Solved Sheet1 Surface Area of Composite Figures Find the | Chegg.com - Free Printable

Let's solve each of the composite figures step by step to find their surface areas, rounding answers to two decimal places and using π = 3.14.

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1) Hemisphere on top of a Cone

- Radius $ r = 2 $ ft
- Slant height of cone $ l = 12 $ ft

Surface Area = Surface area of hemisphere (curved only) + Lateral surface area of cone

> Note: The base of the hemisphere is attached to the cone, so we don’t include the circular base of the hemisphere or the base of the cone (since they are internal).
> But the cone’s lateral surface and hemisphere’s curved surface are exposed.

#### Hemispherical part:
Curved surface area of a hemisphere = $ 2\pi r^2 $

$$
= 2 \times 3.14 \times (2)^2 = 2 \times 3.14 \times 4 = 25.12 \text{ ft}^2
$$

#### Cone lateral surface area:
$ \pi r l = 3.14 \times 2 \times 12 = 75.36 \text{ ft}^2 $

#### Total Surface Area:
$$
25.12 + 75.36 = 100.48 \text{ ft}^2
$$

✔ Answer: 100.48 ft²

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2) Cone on top of a Cylinder

- Radius $ r = 11 $ yd
- Height of cylinder $ h_{\text{cyl}} = 13 $ yd
- Slant height of cone $ l = 9 $ yd

We need:
- Lateral surface area of cylinder
- Lateral surface area of cone
- Top of cylinder is covered by cone, so no top base.
- Base of cylinder is exposed → include bottom base
- Cone has no base (attached to cylinder), so only lateral surface

#### Cylinder lateral surface area:
$ 2\pi r h = 2 \times 3.14 \times 11 \times 13 = 907.48 \text{ yd}^2 $

#### Cone lateral surface area:
$ \pi r l = 3.14 \times 11 \times 9 = 310.86 \text{ yd}^2 $

#### Cylinder base (bottom):
$ \pi r^2 = 3.14 \times (11)^2 = 3.14 \times 121 = 380.14 \text{ yd}^2 $

#### Total Surface Area:
$$
907.48 + 310.86 + 380.14 = 1598.48 \text{ yd}^2
$$

✔ Answer: 1598.48 yd²

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3) Sphere with a cylindrical hole through the center? Wait — looks like a sphere split into two hemispheres around a cylinder.

Wait — this is actually a sphere with a cylinder removed from the middle, but looking at the diagram:

It shows a cylinder of height 10 ft and radius 6 ft, with a hemisphere on top and bottom.

So it's: Two hemispheres + one cylinder → makes a sphere-like shape with a cylinder in the middle?

Wait — actually, it's a cylinder with a hemisphere on top and bottom, forming a spherocylinder.

But since the radius is same for all parts (6 ft), and height of cylinder is 10 ft.

But note: the top and bottom of the cylinder are covered by hemispheres, so we don't count the bases of the cylinder.

Also, the hemispheres have no flat bases because they're glued to the cylinder.

So total surface area:
- Lateral surface area of cylinder
- Curved surface area of top hemisphere
- Curved surface area of bottom hemisphere

But since both hemispheres are full curved surfaces, and they meet the cylinder at the edges, we only add their curved parts.

So:

#### Cylinder lateral surface:
$ 2\pi r h = 2 \times 3.14 \times 6 \times 10 = 376.8 \text{ ft}^2 $

#### Top hemisphere curved SA:
$ 2\pi r^2 = 2 \times 3.14 \times 36 = 226.08 \text{ ft}^2 $

#### Bottom hemisphere curved SA:
Same as top → $ 226.08 \text{ ft}^2 $

#### Total:
$$
376.8 + 226.08 + 226.08 = 828.96 \text{ ft}^2
$$

✔ Answer: 828.96 ft²

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4) Cone on top of a Hemisphere

- Radius $ r = 5 $ m
- Slant height of cone $ l = 12 $ m

This is similar to #1, but now hemisphere is below, cone on top.

But again, the base of the cone is attached to the flat face of the hemisphere, so we don't include that area.

We include:
- Curved surface of hemisphere (only outer curve)
- Lateral surface of cone
- No base of cone, no base of hemisphere (they’re joined)

But wait: the hemisphere has a flat base — is it exposed? In this case, the figure seems to be standing on the flat base of the hemisphere.

So we must decide: Is the flat base of the hemisphere included?

Looking at the diagram: It appears the figure is sitting on the flat base of the hemisphere, so yes, we should include that.

But let's check carefully.

In many such problems, if the figure is resting on a base, the base is included unless stated otherwise.

But here, the cone is on top, so the hemisphere's flat base is on the ground → include it.

So surface area includes:
- Curved surface of hemisphere
- Lateral surface of cone
- Base of hemisphere (circle)

#### Curved surface of hemisphere:
$ 2\pi r^2 = 2 \times 3.14 \times 25 = 157.0 \text{ m}^2 $

#### Lateral surface of cone:
$ \pi r l = 3.14 \times 5 \times 12 = 188.4 \text{ m}^2 $

#### Base of hemisphere (circle):
$ \pi r^2 = 3.14 \times 25 = 78.5 \text{ m}^2 $

#### Total:
$$
157.0 + 188.4 + 78.5 = 423.9 \text{ m}^2
$$

✔ Answer: 423.90 m²

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5) Two cones joined at their bases (like a double cone)

- Each cone has:
- Slant height $ l = 10 $ yd
- Base diameter = 19 yd → radius $ r = 9.5 $ yd

They are joined at the bases, so the bases are not exposed.

So total surface area = lateral surface area of both cones

Each cone:
$ \pi r l = 3.14 \times 9.5 \times 10 = 298.3 \text{ yd}^2 $

Two cones:
$ 2 \times 298.3 = 596.6 \text{ yd}^2 $

✔ Answer: 596.60 yd²

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6) Cylinder with a hemisphere on top

- Radius $ r = 4 $ ft
- Height of cylinder $ h = 5 $ ft

The hemisphere is on top, so:
- Cylinder lateral surface area
- Cylinder base (bottom)
- Hemisphere curved surface (top)
- Hemisphere base is attached to cylinder, so not exposed
- Cylinder top is covered, so not included

So:
- Lateral surface of cylinder
- Base of cylinder
- Curved surface of hemisphere

#### Cylinder lateral surface:
$ 2\pi r h = 2 \times 3.14 \times 4 \times 5 = 125.6 \text{ ft}^2 $

#### Cylinder base:
$ \pi r^2 = 3.14 \times 16 = 50.24 \text{ ft}^2 $

#### Hemisphere curved surface:
$ 2\pi r^2 = 2 \times 3.14 \times 16 = 100.48 \text{ ft}^2 $

#### Total:
$$
125.6 + 50.24 + 100.48 = 276.32 \text{ ft}^2
$$

✔ Answer: 276.32 ft²

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✔ Final Answers:

| Problem | Surface Area |
|--------|--------------|
| 1) | 100.48 ft² |
| 2) | 1598.48 yd² |
| 3) | 828.96 ft² |
| 4) | 423.90 m² |
| 5) | 596.60 yd² |
| 6) | 276.32 ft² |

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