Equilateral Triangle

Here we will learn about equilateral triangles, including what an equilateral triangle is and how to solve problems involving the sides and angles of equilateral triangles.

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

What is an equilateral triangle?

An equilateral triangle is a type of triangle that has the following properties,

• All sides are equal length.

• All angles are equal size $(60^{\circ})$.

An equilateral triangle is the simplest regular polygon.

All equilateral triangles are similar to (an enlargement of) each other as they have the same angles.

Equilateral triangles are only congruent if the lengths of the sides are the same.

An equilateral triangle is a special case of an acute triangle, where all the angles are acute angles.

Properties of equilateral triangles and other types of triangles are very important within geometry.

They are one of the most common shapes to recognise for angles in parallel lines, circle theorems, interior angles, trigonometry, Pythagoras’ theorem and many more. You therefore must be familiar with their individual properties.

What is an equilateral triangle?

Symmetry

An equilateral triangle has three lines of symmetry. The lines go from one vertex to the middle of the opposite side.

The lines of symmetry are angle bisectors of the vertices.

The lines of symmetry are also perpendicular bisectors of the sides of the equilateral triangle.

An equilateral triangle has an order of rotational symmetry of $3$ as it looks the same three times in a full turn.

In other words, if you rotate an equilateral triangle about its centre, it looks like the original every $120^{\circ}$ rotation clockwise.

Area of an equilateral triangle

The area of an equilateral triangle can be found by using the formula $A=\cfrac{1}{2} \, bh,$ where $b$ is the base length and $h$ is the perpendicular height of an equilateral triangle. Sometimes these values need to be calculated.

Remember that an area uses square units.

For example, calculate the area of the equilateral triangle $ABC$ below, where $M$ is the midpoint of the line $AC$, and angle $AMB = 90^{\circ}$.

Here, the base $AC$ is double the length of $AM$ and so $AC = 8{~cm}$.

Substituting the base length $b = 8$ and the perpendicular height $h = 6.92$ into the formula for the area of a triangle, we have

$A=\cfrac{1}{2}\times{8}\times{6.92}= 27.68\text{~cm}^{2}.$

The area of the triangle is $27.68 \, cm^2.$

Step-by-step guide: Area of an equilateral triangle

3D shapes

A regular tetrahedron can be made by using $4$ equilateral triangles.

An octahedron can be made by using $8$ equilateral triangles.

An icosahedron can be made by using $20$ equilateral triangles.

Step-by-step guide: 3D shape names

How to solve problems involving equilateral triangles

In order to solve problems involving equilateral triangles:

1. Locate known angles and calculate any necessary unknown angles.
2. Locate known sides and calculate any necessary unknown side lengths.
3. Solve the problem using any necessary values.

Explain how to solve problems involving equilateral triangles

Related lessons on triangles

Equilateral triangle is part of our series of lessons to support revision on polygons and triangles. You may find it helpful to start with the main triangles 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:

• Polygons
• Triangles
• Right angle triangle
• Types of triangle
• Scalene triangle
• Isosceles triangle

Equilateral triangle examples

Example 1: perimeter of an equilateral triangle

Find the perimeter of an equilateral triangle with side $7{~cm}$.

1. Locate known angles and calculate any necessary unknown angles.

We do not need to calculate any angles as the perimeter is a measure of length and we can determine the side lengths of the triangle from the information provided.

2Locate known sides and calculate any necessary unknown side lengths.

As the triangle is an equilateral triangle, all three side lengths are the same and are equal to $7{~cm}$.

3Solve the problem using any necessary values.

We can calculate the perimeter by multiplying one of the side lengths by 3 (or adding the three side lengths), $3\times 7=21$.

The perimeter of the equilateral triangle is $21{~cm}$.

Example 2: perimeter of a 2D compound shape

Find the perimeter of this trapezium made from congruent equilateral triangles with side $20{~cm}$.

Locate known sides and calculate any necessary unknown side lengths.

Show step

As each side length of an equilateral triangle is the same, each side can be labelled as $20{~cm}$ although, we do not need to label the two lines within the shape as these are not needed for the perimeter.

