Surface area worksheet for composite figures with diagrams and space for calculations.

Let's solve each of the composite figures step by step to find their surface areas. We are instructed to use π ≈ 3.14.

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Problem 1: Rectangular Prism with a Cylinder Removed

Figure: A rectangular prism with a cylindrical hole through its center (like a tunnel).

- Dimensions:
- Prism: Length = 10 cm, Width = 8 cm, Height = 5 cm
- Cylinder: Diameter = 4 cm → Radius = 2 cm, Height = 5 cm (same as prism)

#### Step-by-step:

1. Surface Area of Rectangular Prism (without holes):
$$
SA_{\text{prism}} = 2(lw + lh + wh)
$$
$$
= 2(10 \times 8 + 10 \times 5 + 8 \times 5) = 2(80 + 50 + 40) = 2(170) = 340 \text{ cm}^2
$$

2. Subtract the area of the two circular ends (since they're removed):
- Each circle: $ A = \pi r^2 = 3.14 \times 2^2 = 3.14 \times 4 = 12.56 $
- Two circles: $ 2 \times 12.56 = 25.12 \text{ cm}^2 $

3. Add the lateral surface area of the cylinder (the inner curved surface):
$$
A_{\text{lateral}} = 2\pi rh = 2 \times 3.14 \times 2 \times 5 = 62.8 \text{ cm}^2
$$

4. Total Surface Area:
$$
SA = 340 - 25.12 + 62.8 = 377.68 \text{ cm}^2
$$

✔ Answer for Problem 1: 377.68 cm²

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Problem 2: Cube with a Triangular Pyramid Cut Out

Figure: A cube with a triangular pyramid (tetrahedron) cut out from one corner.

- Cube: Side = 12 in
- Triangle base: Right triangle with legs 12 in and 12 in → Hypotenuse = $ \sqrt{12^2 + 12^2} = \sqrt{288} \approx 16.97 $ in
- Height of triangle = 12 in (from vertex to base)

But wait — looking at the figure: The cut is a triangular face on one side, and we’re removing a pyramid from the cube.

Actually, this appears to be a cube with a triangular prism removed, but the diagram shows a cut-out wedge forming a right triangular pyramid.

But upon closer inspection: The shaded region is a triangle on the top face, and it’s a cutout from the cube.

Wait — actually, the figure shows a cube with a triangular pyramid (a tetrahedron) removed from one corner. But the triangle has sides 12, 12, and 12√2? No — the triangle has legs 12 and 12, and hypotenuse is shown as 12√2, so it's a right triangle.

But the surface area change:

- We remove a triangular face (but that was internal), but actually, when you cut out a pyramid, you:
- Remove one face (the triangular cut), but add three new faces (the three triangular faces of the pyramid).
- Wait — no: if you cut out a pyramid from a corner, you remove the corner and expose three new triangular faces.

But in this case, the figure shows a square face with a right triangle removed — so maybe it's not a full pyramid?

Wait — the diagram shows a cube with a triangular section removed, but the cut is along a diagonal of the face.

Actually, looking carefully: It’s a cube of side 12 inches, and a right triangular pyramid is cut out from one corner, such that the base is a right triangle with legs 12 in and 12 in, and height 12 in.

So the pyramid has:
- Base: Right triangle (legs 12, 12)
- Height: 12 in (depth into the cube)

But since it's a corner cut, we must consider:

- Remove the triangular face that was part of the cube’s face (area of triangle = $ \frac{1}{2} \times 12 \times 12 = 72 $ in²)
- Add three new triangular faces:
1. One with legs 12 and 12 → area = 72
2. One with legs 12 and 12 → area = 72
3. One with legs 12 and 12 → area = 72? No — actually, the three faces are all right triangles with legs 12 and 12.

Wait — no: the three faces of the pyramid are:
- Face 1: On front face — triangle with legs 12 and 12 → area = 72
- Face 2: On side face — same → 72
- Face 3: On top face — same → 72

But the base (the triangle on the outer face) is removed, so we subtract that area, but add the other three.

