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Quadrilaterals

Parallel lines

This topic is relevant for:

Here we will learn about **polygons**, including regular polygons, angles in polygons, and complex polygons.

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

A **regular polygon** is a 2D shape where the sides are all **straight line segments** of **equal length** and **each interior angle in the shape is equal**.

You need to be able to classify geometric shapes based on their properties, and find unknown angles and side lengths in any triangle, quadrilateral and regular polygon.

E.g.

Equilateral triangle | Square | Regular pentagon |

To do this, we need to look closely at the properties of these shapes.

- Regular polygons have
**sides of equal length**and**angles of equal size.** - The table below gives the name of several regular polygon.

Polygon | Number of sides | Image (regular polygon) |

Equilateral triangle | 3 | |

Square (quadrilateral) | 4 | |

Regular pentagon | 5 | |

Regular hexagon | 6 | |

Regular heptagon | 7 | |

Regular octagon | 8 | |

Regular nonagon | 9 | |

Regular decagon | 10 | |

Regular hendecagon | 11 | |

Regular dodecagon | 12 | |

Regular heptadecagon | 17 | |

Regular icosagon | 20 | |

N-gon | n |

- The sum of
**interior angles**can be calculated using the formula:

**Sum of interior angles = (n-2) × 180^{\circ} **

where n represents the number of sides.

As the size of each angle is equal, we can determine the size of each angle by **dividing the sum of the interior angles by the number of sides**:

**Exterior angles**are supplementary to the interior angle:

**Sum of exterior angles of a polygon = 360° **

We can use this property to find either the interior angle, or exterior angle at a vertex.

As the sum of exterior angles is always 360° , for any regular polygon we can divide ** 360 by the number of sides** to work out an exterior angle.

- All regular polygons can be
**inscribed (enclosed) in a circle**.

- All regular polygons are known as
**convex polygons**as all of the interior angles are less than 180° .

E.g.

Convex hexagon | Concave hexagon |

All interior angles are either acute or obtuse. | At least one angle is greater than 180°. |

In order to classify a regular polygon:

**State/calculate the number of sides of the polygon.****Determine the size of the angles/side lengths within the polygon.****Recognise the other properties of the polygon.**

Get your free regular polygon worksheet of 20+ questions and answers. Includes reasoning and applied questions.

COMING SOONGet your free regular polygon worksheet of 20+ questions and answers. Includes reasoning and applied questions.

COMING SOONGiven that ABD is an isosceles triangle, and ADE is a right angle triangle, classify the polygon BCD :

**State/calculate the number of sides of the polygon**.

BCD has 3 sides and so it is a triangle.

2**Determine the size of the angles and/or side lengths within the polygon**.

As ABD is an isosceles triangle, BAD = BDA = (180 - 120) ÷ 2 = 30° .

DAE = 180 - (90+60) = 30° as angles in a triangle total 180° .

Angle ACE = 180 - (30+30+60) = 60°.

Angle CBD is corresponding to angle BAE so angle CBD = 60°.

This is the same for angle CDB as it is corresponding to angle DEA.

3**Recognise the other properties of the polygon**.

All the angles in BCD are equal to 60° .

The polygon is an **equilateral triangle**, which is a regular polygon.

The polygon ABCD has two pairs of parallel sides. The diagonals are perpendicular to each other with AC = BD . Classify the quadrilateral.

**State/calculate the number of sides of the polygon**.

The polygon has four sides so we are looking at a quadrilateral.

**Determine the size of the angles/side lengths within the polygon**.

As AC = BD , the lengths of OA, OB, OC and OD are all equal as O is the centre of the polygon.

Each triangle within the polygon is therefore an isosceles triangle, or equilateral.

As the angle AOD = 90° (perpendicular lines meet at 90° ), the two other angles in the triangle are equal to (180 - 90) ÷ 2 = 45°.

**Recognise the other properties of the polygon**.

The angle at each vertex is equal to 45 + 45 = 90°.

ABCD is a **square**.

The polygon below can be split into eight congruent triangles. One of the angles AOB = 45° . Classify the polygon.

**State/calculate the number of sides of the polygon**.

The polygon has eight sides so it is an octagon.

**Determine the size of the angles and/or side lengths within the polygon**.

As the angle at O for each isosceles triangle is equal to 45° , the other two angles must be equal to (180 - 45) ÷ 2 = 67.5°.

Two adjacent angles are therefore equal to 67.5 × 2 = 135°.

**Recognise the other properties of the polygon**.

As all the interior angles are equal, the polygon is a **regular octagon**.

Two isosceles trapezia are joined together on their longest side to form another polygon. AB = BC . Given the information in the diagram below, determine the classification of the new polygon.