Example 3: external angles of an equilateral triangle

Calculate the size of angle $x$.

Locate known angles and calculate any necessary unknown angles.

Show step

Each side length of the triangle contains a dash and so the triangle is equilateral. This means that all interior angles are $60^{\circ}$.

Example 4: similar equilateral triangles problem solving

An equilateral triangle $ABC$ has a perimeter of $30{~cm}$.

Points $M$ and $N$ are midpoints of the lines $AB$ and $AC$ respectively where $MN$ is parallel to $BC$. Show that the line $MN = 5{~cm}$.

Locate known angles and calculate any necessary unknown angles.

Show step

As $ABC$ is an equilateral triangle, each interior angle is equal to $60^{\circ}$.

As corresponding angles in parallel lines are equal, and angle $AMN$ is corresponding to angle $ABC$, angle $AMN = 60^{\circ}$.

Also, as angle $ANM$ is corresponding to angle $ACB$, angle $ANM = 60^{\circ}$.

We can now see that triangle $AMN$ is also equilateral.

Example 5: equilateral triangles and regular hexagons problem solving

Six congruent equilateral triangles are placed next to each other to form a hexagon $ABCDEF$. The side length of a triangle is $3{~cm}$. Show that the hexagon is a regular hexagon.

Locate known angles and calculate any necessary unknown angles.

Show step

As each triangle is equilateral, each angle within the triangles is $60^{\circ}$. Two angles that meet at the perimeter of the hexagon must therefore have a sum of $60+60=120^{\circ}$. Labelling the diagram with this information, we have

Locate known sides and calculate any necessary unknown side lengths.

Show step

We are also told that the side length of each equilateral triangle is $3{~cm}$, and so labelling each external side length of $ABCDEF$, we have

Example 6: bearings problem with equilateral triangles problem solving

Three planes set off from an airfield at the same height.

Plane $A$ travels at a bearing of $020^{\circ}$, Plane $B$ travels at a bearing of $140^{\circ}$.

Plane $C$ travels at a bearing of $260^{\circ}$. Each plane travels at the same speed.

After $30$ minutes, each plane is exactly $D{~km}$ from the airport.

A diagram of this is shown below.

Without calculating the exact value, show that the planes are all the same distance apart from each other

Locate known angles and calculate any necessary unknown angles.

Show step

Bearings are always taken from North in a clockwise direction. This means that we can work out the exact angle the planes are on with respect to the airfield.

The angle $AOB = 140-20=120^{\circ}.$

The angle $BOC = 260-140=120^{\circ}.$

The angle $AOC = (360-260)+20=120^{\circ}.$

Sketching these onto a separate diagram, we have

Solve the problem using any necessary values.

Show step

Let us assume that the distance $BC = x{~km}$.

This means that $OBC$ is an isosceles triangle as the lines $OB$ and $OC$ are the same length. The angles $OBC$ and $OCB$ must now be equal as they are opposite the two equal side lengths.

As the sum of angles in a triangle is $180^{\circ}$, the remaining two angles are half of the difference between $120^{\circ}$ and $180^{\circ}$.

This means that the angles $OBC$ and $OCB$ are equal to $(180-120)\div{2}=30^{\circ}.$

Repeating this process for triangles $AOB$ and $AOC$, we have exactly the same result.

Looking at the angles $ABC, BCA$ and $CAB$, each of these is equal to $30+30=60^{\circ}$ and as the value of $x{~km}$ is the distance between each plane, $ABC$ is an equilateral triangle, and so all of the planes are the same distance away from each other.

Common misconceptions

• Angles in polygons
  Make sure you know your angle properties. Getting these confused causes quite a few misconceptions.
  • Angles in a triangle total $180^{\circ}$.
  • Angles in a quadrilateral total $360^{\circ}$.

• Angle facts
  Make sure you know your angle facts.
  • Adjacent angles on a straight line add up to $180^{\circ}$.
  • Vertically opposite angles are equal.
  • Alternate angles are equal.
  • Corresponding angles are equal.

The next lessons are

• Symmetry
• Area
• How to work out perimeter

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