One to one maths interventions built for KS4 success

Weekly online one to one GCSE maths revision lessons now available

In order to access this I need to be confident with:

Faces, edges and vertices Substitution Area of a triangle Area of quadrilaterals Pythagoras’ theoremThis topic is relevant for:

Here we will learn about the surface area of a pyramid, including how to find the area of the base and the triangular faces of a pyramid and how to structure a logical solution to calculate the surface area of a pyramid.

There are also surface area of a pyramid* *worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

The **surface area of a pyramid** is the **total sum of the area of each face** of the pyramid.

For example, a square based pyramid consists of a square, and four triangular faces.

**Note:**

There are many types of pyramids

- A rectangular based pyramid consists of a rectangle and two pairs of identical triangles
- A triangular based pyramid consists of 4 triangular faces.
- The base could be any regular polygon or irregular polygon e.g pentagonal or hexagonal pyramids. You need to make sure you have found the area of all lateral faces and added it to the area of the base of the pyramid.

To calculate the total surface area of a pyramid, we need to calculate the area of each face of a pyramid, and then add these areas together. This is exactly the same approach if we were calculating the area of a compound shape.

E.g. Work out the surface area of the pyramid.

\begin{aligned} &\text{Area of the base }=5\times{5}=25\text{ cm}^{2} \\\\ &\text{Area of a triangular face }=\frac{1}{2}\times{8}\times{5}=20\text{ cm}^{2} \\\\ &\text{Surface area }=25+20+20+20+20=25+(4\times{20})=105\text{ cm}^{2} \end{aligned} .

As one single calculation, we can say:

\begin{aligned} \text{Surface area of pyramid}&= \text{Area of base} + \text{Area of triangular faces}\\\\ &=5^2+4\times (\frac{1}{2}\times 5 \times 8)\\\\ &=25+80\\\\ &=105 \ cm^2 \end{aligned}Visualising the individual faces of a pyramid is a key skill.

The **apex** is the convergent point to which all the slanted edges of the pyramid meet.

The apex is located above the **base** of a pyramid, and the **height** of the pyramid is the **perpendicular** distance between the base and the apex.

Let us have a look to see what happens to the triangular faces when we move the apex for a square based pyramid.

*For each pyramid, the base is a congruent square.

**Note:** It is likely that a pyramid will be regular, nevertheless, you may still have to find the surface area of an oblique pyramid (see example 6 below). If you follow the steps carefully, you should be able to find the surface area of any pyramid.

In order to find the surface area of a pyramid:

**Calculate the area of the base.****Calculate the area of all of the triangular faces.****Calculate the total surface area.****Write the answer, including the units.**

Get your free surface area of a pyramid worksheet of 20+ volume and surface area of pyramids questions and answers. Includes reasoning and applied questions.

COMING SOONGet your free surface area of a pyramid worksheet of 20+ volume and surface area of pyramids questions and answers. Includes reasoning and applied questions.

COMING SOONThe apex of a square-based pyramid is directly above the centre of the base. Using the diagram below, calculate the surface area of the pyramid.

**Calculate the area of the base.**

The base is a square with the area 6\times{6}=6^{2}=36\text{ cm}^2 .

2**Calculate the area of all of the triangular faces.**

As the apex of the pyramid is directly above the centre of the base, all four triangular faces are identical and so we can calculate the area of one of these triangles, and multiply the solution by 4 .

The area of a triangle is calculated using the formula A=\frac{1}{2}bh where b is the base of the triangle, and h is the vertical height of the triangle. We therefore have the area of each triangle to be:

\begin{aligned} A&=\frac{1}{2}\times{b}\times{h}\\\\ &=\frac{1}{2}\times{10}\times{6}\\\\ &=30\text{ cm}^{2} \end{aligned} 4A=30\times{4}=120\text{ cm}^{2}.3**Calculate the total surface area.**

Adding the values for the area of the base and the area of the four triangles, we get:

SA=36+120=1564**Write the answer, including the units.**

A softbox light takes the form of a solid, square-based pyramid. The screen has a width of 1m, and the sides of the softbox light have a perpendicular height of 1.2m.

