ = 35^\circ$ (Alternative: $\angle WYZ = \angle YXZ$ because both are complementary to $\angle Z$).
• Marking: 1 mark for finding $\angle Z$ or setting up relation, 1 mark for final answer.

5. Answer: $108^\circ$

• Concept: Sum of interior angles of an $n$-sided polygon is $(n-2) \times 180^\circ$. For a regular polygon, divide by $n$. Note: While the formula $(n-2) \times 180$ is secondary, P6 students are taught that a regular pentagon can be split into 3 triangles from one vertex, or they memorize the interior angle of common regular polygons.
• Working: Sum of interior angles $= 3 \times 180^\circ = 540^\circ$. One interior angle $= 540^\circ \div 5 = 108^\circ$.
• Marking: 1 mark for sum ($540^\circ$), 1 mark for division.

Section B: Circles and Composite Figures

6. Answer: $44 \text{ cm}$

• Concept: Circumference $C = \pi d$.
• Working: $C = \frac{22}{7} \times 14$ $C = 22 \times 2 = 44 \text{ cm}$
• Marking: 1 mark for formula/substitution, 1 mark for answer.

7. Answer: $77 \text{ cm}^2$

• Concept: Area of semi-circle $= \frac{1}{2} \pi r^2$.
• Working: $\text{Area} = \frac{1}{2} \times \frac{22}{7} \times 7 \times 7$ $\text{Area} = \frac{1}{2} \times 22 \times 7$ $\text{Area} = 11 \times 7 = 77 \text{ cm}^2$
• Marking: 1 mark for $\pi r^2$ calculation, 1 mark for halving.

8. Answer: $60 \text{ cm}$

• Concept: Dimensions of bounding rectangle.
• Working: Radius $r = 5 \text{ cm}$. Diameter $d = 10 \text{ cm}$. Width of rectangle $= d = 10 \text{ cm}$. Length of rectangle $= 2 \times d = 20 \text{ cm}$. Perimeter $= 2 \times (20 + 10) = 2 \times 30 = 60 \text{ cm}$.
• Marking: 1 mark for identifying L and W, 1 mark for perimeter calc.

9. Answer: $90^\circ$

• Concept: Angle in a semi-circle is a right angle.
• Working: Since $AC$ is the diameter, the angle subtended at the circumference ($\angle ABC$) is $90^\circ$.
• Marking: 2 marks for correct answer (knowledge-based).

10. Answer: $36 \text{ cm}$

• Concept: Perimeter of quadrant $= \text{Arc length} + 2 \times \text{radius}$.
• Working: Arc length $= \frac{1}{4} \times 2 \pi r = \frac{1}{4} \times 2 \times \frac{22}{7} \times 14$. $\text{Arc} = \frac{1}{2} \times 22 \times 2 = 22 \text{ cm}$ Two radii $= 14 + 14 = 28 \text{ cm}$. Total Perimeter $= 22 + 28 = 50 \text{ cm}$. Wait, let me re-calculate. $\frac{1}{4} \times \frac{22}{7} \times 28 = \frac{1}{4} \times 88 = 22$. Correct. $22 + 14 + 14 = 50$. Correction in Answer Key: The previous draft said 36, which was incorrect. Correct Answer: $50 \text{ cm}$
• Marking: 1 mark for arc length, 1 mark for adding radii.

11. Answer: $42 \text{ cm}^2$

• Concept: Area of square minus area of 4 quadrants (which make 1 full circle).
• Working: Side of square $= 14 \text{ cm}$. Area of square $= 14 \times 14 = 196 \text{ cm}^2$. Radius of each quadrant $= 14 \div 2 = 7 \text{ cm}$. 4 Quadrants $= 1$ Full Circle. Area of Circle $= \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154 \text{ cm}^2$. Unshaded Area $= 196 - 154 = 42 \text{ cm}^2$.
• Marking: 1 mark for square area, 1 mark for circle area, 1 mark for subtraction.

12. Answer: $139.25 \text{ cm}^2$

• Concept: Area of triangle + Area of semi-circle.
• Working: Triangle Base $= 10 \text{ cm}$, Height $= 12 \text{ cm}$. Area of Triangle $= \frac{1}{2} \times 10 \times 12 = 60 \text{ cm}^2$. Semi-circle Radius $= 10 \div 2 = 5 \text{ cm}$. Area of Semi-circle $= \frac{1}{2} \times 3.14 \times 5 \times 5 = \frac{1}{2} \times 3.14 \times 25$. $3.14 \times 25 = 78.5$ $\text{Semi-circle Area} = 39.25 \text{ cm}^2$ Total Area $= 60 + 39.25 = 99.25 \text{ cm}^2$. Wait, re-reading Q12. "Take $\pi = 3.14$". Calculation: $0.5 \times 3.14 \times 25 = 39.25$. Total $= 60 + 39.25 = 99.25$. Correct Answer: $99.25 \text{ cm}^2$
• Marking: 1 mark for triangle area, 1 mark for semi-circle area, 1 mark for total.

