Perimeter Worksheets, Questions and Revision

Perimeter Worksheets, Questions and Revision

GCSE 4 - 5KS3AQAEdexcelOCRWJECFoundationAQA 2022Edexcel 2022OCR 2022WJEC 2022

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.

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Level 4-5 GCSE KS3 AQA Edexcel OCR WJEC

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.

perimeter of a rectangle example

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 GCSE KS3 AQA Edexcel OCR WJEC
perimeter of compound shapes example

Perimeter of Compound Shapes

ABCDEF is a composite shape made up of rectangles.

Calculate the perimeter of shape ABCDEF.

perimeter of compound shapes example

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 GCSE KS3 AQA Edexcel OCR WJEC
perimeter compound shapes example semicircle rectangle

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]

perimeter compound shapes example semicircle rectangle

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)

Level 4-5 GCSE KS3 AQA Edexcel OCR WJEC
perimeter of right triangle

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}

perimeter of right triangle
Level 4-5 GCSE KS3 AQA Edexcel OCR WJEC

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,

 

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}

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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