**State/calculate the number of sides of the polygon**.

The new polygon has 2 joining sides so we need to reduce the total number of sides by 2 :

8 - 2 = 6.The polygon is a type of hexagon.

**Determine the size of the angles and/or side lengths within the polygon**.

As the lines AB and CD are parallel, and we know the original polygon is an isosceles trapezia, we can say that the angle at B is 120° , the angle at C and D are equal to 60° as they are co-interior angles to the angles at A and B respectively.

This means that when the two tapezia are placed together the angles in the new image are a reflection of the original. The angles at C and D are double their original so all of the angles in the hexagon are equal to 120° .

**Recognise the other properties of the polygon**.

As all the interior angles are equal to 120° , and with AB = BC , all the side lengths are equal so the new polygon is a **regular hexagon**.

A polygon has n sides. Each exterior angle is equal to 30° . All the sides are the same length. Determine the classification of the polygon.

**State/calculate the number of sides of the polygon**.

As each exterior angle is equal to 30° , we can calculate the number of sides of the polygon using the formula E_{n} = 360 ÷ n where n is the number of sides and E is the exterior angle:

30 = 360 ÷ n

30 × n = 360

n = 360 ÷ 30

n = 12

**Determine the size of the angles and/or side lengths within the polygon**.

As we know the size of the exterior angle, we can calculate the interior angle at each vertex by using the formula I_{n} + E_{n} = 180° where I is the interior angle for the number of sides n :

I_{n}+ 30 = 180

I_{n}= 150°.

Each interior angle is equal to 150° .

**Recognise the other properties of the polygon**.

As all of the interior angles are equal, and the polygon has 12 sides, the polygon is a **regular dodecagon**.

The interior angles of a polygon are: 3x + 15, 2x + 30 and 5x - 15 . Classify the polygon.

**State/calculate the number of sides of the polygon**.

As there are 3 interior angles, the polygon is a type of triangle.

**Determine the size of the angles and/or side lengths within the polygon**.

As the sum of angles in a triangle total 180° ,

3x + 15 + 2x + 30 + 5x - 15 = 180

10x+30=180

10x = 150

x = 15

As x = 15

3x + 15 = 3\times15 + 15 = 60°

2x + 30 = 2\times15 + 30 = 60°

5x - 15 = 5\times15 - 15 = 60°

**Recognise the other properties of the polygon**.

As all the angels in the polygon are equal to 60° , the triangle is an** equilateral triangle **(a regular polygon).

**Angles in polygons**

Make sure you know your angle properties. Getting these confused causes quite a few misconceptions.

- Angles in a triangle total 180°.
- Angles in a quadrilateral total 360°.
- Angles on a straight line total 180°.

**Incorrect quadrilateral classification**

There are many quadrilaterals and it is common to confuse the properties, especially for a rhombus, parallelogram, or trapezium.

As well as this, stating that the polygon is a quadrilateral is not enough information for a classification.

**Incorrect assumptions for triangles**

Assuming a triangle is isosceles or equilateral can have an impact on the size of angles within the rest of the polygon, so make sure you can explain why you have chosen a specific type of triangle.

As well as this, stating that the polygon is a triangle is not enough information for a classification. You must state whether it is isosceles/ equilateral etc.

1. The triangle ACE is constructed using 4 congruent triangles. ACE has a rotational symmetry of order 3 . Classify the polygon BDF.

Right angle triangle

Equilateral triangle

Isosceles triangle

Scalene triangle

As the triangle has a rotational symmetry of 3 , the interior angles at A, C, and E are equal to 60^{\circ} .

As the four triangles are congruent, each interior angle is equal to 60^{\circ} . The triangle BDF is equilateral.

2. A quadrilateral is inscribed in a circle with centre O . Use circle theorems to determine the classification of the quadrilateral.

Square

Irregular quadrilateral

Trapezium

Kite

Each angle at A, B, C, and D is equal to 90^{\circ} as each diagonal passes through the centre of the circle so the angle in the semicircle is equal to 90^{\circ} .

3. The polygon below is made up of four congruent quadrilaterals and two congruent triangles. By calculating the value of x , classify the polygon.

Regular octagon

Regular triangle

Regular quadrilateral

Regular nonagon

Looking at the quadrilateral AOMH , we can calculate the size of angle x by using the fact that angles in a quadrilateral total 360^{\circ} :

2x+135+90=360

2x=135

x=67.5^{\circ}

Looking at the triangle, as the line MO is parallel to the line AB , angle OAB is alternate to the angle AOM and so angle OAB = x = 67.5^{\circ} . This means that triangle ABO is isosceles as the angle at A and the angle at B are both equal.

