Solutions — Surface Area of Prisms and Cylinders

1. TSA of rectangular prisms. Fluency

   • (a) l=6, w=4, h=3: TSA = 2(6×4 + 6×3 + 4×3) = 2(24+18+12) = 2×54 = 108 cm².
   • (b) l=10, w=7, h=2: TSA = 2(70+20+14) = 2×104 = 208 m².
   • (c) Cube, side 5: TSA = 6 × 5² = 6 × 25 = 150 cm².
   • (d) l=12, w=8, h=5: TSA = 2(96+60+40) = 2×196 = 392 mm².

2. TSA of cylinders. Fluency

   • (a) r=3, h=8: TSA = 2π(9) + 2π(3)(8) = 18π + 48π = 66π ≈ 207.35 cm².
   • (b) r=5, h=12: TSA = 2π(25) + 2π(5)(12) = 50π + 120π = 170π ≈ 534.07 m².
   • (c) d=10 so r=5, h=6: TSA = 2π(25) + 2π(5)(6) = 50π + 60π = 110π ≈ 345.58 cm².
   • (d) r=2.5, h=9: TSA = 2π(6.25) + 2π(2.5)(9) = 12.5π + 45π = 57.5π ≈ 180.64 mm².

3. TSA of triangular prisms. Fluency

   • (a) Legs 5, 12, hyp 13, length 10: 2 triangles: 2×(½×5×12) = 60 cm². Rectangles: (5×10)+(12×10)+(13×10) = 50+120+130 = 300 cm². TSA = 360 cm².
   • (b) Equilateral side 6, length 8: Triangle height = √(6²−3²) = √27 = 3√3. Triangle area = ½×6×3√3 = 9√3. 2 triangles: 18√3 ≈ 31.18 cm². 3 rectangles: 3×(6×8) = 144 cm². TSA ≈ 175.18 cm².
   • (c) Legs 8, 15, hyp 17, length 20: 2 triangles: 2×(½×8×15) = 120 m². Rectangles: (8+15+17)×20 = 40×20 = 800 m². TSA = 920 m².
   • (d) Isosceles: base 8, sides 5, height 3, length 15: 2 triangles: 2×(½×8×3) = 24 cm². Rectangles: (8×15)+(5×15)+(5×15) = 120+75+75 = 270 cm². TSA = 294 cm².

4. Find the missing dimension. Fluency

   • (a) TSA=148, l=6, w=4. Find h: 148 = 2(24 + 6h + 4h) ⇒ 74 = 24 + 10h ⇒ 10h = 50 ⇒ h = 5 cm.
   • (b) TSA=54π, r=3. Find h: 54π = 2π(9) + 2π(3)h = 18π + 6πh ⇒ 36π = 6πh ⇒ h = 6 cm.
   • (c) Cube TSA=294. Find side: 6s² = 294 ⇒ s² = 49 ⇒ s = 7 cm.
   • (d) TSA=100π, h=8. Find r: 100π = 2πr² + 2πr(8) = 2πr² + 16πr ⇒ r² + 8r − 50 = 0 ⇒ r = (−8 + √(64+200))/2 = (−8+√264)/2 ≈ (−8+16.25)/2 ≈ 4.12 m.

5. Surface area from the diagram. Understanding

   • (a) Faces: 5 faces total: 2 right triangles (legs 6 cm and 8 cm) + 3 rectangles: 6×12 cm, 8×12 cm, 10×12 cm.
   • (b) Area of each triangular face: ½ × 6 × 8 = 24 cm² each. Total for both: 48 cm².
   • (c) Area of each rectangular face: 6×12 = 72 cm²; 8×12 = 96 cm²; 10×12 = 120 cm².
   • (d) Total surface area: 48 + 72 + 96 + 120 = 336 cm².

6. Open containers and partial surfaces. Understanding

   • (a) Open box 40×30×20: Base: 40×30 = 1200 cm². Two long walls: 2×(40×20) = 1600 cm². Two short walls: 2×(30×20) = 1200 cm². Total = 1200+1600+1200 = 4000 cm².
   • (b) Open cylinder r=6, h=10: One base: π(36) = 36π. Curved: 2π(6)(10) = 120π. Total = 156π ≈ 490.09 cm².
   • (c) Half-cylinder r=5, h=12: Flat rectangular base: 2r×h = 10×12 = 120 cm². Two semicircular ends: 2×(½πr²) = π×25 = 25π cm². Curved half-surface: ½×2πrh = πrh = 5π×12 = 60π cm². Total = 120 + 25π + 60π = 120 + 85π ≈ 386.92 cm².
   • (d) Tube r=4, h=15 (no ends): Curved surface only: 2π(4)(15) = 120π ≈ 376.99 cm².

