# Nets and Surface Area

KS3AQAEdexcelOCRWJEC

## Nets and Surface Area

A net is a deconstructed 3D shape folded out flat. Nets can be helpful when we want to calculate the surface area of a 3D shape.

## Skill 1: Nets of Cubes

A net shows us each face of a shape when it is laid out flat, and there are often many different nets for a 3D shape. Below are some examples of nets of cubes.

Note: these are only some of the nets of a cube – there are many more

Each of the nets above can be folded up to construct a cube, like so:

Cubes are the simplest nets you will encounter, so make sure you are comfortable drawing them before moving on to harder shapes.

Level 1-3KS3

## Skill 2: Nets of Other Shapes

There are some other common shapes that you should familiarise yourself with, including prisms and pyramids.

A prism is like a 2D shape ‘stretched out’ with the original 2D at either end. This makes drawing a net of a prism fairly simple, as it will feature the 2D shape at both ends with rectangles between them that form the ‘stretched out’ section. A cuboid is also a type of prism, so the net of a cuboid follows the same pattern.

A pyramid always features a 2D shape as its base, with each edge linked to a triangular face. These triangular faces all converge to a central point above the centre of the base. The nets of pyramids are easy to draw – just draw the shape of the base with triangles attached to each side.

A cylinder can be thought of as a circular prism. Its net follows the same pattern as the nets of prisms.

Some examples of nets of 3D shapes are shown below:

Level 1-3KS3

## Skill 3: Finding the Surface Area

We can use nets to find the surface area of a 3D shape – the combined area of all the faces.

Consider the following net of a cuboid.

We can use the net to work out the surface area. There are $3$ ‘pairs’ of faces that are the same – if we calculate the areas of these individual faces we can add them all up to get the surface area of the cuboid.

The total surface area is therefore:

$2 \times48 + 2 \times32 + 2 \times 24 = 208$ cm$^2$

Note: remember that the units of area are cm$^2$, m$^2$ etc.

Level 1-3KS3

## Example 1: Surface Area of a Cylinder

Calculate the surface area of cylinder $A$ from the net shown.

[3 marks]

The diameter of the circular face is $6$ cm, meaning the radius of the circle is $3$ cm. We can calculate the area of one circular face using the formula:

Area of a circle $= \pi r^2$

So the area of both circular faces on the cylinder is:

$2\times \pi \times 3^2=18\pi = 56.55$ cm$^2$

To calculate the area of the rectangle, we need to know the length – this is the same as the circumference of the circle ($2\pi r$).

The length of the rectangle is therefore:

$2\pi r = 6 \pi$ cm

We can now calculate the area of the rectangle face:

$6 \pi \times 6 = 36 \pi = 113.10$ cm$^2$

The total surface area of the cylinder is therefore:

$\text{Area} = 56.55 + 113.10=169.65$ cm$^2$

Level 4-5KS3

## Note:

There is a simple formula for calculating the surface area of a cylinder:

Surface area of a cylinder $=2\pi rh + 2\pi r^2$

Where $r$ is the radius of the circle face and $h$ is the height of the cylinder.

Substituting values into this formula is a quick way of performing the calculations shown in the example above.

## Example Questions

Only nets A and C will form a cube.

The faces on net B will overlap when folded over and net D has too many faces.

Your completed sketch should look something like this:

We need to use the following formula:

Area of a triangle $=\dfrac{1}{2} \times \text{base} \times \text{height}$

So the area of both triangular faces is:

$2\times \dfrac{1}{2}\times4\times3 = 12$ cm$^2$

The area of the three rectangular faces is:

$3\times10\times4 = 120$ cm$^2$

The total surface area is:

$12+120=132$ cm$^2$

Use the formula and substitute in the values of $r=2$ and $h= 8$:

\begin{aligned}\text{Surface area} &= 2\pi r^2 + 2\pi rh \\ &=(2 \times \pi \times 2^2) + (2\times \pi \times 2 \times 8) \\ &= 8\pi + 32\pi \\ &= 25.13+100.53\\ &=125.66 \text{ cm}^2\end{aligned}

The surface area of the pyramid is given by:

$\text{Surface area}= 4\times \text{area of triangle }+ \text{area of square base}$

Use the formula for the area of the triangle:

$\text{Area of a triangle}=\dfrac{1}{2} \times \text{base} \times \text{height}$

Substitute in the values from the net (base $=4$ cm and height $=9$ cm) to calculate the area:

\begin{aligned}\text{Surface area} &= (4\times \dfrac{1}{2}\times4\times9) + (4\times4) \\[1.5em] &= 72+16 \\[1.5em] &= 88 \text{ cm}^2 \end{aligned}

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