2D Shapes

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GCSE Maths Geometry and Measure

2D Shapes

Here we will learn about 2D shapes, including symmetry, perimeter, area, circles, sectors, arcs and angles in polygons.

There are also 2D shapes worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What are 2D shapes?

2D shapes are flat shapes which only have two dimensions; length and width.

Some examples of common 2D shapes names are triangles, rectangles, pentagons, hexagons, heptagons, octagons, nonagons, decagons and circles.

We can solve problems involving 2D shapes using a variety of methods.

What are 2D shapes?

Polygons

Polygons are 2D shapes made from straight lines. You will deal with two different types of polygons.

Regular polygons

Regular polygons have specific properties. They have all sides of equal length and all interior angles are equal.

Examples of basic 2D shapes that are regular polygons are equilateral triangles and squares.

Irregular polygons

Irregular shapes (or polygons) do not have all equal sides and do not have all equal angles. When the number of sides is unknown, we describe this shape as an $\textbf{n-gon}$ where the number of sides is given as $\textbf{n}.$

You will also need to be able to solve problems involving angles in polygons. The angles in a polygon can help determine whether a polygon is regular, a specific type of polygon or to determine how many sides a polygon has. You need to be familiar with interior and exterior angles of polygons.

Interior angles are angles that are contained within the polygon.

The sum of interior angles of any polygon can be calculated using the formula,

Sum of interior angles = $(n-2) \times 180$

where $n$ represents the number of sides.

Exterior angles are supplementary to the interior angle. This means that the sum of the interior and exterior angles at a vertex always equals $180^{o}.$ We can use this property to find either the interior angle, or exterior angle at a vertex.

For any polygon the sum of exterior angles of a polygon $360^{o}.$

Step-by-step guide: Polygons

Symmetry

Symmetry is when a line is drawn through a shape to make one side of the line a reflection of the other. It is a property of a 2D polygon or 3D polyhedron.

There are two different types of symmetry that you need to be aware of. Lines of symmetry and rotational symmetry.

For example, a rectangle has two lines of symmetry

and has order $2$ rotational symmetry.

Step-by-step guide: Symmetry

Area

Area is a measure of how much space there is inside of a $2$ dimensional shape.

Here are formulae we need to remember to calculate the area of certain 2D shapes.

These are seen in the table below.

Step-by-step guide: How to work out area

Perimeter

The perimeter is the total distance around the outside of a 2D shape.

For example, let’s find the perimeter of the triangle.

For example, let’s find the perimeter of the rectangle.

The opposite sides of a rectangle are equal, $a$ and $b$ are the side lengths.

For example, let’s find the perimeter of the circle.

The perimeter of a circle is known as the circumference of a circle.

The circumference of this circle is

\begin{aligned} \text { Circumference } &=2 \times \pi \times r \\\\ &=2 \times \pi \times 12 \\\\ &=24 \pi \ \mathrm{ cm} \end{aligned}

Step-by-step guide: How to work out perimeter

Circles, sectors and arcs

Circles are round plane figures whose boundaries consist of points equidistant from a fixed point (the centre of the circle).

As well as the area and circumference of a circle we can also work out the following.

Step-by-step guide: Circles, sectors and arcs

Parts of a circle

The parts of a circle have specific names and properties which you need to know for all circle related questions. An important fact to remember is that the radius of a circle is half of its diameter.

2D shapes circles, sectors and arcs image 2

Step-by-step guide: Parts of a circle

Area of a sector

The area of a sector is part of the area of a circle.

It can be found by using the formula $\frac{\theta}{360} \times \pi r^{2}.$

2D shapes circles, sectors and arcs image 3

For example,

2D shapes circles, sectors and arcs image 4

The area of this sector is

\begin{aligned} &=\frac{\theta}{360} \times \pi r^{2} \\\\ &=\frac{60}{360} \times \pi \times 4^{2} \\\\ &=\frac{1}{6} \pi \end{aligned}

Step-by-step guide: Area of a sector

See also: Sector of a circle

Arc length

The arc of a circle is part of the circle’s circumference.

