Solutions — Volume of Solids

1. Volume of prisms. Fluency

   • (a) 9×6×4: 216 cm³.
   • (b) Triangular prism legs 5,12, length 8: A = ½×5×12 = 30 m². V = 30×8 = 240 m³.
   • (c) Trapezoidal prism: parallel sides 4,7, height 5, length 10: A = ½(4+7)×5 = 27.5 cm². V = 27.5×10 = 275 cm³.
   • (d) Hyp=13, leg=5 ⇒ other leg=12. Length 6: A = ½×5×12 = 30 cm². V = 30×6 = 180 cm³.

2. Volume of cylinders. Fluency

   • (a) r=4, h=9: V = π(16)(9) = 144π ≈ 452.39 cm³.
   • (b) r=7, h=3: V = π(49)(3) = 147π ≈ 461.81 m³.
   • (c) d=12 so r=6, h=15: V = π(36)(15) = 540π ≈ 1696.46 mm³.
   • (d) r=2, h=20: V = π(4)(20) = 80π ≈ 251.33 cm³ = 251.33 mL.

3. Volume of pyramids and cones. Fluency

   • (a) Square pyramid, base 6, h=8: V = ⅓×36×8 = 96 cm³.
   • (b) Rect pyramid, base 5×9=45, h=6: V = ⅓×45×6 = 90 m³.
   • (c) Cone r=5, h=12: V = ⅓π(25)(12) = 100π ≈ 314.16 cm³.
   • (d) Cone d=8 so r=4, h=15: V = ⅓π(16)(15) = 80π ≈ 251.33 m³.

4. Volume of spheres and hemispheres. Fluency

   • (a) Sphere r=6: V = &frac43;π(216) = 288π ≈ 904.78 cm³.
   • (b) Sphere d=10, r=5: V = &frac43;π(125) = (500/3)π ≈ 523.60 m³.
   • (c) Hemisphere r=9: V = ⅔π(729) = 486π ≈ 1526.81 cm³.
   • (d) V=972π. Find r: &frac43;πr³ = 972π ⇒ r³ = 729 ⇒ r = 9 cm.

5. Volume of a composite solid. Understanding

   • (a) Hemisphere formula: V = ⅔πr³.
   • (b) Hemisphere r=4: V = ⅔π(64) = (128/3)π cm³.
   • (c) Cylinder r=4, h=9: V = π(16)(9) = 144π cm³.
   • (d) Total: Total = (128/3)π + 144π = (128/3 + 432/3)π = (560/3)π ≈ 186.67π ≈ 586.43 cm³.

6. Find the missing dimension. Understanding

   • (a) Cyl V=200π, r=5. Find h: π(25)h = 200π ⇒ h = 8 cm.
   • (b) Prism V=360, l=10, w=6. Find h: 10×6×h = 360 ⇒ 60h = 360 ⇒ h = 6 cm.
   • (c) Cone V=100π, h=12. Find r: ⅓πr²(12) = 100π ⇒ 4πr² = 100π ⇒ r² = 25 ⇒ r = 5 cm.
   • (d) Sphere V=288π. Find r: &frac43;πr³ = 288π ⇒ r³ = 216 ⇒ r = 6 cm.

7. Compare containers. Understanding

   • (a) Cyl A (r=5, h=10) vs Cyl B (r=10, h=5): V_A = π(25)(10) = 250π. V_B = π(100)(5) = 500π. Cylinder B holds twice as much. Doubling the radius quadruples the base area; halving the height only halves it back — net doubling.
   • (b) Cube 8 cm vs sphere d=10 (r=5): Cube: 8³ = 512 cm³. Sphere: &frac43;π(125) ≈ 523.6 cm³. The sphere is slightly larger (by about 11.6 cm³).
   • (c) Cone vs cylinder (same r, h): One third. V_cone = ⅓V_cylinder.
   • (d) Spheres radius r and 2r: V₁ = &frac43;πr³. V₂ = &frac43;π(2r)³ = &frac43;π(8r³) = 8 × V₁. Ratio = 1 : 8. Doubling the radius multiplies the volume by 2³ = 8.

8. Liquid capacity problems. Understanding

   • (a) Cyl tank r=1.2 m, h=2 m: V = π(1.44)(2) = 2.88π ≈ 9.047 m³ = 9047 L.
   • (b) Pool 25×10×1.8: V = 25×10×1.8 = 450 m³ = 450 kL.
   • (c) Cone r=5, h=12: V = ⅓π(25)(12) = 100π ≈ 314.16 cm³ = 314.16 mL.
   • (d) Drain at 20 mL/s: Time = 314.16/20 ≈ 15.71 seconds.

9. Sphere and cylinder relationship. Problem Solving

   • (a) Cylinder dimensions: The sphere touches the sides ⇒ cylinder radius = r. The sphere touches top and bottom ⇒ cylinder height = 2r (diameter of sphere).
   • (b) Cylinder volume: V_cyl = πr²(2r) = 2πr³.
   • (c) Sphere volume: V_sphere = &frac43;πr³.
   • (d) Fraction: V_sphere/V_cyl = (&frac43;πr³)/(2πr³) = &frac43; ÷ 2 = 2/3. The sphere occupies exactly two-thirds of the enclosing cylinder — Archimedes’ tombstone bore a sphere inscribed in a cylinder to celebrate this discovery.

10. Concrete footings. Problem Solving

    • (a) One footing volume: Prism: 0.4×0.4×0.6 = 0.096 m³. Hole: π(0.08)²×0.6 = π×0.0064×0.6 = 0.003840π ≈ 0.012064 m³. Net = 0.096 − 0.012064 ≈ 0.0839 m³.
    • (b) 16 footings: 0.0839 × 16 ≈ 1.3424 m³.
    • (c) Cost: 1.3424 × $180 ≈ $241.63.
    • (d) Steel volume (16 posts): One post: π(0.08)²×0.6 ≈ 0.012064 m³. 16 posts: 0.012064×16 ≈ 0.1930 m³.