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Lesson 7.1.3

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Surface Areas of Cylinders

and Prisms

Lesson 7.1.3

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

California Standards:

Measurement and Geometry 2.1

Use formulas routinely for findingthe perimeter and area of basic two-dimensional figures and the surface areaand volume of basic three-dimensional figures, including rectangles, parallelograms, trapezoids, squares, triangles, circles, prisms, and cylinders.

Measurement and Geometry 3.5

Construct two-dimensional patterns for three-dimensional models, such as cylinders, prisms, and cones.

Mathematical Reasoning 1.3

Determine when and how to break a problem into simpler parts.

What it means for you:

You’ll see how to work out the surface area of 3-D shapes like cylinders and prisms.

- Key words:
- net
- surface area
- cylinder
- prism

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Nets are very useful for finding the surface areaof 3-D shapes. They change a 3-D problem into a 2-D problem.

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Draw a Net to Work Out the Surface Area

The surface area of a three-dimensional solid is the total area of all its faces — it’s the area you’d paint if you were painting the shape.

The net of a three-dimensional solid can be folded to make a hollow shape that looks exactly like the solid.

So one way to work out the surface area of the solid is to work out the surface area of the net.

8 in

8 in

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Example 1

What is the surface area of this cube?

Solution

The net of the cube is six squares. So the surface area of the cube is equal to the area of six squares.

The area of each square is 8 × 8 = 64 in2.

So the surface area of the entire cube is 6 × 64 = 384 in2.

Solution follows…

20 cm

10 cm

10 cm

8.7 cm

10 cm

10 cm

8.7 cm

10 cm

10 cm

20 cm

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Example 2

What is the surface area of this prism?

Solution

The net of this prism has three identical rectangles.

The area of each rectangle is 10 × 20 = 200 cm2.

600 cm2

So the total surface area of the three rectangles is 3 × 200 = 600 cm2.

Solution continues…

Solution follows…

20 cm

10 cm

10 cm

8.7 cm

10 cm

10 cm

8.7 cm

10 cm

87 cm2

10 cm

20 cm

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Example 2

What is the surface area of this prism?

Solution (continued)

There are also two identical triangles.

Each has a base of 10 cm and a height of 8.7 cm. The area of each triangle is 0.5× 10 × 8.7 = 43.5 cm2.

So the surface area of both the triangles together is 2 × 43.5 = 87 cm2.

So the total surface area of the prism is 600 + 87 = 687 cm2.

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Guided Practice

Work out the surface area of the shapes shown in Exercises 1–3.

1.2.

3.

8 in

6 in

9 cm

9 cm

5 in

3 cm

10 cm

96 + 60 + 80 = 236 in2

Height = 8 cm

54 + 30 + 80 = 164 cm2

30 m

2 m

7 m

28 + 120 + 420 = 568 m2

Solution follows…

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Finding the Surface Area of Cylinders

The net of a circular cylinder has a rectangle

and two circles.

So you need to use the formula for the area of a circleto find its surface area.

3 ft

5 ft

3 ft

9.42 ft

5 ft

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Example 3

What is the surface area of this cylinder? Use p = 3.14.

Solution

The net of the cylinder has one rectangle and two identical circles.

To work out the area of the rectangle, you need to know its length.

It’s the same as the circumference of the circles, so it is 3 × p = 9.42 ft.

Solution continues…

Solution follows…

3 ft

5 ft

3 ft

7.065 ft2

9.42 ft

5 ft

7.065 ft2

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Example 3

What is the surface area of this cylinder? Use p = 3.14.

Solution (continued)

So the area of the rectangle is 9.42 × 5 = 47.1 ft2.

The circles have a diameter of 3 feet. So they have a radius of 1.5 feet.

47.1 ft2

The area of each circle is p × 1.52 = p × 2.25 = 7.065 ft2.

Solution continues…

3 ft

5 ft

3 ft

7.065 ft2

9.42 ft

5 ft

7.065 ft2

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Example 3

What is the surface area of this cylinder? Use p = 3.14.

Solution (continued)

Together the two circles have a surface area of 2 × 7.065 = 14.13 ft2.

47.1 ft2

So the total surface area of the cylinder is 47.1 + 14.13 = 61.23 ft2.

1 yd

2 in

3 ft

10 in

9 yd

3 ft

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Guided Practice

Find the surface areas of the cylinders in Exercises 4–6. Use p = 3.14.

4. 5. 6.

2 × 3.14 = 6.28

6.28 × 9 = 56.52

3.14 × 12 = 3.14

3.142 × 2 = 6.28

56.52 + 6.28 = 62.8 yd2

3 × 3.14 = 9.42

9.42 × 3 = 28.26

3.14 × 1.52 = 7.065

7.065 × 2 = 14.13

28.26 + 14.13 = 42.39 ft2

2 × 3.14 = 6.28

6.28 × 10 = 62.8

3.14 × 12 = 3.14

3.14 × 2 = 6.28

62.8 + 6.28 = 69.08 in2

Solution follows…

Two bases

Two bases

Lateral area

Lateral area

Area = (2 × base) + lateral area

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Use Formulas For Prism and Cylinder Surface Areas

The way you work out the surface area of a cylinder, and the way you work out the surface area of a prism are similar.

The surface area of either is twice the area of the base plus the area of the part between the bases of the net.

The part between the bases is sometimes called the lateral area.

6 cm

1 cm

20 in

5 cm

30 in

7 in

10 ft

10 ft

7 in

30 ft

8 ft

Vertical height = 6.8 feet

7 in

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Independent Practice

Work out the surface areas of the shapes shown in Exercises 1–4. Use p = 3.14.

1. 2.

3. 4.

82 cm2

2512 in2

894.4 ft2

294 in2

Solution follows…

3 ft

6 ft

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Independent Practice

5. A statue is to be placed on a marble stand, in the shape of a regular-hexagonal prism.

Find the area of the stand’s base, given that the stand has a surface area of 201.5 square feet and dimensions as shown.

46.75 ft2

Solution follows…

1 m

6 m

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Independent Practice

The inside of a large tunnel in a children’s play area is to be painted. The tunnel is 6 meters long and 1 meter tall. It is open at each end.

6. What is the area to be painted?

7. Cans of paint each cover 5 m2. How many cans do they need to buy?

18.84 m2

4

Solution follows…

Lesson

7.1.3

Surface Areas of Cylinders and Prisms

Round Up

Working out the surface area of a 3-D shape means adding together the area of every part of the outside.

One way to do that is to add together the areas of different parts of the net.

Just make sure you can remember the triangle, rectangle, and circle area formulas.