## Perimeter

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

– Circles

## 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

## 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

## Example 1: Compound Shapes

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

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=\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)

## 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

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=AC^2

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

AC^2=64+36=100

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}

### Take an Online Exam

#### Perimeter Online Exam

### Example Questions

**Question 1:** ABCD is square with an area of 64\text{m}^2.

Calculate the perimeter of ABCD.

**[2 marks]**

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,

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}

**Question 2: **ABCDEF is a regular hexagon. Its perimeter is 21 cm.

Calculate the length of one side.

**[2 marks]**

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}

**Question 3:** The semi-circle below has centre O and a radius of 5 cm.

Find the perimeter of the semi-circle. Give your answer to 1 decimal place.

**[2 marks]**

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

**Question 4: **The diagram below shows a rectangle with a section missing.

Work out the perimeter of the shape.

**[3 marks]**

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

**Question 5:** ABC is an isosceles triangle. The perimeter of ABC is 45 cm.

Work out the value of x.

**[3 marks]**

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}

### Worksheets and Exam Questions

#### (NEW) Perimeter Exam Style Questions - MME

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