Checkpoint Quiz 2
Maths Standard Year 11 · Module 2: Measurement
Lesson 6 — Sectors, Annuli & Composite Shapes
Lesson 7 — SA of Prisms & Cylinders
Lesson 8 — SA of Pyramids, Cones & Spheres
Lesson 9 — Volume of Prisms & Cylinders
Lesson 10 — Volume of Pyramids, Cones & Spheres
10 MC
3 SA
~25 min
0/10
Multiple Choice Score
Answer questions to see your score.
Part A — Multiple Choice (1 mark each)
A $75.40 \text{ cm}^2$
B $150.80 \text{ cm}^2$
C $37.70 \text{ cm}^2$
D $100.53 \text{ cm}^2$
A — $75.40 \text{ cm}^2$. $A = \dfrac{\theta}{360°} \times \pi r^2 = \dfrac{135}{360} \times \pi \times 64 = \dfrac{3}{8} \times 64\pi = 24\pi \approx 75.40\text{ cm}^2$.
A $201.06 \text{ m}^2$
B $100.53 \text{ m}^2$
C $452.39 \text{ m}^2$
D $125.66 \text{ m}^2$
A — $201.06 \text{ m}^2$. $A = \pi(R^2 - r^2) = \pi(10^2 - 6^2) = \pi(100 - 36) = 64\pi \approx 201.06\text{ m}^2$.
A $60 \text{ m}^2$
B $94 \text{ m}^2$
C $120 \text{ m}^2$
D $47 \text{ m}^2$
B — $94 \text{ m}^2$. $SA = 2(lw + lh + wh) = 2(5 \times 4 + 5 \times 3 + 4 \times 3) = 2(20 + 15 + 12) = 2 \times 47 = 94\text{ m}^2$.
A $226.19 \text{ cm}^2$
B $188.50 \text{ cm}^2$
C $245.04 \text{ cm}^2$
D $282.74 \text{ cm}^2$
C — $245.04 \text{ cm}^2$. $SA = 2\pi r^2 + 2\pi r h = 2\pi(9) + 2\pi(3)(10) = 18\pi + 60\pi = 78\pi \approx 245.04\text{ cm}^2$.
A $13 \text{ cm}$
B $17 \text{ cm}$
C $11.31 \text{ cm}$
D $7 \text{ cm}$
A — $13 \text{ cm}$. $l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}$. The Pythagorean triple $5, 12, 13$ applies here.
A $1436.76 \text{ cm}^2$
B $307.88 \text{ cm}^2$
C $615.75 \text{ cm}^2$
D $153.94 \text{ cm}^2$
C — $615.75 \text{ cm}^2$. $SA = 4\pi r^2 = 4\pi(7)^2 = 4 \times 49\pi = 196\pi \approx 615.75\text{ cm}^2$.
A $452.39 \text{ m}^3$
B $226.19 \text{ m}^3$
C $904.78 \text{ m}^3$
D $144.51 \text{ m}^3$
A — $452.39 \text{ m}^3$. $V = \pi r^2 h = \pi(4)^2(9) = 144\pi \approx 452.39\text{ m}^3$.
A $144 \text{ m}^3$
B $96 \text{ m}^3$
C $120 \text{ m}^3$
D $240 \text{ m}^3$
C — $120 \text{ m}^3$. Cross-section area $= \tfrac{1}{2}(4+6) \times 3 = 15\text{ m}^2$. Volume $= 15 \times 8 = 120\text{ m}^3$.
A $1130.97 \text{ cm}^3$
B $376.99 \text{ cm}^3$
C $188.50 \text{ cm}^3$
D $753.98 \text{ cm}^3$
B — $376.99 \text{ cm}^3$. $V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi(36)(10) = 120\pi \approx 376.99\text{ cm}^3$.
A $314.16 \text{ cm}^3$
B $261.80 \text{ cm}^3$
C $1047.20 \text{ cm}^3$
D $523.60 \text{ cm}^3$
D — $523.60 \text{ cm}^3$. $V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi(125) = \dfrac{500\pi}{3} \approx 523.60\text{ cm}^3$.
Part B — Short Answer (show all working)
▼ Show solution
Formula: $A = \pi(R^2 - r^2)$
Substitute: $A = \pi(50^2 - 44^2) = \pi(2500 - 1936) = 564\pi$
Answer: $564\pi \approx 1772.12\text{ m}^2$
▼ Show solution
Faces: 1 circular base + curved surface (no top)
Base: $\pi r^2 = \pi(25) = 25\pi$
Curved surface: $2\pi r h = 2\pi(5)(12) = 120\pi$
Total: $25\pi + 120\pi = 145\pi \approx 455.53\text{ cm}^2$
▼ Show solution
(a) $V_{\text{cyl}} = \pi r^2 h = \pi(16)(6) = 96\pi \approx 301.59\text{ m}^3$
(b) $V_{\text{hemi}} = \tfrac{1}{2} \times \tfrac{4}{3}\pi r^3 = \tfrac{2}{3}\pi(64) = \dfrac{128\pi}{3} \approx 134.04\text{ m}^3$
(c) $V_{\text{total}} = 96\pi + \dfrac{128\pi}{3} = \dfrac{288\pi + 128\pi}{3} = \dfrac{416\pi}{3} \approx 435.63\text{ m}^3$
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