Surface Area of Composite Figures worksheet with diagrams and instructions for calculating surface area.

You’re working on a worksheet titled “Surface Area of Composite Figures”, and you need to find the surface area of each composite 3D shape, rounding to two decimal places. The key is to identify which parts of each shape are exposed (i.e., contribute to the total surface area) and which are hidden or glued together — those hidden parts should not be included.

Let’s go through each problem one by one.

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🔹 Problem 1: Cone + Hemisphere (bottom)

- Cone: radius = 2, slant height = 5
- Hemisphere: radius = 2 (attached to base of cone)

✔ Important: The flat circular base of the hemisphere is attached to the cone’s base — so it’s not part of the external surface area. We only include:
- Lateral surface area of the cone
- Curved surface area of the hemisphere

#### Formulas:
- Lateral SA of cone = π × r × l = π × 2 × 5 = 10π
- Curved SA of hemisphere = 2πr² = 2π(2)² = 8π

✔ Total SA = 10π + 8π = 18π ≈ 56.55

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🔹 Problem 2: Cylinder + Cone (on top)

- Cylinder: radius = 4, height = 7
- Cone: radius = 4, slant height = 9

✔ Important: The base of the cone is glued to the top of the cylinder — so we exclude that circular area from both shapes.

We include:
- Lateral SA of cylinder = 2πrh = 2π×4×7 = 56π
- Lateral SA of cone = πrl = π×4×9 = 36π
- Bottom base of cylinder (since top is covered) = πr² = π×16 = 16π

✔ Total SA = 56π + 36π + 16π = 108π ≈ 339.29

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🔹 Problem 3: Cylinder + Hemisphere (on top)

- Cylinder: radius = 3, height = 6
- Hemisphere: radius = 3 (sitting on top of cylinder)

✔ Important: The flat base of the hemisphere is glued to the top of the cylinder — so we exclude that circle.

We include:
- Lateral SA of cylinder = 2πrh = 2π×3×6 = 36π
- Bottom base of cylinder = πr² = π×9 = 9π
- Curved SA of hemisphere = 2πr² = 2π×9 = 18π

✔ Total SA = 36π + 9π + 18π = 63π ≈ 197.92

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🔹 Problem 4: Cone + Cylinder (cone on top)

- Cone: radius = 6, slant height = 10
- Cylinder: radius = 6, height = 8

✔ Important: The base of the cone is glued to the top of the cylinder — so we exclude that circle.

We include:
- Lateral SA of cone = πrl = π×6×10 = 60π
- Lateral SA of cylinder = 2πrh = 2π×6×8 = 96π
- Bottom base of cylinder = πr² = π×36 = 36π

✔ Total SA = 60π + 96π + 36π = 192π ≈ 603.19

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🔹 Problem 5: Two Pyramids (joined at bases)

- Each pyramid: square base = 8×8, slant height = 10
- They are joined at their bases → so the two square bases are internal and not part of surface area.

✔ We only include the lateral surfaces of both pyramids.

Each pyramid has 4 triangular faces.

Area of one triangle = ½ × base × slant height = ½ × 8 × 10 = 40

So for one pyramid: 4 × 40 = 160
For two pyramids: 2 × 160 = 320

✔ Total SA = 320.00

*(Note: No π here — these are pyramids, not curved shapes.)*

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🔹 Problem 6: Cylinder + Hemisphere (on bottom)

- Cylinder: radius = 5, height = 12
- Hemisphere: radius = 5 (attached to bottom of cylinder)

✔ Important: The flat base of the hemisphere is glued to the bottom of the cylinder — so we exclude that circle.

We include:
- Lateral SA of cylinder = 2πrh = 2π×5×12 = 120π
- Top base of cylinder = πr² = π×25 = 25π
- Curved SA of hemisphere = 2πr² = 2π×25 = 50π

✔ Total SA = 120π + 25π + 50π = 195π ≈ 612.61

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## ✔ Final Answers (rounded to two decimal places):

1. 56.55
2. 339.29
3. 197.92
4. 603.19
5. 320.00
6. 612.61

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📌 Key Tip: Always ask yourself — “Which surfaces are visible?” Any surface where two solids are glued together is not part of the external surface area. Only count the outer surfaces!

Let me know if you want step-by-step diagrams or explanations for any specific problem!