Surface area calculation problems for composite figures using π = 3.14.

Let's solve each of the six composite figures step by step, calculating their surface areas. We'll use π ≈ 3.14, and round answers to two decimal places.

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1) Cone with a hemisphere on top

This is a cone with a hemisphere (half-sphere) attached to its base.

- Radius $ r = 10 \text{ m} $
- Height of cone $ h = 14 \text{ m} $
- The hemisphere sits on the base of the cone, so we don't double-count the circular base.

#### Step 1: Surface area of the cone (excluding base)
$$
\text{Lateral surface area of cone} = \pi r l
$$
We need slant height $ l $:
$$
l = \sqrt{r^2 + h^2} = \sqrt{10^2 + 14^2} = \sqrt{100 + 196} = \sqrt{296} \approx 17.20 \text{ m}
$$

So,
$$
\text{Cone lateral SA} = 3.14 \times 10 \times 17.20 = 539.84 \text{ m}^2
$$

#### Step 2: Surface area of hemisphere (only curved part, since base is glued to cone)
$$
\text{Curved surface area of hemisphere} = 2\pi r^2 = 2 \times 3.14 \times 10^2 = 628.00 \text{ m}^2
$$

> Note: The flat base of the hemisphere is not exposed (attached to cone), so we don't include it.

#### Total surface area:
$$
539.84 + 628.00 = \boxed{1167.84} \text{ m}^2
$$

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2) Rectangular prism with a triangular prism on top

The shape is a rectangular box with a triangular prism on top.

Dimensions:
- Rectangular prism: 15 ft × 11 ft × 5 ft
- Triangular prism on top: right triangle with legs 5 ft and 12 ft, hypotenuse 13 ft (since $ 5^2 + 12^2 = 13^2 $)
- The base of the triangle rests on the top face of the rectangular prism (so only one triangular face is visible)

But wait — the triangular prism has height 12 ft? Wait — let’s read carefully:

From diagram:
- Triangle: 5 ft and 12 ft legs, hypotenuse 13 ft.
- The base of the triangle is placed on the rectangle, so the length of the triangular prism is 11 ft (same as width of the box).

So the triangular prism has:
- Base triangle: 5 ft, 12 ft, 13 ft
- Length (depth): 11 ft

We need total surface area of the whole figure, subtracting overlapping areas.

#### Step 1: Surface area of rectangular prism
- Dimensions: 15 ft (length), 11 ft (width), 5 ft (height)
- SA = $ 2(lw + lh + wh) $
$$
= 2(15×11 + 15×5 + 11×5) = 2(165 + 75 + 55) = 2(295) = 590 \text{ ft}^2
$$

But the top face of the rectangular prism is partially covered by the triangular prism.

Area of top face: $ 15 \times 11 = 165 \text{ ft}^2 $

Area of triangle on top: $ \frac{1}{2} \times 5 \times 12 = 30 \text{ ft}^2 $

So exposed top area of box: $ 165 - 30 = 135 \text{ ft}^2 $

Wait — but when computing total surface area, we must exclude the parts that are glued together.

So instead, better approach:

- Compute full SA of both solids
- Subtract twice the area of the overlapping region (because both surfaces are hidden)

But actually: Only the bottom face of the triangular prism is glued to the top of the box.

So we subtract once the area of the triangle from both shapes? No — we just don’t count the glued faces.

So:
- Full SA of box: 590 ft²
- But we remove the area of the triangle (30 ft²) from the top face → so top face becomes $ 165 - 30 = 135 $
- So net box SA: $ 590 - 30 = 560 \text{ ft}^2 $? Wait — no.

Actually, original box SA includes entire top face (165). Since 30 ft² is covered, we subtract only 30 ft² from the box’s SA.

So box contributes: $ 590 - 30 = 560 \text{ ft}^2 $

Now add triangular prism:

Triangular prism has:
- Two triangular faces: each $ \frac{1}{2} \times 5 \times 12 = 30 \text{ ft}^2 $
- Three rectangular faces:
- Along 5 ft edge: $ 5 \times 11 = 55 \text{ ft}^2 $
- Along 12 ft edge: $ 12 \times 11 = 132 \text{ ft}^2 $
- Along 13 ft edge: $ 13 \times 11 = 143 \text{ ft}^2 $

Total SA of prism: $ 2 \times 30 + 55 + 132 + 143 = 60 + 330 = 390 \text{ ft}^2 $

But the bottom face (triangle, 30 ft²) is glued to the box — so we don’t include it.