Wait — actually, the base of the pyramid is internal — so it’s not part of the original surface.

But the original cube has 6 faces.

When we cut out a pyramid from the corner, we:
- Remove a triangle from one face (so reduce that face area)
- Add three new triangular faces (the lateral faces of the pyramid)

So:

1. Original cube surface area:
$$
6 \times 12^2 = 6 \times 144 = 864 \text{ in}^2
$$

2. Subtract the area of the triangle removed from one face:
$$
\frac{1}{2} \times 12 \times 12 = 72 \text{ in}^2
$$

3. Add the areas of the three new triangular faces:
Each is a right triangle with legs 12 and 12 → area = 72
So total added = $ 3 \times 72 = 216 \text{ in}^2 $

4. Total surface area:
$$
864 - 72 + 216 = 1008 \text{ in}^2
$$

Wait — that seems high. Let's reconsider.

Actually, the three new faces are:
- One face: triangle with legs 12 and 12 → area = 72
- Second: same → 72
- Third: same → 72

Yes, so total added = 216

But we only subtract 72 (the area of the cut on the outer face)

So final SA = $ 864 - 72 + 216 = 1008 $

But let’s double-check: Is the cut really removing a triangle from the face and adding three new ones?

Yes — correct.

✔ Answer for Problem 2: 1008 in²

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Problem 3: Cylinder on Top of a Rectangular Prism

Figure: A cylinder sitting on top of a rectangular prism.

- Prism: 10 cm × 8 cm × 6 cm
- Cylinder: Diameter = 8 cm → Radius = 4 cm, Height = 6 cm

Note: The cylinder sits exactly on the top face of the prism.

#### Steps:

1. Surface Area of Prism (without top):
- Normally: $ 2(lw + lh + wh) = 2(10×8 + 10×6 + 8×6) = 2(80+60+48)=2(188)=376 $
- But top face is covered by cylinder → so do not include top face
- So subtract top area: $ 10×8 = 80 $
- Also, bottom face is included: $ 80 $
- So SA_prism = $ 376 - 80 = 296 $ cm² (excluding top)

But wait — standard way: total SA minus the top face that’s covered.

Alternatively: calculate lateral and bottom.

- Lateral faces: $ 2(lh + wh) = 2(10×6 + 8×6) = 2(60+48)=2(108)=216 $
- Bottom: $ 10×8 = 80 $
- Total prism contribution: $ 216 + 80 = 296 $ cm²

2. Cylinder:
- Lateral surface area: $ 2\pi rh = 2 \times 3.14 \times 4 \times 6 = 150.72 $ cm²
- Top: $ \pi r^2 = 3.14 \times 16 = 50.24 $ cm²
- Bottom: Not exposed (on top of prism), so do not include
- So cylinder contributes: $ 150.72 + 50.24 = 200.96 $ cm²

3. Total Surface Area:
$$
296 + 200.96 = 496.96 \text{ cm}^2
$$

✔ Answer for Problem 3: 496.96 cm²

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Problem 4: Triangular Prism with a Sphere Cut Out

Figure: A triangular prism with a sphere (or hemisphere?) cut out.

Looking closely:
- Triangular prism: Base triangle with legs 12 and 12, hypotenuse 12√2, height of prism = 20
- A sphere is cut out from the center? But it says "radius = 3"

Wait — it shows a circle inside the triangle, radius 3, so likely a cylinder or hemisphere?

But the figure shows a circular hole going through the prism, with radius 3.