The apex of the pyramid is perpendicular to the centre of the screen.

Calculate the surface area of the softbox light.

**Calculate the area of the base.**

Here, the screen is the base. As the screen is a square, the area of the square is 1\times{1}=1\text{ m}^{2} .

**Calculate the area of all of the triangular faces.**

As the apex is perpendicular to the centre of the screen, each side of the softbox light is an identical, isosceles triangle.

\begin{aligned} A&=\frac{1}{2}\times{b}\times{h}\\\\ &=\frac{1}{2}\times{1}\times{1.2}\\\\ &=0.6\text{ m}^{2} \end{aligned}

4A=4\times{0.6}=2.4\text{ m}^2

**Calculate the total surface area.**

SA=1+2.4=3.4

**Write the answer, including the units.**

SA=3.4\text{ m}^{2}

A regular tetrahedron is a special type of triangular-based pyramid as all faces of a regular tetrahedron are congruent, equilateral triangles. Using the diagram below, calculate the surface area of the regular tetrahedron.

**Calculate the area of the base.**

As each face of the tetrahedron is the same equilateral triangle, the area of the triangular base is the same as the area of each of the other triangular faces.

\begin{aligned}
A&=\frac{1}{2}\times{b}\times{h}\\\\
&=\frac{1}{2}\times{8}\times{4\sqrt{3}}\\\\
&=16\sqrt{3}
\end{aligned}

**Calculate the area of all of the triangular faces.**

Again the area of each face is the same and so the area of the other three triangular faces is also equal to 16\sqrt{3} .

**Calculate the total surface area.**

Adding the area of each face together, we get:

SA=16\sqrt{3}+16\sqrt{3}+16\sqrt{3}+16\sqrt{3}=4\times{16}\sqrt{3}=64\sqrt{3}

**Write the answer, including the units.**

SA=64\sqrt{3}\text{ cm}^{2}

ABCDE is a rectangular-based pyramid. The vertex E is directly above the centre of the base ABCD . Calculate the surface area of the pyramid. Write your answer in square centimetres.

**Calculate the area of the base.**

The units for the base are not the same, so we need to make sure that they are for the calculations to be accurate. As the question asks for the answer to be written in square centimetres, we need to convert the units to centimetres first.

1.2m=120cm.

The area of the base is therefore:

A=120\times{90}=10800\text{ cm}^2

**Calculate the area of all of the triangular faces.**

The pyramid has two pairs of congruent faces: ADE=BCE and CDE=ABE.

Calculating the area of ADE , we have

A=\frac{1}{2}\times{120}\times{380}=22800\text{ cm}^2

Triangle ADE=22800\text{ cm}^2

Triangle BCE=22800\text{ cm}^2

Calculating the area of CDE , we have

A=\frac{1}{2}\times{90}\times{382}=17190\text{ cm}^2

Triangle CDE=17190\text{ cm}^2

Triangle ABE=17190\text{ cm}^2

**Calculate the total surface area.**

Adding all of the areas together, we have:

\begin{aligned}
SA&=10800+22800+22800+17190+17190\\\\
&=10800+2(22800)+2(17190)\\\\
&=90780
\end{aligned}

**Write the answer, including the units.**

SA=90780\text{ cm}^2

Using the net below, calculate the surface area of the square-based pyramid. Write your answer to 1 decimal place.