It’s length can be found using the formula $\frac{\theta}{360} \times \pi d.$

2D shapes circles, sectors and arcs image 5

For example,

2D shapes circles, sectors and arcs image 6

The arc length of this sector is

\begin{aligned} &=\frac{\theta}{360} \times \pi d \\\\ &=\frac{60}{360} \times \pi \times 8 \\\\ &=\frac{60}{360} \times \pi \times 8 \\\\ &=\frac{4}{3} \pi \end{aligned}

Step-by-step guide: Arc length

See also: Arc of a circle

Perimeter of a sector

To find the perimeter of a sector you need to find the arc length and then add it to the two straight sides which are both radii (i.e. the length around the outside of the sector).

2D shapes circles, sectors and arcs image 7

For example,

2D shapes circles, sectors and arcs image 8

The perimeter of this sector is

\begin{aligned} &=\frac{60}{360} \times \pi \times 8+4+4 \\\\ &=\frac{4}{3} \pi+8 \end{aligned}

Step-by-step guide: Perimeter of a sector

Segment of a circle

A segment of a circle is created by an arc length and a chord.

You may have to find its area using a combination of mathematical rules such as trigonometry or Pythagoras’ theorem.

2D shapes circles, sectors and arcs image 9

Step-by-step guide: Segment of a circle

Equation of a circle

The equation of a circle (at GCSE) can be given in the form below

x^{2}+y^{2}=r^{2}.

2D shapes circles, sectors and arcs image 10

Step-by-step guide: Equation of a circle

How to use 2D shapes

We can use 2D shapes in lots of different ways.

We will learn about:

Polygons
Symmetry
Area
Perimeter
Circles, sectors and arcs

Explain how to use 2D shapes

Area of 2D shapes worksheet

Get your free 2D shapes worksheet of 20+ area of 2D shapes questions and answers. Includes reasoning and applied questions.

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Area of 2D shapes worksheet

Get your free 2D shapes worksheet of 20+ area of 2D shapes questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE

2D shapes examples

Example 1: regular / irregular polygons

The shape $ABCDE$ below is made from two scalene triangles, and one isosceles triangle. $M$ is a midpoint on the line $CD$ . $AM$ is a line of symmetry.

Determine what type of polygon $ABCDE$ is and determine if it is regular or irregular.

State / calculate the number of sides of the polygon.

The polygon $ABCDE$ has $5$ sides so it is a type of pentagon.

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

As triangle $ACD$ is isosceles and $AM$ is a line of symmetry, angle $ACD =$ angle $ADC = 52^{o}.$

As the sum of angles in a triangle is $180^{o},$

angle $CAD = 180-(52+52)=76^{o}.$

As $AM$ is a line of symmetry, triangles $ABC$ and $ACD$ must be congruent, sharing the same angles and side lengths. This means that,

Angle $BCA =$ angle $ADE =56^{o}$
Angle $DAE =$ angle $CAB =16^{o}$

Again, as the sum of angles in a triangle total $180^{o},$

angle $AED = 180-(56+16)=108^{o}.$

This is the same for angle $ABC$ as it is symmetrical to angle $AED$ and therefore equal.

We now have all of the following angles,

3Recognise the other properties of the polygon.

For a pentagon to be regular, all of the interior angles must be the same and side lengths must be the same.

Furthermore, as the interior angle sum of a pentagon is $540^{o},$ each interior angle of a regular pentagon is equal to

540\div{5}=108^{o}.

Adding the angles at each vertex together, we have

Each interior angle of the polygon $ABCDE$ is equal to $108^{o}.$

However, the triangles $AED$ and $ABC$ are scalene. Therefore, $AB$ has a different length to $BC$ .

The interior angles are equal but the side lengths are not equal.

The polygon is an irregular pentagon.

Example 2: types of quadrilateral

Determine what type of quadrilateral $ABCD$ is below.

Determine the size of the angles / side lengths within the quadrilateral.

Show step

All four side lengths are equal to $9cm$ with two pairs of parallel sides.

Angle $BCD = 48+48=96^o$ and so $ABCD$ cannot be a square.

