Year 10 Mathematics | Victorian Curriculum 2.0
Surface area and volume of composite objects
Topic 09 | Measurement | Answer key

Tier 1

    1. Cube volume =1000= 1000 cm3^3; cylinder volume =π(16)(10)=160π≈502.65= \pi(16)(10) = 160\pi \approx 502.65 cm3^3. Total ≈1502.65\approx 1502.65 cm3^3.
    2. Prism =240= 240 cm3^3; hole =π(4)(5)=20π≈62.83= \pi(4)(5) = 20\pi \approx 62.83 cm3^3. Remaining ≈177.17\approx 177.17 cm3^3.
    3. Volume =72+96=168= 72 + 96 = 168 cm3^3. The joined face (4×3=124 \times 3 = 12 cm2^2) is hidden. SA of each prism: 2(24+18+12)=1082(24+18+12) = 108 and 2(32+24+12)=1362(32+24+12) = 136. Exposed SA =108+136−2(12)=220= 108 + 136 - 2(12) = 220 cm2^2.
    4. Cylinder =π(9)(7)=63π= \pi(9)(7) = 63\pi; hemisphere =23π(27)=18π= \frac{2}{3}\pi(27) = 18\pi. Total =81π≈254.47= 81\pi \approx 254.47 cm3^3.
    5. Cylinder curved SA =2π(3)(7)=42π= 2\pi(3)(7) = 42\pi; cylinder base =9π= 9\pi; hemisphere curved =2π(9)=18π= 2\pi(9) = 18\pi. Total =(42+9+18)π=69π≈216.77= (42 + 9 + 18)\pi = 69\pi \approx 216.77 cm2^2. (Top circle of cylinder is replaced by hemisphere, so not counted.)
    6. Full prism =240= 240; removed =48= 48. Volume =192= 192 cm3^3.
    7. Prism =720= 720 cm3^3; half-cylinder =12π(25)(12)=150π≈471.24= \frac{1}{2}\pi(25)(12) = 150\pi \approx 471.24 cm3^3. Total ≈1191.24\approx 1191.24 cm3^3.
    8. Cube full SA =600= 600; cylinder full SA =2π(16)+2π(4)(10)=32π+80π=112π≈351.86= 2\pi(16) + 2\pi(4)(10) = 32\pi + 80\pi = 112\pi \approx 351.86; shared circle =16π≈50.27= 16\pi \approx 50.27. Exposed SA ≈600+351.86−2(50.27)≈851.32\approx 600 + 351.86 - 2(50.27) \approx 851.32 cm2^2.

Tier 2

    1. Curved SA =2π(0.5)(1.8)=1.8π= 2\pi(0.5)(1.8) = 1.8\pi; top =π(0.25)=0.25π= \pi(0.25) = 0.25\pi. Total area =2.05π≈6.44= 2.05\pi \approx 6.44 m2^2. Cost =6.44×18≈= 6.44 \times 18 \approx $115.92.
    2. Cross-section is a trapezium with parallel sides 1.21.2 and 2.42.4, width 1010. Area =12(1.2+2.4)(10)=18= \frac{1}{2}(1.2 + 2.4)(10) = 18 m2^2. Volume =18×5=90= 18 \times 5 = 90 m3^3.
    3. (a) Base =4= 4 m3^3; column =π(0.16)(8)=1.28π≈4.02= \pi(0.16)(8) = 1.28\pi \approx 4.02 m3^3. Total ≈8.02\approx 8.02 m3^3. (b) Base: bottom =4= 4, four sides =4(2×1)=8= 4(2 \times 1) = 8, top exposed =4−π(0.16)≈3.50= 4 - \pi(0.16) \approx 3.50. Column: curved =2π(0.4)(8)=6.4π≈20.11= 2\pi(0.4)(8) = 6.4\pi \approx 20.11. (Open top, so no top circle.) Total SA ≈4+8+3.50+20.11=35.61\approx 4 + 8 + 3.50 + 20.11 = 35.61 m2^2.
    4. Box =9000= 9000 cm3^3; can =π(25)(15)=375π≈1178.10= \pi(25)(15) = 375\pi \approx 1178.10 cm3^3. Wasted ≈9000−1178.10=7821.90\approx 9000 - 1178.10 = 7821.90 cm3^3. Percentage ≈86.9%\approx 86.9\%.
    5. Large cylinder SA: curved =2π(6)(10)=120π= 2\pi(6)(10) = 120\pi; base =36π= 36\pi; top annulus (ring) =36π−9π=27π= 36\pi - 9\pi = 27\pi. Small cylinder: curved =2π(3)(8)=48π= 2\pi(3)(8) = 48\pi; top =9π= 9\pi. Total exposed =(120+36+27+48+9)π=240π≈753.98= (120 + 36 + 27 + 48 + 9)\pi = 240\pi \approx 753.98 cm2^2.
    6. Cylinder =π(16)(10)=160π= \pi(16)(10) = 160\pi; hemisphere =23π(64)=128π3= \frac{2}{3}\pi(64) = \frac{128\pi}{3}. Remaining =160π−128π3=352π3≈368.61= 160\pi - \frac{128\pi}{3} = \frac{352\pi}{3} \approx 368.61 cm3^3.

