Nets and Surface Area

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:

cube construction from a net

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

Level 1-3KS3AQAEdexcelOCRWJEC

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:

nets of 3D shapesLevel 1-3KS3AQAEdexcelOCRWJEC

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.

surface area 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.

cuboid surface area nets

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 1-3KS3AQAEdexcelOCRWJEC

Example 1: Surface Area of a Cylinder

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

[3 marks]

 

cylinder net surface area

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).

 

 

cylinder net surface area
cylinder net surface area

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 4-5KS3AQAEdexcelOCRWJEC

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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