CDE is a triangle. As the sum of angles in a triangle total $180^{o},$

angle $CED = 180-(42+48)=90^{o}.$

Labelling this on the diagram, we have

Recognise the other properties of the polygon.

Show step

As vertically opposite angles are equal, angle $AED =$ angle $CED = 90^{o}.$

As the sum of angles on a straight line total $180^{o},$

Angle $AED = 180-90=90^{o}.$

This is also true for angle $BEC$ .

We therefore have the diagonals intersecting at $90$ degrees.

$ABCD$ is a rhombus.

Remember: The properties of a rhombus are,

Four equal sides
Opposing sides are parallel
Two opposing pairs of equal angles
Diagonals bisect each other at $90$ degrees
Two lines of symmetry
Rotational symmetry of order $2$

Example 3: symmetry

State the number of lines of symmetry for a regular hexagon.

Locate the centre of the 2D shape.

Show step

To locate the centre of a shape with an even number of vertices, draw a pair of straight lines connecting two opposing vertices.

Use a ruler to visualise a horizontal and/or vertical line of symmetry through the centre of the shape.

Show step

The regular hexagon has a vertical line of symmetry.

The regular hexagon has a horizontal line of symmetry.

Continue to rotate the ruler around $180$ degrees over the centre point to cover all sides and vertices.

Show step

Line of symmetry $3.$

Line of symmetry $4.$

Line of symmetry $5.$

Line of symmetry $6.$

A regular hexagon has $6$ lines of symmetry.

Example 4: area of an isosceles triangle

Calculate the area of the triangle below.

Identify the base and perpendicular height of the triangle.

Show step

The two values that are perpendicular to one another are the $12cm$ along the base, and the vertical height of $8cm.$

The $10cm$ is not required to find the area of this triangle.

Write the area formula.

Show step

The area of a triangle formula is,

$\text{Area of a triangle}=\cfrac{\text{base }\times\text{height}}{2}.$

Substitute known values into the area formula.

Show step

As the base $= 12cm$ and the height $= 8cm,$ we have

$\text{Area of a triangle }=\cfrac{12\times{8}}{2}.$

Solve the equation.

Show step

\text{Area of a triangle }=\cfrac{96}{2}=48.

Write the answer, including the units.

Show step

As the units of length are in centimetres, the units of area are square centimetres.

The area of the triangle is $48cm^{2}.$

Example 5: circumference of a circle

Calculate the circumference of a circle with a radius of $4cm.$ Write your answer correct to $2$ decimal places.

Find the radius or diameter of the circle.

Show step

The radius of the circle is $4cm.$

Use the relevant formula to calculate the circumference of the circle.

Show step

The formula for the circumference of a circle $(C)$ in terms of the radius $(r)$ is

$C=2\pi{r}.$

Substituting $r=4$ into the formula, we have

$\begin{aligned} C&=2\times\pi\times{4} \\\\ &=8\times\pi \\\\ &=8\pi \\\\ &=25.13274123 \\\\ &=25.13\text{ (2dp)} \end{aligned}$

Give your answer clearly with the correct units.

Show step

The circumference of the circle is $25.13cm \ (2dp).$

Example 6: perimeter of compound shapes – real life problem

A car park needs a new boundary fence installing. A sketch of the car park is given below.

Determine the length of the boundary wall (the car park entrance/exit must not be included in this value).

Add all the side lengths.

Show step

A few of the side lengths are missing and so we need to calculate these lengths first.

As $AF = 8.8m$ and $DE = 2.9m,$ the length of $BC$ can be calculated by subtracting $DE$ from $AF$ ,

$8.8-2.9 = 5.9m.$

As $AB = 12.4m$ and $CD = 3.5m,$ the length of $EF$ can be calculated by adding $AB$ and $CD$ together,

$12.4+3.5=15.9m.$

Writing these two measurements onto the diagram, we have

The perimeter is the sum of the side lengths and so we have the sum,

$12.4+5.9+3.5+2.9+15.9+8.8=49.4m.$

Note: For an L shape, this is the same as doubling the sum of the height and the width $((8.8+15.9)\times{2}=49.4m).$

Write the final answer with the correct units.

