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Substitution Maths formulas Area of a quadrilateral Rearranging equations Pythagoras’ Theorem TrigonometryThis topic is relevant for:

Here we will learn about the **area of an equilateral triangle** including how to find the area of an equilateral triangle with given lengths and how to calculate those lengths if they are not given.

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

An **equilateral triangle** is a triangle with all sides equal and all angles equal.

As the angles in a triangle add up to

Since all sides are the same length it is a regular polygon.

We can identify a triangle by putting a capital letter on each vertex (corner).

We can then refer to each of the sides of the triangle by using two letters to describe where the line starts and ends.

We can refer to the entire triangle by using all three letters.

E.g.

Name of sides:**side AB**,

Name of triangle:**triangle ABC**

We use small lines on the sides to indicate that there are three equal sides in an equilateral triangle.

In order to find the area of a triangle, we need to start with the area of a rectangle.

To find the area of a rectangle you must multiply **adjacent **sides together.

The area of the rectangle below would be calculated by multiplying the base

We can split a rectangle into **congruent** (identical) triangles.

So the area of each triangle is exactly half the area of the rectangle.

The formula to find the area of any triangle is:

\[\text { Area of a triangle }= \frac{\text { base } \times \text { height }}{2}\]

This can be shortened to:

\[A=\frac{1}{2} b h\]

where

Your final answer must be given in units^{2}^{2}, m^{2}, mm^{2}

To find the area of an equilateral triangle when only the side length has been given, we will need to calculate the height of the triangle using Pythagoras’ Theorem or Trigonometry (SOHCAHTOA).

**Finding the height using Pythagoras’ Theorem**

Let the side length of the equilateral triangle represent the base, b.

Drawing a perpendicular line from the top vertex to the base forms the height, h, and also forms a right angled triangle with hypotenuse b and short sides h and \frac{1}{2}b.

\begin{aligned} h &=\sqrt{{{b}^{2}}-{{\left( \frac{1}{2}b \right)}^{2}}} \\\\ & =\sqrt{\frac{3}{4}{{b}^{2}}} \\\\ & =\frac{\sqrt{3}}{2}b \end{aligned}**Finding the height using SOHCAHTOA**

Let the side length of the equilateral triangle represent the base, b.

Drawing a perpendicular line from the top vertex to the base forms the height, h, and also forms a right angled triangle with hypotenuse b and angle 60^{\circ}.

\sin \left( 60 \right)=\frac{h}{b}So, h=b\sin \left( 60 \right)

We can also find the area of the equilateral triangle using the area of a triangle formula

\text{Area }=\frac{1}{2}ab\sin CThe formula would give the area as

\text{Area }=\frac{1}{2}{{b}^{2}}\sin \left( 60 \right)Using the exact value of \sin \left( 60 \right)=\frac{\sqrt{3}}{2}, this gives us

Area of an equilateral triangle =\frac{\sqrt{3}}{4}{{b}^{2}}

In order to find the area of an equilateral triangle:

**Identify the height and base length of your triangle if given.**

(You might need to calculate one of these values).**Write the appropriate area formula**

A=\frac{1}{2} b h or A=\frac{1}{2}{{b}^{2}}\sin \left( 60 \right)**Substitute the values into the formula****Calculate**

Get your free area of an equilateral triangle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free area of an equilateral triangle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**Area of an equilateral triangle** is part of our series of lessons to support revision on **area**. You may find it helpful to start with the main area 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:

- Area
- Area of a circle
- Area of a quadrilateral
- Area of a trapezium
- Area of a parallelogram
- Area of compound shapes
- Pi r squared
- Area of a rhombus
- Area of a right angled triangle
- Area of an isosceles triangle
- Area of a triangle
- How to work out area
- Area of a rectangle
- Area of a hexagon
- Area of a pentagon

**Find the area of the triangle below:**

**Identify the height and base length of the triangle****if given**.

**2Write down the appropriate formula.**

\[A=\frac{1}{2} b h\]

**3Substitute the values into the formula**.

\[\begin{aligned}
A &=\frac{1}{2} b h \\\\
&=\frac{1}{2}(3)(2.6)
\end{aligned}\]

