Review Solutions — Measurement: Surface Area and Volume

1. Surface area basics. Fluency

   • (a) Prism 8×5×4: 184 cm².
   • (b) Cylinder r=6, h=10: 192π cm².
   • (c) Sphere r=5: 100π ≈ 314.16 cm².
   • (d) Cone r=9, h=12: l=15. 216π ≈ 678.58 cm².

2. Volume basics. Fluency

   • (a) Tri prism: 360 m³.
   • (b) Cone r=4, h=9: 48π cm³.
   • (c) Sphere d=14: (1372/3)π ≈ 1436.76 cm³.
   • (d) Pyramid 10×10, h=6: 200 m³.

3. Unit conversions. Fluency

   • (a) 4.8 m²: 48 000 cm².
   • (b) 750 000 mm³: 750 cm³.
   • (c) 2.4 m³: 2400 L.
   • (d) 65 000 m²: 6.5 ha.

4. Find the missing dimension. Fluency

   • (a) Cube TSA=216: s = 6 cm.
   • (b) Cyl V=320π, h=5: r = 8 cm.
   • (c) Sphere TSA=576π: r = 12 cm.
   • (d) Cone V=75π, r=5: h = 9 cm.

5. Composite solid (cylinder minus cone, r=5). Understanding

   • (a) Cylinder formula: V = πr²h = π(25)(14) = 350π cm³.
   • (b) Cone formula: V = ⅓πr²h = ⅓π(25)(10) = (250/3)π cm³.
   • (c) Remaining volume: 350π − (250/3)π = (1050/3 − 250/3)π = (800/3)π ≈ 837.76 cm³.
   • (d) Mass (density 7.8 g/cm³): 837.76 × 7.8 ≈ 6534 g ≈ 6.53 kg.

6. Composite surface area (trophy). Understanding

   • (a) Slant heights: From 8-side (half-width=3): l⊂1; = √(9+25) = √34 ≈ 5.83 cm. From 6-side (half-length=4): l⊂2; = √(16+25) = √41 ≈ 6.40 cm.
   • (b) Four pyramid faces: 2×(½×8×5.83) + 2×(½×6×6.40) = 46.64 + 38.40 ≈ 85.04 cm².
   • (c) Prism without top: 2(8×12)+2(6×12)+8×6 = 192+144+48 = 384 cm².
   • (d) Total: 384 + 85.04 ≈ 469.04 cm².

7. Scaling and comparison. Understanding

   • (a) Sphere doubled radius: 288π cm³.
   • (b) Cone as % of cylinder: 33.3%.
   • (c) Party can vs regular: ≈ 3.06 times.
   • (d) Scale 1.5 ⇒ curved SA: Scales by 2.25.

8. Practical applications. Understanding

   • (a) Pipe plastic volume: 1800π ≈ 5655 cm³.
   • (b) Hemispherical bowl r=12: 1152π ≈ 3619 cm³.
   • (c) Wrapping paper +10%: ≈ 663.5 cm².
   • (d) Cheese cubes: 200 cubes.

9. Water tank design. Problem Solving

   • (a) Cylinder r=h radius: r ≈ 1.47 m.
   • (b) Cylinder TSA: ≈ 27.18 m².
   • (c) Sphere radius and TSA: r ≈ 1.336 m. TSA ≈ 22.44 m².
   • (d) Conclusion: Sphere uses less material, but is harder to build and install.

10. Recycling a sphere into a cone. Problem Solving

    • (a) Sphere r=9: 972π ≈ 3053.63 cm³.
    • (b) Cone height: h = 81 cm.
    • (c) Slant height: l ≈ 81.22 cm.
    • (d) TSA change: Sphere 324π ≈ 1017.9 cm². Cone ≈ 1643.8 cm². +61.5%.

11. Composite silo (cylinder + cone + hemisphere). Problem Solving

    • (a) Total volume: (100/3)π ≈ 104.72 m³.
    • (b) Total outer surface area: ≈ 123.21 m².
    • (c) 80% full, density 750 kg/m³: ≈ 62 830 kg.
    • (d) Pure cylinder same volume, r=2: h = 25/3 ≈ 8.33 m.

12. Optimisation: minimum material tin. Problem Solving

    • (a) h in terms of r: h = 1000/(πr²).
    • (b) TSA as function of r: TSA = 2πr² + 2000/r.
    • (c) Minimum near r=?: Minimum near r = 5 cm (TSA ≈ 557 cm²).
    • (d) Cost per tin: ≈ $11.14.