**Calculate the area of the base.**

The base is a square and so the area of the base is 5.6\times{5.6}=31.36\text{ cm}^2

**Calculate the area of all of the triangular faces.**

To calculate the area of a triangle, we need to know the height of the triangle. To do this, we need to use Pythagoras theorem (remember the base of the right angle triangle is half the width of the isosceles triangle in the net):

h=\sqrt{7.8^{2}-2.8^{2}}=\sqrt{53}

The area of the triangle above is therefore:

\begin{aligned} A&=\frac{1}{2}\times{5.6}\times{\sqrt{53}} \\\\ A&=20.38430769… \end{aligned}

The area of all four of these triangles is therefore:

4A=4\times{20.38430769}=81.53723076\text{ cm}^2

**Calculate the total surface area.**

Adding all of the areas together, we have:

\begin{aligned}
SA&=31.36+81.53723076\\\\
&=112.8972308
\end{aligned}

**Write the answer, including the units.**

SA=112.9\text{ cm}^2\text{ (1dp)}

Below is the oblique, square-based pyramid ABCDE. The apex E is directly above the vertex D . Given that AB=8cm , \ BC=4cm, \ DE=6cm, \ CE=10cm, and AE=2\sqrt{13}\text{ cm} , calculate the surface area of the square-based pyramid.

**Calculate the area of the base.**

The base is a rectangle. The area of the rectangle is 8\times{4}=32\text{ cm}^2

**Calculate the area of all of the triangular faces.**

Let’s work through each triangular face separately as they are all different sizes.

\begin{aligned} &ABE=\frac{1}{2}\times{8}\times{2\sqrt{13}}=8\sqrt{13}\text{ cm}^{2} \\\\ &CDE=\frac{1}{2}\times{8}\times{6}=24\text{ cm}^{2} \\\\ &ADE=\frac{1}{2}\times{4}\times{6}=12\text{ cm}^{2} \end{aligned}

BCE=\frac{1}{2}\times{4}\times{10}=20\text{ cm}^2

**Calculate the total surface area.**

Adding up all of the values for each face of the square-based pyramid, we have the surface area to be:

SA=32+8\sqrt{13}+24+12+20=116.8444102…

**Write the answer, including the units.**

SA=116.8\text{ cm}^2

**The height of the triangle is the perpendicular height**

The height of the triangle needs to be the perpendicular height. Then we can use the perpendicular height of the triangle to find its area.

**Surface area has square units**

The surface area will have square units such as square centimetres ( cm^2 ) or square metres ( m^2 ).

**Be accurate**

When there are two or more steps in your workings, do not round your workings. For example – do not round the area of the triangular faces. Only round at the end of the question so that your answer is accurate.

**Take care with rounding**

At the end of the question, make sure you round your answer to the correct number of decimal places or significant figures.

Surface area of a pyramid is part of our series of lessons to support revision on pyramids. You may find it helpful to start with the main pyramid lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

1. Calculate the surface area of the pyramid below with the apex directly above the centre of the base:

384 \text{ cm}^2

260 \text{ cm}^2

140 \text{ cm}^2

224 \text{ cm}^2

\begin{aligned}
\text{Surface area of pyramid}&= \text{Area of base} + \text{Area of triangular faces}\\\\
&=8^2+4\times (\frac{1}{2}\times 8\times 10)\\\\
&=64+160\\\\
&=224 \ cm^2
\end{aligned}

2. Find the surface area of the pyramid.

781 \text{ m}^2

330 \text{ m}^2

451 \text{ m}^2

555 \text{ m}^2

\begin{aligned}
\text{Surface area of pyramid}&= \text{Area of base} + \text{Area of triangular faces}\\\\
&=11^2+4\times (\frac{1}{2}\times 11\times 15)\\\\
&=121+330\\\\
&=451 \ m^2
\end{aligned}

3. The apex E lies perpendicular to the centre of the base ABCD of the pyramid below. Calculate the surface area of the pyramid.

7280 \text{ cm}^2

4788 \text{ cm}^2

1366 \text{ cm}^2

3077 \text{ cm}^2

\begin{aligned}
\text{Surface area of pyramid}&= \text{Area of base} + \text{Area of triangular faces}\\\\
&=(42\times{10})+2\times (\frac{1}{2}\times{42}\times{52})+2\times (\frac{1}{2}\times{10}\times{47.3})\\\\
&=420+2184+473\\\\
&=3077 \text{ cm}^2
\end{aligned}

4. The apex of the pyramid below is directly above the centre of the base. Use the diagram below to calculate the surface area of the pyramid. Write your answer to 1 decimal place.