Tier 3

    1. Each component’s full SA includes the shared face as part of its own total. When the two solids join, that face is hidden on both solids — it disappears from the outside of solid 1 and from the outside of solid 2. Since it was counted once in each SA calculation, you must subtract it twice to get the correct exposed SA.
    2. Cylinder =π(1)2(2)=2π= \pi(1)^2(2) = 2\pi m3^3; cone =13π(1)2(0.5)=π6= \frac{1}{3}\pi(1)^2(0.5) = \frac{\pi}{6} m3^3. Total =13π6≈6.81= \frac{13\pi}{6} \approx 6.81 m3≈6807^3 \approx 6807 L. The cone adds only π6≈0.52\frac{\pi}{6} \approx 0.52 m3^3 (≈524\approx 524 L), about 7.7%7.7\% of the total, because the cone formula includes the 13\frac{1}{3} factor and the cone height is small.
    3. Design A (cylinder): choose rr and hh with πr2h=2000\pi r^2 h = 2000. For example r=6r = 6 cm gives h=200036π≈17.68h = \frac{2000}{36\pi} \approx 17.68 cm; SA =2π(36)+2π(6)(17.68)≈226.19+666.18≈892.4= 2\pi(36) + 2\pi(6)(17.68) \approx 226.19 + 666.18 \approx 892.4 cm2^2. Design B (cube + hemisphere): cube side ss with hemisphere r=s/2r = s/2; volume =s3+23π(s/2)3=s3(1+π12)≈1.262s3=2000= s^3 + \frac{2}{3}\pi(s/2)^3 = s^3(1 + \frac{\pi}{12}) \approx 1.262 s^3 = 2000, so s≈11.67s \approx 11.67 cm. SA =5s2+2π(s/2)2=5s2+πs22≈6.571s2≈894.7= 5s^2 + 2\pi(s/2)^2 = 5s^2 + \frac{\pi s^2}{2} \approx 6.571 s^2 \approx 894.7 cm2^2. Both designs use roughly similar material; the optimal choice depends on exact dimensions.
    4. V=a×a×2a−π(a4)2(2a)=2a3−πa38=a3(2−π8)V = a \times a \times 2a - \pi\left(\frac{a}{4}\right)^2(2a) = 2a^3 - \frac{\pi a^3}{8} = a^3\left(2 - \frac{\pi}{8}\right).

Challenge

    1. (a) Cylinder =π(9)(10)=90π= \pi(9)(10) = 90\pi; hemisphere =23π(27)=18π= \frac{2}{3}\pi(27) = 18\pi. Total =108π≈339.29= 108\pi \approx 339.29 m3^3. (b) Base circle =9π= 9\pi; curved cylinder =2π(3)(10)=60π= 2\pi(3)(10) = 60\pi; hemisphere curved =2π(9)=18π= 2\pi(9) = 18\pi. Total SA =(9+60+18)π=87π≈273.32= (9 + 60 + 18)\pi = 87\pi \approx 273.32 m2^2. Grain volume =34(108π)=81π≈254.47= \frac{3}{4}(108\pi) = 81\pi \approx 254.47 m3^3.
    2. Base prism =128= 128 cm3^3; cylinder =π(4)(12)=48π= \pi(4)(12) = 48\pi; sphere =43π(27)=36π= \frac{4}{3}\pi(27) = 36\pi. Total volume =128+84π≈391.88= 128 + 84\pi \approx 391.88 cm3^3. SA: prism bottom =64= 64; prism four sides =4(16)=64= 4(16) = 64; prism top annulus =64−4π= 64 - 4\pi; cylinder curved =2π(2)(12)=48π= 2\pi(2)(12) = 48\pi; sphere =4π(9)=36π= 4\pi(9) = 36\pi (but a circle π(4)\pi(4) is hidden where sphere meets cylinder, subtract 2π(4)=8π2\pi(4) = 8\pi). Sphere sits on top so small circle hidden; actually sphere r=3>r=3 > cylinder r=2r=2, so the sphere rests on the cylinder rim; exposed sphere SA =36π−π(4)=32π= 36\pi - \pi(4) = 32\pi. Total SA ≈64+64+(64−4π)+48π+32π≈192+76π≈430.73\approx 64 + 64 + (64 - 4\pi) + 48\pi + 32\pi \approx 192 + 76\pi \approx 430.73 cm2^2.
    3. Outer volume =π(0.15)2(50)=1.125π= \pi(0.15)^2(50) = 1.125\pi m3^3; inner volume =π(0.12)2(50)=0.72π= \pi(0.12)^2(50) = 0.72\pi m3^3. Wall volume =(1.125−0.72)π=0.405π≈1.272= (1.125 - 0.72)\pi = 0.405\pi \approx 1.272 m3^3.
    4. Cube has 5 exposed faces (top replaced by pyramid base): 5×36=1805 \times 36 = 180 cm2^2. Pyramid lateral area =12(24)(5)=60= \frac{1}{2}(24)(5) = 60 cm2^2. Total exposed SA =180+60=240= 180 + 60 = 240 cm2^2.
Year 10 Mathematics study companion | Answer key