**4Calculate.**

\[\begin{aligned}
A &=\frac{1}{2} b h \\\\
&=\frac{1}{2}(3)(2.6)\\\\
&=3.9cm^2
\end{aligned}\]

**Remember: Your final answer must be in units squared**.

**Find the area of the triangle below:**

**Identify the height and base length of the triangle if given.**

*Note: You have 2 different units here. You must convert them to a common unit*:

\[\begin{array}{l}
h = 4.3m\\\\
b = 5m
\end{array}\]

**Write down the appropriate formula**.

\[A=\frac{1}{2} b h\]

**Substitute the values into the formula.**

\[\begin{aligned}
A &=\frac{1}{2} b h \\\\
&=\frac{1}{2}(5)(4.3)
\end{aligned} \]

**Calculate**.

\[\begin{aligned}
A &=\frac{1}{2} b h \\\\
&=\frac{1}{2}(5)(4.3)\\\\
&=10.75m^2
\end{aligned} \]

**Remember: Your final answer must be in units squared.**

**Find the area of triangle ABC below:**

**Identify the height and base length of the triangle if given.**

In this question, we are given one side length of the triangle. We are missing the height of the equilateral triangle.

To calculate the height of an equilateral triangle using the Pythagorean Theorem but we must split the equilateral triangle into two equal right triangles with one of the sides of the equilateral triangle as the hypotenuse.

Draw a straight line from the top vertex to the midpoint of the base. This will create two congruent right triangles with a hypotenuse of

Using one of the right angled triangles we can apply Pythagorean Theorem:

\begin{array}{l}
h^{2}&=8^{2}-4^{2} \\\\
h^{2}&=64-16 \\\\
h^{2}&=48 \\\\
h&=\sqrt{48} \\\\
h &= 6.92820… \mathrm{~mm}
\end{array}

**Write down the appropriate formula**.

\[A=\frac{1}{2} b h\]

**Substitute the values into the formula.**

\[\begin{aligned}
A &=\frac{1}{2} b h \\\\
&=\frac{1}{2}(8)(6.928…)
\end{aligned}\]

**Calculate**.

\[\begin{aligned}
A &=\frac{1}{2} b h \\\\
&=\frac{1}{2}(8)(6.928…)\\\\
&=27.7mm^2 (1.d.p)
\end{aligned}
\]

**Remember: Your final answer must be in units squared.**

Below is an equilateral triangle with side length 9 \, cm.

Find the area of the triangle.

**Identify the height and base length of your triangle if given. (You might need to calculate one of these values)**

The base of the triangle is 9 \, cm. We can calculate the height using SOHCAHTOA

\begin{aligned} & h=9\sin \left( 60 \right) \\\\ & =7.794... \, cm \end{aligned}

**Write down the appropriate formula.**

\[A=\frac{1}{2} b h\]

**Substitute the values into the formula.**

\[\begin{aligned}
A &=\frac{1}{2} b h \\\\
&=\frac{1}{2}(9)(7.79)
\end{aligned}
\]

**Calculate**.

\[\begin{aligned}
A &=\frac{1}{2} b h \\\\
&=\frac{1}{2}(9)(7.79)\\\\
&=35.07 \, cm^2
\end{aligned}
\]

Below is the layout for a new garden plot that needs to be filled with soil. Each bag of soil costs ^{2}

**Identify the height and base length of the triangle if given.**

Split the plan into

For the triangle:

**Write down the appropriate formula.**

\[A=\frac{1}{2} b h\]

**Substitute the values into the formula.**

\[\begin{aligned}
A &=\frac{1}{2} b h \\\\
&=\frac{1}{2}(7.5)(6.5)
\end{aligned}
\]

**Calculate.**

\[\begin{aligned}
A &=\frac{1}{2} b h \\\\
&=\frac{1}{2}(7.5)(6.5)\\\\
&=24.4m^2 (1.d.p)
\end{aligned}
\]

**Now you must find the area of the rectangle:**

\[\begin{aligned}
\text { Area of a rectangle }&=l \times w \\\\
&=5 \times 7.5 \\\\
&=37.5 m^{2} \\\\
\end{aligned}
\]

Total Area ^{2}

Now divide ^{2}

Now to work out the cost:

The total cost to fill the plot with soil is

Below is an equilateral triangle with side length 15 \, cm.

