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9. 2D Measurement
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Level 6

9.06 Perimeter and area

Lesson
Practice
Lesson

Are you ready?

Can you find the area of a rectangle by using the length and width?

Examples

Example 1

Find the area of the rectangle shown.

A rectangle of length 12 centimetres and width 2 centimetres.
Worked Solution
Create a strategy

Use the area of a rectangle formula: \text{Area}=\text{Length} \times \text{Width}

Apply the idea

We can see that length is 12 \text{ cm} and the width is 2 \text{ cm}.

\displaystyle \text{Area}\displaystyle =\displaystyle \text{Length} \times \text{Width}Use the formula
\displaystyle =\displaystyle 12 \times 2 Substitute the length and width
\displaystyle =\displaystyle 24 \text{ cm}^2Double 12
Idea summary

The area of a rectangle is given by the formula: \text{Area}=\text{Length} \times \text{Width}

Break up shapes to work out area

By breaking a shape into smaller rectangles, we can work out the area of those rectangles first. We can then add those two values together, to work out the total area of our shape.

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Examples

Example 2

Find the area of the given shape.

A shape made from two rectangles. Ask your teacher for more information.
Worked Solution
Create a strategy

The shape can be broken down into two rectangles as shown below.

An image showing 2 rectangles with a plus sign between them.

So the area of the shape can be found by adding the areas of the two rectangles: \text{Area} = \text{Area of top rectangle} + \text{Area of bottom rectangle}

Apply the idea

For the top rectangle:

\displaystyle \text{Area}\displaystyle =\displaystyle 5 \times 5Multiply length and width
\displaystyle =\displaystyle 25\text{ cm}^2

For the bottom rectangle:

\displaystyle \text{Area}\displaystyle =\displaystyle 9 \times 4Multiply length and width
\displaystyle =\displaystyle 36\text{ cm}^2
\displaystyle \text{Area}\displaystyle =\displaystyle 25 + 36Add the areas
\displaystyle =\displaystyle 61 \text{ cm}^2
Idea summary

We can find the area of more complicated shapes by breaking the shape up into rectangles, and adding the area of each rectangle.

Distributive property

Once we know how to work out the area of rectangles, there are some handy things we can do. The distributive property of area means we can work out the area of a rectangle by breaking it into smaller rectangles.

To see how we can do this watch this video.

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Examples

Example 3

The rectangle below has been split in to two rectangles. We want to work out the area.

A large rectangle of height 7 and length 11 metres made of purple and green squares. Ask your teacher for more information.
a

What is the area of the purple rectangle?

Worked Solution
Create a strategy

Consider the purple rectangle shown below and find the area using the formula: \text{Area} = \text{length} \times \text{width}

Apply the idea

The length of the purple rectangle is 7 metres and the width is 4 metres.

\displaystyle \text{Area}\displaystyle =\displaystyle 7 \times 4Substitute the length and width
\displaystyle =\displaystyle 28 \text{ m}^2
b

What is the area of the green rectangle?

Worked Solution
Create a strategy

Consider the green rectangle shown below and find the area using the formula: \text{Area} = \text{length} \times \text{width}

Apply the idea

The length of the green rectangle is 7 metres and the width is 7 metres.

\displaystyle \text{Area}\displaystyle =\displaystyle 7 \times 7Substitute the length and width
\displaystyle =\displaystyle 49 \text{ m}^2
c

What is the area of the whole rectangle?

Worked Solution
Create a strategy

Consider the whole rectangle and find the area using the formula: \text{Area} = \text{length} \times \text{width}

Apply the idea

The length of the whole rectangle is 7+4 metres and the width is 7 metres.

\displaystyle \text{Area}\displaystyle =\displaystyle 7 \times (4 + 7)Substitute the length and width
\displaystyle =\displaystyle 7 \times 11Add the numbers inside the brackets
\displaystyle =\displaystyle 77 \text{ m}^2
d

Is the area of the large rectangle equal to the sum of the two smaller rectangles? In other words, is the following statement true: 7 \times 4 + 7 \times 7 =7 \times 11

Worked Solution
Create a strategy

Add the answer in part (a) and (b) and compare it to the answer in part (d).

Apply the idea
\displaystyle \text{Area of smaller rectangles}\displaystyle =\displaystyle 7\times 4 + 7\times 7Add the areas from (a) and (b)
\displaystyle =\displaystyle 28 + 49Find each area
\displaystyle =\displaystyle 77 \text{ m}^2Add the areas

So the sum of the areas of the two smaller rectangles is equal to the area of the large rectangle.

Reflect and check

What we have shown is that the distributive property is true: 7\times 4 + 7\times 7=7\times (4 + 7)

Idea summary

Make sure to add all the smaller shapes together to get the total area of the shape.

We can use the area of a rectangle to show that the distributive property is true. For example:7\times 4 + 7\times 7=7\times (4 + 7)

Perimeter and area

You may have noticed already that shapes with the same perimeter don't always have the same area, as shown in the rectangles below. Similarly, shapes with the same area don't always have the same perimeter.

An array with 1 row and 4 columns of squares.
This rectangle has a perimeter of 10 units and an area of 4 square units.
An array with 2 rows and 3 columns of squares.
This rectangle has a perimeter of 10 units and an area of 6 square units.

Let's look more at the relationship between perimeter and area now.

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Examples

Example 4

Which of these rectangles has an area of 24 \text{ cm}^2 and a perimeter of 28 cm?

(Note: Diagrams are not to scale.)

A
A rectangle of length 12 centimetres and width 2 centimetres.
B
A rectangle of length 10 centimetres and width 4 centimetres.
C
A rectangle of length 5 centimetres and width 4 centimetres.
D
A rectangle of length 8 centimetres and width 3 centimetres.
Worked Solution
Create a strategy

Find the area and perimeter of each of the rectangles using the rules:

\text{Area}=\text{length} \times \text{width}

\text{Perimeter}=\text{Sum of all the sides}

Apply the idea

Let's find the areas of each rectangle first.

Option A:

\displaystyle \text{Area}\displaystyle =\displaystyle \text{Length} \times \text{Width}Use the formula
\displaystyle =\displaystyle 12 \times 2Multiply the length and width
\displaystyle =\displaystyle 24\, \text{cm}^2

Option B:

\displaystyle \text{Area}\displaystyle =\displaystyle 10 \times 4Multiply the length and width
\displaystyle =\displaystyle 40\, \text{cm}^2

Option C:

\displaystyle \text{Area}\displaystyle =\displaystyle 5 \times 4Multiply the length and width
\displaystyle =\displaystyle 20\, \text{cm}^2

Option D:

\displaystyle \text{Area}\displaystyle =\displaystyle 8 \times 3Multiply the length and width
\displaystyle =\displaystyle 24\, \text{cm}^2

The areas of options A and D are 24\text{ cm}^2. So now we will find their perimeters.

Option A:

\displaystyle \text{Perimeter}\displaystyle =\displaystyle 12+12+2+2Add the sides
\displaystyle =\displaystyle 28\text{ cm}

Option D:

\displaystyle \text{Perimeter}\displaystyle =\displaystyle 8+8+3+3Add the sides
\displaystyle =\displaystyle 22\text{ cm}

Option A has the correct perimeter. The answer is option A.

Idea summary

Rectangles can have the same perimeter but different areas.

Outcomes

VCMMG224

Solve problems involving the comparison of lengths and areas using appropriate units.

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