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Australia
Year 10

7.01 Surface area of prisms and cylinders

Lesson

Use of net to find the surface area

The surface area of a prism is the sum of the areas of all the faces.

To find the surface area of a prism, we need to determine the kinds of areas we need to add together.

Consider this cube:

A cube with side length of 4. We can see the front, side and top faces.

From this angle we can see three square faces with side length 4, and the area of these faces will contribute to the surface area. But we also need to consider the faces we can't see from this view.

A net is has 6 identical squares: a row of 4 squares, 1 square at the bottom, and 1 square at the top of the second square.

By drawing the net of the cube we can see all the faces at once.

Now we know that the surface is made up of six identical square faces, and finding the surface area of the cube is the same as finding the area of a square face and multiplying that by 6: A=6 \times 4^{2}=96

Examples

Example 1

A rectangular prism with width of 5 metres, length of 7 metres, and height of 15 metres.

Consider the rectangular prism with a width, length and height of 5\text{ m},\,7\text{ m} and 15\text{ m} respectively.

Find the surface area.

Worked Solution
Create a strategy

We need to add the areas of the 3 pairs of equal rectangular faces: Front and back, the left and right sides, and the top and bottom.

4 rectangular prisms showing the pairs of equal sides that make up the surface area. Ask your teacher for more information.
Apply the idea
\displaystyle \text{Front and back}\displaystyle =\displaystyle 2 \times 15 \times 5Multiply the area by 2
\displaystyle =\displaystyle 150\text{ m}^2Evaluate
\displaystyle \text{Sides}\displaystyle =\displaystyle 2 \times 15 \times 7Multiply the area by 2
\displaystyle =\displaystyle 210\text{ m}^2Evaluate
\displaystyle \text{Top and bottom}\displaystyle =\displaystyle 2 \times 7 \times 5Multiply the area by 2
\displaystyle =\displaystyle 70\text{ m}^2Evaluate
\displaystyle \text{Surface area}\displaystyle =\displaystyle 150+210+70Add the areas
\displaystyle =\displaystyle 430\text{ m}^{2}Evaluate
Idea summary

The surface area of a prism is the sum of the areas of all the faces.

Net of a cylinder

We can use a similar method for a cylinder. By "unwrapping" the cylinder we can treat the curved surface as a rectangle, with one side length equal to the height of the cylinder, and the other the perimeter (circumference) of the base circle. This is given by 2\pi r, where r is the radius.

The image shows a cylinder and its net with with dimensions. Ask your teacher for more information.

This means the surface area of the curved part of a cylinder is 2\pi rh, where r is the radius and h is the height.

Exploration

We can see how the cylinder unrolls to make this rectangle in the applet below:

Loading interactive...

The length of the rectangle that wraps around the cylinder is equal to the circumference of the face of the cylinder.

To find the surface area of the whole cylinder, we need to add the area of the top and bottom circles to the area of the curved part. Both of these circles have an area of \pi r^{2}, so the surface area of a cylinder is: \text{Surface area of a cylinder}=2\pi r^{2}+2\pi rh, where r is the radius and h is the height of the cylinder.

Examples

Example 2

Consider the following cylinder.

A cylinder with radius of 3 metres and a height of 8 metres.
a

Find the curved surface area of the cylinder to two decimal places.

Worked Solution
Create a strategy

Use the formula of the curved surface area of the cylinder given by: 2\pi r h.

Apply the idea

We are given: r=3 and h=8.

\displaystyle \text{Curved surface area}\displaystyle =\displaystyle 2\pi r hUse the given formula
\displaystyle =\displaystyle 2\pi \times 3 \times 8Substitute r and h
\displaystyle =\displaystyle 150.80\text{ m}^{2}Evaluate and round
b

Using the result from part (a) or otherwise, find the total surface area of the cylinder. Round your answer to two decimal places.

Worked Solution
Create a strategy

Use the formula of the surface area of the cylinder: 2\pi r^2+2\pi rh.

Apply the idea

We are given: r=3 and h=8.

\displaystyle \text{Surface Area}\displaystyle =\displaystyle 2\pi r^2+2\pi rhUse the given formula
\displaystyle =\displaystyle 2\pi \times(3)^2+2\pi\times3\times8Subsitute the given values
\displaystyle =\displaystyle 207.35\text{ m}^2Evaluate

Example 3

Find the surface area of the figure shown.

Two parallelogram faces and four rectangular faces, where the opposite faces of the prism are equal. Ask your teacher for more information.
Worked Solution
Create a strategy

Use the formula on area of a parallelogram given by: A=bh

Apply the idea

The front and back are parallelograms with bases of b=9\text{ cm} and perpendicular heights of h=10\text{ cm}.

The four rectangular sides can be unwrapped to form a large rectangle with length 11+9+11+9=40\text{ cm}, and width of 5\text{ cm}.

\displaystyle \text{Surface Area}\displaystyle =\displaystyle \text{Area of }2\text{ parallelograms}+ \text{Area of unwrapped rectangle}
\displaystyle =\displaystyle 2\times 9\times 10+ 5\times 40
\displaystyle =\displaystyle 380 \text{ cm}^2
Idea summary
\displaystyle \text{Surface area of a cylinder}=2\pi r^{2}+2\pi rh
\bm{r}
is the radius
\bm{h}
is the height of the cylinder

Outcomes

ACMMG242

Solve problems involving surface area and volume for a range of prisms, cylinders and composite solids

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