Angle AOB = 45^{\circ} as angles in a triangle total 180^{\circ} .

Angle BAH = 67.5 + 67.5 = 135^{\circ} .

Angle ABC = 67.5 + 67.5 = 135^{\circ} and Angle BCN = 135^{\circ} as shape BCNO is congruent to shape AOMH .

The line MN bisects the shape into two equal halves and so the angles on the opposite side of this line are mirrored. All of the interior angles of the octagon are equal to 135^{\circ} and so the polygon is a regular octagon.

4. 6 congruent triangles are joined together to form a 6 sided polygon. Each line segment within the polygon is the same length and all the interior angles of the polygon are obtuse. Classify the new polygon.

Hexagon

Regular hexagon

Equilateral triangle

Trapezium

As each line segment is the same length, each triangle must be equilateral. The interior angles are all obtuse. The only way to arrange the triangles to produce a 6 sided shape with all obtuse angles is:

Each interior angle is equal to 120^{\circ} as two adjacent angles of two equilateral triangles are 60 + 60 = 120^{\circ} .

5. A polygon has n sides. Each exterior angle is equal to 45^{\circ} . All the side lengths are equal. Determine the classification of the polygon.

Regular nonagon

Regular heptagon

Regular quadrilateral

Regular octagon

360 ÷ 45 = 8 sides. All the side lengths are the same as stated in the question.

6. The interior angles of a polygon are: 10x + 8, 12x – 12, 9x + 18, 6x + 48, and 13x – 22 . The sum of the angles is equal to 540^{\circ} . Classify the polygon.

Pentagon

Hexagon

Regular pentagon

Regular hexagon

10x+8+12x-12+9x+18+6x+48+13x-22=540

50x+40=540

50x=500

x=10

As x=10

10x+8=10\times10+8=108^{\circ}

12x-12=12\times10-12=108^{\circ}

9x+18=9\times10+18=108^{\circ}

6x+48=6\times10+48=108^{\circ}

13x-22=13\times10-22=108^{\circ}

All 5 of the angles are equal to 108^{\circ} .

1. (a) Two regular polygons with the same number of sides join down one side. The angle between the polygons is equal to 120^{\circ} . What type of polygons are they?

(b) Calculate the number of sides of a polygon with an internal angle of 156^{\circ} .

**(4 marks)**

Show answer

(a)

(360 – 120)\div2 = 120^{\circ}

**(1)**

Regular hexagons

**(1)**

(b)

180 – 156 = 24^{\circ}

**(1)**

360\div24 = 15 sides

**(1)**

2.

(a) The polygon ABCDE is made from 2 congruent isosceles triangles and two congruent right angle triangles.

The point M is the midpoint of the line DE .

Angle CDE and BAE = 108^{\circ} .

Angle BME = 90^{\circ} and BM is a line of symmetry.

Classify the following polygon ABCDE .

(b) Does this shape tessellate? Explain your answer.

**(7 marks)**

Show answer

(a)

AED = BCD = 108^{\circ} due to symmetrical properties

**(1)**

Angles in a pentagon total 540^{\circ} or (180\div(n-2))

**(1)**

Angle ABC = 540 – (108\times4) = 108^{\circ}

**(1)**

All sides equal as the polygon has a line of symmetry

**(1)**

Regular pentagon

**(1)**

(b)

No

**(1)**

360\div108 is not an integer

**(1)**

3. The pentagram below is made up of 5 isosceles triangles and a regular pentagon. Calculate the interior angles for the pentagram labelled x, y, and z .

Give reasons for your answer.

**(6 marks)**

Show answer

x=108^{\circ}

**(1)**

Interior angle of a regular pentagon is 108^{\circ}

**(1)**

y=72^{\circ}

**(1)**

Angles on a straight line total 180^{\circ}

**(1)**

z=36^{\circ}

**(1)**

Angles in an isosceles triangle

**(1)**

You have now learned how to:

- Plot specified points and draw sides to complete a given polygon
- Distinguish between regular and irregular polygons based on reasoning about equal sides and angles
- Compare and classify geometric shapes based on their properties and sizes and find unknown angles in any triangles, quadrilaterals, and regular polygons
- Describe, sketch and draw using conventional terms and notations: points, lines, parallel lines, perpendicular lines, right angles, regular polygons, and other polygons that are reflectively and rotationally symmetric
- Derive and use the sum of angles in a triangle and use it to deduce the angle sum in any polygon, and to derive properties of regular polygons

- Irregular polygons
- Types of quadrilaterals
- Types of triangles

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