But the prism has a triangular cross-section, so if a cylinder of radius 3 is drilled through, then:

- We need to compute:
- Surface area of the prism
- Subtract the area of the two circular ends (if fully drilled)
- Add the lateral surface area of the cylinder

But the cylinder goes through the entire prism (length = 20), so:

#### Step 1: Surface Area of Triangular Prism

- Triangle: Right triangle with legs 12 and 12 → hypotenuse = $ \sqrt{12^2 + 12^2} = \sqrt{288} \approx 16.97 $
- Area of triangle: $ \frac{1}{2} \times 12 \times 12 = 72 $
- Perimeter: $ 12 + 12 + 16.97 = 40.97 $
- Lateral surface area: $ \text{perimeter} \times \text{height} = 40.97 \times 20 = 819.4 $
- Two bases: $ 2 \times 72 = 144 $
- Total prism SA: $ 819.4 + 144 = 963.4 $

But now, a cylinder of radius 3 is drilled through the length (20 units), so:

- Remove two circular areas from the two triangular faces: $ 2 \times \pi r^2 = 2 \times 3.14 \times 9 = 56.52 $
- Add lateral surface area of cylinder: $ 2\pi r h = 2 \times 3.14 \times 3 \times 20 = 376.8 $

So total surface area:
$$
963.4 - 56.52 + 376.8 = 1283.68 \text{ cm}^2
$$

✔ Answer for Problem 4: 1283.68 cm²

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Problem 5: Rectangular Prism with a Pyramid on Top

Figure: A rectangular prism with a square pyramid on top.

- Prism: 10 ft × 10 ft × 8 ft
- Pyramid: Square base 10 ft × 10 ft, height = 6 ft

We assume the pyramid sits perfectly on top.

#### Step 1: Surface Area of Prism (without top)

- Lateral faces: $ 4 \times (10 \times 8) = 320 $
- Bottom: $ 10 \times 10 = 100 $
- Total prism: $ 320 + 100 = 420 $ ft²

Top face is covered by pyramid → not included

#### Step 2: Pyramid

- Base: 10×10 = 100 → but not exposed, so don't include
- Lateral faces: 4 triangular faces

Each triangle:
- Base = 10 ft
- Slant height: need to find

From pyramid: height = 6 ft, base = 10 ft → distance from center to edge = 5 ft

So slant height $ l = \sqrt{6^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61} \approx 7.81 $ ft

Area of one triangle: $ \frac{1}{2} \times 10 \times 7.81 = 39.05 $

Four triangles: $ 4 \times 39.05 = 156.2 $

#### Total Surface Area:
$$
420 + 156.2 = 576.2 \text{ ft}^2
$$

✔ Answer for Problem 5: 576.2 ft²

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Problem 6: Cone on Top of a Cylinder

Figure: A cone sitting on top of a cylinder.

- Cylinder: Diameter = 8 cm → radius = 4 cm, height = 10 cm
- Cone: Same base, height = 5 cm

So the cone is glued on top of the cylinder.

#### Step 1: Cylinder

- Lateral surface area: $ 2\pi r h = 2 \times 3.14 \times 4 \times 10 = 251.2 $
- Bottom: $ \pi r^2 = 3.14 \times 16 = 50.24 $
- Top: Covered by cone → not included

So cylinder contributes: $ 251.2 + 50.24 = 301.44 $

#### Step 2: Cone

- Lateral surface area: $ \pi r l $
- Slant height $ l = \sqrt{r^2 + h^2} = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \approx 6.403 $
- LSA = $ 3.14 \times 4 \times 6.403 \approx 3.14 \times 25.612 \approx 80.43 $
- Base: covered → not included

So cone contributes: 80.43

#### Total Surface Area:
$$
301.44 + 80.43 = 381.87 \text{ cm}^2
$$

✔ Answer for Problem 6: 381.87 cm²

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✔ Final Answers:

| Problem | Surface Area |
|--------|--------------|
| 1 | 377.68 cm² |
| 2 | 1008 in² |
| 3 | 496.96 cm² |
| 4 | 1283.68 cm² |
| 5 | 576.2 ft² |
| 6 | 381.87 cm² |

Let me know if you'd like these rounded to nearest tenth or whole number!