7. Composite shapes. Understanding

   • (a) Small cube (3 cm) on large cube (8 cm): Large cube TSA = 6×64 = 384. Remove one 3×3 face from top of large cube (where small cube sits): −9. Remove bottom face of small cube: −9. Small cube remaining 5 faces: 5×9 = 45. Total = 384 − 9 + 45 = 420 cm².
   • (b) Box 10×6×4 with cylindrical hole r=1 through length: Box TSA = 2(10×6+10×4+6×4) = 2(60+40+24) = 248. Remove 2 circles (ends of hole): −2π(1)² = −2π. Add cylinder interior: 2π(1)(10) = 20π. Total = 248 − 2π + 20π = 248 + 18π ≈ 248 + 56.55 = 304.55 cm².
   • (c) Shed: rectangular prism (4×10×3) + triangular prism roof (equilateral side 4, length 10): Roof triangle height = √(4−4) ... equilateral side 4: h = √(16−4) = √12 = 2√3. 2 gable triangles: 2×(½×4×2√3) = 8√3 ≈ 13.86 m². 2 sloping roof rectangles: 2×(4×10) = 80 m². 4 walls: 2(4×3)+2(10×3) = 24+60 = 84 m². (No floor; roof sits flush so top of walls excluded.) Total outer = 13.86 + 80 + 84 ≈ 177.86 m².
   • (d) Cylinder (r=3, h=8) on prism (10×10×5): Prism TSA = 2(100+50+50) = 400. Remove cylinder base circle: −π(9) = −9π. Cylinder: TSA = 2π(9)+2π(3)(8) = 18π+48π = 66π; remove one base circle: 66π−9π = 57π. Total = 400 − 9π + 57π = 400 + 48π ≈ 400 + 150.80 = 550.80 cm².

8. Spheres and cones. Understanding

   • (a) Sphere r=7: TSA = 4π(49) = 196π ≈ 615.75 cm².
   • (b) Sphere TSA=324π. Find r: 4πr² = 324π ⇒ r² = 81 ⇒ r = 9 cm.
   • (c) Cone r=6, h=8: l = √(36+64) = √100 = 10 cm. TSA = π(36) + π(6)(10) = 36π + 60π = 96π ≈ 301.59 cm².
   • (d) Ice-cream cone (open, no base) r=4, l=11: Curved surface only: πrl = π(4)(11) = 44π ≈ 138.23 cm².

9. Painting and wrapping. Problem Solving

   • (a) Gable triangle (base 3 m, slant sides 2 m): Height = √(2² − 1.5²) = √(4−2.25) = √1.75 ≈ 1.322 m. Area = ½ × 3 × 1.322 ≈ 1.98 m².
   • (b) Four walls (no floor, no gable): Two long walls (5×2.4): 2×12 = 24 m². Two short walls (3×2.4): 2×7.2 = 14.4 m². Total = 38.4 m².
   • (c) Two sloping roof rectangles (2×5 each): 2 × (2×5) = 20 m².
   • (d) Cost: Roof (2 gables + 2 slopes): 2×1.98 + 20 = 23.96 m² × $12 = $287.52. Walls: 38.4 m² × $8 = $307.20. Total = $594.72.

10. Packaging design. Problem Solving

    • (a) Box dimensions: The cylinder diameter = 10 cm, so the box is 10 cm × 10 cm × 18 cm.
    • (b) Canister TSA (r=5, h=18): TSA = 2π(25) + 2π(5)(18) = 50π + 180π = 230π ≈ 722.57 cm².
    • (c) Box TSA (10×10×18): TSA = 2(100 + 180 + 180) = 2×460 = 920 cm².
    • (d) Extra as percentage: Extra = 920 − 722.57 = 197.43 cm². Percentage = 197.43/722.57 × 100 ≈ 27.3% more.