So prism contributes: $ 390 - 30 = 360 \text{ ft}^2 $

Total surface area:
$$
\text{Box (adjusted)} + \text{Prism (adjusted)} = 560 + 360 = \boxed{920.00} \text{ ft}^2
$$

Wait — but is that correct?

Alternative method: Add all outer surfaces.

- Box: full SA = 590
- But top face has 30 ft² covered → so reduce by 30
- Prism: has 3 sides and 1 triangle (the top one), but bottom triangle is hidden → so include only:
- 2 triangles? No — only the top triangle is exposed?

Wait! The triangular prism has two triangular ends. But in this case, the bottom triangle is glued to the box, so only top triangle is exposed.

And the three rectangular faces are all exposed.

So prism contributes:
- Top triangle: 30 ft²
- 3 rectangles: 55 + 132 + 143 = 330 ft²
- Total: 360 ft²

Box:
- All faces except top is partially covered.
- Top face: $ 15 \times 11 = 165 $, minus 30 ft² covered → so 135 ft² exposed
- Other faces: front, back, left, right, bottom — all exposed
- So box SA = $ 2(15×5 + 11×5) + 15×11 $ (bottom) + 135 (top)
- $ 2(75 + 55) = 2(130) = 260 $
- Bottom: 165
- Top: 135
- Total: $ 260 + 165 + 135 = 560 $

Then prism adds: 360

Total: $ 560 + 360 = \boxed{920.00} \text{ ft}^2 $

✔ Correct.

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3) Two rectangular prisms stacked

Bottom prism: 18 yd × 12 yd × 7 yd
Top prism: 5 yd × 5 yd × 3 yd (stacked on top center)

Assume top prism is centered on bottom.

We compute total surface area, subtracting the overlapping area.

#### Step 1: SA of bottom prism
$$
SA = 2(lw + lh + wh) = 2(18×12 + 18×7 + 12×7) = 2(216 + 126 + 84) = 2(426) = 852 \text{ yd}^2
$$

#### Step 2: SA of top prism
$$
SA = 2(5×5 + 5×3 + 5×3) = 2(25 + 15 + 15) = 2(55) = 110 \text{ yd}^2
$$

Now, the bottom face of top prism is glued to the top of bottom prism.

Area of contact: $ 5 \times 5 = 25 \text{ yd}^2 $

So:
- Remove 25 from top prism (its bottom is hidden)
- Remove 25 from bottom prism (its top is partially covered)

So total SA:
$$
852 + 110 - 2 \times 25 = 962 - 50 = \boxed{912.00} \text{ yd}^2
$$

✔

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4) Rectangular prism with a cone on top

Rectangular prism: 5 in × 3 in × 6 in
Cone on top: radius 2 in, height 6 in

But note: the cone is sitting on the top face of the prism.

So the base of the cone is glued to the prism.

We need:
- SA of prism (minus the area where cone is attached)
- SA of cone (minus base)
- Add them

#### Step 1: SA of rectangular prism
$$
SA = 2(5×3 + 5×6 + 3×6) = 2(15 + 30 + 18) = 2(63) = 126 \text{ in}^2
$$

Top face area: $ 5 \times 3 = 15 \text{ in}^2 $

Area covered by cone base: $ \pi r^2 = 3.14 \times 2^2 = 12.56 \text{ in}^2 $

So exposed top of prism: $ 15 - 12.56 = 2.44 \text{ in}^2 $

But we don’t need to compute separately — better to do:

- Full prism SA: 126
- Subtract area of circle (since it's covered): 12.56
- So prism contributes: $ 126 - 12.56 = 113.44 \text{ in}^2 $

#### Step 2: SA of cone (lateral only, since base is glued)

Slant height $ l = \sqrt{r^2 + h^2} = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} \approx 6.32 \text{ in} $

Lateral SA of cone: $ \pi r l = 3.14 \times 2 \times 6.32 = 3.14 \times 12.64 \approx 39.70 \text{ in}^2 $

#### Total surface area:
$$
113.44 + 39.70 = \boxed{153.14} \text{ in}^2
$$

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5) Rectangular prism with a hemisphere on top

Prism: 5 ft × 10 ft × 5 ft
Hemisphere on top: diameter = 5 ft → radius $ r = 2.5 $ ft

Note: Hemisphere sits on top of prism, so its flat base is glued to the prism.

