What you need to know

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.

Here we will look at working with perimeters of various kinds of shapes, including composite shapes – shapes made by combining 2 or more shapes together – which means that you may have to utilise all kinds of geometry tools and knowledge to answer a question. Make sure you’re at least familiar with common types of shapes.

Example: Below is a rectangle with length 11cm and width 3cm. Calculate its perimeter.

Here, the important fact about a rectangle is that its opposite sides are equal in length. So, the two sides that are not labelled must also be 3cm and 11cm. Therefore, we get

\text{perimeter }=3+3+11+11=28\text{ cm}

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

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

a^2+b^2=c^2,

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

8^2+6^2=c^2

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

c^2=64+36=100

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

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

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

Example: ABCDE is a shape made up of a rectangle and a semi-circle. AB = 6,\,BC=15. Calculate the perimeter of this shape to 3 significant figures.

For this question, you should be familiar with calculating the circumference (the fancy name for the perimeter of a circle). Click here (https://mathsmadeeasy.co.uk/gcse-maths-revision/areas-circles-circle-segments-gcse-maths-revision-worksheets/) to find out more on this topic if you aren’t sure.

Working with composite shapes often means breaking the shape up into parts that we are familiar with. In this case, we’ll deal with the rectangle first and then the semi-circle.

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

6+13+6=25

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=(\pi \times 13)=20.420...

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

25+20.420...=45.4\text{ (3sf)}

Example: PQR is an isosceles triangle. PQ=QR. The perimeter of PQR is 45. Work out the value of x.

The difference in this question is that we are given the perimeter, whilst the side-lengths are expressed in terms of some unknown x.

We know that if we add up all the sides, that must equal 45. Before we can do that, we need to have some expression for the other side: QR. Since this triangle is isosceles, it is the same as PQ, so we have that

QR=x+5

Now, we can’t add these side-lengths us as numbers, but we can add them together as algebra. Doing so, we get

x+5\,\,+\,\,x+5\,\,+\,\,3x=5x+10

Now, since the perimeter is equal to all the sides added up, we can make this equal to 45:

5x+10=45

Now, to find x we must solve this equation. Subtract 10 from both sides to get

5x=45-10=35

Then, dividing both sides by 5 we get the answer to be

x=\dfrac{35}{5}=7

Example Questions

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}

To find the perimeter, we need to find the length of one side (all sides the same, since it’s a square). If we say that x is the length of one side, then we find the area (which is given to be 64) by finding x^2. In other words, we know

 

x^2=64

 

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}

The perimeter is the distance around the outside of the shape, so we know that if we add all the side-lengths together, the result must be equal to 82. We’re going to add all the side-lengths together and make this expression for the perimeter equal to 82. This will give us an equation we can solve for x.

 

We don’t currently know all the sides, but we can work them out (or at least an expression for them). Looking at the shape, we can see that the sum of the lengths of FE and DC must equal the length of AB, which is 12. So, we get

 

DC=AB-FE=12-3=9

 

Now, the only side left is AF. Taking a similar approach, we can see on the picture that AF is equal to the sum of the lengths of CB (5) and DE (2x). So, we get

 

FA=CB+DE=5+2x

 

Now, marking these new side-lengths on the picture, we now have

 

 

So, to find an expression for the perimeter, we add up all these sides lengths:

 

12+5+9+2x+3+5+2x

 

Collecting terms, this becomes

 

4x+34

 

Making this equal to 82 and solving for x, we get

 

\begin{aligned}4x+34 &=82 \\ 4x &=48 \\ x &= \dfrac{48}{4}=12\end{aligned}

Revision and Worksheets

Coming Soon....

Need some extra help? Find a Maths tutor now

Or, call 020 3633 5145