27.5 \text{ cm}^2

55.0 \text{ cm}^2

38.8 \text{ cm}^2

25.9 \text{ cm}^2

The pyramid is a square-based pyramid as the width is 1.95 \times 2=3.9cm.

\begin{aligned} \text{Surface area of pyramid}&= \text{Area of base} + \text{Area of triangular faces}\\\\ &=3.9^2+4\times (\frac{1}{2}\times 3.9\times 5.1)\\\\ &=15.21+39.78\\\\ &=54.99\\\\ &=55.0\text{ cm}^{2}\text{ (1dp)} \end{aligned}

5. Here is the net of a square-based pyramid where the apex is directly above the centre of the base. Calculate the surface area of the pyramid:

336 \text{ cm}^2

240 \text{ cm}^2

264 \text{ cm}^2

384 \text{ cm}^2

First we need to find the height of the triangular face.

h=\sqrt{10^{2}-6^{2}}=8\text{ cm}

\begin{aligned} \text{Surface area of pyramid}&= \text{Area of base} + \text{Area of triangular faces}\\\\ &=12^2+4\times (\frac{1}{2}\times 12\times 8)\\\\ &=144+192\\\\ &=336 \text{ cm}^2 \end{aligned}

6. Here is the net of a square-based pyramid where the apex is directly above a vertex of the base. Calculate the surface area of the pyramid.

48 \text{ cm}^2

68.8 \text{ cm}^2

112 \text{ cm}^2

64 \text{ cm}^2

The area of the base is 4\times{4}=16\text{ cm}^2

The area of two of the triangles is:

(\frac{1}{2}\times{6}\times{4})\times{2}=24\text{ cm}^2

The height of the other two triangular faces is (using Pythagoras theorem):

h=\sqrt{4^{2}+6^{2}}=2\sqrt{13}

The area of the other two triangles is now equal to:

(\frac{1}{2}\times{2\sqrt{13}}\times{4})\times{2}=8\sqrt{13}\text{ cm}^2

The total surface area is therefore:

SA=16+24+8\sqrt{13}=68.8\text{ cm}^2\text{ (1dp)}

1. The diagram shows a pyramid. The apex is perpendicular to the centre of the square base.

Calculate the surface area of the pyramid.

**(3 marks)**

Show answer

\frac{1}{2}\times 9 \times 12=54

**(1)**

9^2+4\times 54

**(1)**

297 \ cm^2

**(1)**

2. The diagram shows a square-based pyramid.

Calculate the surface area of the pyramid.

Give your answer in cm^2.

**(4 marks)**

Show answer

1.3 \text{ m} = 130 \text{ cm}

**(1)**

\frac{1}{2}\times 70 \times 130=4550

**(1)**

70^2+4\times 4550

**(1)**

23100

**(1)**

3. Here is the net of a square-based pyramid.

Calculate the surface area of the pyramid.

Give your answer in cm^2.

**(4 marks)**

Show answer

\sqrt{13^2-5^2}=12

**(1)**

\frac{1}{2}\times 10 \times 12=60

**(1)**

10^2+4\times 60

**(1)**

340

**(1)**

4. (a) A fair, tetrahedral dice has a side length of 12mm . Calculate the perpendicular height of each face of the tetrahedron. Write your answer as a simplified surd.

(b) Use your answer to part (a) to calculate the surface area of the dice. Write your answer in the form a\sqrt{b}\text{ mm}^{2} where a and b are integers to be found.

**(4 marks)**

Show answer

(a)

\sqrt{12^2-6^2}

**(1)**

6\sqrt{3}\text{ cm}

**(1)**

(b)

\frac{1}{2}(12\times{6\sqrt{3}})\times{4}

**(1)**

144\sqrt{3}\text{ mm}^{2}

**(1)**

You have now learned how to:

- Calculate surface areas of pyramids

Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.

Find out more about our GCSE maths tuition programme.