Area of Composite Shapes II

We have already had a look at combining 2D shapes together and finding the area of composite shapes.  But since then we have learnt about the areas of a whole lot more shapes.  

Let's just look at a summary of areas of 2D shapes we have looked at so far with our shape robot:

Strategies

The key skills you need to remember when thinking about composite shapes are to consider

• can the shape be considered as a larger easier shape with smaller bits missing
• can the shape be considered the sum of a number of smaller shapes

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

$A=L\times W$A=L×W

$\text{Area of a Square }=\text{Side }\times\text{Side }$Area of a Square =Side ×Side

$A=S\times S$A=S×S

$\text{area of a triangle }=\frac{1}{2}\times\text{base }\times\text{height }$area of a triangle =12​×base ×height

$A=\frac{1}{2}bh$A=12​bh

$\text{Area of a Parallelogram }=\text{Base }\times\text{Height }$Area of a Parallelogram =Base ×Height

$A=b\times h$A=b×h

$\text{Area of a Trapezium}=\frac{1}{2}\times\left(\text{Base 1 }+\text{Base 2 }\right)\times\text{Height }$Area of a Trapezium=12​×(Base 1 +Base 2 )×Height

$A=\frac{1}{2}\times\left(a+b\right)\times h$A=12​×(a+b)×h

$\text{Area of a Kite}=\frac{1}{2}\times\text{diagonal 1}\times\text{diagonal 2}$Area of a Kite=12​×diagonal 1×diagonal 2

$A=\frac{1}{2}\times x\times y$A=12​×x×y

$\text{Area of a Rhombus }=\frac{1}{2}\times\text{diagonal 1}\times\text{diagonal 2}$Area of a Rhombus =12​×diagonal 1×diagonal 2

$A=\frac{1}{2}\times x\times y$A=12​×x×y

$\text{Area of a circle }$Area of a circle

$A=\pi r^2$A=πr2

$\text{Area of a sector }$Area of a sector

$A=\frac{\text{angle of sector}}{360}\times\pi r^2$A=angle of sector360​×πr2

Worked Examples

QUESTION 1

Find the total area of the figure shown.

Reveal Solution

QUESTION 2