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9.01 Surface area of prisms and cylinders

Lesson
Summary

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

The surface area of a cylinder of radius $r$r and height $h$h is $A=2\pi rh+2\pi r^2$A=2πrh+2πr2.

 

Finding surface area using the net

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

Exploration

Consider this cube:

From this angle we can see three square faces with side length $4$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.

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$6:

$A=6\times4^2$A=6×42$=$=$96$96

Using the net is useful for seeing exactly what areas need to be added together, but it isn't always this easy to find.

 

Using the 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$2πr, where $r$r is the radius.

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

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

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 $\pi r^2$πr2, so the surface area of a cylinder is:

$A=2\pi rh+2\pi r^2$A=2πrh+2πr2

 

Another way to find surface area

Much like how the curved surface of the cylinder unwrapped to become a rectangle, the non-base faces of a prism can be unfolded to make a large rectangle.

Worked example

Find the surface area of the trapezoidal prism.

Think: To find the surface area, we can add the areas of the two base faces to the large folded rectangul joining the two faces. Here is the net of the prism:

Do: Find the area of the base face using the area formula for a trapezium:

$\text{Base area}=\frac{1}{2}\left(9+12\right)\times5$Base area=12(9+12)×5

Since the perimeter of the base is $6+9+7+12$6+9+7+12, the area of the large rectangle connecting the base faces is:

$\text{Rectangular area}=\left(6+9+7+12\right)\times15$Rectangular area=(6+9+7+12)×15

Adding two base areas together with the rectangular area gives us the total surface area:

$\text{Surface area}=2\times\frac{1}{2}\left(9+12\right)\times5+\left(6+9+7+12\right)\times15$Surface area=2×12(9+12)×5+(6+9+7+12)×15 $=$= $615$615

Reflect: We could have calculated the area of each of the four rectangular faces individually, but by using the perimeter of the base we saved on calculation time.

 

Practice questions

Question 1

Consider the following rectangular prism with a width, length and height of $5$5 m, $7$7 m and $15$15 m respectively. Find the surface area.

Question 2

Find the surface area of the figure shown.

Question 3

Consider the following cylinder.

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

  2. Using the result from part (a) or otherwise, find the total surface area of the cylinder.

    Round your answer to two decimal places.

Question 4

Find the surface area of the figure shown.

Round the answer to two decimal places.

Outcomes

VCMMG343

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

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