Show step

As we need to calculate the boundary excluding the entrance/exit, we need to subtract the width of the entrance from the total perimeter.

$49.4-2.4=47m$

The perimeter of the car park excluding the entrance/exit is $47m.$

Example 7: area of a sector

A circular cake is cut into $8$ equal slices as shown below.

Calculate the area of the top of one slice of cake.

Find the length of the radius $\textbf{r}$ .

Show step

The radius $r = 9.5cm.$

Find the size of the angle creating the sector

Show step

The angle of the sector is one eighth of a full turn. As a full turn is $360$ degrees,

$360\div{8}=45.$

The angle of the sector is $45^{o}.$

Substitute the value of the radius and the angle into the formula for the area of a sector.

Show step

The formula for the area of the sector is

$\text{Area of a sector }=\frac{\theta}{360}\times\pi{r}^{2}$

where $\theta$ is the angle of the sector, and $r$ is the radius of the overall circle.

Substituting $\theta=45^{o}$ and $r = 9.5cm,$ we have

$\begin{aligned} \text{Area of a sector }&=\frac{45}{360}\times\pi\times{9.5}^{2} \\\\ &=\frac{1}{8}\times\pi\times{90.25} \\\\ &=11.28125\pi \\\\ &=35.44109212 \\\\ &=35.44\text{cm}^{2}\text{ (2dp)} \end{aligned}$

Clearly state your answer.

Show step

The area of the top of the cake slice is $35.44cm^{2}.$

Example 8: equation of a circle

Determine the equation of the unit circle, centred at the origin.

Write the general equation of a circle.

Show step

The general equation of a circle is $x^{2}+y^{2}=r^{2}.$

State any variables you know.

Show step

The radius of the unit circle is $1.$ We can also see this as the centre of the circle lies at the point $(0,0)$ and a coordinate on the circumference of the circle is $(1,0),$ giving us a radius of $1.$

Substitute any values you know into the equation.

Show step

Substituting $r=1$ into the equation of a circle, we have

$x^{2}+y^{2}=1^{2}.$

Use the information you have to solve the problem.

Show step

Evaluating $1^{2}$ we get $1^{2}=1 \times 1=1.$

Clearly state the answer.

Show step

The equation of the unit circle, centred at the origin is $x^{2}+y^{2}=1.$

Common misconceptions

Rotational symmetry/lines of symmetry

Lines of symmetry are often confused with rotational symmetry. A line of symmetry on a two-dimensional shape divides the shape equally into two symmetrical pieces.

Rotational symmetry is the number of times a shape fits into itself when rotated around its centre.

Confusing perimeter with area

Remember, perimeter is distance around the outside of a shape, whilst area is the space inside the shape.

Angles properties

Ensure you have a good understanding of the different types of angles (for example, acute angles and right angles) along with how to calculate angles in polygons and angles in parallel lines.

Getting these confused can lead to misconceptions when problem solving.

Practice 2D shapes questions

Parallelogram

Rhombus

Square

Kite

A parallelogram has no lines of symmetry and rotational symmetry order $2.$

A square has $4$ lines of symmetry and rotational symmetry order $4.$

A kite has $1$ line of symmetry and no rotational symmetry.

2D shapes practice question 1

4cm

11cm

24cm

6cm

Form an equation for the perimeter.

2x+x+9+2x+x+9=18

6x+18=42

Solve the equation, to find $x=4.$

60cm^2

120cm^2

96cm^2

48cm^2

This is an isosceles triangle. We can split it vertically to form two identical triangles with base $6cm,$ and hypotenuse $10cm.$

We then use Pythagoras’ theorem to find the height of the triangle.

\begin{aligned} &a^{2}+b^{2}=c^{2} \\\\ &6^{2}+\text { height }{ }^{2}=10^{2} \\\\ &\text { height }{ }^{2}=100-36 \\\\ &\text { height }{ }^{2}=64 \\\\ &\text { height }=8cm \end{aligned}

Then we can use the formulae for the area of a triangle, and multiply the height by the base (remembering to use the base as $12cm$ for the area of the whole triangle) then divide by $2.$

\begin{aligned} \text { Area of a triangle } &=\frac{1}{2} \times \text { base } \times \text { height } \\\\ &=\frac{1}{2} \times 12 \times 8 \\\\ &=48cm^{2} \end{aligned}

3 \pi \ cm

18+18 \pi \ cm

9+3 \pi \ cm

18+3 \pi \ cm

To find the perimeter of the sector we need to calculate the arc length. To do this we need to know the diameter of the sector.

