Perimeter Worksheets | Questions and Revision | MME

Perimeter Worksheets, Questions and Revision

Level 1-3
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The perimeter of a 2D shape is the distance around the edges of the shape. It is found by adding up the lengths of each individual edge of the shape. This can be easy for simple shapes and harder for composite shapes. 

Make sure you are happy with the following the following topics:

2D Shapes and Quadrilaterals 


Perimeter of Simple Shapes

The rectangle shown below has a length of 11cm and a width of 3cm.

Calculate the perimeter of the rectangle shown. 

The opposite sides of a rectangle are equal in length.

So, the two sides that are not labelled must also be 3 cm and 11 cm.

Therefore, we get

Perimeter = 3+3+11+11=28 cm

Level 1-3

Perimeter of Compound Shapes 

ABCDEF is a composite shape made up of rectangles.

Calculate the perimeter of shape ABCDEF.

Usually, with compound shapes there will be some missing sides which need to be calculated. 

First we need to calculate ED.

We can see that FA = ED + CB, so we can calculate:

7 = 5 + ED

ED = 2

Next, we do the same with DC.

AB = FE + DC

12 = 3+DC

DC = 9

Finally, we can calculate the perimeter by adding up all the side lengths. 

3 + 2+ 9 + 5+ 12+ 7 = 38

Level 4-5

Example 1: Circles 

ABCDE is a shape made up of a rectangle and a semi-circle.
AB = 6,\,BC=13.

Calculate the perimeter of this shape to 3 significant figures.

[2 marks]

With compound shapes we first need to break it up into two parts. 

1) Rectangle – We know that rectangles have two pairs of equal sides, so we know that DC=6 and AD=13. However, since it is not on the outside of the shape, we won’t be counting AD with the perimeter. Therefore, the perimeter so far (sides AB, BC, and CD) is


2) Semi-circle – The formula for the circumference of a circle is \pi d, where d is the diameter. Because it is a semi-circle we’re dealing with, we will half the result. Doing this, we get

AED=\dfrac{(\pi \times 13)}{2}=20.420...

Now, adding the values together we get the total perimeter to be

25+20.420...=45.4 (3 sf)

Level 4-5

Example 2: Perimeter of Triangles 

ABC is a triangle. AB = 8 cm, BC = 6 cm. Angle ABC is a right-angle. Calculate the perimeter of triangle ABC.

[2 marks]

In this case, we are given two sides of the triangle but will have to work out the third if we want to find the perimeter. Since this is a right-angled triangle, we can use Pythagoras! Pythagoras’ theorem says


Where c is the hypotenuse and a and b are the other two sides. So, the equation becomes


Evaluating the left-hand side, and then square rooting, we get


AC=\sqrt{100}=10\text{ cm}

Now we have all three sides, simply add them to get the perimeter:

6+8+10=24\text{ cm}

Level 4-5
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Example Questions

To find the perimeter, we need to find the length of one side (all sides are the same, since it’s a square). If we say that x is the length of one side, then the area is,




So, if we square root both sides, we find that x=8. Therefore, the perimeter of the square is,


8+8+8+8=32\text{ m}

This hexagon is regular, so all 6 of its sides must be the same length. Since the perimeter is the result of adding all the sides together, we get:


\text{Length of one side}=21 \div 6=3.5\text{ cm}

Perimeter of semi-circle arc can be calculated by finding half of the circumference of a complete circle:


\dfrac{1}{2}\times\pi\times10=5\pi cm


Length of base = 10 cm

Total Perimeter = 10 +5\pi = 25.7 cm

To find the perimeter, we first need to find the missing lengths: 


120-55=65 cm

195-70=125 cm


Hence adding all the lengths together: 


Total Perimeter =120+70+65+125+55+195=630 cm

Here we are given the perimeter, whilst the side-lengths are expressed in terms of x. Since this triangle is an isosceles triangle, we know that, 


AB = BC = x+5


Adding together the three sides:

(x+5) + (x+5) + 3x= 45 \text{cm}


Now, to find x we must solve this equation,


\begin{aligned}5x+10&=45 \\ 5x&=35 \\ x&= 35\div5=7\text{cm}\end{aligned}

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