We calculate:
- SA of prism (minus top area covered)
- SA of hemisphere (curved part only, since base is glued)

#### Step 1: SA of prism
$$
SA = 2(5×10 + 5×5 + 10×5) = 2(50 + 25 + 50) = 2(125) = 250 \text{ ft}^2
$$

Top face area: $ 5 \times 10 = 50 \text{ ft}^2 $

Area covered by hemisphere base: $ \pi r^2 = 3.14 \times (2.5)^2 = 3.14 \times 6.25 = 19.625 \text{ ft}^2 $

So exposed top: $ 50 - 19.625 = 30.375 $, but again, better to subtract from total.

Prism contributes: $ 250 - 19.625 = 230.375 \text{ ft}^2 $

#### Step 2: Curved surface area of hemisphere
$$
= 2\pi r^2 = 2 \times 3.14 \times 6.25 = 39.25 \text{ ft}^2
$$

#### Total:
$$
230.375 + 39.25 = \boxed{269.62} \text{ ft}^2
$$

(Rounded to two decimals: 269.62)

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6) Sphere with cylinder removed? Or is it a sphere with a cylindrical hole?

Wait — looks like a sphere with a cylindrical hole through it, but the diagram shows a cylinder inside a sphere, or perhaps a spherical cap?

But looking at the image: It says "2.4" radius, and height 10 yd.

Wait — actually, the shape appears to be a sphere with radius 2.4 yd, and height 10 yd? That can't be — height of sphere should be diameter.

Wait — maybe it's a cylinder with hemispheres on both ends?

Yes! This is a capsule-shaped object: a cylinder with two hemispheres on the ends.

But the diagram shows:
- A cylinder with length 10 yd
- Two hemispheres on ends, each with radius 2.4 yd

So total shape: cylinder + two hemispheres = sphere + cylinder

But since two hemispheres = one sphere, so it's a sphere of radius 2.4 yd, but with a cylindrical "neck"?

No — actually, if the cylinder has radius 2.4 yd, and hemispheres have same radius, then it's a cylinder with hemispheres on both ends, forming a capsule.

But the total length is 10 yd.

Let’s see:
- Each hemisphere has diameter = $ 2 \times 2.4 = 4.8 $ yd
- So two hemispheres take up $ 4.8 $ yd
- Remaining length for cylinder: $ 10 - 4.8 = 5.2 $ yd

So:
- Cylinder: radius $ r = 2.4 $ yd, height $ h = 5.2 $ yd
- Two hemispheres → one full sphere

But surface area:
- Cylinder: lateral surface only (ends are covered by hemispheres)
- Hemispheres: curved surfaces only (bases are glued to cylinder)

So total SA:
- Lateral SA of cylinder: $ 2\pi r h = 2 \times 3.14 \times 2.4 \times 5.2 $
$$
= 6.28 \times 2.4 \times 5.2 = 6.28 \times 12.48 = 78.45 \text{ yd}^2
$$

- Surface area of sphere: $ 4\pi r^2 = 4 \times 3.14 \times (2.4)^2 = 4 \times 3.14 \times 5.76 = 4 \times 18.0864 = 72.3456 \text{ yd}^2 $

Total SA:
$$
78.45 + 72.3456 = \boxed{150.79} \text{ yd}^2
$$

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✔ Final Answers:

1) $ \boxed{1167.84} $ m²
2) $ \boxed{920.00} $ ft²
3) $ \boxed{912.00} $ yd²
4) $ \boxed{153.14} $ in²
5) $ \boxed{269.62} $ ft²
6) $ \boxed{150.79} $ yd²

Let me know if you'd like diagrams or clarifications!