Find the area of the triangle.

We know all sides are of length 15 \, cm and each angle is 60^{\circ}, so we can use the A=\frac{1}{2}{{b}^{2}}\sin \left( 60 \right) formula.

**Write down the appropriate formula.**

\[A=\frac{1}{2}{{b}^{2}}\sin \left( 60 \right)\]

**Substitute the values into the formula.**

\[A=\frac{1}{2}{{15}^{2}}\sin \left( 60 \right) \]

**Calculate**.

\[\begin{aligned}
A& =\frac{1}{2}{{\left( 15 \right)}^{2}}\sin \left( 60 \right) \\\\
& =97.43 \, cm^{2}
\end{aligned}
\]

**The height of an equilateral triangle is not the same as the side length**

When dealing with an equilateral triangle’s area it is important to note that even though the length of all sides are equal, the height is not the same as well. It will either be given in the question or you will have to calculate it.

**Units**

It is common to forget the units for area in the final answer. When calculating area, your answer must always have units squared.

1. Find the area of the triangle below:

55.2 cm^2

27.6 cm^2

13.8 cm^2

110.4 cm^2

The area of the triangle is given by

\frac{1}{2} \times 8 \times 6.9 = 27.6cm^{2}

2. Find the area of the triangle below:

87m^2

43.5m^2

4350m^2

4.35m^2

We convert the height into metres and then use the formula,

\frac{1}{2} \times 10 \times 8.7 = 43.5m^{2}

3. Shown below is a rhino enclosure. Each rhino needs a minimum of 9m^2 to roam around. What is the maximum number of rhinos that can fit into this enclosure?

9 rhinos

10 rhinos

5 rhinos

18 rhinos

The shape can be split into a rectangle and triangle.

The area of the rectangle is:

8.5 \times 6 = 51m^{2}

The area of the triangle is:

\frac{1}{2} \times 8.5 \times 7.4 = 31.45m^{2}

This means the total area is:

51 + 31.45=82.45m^{2}

By considering multiples of 9 , we conclude that 9 rhinos will fit.

4. Find the area of the triangle below:

72cm^2

36cm^2

62.4cm^2

31.2cm^2

The height can be calculated using Pythagoras’ Theorem:

Height = \sqrt{12^{2}-6^{2}}

Height = 10.392…

Then using the formula with measures for the base and height:

\frac{1}{2} \times 12 \times 10.392 = 62.4 (1.d.p)

1. This regular hexagon is made using six equilateral triangles.

Find the area of the hexagon.

**(4 marks)**

Show answer

20.8 \div 2 = 10.4

**(1)**

One triangle:

A= \frac{1}{2} \times 12 \times 10.4

**(1)**

**(1)**

Total area:

6 \times 62.4=374.4cm^{2}

**(1)**

2. Lily wants to hire a plasterer to plaster the following wall.

Brian charges \pounds 20 + \pounds 2 per square metre.

Natalie charges a flat rate of \pounds 150 .

Which plasterer would be the cheapest?

**(5 marks)**

Show answer

Area of rectangle:

\begin{aligned}A&=6 \times 8\\\\A&=48m^{2}\end{aligned}

**(1)**

Area of triangle:

\begin{aligned} A&=\frac{1}{2} \times 6 \times 5.2\\\\ A&=15.6m^{2} \end{aligned}

**(1)**

Total area:

48+15.6=63.6m^{2}

**(1)**

Brian:

\pounds 20 + 63.6 \times \pounds 2 = \pounds 147.20

**(1)**

Brian would be cheaper

**(1)**

3. Find the area of an equilateral triangle with sides of length 10cm .

**(4 marks)**

Show answer

Find height using Pythagoras Theorem:

h^{2}=10^{2}-5^{2}

**(1)**

**(1)**

**(1)**

**(1)**

You have now learned how to:

- Apply a formula to calculate and solve problems involving the area of triangles
- Use Pythagoras’ Theorem to solve problems involving triangles

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