Lesson

Area is a measure of the space inside a two-dimensional shape. To find the area of a rectangle or square, we could fill the shape with unit squares, and count how many unit squares there are. Another way though, is to use a rule:

Area of a rectangle or square

$\text{Area of a Rectangle}=\text{length}\times\text{width}$Area of a Rectangle=length×width

$\text{Area of a Square}=\text{side}^2$Area of a Square=side2

The fact we are using here is that rectangles have two pairs of equal sides, so we only need two dimensions, a length and a width. A square is a special type of rectangle where **every **side is equal, therefore we only need one dimension, the side length of the square.

Find the area of the rectangle shown.

If we know the area of a rectangle, and only one other dimension, we can use division to work out the value of the missing dimension. In fact, since a square has $4$4 equal sides, if we know its area, we can calculate the length of the sides.

The following applet allows us to change the length and/or width of a rectangle to see how the area changes. By hiding the length or the width, we can have a go at trying to work out the missing dimension ourselves.

Find the perimeter of a square whose area is $49$49cm^{2}.

Once we know how to calculate the area of a rectangle, or find the value of an unknown dimension, we can begin solving problems involving rectangles and squares, such as:

- Calculating how much paint or material is required to cover a surface
- The amount of rainwater that could be collected from the roof of a house

A rectangular driveway is $8$8 m long and $3$3 m wide.

What is the area of the driveway?

A kitchen floor is tiled with the tiles shown in the picture. If $30$30 tiles are needed to tile the floor, what is the total area of the floor? Give your answer in square centimetres.

We know the area of a rectangle can be calculated by multiplying the length by the width. The area of a triangle is exactly half of the area of a rectangle that has the same base dimension and height. The following video explains why this is true:

The rule for the area of a triangle is given by:

Area of a triangle

$\text{Area}$Area | $=$= | $\frac{1}{2}\times\text{base}\times\text{height}$12×base×height |

$A$A |
$=$= | $\frac{1}{2}bh$12bh |

The important thing is that we choose one of the triangle's sides as the base. The height will then be the perpendicular height from the base to the opposite corner (vertex).

For a right-angled triangle, this is relatively simple: we choose the base to be one of the sides adjacent to the right angle. The perpendicular height is the length of the other side adjacent to the right-angle.

For all other triangles, the perpendicular height is the line from the base to the opposite vertex, so that a right angle is formed at the base. This line may even be constructed outside the triangle, like the image on the right below.

In the following applet, we can change the lengths of the base and the height of a triangle, to see how the area changes.

If we know the area of a triangle and only one of its dimensions, we can work out the value of the unknown dimension, using the method explained in the following video.

In the following applet, we can change the dimensions of the triangle, to change its area. We can also choose to hide the base or height and then work out the value of the missing dimension.

Use the picture to answer the following questions about the area of the rectangle and the triangle.

Firstly, find the area of the entire rectangle.

Now find the area of the triangle.

Area of triangle $=$= $\frac{1}{2}$12 $\times$× area of rectangle m ^{2}Area of triangle $=$= $\frac{1}{2}\times\editable{}$12× m ^{2}(Fill in the value for the area of the rectangle.) Area of triangle $=$= $\editable{}$ m ^{2}(Complete the multiplication.)

Find the value of $h$`h` in the triangle with base length $6$6 cm if its area is $54$54 cm^{2}.

A gutter running along the roof of a house has a cross-section in the shape of a triangle. If the area of the cross-section is $50$50 cm^{2}, and the length of the base of the gutter is $10$10 cm, find the perpendicular height $h$`h` of the gutter.

A parallelogram is a four-sided shape where the opposite sides are parallel. It is similar to a rectangle, but doesn't necessarily have right angles. This means a rectangle is a special type of parallelogram.

The good news is, we can use the commonalities between parallelograms and rectangles to calculate the area of a parallelogram. The following applet explores these commonalities further.

Because a parallelogram might not have right angles, we need to make sure we use the perpendicular height, when we work out the area. It doesn't matter which side we use as the base, as long as the height is the perpendicular height.

Area of a parallelogram

$\text{Area}$Area | $=$= | $\text{base}\times\text{height}$base×height |

$A$A |
$=$= | $bh$bh |

Find the area of the parallelogram shown.

Find the unknown measure in the parallelogram pictured, given that its area is $255$255 square units.

Some car parks require the cars to park at an angle as shown.

The dimensions of the car park are as given, where each individual parking space has a length of $4.9$4.9 m and a width of $4.4$4.4 m.

What area is needed to create an angled carpark suitable for $7$7 cars?

Trapeziums (or "trapezia") are related to parallelograms. That's important, because it means we can use the area of a parallelogram to help us work out the area of a trapezium.

The following applets explores how a trapezium is exactly half of its related parallelogram or rectangle.

Area of a trapezium

$\text{Area}$Area | $=$= | $\text{half the perpendicular height}\times\left(\text{sum of the lengths of the parallel sides}\right)$half the perpendicular height×(sum of the lengths of the parallel sides) |

$A$A |
$=$= | $\frac{h}{2}\left(a+b\right)$h2(a+b) |

The sides $a$`a` and $b$`b` are the lengths of the parallel sides, and $h$`h` is the perpendicular height.

Two identical trapezia are placed to make a parallelogram.

Use the picture to answer the following questions about the area of the parallelogram and one of the trapezia.

Find the area of the entire parallelogram.

Find the area of one of the trapezia.

area of trapezium $=$= $\frac{1}{2}$12 $\times$× area of parallelogram mm ^{2}area of trapezium $=$= $\frac{1}{2}\times\editable{}$12× mm ^{2}Fill in the value for the area of the parallelogram. area of trapezium $=$= $\editable{}$ mm ^{2}Complete the multiplication

Find the area of the trapezium shown.

Find the height $\left(h\right)$(`h`) if the area of the trapezium shown is $36$36 cm^{2}.

Start by substituting the given values into the formula for the area of a trapezium.

$A=\frac{1}{2}\left(a+b\right)h$

`A`=12(`a`+`b`)`h`

A harbour has a trapezoidal pier with a perpendicular height of $8$8m. One base of the pier has a length of $9$9m and the other has a length of $5$5m. What is the area of the pier?

Area of a rectangle

$\text{Area of a Rectangle}=\text{length}\times\text{width}$Area of a Rectangle=length×width

$\text{Area of a Square}=\text{side}^2$Area of a Square=side2

$\text{Area of a Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$Area of a Triangle=12×base×height

$\text{Area of a Parallelogram}=\text{base}\times\text{height}$Area of a Parallelogram=base×height

$\text{Area of a Trapezium}=\frac{1}{2}\times\text{base+top}\times\text{height}$Area of a Trapezium=12×base+top×height