We can work backwards using the formula for the area of a sector to find the radius to help us find the diameter.

\text { Area of a sector }=\frac{\theta}{360} \times \pi \times r^{2}

Substituting in what we know,

\begin{aligned} &\frac{60}{360} \times \pi \times r^{2}=\frac{27}{2} \pi \\\\ &\frac{60}{360} r^{2}=\frac{27}{2} \\\\ &r^{2}=81 \\\\ &r=9cm \end{aligned}

Now we can calculate the arc length. If the radius is $9cm,$ the diameter must be $18cm.$

\text { Arc length }=\frac{\theta}{360} \times \pi \times d

Substituting in what we know,

\begin{aligned} &=\frac{60}{360} \times \pi \times 18 \\\\ &=3 \pi \ cm \end{aligned}

For the total perimeter we need to add the two radii to the arc length.

\text { Perimeter }=3 \pi+18cm

360^{\circ}

540^{\circ}

720^{\circ}

900^{\circ}

\begin{aligned} \text { Sum of the interior angles of a pentagon } &=(5-2) \times 180 \\\\ &=3 \times 180 \\\\ &=540^{\circ} \end{aligned}

24

15

12

36

The interior and exterior angle of a polygon sum to $180^{\circ}.$ We can find the exterior angle using this fact.

\begin{aligned} &\text{Exterior angle }=180-\text{interior angle} \\\\ &\text{Exterior angle }=180-165=15^{\circ} \end{aligned}

To find out the number of sides we then do,

\text { Number of sides }=360 \div 15=24.

2D shapes GCSE questions

1. The hexagon $ABCDEF$ has one line of symmetry.

2D shapes gcse question 1

Angle $FAB =$ angle $ABC = 110^{\circ}.$

Angle $AFE =$ angle $BCD$ .

Angle $FED =$ angle $CDE$ .

Angle $CDE :$ angle $BCD = 3 : 2.$

Find the size of angle $AFE$ .

(5 marks)

Show answer

Indicating sum of angles is $720^{\circ}$ or $540^{\circ}$ if line of symmetry used to form a pentagon.

(1)

Finding the sum of $AFE$ and $FED$ is $250^{\circ}.$

(1)

Use of ratio $3:2$ or sight of $3x$ and $2x.$

(1)

Finding $x=50.$

(1)

Angle $AFE = 100^{\circ}.$

(1)

2. The perimeter of the rectangle is twice the perimeter of the isosceles triangle.

2D shapes gcse question 2

Find the area of the rectangle.

(5 marks)

Show answer

Correct expression for perimeter of rectangle or triangle,

6x+2 \ , \ 2x+5.

(1)

Forming an equation linking the perimeters.

For example, $6x+2=2(2x+5)$ or equivalent.

(1)

Solving equation to get $x=4.$

(1)

Correctly substituted into length and width of rectangle.

(1)

Area given as $42cm^{2}.$

(1)

3. The sector has a perimeter $25cm$ and radius $6cm.$

2D shapes gcse question 3

Find the size of angle $x.$ Give your answer to $3$ significant figures.

(3 marks)

Show answer

Sight of arc length given as $13cm.$

(1)

Process to use formula $\frac{x}{360} \times 2 \times \pi \times 6=13$

(1)

x=124^{\circ}

(1)

Learning checklist

You have now learned how to:

Identify lines of symmetry and order of rotational symmetry in 2D shapes

How to work out the perimeter

Work out the area of triangles, quadrilaterals and circles

Work out the area of compound shapes

Identify and apply circle definitions and properties

Calculate the circumference of a circle

Calculate the area and arc length of a sector

Distinguish between regular and irregular polygon

The next lessons are

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