Surface Area of Prisms and Cylinders — Solutions

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1. SA of rectangular prisms

   1. SA = 2(8×5 + 8×3 + 5×3) = 2(40+24+15) = 2(79) = 158 cm²
   2. SA = 2(10×4 + 10×6 + 4×6) = 2(40+60+24) = 2(124) = 248 m²
   3. Cube: SA = 6 × 7.5² = 6 × 56.25 = 337.5 cm²
   4. SA = 2(12×9 + 12×2 + 9×2) = 2(108+24+18) = 2(150) = 300 mm²
   5. SA = 2(1.2×0.8 + 1.2×0.5 + 0.8×0.5) = 2(0.96+0.60+0.40) = 2(1.96) = 3.92 m²

2. SA of cylinders

   1. SA = 2π(3)² + 2π(3)(10) = 18π + 60π = 78π ≈ 245.0 cm²
   2. SA = 2π(7)² + 2π(7)(2) = 98π + 28π = 126π ≈ 395.8 m²
   3. SA = 2π(4.5)² + 2π(4.5)(8) = 40.5π + 72π = 112.5π ≈ 353.4 cm²
   4. SA = 2π(1.5)² + 2π(1.5)(5) = 4.5π + 15π = 19.5π ≈ 61.3 m²
   5. r = 10 cm. SA = 2π(10)² + 2π(10)(15) = 200π + 300π = 500π ≈ 1570.8 cm²

3. SA of triangular prisms

   1. Triangle area = ½×6×8 = 24 cm². Rectangular faces: 6×15=90, 8×15=120, 10×15=150. SA = 2(24)+90+120+150 = 48+360 = 408 cm²
   2. Equilateral triangle area = (√3/4)×36 = 9√3 m² ≈ 15.588 m². Three rectangular faces each: 6×10=60 m². SA = 2(9√3)+3(60) = 18√3+180 ≈ 31.18+180 = 211.2 m²
   3. Triangle area = ½×9×4 = 18 cm². Rectangular faces: 9×12=108, 5×12=60, 5×12=60. SA = 2(18)+108+60+60 = 36+228 = 264 cm²

4. Cylinder net (r = 6 cm, h = 10 cm)

   1. Net: two circles of radius 6 cm, and a rectangle of width 2π(6) = 12π cm and height 10 cm.
   2. Area of each circle = π(6)² = 36π ≈ 113.10 cm²
   3. Curved surface area = 2π(6)(10) = 120π ≈ 376.99 cm²
   4. Total SA = 2(36π) + 120π = 72π + 120π = 192π ≈ 603.19 cm²
   5. Open at one end: remove one circular face. SA = 36π + 120π = 156π ≈ 490.09 cm²

5. Comparing SA of cube vs cylinder

   1. Cube SA = 6 × 10² = 600 cm²
   2. Cylinder: r = 5 cm, h = 10 cm. SA = 2π(25) + 2π(5)(10) = 50π + 100π = 150π ≈ 471.2 cm²
   3. Cube is greater. Difference = 600 − 471.2 = 128.8 cm²
   4. In this case the cube has more SA. However, the claim is not universally true — a very tall thin cylinder could exceed a cube of the same height if the diameter is large enough. The comparison depends on both height and the cross-sectional dimensions.

6. Finding missing dimensions

   1. 100π = 2π(5)² + 2π(5)h = 50π + 10πh. So 50π = 10πh, h = 5 cm
   2. 294 = 6s², s² = 49, s = 7 cm
   3. SA = 2(lw+lh+wh): 158 = 2(8×3 + 8h + 3h) = 2(24+11h). 79 = 24+11h, 55 = 11h, h = 5 cm

7. Unit conversion and surface area

   1. SA = 2(2×1.5 + 2×0.8 + 1.5×0.8) = 2(3+1.6+1.2) = 2(5.8) = 11.6 m². In cm²: 11.6 × 10 000 = 116 000 cm²
   2. r = 1.5 cm, length = 200 cm. Curved SA = 2π(1.5)(200) = 600π ≈ 1884.96 cm²
   3. 1 m = 100 cm, so 1 m² = 100² cm² = 10 000 cm². 11.6 × 10 000 = 116 000 cm² ✓

8. Triangular prism shed painting problem

   1. Area of triangular end = ½ × 6 × 4 = 12 m²
   2. Each sloping rectangular roof panel: 5 m (slant side) × 10 m (length) = 50 m². Two panels: 100 m²
   3. Total area = 2(12) + 100 = 24 + 100 = 124 m²
   4. Litres needed = 124 ÷ 12 = 10.33… Round up: 11 litres
   5. Total cost = 11 × $18.50 = $203.50

9. Tin can labelling cost

   1. r = 4 cm, h = 11 cm. SA = 2π(4)² + 2π(4)(11) = 32π + 88π = 120π ≈ 376.99 cm²
   2. Label height = h = 11 cm. Label length = circumference + overlap = 2π(4) + 0.5 = 8π + 0.5 ≈ 25.63 + 0.5 = 26.13 cm wide, 11 cm tall (label wraps around so “length” = 26.13 cm, “width” = 11 cm)
   3. Label area = 26.13 × 11 = 287.43 cm². Cost = 287.43 × 0.003 = $0.862 ≈ $0.86
   4. 10 000 × $0.86 = $8 600

10. Composite solid surface area (cylinder on rectangular prism)

    1. Sketch: rectangular prism 8×8×5 with a cylinder (r=3, h=4) on the centre of the top face.
    2. Full prism SA = 2(8×8 + 8×5 + 8×5) = 2(64+40+40) = 288 cm². Subtract the circular footprint of the cylinder: π(3)² = 9π ≈ 28.27 cm². Prism contribution = 288 − 9π ≈ 259.73 cm²
    3. Cylinder without base: top circle + curved surface = π(3)² + 2π(3)(4) = 9π + 24π = 33π ≈ 103.67 cm²
    4. Total SA = (288 − 9π) + 33π = 288 + 24π ≈ 288 + 75.4 = 363.4 cm²