Section C: Complex Geometry and Problem Solving

13. Answer: $128 \text{ cm}^2$

• Concept: Area of trapezium $= \frac{1}{2} (a + b) h$.
• Working: $\text{Area} = \frac{1}{2} \times (12 + 20) \times 8$ $\text{Area} = \frac{1}{2} \times 32 \times 8$ $\text{Area} = 16 \times 8 = 128 \text{ cm}^2$
• Marking: 1 mark for sum of parallel sides, 1 mark for formula application, 1 mark for answer.

14. Answer: $120^\circ$

• Concept: Exterior angle of a regular polygon / Angles on a straight line.
• Working: Interior angle of equilateral triangle $= 60^\circ$. $\angle ACB = 60^\circ$. $BCD$ is a straight line, so $\angle ACB + \angle ACD = 180^\circ$. $\angle ACD = 180^\circ - 60^\circ = 120^\circ$
• Marking: 1 mark for identifying $60^\circ$, 1 mark for subtraction, 1 mark for answer.

15. Answer: $216 \text{ cm}^2$

• Concept: Surface area of a cube remains unchanged when a corner cube is removed (the 3 outer faces removed are replaced by 3 inner faces of the same area).
• Working: Original Surface Area $= 6 \times (6 \times 6) = 6 \times 36 = 216 \text{ cm}^2$. Removing a corner cube removes 3 faces of area $2 \times 2$ but exposes 3 new internal faces of area $2 \times 2$. Net change $= 0$. New Surface Area $= 216 \text{ cm}^2$.
• Marking: 1 mark for original SA calc, 2 marks for reasoning that SA is unchanged.

16. Answer: $124 \text{ cm}^2$

• Concept: Principle of Inclusion-Exclusion. Area($A \cup B$) = Area($A$) + Area($B$) - Area($A \cap B$).
• Working: Area of Large Square $= 10 \times 10 = 100 \text{ cm}^2$. Area of Small Square $= 6 \times 6 = 36 \text{ cm}^2$. Overlap $= 12 \text{ cm}^2$. Total Area $= 100 + 36 - 12 = 124 \text{ cm}^2$.
• Marking: 1 mark for individual areas, 1 mark for subtraction of overlap, 1 mark for answer.

17. Answer: $410.67 \text{ cm}^2$ (or $410 \frac{2}{3}$)

• Concept: Area of sector. Angle at center.
• Working: $\angle AOB$ is a straight line ($180^\circ$). $\angle BOC = 60^\circ$. $\angle AOC = 180^\circ - 60^\circ = 120^\circ$. Fraction of circle $= \frac{120}{360} = \frac{1}{3}$. Area of Circle $= \pi r^2 = \frac{22}{7} \times 14 \times 14 = 22 \times 2 \times 14 = 616 \text{ cm}^2$. Area of Sector $AOC = \frac{1}{3} \times 616 = \frac{616}{3} = 205.33...$ Wait, calculation check: $\frac{22}{7} \times 196 = 22 \times 28 = 616$. Correct. $616 / 3 = 205.33$. Correct Answer: $205.33 \text{ cm}^2$ (or $205 \frac{1}{3}$)
• Marking: 1 mark for finding angle $120^\circ$, 1 mark for area of full circle, 1 mark for fraction calculation, 1 mark for final answer.

18. Answer: $7200 \text{ cm}^3$

• Concept: Volume of displaced water = Volume of stone.
• Working: Rise in water level $= 18 \text{ cm} - 15 \text{ cm} = 3 \text{ cm}$. Base Area of tank $= 60 \times 40 = 2400 \text{ cm}^2$. Volume of stone $= \text{Base Area} \times \text{Rise in height}$. $V = 2400 \times 3 = 7200 \text{ cm}^3$
• Marking: 1 mark for height difference, 1 mark for base area, 1 mark for multiplication, 1 mark for answer.

19. Answer: $76 \text{ cm}$

• Concept: Perimeter of composite shape.
• Working: The shape consists of:
  1. Three sides of the rectangle: Two lengths ($20 \text{ cm}$ each) and one width ($14 \text{ cm}$). The other width is internal. Sum $= 20 + 20 + 14 = 54 \text{ cm}$.
  2. The arc of the semi-circle. Diameter $= 14 \text{ cm}$. Arc Length $= \frac{1}{2} \times \pi \times d = \frac{1}{2} \times \frac{22}{7} \times 14 = 22 \text{ cm}$. Total Perimeter $= 54 + 22 = 76 \text{ cm}$.
• Marking: 1 mark for straight sides sum, 1 mark for arc length, 1 mark for addition, 1 mark for answer.

20. Answer: $112 \text{ cm}^2$

• Concept: Area of overlap of two quadrants in a square.
• Working: Area of one quadrant $= \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{4} \times 616 = 154 \text{ cm}^2$. Area of two quadrants $= 154 \times 2 = 308 \text{ cm}^2$. Area of Square $= 14 \times 14 = 196 \text{ cm}^2$. The two quadrants cover the square, but the overlapping region is counted twice. Area of Overlap $= (\text{Area of 2 Quadrants}) - (\text{Area of Square})$. $\text{Overlap} = 308 - 196 = 112 \text{ cm}^2$
• Marking: 1 mark for area of one quadrant, 1 mark for sum of two quadrants, 1 mark for subtraction of square area, 1 mark for answer.