8.03 Area of composite shapes
A composite shape is a shape that can be broken up into smaller more recognisable shapes. For example, this shape is a square and a triangle combined.
Finding areas of composite shapes requires us to be able to break up the shape into recognisable components.
Cutting up shapes
It may be easier to see small shapes that make up a large one ("adding" shapes together):
Sometimes it's easier to see a large shape with a bit missing ("subtracting" shapes). It's good to get practice at both. So, the same shape could be a large rectangle with a small rectangle cut out of it.
What shapes can you see in these pictures? Practice breaking the composite shape up into smaller parts, or looking for a larger shape with a piece cut out of it. Remember to look for shapes already studied such as rectangles, squares, triangles or parallelograms.
Finding areas of composite shapes
To find the areas of composite shapes, being able to identify the shapes is only the first step. The next is calculating the areas of the parts. Consider the following two examples from the shapes above.
Worked examples
example 1
Find the area of the composite shape below.
Think: What shapes can we see? We can break up the composite shape into two triangles.
Do: The first triangle is given below.
The base measurement is $3-1=2$3−1=2 cm and the height is $3.2$3.2 cm. So the area of the triangle is:
| Area of triangle | $=$= | $\frac{1}{2}\times b\times h$12×b×h |
| $=$= | $\frac{1}{2}\times2\times3.2$12×2×3.2 | |
| $=$= | $3.2$3.2 cm2 |
The other triangle is given below.
The base measurement is $3$3 cm and the height measurement is $3$3 cm. So the area of the triangle is:
| Area of triangle | $=$= | $\frac{1}{2}\times b\times h$12×b×h |
| $=$= | $\frac{1}{2}\times3\times3$12×3×3 | |
| $=$= | $4.5$4.5 cm2 |
So the total area of the composite shape is $3.2+4.5=7.7$3.2+4.5=7.7 cm2.
example 2
A backyard garden needs to have turf laid. The shape and dimensions of the garden are indicated in the picture below. Find the area of the turf required.
Think: What shapes can we see? We can think of the turf as a rectangle with a triangular piece taken out as shown below.
Do:
The outside rectangle has area:
| Area | $=$= | $l\times w$l×w |
| $=$= | $12\times9$12×9 | |
| $=$= | $108$108 m2 |
The corner triangle has area:
| Area | $=$= | $\frac{1}{2}\times b\times h$12×b×h |
| $=$= | $\frac{1}{2}\times\left(12-5\right)\times\left(9-4\right)$12×(12−5)×(9−4) | |
| $=$= | $\frac{1}{2}\times7\times5$12×7×5 | |
| $=$= | $17.5$17.5 m2 |
So the total area is the area of the rectangle minus the area of the triangle:
| Total area | $=$= | $108-17.5$108−17.5 |
| $=$= | $90.5$90.5 m2 |
Practice questions
Question 1
Find the total area of the figure shown.
Question 2
Find the area of the shaded region in the figure shown.
Annulus
An annulus is a doughnut shape where the area is formed by two circles with the same centre.
$\text{Area of an annulus}=\text{area of larger circle}-\text{area of smaller circle}$Area of an annulus=area of larger circle−area of smaller circle
Practice question
Question 3
Find the area of the shaded region in the following figure, correct to one decimal place.
Sectors
$\text{Area of a circle}=\pi r^2$Area of a circle=πr2
$\text{Area of a semi-circle}=\frac{1}{2}\pi r^2$Area of a semi-circle=12πr2
$\text{Area of a quarter-circle}=\frac{1}{4}\pi r^2$Area of a quarter-circle=14πr2
Always consider the fraction of the circle. For example consider a quarter circle–the angle of this sector is $90^\circ$90°. The fraction of the circle is $\frac{90}{360}=\frac{1}{4}$90360=14. Therefore to calculate the area of a quarter of a circle, we calculate the whole circle and divide by $4$4.
Practice questions
Question 4
Calculate the area of the following figure, correct to one decimal place.
Question 5
Find the area of the shaded region in the following figure, correct to one decimal place.
Question 6
Find the area of the shaded region in the following figure, correct to one decimal place.
